CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 14 Question Answer Organisms and Environment

Organisms and Environment Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Multiple choice questions

Question 1.
A population is a group of
(a) individuals in a species
(b) species’in a community
(c) individuals in a family
(d) communities in an ecosystem
Answer:
(a) individuals in a species

Question 2.
Exponential growth occurs when there is
(a) a great environmental resistance
(b) a fixed carrying capacity
(c) no biotic potential
(d) no environmental resistance
Answer:
(d) no environmental resistance

Question 3.
In a population, unrestricted reproductive capacity is called as
(a) carrying capacity
(b) birth rate
(c) biotic potential
(d) fertility rate
Answer:
(b) birth rate

CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment

Question 4.
Two opposite forces operate in the growth and development of a population. One of them is related to the ability to reproduce at a given rate. The force opposite to it is called
(a) environmental resistance
(b) mortality
(c) fecundity
(d) biotic control
Answer:
(b) mortality

Question 5.
The carrying capacity of a population is determined by its
(a) population growth rate
(b) mortality
(c) limiting resources
(d) natality
Answer:
(c) limiting resources

Question 6.
Which of the following at a conduit for energy transfer across trophic levels?
(a) mutualism
(b) parasitism
(c) protocooperation
(d) predation
Answer:
(d) predation

Question 7.
Phenomenon of inhibition of growth of one species by other species through secretion of some chemicals is termed as
(a) commensalism
(b) allelopathy
(b) mutualism
(d) predation
Answer:
(b) allelopathy

Question 8.
Predation performs all, except
(a) transfer of energy
(b) loss of sense organs
(c) keeps prey population under control
(d) maintains species diversity
Answer:
(b) loss of sense organs

Question 9.
Two important factors that influence the life of organisms are
(a) soil, temperature
(b) soil, light
(c) light, water
(d) water, temperature
Answer:
(c) light, water

Question 10.
Ecology describes
(a) interactions between living organisms only
(b) intraspecific competitions only
(c) interactions between members of single species
(d) interactions of organisms and abiotic components around
Answer:
(d) interactions of organisms and abiotic components around

CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment

Express in one or two words

Question 1.
Study of interrelationship between the environment and a plant species.
Answer:
Ecology

Question 2.
Amount of water vapours actually present in the air at any given time.
Answer:
Humidity

Question 3.
The total amount of water in the soil, except the gravitational water.
Answer:
Holard

Question 4.
Association of fungi and algae.
Answer:
Lichen

Question 5.
The study of soil.
Answer:
Pedology

Question 6.
Vegetation where the annual rainfall is more than 50 inches.
Answer:
Grassland

CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment

Correct the sentences changing the underlined word only

1. Plants those grow in soil and mud are xerophytes.
Answer:
mesophytes

2. Sunken stomata is a characteristic of hydrophytes.
Answer:
xerophytes

3. Air pockets are found in mesophytes.
Answer:
hydrophytes

4. The pre-reproductive mass is found more in urn- shaped pyramid.
Answer:
triangular

5. Population consists of different kinds of species.
Answer:
Community

Fill in the blanks

1. Shallow water region present on the edge of lakes is called …………. .
Answer:
littoral zone

2. The most relevant ecological factor is …………….. .
Answer:
light

3. Mortality and …………. contribute to a decrease in population density.
Answer:
emigration

4. J-shaped curve represents ………… growth.
Answer:
exponential

CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment

5. Geometric representation of age structure is a characteristic of ………..
Answer:
population

Short Answer Type Questions

Question 1.
Camouflage
Answer:
Camouflage It is the act of hiding the identify either by colour changes or by making animals or objects hard to see.
Some species of insects and frogs are cryptically coloured (camouflaged) to avoid being detected easily by predators.

Question 2.
Edaphic factor
Answer:
Edaphic factor An edaphic factor relating to physical or chemical composition of the soil found in a particular area, e.g. soil pH, porosity, water holding capacity, etc.

Question 3.
Habitat
Answer:
Environment is termed as sum total of all external conditions which influence the organisms in term of survival and reproduction. Habitat is a natural abode or locality where a plant/ animal grow, based on the environment, the differences in the vegetation and species of different places are observed.

Each organism plays an important role in its surrounding. Niche is the role of an organism plays in its ecosystem. In other words, we can say that niche is how an organism makes a living, interacts with other organisms and helps in cycling of nutrients.

Question 4.
Temperature
Answer:
Temperature
It is the most ecologically significant environmental factor. It varies seasonally on land and decreases progressively from the equator towards the poles and from plains to the mountain tops. It ranges from sub-zero levels in polar areas and high altitudes to > 50°C in tropical deserts in summer. There are also certain unique habitats such as thermal springs, deep sea hydrothermal vents where the average temperature exceeds 100°C.

Question 5.
Biomes
Answer:
Biomes The large unit of ecology which consist of major vegetation type and associated fauna in a particular climatic zone is called biome, e.g. sea coast, deserts. Deciduous forest are the major biomes of India.

Question 6.
Population
Answer:
The term ‘population’ can be defined as a group or assemblage of organisms of the same species living in a particular area at a given time.
For example, the lion population of the Gir forest or the peacock population of India, etc.

The population may be subdivided into demes or local populations, which are a group of interbreeding organisms. It is the smallest collective unit of a plant or animal population.
The other is metapopulation, which consists of whole set of local populations connected by dispersing individuals.

CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment

Question 7.
Competition
Answer:
It is generally believed to occur when closely related the two individuals of species compete for the same resources that are limiting. This can happen between the members of the same species, i.e., intraspecific or different species, i.e. interspecific.

Question 8.
Abiotic factors
Answer:
Abiotic factors The non-living factors which influence an ecosystem and the organisms are called abiotic factors. It can determine which species of organism will survive in an given environment.

Question 9.
Population density
Answer:
Population Density
This refers to the size of population to a unit space at a particular time. This can be measured is several ways an mentioned below

  • Abundance (absolute number in population).
  • Numerical density (number individuals per unit area).
  • Biomass density (biomass per unit area).

Question 10.
Necessity of adaptations
Answer:
Necessity of adaptations Adaptations take a lot of time to evolve. They are necessary because they help the organism to fit in to its niches. Also, it is important for’ survival of both, the individual and the species.

Differentiate between the following

Question 1.
Habitat and Niche
Answer:
Differences between habitat and niche are as follows

Habitat Niche
A place or part of an ecosystem, occupied by a particular organism. A functiohal description of the role, a species plays in a community.
It can have number of niches. It does not have any components.
A variety of environmental variables are present in a habitat. Every niche has its own specific environment.

Question 2.
Mutualism and Parasitism
Answer:
Differences between mutualism and parasitism are as follows

Mutualism Parasitism
It type of interaction in which both the interacting species gets benefits. The mode of interaction in two species in which one (parasites) is dependent on another for benefits there by damaging the other one (host).
Mycorrhiza show mutual association between fungi and roots of higher plants. Human liver-fluke depends on two intermediate hosts to complete its life cycle.

CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment

Question 3.
Hydrophytes and Xerophytes
Answer:
Differences between hydrophytes and xerophytes are as follows

Hydorphytes Xerophytes
The plants which are adapted to live in abundance of water are called hydrophytes. The plants which are adapted to live in condition of water scarcity and dry habitat are called xerophytes.
Root are poorly developed. Roots are very well- developed.
Leaves are well-developed. Leaves are modified in various structure.
Stomata are present abundantly. Sunken stomata are present.
e.g. Hydrilla, Pistia. e.g. Opuntia, Asparagus, etc.

Question 4.
Birth rate and Death rate
Answer:
Differences between birth rate and death rate are as follows

Birth rate Death rate
It is the number of birth of new individuals per unit of population per unit time. It is the number of loss of individuals per unit of population per unit time.
It increases the size of population. It decreases the size of population.

Question 5.
Fertility and Fecundity
Answer:
Differences between fertility and fecundity are as follows

Fertility Fecundity
Fertility is the natural ability to reproduce and is defined as the offspring per couple. Fecundity is the actual reproductive rate of an organism or population measured by number of gametes, (seed sets or asexual propagules).

Question 6.
Logarithmic and Exponential growth
Answer:
Differences between logarithmic and exponential growth are as follows

Logarithmic growth Exponential growth
Logistic or Logarithmic occurs rapidly and then slow down. The kind of growth which is slow initially but, it increases as the population growth takes place.
It considers factors like competition and limited resources. It requires specific ideal conditions to occur.
It occurs when the resources are limited. It occurs when the resources are abundant.
It has four phases-lag, log, decelearatlon and steady. It has two phases-lage and log.
It is more common, in human, etc. It occurs in algal bloom, etc.

CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment

Long Answer Type Questions

Question 1.
Explain, what is population? Describe the different characteristics of population.
Answer:
The term ‘population’ can be defined as a group or assemblage of organisms of the same species living in a particular area at a given time.
For example, the lion population of the Gir forest or the peacock population of India, etc.

The population may be subdivided into demes or local populations, which are a group of interbreeding organisms. It is the smallest collective unit of a plant or animal population.
The other is metapopulation, which consists of whole set of local populations connected by dispersing individuals.

Population Interactions:
In nature, living organisms such as animals, plants and microbes, cannot live in isolation and therefore, interact in various ways to form a biological community. Interspecific interactions occur between the populations of two different species living together within a community.

These interactions could be beneficial (+), detrimental (-) or neutral (0), as shown in table below
Population Interactions and their Effects

Name of interaction Effect on species-A Effect on species-B
Mutualism + +
Competition
Predation +
Parasitism +
Commensalism + 0
Ammensalism 0

Mutualism:
It is an interaction that confers benefits to both the interacting species. Some examples of mutualism are
1. Lichens represent an intimate mutualistic relationship between a fungus (mycobiont) and photosynthesising algae (phycobiont) or cyanobacteria. Here, the fungus helps in the absorption of nutrients and provides protection, while algae prepares the food.

2. Mycorrhiza show dose mutual association between fungi and the roots of higher plants. Fungi help the plant in absorption of nutrients, while the plant provides food for the fungus, e.g. many members of genus -Glomus.

3. Plants need help from animals for pollination and dispersal of seeds. In return, plants provide nectar, pollens and fruits to them.

4. Co-evolution to safeguard the mutually beneficial system, plant-animal interactions involve co-evolution of the mutualists, i.e. the evolution of the flower and its pollinator species are tightly linked with one another, e.g.

(i) Fig and its partner wasp species, the female wasp uses the fruit not only as an oviposition (egg-laying) site, but uses the developing seeds from the fruit for nourishing its larvae. In return, the wasp pollinates the fig inflorescence, while searching for suitable egg-laying sites.

(ii) Mediterranean orchid Ophrys employs ‘sexual deceit’ to get pollinated by a species of bee. One petal of its flower bears an uncanny resemblance to the female of the bee in size, colour and markings. The male bee is attracted to what it perceives as a female and ‘pseudocopulates’ with the flower. During this process, pollen are dusted from the flower onto the male bee. When the same bee pseudocopulates with another flower, it transfers pollen to it and thus pollinates the flower.

Competition:
It is generally believed to occur when closely related the two individuals of species compete for the same resources that are limiting. This can happen between the members of the same species, i.e., intraspecific or different species, i.e. interspecific. These are as follows

(i) Intraspecific competition occur for the resources which are short in supply such as food, space or mate. There are of two types

  1. Content competition Where each organism claims a part of the resource, but due to competition, some are successful to get, but some fail.
  2. Scramble competition In this case, the resource gets divided to many other smaller parts to which all have access. Individual organism scramble for resources. Each individual succeeds to get some of the resource available to survive.

(ii) Interspecific competition It occurs between members of different species, e.g. competition is likely to have two outcomes. These are as follows
1. Competitive exclusion According to Gause’s competitive exclusion principle two closely related species competing for same resources cannot co-exist indefinitely and the competitively inferior species will be eliminated eventually two species of beetles (Tribolium castaneum and T. confusum) feed on stored flour, one species would survive and other will die.

2. Co-existence Species facing competition might evolve a mechanism to live in the same niche by changing the feeding time or foraging pattern. This is called resource partitioning.

Question 2.
What do you understand by population? Explain the different attributes of the population.
Answer:
The term ‘population’ can be defined as a group or assemblage of organisms of the same species living in a particular area at a given time.
For example, the lion population of the Gir forest or the peacock population of India, etc.

The population may be subdivided into demes or local populations, which are a group of interbreeding organisms. It is the smallest collective unit of a plant or animal population.
The other is metapopulation, which consists of whole set of local populations connected by dispersing individuals.

Population Attributes:
Population is a unit of ecosystem through which the energy flows and nutrients get cylced which helps in maintaining its stability. A population has some features such as birth rate, death rate, growth form, age structure, density, etc. which are discussed below

1. Growth
When a few organisms are introduced to a particular unoccupied area, the population show growth (increase in size) in sigmoid or ‘S’-shaped logistic fashion.

The few organisms are introduced to a particular unoccupied area, the growth of population initially is slow (positive acceleration phase). Then suddenly there is sharp increase and growth becomes vary rapid (logarithmic phase). At last stage it finally slow and down because of the increased environmental resistance (negative acceleration phase). It is dipicted in the given [fig. 14.19(a)], The increase of population size does not occure beyond a certain saturation limit. This is known as carrying capacity .
CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 1
The sigmoid and exponential growth curves

Another kind of population curve in J-shaped this the density of organism increase rapidly, but stops suddenly due to environmental resistance. It is called exponential growth curve.

(i) Growth Rate:
The growth rate of a population can be determined by the following formula
CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 2

Population size for a given species is not static parameter, but it is ever changing based on different factors including food availability, predation pressure and weather diversity. The population thickness changes because of the following four changes

  • Natality It refers to the number of “births during a given period in the population that are added to the initial density.
  • Mortality It is the number of deaths in the population during a given period.
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 3
    Factors influencing population density

2. Birth Rate and Death Rate:
The increased and decreased rate of population depends on birth and death natality is related new homes, i.e. the reproduction capacity of individuals. It is related with two aspects of reproduction.

Thus, natality or birth rate of a population can be expressed as
CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 4
Where, b = Natality per unit time, d – Changing value of the entity, N = Initial number of individuals in population, Nn – Number of new individuals added and t = Unit time.

Similarly, the death rate or mortality rate is the death or loss of individuals from the population in unit time and can be expressed as
CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 5
The death rate has a relation with the natality because of over crowding, predation and spread of disease.

3. Sex Ratio:
It represents the ratio of organisms of different sexes of the population. In animals, sex is either male or female which creates difference in the characteristic of the population.

4. Age Distribution:
A population at a given time is composed of individuals of different age groups such as pre-reproductive, reproductive and post-reproductive. The population age distribution is related to the growth rate of the population and this can be used to calculate whether the population is expanding or contracting.

Ordinarily, a rapidly expanding population would have a large proportion of young individuals, a stationary population with even distribution of age groups and declining population contain a large proportion of old individuals.

These age groups of the population can be portrayed through the graphical age pyramid representations.In human population, the age pyramids generally express age distribution of males and females in a combined diagram. The shapes of the pyramids reflect the growth status of the population.

The pyramids can be of three difference types as follows
(i) Expanding (Triangular) This is a type of a growing population representation is like a triangle.
The population carries a high proportion of pre-reproductive individuals followed by reproductive individuals and post-reproductive individuals. Because of the very large number of pre-reproductive individuals, more and more of them enter reproductive phases and rapidily increases the size of the population.

(ii) Stable (Bell-shaped) This type of pyramid will represent a stationary or stable population having an equal number of young and middle aged class of individiuals.

(iii) Declining (Urn-shaped) This group has a small number of pre-reproductive individuals followed by a large number of reproductive individuals. As, there is less number of individuals in pre-reproductive groups.
CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 6

5. Population Density
This refers to the size of population to a unit space at a particular time. This can be measured is several ways an mentioned below

  1. Abundance (absolute number in population).
  2. Numerical density (number individuals per unit area).
  3. Biomass density (biomass per unit area).

The density of population can be expressed in the following manner
(a) Crude density This is the density of a species population with reference is the total area. It varies according to the season, weather, food supply rate of reproduction, etc.
(b) Ecological density This is the density of a species with reference to the actual area of habitat available to the species.

Question 3.
Explain, how different organisms interact in a population emphasising on the possibilities of various relationships.
Answer:
The term ‘population’ can be defined as a group or assemblage of organisms of the same species living in a particular area at a given time.
For example, the lion population of the Gir forest or the peacock population of India, etc.

The population may be subdivided into demes or local populations, which are a group of interbreeding organisms. It is the smallest collective unit of a plant or animal population.
The other is metapopulation, which consists of whole set of local populations connected by dispersing individuals.

Predation
It is an interspecific interaction, where an animal called predator kills and consumes the other weaker animal called prey. This is a biological control method. It is the nature’s way of transferring energy to higher trophic levels, which is fixed by plants at the first trophic level, e.g. tiger (predator) and deer (prey).

Important roles of predators are as follows
1. In the absence of predators, prey species could achieve very high population densities and cause instability. So, besides acting as ‘conduits’ for energy transfer across trophic levels, predators play very important role in providing population stability.

2. They help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species, e.g. predator starfish Pisaster in the rocky intertidal communities of American Pacific Coast. In a field experiment, when all the starfish were removed from the area, more than 10 species of invertebrates became extinct within a year, because of interspecific competition.

3. When certain exotic species are introduced into a geographical area, they become invasive and start spreading fast because the invaded land does not have natural predators, e.g. prickly pear cactus introduced in Australia in early 1920s was brought under control by introducing its predator (i.e. a moth) in the country.

If a predator is too efficient and over exploits its prey, then the prey might become extinct.
Following it, the predator will also become extinct because of the lack of food. This is why predators in nature are prudent.
Prey species have evolved various defence mechanisms to lessen the impact of predation. These are as follows

In Animals:

  1. Camouflage Some species of insects and frogs are cryptically coloured (camouflaged) to avoid being detected easily by the predator. Some are poisonous and therefore, avoided by the predators.
  2. Chemical emission Monarch butterfly is highly distasteful to its predators (birds) because of a special chemical present in its body. The butterfly acquires this chemical during its caterpillar stage by feeding on a poisonous weed.
  3. Mimicry It refers to the resemblance of an organism to their natural surroundings which hides them from eyes of predators.

In Plants
Nearly 25% of all insects are known to be phytophagous (feeding on plant sap and other parts of plants) apart from other herbivores. So, plants have evolved various defences against them, e.g., thorns of Acacia and cactus are the most common morphological means of defence.

Some plants produce highly poisonous chemicals like cardiac glycosides, e.g., weed Calotropis that makes the herbivore sick, etc.
Chemicals like nicotine, caffeine, quinine, strychnine, opium, etc., are actually defence mechanisms against grazers and browsers.

Parasitism:
It is the mode of interaction between two species in which one species (parasite) depends on the other species (host) for food and shelter and damages the host. In this process, one organism is benefitted (parasite), while the other is being harmed (host).

(i) Adaptation methods of a parasite are
Parasite is host-specific in a way that both host and parasite tend to co-evolve. According to its lifestyle, a parasite, evolved special adaptations as

  • Loss of unnecessary sense organs.
  • The presence of adhesive organs or suckers for clinging on to host.
  • Loss of digestive system.
  • High reproductive capacity.

(ii) The life cycles of parasites are often complex, involving one or two intermediate hosts or vectors to facilitate parasitisation of its primary host,, e.g.

  • Human liver fluke (a trematode parasite) depends on two intermediate hosts (a snail and a fish) to complete its life cycle.
  • Malarial parasite (Plasmodium) needs a vector (mosquito) to spread to other hosts.

(iii) Majority of parasites harm the host. The harm is done in the following ways

  • They reduce the survival, growth and reproductive ability of the host.
  • They reduce its population density.
  • They might render the host more vulnerable to predation by making it physically weak.

Types of Parasites
Parasites are broadly divided into following main types

  1. Ectoparasites depend on the external surface of the host organism for food and shelter, e.g., lice on humans, ticks on dogs, copepods in marine fishes and Cuscuta, a parasitic plant that grows on hedge plants.
  2. Endoparasites live inside the host’s body at different sites like liver, kidney, lungs, etc., for food and shelter, e.g. tapeworm, liver fluke, Plasmodium, etc. The life cycles of endoparasites are more complex because of their extreme specialisation.
  3. Brood parasitism is an example of parasitism in which one organism (parasite) lays its eggs in the nest of another organism (host) for the later to incubate them.

The eggs of parasitic birds have evolved to resemble the host’s egg to reduce the chances of host bird from detecting and ejecting the parasitic eggs from nest, e.g., cuckoo (koel, parasite) and crow (host) during breeding season (spring to summer).

Effects of Parasites on Host
There always exist some kind of host parasite relationship which has following characteristics parasites do not kill the host immediately, but they make the host to suffer, they damage their body and in extreme case they may cause their death. These reduce the growth of the host and also afffect their reproductive potential. They also reduce the size of the population of host.

Question 4.
What is habitat? Describe the different types of abiotic factors present in the habitat.
Answer:
Habitat:
Environment is termed as sum total of all external conditions which influence the organisms in term of survival and reproduction. Habitat is a natural abode or locality where a plant/ animal grow, based on the environment, the differences in the vegetation and species of different places are observed.

Abiotic Factors:
These include climatic, edaphic and topographic factors. Climatic factors are light, temperature, precipitation, humidity, wind. Edaphic factors are factors related to soil whereas topographic factors are physical factors related to slope, altitude and others concerned with earth surface. Some of the major abiotic factors are discussed below
1. Light
Sunlight is the primary source of light. It plays an important role in almost all ecosystems. The entire food chain starts with the organisms that are photosynthetic (producers).

So without sunlight, all life excluding some microbes would perish, not just the plants. The total amount of light that falls on the earth varies according to the season, latitude, altitude and conditions of the atmosphere. The significance of light lies in the fact that all autotrophs depend upon light as a source of energy for preparing their food by photosynthesis and release oxygen during the process.

Therefore, it is an important factor for life to exist on earth. Small herbs and shrubs growing in forests are adapted to photosynthesis under very low light intensities, because they are overshadowed by the tall, canopied trees. Most plants depend on sunlight to meet their photoperiodic requirement for flowering also.

The spectral quality of solar radiation as shown in figure is also important for life. The UV component of light is harmful for many organisms. Different components of the visible spectrum are available for marine plants living at different depths of the ocean.
CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 1
This is why different types of algae, i.e., green, brown and red algae occur at different depths in sea in the upper, middle and deep layers of water respectively.
Plants which grow in bright light are called sun plants or heliophytes while plants growing in shade or low intensity light are called sciophytes.
Depending upon the availability of light aquatic body is divided into the following zones

  1. Littoral zone ft adjoins the shore and extends to the point water body when producers (plants) show the light compensation level. The point indicate the rate of photosynthesis equalising with the rate of respiration.
  2. Limnetic zone It is the major open area of waterbody next to the littoral zone. The O2 availability to the organisms living in this area is from the photosynthetic activity of phytoplanktons and in the atmosphere immediately over the water or lake’s surface.
    In the limnetic zone, the region which gets the maximum light above the light compensation point is called euphotic zone.
    The region where light is received less and diffused and is below the light compensation point is disphotic zone.
  3. Profundal zone This is the bottom area of the pond, where respiration is greater than production.
  4. Benthic zone It is the zone of darkness. No light, penetrate in the deeper layers of waterbody. The organism living here in this zone have poorly developed eye sight. The producers are mainly chemosynthetic bacteria.
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 15
    Zonation in deep lake showing gradient of light and oxygen

2. Temperature
It is the most ecologically significant environmental factor. It varies seasonally on land and decreases progressively from the equator towards the poles and from plains to the mountain tops. It ranges from sub-zero levels in polar areas and high altitudes to > 50°C in tropical deserts in summer. There are also certain unique habitats such as thermal springs, deep sea hydrothermal vents where the average temperature exceeds 100°C.

Physiological functions as well as geographical distribution of plants and animals are governed by the temperature conditions and their thermal tolerance. Organisms which can tolerate and thrive in a wide range of temperatures are called eurythermal, e.g., most mammals and birds while organisms which can tolerate a narrow range of temperatures are called stenothermal, e.g., polar bear, amphibians.

3. Precipitation
Rainfall and hail storm are two major forms of precipitation. The only available form of water to the plant is the soil water. So, precipitation is indirect form of water which affect the plant life and water content of ‘ the soil. The vegetation of an area is directly or indirectly affected by the precipitation received by that area.

4. Soil
The nature and properties of soil in different places vary significantly. It is dependent mainly on the following factors

  1. Climate
  2. Weathering process
  3. Whether soil is transported or sedimentary
  4. Soil development process

Water holding capacity and percolation of the soil is determined by its various characteristics, such as soil composition, soil particle size and aggregation of soil particles.

These characteristics of soil along with its pH, mineral composition, topography, etc., determine the type of plants that can grow in a particular habitat and the type of animals that can feed on them. In aquatic environment also, the bottom sediments and its characteristics determine the type of benthic animals that can live there. Thus, the key abiotic factors affecting the organisms.

Besides these prominent abiotic factors, biotic factors are also forming a habitat and are the part of ecological environment.

Question 5.
What are the various adaptations different plants adapt for their survival in different habitats?
Answer:
Adaptations in Hydrophytes
Adaptations in hydrophytes can be discussed under three headings, i.e. morphological, anatomical and physiological.
1. Morphological Adaptations
Hydrophytes show various kinds of structural adaptations in their roots, stems and leaves.
(i) Root

  • Roots may be entirely absent, e.g. Woljfia, Salvinia or poorly developed, e.g. Hydrilla.
  • Roots are well-developed with distinct root caps, e.g. Ranunculus (emergent hydrophytes), aerenchymapresent.
  • In Eichhornia root caps are replaced by root pockets.
  • Some plants, i.e. Jussiaea have two types of roots, one is normal type and other is spongy and negatively geo trophic.

(ii) Stem

  • In Hydrilla, Potamogeton, the stems are slender spongy flexible.
  • Horizontal stems are found in floating hydrophytes like Azolla, Pistia or Eichhornia.
  • In rooted hydrophytes like Sagittaria, Cyperus, Scirpus, the stem is rhizome or stolon.
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 7
    Stolon stems in marshy plants

(iii) Petioles
Some hydrophytes show special features in the petioles.

  • Petioles in submerged plants, with free-floating leaves like Nymphaea and Nelumbium, are long, slender and spongy.
  • In the free-floating hydrophyte Eichhornia, stems are long slender and spongy.
  • In the free-floating hydrophyte Eichhornia, the petiole is swollen have big air spaces in side tissues and helps in floating.
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 8
    TS of petiole of Eichhornia

(iv) Leaves
The structural details of hydrophytes show number of variations, which can be summerised as follows

  • In Vallisneria the leaves are long and narrow.
  • These are finely dissected in Utricularia, Myriophyllum and Ceratophyllum. This helps aquatic plants to provide little resistance against water.
  • The free-floating hydrophytes have wax coating on than, these are shiny and smooth. Wax coating present dessication of leaves in water scarce condition.
  • In Nelumbium and Nymphaea the leaves remain in touch with water surface and upper layer is exposed to the air.
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 9

The amphibions hydrophytes exhibit the phenomenon of heterophylly (two types of leaves). The submerged leaves are dissected to go with’water currents while above water are broad, e.g. Sagittaria, Ranunculus, Limnophylla heterophylla.

2. Anatomical Adaptations
Hydrophytes show the following anatomical features
(i) Reduction in Protecting Structures

  • Cuticle is absent in submerged portion.
  • Epidermis has chloroplast, used as photosynthesising organ.
  • Hypodermis is poorly developed.

(ii) Reduction of Mechanical Tissue

  • Sclerenchyma is absent or poorly developed in submerged portions.
  • Asterosclereids is present that provide mechanical support in case sclerenchyma is absent.
  • Sclerenchyma present only in aerial tissues.

(iii) Reduction of Conducting Tissue

  • Vascular bundles are reduced to few or even one, e.g Hydrilla.
  • Xylem cells are very few.
  • Phloem is usually poorly developed, but in some cases it is well-developed.
  • Secondary vascular tissue is totally absent.

(iv) Increase in Aeration

  • Stomata are totally absent or poorly developed in submerged parts.
  • If present, stomata are confined to upper surface leaves.
  • In amphibious plants, stomata are scattered on the aerial portions.
  • Roots, stems and leaves of most hydrophytes have parenchymatous tissue with air chambers. These chambers store gases like CO2 and O2 and help in respiration and photosynthesis. These are hence, called aerenchyma. Besides, the air chambers help in buoyancy and provide mechanical support.
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 10
    TS of root of Typha

3. Physiological Adaptations
Besides their adaptations in the morphological and anatomical characters, hydrophytes also show physiological adaptations.

  • Osmotic concentrations of cell sap is low.
  • Photosynthetic and respiratory gases are retained in air chambers for further use.
  • No transpiration occurs in submerged plants.
  • In hydrophytes, mostly vegetative reproduction.

Xerophytes:
These are the plants which are adapted to drier regions and have high rate of transpiration than absorption of water.

Types of Xerophytes
Xeric habitats are of two types
(i) Physically dry habitats are those in which water cannot be retained (deserts, rock surface).
(ii) Physiologically dry habitats have plants of water, but the water is not available to the plant.

Based on their adaptation to water scarcity or drought conditions, xerophytes are of three types

  1. Drought resistant plants are such that they can survive in extreme conditions, drought enduring plants can tolerate drought though they may hot have adaptation.
  2. Drought enduring plants these do not have distinct adaptation.
  3. Drought escaping plants these are short lived plants, complete the life cycle before the arrival of dry condition, e.g. Artemisia, Astragalus.
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 11
    Succulents xerophytes

Based on their capacity to store water, xerophytes are classified as succulents and non-succulents. Succulents like Opuntia, have their organs swollen due to accumulation of water, whereas non-succulents are considered as true xerophytes.
CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 12

Adaptation in Xerophytes:
These show varied adaptions in the morphology, anatomy and physiology which are as follows
1. Morphological Adaptations
Xerophytes exhibit a number of special features in their body organs as given below

(i) Roots

  • Roots are very extensive, long tap roots, with branching spread over wide areas.
  • Root hairs and root caps are very well-developed.

(ii) Stem

  • Stems are stunted, woody, dry, hard and covered with thick waxy cuticle.
  • In Opuntia stem becomes green and fleshy (phylloclade).
  • On stems and leaves, there are hairs and waxy coatings.
  • Succulents have their stem modified into leaf-like structures called cladodes as in Asparagus.

(iii) Leaves
The leaves of xerophytes are reduced to spines to various structures. This helps plants to reduce rate of respiration. The following types of leaf conditions are seen

  • Microphyllous when leaves are scaly (Casuarina) and needle-like Pinus.
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 13
  • Trichophyllous when leaves are covered with hairs (.Nerium, Calotropis).
    CHSE Odisha Class 12 Biology Solutions Chapter 14 Organisms and Environment 14
  • Macrophyllous when leaves are soft and flesh (Begonia).
  • Sclerophyllous plants showing leaves which are tough and hard.
  • Cadueous when leaves fall early, i.e. plants with no leaves.
  • Rolling leaves in ammophilis stomata are directed in words.

Anatomical Adaptations:
These adaptations can be conveniently discussed under the headings, i.e. epidermis, hypodermis and vascular tissue.
(i) Epidermis

  • Some xerophytes have multiple epidermis, e.g. Nerium.
  • It has thick cuticle and deposition of waxes, resins, etc.
  • Epidermal hairs are present in grooves.
  • Some leaves have bulliform cells that help in rolling.
  • The stomata are present in sunken pits to reduce transpiration rate.
  • Stomatal frequency is very low in xerophytes.

(ii) Hypodermis
It is thick and well-developed and is made up of parenchymatous cells.

(iii) Ground Tissue

  • In stems, there is abundant mechanical tissue in the form of sclerenchyma, e.g. Casuarina.
  • Since, leaves are reduced, the stems usually have chlorenchyma.
  • In succulent plants, cortex is filled with water, mucilage, latex, etc.
  • In plants that have leaves, palisade parenchyma is well-developed.
  • In Pinus, mesophyll cells are modified.
  • Intercellular spaces are greatly reduced.

(iv) Conducting Tissue
Vascular tissue (xylem and phloem) .is very well-developed in xerophytes.

3. Physiological Adaptations

  • Osmotic concentration of the cell sap is very high.
  • Succulents have high pentosan (chemicals derived from polysaccharides) resulting in accumulation of water.

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 13 Question Answer Applications of Biotechnology

Applications of Biotechnology Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Multiple Choice Questions

Question 1.
Golden rice is produced by rice plant having a transgene encoding an enzyme in biosynthetic pathway of
(a) P-carotene
(b) luciferin
(c) glyphosate
(d) Bt protein
Answer:
(a) P-carotene

Question 2.
Fruit ripening is delayed by preventing the expression of the enzyme
(a) luciferase
(b) polygalacturonase
(c) nitrogenase
(d) adenosine deaminase
Answer:
(b) polygalacturonase

Question 3.
Humulin is manufactured by
(a) Pfizer
(b) Hoechst
(c) Eli Lilly
(d) Aventis
Answer:
(c) Eli Lilly

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Question 4.
Genetic correction of inflicted cells is made in vitro and then reimplanted into its natural environment. This therapy is known as
(a) ex vivo gene therapy
(b) in vivo therapy
(c) in vitro therapy
(d) in toto therapy
Answer:
(a) ex vivo gene therapy

Question 5.
The first genetic disorder treated by gene replacement therapy is
(a) Familial Hypercholesterolemia (FH)
(b) Cystic Fibrosis (CF)
(c) Duchenne Muscular Dystrophy (DMD)
(d) Severe Combined Immunodeficiency (SCID)
Answer:
(d) Severe Combined Immunodeficiency (SCID)

Question 6.
Patent is not granted for
(a) a novel invention
(b) an invention having an industrial application
(c) a discovery made by previously existing knowledge
(d) an invention having an inventive step
Answer:
(c) a discovery made by previously existing knowledge

Question 7.
Which of the following is not related to biosafety?
(a) Convention on Biological Diversity
(b) Cartagena Protocol
(c) World Trade Organisation
(d) UNICEF
Answer:
(d) UNICEF

Question 8.
Which of the following patent cases, India is not directly or indirectly connected with?
(a) Soybean patent case
(b) Neem patent case
(b) Turmeric patent case
(d) Chakraborty patent case
Answer:
(a) Soybean patent case

Question 9.
The supermouse is a genetically modified animal with
(a) insulin transgene
(b) lipid biosynthesis transgene
(c) growth hormone transgene
(d) steroid hormone transgene
Answer:
(c) growth hormone transgene

Question 10.
Which is the nodal centre for Indian biosafety network?
(a) Department of Biotechnology
(b) Department of Science and Technology
(c) Indian Agricultural Research Institute
(d) Department of Forest and Environment
Answer:
(a) Department of Biotechnology

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Fill in the blanks

Question 1.
The mass of undifferentiated plant cells in a plant tissue culture media is known as ……………..
Answer:
Callus

Question 2.
Herbicide resistant plants are generated by plant tissue culture technique by transferring …………… gene of a bacterium into a plant protoplast.
Answer:
glyphosate

Question 3.
A bacterium species of ……………. genus is genetically engineered to prevent frost formation in plants.
Answer:.
Pseudomonas

Question 4.
A bioluminescent plant is generated by transferring ………….. gene of a firefly into plant protoplasts.
Answer:
luciferase

Question 5.
Golden rice producing plant is a transgenic plant, whose cells have a transgene encoding …………
Answer:
ß-carotene

Question 6.
Delayed ripening in tomato is due to the inhibition of expression of an enzyme ……………..
Answer:
polygalacturonase

Question 7.
The first recombinant human vaccine produced and marketed is …………. vaccine.
Answer:
hepatitis-B

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Question 8.
Recombinant insulin in the trade name of Humulin is manufactured by …………….
Answer:
Eli Lilly Corporation

Question 9.
Monoclonal antibody is synthesised and secreted by a cell known as ……….
Answer:
B-lymphocytes

Question 10.
Severe Combined Immunodeficiency (SCID) is expressed due to the absence of an enzyme, …………… .
Answer:
adenosine deaminase

Question 11.
A forensic analysis of DNA for establishing the identity of a person is known as …………….. .
Answer:
DNA fingerprinting

Question 12.
An immunological technique, applied to detect the presence of very minute quantity of antigen in the serum is known as …………. .
Answer:
ELISA

Question 13.
A biopesticide, known as Bt protein is expressed by a bacterial species, ………… .
Answer:
Bacillus thuringiensis

Question 14.
A legal right, privilege and authority granted to a person for a limited period for an invention is known as …………. .
Answer:
patent

Question 15.
The use of novel biological resource of a sovereign country without its due permission is known as …………….. .
Answer:
biopiracy

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Express in one or two word(s)

Question 1.
The tomato plant variety that bears tomatoes exhibiting delayed ripening.
Answer:
Flaw savr

Question 2.
The somatic hybrid cell, which produces monoclonal antibodies.
Answer:
Hybridoma

Question 3.
Genetically engineered rice, rich in vitamin-A.
Answer:
Golden rice

Question 4.
An insecticidal protein, produced by Bacillus thuringiensis.
Answer:
Cry protein

Question 5.
A broad spectrum herbicide that is used world over.
Answer:
Glyphosate

Question 6.
The biotech company, which commercially manufactured the first recombinant human insulin.
Answer:
Genentech

Question 7.
The first genetic disorder that was treated by gene therapy.
Answer:
SCID

Question 8.
The gene transfer into the mammalian fertilised egg with a micropipette.
Answer:
Microinjection

Question 9.
The gene transfer method practiced by passing intermittent pulses of electric current through the medium containing plant protoplasts.
Answer:
Electroporation

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Question 10.
The corn was genetically engineered by transferring Bt protein gene into plant protoplasts. The brand was marketed and later was withdrawn due to safety reasons.
Answer:
Star Link corn

Question 11.
The biosafety protocol that was drafted in 1995 and adopted in 2000.
Answer:
Cartagena Protocol

Short Answer Type Questions

Answer each of the following within 50 words

Question 1.
What is golden rice?
Answer:
Golden rice is a transgenic variety of rice with an elevated level of ß-carotene (provitamin-A), a precursor of vitamin-A. The genes encoding the enzymes of the ß-carotene biosynthetic pathway are introduced into rice plant cells in culture. The transgenic rice plants generated produce rice with ß-carotene.

Question 2.
What is Flavr Savr tomato?
Ans.
Fruit ripening in tomato and other fruits and vegetables are delayed by manipulating a gene, involved in softening and ripening. A variety of tomato plant has been successfully engineered, which bears tomatoes, known as Flavr Savr tomatoes. This variety exhibits delayed ripening.

Question 3.
What does the recombinant hepatitis-B vaccine contain?
Answer:
Hepatitis-B vaccine contains surface antigen proteins (HBs Ag) extracted from Hepatitis-B Virus (HBV).

Question 4.
What do you understand by ex vivo gene therapy?
Answer:
Ex vivo gene therapy The affected cells are removed from the body and transformed by the remedial gene in vitro. The transformed cells are grown in a cell culture medium to a sufficient number and then returned to the body by transfusion or transplantation.

Question 5.
What do you mean by a biopesticide? Give an example.
Answer:
Biopesticides are the type of pesticides produced from an organism. They are equally potent but do not inflict a damage on the environment, e.g. a species of bacteria with insecticidal properties is Bacillus thuringiensis. It produces insecticidal cry protein or Bt protein.

Question 6.
What is a supermouse?
Answer:
R L Brinster and R Palmiter (1982) successfully created the first transgenic mouse by transferring the rat growth hormone gene into the fertilised mouse egg by microinjection. This act was carried out in vitro.

Following the transfer, the fertilised egg was implanted into the uterus of a pseudopregnant mouse.
The mouse gave birth to mice that were relatively larger in size, possibly due to an increased synthesis of growth hormone directed by the rat growth homone transgene. This mouse was called supermouse because of its abnormal growth.

Question 7.
Explain biopiracy.
Answer:
Illegal transfer of biological resources has been termed as biopiracy. It describes a practice, in which indigenous knowedge and processes used by indigenous people of a region is used by others for profit without permission from and with little or no compensation or recognition to the indigenous people themselves. This is an illegal practice and enforceable in the court of law.

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Question 8.
Ennumerate and explain in brief two biosafety issues, biotechnology is confronted with.
Answer:
Biosafety in a broad sense, refers to the prevention of loss of biological integrity of biological processes and products, harvested by using living organisms. For example, recombinant insulin was manufactured in a complex biological process putting in thought, knowledge, skill and execution method of the inventor. Secondly, a lot of energy and money was spent in the successful execution of the process. Therefore, the right of the inventor needs to be protected by law considering it as a property.

On the other hand, insulin that is manufactured a prescribed trial process to prove that it is suitable for human use. Another potential hazard was the release of Genetically Modified Organisms (GMOs) into the wild. There was a threat that it might sexually reproduce with organisms of its own species and exchange genes, consequently changing the structure of the gene pool.
This might have produced an adverse effect on organic evolution. Thus, normal biological diversity might be destabilised.

Question 9.
Describe the evolution of Indian Patent Act.
Answer:
India enacted the Patent Act in 1970. The Act has undergone amendments in 1999, 2002, 2005 and 2006. The headquarter for the same is in KolKata, West Bengal. The nodal centre for Indian biosafety network is the Department of Biotechnology, Government of India.

Question 10.
Describe the neem patent case.
Answer:
Neem Patent Case:
The multinational agribusiness company, WR Grace of New York and United States Department of Agriculture, Washington DC filed, for a European patent for the fungicidal use of neems oil in the European Patent Office (EPO). It was stated that neem oil controlled fungal growth on plants.

The plea was accepted and a patent was granted. However, following the publication, Dr. Vandana Shiva of Research Foundation for Science and Technology and Natural Resource Policy, New Delhi and others filed a legal opposition to the grant of patent in the EPO.

Write brief notes on the following

Question 1.
Flerbicide resistant plants
Answer:
Herbicide resistant plants Herbicide resistant transgenic plants are generated by transferring bacterial herbicide resistant genes into plant cells grown in culture. Glyphosate is the most widely used broad-spectrum herbicide world over.
A glyphosate resistant gene from Petunia plant is transferred into isolated plant cell§ in culture and glyphosate resistant plants are generated.

Question 2.
Humulin
Answer:
Genentech is the first biotech company to manufacture recombinant human insulin in 1978 using bacteriophage vector and E. coli as the cloning and expression host cell. Later, this technology was licenced to Eli Lilly Corporation of USA.

The recombinant human insulin was termed as Humulin. It was approved as the first recombinant drug by the Food and Drug Administration (FDA), USA for human use. Since then, several companies all over the world have been manufacturing recombinant human insulin on a commercial basis.

Some noteworthy companies are Novo Nordisk of Denmark; Hoechst and Aventis of Germany and Pfizer of USA. Wokhardt Limited, a pharmaceutical company has been manufacturing human insulin in India under the trade name of Wosulin.

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Question 3.
Recombinant vaccine
Answer:
Recombinant Vaccines
An antigcnic agent, which after being administered into an animal, generates an active acquired immune response is called as vaccines. The antigenic agent involved in vaccine production varies from vaccine to vaccine.

They generally belong to three classes-attenuated (inactivated) whole organisms isolated antigenic proteins such as coat proteins of viruses and inactivated touns. The latter two fall under the subunit vaccine class, wherein part of the organism possessing antigenic property is used in the vaccine production.

In these vaccines, a DNA insert encoding an antigen (like bacterial surface proteins) is introduced into a less virulanc host. These elicit an immune response expressing the antigens but do cause infection. The expressed antigens are isolated and puri&d and injectedinto the human hosts as a vaccine. It is called recombinant vaccine.

Question 4.
Gene therapy
Answer:
Gene therapy is a therapy or treatment of a gene, which has been mutated. It is a therapy to correct the damage. A genetic disorder is expressed, when a particular gene is mutated.

The mutant gene, as it is known, encodes a different polypeptide other than a normal. This polypeptide is the root cause of the expression of symptoms of a genetic disorder.
Attempts have been made to rectify or replace the mutant gens, so that they express normally. This replacement process is called gene therapy.

Question 5.
Biopesticide
Answer:
Bacillus thuringiensis is known to produce endotoxins or insecticidal crystalline protein or Cry protein or Bt protein.
Bt protein is hydrolysed by an alkali into 250 kD (kilodalton) units, known as protoxins. Each protoxin consists of two 130 kD polypeptides. The 130 kD polypeptide is digested into a 68 kD toxin polypeptide in an alkaline pH. When catepillars eat the leaves of crop plants, on which the bacterial spores are deposited, they ingest the spores.

The spores germinate in the alimentary canal, the bacteria grow in size and produce Bt protein. This protein is digested into 68kD toxin polypeptides in the intestine of the larva. The action of the poypeptide, eventually kills the larva. The alimentary canal of mammals, including human, produces an acid, which degrades the Bt protein. Thus, it is apparently harmless to human and other mammals. Since, this pesticide is produced from an organism, it has been identified as biopesticide.

Question 6.
Transgenic animals
Answer:
Transgenic animals are those which can grow faster, yield more milk, lay bigger eggs and so on. They are produced by combining traditional breeding with gene technology, which yield encouraging results. It involves selecting, isolating, purifying and transferring beneficial genes of one species to another to harvest a beneficial effect. These animals are the transfer of the beneficial gene (genes) developed. The gene that is transferred is known as a transgene. The transgene is transferred by using one of the several methods of gene transfer in practice. Microinjection is found to be most suitable for animal cells.

CHSE Odisha Class 12 Biology Solutions Chapter 13 Applications of Biotechnology

Question 7.
Patent
Or What is patent?
Answer:
Patent is an open letter granting legal right, privilege and authority by a sovereign state to a person or an institution for a limited period of time. It is given for an invention using scientific and technical knowledge. All sovereign countries have enacted their own Patent Acts to regulate the use of such properties.

An invention involves new knowledge, while a discovery is an application of the knowledge. For example, the double helical model proposed by Watson and Crick was a discovery and hence, does not qualify to be patented, while new forms of DNA, such as recombinant DNAs have been patented.

Question 8.
Biopiracy
Answer:
Biopiracy is theft or robbery or exploitation of biological and genetic resources indigenous to a country. These biological resources are often the main targets of enterprising businessmen because of their many uses in agriculture, healthcare and chemical industries. The process of biopiracy involves collection of samples of biological resources, which then undergo product development for their use on a commercial scale.

Biopiracy begins with biodiversity prospecting, which is exploration of wild plants and animals for commercially viable genetic and biochemical resources. Genetic resources are the genes found in plants and animals that are of actual or potential value to people.

Through the use of new biotechnologies, genes/germplasm from any plant or animal can be transferred to another. Such genetically engineered organisms (plants, animals and microorganisms) are being used for new industrial applications, pharmaceuticals, farming, cattle breeding and poultry farming.

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 12 Question Answer Principles and Processes of Biotechnology

Principles and Processes of Biotechnology Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Multiple choice questions

Question 1.
The double helical structure of DNA was proposed by
(a) Jacob and Monod
(b) Sanger and Gilbert
(c) Watson and Crick
(d) Beadle and Tatum
Answer:
(c) Watson and Crick

Question 2.
Polymerase chain reaction was discovered by
(a) H G Khorana
(b) K Mullis
(c) R Holley
(d) M Nirenberg
Answer:
(b) K Mullis

Question 3.
Exonuclease is an enzyme that
(a) makes internal cuts in polynucleotide
(b) polymerises nucleotides
(c) joins two polynucleotide fragments
(d) removes nucleotides from the termini one after another
Answer:
(d) removes nucleotides from the termini one after another

Question 4.
DNA ligase is commonly known as
(a) molecular scissors
(b) molecular marker
(c) molecular glue
(d) molecular probe
Answer:
(c) molecular glue

Question 5.
During electrophoresis, DNA fragments move from
(a) anode to cathode
(b) remain static
(c) move randomly
(d) cathode to anode
Answer:
(d) cathode to anode

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Question 6.
The blotting of protein molecules to a nylon membrane is known as
(a) Southern blotting
(b) Western blotting
(c) Northern blotting
(d) Eastern blotting
Answer:
(b) Western blotting

Question 7.
Detection of a desired DNA fragment by using radioactive emission is known as
(a) hybridisation
(b) denaturation
(c) autoradiography
(d) electrophoresis
Answer:
(c) autoradiography

Question 8.
Choose the incorrect statement.
(a) A plasmid is small, double-stranded circular DNA
(b) A plasmid contains an origin of replication
(c) A plasmid has several restriction sites
(d) A plasmid has telomeres
Answer:
(d) A plasmid has telomeres

Question 9.
A cosmid is a
(a) plasmid phage hybrid vector
(b) DNA bacteriophage vector
(c) expression vector
(d) viral vector
Answer:
(a) plasmid phage hybrid vector

Question 10.
The example of a plant cell compatible vector is
(a) fertility plasmid
(b) colicinogenic plasmid
(c) tumour inducing plasmid
(d) resistance plasmid
Answer:
(c) tumour inducing plasmid

Question 11.
Amplification of DNA by PCR uses a DNA polymerase called
(a) Taq DNA polymerase
(b) RNA polymerase
(c) DNA polymerase-III
(d) Reverse transcriptase
Answer:
(a) Taq DNA polymerase

Fill in the blanks

Question 1.
The phenomenon of fermentation was demonstrated by ……………
Answer:
Louis Pasteur

Question 2.
The word ‘biotechnology’ was coined by …………..
Answer:
Karl Ereky

Question 3.
Class II restriction endonucleases (enzymes) recognise specific nucleotide sequence in DNA called ……………
Answer:
palindromic sequence

Question 4.
Cohesive ends in the DNA fragments are generated by …………… cutting.
Answer:
staggered

Question 5.
The anionic detergent, used in polyacrylamide gel electrophoresis is known as
Answer:
Sodium Dodecyl Sulphate (SDS)

Question 6.
The transfer of separated protein molecules from the gel into a nylon membrane is known as ………….
Answer:
Western blotting

Question 7.
Detection of desired DNA fragment by the emission of ionising radiation is known as ……….
Answer:
autoradiography

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Question 8.
The conjoint structure formed by the joining of the vector DNA and the target DNA fragment is known as …………..
Answer:
recombinant DNA

Question 9.
The uptake of the recombinant DNA by the bacterial host cell is known as ………….
Answer:
transformation

Question 10.
The delivery of a foreign DNA fragment into the fertilised egg with a micropipette is known as …………
Answer:
microinjection

Express in one or two word(s)

1. The technique of separation of DNA fragments based on their molecular weight and electrical charge.
Answer:
Electrophoresis

2. The restriction endonuclease isolated from Escherichia coli.
Answer:
Eco RI

3. The DNA digesting enzyme that removes nucleotides from the termini.
Answer:
Exonucleases

4. The enzyme that catalyses the synthesis of RNA on a DNA template.
Answer:
RNA polymerases

5. The enzyme that catalyses the replication of DNA.
Answer:
DNA polymerases

6. The enzyme that catalyses the synthesis of a complementary DNA strand on an RNA template.
Answer:
Reverse transcriptase

7. The fluorescent dye used in agarose gel electrophoresis.
Answer:
Ethidium bromide

8. Transfer of DNA fragments from the agarose gel to a nylon membrane.
Answer:
Southern blotting

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

9. Breaking of hydrogen bonds in a duplex so as to make it single-stranded.
Answer:
Denaturation

10. The DNA that helps carry the target DNA fragment to the host cell for cloning.
Answer:
Vector

11. The plasmid phage hybrid cloning vector.
Answer:
Cosmid

12. A plant cell, whose cellulose cell wall is digested.
Answer:
Protoplast

13. The plasmid present in Agrobacterium tumefaciens.
Answer:
Tumour inducing (Ti) plasmid

14. Transfer of a DNA fragment into a host cell in a medium by passing brief pulses of electric current through the medium.
Answer:
Electroporation

15. The instrument used in PCR amplification.
Answer:
Thermocycler

Match the words of group ‘A with those of group ‘B’ to make meaningful pairs

Question 1.

Group-A Group-B
Restriction endonuclease End modifying enzyme
DNA ligase RNA dependent DNA polymerase
Exonuclease Molecular scissors
Reverse transcriptase Removes nucleotides from both ends
RNA polymerase DNA dependent DNA synthesis
DNA polymerase Molecular glue
Alkaline phosphatase DNA dependent RNA synthesis

Answer:

Group-A Group-B
Restriction endonuclease Molecular scissors
DNA ligase Molecular glue
Exonuclease Removes nucleotides from both ends
Reverse transcriptase RNA dependent DNA polymerase
RNA polymerase DNA dependent RNA synthesis
DNA polymerase DNA dependent DNA synthesis
Alkaline phosphatase End modifying enzymes

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Question 2.

Group-A Group-B
Polyacrylamide Agrobacterium tumefaciens
Southern blotting Thermocycler
Plasmid Protein electrophoresis
Agarose gel Denaturation
Cos Taq polymerase
Tumour inducing plasmid Cloning vehicle
Recombinant DNA Molecular marker
DNA coated tungsten particles Nucleic acid electrophoresis
insertional inactivation Chimeric DNA
Breaking of interchain hydrogen bonds Particle gun
Equipment for PCR amplification Screening of clones
Ampicillin resistance gene Blotting of DNA
Thermophilus aquaticus Cohesive site

Answer:

Group-A Group-B
Polyacrylamide Protein electrophoresis
Southern blotting Nucleic acid electrophoresis
Plasmid Cloning vehicle
Agarose gel Blotting of DNA
Cos Cohesive site
Tumour inducing plasmid Agrobacterium tumetaciens
Recombinant DNA Chimeric DNA
DNA coated tungsten particles Particle gun
insertional inactivation Screening of clones
Breaking of interchain hydrogen bonds Denaturation
Equipment for PCR amplification Thermocycler
Ampicillin resistance gene Molecular marker
Thermophilus aquaticus Taq polymerase

Short Answer Type Questions

Answer each of the following in 50 words

Question 1.
Define biotechnology.
Answer:
‘Biotechnology is an application of knowledge and technique of biochemistry, microbiology, genetics, immunology, tissue and cell culture, molecular biology, chemical engineering and computer science to living systems or parts therefore, for harvesting beneficial products and/or services for mankind.

Question 2.
Define gene cloning.
Answer:
The process of making a number of identical copies of a beneficial gene is known as gene cloning. Microorganisms, especially bacteria are chosen as host cells for this work, which provide a suitable environment for replication (amplification) of genes.

The gene is introduced into a bacterium by a process, known as transformation. For transformation, a gene is delivered into a bacterial host conjointly with a carrier, also DNA, known as a vector.

Question 3.
What is a restriction endonuclease (restriction enzyme)? Why is the word restriction used to designate these?
Answer:
These enzymes recognise specific recognition sequences on DNA and make the cut either within that sequence only or at a variable distance from that sequence. The site where they cleave or cut the DNA is called recognition site.
In rDNA technology, class-II REs are used.

Question 4.
Describe two types of cutting of DNA, executed by restriction endonucleases.
Answer:
Restriction Endonucleases
Each restriction endonuclease functions by ‘inspecting’ the length of a DNA sequence.
Once, it finds its specific recognition sequence, it will bind to the DNA and cut each of the two strands of the double helix at specific points in their sugar phosphate backbones.

Each restriction endonuclease recognises a specific palindromic nucleotide sequences in the DNA. Palindromes are group of letters that form the same words when read both forward and backward, e.g., ‘MALAYALAM’. Therefore, the palindrome in DNA is a base pair sequence that is the same when read forward or background, e.g., the following sequences read the same on the two strands in 5′ → 3′ direction as well as 3′ → 5′ direction.
5′- GAATTC – 3′
3′ – CTTAAG – 5′
Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands.

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Question 5.
What is electrophoresis? How many types of electrophoresis you have studied?
Answer:
The cutting of DNA by restriction endonucleases results in the generation of DNA fragments which are then separated by a technique known as gel electrophoresis.
CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology 1
The DNA fragments are resolved (separate) according to their size through sieving effect provided by the gel when the potential difference is applied.
Two types of gels used in molecular separation are

  • Agarose Used for nucleic acids (RNA and DNA).
  • Polyacrylamide Used for proteins.

Question 6.
What is a palindrome? Give an example.
Answer:
Each restriction endonuclease recognises a specific palindromic nucleotide sequences in the DNA. Palindromes are group of letters that form the same words when read both forward and backward, e.gy‘MALAYALAM’. Therefore, the palindrome in DNA is a base pair sequence that is the same when read forward or background, e.g., the following sequences read the same on the two strands in 5′ → 3′ direction as well as 3′ →5′ direction.
5′- GAATTC – 3′
3′ – CTTAAG – 5′

Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands.

Question 7.
What is a polymerase? How-many types of polymerases you have studied?
Answer:
Polymerases
These enzymes catalyse the process of copying the nucleic acid molecules. These are of following types
(i) RNA polymerase These are DNA dependent RNA polymerase.
(ii) DNA polymerase They copy DNA strand into another complementary strand through replication. These are also DNA dependent enzymes.
(iii) Reverse transcriptase They catalyse the synthesis of complementary DNA strand on RNA template. These are found in retroviruses and are RNA dependent DNA polymerase.
(iv) RNA polymerase They catalyse the process of strand copying of DNA into RNA through transcription.

Question 8.
Why is DNA ligase called molecular glue?
Answer:
DNA Ligases
These enzymes help to join or seal 3′-OH end of one nucleic acid fragment with the 5′-P end of another nucleic acid fragment by forming a phosphodiester bond. As they are able to stick DNA fragments together, they are known as molecular glue. The widely used ligase enzyme is T4 DNA ligase. It is purified from E. coli cells that are infected by T4 bacteriophages.

Question 9.
What is Southern blotting?
Answer:
Southern Blotting:
When the blotting is applied to DNA it is known as Southern blotting (named so after its discover EM Southern in 1975).
Following this, complementary DNA or RNA sequence (probe) labelled with 32 P (radioactive phosphorus) are, hybridised.
The probe binds its complementary target DNA sequence, by forming hydrogen bonds and a duplex is formed. This process is known as molecular hybridisation. The procedure that combines Southern blotting and molecular hybridisation is known as Southern blot hybridisation. After this, the target DNA sequence is identified using autoradiography.
CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology 2

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Question 10.
Why is SDS used in polyacrylamide gel electrophoresis?
Answer:
A protein molecule contains many amino acids and all the amino acids do not have similar acid-base properties. Hence, a protein molecule cannot be conferred with uniform negative charges in a buffer solution.
It is therefore, treated with Sodium Dodecyl Sulphate (SDS), an anionic detergent that coats a protein molecule and confers negative charges uniformly.

Question 11.
What is autoradiography?
Answer:
Autoradiography:
The separated fragments in the gel are treated with an alkali to render the double-stranded fragments into single-stranded. This phenomenon is known as denaturation. These single-stranded fragments are then transferred onto a nitrocellulose filter paper or nylon membrane by a process known as blotting.

Question 12.
Enumerate the features of a suitable cloning plasmid.
Answer:
Plasmids:
These are small, autonomously replicating usually circular, extrachromosomal double-stranded DNA molecules that occur in many bacteria and some yeasts, but naturally occurring plasmids are not suitable for gene cloning. Therefore, they are genetically engineered so to make them compatible for cloning. These are then called as cloning vectors and, they possess the following additional properties other than these found in an ideal vector

  • They are non-conjugative.
  • They have relaxed replication and a high copy number.
  • They should be able to clone a gene in the range of 5-10Kb

The most commonly used cloning vector is pBR322 where p = plasmid; BR = Name of its engineers
[B = Boliver and R= Rodriguez]; 322 = Number of relevant workers who worked out this plasmid. pBR322 has the following features

• It is small, ds circular DNA molecule having ori.
• It has two antibiotic marker genes namely, ampicillin resistance gene ampR and tetracyclin resistance gene (tetR). In between these two genes, Eco RI restriction site is present.
• ampR gene contains Pvu I and Pst I restriction sites.
• tetR gene contains Bam HI and Sal I restriction sites.
• The target DNA and the vector DNA fragments should be compatible for a successful RDT process.
CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology 3

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Question 13.
What is a recombinant DNA?
Answer:
The heterogeneous combination of two different DNAs in which the gene of interest is present is known as recombinant DNA or chimeric DNA and the technology is called recombinant DNA technology.

Question 14.
What is microinjection?
Answer:
Microinjection In this method, foreign DNA is directly injected into the nucleus of animal cell-or plant cell by using microneedles or micropip’ettes without any use of a vector DNA. It is used to v deliver a transgene or foreign DNA into the egg. Microinjection is especially used for transferring the foreign DNA in animal cells. The transgene is injected into the male pronuclei because it is larger and traceable with a dissecting microscope.

The male pronucleus is then fused with female pronucleus to produce zygote in vitro. In plant cells, the cellulose cell wall is first removed by enzymatic digestion before micro injection. This process is very tedious and involves a lot of precision.
CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology 3

Question 15.
Describe briefly electroporation.
Answer:
Electroporation It is the formation of temporary pores in the plasma membrane of host cells by using lysozyme or calcium chloride under high voltage of electric current. These pores are used for introduction of foreign DNA.
It can be used to transform plant protoplast, i.e. the plant cell with removed cell wall.

Question 16.
What is Polymerase Chain Reaction (PCR)?
Answer:
Polymerase Chain Reaction (PCR)
It is one of the important technique that serves the purpose of amplification of nucleic acid without using a host cell. PCR was discovered by Kary B Mullis in 1983.

Write brief notes on the following

Question 1.
Restriction endonuclease
Answer:
Restriction endonuclease These are specific enzymes, which recognise specific sequences called recognition sequences on DNA and make double-stranded cuts either within the recognition sequence or at a variable distance from the recognition sequence.

They act as scissors and therefore, often are referred to as molecular scissors. The point of cleavage is known as a restriction site. There will be as many restriction sites as the number of recognition sites. The REs that are used in the recombinant DNA technology fall under class-II. These make double-stranded symmetrical cuts within recognition sequences, generating cohesive or blunt ends.

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Question 2.
DNA Ligases
Answer:
DNA Ligases
These enzymes help to join or seal 3-OH end of one nucleic acid fragment with the 5′-P end of another nucleic acid fragment by forming a phosphodiester bond. As they are able to stick DNA fragments together, they are known as molecular glue. The widely used ligase enzyme is T4 DNA ligase. It is purified from E. coli cells that are infected by T4 bacteriophages.

Question 3.
DNA polymerase
Answer:
DNA polymerase enzyme polymerises the DNA synthesis on DNA template or complementary DNA (cDNA). It was discovered by A. Kornberg and co-workers in E. coli in 1956 in the presence of a preformed DNA template, it produces a parallel strand in the presence of ATP.

Question 4.
Southern blotting
Answer:
When the blotting is applied to DNA it is known as Southern blotting (named so after its discover EM Southern in 1975).
Following this, complementary DNA or RNA sequence (probe) labelled with 32 P (radioactive phosphorus) are, hybridised.
The probe binds its complementary target DNA sequence, by forming hydrogen bonds and a duplex is formed. This process is known as molecular hybridisation. The procedure that combines Southern blotting and molecular hybridisation is known as Southern blot hybridisation. After this, the target DNA sequence is identified using autoradiography.

Question 5.
Agarose gel electrophoresis
Answer:
Agarose Gel Electrophoresis:
The molecules of nucleic acids are given uniform negative charge in alkaline buffer solution. These molecules then migrate towards the anode. The smaller molecules migrate faster than larger molecules because the rate of migration is inversely proportional to the molecular weight of separating molecules and directly proportional to the strength of electric field.

Question 6.
Cloning plasmid
Answer:
Plasmids were discovered by William Hayes and Joshua Lederberg (1952). These are extrachromosomal, self-replicating, usually circular, double-stranded DNA molecules, found naturally in many bacteria and also in some yeast. Although plasmids are usually not essential for normal cell growth and division, they often confer some traits on the host organism. For example, resistance to certain antibiotics or toxins that can be a selective advantage under certain conditions.

The plasmid molecules may be present as 1 or 2 copies or in multiple copies (500-700) inside the host organism. These naturally occurring plasmids have been modified to serve as vectors in the laboratory. The most widely used, versatile, easily manipulated vector pBR 322 is an ideal plasmid vector.

Question 7.
Cosmid
Answer:
Cosmid (cos + plasmid) vectors The term cosmid is a combination of two words. COS + MID. COS is taken from COS site of Lambda phage and MID is taken from plasmid DNA. Cosmid was developed for the first time by Collins and Honn (1978). The simplest cosmid vector contains a plasmid’s origin of replication, a selectable marker, suitable restriction enzyme sites and the lambda ‘cos’ site. Cosmids can be used to clone DNA fragments of upto 45 kb length.

CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology

Question 8.
Recombinant DNA
Answer:
Recombinant DNA is the DNA that has formed artificially by genetic engineering. It contain the gene of interest and it becomes the part of host’s genetic makeup where it further replication. The process of constructing recombinant DNA is called Recombinant DNA Technology (RDT).

Question 9.
Polymerase chain reaction
Answer:
A gene can also be cloned or amplified without the assistance of a host cell by a specific reaction, known as Polymerase Chain Reaction (PCR). PCR was discovered by Kary B. Mullis in 1983. The reaction is carried out in a thermocycler having a thermal cycling (heating and cooling) programme.

The target gene is put along with other substrates in the thermocycler. The DNA fragment, doubles at the end of the first cycle and quadruples after the second cycle and so on. In this amplification process, no host cell is used as the supporting system. It is carried out in three steps-denaturation, primer annealing and extension.

Question 10.
Micro injection
Answer:
Microinjection In this method, foreign DNA is directly injected into the nucleus of animal cell-or plant cell by using microneedles or micropip’ettes without any use of a vector DNA. It is used to v deliver a transgene or foreign DNA into the egg. Microinjection is especially used for transferring the foreign DNA in animal cells. The transgene is injected into the male pronuclei because it is larger and traceable with a dissecting microscope.

The male pronucleus is then fused with female pronucleus to produce zygote in vitro. In plant cells, the cellulose cell wall is first removed by enzymatic digestion before micro injection. This process is very tedious and involves a lot of precision.
CHSE Odisha Class 12 Biology Solutions Chapter 12 Principles and Processes of Biotechnology 3

Long Answer Type Questions

Question 1.
Describe briefly about the making of a recombinant DNA.
Or Describe briefly recombinant DNA technology.
Answer:
Recombinant DNA Technology
Genetic engineering is alternately called as recombinant DNA technology or gene cloning. Paul Berg (1972) was ‘ awarded Nobel Prize in 1980 and is considered as father of genetic engineering. The first recombinant DNA (rDNA) was constructed by Stanley Cohen and Herbert Boyer in 1972. Recombinant DNA technology involves the following steps

1. Isolation of the genetic material (DNA) is carried out as follows
(a) DNA is enclosed within the membranes. To release DNA along with other macromolecules -such as RNA, proteins, polysaccharides and lipids, bacterial cells/plant or animal tissue are treated with enzymes such as lysozyme (bacteria), cellulase (plant cells), chitinase(fungus).
(b) Other molecules can be removed by appropriate treatments and purified DNA ultimately precipitates out as fine threads in the suspension after the addition of chilled ethanol.

2. Cutting of DNA at specific locations is done by using restriction enzymes. The purified DNA is incubated with the specific restriction enzyme at conditions optimum for the enzyme to act.

3. Isolation of desired DNA fragment is carried out using agarose gel electrophoresis.

4. Amplification of gene of interest using Polymerase Chain Reaction (PCR) is a reaction in which multiple copies of specific DNA (gene of interest) sequence are made (amplification) in-vitro.

5. Ligation of DNA fragment into a vector requires a vector DNA and source DNA.
(a) These are cut with endonuclease to obtain sticky ends.
(b) Both are then ligated by mixing vector DNA, gene of interest and enzyme DNA ligase to form recombinant DNA.

6. Insertion of recombinant DNA into the host cell/organism occurs by several methods, before which the recipient cells are made competent to receive the DNA.
(a) If recombinant DNA carrying antibiotic resistance gene (e.g. ampicillin), is transferred into E. coli cells, the host cell is transformed into ampicillin resistant cell.
(b) The ampicillin resistant gene can be called a selectable marker.
(c) When transformed cells are grown on agar plates containing ampicillin, only transformants will grow and others will die.

7. Culturing the host cells The cell containing the foreign gene is cultured on an appropriate medium at optimal conditions. The DNA gets multiplied.

8. Extraction of desired gene product is carried out in the following steps
(a) A protein encoding gene expressed in a heterologous host is called recombinant protein.
(b) Cells having genes of interest can be grown on a small or on a large scale.

CHSE Odisha Class 12 Biology Solutions Chapter 11 Microbes in Human Welfare

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 11 Microbes in Human Welfare Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 11 Question Answer Microbes in Human Welfare

Microbes in Human Welfare Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Multiple choice questions

Question 1.
In curd making, …………. is useful in coagulation of milk protein. (Lactobacillus, Saccharomyces, Penicillium, Aspergillus)
Answer:
Lactobacillus

Question 2.
Antibiotic streptomycin is obtained from ……….. (Streptomyces griseus, S. aureofaciens, S. nouresi, Saccharomyces cerevisiae)
Answer:
Streoptomyces griseus

Question 3.
Citric acid is produced when fermentation is caused by …………. (Lactobacillus, Aspergillus sp., Penicillium sp., Acetobacter sp.)
Answer:
Aspergillus sp.

Question 4.
Lipase enzyme is produced by the activity of …………. (Trichoderma viride, Rhizopus sp., Aspergillus sp., Saccharomyces cerevisiae)
Answer:
Rhizopus sp.

CHSE Odisha Class 12 Biology Solutions Chapter 11 Microbes in Human Welfare

Question 5.
In pest control of crop plants ……………. has pesticidal properties, (baculovirus, papilloma virus, pox virus, Rhizobium)
Answer:
baculovirus

Express in one word only

Question 1.
What is called the process of heating and cooling of milk for inactivation of bacteria?
Answer:
Pasteurisation

Question 2.
What is called the secretions of microorganisms which are toxic to pathogenic bacteria?
Answer:
Antibiotics

Question 3.
What is the commercial name of acetic acid?
Answer:
Vinegar

Question 4.
What is called the accumulated microorganisms and organic matter in the treatment of sewage?
Answer:
Sludge

Question 5.
What is the major component of biogas?
Answer:
Methane

Question 6.
What can be called the natural pest killing agent other than artificial chemical?
Answer:
Biopesticides

Question 7.
What is called the association between Rhizobium in the root system of legumes?
Answer:
Symbiotic association

CHSE Odisha Class 12 Biology Solutions Chapter 11 Microbes in Human Welfare

Correct the sentences, if required, by changing the underlined word(s)

Question 1.
Antibiotic tetracyclin is obtained from Penicillium notatum.
Answer:
penicillin

Question 2.
In biogas, methane is produced due to the activities of nitrogen-fixing bacteria.
Answer:
methanogenic

Question 3.
The first antibiotic extracted from bacterial culture is nystatin.
Answer:
streptomycin

Question 4.
Industrial production of organic acids through microbial cultures is due to the oxidation process by bacteria.
Answer:
fermentation

Question 5.
Acetic acid is produced by Lactobacillus sp.
Answer:
Lactic acid

CHSE Odisha Class 12 Biology Solutions Chapter 11 Microbes in Human Welfare

Fill in the blanks

1. In biogas production ………… bacteria are used.
Answer:
methanogenic

2. BGA used in biological nitrogen-fixation are called ………….. bacteria.
Answer:
cyano

3. Ethanol obtained by ………….. fermentation is used in industry.
Answer:
microbial

4. Acetobacter converts ……………. to vinegar by aerobic fermentation of legumes.
Ans.
ethyl alcohol

Short Answer Type Questions

Write notes on the following with atleast 3 valid points

Question 1.
Biogas
Answer:
It is a complex mixture of gasese like CH4,CO2, H2, etc., that is produced by anaerobic digestion of biomass. It is used as fuel.

Question 2.
Biopesticides
Answer:
These are biodegradable, highly pest specific biological agents that are used to control pest population without harming the environment.

Question 3.
Biofertilisers
Answer:
These are the preparations containing microorganisms that help the plants to uptake various nutrients in utilisable form, e.g. Rhizobium, BGA, etc.

Question 4.
Microbes in industry.
Answer:
Industrial use of microbes includes the production of beverages, antibiotics, etc., that are useful for the human. The large scale production of these products is carried out in bioreactors using the appropriate microbes, e.g. butyric acid is derived from Clostridium and terramycin is derived from Streptomyces rimosus.

CHSE Odisha Class 12 Biology Solutions Chapter 11 Microbes in Human Welfare

Question 5.
Microbes in antibiotics production.
Answer:
The term ‘antibiotics’ was coined by Waksman (1942), and it is derived from the Greek words Anti-against and bios-Yiit, together they mean ‘against life’ (with reference to disease causing organisms). Antibiotics are the chemical substances, produced by some microbes that can kill or retard the growth of other disease causing microbes.

The first antibiotic was obtained from the species of Penicillium notatum in 1928 by Sir Alexander Fleming and it was named Penicillin.

Question 6.
Microbes in sewage treatment.
Answer:
Sewage refers to the municipal waste water generated everyday in cities and towns. Human excreta is the major component of it. It contains large amounts of organic matter and microbes, out of which many are pathogenic. So, it cannot be discharged directly into natural water bodies like rivers, streams, etc.

Differentiate between two words in the following pairs of words

Question 1.
Chemical fertilisers and Biofertilisers.
Answer:
Differences between chemical fertilisers and biofertilisers are as follows

Chemical fertilisers Biofertilisers
These are industry made products which are used to increase the output of a crop plant. These are the microorganisms which increase the nutrient level of soil.
These are harmful and cause pollution to water bodies as well as ground water. These are not harmful as they lead to nutrient enrichment in an organic way.

Question 2.
Synthetic pesticides and Biopesticides.
Answer:
Differences between synthetic pesticides and biopesticides are as follows

Synthetic pesticides Biopesticides
These are not very specific, so harm non-targeted species. These are highly specific, so do not harm non-targeted species.
They cause pollution. They do not cause pollution.
Insects may become resistant, e.g. Heliothis, has become resistant to most insecticides. Insects are not expected to develop resistance to biopesticides.
Harmful residues may often remain in food, fodder and fibres. No harmful residues remain in food, fodder and fibres.

Question 3.
Baker’s and Brewer’s yeast.
Answer:
Differences between baker’s and brewer’s yeast are as follows

Baker’s yeast Brewer’s yeast
Baker’s yeast is Saccharomyces cerevisiae. It is used in fermentation to prepare dough that is used to make bread, idli, dosa, etc. Brewer’s yeast is also Saccharomyces cerevisiae, but a different strain. It is used to produce various alcoholic drinks by fermenting malted cereals and fruit juices.
In this, CO2 released during the process of fermentation gives the fluffy appearance. In this, CO2 released does not cause any fluffy appearance.

Question 4.
Symbiotic nitrogen-fixation and Mycorrhizal nitrogen-fixation.
Answer:
Differences between symbiotic nitrogen-fixation and mycorrhizal nitrogen-fixation are as follows

Symbiotic nitrogen-fixation Mycorrhizal nitrogen-fixation
It is a mutually beneficial association of bacteria with the plants for food and shelter. It is a mutually beneficial association of fungus with the root of higher plants.
The most common is Rhizobium which resides in root nodules of leguminous plants and fixes nitrogen. Members of genus Glomus form mycorrhizal association.

CHSE Odisha Class 12 Biology Solutions Chapter 11 Microbes in Human Welfare

Long Answer Type Questions

Question 1.
Give a detailed account of industrial application of microbes.
Answer:
Industrial Products:
A variety of microbes are used to synthesise a number of products in large scale that are valuable to human beings, e.g., beverages, antibiotics, etc.
A microbe should have following characteristics for its application in industrial fermentation.

  • It should be non-pathogenic and its raw materials should be cheap and easily available.
  • It should have the ability to grow rapidly on suitable nutrients.
  • It should have the ability to high yield of desired products consistently in a reasonable time.
  • It should possess high levels of enzymes for rapid production of the end products.

Now-a-days a number of products are obtained on commercial level with the help of microbes. To accomplish this, microbes are grown in very large vessels called fermentors or bioreactors.
Some of the industrial products obtained using microbes are as follows

Antibiotics:
The term ‘antibiotics’ was coined by Waksman (1942), and it is derived from the Greek words Anti-against and bios-Yiit, together they mean ‘against life’ (with reference to disease causing organisms). Antibiotics are the chemical substances, produced by some microbes that can kill or retard the growth of other disease causing microbes.

The first antibiotic was obtained from the species of Penicillium notatum in 1928 by Sir Alexander Fleming and it was named Penicillin.

It was obtained at commercial scale by growing the microbe in fermentor. The medium used for growing microbe, contained a carbohydrate as energy source, mineral salts and corn steep liquor. The culture was kept under vigorous aeration to obtain the maximal production of ‘penicillin’. It was called the wonder drug as it was used during second World War to give relief to the wounded soldiers from pain and suffering.
Antibiotics and Their Sources

Antibiotic Sources Action
Penicillin Penicillium chrysogenum, P. notatum. Tonsilitis, sore throat, gonorrhoea.
Streptomycin Streptomyces griseus Pneumonia, tuberculosis and local infections.
Erythromycin Streptomyces erythreus Typhoid, diphtheria, whooping cough.
Terramycin Streptomyces rimosus Intestinal and urinary infections.
Tetracyclines Streptomyces aureofaciens Eye infections.
Chloramphenicol S. venezuelae Conjunctivitis.
Nystastin S. nouresi Candida infection.
Polymixin Bacillus polymyxa Antifungal.

The second antibiotic, streptomycin was obtained from the bacterium Streptomyces griseus. The bacteria were grown on culture medium containing glucose, soyameal and mineral salts at pH 7.4 -7-5.

The fermentation was carried out under submerged condition at 25-30°C for 5-7 days. Now-a-days industrial fermentation is used to produce several antibiotics against diseases which have earlier caused widespred destruction in the form of epidemics.

Alcoholic Beverages:
Earlier, people used to produce alcohol by fermentation. Later, another method was used for the same which included catalytic hydration of ethylene. In modern time, again fermentation process is used for the production of ethanol.

It is used for dual purpose, i.e. as chemical and as fuel. Sugar-beet, potatoes, corn, cassava and sugarcane, etc., are used as substrate for the production of ethanol.

Yeasts (like Saccharomyces cerevisiae, S. uvarum, S. carlsbergensis), Candida brassicae, C. utilis and bacteria (Zymomonas mobilis) are used for the production of ethanol at industrial scale. The type of alcoholic drink depends upon the raw material used for its production.

Beer is obtained by the fermentation of barley grains while wine is produced by grapes. This process of alcohol production is known as brewing. In this process, CO2 is produced as a byproduct which is further used in bakery to provide sponginess to breads, cakes, etc.

Production of Organic Acids:
These are produced by the metabolic actions of microbes, i.e. microbial fermentation. Important organic acids producing organisms are listed below

Organic Acid Microbes involved
Citric acid Aspergillus niger and Penicillium sp. (fungi)
Acetic acid Acetobacter aceti (bacteria)
Butyric acid Clostridium butylicum (bacteria)
Lactic acid Lactobacillus (bacteria)
Gluconic acid Aspergillus niger and P. chrysogenum
Fumaric acid Penicillium sp.

The methods of production of some organic acids are as follows

  1. Acetic acid The production of acetic acid or vinegar occurs in two steps-preliminary fermentation and secondary fermentation. Former involves production of ethyl alcohol while later involves the production of acetic acid under aerobic conditions.
  2. Lactic acid It is produced with the help of Lactobacillus. The starchy substances, e.g. corn starch, potato starch, molasses and whey, etc., are initially hydrolysed to obtain simple sugars. It is followed by fermentation under suitable environmental conditions.
  3. Citric acid It is produced by the fermentation of beet molasses, sucrose, commercial glucose, starch hydrates, etc. Many fungi, bacteria and yeasts are used for the same.

Use of Organic Adds:
Organic acids are used as preservatives, flavour enhancers and flavouring agent. They are also used to prevent oxidation and turbidity of food products.

Production of Enzymes:
When microbes are grown in culture medium, they release various substances in the medium including enzymes. These substances can be extracted from the medium and can be used for various purposes like enzymes are used in pharmaceutical, food and textile industries. The quality and quantity of enzymes depend upon the microbial strain and cultural conditions.

Some enzymes and their applications have been discussed below

  1. Lipase produced by Rhizopus sp., used in detergent formulations and helps in removing oily stains from the laundry due to its digestive application.
  2. Pectinase produced by Aspergillus sp., used for clarifying bottled fruit juices.
  3. Proteases They are produced by Aspergillus niger and Bacillus subtilis. They are used as clarifying agents for beer, meat tenderizer, etc.
  4. Amylase produced by Aspergillus sp., used for digestive purpose and in the preparation of glucose syrup.
  5. Cellulase produced by Trichoderma viridie, used for the degradation of cellulose.

Streptokinase produced by Streptococcus and modified by genetic engineering is used as a clot buster for removing clots from blood vessels of patients, who have undergone myocardial infarction leading to heart attack.

Production of Bioactive Molecules:
The bioactive molecules produced by microbes are

  1. Cyclosporin-A It is produced by Trichoderma polysporum (fungus). It is used as an immunosuppressive agent for the patients, who have undergone organ transplantation.
  2. Statins They are produced by Monascus purpureus (yeast) and are used as blood cholesterol lowering agents.

CHSE Odisha Class 12 Biology Solutions Chapter 11 Microbes in Human Welfare

Question 2.
Explain how microbes are useful in pollution control and also in production of alternative source of energy.
Answer:
Microbes are major component of biological world on this earth. Although they are the causal agents of most of the infectious diseases, still they are of great importance to humans. Now, scientist are working on the ways in which with the use of microorganisms pollution problem can be solved. This can be done by bioremediation. In this technique microorganisms are used to neutralise pollutants from a contaminated site by their oxidation. Pollution can be controlled by microbes in two ways

  1. By enhancing the growth and activity of microbes already present an pollutant site.
  2. By adding some new microbes to the pollution site. Pollution control by the application of microbes works best when pollutants are a known mixture of organic compounds that are related to each other in structure and when there is no competition from indigenous microorganisms.

Production of Alternative Source of Energy :

Microbes in Biogas Production:
The excreta of cattle, commonly called gobar, is rich in methanogenic bacteria. Thus, cattle dung can be used for the generation of biogas, commonly called gobar gas. As cattle dung is available in large quantities in rural areas this method of biogas production is mostly functional in those areas.

Economically viable biogas is produced in large vessels called bioreactors.
Biogas plant consists of a concrete tank (10-15 feet deep) in which bio-wastes are collected and slurry of dung is fed. A floating cover is’placed over the slurry, which keeps on rising, as the gas is produced in the tank due to the microbial activity.

Methanobacterium present in the dung acts on the bio-waste to produce biogas. An outlet is also present which connects to a pipe that supplies biogas to the nearby houses.
Img 1

There is another outlet from which spent slurry is removed that can be used as fertiliser. Biogas production technology was developed in India mainly by Khadi and Village Industries Commission (KVIC) and Indian Agricultural Research Institute (LARI).

The production of biogas occurs in following three steps

Solubilisation: Decomposition of lipids, proteins, cellulose, hemicellulose, etc., present in organic matter to monomers by the action of hydrolytic enzymes like lipases, cellulases, proteases, peptidases, etc., secreted by microorganisms.

Acidogenesis: Conversion of monomers to organic acids with the help of fermentative microbes, e.g. Propionibacterium, Acetovibrio. The most common organic acid produced is acetic acid.

Methanogenesis: Biogas production with the action of methanogens, e.g. Methanococcus Methanobacillus.

CHSE Odisha Class 12 Biology Solutions Chapter 10 Improvement in Food Production

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 10 Improvement in Food Production Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 10 Question Answer Improvement in Food Production

Improvement in Food Production Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Fill in the blanks with correct answers from the choices given in the brackets of each bit

1. Physical removal of anthers is done by …………… process.
(introduction, mutation, hybridisation, emasculation)
Answer:
emasculation

2. The cross between two varieties of same crop is called ………… hybridisation. (intervarietal, intravarietal, intrageneric, intergeneric)
Answer:
intervarietal

3. In the process of breeding, genetic makeup of ……….. the concerned organism may be changed, (mutation, interspecific, selection, intraspecific)
Answer:
mutation

4. The plant part used in tissue culture is called …………… .(cells, zygote, explant, gamete)
Answer:
explant

5. To produce haploid plants ………… culture can be made.
(anther, embryo, endosperm, zygote)
Answer:
anther

6. The autotroph ………… is cultured to obtain single cell protein.
(Saccharomyces, Pseudomonas, Spirulina, Chaetomicem)
Answer:
Spirulina

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Multiple choice questions

Question 1.
The cross breed milch breed is
(a) Red Sindhi
(b) Tharparkar
(c) Frieswal
(d) Sahiwal
Answer:
(c) Frieswal

Question 2.
The exotic breed of cattle is
Or Which is an exotic breed of cattle?
(a) Jersey
(b) Sahiwal
(c) Gir
(d) Red Sindhi
Answer:
(a) Jersey

Question 3.
An indigenous breed of cattle is
(a) Red Dane
(b) Jersey
(c) Karan Swiss
(d) Rathi
Answer:
(d) Rathi

Question 4.
Sunandini, a cross breed cattle is produced by crossing.
(a) Brown-Swiss bull with Sahiwal cow
(b) Jersey bull with Red Sindhi cow
(c) Red Dane bull with Sahiwal cow
(d) Holstein-Friesian bull with Rathi cow
Answer:
(a) Brown-Swiss bull with Sahiwal cow

Question 5.
An indigenous milch breed of buffalo is
(a) Haryana
(b) Jaffarabadi
(c) Kankrej
(d) Jamunapuri
Answer:
(b) Jaffarabadi

Question 6.
The indigenous breed of poultry is
(a) Nicobari
(b) Rhode Island Red
(b) Barred Plymouth Rock
(d) New Hampshire .
Answer:
(a) Nicobari

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 7.
Commercial poultry production is done under
(a) free range system
(b) intensive system
(c) semi-intensive system
(d) folding unit system,
Answer:
(b) intensive system

Question 8.
Which of the following is a disease of cattle
(a) Ranikhet disease
(b) Marek’s disease
(c) bacillary white diarrhoea
(d) All of the above
Answer:
(d) All of the above

Express in one or two word(s)

1. What is the nutrient source not required for obtaining single cell protein from autotrophs ?
Answer:
Carbon source

2. What is called to the amorphous mass of loosely arranged thin-walled parenchymatous cells developed in the process of tissue culture?
Answer:
Callus

3. What is called to the remaining part of plant cells when its wall is mechanically or enzymatically removed?
Answer:
Protoplast

4. What is called to the sum total of all the alleles of gene present in a particular species and its allied wild and cultivated varieties?
Answer:
Gene pool

5. What is the process called, where flower buds are artificially enclosed to avoid undesired pollination?
Answer:
Bagging

6. In which process can genetic makeup of concerned organism changed ?
Answer:
Mutation

7. The preservation of semen at ultra low temperature.
Answer:
Cryopreservation

8. The substance the queen bee is fed with.
Answer:
Royal jelly

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

9. Development of haploid eggs without fertilisation.
Answer:
Parthenogenesis

10. The important monosaccharide present in honey.
Answer:
Levulose

11. The repeated breeding between closely related individuals.
Answer:
Upgrading

12. Breeding between unrelated individuals.
Answer:
Cross-breeding

Correct the sentences, if required by changing the underlined word only

1. The process of aseptic transfer of explant from nutrient medium to culture vessels is called micropropagation.
Answer:
inoculation

2. When cytoplasms are fused and one of the two nuclei lost in formation of new organism, it is called a hybrid.
Answer:
cybrid

3. For nuclear fusion, PEG is used.
Answer:
protoplast fusion

4. Cross between different genotypes of same variety is called intrageneric hybridisation.
Answer:
intravarietal

5. When pollens from selected male parents are transferred to stigma, it is called natural pollination.
Answer:
cross

Fill in the blanks

1. The cross between two species of a genus is called …………. hybridisation.
Answer:
intrageneric

2. In selection and testing of superior recombinants, F1-generation offsprings are …………… pollinated.
Answer:
self

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

3. Biofortification is done to enrich crops with micronutrients like minerals and ………. .
Answer:
vitamins

4. Explants sterilised by mercuric chloride or hydrogen peroxide, etc. are known as ………….. sterilisation.
Answer:
surface

5. Through the process of tissue culture, large number of plants raised in a small area and called micropropagation or ……. propagation.
Answer:
clonal

6. Triploids can be raised by ……….. culture.
Answer:
endosperm

7. Milk yielding cattle breeds are known as ………… breeds.
Answer:
milch

8. Foot and mouth disease is a common disease of ………
Answer:
cattle

9. Traditional method of breeding is substituted by artificial ………. .
Answer:
insemination

10. The housing system employed in the commercial poultry farming is known as ……….. farming.
Answer:
intensive system

11. Ranikhet disease is a common disease of …………..
Answer:
poultry

12. Culture of honeybee on a commercial basis is known as ……………
Answer:
apiculture

13. The drones develop from haploid eggs, which are not fertilised. This development is termed as …………….
Answer:
parthenogenesis

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

14. The deserting of the queen bee is known as ……………… .
Answer:
swarming

15. The characteristic flight of the queen bee during fertilisation is known as …………..
Answer:
nuptial flight

16. The juvenile bees are reared in …………. chamber of the honey comb.
Answer:
brood

17. The worker bees develop from fertilised eggs and hence are diploid ………….. .
Answer:
sterile females

Short Answer Type Questions

Answer the following within 50 words each

Question 1.
Name five indigenous breeds of cattle.
Answer:
Indigenous Milch Breeds of Cattle
The indigenous breeds of cattle are classified into

  • Milch breeds These are high milk producers.
  • Dual purpose breeds In these breeds, cow yield average quantity of milk, while males are good working bullocks.
  • Draught breeds These include poor milkers but superior quality bullocks.

These breeds are the highest milk producing catde. These include Sahiwal, Red Sindhi, Gir, Tharparkar and Rathi.

Question 2.
Name three cross-breeds of cattle.
Answer:
Cross-Breed Strains of Cattle:
Some of the cross-breed cows are as follows
(a) Karan Swiss This breed was developed at National Dairy Research Institute ‘Karnal, by the breeding of Sahiwal cows with Brown Swiss bulls imported from USA.
(b) Karan Fries This breed has got its origin at the National Dairy Research Institute, Karnal by the crossing between Tharparkar and Holstein-Friesian.
CHSE Odisha Class 12 Biology Solutions Chapter 10 Improvement in Food Production 1
(c) Sunandini This breed originated in Kerala by crossing the local non-descript cattle with Jersey, Brown Swiss and Holstein-Friesin breeds.
(d) Frieswal This breed is developed by crossing Holstein-Friesian bulls with Sahiwal cows.

Question 3.
Name three exotic breeds of cattle.
Answer:
Exotic Milch Breeds of Cattle
This breed includes high milk producing cattle breeds of other countries which have been cross breed with indigenous breeds for producing high yielding hybrids which can easily adopt to Indian conditions.
The common exotic breeds which are used in such programme include Holstein-Friesian of Netherlands, Brown-Swiss of Switzerland, Jersey of Europe and America and Red Dane of Denmark.
Img 2

Question 4.
What is cryopreservation?
Answer:
Cryopreservation is the preservation of tissues, embryos gametes, etc., at -196°C (liquid nitrogen), the preserved material is revived through special technique when required. Cryopreservation methods seek to reach low temperatures without causing additional damage caused by the formation of ice during freezing. Traditional cryopreservation has relied on coating the material with cryoprotectants.

Question 5.
Describe organic dairy farming
Answer:
Organic Dairy Farming
Though the milk production has increased by leaps and bounds by the use of synthetic chemicals on cattle and buffaloes, there is a chance of contamination of the harvested milk. It has been found that milk and milk products are contaminated by residue components of harmful chemicals.

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 6.
What is free-range poultry farming?
Answer:
The term ‘poultry’ means rearing of domesticated fowl (birds) for food (meat) or their eggs, i.e. chicken, ducks, , geese, turkeys and some varieties of pigeons. Poultry farming upto 1960 was considered under free-range condition.

Free-range system In this system, the birds are allowed to move freely to outside. The farmer provides them with food supplements as requird moving outside.

Question 7.
What is intensive housing system in poultry?
Answer:
Intensive system
This system is practiced where there is a need of a large scale production of meat and eggs. The birds do not have access to outside and are kept in a walled house. There are two types of housing system. Cage system is the system, where birds are kept in cages and it also helps in preventing the spread of diseases. In litter system however, birds are kept on a floor covered with rice husk, saw dust, dried leaf, etc.

Question 8.
Describe the protein source in poultry nutrition.
Answer:
These are required for growth and repair of the body tissues. For protein, the feed is supplemented with soyabean meal, groundnut cake, sunflower cake.
The fish meal prepared from the wastes of fish processing industry and meat meal from the wastes of meat processing industry are also used to feed poultry birds. The skimmed milk is highly nutritive for young chicks and should be given in clean vessels. The green foods such as fresh tender grass, garlic lettuce, onions, etc. are important part of the feed.

Fats Only unsaturated fats could be digested by poultry. This requirement is met by providing groundnut, cake, sunflower cake, etc.

Minerals Some minerals are required in large quantity such as calcium, sodium, etc. Minerals such as zinc, iron, copper, are required in lesser quantities and so, are called as trace minerals.

Vitamins fat soluble (A, D, E, K) and water soluble (B,C) all are essential for normal growth of chickens.
CHSE Odisha Class 12 Biology Solutions Chapter 10 Improvement in Food Production 2

Question 9.
Why is inbreeding harmful?
Answer:
Breeding between the animals of the same breed for 4-6 generation is called inbreeding. It increases homozygosity. Thus, inbreeding is necessary to develop pureline. It also helps in accumulation of superior genes and elimination of less desirable genes. But continued inbreeding reduces fertility and even productivity so, it is harmful also.

Question 10.
What is artificial insemination?
Answer:
Artificial Insemination
It is a method in which the semen collected from a superior male parent is injected into the reproductive tract of the selected female parent by the breeder. This results in development of progeny with superior traits like better growth and increased milk production.
The success rate of artificial insemination is fairly low, even then it is carried out because of the following advantages

  • The semen collected can be used immediately or stored in frozen form for later use.
  • The semen from a desired breed can be easily transported in the frozen form to distant places, where the selected females are present and thus can be used for impregnating the females on a large scale.
  • It helps to overcome several problems of normal mating.

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 11.
What is in vitro fertilisation?
Answer:
In Vitro Fertilisation and Embryo Transfer
In this modern method of animal breeding, eggs of an ovulating livestock animal, e.g. cattle are isolated. Then they are fertilised in- vitro by semen of a bull which possesses desired characters.

This fertilised egg is kept in a suitable medium and is then stimulated to undergo cleavage to a 8-16 called stage embryo which is known as a blastocyst, also in-vitro. This embryo is then finally transferred or implanted into uterus of a surrogate cow and made pseudopregnant.
The cow completes the term and then give birth to calf with the required characters.

Question 12.
What is a transgenic animal?
Answer:
Transgenic Animals:
It involves the transfer of genes intti special cells or embryos. In this method, the unfertilised egg is enucleated by treating with cytochalasin-B and the blastula stage nuclei are obtained from embryo donor. These two are incubated together in’the presence of Polyethylene Glycol (PEG) and transferred into surrogate mother for fusion. The foetus develops into a transgenic animal. Various products like a-antitrypsin, haemoglobin, lacteferrin, iron binding protein, etc., are obtained from transgenic animals. In Japan, gynogenesis is being used to improve fish size.

Question 13.
Describe swarming.
Answer:
Swarming It is the process by which a new honeybee colony is formed when the queen bee leaves the colony with a large group of worker bees.
Swarming is mainly a spring phenomenon, usually within a two or three weeks period depending on the locality but occasional swarms can happen throughout the producing season.

Write notes on with 2/3 valid points

Question 1.
Germplasm collection
Answer:
Collection of Germplasm:
This is the major step acting as the root of any breeding programme. In this step, the pre-existing genetic variability available in purelines, wild varieties, species no longer cultivated and relatives of many crop species are collected and preserved.

Evaluation of their characteristics is a pre-requisite for the effective exploitation of natural genes available in the populations. The entire collection of plants/seeds having all diverse alleles for all genes in a given crop is called germplasm collection. A good germplasm collection is essential for a successful breeding programme.

Question 2.
Emasculation
Answer:
Emasculation
It is the process of removal of stamens of a flower, without affecting the female reproductive organs. Emasculation is usually done in bisexual flowers before the anthers mature and stigma has become receptive. It can be done by various methods, such as, hand emasculation, suction method, hot water emasculation, alcohol treatment, cold treatment and genetic emasculation. Among these methods, hand and suction method are mostly used.

For example, in Triticum (wheat) flowers may be exposed to some chemical like 2,4-dichloro phenoxyacetic acid, maleic hydrozide or a panicle of Sorghum is dipped in lukewarm (50°C) water for 10 minutes, etc. These methods are applied on those cases where the methods of physical nature could not be applied.

Question 3.
Bagging
Answer:
Bagging In this method, the emasculated flower or inflorescence is immediately bagged to avoid . pollination by any foreign pollen. Emasculation bags made up of butter paper, fine cloth or polythene, etc., may be used depending upon the crop.

Question 4.
Artificial pollination
Answer:
Artificial pollination In order to bring about artificial pollination, the collected pollen grains from selected male parents are dusted on the stigma of female plant. The properly labelled flowers are then allowed to cross-pollinate. The crossed flowers are then tagged again.

Question 5.
Breeding for disease resistance
Answer:
Resistance of the host plant is the ability to prevent the pathogen from causing disease and is determined by the genetic constitution of host plant. Crops are required to be disease resistant, as there are a wide range of fungal, bacterial and viral pathogens that affect the yield of cultivated crop species, especially in tropical climates.

Question 6.
Biofortification
Answer:
It is a method of breeding crops with higher levels of vitamins, minerals, healthier fats to improve public health. The objective of breeding for improved nutritional quality is to enhance
(i) Protein, oil content and quality.
(ii) Vitamin content.
(iii) Micronutrients and mineral content

Question 7.
Explant
Answer:
It is the technique of maintaining and growing plant cells, tissues or organs in nutrient media under controlled environmental conditions. The plant part taken out to be grown in a test tube in special nutrient media is called explant.

Explant Selection:
It is the tissue obtained from the plant for the purpose of tissue culture. The most commonly used explant tissues are the meristematic ends of the plants. These include stem tip, auxiliary bud tip and root tip. Meristematic tissues have high rate of cell division. Before the procedure starts, explants are cleaned and surface sterilised with the help of disinfectants and detergents to remove germs.

Question 8.
Tissue culture medium
Answer:
Tissue Culture Techniques and Steps
Plant tissue culture involves producing entire plants from a few plant cells or tissues by growing them in an artificial medium.

1. Explant Selection
2. Sterilisation
3. Preparation of Nutrient or Culture Medium
4. Inoculation
5. Callus Formation and its Culture
6. Organogenesis
7. Somatic Embryogenesis

Question 9.
Totipotency .
Answer:
Totipotency The capacity to generate a whole plant from any cell/ explant is called cellular totipotency in fact, the whole plant can be regenerated from any plant part or cells.

Question 10.
Micropropagation
Answer:
Micropropagation or Clonal Propagation
By the process of plant tissue culture which requires lesser space and lesser time, a large population of plants could be raised. Also since the plants produced are genetically identical, this process is also called as clonal propagation. Examples of plants cutlivated micropropagation include grapes, bamboo, coffee, banana, cardamoms, etc.

Question 11.
Anther culture
Answer:
This technique was developed by Guha and Maheshwari (1946) in Datura innoxia. In this technique, floral buds are opened to remove anthers. These anthers are then cultured for the production of haploid embryoids. The plants produced by haploid culture are sterile. These haploids could be subjected to colchicine treatment in order to double their chromosome number.

Question 12.
Somaclonal variation
Answer:
Somaclonal variation Genetic variation present among plant cells of a culture is called somaclonal variation.
The term somaclonal variation is also used for the genetic variation present in plants regenerated from a single culture.

Question 13.
Synthetic seeds
Answer:
Synthetic Seeds/Artificial Seeds:
Artificial seeds are those seeds in which somatic embryos or plantlets are encapsulated by calcium alginates. This can put a stop to desiccation and they could be used by farmers like normal seeds and are also used for rapid propagation of crop plants.

Question 14.
Secondary metabolites
Answer:
Secondary Metabolites Production:
Cell suspension culture has been employed for commercial production of secondary metabolites like tannin, latex, resin. These cost of secondary metabolites production would be very high if manufactured chemically. So, the plant tissue culture comprising of large scale cell suspension culture has been used.

For example, taxol which is an anticancer drug is obtained from Taxus. The cells of Taxus are cultured which produce a similar chemical which is later chemically modified to taxol. Another example includes Digitalis lantana which is being employed to modify digoxin, to digitoxin, a drug used in cardiac treatment.

Question 15.
Embryo rescue
Answer:
Endosperm Culture:
It is used to produce triploids. Endosperm culture is helpful in producing seedless apple, Citrus which are of better commercial values.

Differentiate between the following

Question 1.
Bagging and Tagging.
Answer:
Differences between bagging and tagging are as follows

Bagging Tagging
It is the step involved in hybridisation. It is also a step of hybridisation.
Emasculated flowers are immediately covered by paper, plastic or polythene bags. Bapqed flowers tagged by writing date, time, male and female, parents.

Question 2.
Chemical pest control and Biological pest control.
Answer:
Differences between chemical pest control and biological pest control are as follows

Chemical pest control Biological pest control
In this method chemicals are used for pest control. In this method biological organisms, i.e predator, parasitoids are used for pest control.
Various types of chemicals are used, i.e herbicides, insecticides, pesticides, etc. Various predators, i.e Trichoderma, B. thuringiensis, etc. are used.
These are expensive. These are cost effective methods.

Question 3.
Callus and Protoplast.
Answer:
Differences between callus and protoplast are as follows

Callus Protoplast
It is a growing mass of unorganised plant parenchyma cells. It refers to entire cell excluding cell wall.
Explants are supplemented with auxin, cytokinin, etc to initiate callus. It is used to study membrane biology, protoplast fusion, etc.

Question 4.
Synthetic seeds and Embryo.
Answer:
Differences between synthetic seeds and embryo are as follows

Synthetic seeds Embryo
These are encapsulated somatic embryo, shoot buds or aggregates of cell or any tissues which has the ability to form a plant in in vitro. It is the part of seed, consisting of precursor tissues for the leaves, stem and roots.
Hybrid plants can be easily propagated using synthetic seeds. Embryo culture is used for the culturing of embryo.

Question 5.
Endosperm culture and Anther culture
Answer:
Differences between endosperm culture and anther culture are as follows

Endosperm culture Anther culture
Endosperm is used for culturing. Anther is used for culturing.
Triploid plants are formed. Haploid plants are formed.
Used in production of seedless fruits. It is useful for the improvement of crop plants.

Question 6.
Hybrid and Cybrid.
Answer:
Differences between hybrid and cybrid are as follows

Hybrid Cybrid
It is the result of combining the qualities of two organisms of different breeds, varieties, species through hybridisation. It is a eukaryotic cell produced by the fusion of two protoplast.
Hybrids are produced through hybridisation. These are produced through somatic hybridisation.

Long Answer Type Questions

Question 1.
Describe the main steps of breeding to develop genetic variability in crop plants.
Answer:
Steps in Plant Breeding:
The major steps in breeding a new genetic variety of a crop are
(i) Collection of Germplasm:
This is the major step acting as the root of any breeding programme. In this step, the pre-existing genetic variability available in purelines, wild varieties, species no longer cultivated and relatives of many crop species are collected and preserved.

Evaluation of their characteristics is a pre-requisite for the effective exploitation of natural genes available in the populations. The entire collection of plants/seeds having all diverse alleles for all genes in a given crop is called germplasm collection. A good germplasm collection is essential for a successful breeding programme.

(ii) Evaluation and Selection of Parents:
It is carried out by evaluating germplasm, to identify plants with desirable combination of characters.
The selected plants are multiplied and hybridised by self-pollination. Purelines are created, whenever desired and possible.

(iii) Cross Hybridisation among Selected Parents
It is possible by cross hybridising the two parents to produce hybrids that genetically combine the desired characters in a single plant. It is known to be a time consuming and tedious process as it involves collection of pollen grains from the desired plants (male parent) and have to be placed on the stigma of the selected flower (female parent) to incorporate desired traits.

It is also not necessary that the hybrids do combine desired characters. The chances of desirable combination is usually only one in few hundred to a thousand crosses carried out.

Some of the objectives of hybridisation are as follows
(a) To produce variations in progeny which are useful. It is achieved by recombination of characters.
(b) To make the use of hybrid vigour which is the superiority of progeny over its parents.
(c) To develop high yielding varieties which are also resistant to diseases.

Depending on the nature of plants involved in the cross, there may be different types of hybridisation such as
(a) Inter-varietal Cross between two varieties of same crop.
(b) Intra-varietal Cross between different genotypes of the same variety.
(c) Intra-generic Cross between two species of a genus.
(d) Inter-generic Cross between two genera.

Techniques of Hybridisation:
The first step in hybridisation is to ensure that pollination can not occur before the intended artificial process. This can be achieved by
(a) Emasculation It is the process of removal of stamens of a flower, without affecting the female reproductive organs. Emasculation is usually done in bisexual flowers before the anthers mature and stigma has become receptive. It can be done by various methods, such as, hand emasculation, suction method, hot water emasculation, alcohol treatment, cold treatment and genetic emasculation. Among these methods, hand and suction method are mostly used. For example, in Triticum (wheat) flowers may be exposed to some chemical like 2,4-dichloro phenoxyacetic acid, maleic hydrozide or a panicle of Sorghum is dipped in lukewarm (50°C) water for 10 minutes, etc. These methods are applied on those cases where the methods of physical nature could not be applied.

(b) Bagging In this method, the emasculated flower or inflorescence is immediately bagged to avoid pollination by any foreign pollen. Emasculation bags made up of butter paper, fine cloth or polythene, etc., may be used depending upon the crop.

(c) Tagging In this process, the emasculated and bagged inflorescence or flowers are tagged and properly labelled.
The labels contain following information

1 Date of emasculation
2 Date of pollination
3 Name of female and male plants. The name of the female parent plant is written first and that of the male parent plant is written later.

(d) Artificial pollination In order to bring about artificial pollination, the collected pollen grains from selected male parents are dusted on the stigma of female plant. The properly labelled flowers are then allowed to cross-pollinate. The crossed flowers are then tagged again.

(e) Selection and testing of superior recombinants:
This step consists of selection of plants among the progeny of the hybrids with desired combination of characters. It yields plants that are superior than both the parents. This is known as hybrid vigour/heterosis. These are self-pollinated for several generations, till they reach a state of uniformity or homozygosity, so that the characters will not segregate in the progeny.

(f) Testing, release and commercialisation of new varieties Evaluation is done for newly selected lines for their yield and other agronomic traits of quality, disease resistance, etc. Selected plants are grown in research fields and their performance is recorded under ideal fertiliser applications, irrigation and other crop management practices.

Testing of hybrid line is done in farmer’s field after evaluation. After testing, the crop is grown at different locations in the country with different agroclimatic zones for atleast three growing seasons. The tested material is evaluated in comparison to the best available local crop cultivar used as reference cultivar. Release of tested material is finally done in bulk after selection and certification.

Examples of Some Improved Varieties
(i) Wheat Kalyan Sona and Sonalika are semi-dwarf, high yielding and resistant to root disease, introduced to wheat growing belt of India.
(ii) Rice Along with the above wheat varieties, rice varieties such as IR – 8 and Taichung and their derivatives Jaya and Ratna varieties were introduced around the same time in India. All these varieties contributed to the quantum jump in food production which is called green revolution.

Question 2.
Describe the techniques of hybridisation.
Answer:
Techniques of Hybridisation:
The first step in hybridisation is to ensure that pollination can not occur before the intended artificial process. This can be achieved by
(a) Emasculation It is the process of removal of stamens of a flower, without affecting the female reproductive organs. Emasculation is usually done in bisexual flowers before the anthers mature and stigma has become receptive. It can be done by various methods, such as, hand emasculation, suction method, hot water emasculation, alcohol treatment, cold treatment and genetic emasculation. Among these methods, hand and suction method are mostly used. For example, in Triticum (wheat) flowers may be exposed to some chemical like 2,4-dichloro phenoxyacetic acid, maleic hydrozide or a panicle of Sorghum is dipped in lukewarm (50°C) water for 10 minutes, etc. These methods are applied on those cases where the methods of physical nature could not be applied.

(b) Bagging In this method, the emasculated flower or inflorescence is immediately bagged to avoid pollination by any foreign pollen. Emasculation bags made up of butter paper, fine cloth or polythene, etc., may be used depending upon the crop.

(c) Tagging In this process, the emasculated and bagged inflorescence or flowers are tagged and properly labelled.
The labels contain following information

1 Date of emasculation
2 Date of pollination
3 Name of female and male plants. The name of the female parent plant is written first and that of the male parent plant is written later.

(d) Artificial pollination In order to bring about artificial pollination, the collected pollen grains from selected male parents are dusted on the stigma of female plant. The properly labelled flowers are then allowed to cross-pollinate. The crossed flowers are then tagged again.

(e) Selection and testing of superior recombinants:
This step consists of selection of plants among the progeny of the hybrids with desired combination of characters. It yields plants that are superior than both the parents. This is known as hybrid vigour/heterosis. These are self-pollinated for several generations, till they reach a state of uniformity or homozygosity, so that the characters will not segregate in the progeny.

(f) Testing, release and commercialisation of new varieties Evaluation is done for newly selected lines for their yield and other agronomic traits of quality, disease resistance, etc. Selected plants are grown in research fields and their performance is recorded under ideal fertiliser applications, irrigation and other crop management practices.

Testing of hybrid line is done in farmer’s field after evaluation. After testing, the crop is grown at different locations in the country with different agroclimatic zones for atleast three growing seasons. The tested material is evaluated in comparison to the best available local crop cultivar used as reference cultivar. Release of tested material is finally done in bulk after selection and certification.

Question 3.
Give an account of techniques and steps of plant tissue culture.
Answer:
Tissue Culture:
It is the technique of maintaining and growing plant cells, tissues or organs in nutrient media under controlled environmental conditions. The plant part taken out to be grown in a test tube in special nutrient media is called explant.
The capacity of producing a whole plant from this explant is called totipotency.

It was Gottilieb Haberlandt (1902) who discovered totipotency. Haberlandt also attempted to cultivate plant leaf cells for the first time in simple nutrient medium. Plant tissue culture technique is a major tool in various areas of crop improvement, experimental biology and fundamental or applied research.

Tissue Culture Techniques and Steps:
Plant tissue culture involves producing entire plants from a few plant cells or tissues by growing them in an artificial medium.

1. Explant Selection
It is the tissue obtained from the plant for the purpose of tissue culture. The most commonly used explant tissues are the meristematic ends of the plants. These include stem tip, auxiliary bud tip and root tip. Meristematic tissues have high rate of cell division. Before the procedure starts, explants are cleaned and surface sterilised with the help of disinfectants and detergents to remove germs.

2. Sterilisation
For plant tissue culture, it is essential that the explants, culture vessel, media and instruments, etc. are free from microbes. For this purpose, explants are treated with specific antimicrobial chemicals like mercury chloride, hydrogen peroxide, etc.
This procedure is called surface sterilisation. The vessels, media and instruments, are suitably treated with steam (in autoclave), dry heat or alcohol or subjected to filtration to make them free from microbes. This is complete sterilisation.

3. Preparation of Nutrient or Culture Medium
The medium on which explants are cultured is known nutrient medium or culture medium or simply medium. The culture media may be solid or liquid. The optimum pFl of the media should be 5.7.
The basic components of culture media are
(i) Inorgnic nutrients These include salts, providing all essential macro and microelements. Macronutrients include salts of calcium, magnesium, nitrogen, phosphorus and potassium. Micronutrients include iron, chloride, copper, zinc, boron and molybdenum.
(ii) Source of carbon These are the sources of energy in the form of sucrose, glucose, fructose or carbohydrates, amino acids, etc.
(iii) Growth hormones and vitamins Auxin (2, 4 -D) and cytokinins (benzyl amino purine), vitamins (Pyridoxine HCL) are commonly used in tissue culture. A media without growth hormones is known as base media. High auxins result in root formation while high cytokinins may yield shoots. After the preparation of medium, agar-agar is added in order to obtain a solid medium. In some cases liquid medium is required, for example in root culture where no agar-agar is added.

4. Inoculation
The process of transfer of explant to suitable nutrient medium contained in culture vessels is called inoculation. It is done under sterile conditions, achieved in an inoculation chamber or under laminar air flow. After the process, temperature and light of vessels is kept at controlled range. The temperature ranges between 18-25°C.

5. Callus Formation and its Culture
In callus culture, cell division in the explant forms a callus. It is an amorphous mass of’loosely arranged thin-walled parenchymatous cells developing frome proliferating cells of the parent tissue (Dodos and Roberts; 1985). This is usually maintained on a medium gelled with agar. If nutrient medium contains auxins cell division occurs and the upper surface of explant gets covered by callus. This callus later develops into normal roots, shoots and finally lead to the formation of plant. The development of callus occurs through three stages, i. e. induction, cell division and differentiation.

The callus formation is affected by the composition of medium, the source from which explant has been taken and the surrounding environmental factors. The first stage (induction) involves stimulation of metabolic rate of cells. With the increase in the metabolic rate, these cells enter the cell division stage. Finally in differentiation stage, the cells produce secondary products by the expression of certain metabolic pathways. It is necessary to subculture the callus in fresh media when the callus is being grbwn for a long time on the nutrient media.

6. Organogenesis
It involves the formation of plant organs, i.e. roots and shoots directly from cultured tissues. Organogenesis begins from the stimulation provided by the chemicals of medium, endogenous compounds produced by the culture and substances that have been carried over from the original explants. The process of formation of roots is known as rhizogenesis while that of shoots is caulogenesis. Organogenesis is highly controlled by the ‘ melatin concentrations of auxin and cytokinin in medium.

It was Skoog and Miller (1957) who demonstrated that a high concentration of auxin promotes rhizogenesis while high concentration of cytokinin promotes caulogenesis.

7. Somatic Embryogenesis
A Somatic Embryo (SE) is an embryo derived from a somatic cell, other than zygote. SEs are obtained usually on culture of the somatic cells in vitro. So, this process is called somatic embryogenesis. The embryo formed is known as embryoids. SEs are bipolar structures, i.e. they have a radicle and a plumule.

It is induced by a relatively high concentration of an auxin, like 2, 4-D. There are two different media for the formation of embryoids. The first medium comprises of auxin that initiates embryogenic cells.

The second medium either lacks or has decreased level of auxins. For the further development of embryonic cells into embryoids and plantlets the embryogenic cells undergo three developmental stages such as globular, heart-shaped and torpedo stage. Examples of plants undergoing embryogenesis in vitro include Nicotiana tabaccum, Loffea arabica, Atropa belladona, Brassica oleracea, etc.
The method used for tissue culture is as given below
Major Methodology of plant Tissue Culture
CHSE Odisha Class 12 Biology Solutions Chapter 10 Improvement in Food Production 3
Events of somatic embryogenesis

Question 4.
Elaborate the application of plant tissue culture.
Answer:
Applications of Plant Tissue Culture
(i) Micropropagation or Clonal Propagation:
By the process of plant tissue culture which requires lesser space and lesser time, a large population of plants could be raised. Also since the plants produced are genetically identical, this process is also called as clonal propagation. Examples of plants cutlivated micropropagation include grapes, bamboo, coffee, banana, cardamoms, etc.

(ii) Production of Virus-Free Plants:
Crop plants that reproduce asexually are susceptible to viral infections which advances through the vegetative organs for propagation like stem, tuber, rhizome, etc. The cambium culture in some plants produces virus-free plants.

(iii) Synthetic Seeds/Artificial Seeds:
Artificial seeds are those seeds in which somatic embryos or plantlets are encapsulated by calcium alginates. This can put a stop to desiccation and they could be used by farmers like normal seeds and are also used for rapid propagation of crop plants.

(iv) Embryo Rescue:
It is a technique in which immature embryos are dissected out from the fruit (seeds). They are then grown in nutrient medium which lead to the formation of plantlets. Embryo rescue technique is done in conditions where the embryo does not develop after initial divisions though the pollination and fertilisation had been successfully completed.
This technique is being used to improve chick-pea, groundnut, etc. at International Crop Research Institute for Semi Arid Tropics (ICRISAT), Hyderabad.

(v) Endosperm Culture:
It is used to produce triploids. Endosperm culture is helpful in producing seedless apple, Citrus which are of better commercial values.

(vi) Secondary Metabolites Production:
Cell suspension culture has been employed for commercial production of secondary metabolites like tannin, latex, resin. These cost of secondary metabolites production would be very high if manufactured chemically. So, the plant tissue culture comprising of large scale cell suspension culture has been used.

For example, taxol which is an anticancer drug is obtained from Taxus. The cells of Taxus are cultured which produce a similar chemical which is later chemically modified to taxol. Another example includes Digitalis lantana which is being employed to modify digoxin, to digitoxin, a drug used in cardiac treatment.

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 9 Question Answer Health and Diseases

Health and Diseases Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Multiple choice questions

Question 1.
Which one of the following diseases are communicable ?
(a) Deficiency diseases
(b) Allergies
(c) Degenerative diseases
(d) Infectious diseases
Answer:
(d) Infectious diseases

Question 2.
The nature of the spread of communicable diseases is termed as
(a) parasitology
(b) immunology
(c) epidemiology
(d) None of these
Answer:
(c) epidemiology

Question 3.
Which one of the following is a sexually transmitted disease ?
(a) Q-fever
(b) Leprosy
(c) Whooping cough
(d) Gonorrhoea
Answer:
(d) Gonorrhoea

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 4.
Gonorrhoea is a
(a) bacterial disease
(b) Veneral disease
(c) STD
(d) All of these
Answer:
(c) STD

Question 5.
Anthrax is caused by
(a) Vibrio
(b) Bacillus
(c) Salmonella
(d) virus
Answer:
(b) Bacillus

Question 6.
Some common diseases caused by bacteria are
(a) measles, mumps and malaria
(b) tetanus, typhoid and tuberculosis
(c) syphilis, smallpox and sleeping sickness
(d) pneumonia, poliomyelitis and psittacosis
Answer:
(b) tetanus, typhoid and mberculosis

Question 7.
Which one of the following disease is spread through wounds ?
(a) Tetanus
(b) Cholera
(c) Plague
(d) Tuberculosis
Answer:
(a) Tetanus

Question 8.
Which of the following is a bacterial disease ?
(a) Measles
(b) Smallpox
(c) Rabies
(d)Tuberculosis
Answer:
(d)Tuberculosis

Question 9.
Causative agent of TB is
(a) Salmonella
(b) Streptococcus
(c) Mycobacterium
(d) Pneumococcus
Ans.
(c) Mycobacterium

Question 10.
BCG vaccine is a preventive measure against
(a) Tuberculosis
(b) Typhoid
(c) AIDS
(d) Cholera
Ans.
(a) Tuberculosis

Question 11.
Which one is not a bacterial disease ?
(a) Tuberculosis
(b) Typhoid
(c) AIDS
(d) Cholera
Ans.
(c) AIDS

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 12.
Mantoux test is for
(a) scarlet fever
(b) diptheria
(c) rheumatoid fever
(d) tuberculosis
Ans.
(d) tuberculosis

Question 13.
Chickenpox is caused by
(a) Varicella virus
(b) adenovirus
(c) SV-40 virus
(d) bacteriophage-T2
Ans.
(a) Varicella virus

Question 14.
Smallpox is due to
(a) virus
(b) bacterium
(c) protozoan
(d) helminth
Ans.
(a) virus

Question 15.
The disease caused by virus is
(a) pneumonia
(b) tuberculosis
(c) smallpox
(d) typhoid
Ans.
(c) smallpox

Question 16.
Polio is caused by
(a) virus with double-stranded DNA
(b) virus with double-stranded RNA
(c) virus with single-stranded DNA
(d) virus with single-stranded RNA
Ans.
(d) virus with single-stranded RNA

Question 17.
Mumps is a
(a) protozoan disease
(b) viral disease
(c) fungal disease
(d) bacterial disease
Ans.
(b) viral disease

Question 18.
Which one is a viral disease ?
(a) Measles
(b) Rickets
(c) Syphilis
(d) Congenital night blindness
Answer:
(a) Measles

Question 19.
Amoebiasis is caused by
(a) Plasmodium vivax
(b) Entamoeba gingivalis
(c) Trypanosoma gambiense
(d) Entamoeba histolytica
Answer:
(d) Entamoeba histolytica

Question 20.
Entamoeba histolytica infection occurs through
(a) mosquito bite
(b) bird droppings
(c) sweat
(d) contaminated food and water
Answer:
(d) contaminated food and water

Question 21.
The infective stage of Entamoeba histolytica is
(a) binucleate form
(b) tetranucleate form
(c) minute form
(d) sporozoite stage
Answer:
(b) tetranucleate form

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 22.
Malaria is transmitted by
(a) male Anopheles
(b) female Anopheles
(c) female Culex
(d) female Aedes
Answer:
(b) female Anopheles

Question 23.
Select the incorrect pair
(a) Pedicuius-Typhoid
(b) Xenopsylla-Plague
(c) Culex-Malaria
(d) Aedes-Yellow fever
Answer:
(a) Pedicuius-Typhoid

Question 24.
Filaria is transmitted by
(a) tse-tse fly
(b) sand fly
(c) Anopheles
(d) Culex
Answer:
(b) sand fly

Question 25.
Culex causes the disease
(a) malaria
(b) filariasis
(c) yellow fever
(d) sleeping sickness
Answer:
(b) filariasis

Question 26.
The disease elephantiasis is caused by
(a) Culex mosquito
(b) Anopheles mosquito
(c) housefly
(d) tse-tse fly
Answer:
(a) Culex mosquito

Question 27.
Microfilariae are found in the peripheral blood of man during
(a) day time
(b) day and night time
(c) night time
(d) None of the above
Answer:
(c) night time

Question 28.
Infection of Ascaris occurs due to
(a) tse-tse fly
(b) mosquito bite
(c) imperfectly cooked pork
(d) contaminated food and water
Answer:
(d) contaminated food and water

Question 29.
A disease caused by nematode parasite
(a) filariasis
(b) leprosy
(c) amoebiasis
(d) poliomyelitis
Answer:
(a) filariasis

Question 30.
AIDS is caused by
(a) HTLV-III
(b) herpes virus
(c) rotavirus
(d) orthomyxovirus
Answer:
(a) HTLV-III

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 31.
Cerebral malaria is caused by Plasmodium
(a) vivax
(b) ovale
(c) falciparum
(d) All of these
Answer:
(c) falciparum

Question 32.
Which of the glands is often referred in relation with AIDS?
(a) Thyroid
(b) Adrenal
(c) Thymus
(d) Pancreas
Ans.
(c) Thymus

Question 33.
AIDS is caused by
(a) virus
(b) fungus
(c) helminth
(d) bacterium
Ans.
(a) virus

Question 34.
AIDS is due to
(a) reduction in number of helper T-cells
(b) lack of interferon
(c) reduction is number of killer T-cells
(d) auto-immunity
Answer:
(a) reduction in number of helper T-cells

Question 35.
AIDS virus has
(a) double-stranded DNA
(b) single-stranded DNA
(c) single-stranded RNA
(d) double-stranded RNA
Ans.
(c) single-stranded RNA

Question 36.
AIDS spreads through
(a) immoral way of life
(b) infected needles and syringes
(c) homosexuality
(d) All of the above
Ans.
(d) All of the above

Question 37.
Cancer is
(a) non-malignant tumour
(b) controlled division of cells
(c) unrestrained division of cells
(d) microbial infection
Ans.
(c) unrestrained division of cells

Question 38.
Cancer cells are damaged by radiations while others are not
(a) being different in nature
(b) being starved
(c) undergoing rapid division
(d) None of the above
Answer:
(c) undergoing rapid division

Question 39.
Sarcoma is the cancer of
(a) epithelial tissues
(b) connective tissues
(c) blood
(d) endodermal tissues
Answer:
(b) connective tissues

Question 40.
Blood cancer is called
(a) leukemia
(b) haemophilia
(c) thrombosis
(d) haemolysis
Answer:
(a) leukemia

Question 41.
The cells affected by leukemia are
(a) plasma cells
(b) erythrocytes
(c) thrombocytes
(d) leucocytes
Answer:
(a) plasma cells

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 42.
Genes involved in cancer are
(a) tumour genes
(b) oncogenes
(c) cancer genes
(d) regulator genes
Answer:
(b) oncogenes

Question 43.
Oncology is the study of
(a) living cells
(b) cancer cells
(c) dead cells
(d) dividing cells
Answer:
(b) cancer cells

Question 44.
The most common cancer in women is
(a) breast cancer
(b) skin cancer
(c) cervix cancer
(d) leukemia
Answer:
(a) breast cancer

Question 45.
Breast cancer is an examle of
(a) adenoma
(b) lymphoma
(c) carcinoma
(d) sarcoma
Answer:
(c) carcinoma

Question 46.
Cancer treatment includes
(a) surgery
(b) radiotherapy
(c) treatment with anticancer drugs
(d) All of the above
Answer:
(d) All of the above

Question 47.
The most common type of cancer in man is
(a) skin cancer
(b) lung cancer
(c) cancer of prostate
(d) cancer of bladder
Answer:
(b) lung cancer

Question 48.
Which one of the following is a cancer causing agent ?
(a) Tobacco
(b) Radiation
(c) Smoking
(d) All of these
Answer:
(d) All of these

Question 49.
Which one of the following is an oncogenic virus ?
(a) Herpes simplex-II
(b) Papilloma
(c) Epstein-Barr
(d) All of these
Answer:
(c) Epstein-Barr

Question 50.
The spread of cancerous cells to distant sites is termed
(a) metamorphosis
(b) metagenesis
(c) metastasis
(d) metachrosis
Answer:
(c) metastasis

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 51.
Adenoma refers to the cancer of
(a) glands
(b) lymph nodes
(c) blood
(d) muscles
Answer:
(a) glands

Question 52.
Which one of the following is an anticancer drug?
(a) Aspirin
(b) Flagyl
(c) Streptomycin
(d) Vincristine
Answer:
(d) Vincristine

Question 53.
Which of the following scientists got Noble prize in 1989 for the studies on the genetic basis of cancer ?
(a) Philip Sharp and Richard Roberts
(b) David Baltimore and Howard Temin
(c) Michael Bishop and Harold Varmus
(d) Stanley B Prusiner
Answer:
(c) Michael Bishop and Harold Varmus

Question 54.
HIV attacks which one of the following ?
(a) B-cells
(b) T- cells
(c) Antigen preventing cell
(d) T-helper cells
Answer:
(d) T-helper cells

Question 55.
Which one of the following is not a component of innate immunity ?
(a) Antibodies
(b) Interferons
(c) Complement proteins
(d) Phagocytes
Answer:
(a) Antibodies

Question 56.
Which of the following is involved in defense mechanism of the body ?
(a) Lymphocytes
(b) Neutrophils
(c) Macrophages
(d) All of these
Answer:
(d) All of these

Question 57.
During allergic reactions, which of the following is secreted ?
(a) Allergens
(b) Histamines
(c) Immunoglobulins
(d) Pyrogens
Answer:
(b) Histamines

Question 58.
Immunoglobulins are
(a) antigen
(b) antibodies
(c) antiseptics
(d) antibiotics
Answer:
(b) antibodies

Question 59.
B-lymphocytes are produced by
(a) liver
(b) thymus
(c) spleen
(d) bone marrow
Answer:
(d) bone marrow

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 60.
Cell-mediated immunity is due to
(a) B-cells
(b) T-cells
(c) T-helper cells
(d) All of these
Answer:
(b) T-cells

Question 61.
The cells which release the antibodies are
(a) helper T- cells
(b) B-cells
(c) plasma cells
(d) T-cells
Answer:
(b) B-cells

Question 62.
Antiviral substances are
(a) antibodies
(b) antibiotics
(c) interferons
(d) vaccines
Answer:
(c) interferons

Question 63.
The major phagocytic cells are
(a) lymphocytes
(b) mast cells
(c) macrophages
(d) plasma cells
Answer:
(c) macrophages

Question 64.
Which immunoglobulin is the largest in size ?
(a) IgA
(b) IgD
(c) IgE
(d) IgM
Answer:
(d) IgM

Question 65.
Vaccine for rabies was first produced by
(a) Louis Pasteur
(b) Edward Jenner
(c) Paul Berg
(d) None of these
Answer:
(a) Louis Pasteur

Question 66.
Vaccination means introduction in our body of
(a) weakened germs
(b) WBCs from other animals
(c) antibodies
(d) All of the above
Answer:
(a) weakened germs

Question 67.
The biochemical basis of vaccination was given by
(a) Louis Pasteur
(b) Salk
(c) Kohler
(d) Edward Jenner
Answer:
(d) Edward Jenner

Question 68.
Against which foreign organism (antigen) antibiotic is effective?
(a) Virus
(b) Bacteria
(c) Fungal infection
(d) Protozoan
Answer:
(b) Bacteria

Fill in the blanks

Question 1.
The immunity, present right from birth is known as ………….. immunity.
Answer:
innate

Question 2.
The immunity generated on exposure to foreign antigens is known as ………… immunity.
Answer:
acquired

Question 3.
Anti Tetanus Serum (ATS) administration generates ……….. immunity in the body.
Answer:
artificial passive

Question 4.
Toxoid is an example of ………. immunity.
Answer:
adaptive

Question 5.
A part of an antigen that evokes an immune response is called antigen …………
Answer:
determinant (epitope)

Question 6.
Antibodies segregate with ………. class of serum proteins.
Answer:
immunogens

Question 7.
The stem of the ‘Y’-shaped immunoglobulin molecule carries out …………. functions.
Answer:
effector

Question 8.
Among all immunoglobulins ………… can cross the placental barrier.
Answer:
IgM

Question 9.
During primary immune response, ………… immunoglobulin is predominant.
Answer:
IgG

Question 10.
Immunoglobulin ……………. is present in the mother’s milk, tear and saliva.
Answer:
IgA

Question 11.
Formation of antibodies against self antigens leads to an ………… disorder.
Answer:
autoimmune

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 12.
……………… released by mast cells causes inflammatory response.
Answer:
Histamine

Question 13.
Humans get AIDS virus from
Answer:
HIV contaminated needle

Question 14.
The tests conducted for determining AIDS and typhoid are test and test respectively.
Answer:
ELISA, Widal

Answer the following in one or a few sentences

Question 1.
What is passive acquired immunity ? Explain.
Answer:
It is the immunity which is acquired by readymade antibodies or sensitised WBC directly injected to a person. It provides immediate relief and is not long lasting. It is classified as
• Natural passive acquired immunity
• Artificial passive acquired immunity

Question 2.
What is an antigenic determinant (epitope)?
Answer:
Epitope is the component or an active site of an antigen which binds to the complementary past of an antibody called paratope. It is also known as antigenic determinant.

Question 3.
Explain humoral immunity.
Answer:
The immunity which is mediated by antibodies present in blood and lymph is known as humoral immunity or immune response.

Question 4.
Explain about the antigen binding sites of an antibody.
Answer:
‘Y’-shaped antibody molecule possesses antigen-binding sites and are known as Fragment antigen binding (Fab). This site has the ability to recognise a complementary antigen and bind to it.

Question 5.
Mention about the effector functions of an antibody.
Answer:
Antibodies have several mechanisms by which they act in body. To combat pathogens which are replicated outside cells, antibodies binds to pathogens to link them together.
It causes them to agglutinate. So, by coating the pathogen, antibodies stimulate effector functions against pathogen.

Question 6.
How do antigens interact with their antibodies?
Answer:
Antigen-Antibody Interaction:
Antigens or immunogens are whole organisms or foreign particles that can evoke immune responses and can bind to antibodies in a specific manner. An antibody interacts with the small specific part of an antigen, called epitope or antigenic determinant. Epitope is the immunologically specific component or active site of an antigen, which hinds to the complementary part of an antibody called paratope.

Question 7.
What is a toxoid ? Name the bacterial diseases against which toxoids are used as vaccines.
Answer:
Some pathogenic bacteria producess exotoxins which are isolated and chemically modified to reduce their toxicity. Such exotoxins are non-toxic immunogenic deterivatives also called as toxoids. Diphtheria and tetanus vaccines are produced from toxoids and treat bacterial diseases.

Question 8.
What is an oral polio vaccine?
Answer:
Polio is an infectious disease caused by a virus. Oral Polio Vaccine or OPV are the predominant vaccine used to eradicate polio. Oral polio vaccine results in vaccine associated paralytic polio.

Question 9.
What is immunosuppression?
Answer:
The reduction of activation of immune system is called immuno suppression. It can either be deliberate or as an adverse effect of any therapeutic agent. The major causes of immunosuppression are diabetes, chronic alcoholism, renal failure, autoimmune disorders or CNS infection.

Question 10.
Explain autoimmune haemolytic anaemia.
Answer:
The condition in which antibodies of a person target their own blood cells and cause them to burst, leading to an insufficient oxygen carrying blood cells in the circulatory system, is called Autoimmune Haemolytic Anaemia (AIHA).

Question 11.
What is an immune deficiency?
Answer:
The state in which immune system’s ability to fight diseases is negligible or completely absent is called immunodeficiency. It usually occurs as a result of extrinsic factors which includes HIV infection, extremes of age or environmental factors.

Question 12.
Explain reticular dysgenesis.
Answer:
Reticular Dysgenesis (RD ) is a rare inherited autosomal recessive disease that results in immune deficiency. A weakened immune system leave patients susceptible to different kinds of infections.

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Short Answer Type Questions

Question 1.
What is the causative organism of filariasis and write a note on its prevention and control.
Answer:
The causative organism of filariasis is Wuchereria bancrofii and Wuchereria malayi and Culex mosquito is the vector. Anti-mosquito measures are taken to eliminate the breeding places for the vectors which causes filariasis and also helps to control the spread.

Question 2.
Write the names of five drugs to control malaria.
Answer:
The drugs which helps to control malaria.
(i) Chloroquine
(ii) Doxycycline
(iii) Resochins
(iv) Paludrine
(v) Daraprim

Question 3.
What are the different species of malarial parasite?
Answer:
Malaria is caused by potozoan parasite Plasmodium. The four different species which cause malaria in humans

  • P. falciparum
  • P. ovale
  • P. malarial
  • P. vivax

Question 4.
What are the causes of non-communicable diseases?
Answer:
The major causes for the occurrence of non-communicable diseases are

  • Air-borne germs These are spread through air to a healthy individuals. Common diseases which spread through air-borne germs are measles, tuberculosis and chicken pox.
  • Direct/ indirect contact With a person suffering from communicable diseases can cause the spread.
  • Food borne/water borne The disease can also be caused by sharing food or water with infected person.

Question 5.
What are the measures taken to control malaria?
Answer:
Control Malaria:
In 1979, WHO expert committee summarised few antimalarial measures. These are

  • Use of mosquito repellents, bed-nets and cleaning of houses.
  • Use of aerosols near domestic area.
  • Destroy mosquito larvae by larvicides by using larvivorous fishes like Gambusia.
  • Manage water fills and digs to prevent the area from the development of larvae.
  • Chemoprophylaxis or little dose of quinine to be administered in malaria prone area.
  • Chemotherapy in which medicines like quinine, paluidine, camoquin, resochin, mepacrine, lavagnin, daraprin, etc., are given to people to prevent them from malarial infection.

Question 6.
Write a short note on tumour and their types.
Answer:
It involves the following common methods

  • Surgery In this primary approach, tumours are removed by surgery to check further spread of cancer cells.
  • Radiotherapy In this technique, tumour cells are irradiated by lethal doses of radiation by protecting the surrounding normal cells.
  • Chemotherapy In this several chemotherapeutic drugs are used to kill cancer cells. But their side effects like hair loss, anaemia are also reported.
  • Immunotherapy In this process of treatment, several biological modifiers like a-interferons are used to activate the immune system and help in destroying the tumour.

Question 7.
What is ascariasis and how it is controlled?
Answer:
Ascariasis:
It is caused by an intestinal endoparasite of human, i.e. Ascaris lumbricoides commonly known as roundworm. It is the most common nematode parasite that occurs worldwide and mostly found in tropical and subtropical areas where hygiene and sanitation are poor. The adult female is about 12 inches in length while male adults are smaller.

Infection is more common in rural areas of South-Eastern part in India. Children get more affected than adults by this disease due to poor sanitation habits.

Control:
Few preventive measures are given below

  • Maintain personal hygiene.
  • Consume thoroughly washed and properly cooked vegetables and fruits.
  • Drink packaged or boiled water.
  • Disposal of fecal matter away from habitation crops and water sources.
  • Do not let children to play in soil.

Question 8.
Write a short note on amoebiasis.
Answer:
It is caused by Entamoeba histolytica which is found inside or outside the intestine. The symptoms of intestinal amoebiasis are amoebic dysentry, non-dysentric colitis, amoeboma and amoebic appendicitis leading to complications like intestinal perforation, peritonitis and haemorrhage. The extra-intestinal amoebiasis can also occur in liver, lungs, brain, spleen and skin. The most common type of amoebiasis is hepatic amoebiasis.
E. histolytica is a monogenetic parasite and its only host is human.

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 9.
What is AIDS? How can it be prevented?
Answer:
AIDS or Acquired Immuno Deficiency Syndrome refers to a disorder in which the immunity of body is decreased due to reduction of T-helper cells that activate other lymphocytes too.
It’s caused due HIV or Human Immuno Deficiency virus.

Prevention of AIDS

  • Sterlise all surgical instruments before use.
  • The transfusion of blood should be subjected to HIV test.
  • Infected mother should avoid pregnancy otherwise, it may also transmit to child.
  • Heterosexual activites should be prohibited.
  • Motivate to use condoms during sexual activities.
  • Proper medical dispose off should be established.

Question 10.
What is diabetes mellitus ? How can it be controlled ?
Answer:
The increased level of blood sugar in human body due to hyposecretion of insulin hormone leads to a condition called hyperglycemia. Prolonged hyperglycemia leads to diabetes mellitus characterised by high sugar, weight loss and production of excess urine. It is an acquired non-communicable disease of humans.
Diabetes can be kept under control by changing the diet to sugar free, administration of insulin hormone through injections, etc.

Question 11.
What are carcinogen?
Answer:
Causes of Cancer:
The agents which cause cancer, are called carcinogens.
Cancer can be induced by either physical, chemical or biological agents.

  1. Physical agents These are ionising radiations like X-rays, Y-rays and the non-ionising rays like UV-rays which can cause DNA damage leading to neoplastic transformation.
  2. Chemical agents These are tobacco smoke, benzene fumes, arsenic, hormones and aflatoxin, etc.
  3. Nutritional agents Deficiency of some nutrients like vitamins, minerals and proteins causes cancer.
  4. Biological agents Many cancers are caused by viruses. The association of Hepatitis-B and primary liver cancer has been established.
  5. Mechanical factors Trauma, irritation and severe friction have been identified to cause malignancy.
  6. Host factors These include age, sex, marital status, race, socio-economic status, customs and habits of the host.
  7. Environmental factors These include radiation, air pollution, diet, drugs and social environment.
  8. Genetic factors All factors have a genetic basis and are caused by genetic transformations of cells.

Question 12.
What STDs stand for? Explain with examples.
Answer:
STD stands for sexually transmitted diseases, which are transmitted through sexual intercourse with infected persons. For example AIDS, syphilis, trichomoniasis.

  1. Syphilis is caused by bacteria named Treponema pallidum. Initially, it leads to ulcers on the genitalia followed by skin lesions, rashes and swollen joints.
    It is cured by taking penicillin or tetracycline as antibiotics.
  2. AIDS is caused by human immunodeficiency virus. It leads to decreased immunity of the patient along with many other symptoms like lethargy, weight loss, nausea, fiver. Although, AIDS is incurable yet a drug zidovudine (AZT) is used to treat this.
  3. Trichomoniasis is a STD caused by Protozoa Trichomonas vaginalis. It infects both male and female causing foul smelling, yellow discharge and burning sensation in females and pain and burning sensation in males. It is usually treated by metronidazole in both the cases.

Question 13.
What is cancer? Give its causes.
Answer:
It is defined as an uncontrolled growth or proliferation of cells without any differentiation. Cancer cells divide repeatedly in an uncontrolled manner. It has an ability to invade other tissues or organs, cause necrosis or programmed cell death, i.e. apoptosis.

In normal cells, cell growth and differentiation is highly controlled and regulated. Normal cell shows a property called contact inhibition by virtue of which contact with other cells stops their uncontrolled growth. Cancerous cells appear to have lost this property. As a result, these cells continue to divide to produce a mass of cells called tumour or neoplasm.

Causes of Cancer
The agents which cause cancer, are called carcinogens.
Cancer can be induced by either physical, chemical or biological agents.

  1. Physical agents These are ionising radiations like X-rays, v-rays and the non-ionising rays like UV-rays which can cause DNA damage leading to neoplastic transformation.
  2. Chemical agents These are tobacco smoke, benzene fumes, arsenic, hormones and aflatoxin, etc.
  3. Nutritional agents Deficiency of some nutrients like vitamins, minerals and proteins causes cancer.
  4. Biological agents Many cancers are caused by viruses. The association of Hepatitis-B and primary liver cancer has been established.
  5. Mechanical factors Trauma, irritation and severe friction have been identified to cause malignancy.
  6. Host factors These include age, sex, marital status, race, socio-economic status, customs and habits of the host.
  7. Environmental factors These include radiation, air pollution, diet, drugs and social environment.
  8. Genetic factors All factors have a genetic basis and are caused by genetic transformations of cells.

Question 14.
Write down different types of cancer.
Answer:
Types of Cancer
On the basis of localisation, cancer is divided into several main types

  1. Carcinoma Cancer of epithelial tissues and their derivatives, e.g. breast cancer, lung cancer.
  2. Sarcoma Cancer of connective tissues, e.g. bone cancer, muscle cancer, cancer of lymph nodes.
  3. Lymphoma Excessive production of lymphocytes by lymph nodes and spleen, e.g. Hodgkins disease, multiple myeloma and other immunoproliferative diseases.
  4. Leukemia Cancer of blood forming tissues like stem cells in bone marrow. There is increase in WBC number which destroys the cells of other organs, commonly known as blood cancer.

Some Other Types of Cancer

  • Adenoma Cancer of glands.
  • Lipoma Cancer of adipose tissue.
  • Glioma Cancer of glial cells of central nervous system.
  • Myoma Cancer of muscular tissue.
  • Melanoma Cancer of pigmented epithelium of skin.

Question 15.
What is the causative agent of gonorrhoea? What are its symptoms and treatment?
Answer:
Gonorrhoea is caused by bacteria Neisseria gonorrhoea. It resides in the genital tube and produces pus-containing discharge, pain around genitalia and burning sensation during urination. It can be cured through appropriate medicines like penicillin or ampicillin.

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 16.
Explain oncogenes.
Answer:
Oncogenes encode oncoproteins which promote the loss of growth control and the transformation of cell to a malignant cells. Cancer causing viruses are called oncoviruses and their genes as v-onc (viral protooncogenes). These viral oncogenes have homologous regions in human genome.

The homologous genes are called cellular protooncogenes (c-onc). Nearly 100 protooncogenes are known and these are involved in cell functions. The sudden-change, i.e. mutation in protooncogenes induces abnormal functioning and tumour formation.
Viral protooncogenes constitutes another class of factors transforming protooncogenes into expression ready cellular oncogenes. Which encodes for abnormal proteins known as oncoproteins.
The name of oncogenes are derived from the names of the host viruses are v-src, c-myc, etc.

Question 17.
Explain tumour suppressor gene or antioncogene.
Answer:
Tumor suppressor gene or antioncogene are the genes which protects a cell form the formation of cancerous cells.
When these genes are characterised by mutations, it leads to reduction in its function and cell becomes prone to cancer along with many genetic changes. Tumor suppressor genes are categorised into caretaker gatekeeper and landscaper genes.

Question 18.
Write a note on parasite.
Answer:
Parasite are the organism which lives on other organisms called host and derive their nutrition from the host. They are dependent on host for their survival and they have to be in host, to live, grow and multiply. The one which lives on the surfaces of earth are called ectoparasite, while which lives in the organisms are called endoparasites.

Question 19.
Explain incubation period of malaria parasite.
Answer:
The incubation period in malaria is defined as the period between infection and beginning of the symptoms. It typically lasts between 10 days to 4 weeks.
The incubation period is affected by the type of Plasmodium parasite responsible for the infection. If a patient is given antimalarial drugs which prevent the spread of disease, it can also increase the incubation period by weeks or months.

Question 20.
What kind of physical changes are characteristic of adolescence?
Answer:
Physical changes Adolescence is a period of active growth and sexual maturity. Growth becomes once apparent with an increase in body size, height and weight due to continued secretion of growth and sex hormones (FGH and LH).

Under the influence of these hormones, the body begins to develop secondary sexual characters in males like beard growth, change in voice pitch, etc., and females such as initiation of menstruation, enhanced breast size, etc.

Question 21.
What kind of psychological changes characterise adolescence?
Answer:
Psychological changes Adolescence shows changes in behaviour, emotions and attitude.
For example, difficulty in accepting parental decisions, coping with studies, competition, increased need for money, keeping bad company, etc.

Question 22.
Which is the most common skin problem that affects the youth in adolescence? What are its causes ?
Answer:
The most common skin problem that occurs during youth or teen years is acne.
During puberty hormone level increases and the skin starts releasing more oil (sebum). When this mixes with dead cells of the skin, it closes the pores and causes swelling, redness and pus.
Few medical conditions such as polycystic ovary syndrome, cushing’s syndrome can also lead to acne.

Question 23.
What is the cause of alcoholism?
Answer:
The dependence on alcohol or when a person becomes addicted to alcohol is called alcoholism. It is a result of combination of genetic, psychological, environmental and social factors described below

  • People become addicted to alcohol to relieve stress and the deal with the pressure in their families or workplace.
  • Disorders like anxiety, depression, bipolars disorders or other medical issue can increase the risk of alcoholism.

Question 24.
What are the effects of alcoholism in the body?
Answer:
Effects of Alcohol

  1. High dose of alcohol, i.e. more than 30 ml acts as an intoxicant and affects the functioning of CNS.
    Alcoholism damages internal organs like liver, as alcohol is converted to acetaldehyde then to fat in liver. This fat begins to deposit in body and cause cirrhosis.
  2. It many also cause hepatitis and liver cancer.
  3. Increased consumption of alcohol per day dilates the blood vessels and it leads to hardening of blood vessels. This causes bradycardia and myocardiopathy.
  4. Alcohol decreases ADH secretion and this may cause dehydration.
  5. Drinking alcohol makes the person unusually aggressive and also, affects this judgement, coordination, alterness, vision and responsiveness.
  6. Excessive intake of alcohol affects the behaviour of an individual.

Question 25.
What are the moral and social implications of drinking?
Answer:
Social and Moral Implications of Addiction

  • Habitual drinking creates differences in the family. The addict cause public misdemeanour and misbehave hence is isolated from society as drinking is considered a social evil.
  • Family status declines due to approaching poverty.
  • Such individuals become violent suicidal, antisocial and- lazy.
  • Drug and alcohol addicts develop habits like stealing, burrowing money for fulfilling their addiction.

Question 26.
What are the reasons of drug abuse by the youth?
Or
Mention the causes of drug abuse.
Answer:
Serious effects of drug abuse are

  1. Academic performance decreases.
  2. Frequent absence from school or college.
  3. Isolation, fatigue, depression and aggressiveness occurs in behaviour.
  4. No coordination with family members and friends.
  5. Frequent fluctuations in weight.
  6. Intravenous drug intake leads to risk of AIDS and hepatitis-B.
  7. Excess use of alcohol or drug damages the nervous system and causes liver cirrhosis or cancer.
  8. During pregnancy in females, drugs affect foetus seriously.

Question 27.
Write briefly on the main classes of drugs in use.
Answer:
Drugs are placed into one of three classes A, B or C under the misuse of drugs act 1971.
A drug class is the set of similar action placed under same group.
Class-A includes heroin (diamorphine), cocaine, methadone, LSD, ecstasy and magic mushrooms.
Class-B includes amphetamines, barbiturates, codeine, cannabis, cathinones and synthetic cannabiroids.
Class-C includes benzodiazepines (transquilisers), ketamine, anabolic steroids and Benzylpiperazines (BZP).

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Question 28.
What are the withdrawal symptoms that are seen after drug abuse?
Answer:
The withdrawal symptoms of alcohol and,drug abuse are-

  1. Hallucination
  2. Fits
  3. Tremors

The treatment of drug abuse includes

  1. Use of detoxifying drugs like diazepan, vitamin-B, chlordizepoxide, apomorphine.
  2. Use of antioxidants like disulfiram, cephalosporin, metronidazole.

Question 29.
What are the social and moral implications of drug abuse?
Answer:
Social and Moral Implications of Addiction

  1. Habitual drinking creates differences in the family. The addict cause public misdemeanour and misbehave hence is isolated from society as drinking is considered a social evil.
  2. Family status declines due to approaching poverty.
  3. Such individuals become violent suicidal, antisocial and- lazy.
  4. Drug and alcohol addicts develop habits like stealing, burrowing money for fulfilling their addiction.

Question 30.
What are the effects of tobacco use in the body?
Answer:
Dried and crushed leaves of Nicotiana tabacum and Nicotiana rustica are used to make tobacco. It can induce lung cancer, bronchitis, emphysema, coronary heart disease, cancer of throat, oral cancer, cancer of urinary bladder, etc.
Smoking leads to the increase in the content of carbon monoxide in the blood which reduces the concentration of haemoglobin bound oxygen. This leads to oxygen deficiency in the body

Question 31.
What kind of diseases affect the body in smoking?
Answer:
Major diseases which affect the body in smoking are

  1. Stroke As smoking affects the arteries of a person, it can trigger stroke.
  2. Lung cancer It is the most common type of cancer caused due to smoking.
  3. Chronic obstructive pulmonary disease It is an obstructive lung disease which leads to difficulty in breathing. It leads to early death.
  4. Asthma Smoking irritates air passages and can trigger sudden and severe asthma attacks.
  5. Cancer Over ten other types of cancer including colon, liver, cervix, stomach and pancreas are caused due to smoking.

Question 32.
What is mental illness?
Answer:
Mental illness is a state of emotional and psychological well being of a person which allows him/her to attain his/ her physical cognitive and emotional capabilities.

Question 33.
What are the causes of mental illness?
Answer:
The causes of mental illness are depression, obsessive, compulsive disorder, mood disorder, attention deficiency disorder, sleeplessness, self destructive actions, loss of memory, etc.

Question 34.
What are the different types of mental disorders seen in man?
Answer:
The different types of mental disorders are

  1. Psychosis
  2. Neurosis
  3. Schizophrenia
  4. Phobia
  5. Epilepsy
  6. Parkinson’s disease
  7. Alzhaimer’s disease.

CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases

Differentiate between the following

Question 1.
Amoeba and Entamoeba.
Answer:
Differences between Amoeba and Entamoeba are as follows

Amoeba Entamoeba
Amoeba are a large class of unicellular organisms that are eukaryotic. Entamoeba are a genus of amoeba that live on other organisms.
They move by means of pseudopodia. They are usually a part of normal fauna and live in symbiotic association.

Question 2.
Filaria and Malaria.
Answer:
Differences between filaria and malaria are as follows

Filaria Malaria
It is a parasitic disease caused by infection of roundworms. Malaria is caused by parasitic protozoan to Plasmodium.
The symptom includes edema which causes thickening of skin and underlying tissues. The symptoms include headache, fever, joint pain, retinal damage, etc.
Wuchereria bancrofti is the parasite which causes filariasis. Female Anopheles mosquito transmits a mature infective form to the host.

Question 3.
Communicable Diseases And Non-Communicable Diseases
Answer:
Differences between communicable diseases and non-communicable diseases are as follows

Communicable diseases Non-communicable diseases
These disease do not remain confined to the person who suffer from them. These disease remain confined to the person who suffer from them.
These are transmitted from infected person to other persons directly or indirectly by any causative organisms. They are not transmitted from infected person to other persons except for genetic transmission in some case.
e.g. viral diseases (influenza, mumps, AIDS, smallpox) bacterial diseases (cholera,typhoid, TB,tetanus,etc.) e.g, diabetes, cancer, arthritis, cardiovascular diseases, etc.

Question 4.
Magna and Minuta stage.
Answer:
Differences between magna stage and minuta stage are as follows

Magna stage Minuta stage
It is the active stage of pathogenic stage of Entamoeba histolytica also called trophozoite. It is the non-pathogenic and non-motile and non-feeding form.
It resembles Amoeba in its active form and cytoplasm is divisible into ectoplasm and endoplasm. It lives in the lumen of intestine and may develop into magna by penetrating intestinal wall.
It measures 20-30 p in diameter. It measures about 12-15 pt in diameter.

Question 5.
Infection and Infestation.
Answer:
Differences between infection and infestation are as follows

Infection Infestation
Infection is caused by microorganisms or germs or viruses. Infestation refers to the infection by larger and more complete organisms like pests or parasites.
The germs or microorganisms which causes infection usually grow inside the body and cause illness. Infestation is commonly used in context to one organism present on external surface.

Question 6.
Carcinoma and Sarcoma.
Answer:
Differences between carcinoma and sarcoma are as follows

Carcinoma Sarcoma
The cancer of epithelial epidermal tissues and their derivatives. The cancer of connective tissues is called sarcoma.
It usually includes cancer of lungs, breast, etc. It includes bone, muscle or cancer of lymph nodes.
They spread throughout the body by blood and lymph. They spread through nodules.
They occur primarily in people over 50 years of age. It affects both young and old people.

Question 7.
Benign tumour and Malignant tumour.
Answer:
Differences between benign tumour and malignant tumour are as follows

Benign tumour Malignant tumour
It remain confined to the site of its origin. It is not confined to the tissues.
It may grow in size but does not spread to other parts of the body. They are carried to other parts of the body by blood or lymph.
It is enclosed in connective tissue. It is not enclosed in any specific tissue.

Question 8.
Sporogony and Gamogony.
Answer:
Differences between sporogony and gamogony are as follows

Sporogony Gamogony
It is an asexual stage which produces haploid sporozoites. It is a sexual phase followed by sporogony which produces a diploid zygote.
The oocyte so formed are liberated into haemolymph of mosquito. The zygote elongates into motile worm-like vermicule which penetrates the stomach wall, enclosed itself in a cyst and grows in size.

Question 9.
Innate immunity and Acquired immunity.
Answer:

Innate immunity Acquired immunity
It is a non-specific type of immunity. It is a pathogen specific immune response.
It is inherited from parents and protects the child since birth. It is acquired after the birth of an individual, during its lifetime.
It provides barrier against the entry of pathogen in the body. It produce B-lymphocytes and T-lymphocytes. It also produces primary and secondary types of immune response.

Question 10.
Cell-mediated immunity and Humoral immunity.
Answer:
Differences between cell-mediated immunity and humoral immunity are as follows

Cell-mediated immunity Humoral immunity
It is the type of immunity which is mediated by T-lymphocytes to produce antibodies. It is mediated by antibodies present in blood and lymph.
It provides immunity against all pathogens including fungi and Protozoa. It provides immunity against virus and bacteria.
It shows reaction against organ transplantation. It does not react against organ transplantation.

Question 11.
Vaccination and Immunisation.
Answer:
Differences between vaccination and immunisation are as follows

Vaccination Immunisation
Vaccination is the process of introducing the body to a form of virus. Immunisation is the process of body building up natural defence against bacteria.
It is injected in the form of drops. It does not require administering as its the natural capacity of body.

Long Answer Type Questions

Question 1.
What are pathogens ? Classify diseases and give a note on this.
Answer:
Pathogens:
Infectious disease causing agents are called pathogens and their disease causing capacity is known as pathogenicity or virulence. Most of the pathogens are parasites. They can cause harm to the organism (host) by either living in (as endoparasites) or on them (as ectoparasite). These disrupt the normal physiology of organisms, either plants or animals and express a number of symptoms. The human body contains many natural defence mechanisms against some common pathogens. Certain pathogens have been found to be responsible for massive casualities.

Despite many medical advances for safeguarding human beings from infections by pathogens through the use of vaccines, antibiotics and fungicides, pathogens continue to threaten human lives. They can enter our body by various means s.uch as air, water, food, etc., and can multiply and interfere with the normal vital activities of the body, thus, resulting in the morphological and functional damage.

Classification of Pathogens:
Major classes of pathogens which cause disease, produce toxin and induce immunosuppression in the most are given below

Classes of Pathogens Examples
Viruses Adenovirus, picorna virus, retrovirus, papovavirus, polyma virus, etc.
Bacteria Mycobacterium, Streptococcus, Shigella and Salmonella.
Fungi Saprophytic pathogenic fungi.
Prions Protein pathogens that cannot contain nucleic acids.
Parasites Protozoan and helminth parasites.

Question 2.
Give the symptoms, infection, prevention and control of typhoid.
Answer:
Typhoid:
It is also known as enteric fever and is caused by bacterium Salmonella typhi. It is common in developing countries, where it affects 21.5 million persons every year (1 million in India).

Symptoms and Diagnosis
The incubation period of parasite is about 1-2 weeks and the duration of illness is about 4-6 weeks. The symptoms of typhoid include fever (39-40°C), lethargy, stomach pain, headache, poor appetite, diarrhoea or constipation and rose spots on abdomen. The intestinal
perforation or bleeding may occur in severe cases, which may lead to death. The reccurrence (relapsing) of disease is observed in 10% of patients. Typhoid is diagnosed by WIDAL test. ”

Infection and Transmission
Salmonella typhi invades human intestine through contaminated water or food, from where they are carried by white blood cells to the liver, spleen and bone marrow. They multiply in these organs and re-enter the bloodstream. At this stage, a person begins to develop symptoms like fever. Through the bloodstream the bacteria further invade various organs like gall bladder, biliary system, lymphatic tissues and ultimately pass into the intestinal tract. From here the bacteria can be diagnosed in cultures of stool.
The disease may be transmitted through carriers also. These are those person who recovered from typhoid but continue to carry the bacterial infection.

Treatment
Treatment includes antibiotics like fluoroquinolones, ceftriaxone and azithromycin. But, Salmonella develops resistance to multiple antibiotics. The emergence of multi-drug resistant typhoid has complicated the treatment procedure, especially in those who have acquired infection from South Asia. Therefore, antibiotic susceptibility test in helpful in deciding an appropriate therapy pathway.

Vaccination
The following vaccines are given in the treatment of typhoid

Vaccine Type Dose
Vi antigen Inactivated vaccine injectable A single dose of 0.5 ml administered intramuscularly on thighs and arms.
Oral ty21a Oral live vaccine Course of 3 capsules given on alternate days.

Prevention and Control
There are several ways through which bacterial infection of typhoid can be avoided

  1. Maintain personal hygiene.
  2. Consume thoroughly washed and properly cooked vegetables and fruits.
  3. Drink packaged water with statutory quality or boiled water. Bottled carbonated water is also used to consume and it is safer than other.
  4. Vaccination against typhoid can be done.

Question 4.
What are acquired and innate immunity?
Discuss the mechanical and chemical barriers of innate immunity.
Answer:
Innate Immunity (Inborn)
It is the type of immunity which is present from birth and is inherited from the parents. That’s why it is also called as natural immunity. It is non-specific in nature as it involves general protective measures against any invasion. Innate immunity provides the early lines of defense against pathogens. The principal components of innate immunity that act as barrier system to prevent the entry of pathogens are given below
1. Mechanical barriers
2. Chemical barriers
3. Phagocytosis
4. Fever
5. Inflammation
6. Acute phase proteins
7. Natural Killer (NK) cells

1. Mechanical or Physical Barriers:
They prevent entry of microorganisms in the body, e.g. skin, mucous coating of epithelium lining the respiratory, gastrointestinal and urogenital tracts. These barriers are also called as first line of defence.

  • Skin It is outer and tough layer of epidermis that consists of insoluble protein called keratin. It prevents the entry of bacteria and viruses. The periodical sheding off process of skin removes any clinging pathogen.
  • Mucous membrane The gastrointestinal tract, urinogenital tract and conjuctiva are lined by mucous membrane.

This membrane secretes mucus which entraps microbes, dust or any foreign particles and finally propelled them out through tears, saliva, coughing and sneezing.

2. Chemical or Physiological Barriers
It includes certain chemicals which dispose off the pathogens.
These are given below

  1. Acid of stomach, kills the ingested microorganisms by secreting acid gastric secretion (pH 1.5 – 2.0).
  2. Low pH of sebum (i.e. 3.0-5.0) forms a protective film over the skin that inhibits growth of many microbes.
  3. Lysozyme is a hydrolytic enzyme present in all mucous secretions like tears, saliva and nasal secretions. It attacks bacteria and dissolves their cell walls.
  4. Gastro and duodenal enzymes secrete proteases and lipases. These enzymes digest a variety of structural and chemical constituents of pathogens, e.g. gastric acids easily inactivate rhinoviruses.
  5. Mothers milk Lactoferrin and neuraminic acid are antibacterial substances present in human milk to fight against Staphylococci.
  6. A group of proteins produced by virus infected cells, i.e. interferons induces a generalised activated state in neighbouring uninfected cells.
  7. Humans and some other animals secrete an number of antimicrobial peptides such as defensins. One micrometre thick biofilm of defensins protects the skin from microbial assault.

3. Phagocytosis
When pathogens or microbes penetrate the skin or mucous membrane certain cell types surge towards the site of infection. These can be neutrophils, monocytes and macrophages which engulf the pathogens to form a large intracellular vesicle called phagosome.
The phagosome fuses with lysosome to form phagolysosome. The secretion of lysosomal enzymes digests bacterial cells. The useful products remains in the cell while the waste is egested out of the cell. Therefore, these phagocytes are also known as second line of defence.

4. Fever
It may be brought about by endotoxins or proteins (cytokines) produce by pathogens called endogenous pyrogens.
When enough pyrogens are produced, then there is rise in temperature which strengthens the defence mechanism to inhibit the growth of microbes. Fever is a symptom of an internal diagnoses of the cause of infections.

5. Inflammation
It is a defensive response of the body to tissue damage.
It is characterised by abrasions, chemical irritations, heat, swelling, redness and pain. Inflammation in a non-specific response of the body to injury. It is an attempt to dispose off microbes, toxins or foreign material at the site of injury by macrophages to prevent their spread to other tissues and to prepare the site for tissue repair. Thus, it helps to restore tissue homeostasis.

Broken mast cells release histamine, bradykinin, etc., which cause dilation of capillaries and small blood vessels. As a result more blood flows in these areas making them red and warm. Therefore, the accumulation of this results into tissue swelling (oedema).
After few days, due to phagocytosis, a cavity containing necrotic tissue and dead bacteria is formed. This fluid mixture is called pus.

6. Acute Phase Proteins
The chemical messenger of immune cells called cytokines are important low molecular weight proteins. These heterogenous proteins stimulate or inhibit the differentiation, proliferation or function of immune cells and also certain viral infections.

7. Natural Killer (NK) Cells
These are non-phagocytic granular lymphocytes which are present in spleen, lymph nodes and bone marrow.

Then can produce perforins or cytolysin which lyses the vi ral infected cells.
These cells can kill a range of tumour cells without any antigen specificity.

Mechanism of Active Aquired immunity
Active acquired immunity is more effective and superior than passive immunity. It occurs in two different forms called cell-mediated and humoral immune responses.
(i) Cell-Mediated Immune Response or Cell-Mediated Immunity (CMI) It is the type of acquired immunity mediated by T-lymphocytes. Activated T-lymphocytes undergo proliferation and differentiate into different types of effector cells, such as T-helper (TH) and T-cytotoxic or killer (Tc) lymphocytes and memory T-lymphocytes (TM). TM confers a long term memory against the invading pathogen. Tc /TK cells directly kill or destroy antigens or antigen bearing pathogens. TH cooperates with B-lymphocyte and triggers its transformation into a plasma cell.

(ii) Humoral response or Antibody-Mediated Immunity (AMI) It is mediated by antibodies present in blood and lymph. Immunoglobulins or antibodies are glycoproteins produced in the body by B-cells in response to an antigen, e.g. IgA. IgG, IgM, IgE and IgD.

B-cells multiple in large number and transform into larger cells called plasma cells or plasmocytes. The transformation into plasma cells in assisted By T-Helper cells (TH). These antibodies destroy antigens by specific antigen-antibody interaction.

Question 5.
Mention the factors causing cancer. Add a note on diagnosis and prevention of cancer.
Answer:
Causes of Cancer
The agents which cause cancer, are called carcinogens.
Cancer can be induced by either physical, chemical or biological agents.

  1. Physical agents These are ionising radiations like X-rays, v-rays and the non-ionising rays like UV-rays which can cause DNA damage leading to neoplastic transformation.
  2. Chemical agents These are tobacco smoke, benzene fumes, arsenic, hormones and aflatoxin, etc.
  3. Nutritional agents Deficiency of some nutrients like vitamins, minerals and proteins causes cancer.
  4. Biological agents Many cancers are caused by viruses. The association of Hepatitis-B and primary liver cancer has been established.
  5. Mechanical factors Trauma, irritation and severe friction have been identified to cause malignancy.
  6. Host factors These include age, sex, marital status, race, socio-economic status, customs and habits of the host.
  7. Environmental factors These include radiation, air pollution, diet, drugs and social environment.
  8. Genetic factors All factors have a genetic basis and are caused by genetic transformations of cells.

Cancer Detection and Diagnosis
Cancer can be detected by the following well-known methods

  1. Blood and bone marrow tests are conducted to know number of cell counts, e.g. WBC count in leukemia.
  2. Biopsy of a piece of suspected tissue is done by cutting thin sections, staining and examining them under the microscope.
  3. Radiography by X-rays is done to detect cancer of the internal organs.
  4. Computed tomography using X-rays is done to generate a 3-D image of internal tissue.
  5. Magnetic Resonance Imaging (MRI) involves the use of non-ionising radiation and strong magnetic field to detect pathological and physiological changes in living tissue.
  6. Monoclonal antibodies against cancer-specific antigens are also used for cancer detection. These are homogenous immunological reagents of defined specificity.
  7. Mammography for detection of breast cancer.
  8. Fine Needle Aspiration Cytology (FNAC)
  9. PAP test (cytological staining) used for detection of cervix cancer.

Prevention
According to WHO, the prevention of cancer is ‘the elimination of or protection against, factors known or believed to be involved in carcinogenesis (formation cancerous tumours) and the treatment of precancerous conditions’.
This implies that cancer cannot be cured at the late-stage so it is better to adopt some preventive measures be for its initiation and progression.

The preventive measures are

  1. Educate people to go for early diagnosis and early treatment for better chance of survival.
  2. Motivate people to know about the oncogenic effects of tobacco.
  3. Prohibit the advertisements of cigarettes and drugs that may increase the chances of cancer.
  4. Maintain personal hygiene.
  5. Control environmental pollution by taking major steps.
  6. Reduce amount of radiation.
  7. Organise occupational health programmes.
  8. Take treatment of cancerous.

Symptoms of Initiation and Progress of Cancer

  1. A hump or hard area in the breast.
  2. A change in wart or mole.
  3. A persistent change in digestive and bowel habit.
  4. A persistent cough or hoarseness, excess loss of blood at the monthly period or loss of blood outside the usual dates.
  5. Blood loss from any natural orifice.
  6. A swelling or sore that does not get better.
  7. Unexplained loss of weight.

Question 6.
Give the structure of HIV. Give an account of infection, control and prevention of AIDS.
Answer:
Structure of HIV:
The virus belongs to retrovirus family and is roughly spherical-shaped with a diameter of 90-120 nm. The core has two single-stranded RNA enveloped with conical capsid mode up of viral proteins P24.
Each RNA fdament is segmented into two identical filaments and associated with nucleo-capsid proteins and enzymes like reverse transcriptase enzyme and integrase.
CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases 1
Reverse transcription in HIV

Between the capsid and matrix of the virus proteases and other proteins are present. The matrix consists of lipid bilayer of host cell membrane and projecting knob like glycoprotein spikes. It contains two proteins called gpl20 and gp 41. These proteins are required for anchoring to the host cell and entering into it.
CHSE Odisha Class 12 Biology Solutions Chapter 9 Health and Diseases 2
Structure of human immunodeficiency virus

Infection:
HIV infects CD 4 T-lymphocytes and dendritic cells. It anchors to the surface of host cell by adsorption assisted by the glycoprotein. This release the capsid and then the begins the replication of retrovirus begins in the host cell. The HTV lefts untreated then it progresses in three stages

  1. Acute infection Patient develops some symptoms like flu, fever, swollen glands, soar throat, rash, muscle and joint pain and headache, within 2-4 weeks and HIV infection. This stage is at higher risk of transmitting virus through coitus and injectable drugs using contaminated needle.
  2. Clinical latency stage In this case, virus replicates in host cell without expressing symptoms. If patient undergoes retroviral therapy he may live for decades. If he is not on the therapy, then the latency stage lasts on an average for ten years.
  3. Typical (AIDS) stage At this stage, HIV infection acquires its full strength and damages the immune system badly. The value of CD 4 lymphocytes fall below 200 per cubic mililitre of blood. The infected person contracted many other bacterial and fungal diseases.

Some of the prevalent symptoms at this stage are

  1. Weight loss and unexplained tireness
  2. Chronic diarrhoea
  3. Pneumonia
  4. Prolonged swelling of the lymph glands of armpit, groin and neck.
  5. Recurring fever with night sweats.
  6. Persistent cough
  7. Mouth and skin problems
  8. Recurrent infections
  9. Sores of the mouth, anus and genitals .
    Without treatment, people who progress to AIDS survive about three years.

Diagnosis:
There are several methods for the diagnosis of HIV infection such as viral culture, Enzyme Linked Iramuno Sorbent Assay (ELISA), PCR test, Western blotting, etc. Out of these, ELISA and Western blotting are widely used.

Prevention:
There is no effective treatment developed to treat AIDS. Therefore, some preventive measures are recommended to prevent its infection

The preventive measures are as follows

  1. Sterlise all surgical instruments before use.
  2. The transfusion of blood should be subjected to HIV test.
  3. Infected mother should avoid pregnancy otherwise, it may also transmit to child.
  4. Heterosexual activites should be prohibited.
  5. Motivate to use condoms during sexual activities.
  6. Proper medical dispose off should be established.

Government of Indian launched national AIDS control board, national AIDS committee, national AIDS control organisation, etc., to create awareness among people about HIV transmission and progression of AIDS.

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 8 Question Answer Evolution

Evolution Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Choose the correct options

Question 1.
Life originated on earth about
(a) 2.5 billion years ago
(b) 3.5 billion years ago
(c) 4.5 billion years ago
(d) 5.5 billion years ago
Answer:
(c) 4.5 billion years ago

Question 2.
Which theory proposes the formation of living beings from non-living things?
(a) Theory of panspermia
(b) Theory of abiogenesis
(c) Theory of biogenesis
(d) Theory of special creation
Answer:
(b) Theory of abiogenesis

Question 3.
Who proposed the chemical evolution of life?
(a) AI Oparin – JBS Haldane
(b) Louis Pasteur – AI Oparin
(c) Francesco Redi – JBS Haldane
(d) Spallanzani – Louis Pasteur
Answer:
(a) AI Oparin – JBS Haldane

Question 4.
Which of the following compounds Miller-Urey used in the experimental synthesis of amino acids?
(a) CH4 NH3,CO2 and H2O
(b) CH4,NH3,H2 and H2O
(c) CH4,CO2 H2 and H2O
(d) CH2,N2,H2 and H2O
Answer:
(b) CH4,NH3,H2 and H2O

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 5.
Hot ocean water containing concentrated of prebiotic organic compounds was known as
(a) colloid
(b) crystalloid
(c) gelatinous mixture
(d) primordial soup
Answer:
(d) primordial soup

Question 6.
Which of the following was formed first?
(a) Virus
(b) Prokaryote
(c) Coacervates
(d) Eukaryote .
Answer:
(c) Coacervates

Question 7.
A paper on ‘natural selection’ and ‘origin of species’ was presented in the Linnaean Society of London in 1858 by
(a) Charles Darwin – Robert Malthus
(b) Charles Darwin – Alfred R Wallace
(c) Hugo de Vries – Robert Malthus
(d) Alfred R Wallace – August Weismann
Answer:
(b) Charles Darwin-Alfred R Wallace

Question 8.
Analogous organs have
(a) different origin and similar function
(b) similar origin and similar function
(c) similar origin and different function
(d) different origin and different- function
Answer:
(a) different origin and similar function

Question 9.
Find out the odd match.
(a) Aerial-Flying
(b) Fussorial-Burrowing
(c) Cursorial-Running
(d) Arboreal-Swimming
Answer:
(d) Arboreal-swimming

Question 10.
Which one of the following sets of organs constitutes vestigial organs?
(a) Appendix, coccyx and plica semilunaris
(b) Appendix, pectoral girdle and caecum
(c) Large intestine, coccyx and ear muscle
(d) Appendix, coccyx and rectum
Answer:
(a) Appendix, coccyx and plica semilunaris

Question 11.
What is the correct ascending order?
(a) Mesozoic, Cenozoic and Palaeozoic
(b) Cenozoic, Mesozoic and Palaeozoic
(c) Palaeozoic, Mesozoic and Cenozoic
(d) Palaeozoic, Cenozoic and Mesozoic
Answer:
(d) Palaeozoic, Cenozoic and Mesozoic

Question 12.
Who is known as the ’Father of Modern Palaeontology’?
(a) Leonardo da Vinci
(b) Karl Ernst von Baer
(c) Ernst Haeckel
(d) Georges Cuvier
Answer:
(d) Georges Cuvier

Question 13.
Find the incorrect match.
(a) Blood group ‘A’-Antigen A
(b) Blood group ‘AB’-No antibody
(c) Blood group ‘O’-Andgen A and B
(d) Blood group ‘B’-Antibody anti-A
Answer:
(c) Blood group-‘0’-antigen A and B

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 14.
Which is not a case of chromosomal aberration?
(a) Recombination
(b) Duplication
(c) Inversion
(d) Translocation
Answer:
(a) Recombination

Question 15.
Which type of natural selection removes individuals from both ends of a phenotypic distribution?
(a) Directional
(b) Disruptive
(c) Stabilising
(d) None of these
Answer:
(c) Stabilising

Question 16.
Which is not a great ape?
(a) Gorilla
(b) Chimpanzee
(c) Orangutan
(d) Macaque
Answer:
(d) Macaque

Question 17.
What is the correct sequence in human evolution?
(a) Homo habilis, H. erectus, H. neanderthalensis, H. sapiens
(b) Homo erectus, H. habilis, H. neanderthalensis, H. sapiens
(c) Homo habilis, H. neanderthalensis, H. sapiens, H. erectus
(d) Homo erectus, H. neanderthalensis, H. habilis, H. sapiens
Answer:
(a) Homo habilis, H. erectus, H. neanderthalensis, H. sapiens

Fill in the blanks

Question 1.
Organic evolution refers to a change in diversity and ……….. in populations of organisms.
Answer:
adaptations

Question 2.
The concept of chemical evolution was proposed by JBS Haldane and a Russian scientist, ………… .
Ans
AI Oparin

Question 3.
Charles Robert Darwin hailed from …………….
Ans
Britain

Question 4.
Charles Robert Darwin went on a voyage on board the ship
Answer:
HMS Beagle

Question 5.
Jean Baptiste de Lamarck wrote a book, entitled ……….., which embodied his theory of inheritance of acquired characters.
Answer:
Philosophic Zoologique

Question 6.
Charles Darwin was inspired by the population theory proposed by …………… .
Answer:
Robert Malthus

Question 7.
Darwin’s contemporary ………… was studying population diversity in the erstwhile East Indies.
Answer:
Alfred Russel Wallace

Question 8.
The mutation theory was proposed by ………….. .
Answer:
Hugo de Vries

Question 9.
All the present day life has originated from a single ancestral form, designated as …………. .
Answer:
protobionts

Question 10.
A gene mutation involving only one nucleotide is called as …………….
Answer:
point mutation

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 11.
Abiogenesis of simple organic molecules was experimentally proved by ………….. and …………..
Answer:
Miller, Urey

Question 12.
The ‘theory of inheritance of acquired characters was proposed by …………. who hailed from …………. .
Answer:
Lamarck, France

Question 13.
August Weismann’s ……….. theory gave a thunder blow to Lamarckism.
Answer:
germplasm

Question 14.
Charles Darwin studied the diversity of a class of birds, commonly known as ………….. in the Galapagos archipelago.
Answer:
Darwin’s finches

Question 15.
The original title of Darwin’s book was ……………. .
Answer:
‘On the origin of species by means of natural selection’.

Question 16.
Natural selection in action was demonstrated by …………. moth.
Answer:
peppered

Question 17.
The earliest form of horse was ………….. that was living in the plains of North America.
Answer:
Eohippus

Question 18.
The fossil of ……….. discovered from the sedimentary rocks of Bavaria, Germany is the missing link between reptiles and birds.
Answer:
Archaeopteryx

Question 19.
Digits II and IV persist in modern horse as reduced structures, known as ……….. bones.
Answer:
splint

Question 20.
Modifications of the basic pentadactyl limb plan in vertebrates to meet their needs is known as ……….
Answer:
homology

Question 21.
The arrangement of different eras, periods and epochs in their ascending order of time constitutes the …………..
Answer:
geological time scale

Question 22.
Peripatus is connecting link between …………. and …………….
Answer:
Annelida, Arthropoda

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 23.
………… era is known as the era of reptiles.
Answer:
Mesozoic

Question 24.
Sudden reappearance of some ancestral characters in the present organisms is called as ……………
Answer:
atavism

Question 25.
The effect of …………. is larger in small populations and smaller in large populations.
Answer:
genetic drift

Answer each of the following in one word or more words, wherever necessary

Question 1.
The theory that explains that life originated on this planet from non-living chemical constituents.
Answer:
Theory of abiogenesis

Question 2.
The ocean water that contained concentrated amount of prebiotic organic compounds.
Answer:
Primordial soup or prebiotic

Question 3.
The droplets formed by the separation of high molecular weight organic compounds in a colloidal solution.
Answer:
Coacervates

Question 4.
Protenoids, when dissolved in water by boiling and then cooling, organised structures are formed.
Answer:
Microspheres

Question 5.
Buffon, Erasmus Darwin and Lamarck proposed theories on organic evolution, which had one thing in common.
Answer:
Inheritance of acquired characters

Question 6.
Name the naturalist, who proposed that ontogeny recapitulates phylogeny.
Answer:
Ernst Haeckel

Question 7.
Name the theory, which explains about the origin of amphibians from aquatic fish-like ancestors.
Answer:
Recapitulation theory

Question 8.
DNA → RNA → Protein concept.
Answer:
Central dogma

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 9.
Genetic recombination occurs in cell division. Name the cell division.
Answer:
Meiosis

Question 10.
Hugo de Vries proposed mutation theory on his observations on the morphological features of a plant. Name the plant.
Answer:
Evening primrose (Oenothera lamarckiana)

Question 11.
Breakage, exchange and rejoining of homologous chromosomal segments.
Answer:
Genetic recombination

Question 12.
A single nucleotide substitution in the nucleotide sequence of a gene.
Answer:
Point mutation

Question 13.
The collection of all genes of a population of species.
Answer:
Gene pool

Question 14.
A sudden change in the genetic make up that ends up in a new expression.
Answer:
Mutation

Write whether the following statements are ‘True’or ‘False’

Question 1.
The primitive atmosphere was reducing.
Answer:
True

Question 2.
Heterotrophic organisms with aerobic respiration evolved prior to anaerobic ’ organisms.
Answer:
False

Question 3.
Continuous genetic variation originates through mutation.
Answer:
False

Question 4.
Serum proteins of closely related animals are similar in their amino acid sequences to a greater extent.
Answer:
True

Question 5.
Reptiles flourished in the Palaeozoic era.
Answer:
False

Question 6.
Close similarity in the nucleotide sequence between two organisms depicts close relationship between them. ‘
Answer:
True

Question 7.
Numerical changes, involving one or both chromosomes of a homologous pair, are known as euploidy.
Answer:
True

Question 8.
Genetic drift is the main driving force of evolution in a large randomly breeding population.
Answer:
True

Question 9.
Discontinuous variation is the product of mutation.
Answer:
True

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Short Answer Type Questions

Question 1.
Explain the theory of spontaneous generation.
Answer:
Theory of spontaneous generation It states that life originated from non-living matter automatically. This theory is also known as theory of abiogenesis or autobiogenesis. It was also supported by von Helmont (1642), who claimed that the mice were formed in 21 days from a dirty, sweat-soaked shirt put in a wheat barn in the dark. Abiogenesis was continued to be believed till the 17th century.

Question 2.
What do you mean by chemical evolution?
Answer:
This theory was given by AI Oparin (1923) and JBS Haldane (1928). According to them, the first form of life came from pre-existing, non-living organic molecules (like RNA, protein, etc.) and chemical evolution was followed by the formation of life, i.e. formation of diverse organic molecules from inorganic constituents.

The conditions on the earth favouring chemical evolution were high temperature, volcanic storms and reducing atmosphere containing CH4, NH3, etc.

Question 3.
Describe Miller-Urey experiment.
Answer:
Stanley Miller and Harold Urey in 1953 performed an experiment to demonstrate that ultraviolet radiations or electrical discharges or a combination of these can produce complex organic compounds from a mixture of CH4, NH3, H2 and water vapour (H20) at 800° C.

Electric discharge was created in a closed glass flask containing CH4, NH3, H2 in the ratio of 2 : 1 : 2 as shown in figure. The conditions were set similar to those of the primitive atmosphere in the laboratory.

They observed the formation of amino acids while in similar experiments performed by other scientists the presence of complex molecules like sugar, pigments, nitrogen bases and fats in the flask were also observed. Further reactions occurred in the aqueous medium of flask resulted in the formation of complex organic molecules at the bottom of flask, such as polysaccharides, fats, proteins, nucleotides and later nucleic acids (DNA and RNA).

Hence, it was concluded that during the beginning of life on earth; complex organic compounds could have formed simpler inorganic precursors.
CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 1

Question 4.
Explain prebiotic or primordial soup.
Answer:
Formation of molecular aggregates and cell-like structures Aggregates of complex organic molecules were formed in the oceans of the early earth which was termed the ‘Hot dilute soup or prebiotic or primordial soup’ by JBS Haldane (1920). These were considered as precursors of colloidal particles which could grow and divide. These are small complex molecules which are spherical and are covered by external mambranes.

Question 5.
What is prodigality of reproduction ? Give an example.
Answer:
Prodigality of Reproduction (Overproduction):
All organisms possess enormous fertility. They multiply in a geometric proportion with some organisms producing very large number of species. Despite of this high rate of reproduction of a species, its number remains constant under fairly stable environment. The production of more offsprings by some organisms and fewer by others is termed as differential reproduction.

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 6.
Write three criticisms on Darwinism.
Answer:
Criticism to Darwinism
Darwin’s theory was widely accepted, but Sir Richard Owen and Adam Sedgewick criticised it due to following • reasons
(i) Darwin emphasised on inheritance of useful variations,. However, sometimes inheritance of small variations, which are not useful to individuals are also seen.
(ii) He could not explain the presence of vestigial organs and concept of use and disuse of organs.
(iii) Darwinism failed to explain the arrival of the fittest.
(iv) Darwinism failed to differentiate between the somatic and germinal variations and considered all types of variations as heritable.

Question 7.
Explain how homologous organs reflect organic evolution?
Answer:
Homologous Organs:
It is the relation among the organs of different groups of organisms, that show similarity in the basic structure and embryonic development, but have different functions. Homology in organs indicates common ancestry.
It is based on divergent evolution which leads to the formation of homologous organs.

Question 8.
Describe homology in early embryonic development.
Answer:
Homologous Organs and Homology:
It is the relation among the organs of different groups of organisms, that show similarity in the basic structure and embryonic development, but have different functions. Homology in organs indicates common ancestry.
It is based on divergent evolution which leads to the formation of homologous organs.

In divergent evolution, a same basic organ gets specialisation to perform different functions, in order to adapt to the different environmental conditions prevailing in the habitat, e.g. forelimbs of vertebrates. Examples of homology are as follows

  • Structural organisation of vertebrate’s heart, brain, kidney, muscles, skull, etc.
  • Different mouthparts of some insects.
  • Forelimbs of animals like – whales, bats, cheetah and mammals (e.g. humans).
    CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 2

Question 9.
Explain the theory of recapitulation.
Answer:
Recapitulation in embryos Von Baer stated that during the embryo development, distantly related animals depart more and more than do closely animals. Ernst Haeckel (1905) reinterpreted Baer’s law in the form of recapitulation theory in the light of evolution. The theory of recapitulation or biogenetic law states that ontogeny (development of embryo) recapitulates phylogeny (ancestral sequence).

Question 10.
How do fossils support organic evolution?
Answer:
The study of fossil in different sedimentary layers indicates the geological period in which they existed. It also shows that the life forms varied over time and certain life forms are restricted to certain geological time scale. Hence, new forms of life have evolved at different times in the history of earth. All this is called palaeontological evidence.

Question 11.
Why do you call Archaeopteryx as a connecting link between reptiles and birds ?
Answer:
Archaeopteryx is a connecting link between reptiles and birds. It was of the size of crow and had both reptilian and avian characters. Presence of teeth in jaws, fingers having claws, long tail with free caudal vertebrae are the reptilian characters. While, presence of feathers on the body, and forelimbs modified into wings are the avian characters.

Question 12.
What do you mean by a geological time scale?
Answer:
Geological. Time Scale:
It covers the whole span of the earth’s history to correlate the evolutionary events in a proper sequence of ascending order of time. On the basis of time, the geological history of the earth has been divided into five eras namely, Archacozoic, Proterozojc, Palaeozoic, Mesozoic and Coenozoic.

Each era includes several periods and each period is further divided into epochs.
The most primitive era, i.e. Archaeozoic is placed at the bottom and the most recent era, i.e. Cenozoic is placed at the top.

Question 13.
Explain serological test.
Answer:
Serological tests These are tests done to study serum and other bodily fluids in animals. In serological tests, close similarity is found in species of a single genus, while genera of the same family show moderate reaction and families of the same order show slight, but detectable similarity. This indicates that closely related organisms show more similarity in protein structure and function.

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 14.
What is industrial melanism?
Answer:
Industrial Melanism:
Industrial melanism in peppered moth (Biston betularia) in Manchester city is an example of natural selection. In England, 1850s before industrialisation there were more white-winged moths on trees than dark-winged or melanic forms. But after industrialisation, i.e. in 1920, dark-winged moths became more in number than white-winged moths. This is because during industrialisation, the tree trunks became dark due to air pollution (dust and soot particles).

Question 15.
Explain genetic drift.
Answer:
Random Genetic Drift:
Random genetic drift refers to a random change in gene frequency of a population. In small population, frequencies of particular alleles may change drastically by a single change alone. These changes usually occur randomly.

Sewall Wright recognised this phenomenon as genetic drift (also known as Sewall Wright effect.) In large population, this phenomenon is rare because only favourable variations are selected by nature and unfit variations are eliminated.
Thus, genetic drift acts as the driving force for evolution in large populations.

Question 16.
What is speciation ?
Answer:
Speciation:
It is the process of formation of one or more new species from an existing species due to the accumulation of inversible adaptive change in their structure. A species is a collection of a group of populations with common gene pool (i.e total collection of all genes and its allele in a population). The factors which influence speciation, include mutation, recombination, natural selection, hybridisation, genetic drift, polyploidy (i.e increase in number of chromosomal set) and isolation.

Question 17.
What is bottleneck effect ?
Answer:
Bottleneck Effect:
It occurs when there is a disaster of some sort that reduces the size of a large population to an insignificance. This leaves smaller variation among the small number of surviving individuals, which disables natural selection to operate. In this situation, random genetic drift becomes main driving force of evolution.

Question 18.
What is Hardy-Weinberg’s principle?
Answer:
Mutations introduce new genes into a species resulting in the change in gene frequencies. In 1908, GH Hardy and W Weinberg established a simple mathematical relationship to the study these gene frequencies and gave Hardy-Weinberg principle.

This principle states that the allele frequencies in a . population are stable and is constant from generation to generation, i.e. gene pool remains constant. This is .called genetic equilibrium or Hardy-Weinberg equilibrium.

Differentiate between the following

Question 1.
Abiogenesis and Biogenesis.
Answer:
Differences between abiogenesis and biogenesis are as follows

Abiogenesis Biogenesis
A theory which describes the origin of life on the earth from non-living thing is called Abiogenesis. A theory which describes  the origin of life on the earth from pre-existing living organisms is called Biogenesis.
It is based on observations and thoughts. It was based on practical experiments and ffiaterial evidence.
It was supported by the fungus of bread and production of frogs in the mud. It was supported by the experiments performed by Redi and Pasteur.
It gives no scientific reasoning about the production of life. It describes the process of reproduction as an essential ability of living organism.

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 2.
Chemoautotrophs and Photoautotrophs
Answer:
Differences between chemoautotrophs and photoautotrophs are as follows

Chemoautotrophs Photoautotrophs
These are the organisms which obtain their energy by oxidising electron donor. Phototrophs are the organisms that capture protons in order to acquire energy.
Energy source is the oxidizing energy of chemical compounds. Energy source is mainly sunlight.
Classified as chemoorganotrophs and chemolithotrophs. Classified as photoautotrophs and photoheterotrophs.
Examples include nitrifying bacteria like Nitrosomonas, sulfur bacteria like Thiothrix, etc. Some examples are plants algae, cyanobacteria and phytoplanktons.

Question 3. Chemical evolution and Biological evolution.
Answer:
Differences between chemical evolution and biological evolution are as follows

Chemical evolution Chemical evolution
It occurs due to changes in the structure of molecules from complex molecules to simple molecules with the passage of time. It occurs due to changes in the structure of molecules from complex molecules to simple molecules with the passage of time.
It is relatively fast process and hence, it is possible to prove it in a laboratory. It is relatively fast process and hence, it is possible to prove it in a laboratory.
It involves the evolution of chemical such as water vapour, methane, ammonia and hydrogen into organic molecules such as sugars which later combined to form big molecules such as proteins, RNA and DNA. It involves the evolution of chemical such as water vapour, methane, ammonia and hydrogen into organic molecules such as sugars which later combined to form big molecules such as proteins, RNA and DNA.
Chemical evolution came into act before organic evolution. Organic evolution is a consequence of biological evolution.

Question 4.
Homologous organs and Analogous organs.
Answer:
Differences between homologous organs and analogous organs

Homologous organs Analogous organs
They have same basic structural plan. They have totally different structural plan.
They are found in closely related organisms which arise from some common ancestor. These organs found in totally unrelated organisms.
They differ in appearance. They have similar appearance.
They are modified to carry out different functions. These organs carry out the same function.
They lead to adaptive radiation or divergent evolution. They lead to convergent evolution or adaptive convergence.

Question 5.
Moulds and Casts.
Answer:
Differences between moulds and casts are as follows

Moulds Casts
A mold is a cavity left behind when the organic material is dissolved away. Only the external impressions remain during fossilisation. A cast is usually a very finely preserved representation of the surface features of the organism during fossilisation.

Question 6.
Genetic recombination and Mutation.
Answer:
Differences between genetic recombination and mutation are as follows

Genetic recombination Mutation
It is the production of offspring with combinations of traits that differ from those found in either parents. It is random nucleotide alterations such as copying errors or changes induced by external mutagens.
It is performed by the cell during the preparation of gametes (sperm, egg, pollen) which are used for sexual reproduction. It may take place both in somatic cells and germline cells.
Genetic recombination is heritable. Mutation can be heritable as well as non-heritable.

Question 7.
Somatic variation and Germinal variation.
Answer:
Differences between somatic variations and germinal variations are as follows

Somatic variations Germinal variations
Somatic variations are the variations in the somatic cells of an organism which may be acquired by them in their life. Germinal variations are the variations in the germ cells of an organism.
These are not passed on to their progenies. These are passed on to their progenies.

Question 8.
Chromosomal aberration and Gene mutation.
Answer:
Differences between chromosomal aberration and gene mutation are as follows

Chromosomal aberration Gene mutation
Chromosomal aberration is any change in the number and structure of chromosomes in an organism. Gene mutation is an alteration that occurs in the DNA base sequence of a gene.
It can change the total number of chromosomes in an organism. Gene mutation does not cause changes in the total number of chromosomes in an organism.
It may include many gene alterations. Gene mutation commonly refers to a single gene alteration.
Damages due to chromosomal aberration are large scale compared to gene mutation. Nucleotide damage is small in scale compared to chromosomal aberration. However, it can cause serious health problems.

Question 9.
Euploidy and Aneuploidy.
Answer:
Differences between euploidy and aneuploidy are as follows

Euploidy Aneuploidy
Euploidy is a variation in a chromosomal set of a cell or organism. Aneuploidy is a variation in total chromosome number of a cell or organism.
The number of chromosome sets is changed. The number of chromosome sets is not changed.
Cells have states of 3n, An, etc. Ceils are in the states of 2n+1, 2n-1, n-1,n+1, etc.
Euploidy occurs due to fertilisation of one ovum with two sperms, etc. Aneuploidy arises due to non-disjunction in meosis I and II and mitosis.
Euploidy is not seen in humans. Aneuploidy is seen in humans.

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 10.
Natural selection and Genetic drift.
Answer:
Differences between natural selection and genetic drift are as follows

Natural selection Genetic drift
It is a process where more adaptive species are selected in response to the environmental challenges. It is a random selection.
It occurs due to environmental challenges. It does not occur due to environmental challenges.
It ends up with selecting the more successive trait over the detrimental trait. Due to genetic drift important alleles may disappear completely.
It increases the frequency of the trait more adaptive to the environment. Genetic drift rarely results in more adaptive species to the environment.
It increases genetic variation. It does not increase genetic variation. Infact, sometimes it causes some variants to be extinct completely.

Question 11.
Convergent evolution and Divergent evolution
Answer:
Differences between convergent evolution and divergent evolution are as follows

Convergent evolution Divergent evolution
It is supported by the analogous structures. It is supported by the homologous structures.
It occurs in organisms which are not closely phylogenetically related. It occurs in phylogenetically related organisms.
Example insects, birds, pterosaus and bats. All have developed the similar nature of the flight/wings. Example Darwin’s finches.

Long Answer Type Questions

Question 1.
Give an account of the chemical basis of origin of life.
Answer:
Life on earth appeared 500 million years after its formation. In order to explain origin of life on the earth, different theories were given by different thinkers and scientists. They are
1. Theory of special creation It states that God has created life by his divine act of creation, i.e. the earth, light, plants and animals are all being created by the supernatural power.

2. Theory of catastrophism It states the creation of new life forms occurred after each catastrophe on earth.

3. Cosmozoic theory or Theory of panspermia It was given by early Greek thinkers, which states that the life on earth arose from the spores or panspermia, which came from outer space and developed into living forms.
The above three theories have been discarded due to the lack of logical explanation. Later on few more theories were proposed to explain the orgin of life.
These are
(i) Theory of spontaneous generation It states that life originated from non-living matter automatically. This theory is also known as theory of abiogenesis or autobiogenesis. It was also supported by von Helmont (1642), who claimed that the mice were formed in 21 days from a dirty, sweat-soaked shirt put in a wheat barn in the dark. Abiogenesis was continued to be believed till the 17th century.

(ii) Theory of biogenesis According to this theory, life originates from pre-existing life. This theory was supported by some scientists like Francisco Reddi (1668), Lazzaro Spallanzani (1767), Louis Pasteur (1862), Harvey and Huxley.

Chemical Evolution:
This theory was given by AI Oparin (1923) and JBS Haldane (1928). According to them, the first form of life came from pre-existing, non-living organic molecules (like RNA, protein, etc.) and chemical evolution was followed by the formation of life, i.e. formation of diverse organic molecules from inorganic constituents.
The conditions on the earth favouring chemical evolution were high temperature, volcanic storms and reducing atmosphere containing CH4, NH3, etc.
Chemical Evolution has Occurred in Following Steps

(i) Primitive earth had no atmosphere and it was anaerobic. Initially, the earth was a ball of hot gaseous mass.

(ii) Formation of Early Molecules
(a) Earth gradually began to cool and condense into a solid form. As a result, the free atoms or elements present on earth began to segregate into three concentric masses according to their weight.

(b) The heavier elements like nickel, iron, etc., moved to the core of the earth, intermediate one like silicon and aluminium moved to the middle and the lighter elements like oxygen, nitrogen, hydrogen, carbon, etc., remained on the surface and formed the early atmosphere of earth.

(c) With further cooling of the earth lighter elements such as water (H2O), methane (CH4), ammonia (NH3), hydrogen cyanide (HCN) . and oxides of carbon were formed. The primary atmosphare was devoid of free molecular oxygen (O2) due to which it was of reducing type.
Hot oceans, seas, lakes and other water bodies were formed due to the accumulation of heavy rainwater in the depressions of the earth.

(iii) Formation of simple organic molecules The high concentration of simple inorganic compounds (CH4,HCN,NH3) in water bodies induces them to react with each other so as to produce some unsaturated hydrocarbons.

As a result of further interaction of these hydrocarbons, some simple organic compounds such as simple sugars (e.g. glucose, ribose, deoxyribose, etc.), nitrogenous bases (purines like adenine, guanine and pyrimidines like thymine, cytosine and uracil), amino acids, fatty acids and glycerols were formed.

The energy required for these reactions came from the following sources

  • Solar radiations like ultraviolet light (UV) rays, cosmic rays and X-rays, etc.
  • Electrical discharge from lightning. High energy radiations from radioactive unstable isotopes on primitive earth.

(iv) Formation of complex organic molecules A number of hydrocarbons, purines and pyrimidine bases, amino acids, fatty acids, sugars and other organic compounds were accumulated in the primitive seas. Further reactions like polymerisation lead to the formation of larger organic molecules which later on formed complex organic molecules like polysaccharides, fats, proteins, nucleotides and then nucleic acids (DNA and RNA).

(v) Formation of molecular aggregates and cell-like structures Aggregates of complex organic molecules were formed in the oceans of the early earth which was termed the ‘Hot dilute soup or prebiotic or primordial soup’ by JBS Haldane (1920). These were considered as precursors of colloidal particles which could grow and divide. These are small complex molecules which are spherical and are covered by external mambranes.
These colloidal aggregrates were called coacervates by Oparin and microspheres by Sydney Fox (1965).

The firest non-cellular cells were believed to contain nucleoproteins and other macromolecules like polypeptides, lipids, etc. These cells were called protocells or protobionts or eubionts and they in the ancient ocean represented the beginning of life approx. 3.5 billion years ago. Probably viruses were evolved at the same time.

The first cellular life forms are believed to be evolved approximately 2 billion years back by the aggregation of various non-living molecules. From there, the evolution of diverse species of organisms occurred in course of time.

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 2.
Discuss the evidences of organic evolution from comparative anatomy and morphology.
Answer:
Evidences from Comparative Anatomy and Morphology
These evidences help to identify the similarities and differences among the organisms of today and those that existed years ago. Comparative study of external and internal structure can be used to understand the occurrence of organic evolution.

These can be determined by the following types
Homologous Organs and Homology
It is the relation among the organs of different groups of organisms, that show similarity in the basic structure and
embryonic development, but have different functions. Homology in organs indicates common ancestry.
It is based on divergent evolution which leads to the formation of homologous organs.

In divergent evolution, a same basic organ gets specialisation to perform different functions, in order to ‘ adapt to the different environmental conditions prevailing in the habitat, e.g. forelimbs of vertebrates. Examples of homology are as follows

  • Structural organisation of vertebrate’s heart, brain, kidney, muscles, skull, etc.
  • Different mouthparts of some insects.
  • Forelimbs of animals like – whales, bats, cheetah and mammals (e.g. humans).
    CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 2
    Flomologous organs as exhibited by the forelimbs of vertebrates; (a) Fluman, (b) Bat, (c) Whale, (d) Horse

Adaptive Radiation (Divergent Evolution):
It is the diversification of the organisms of a population into a number of new groups with adaptive characters suiting their need for survival.

Thus, it can be concluded that adaptive radiation and divergent evolution are interrelated and based on the modification of homologous structures. This can be proved studying the basic pattern of the pentadactyl limb which has undergone adaptive modifications in vertebrates,

All these animals have five digits (pentadactyl) in their forelimbs. All these digits possess the same number of skeletal elements that are arranged in same order (i.e. proximal to distal) along with similar muscle, nerve fibres, blood vessels, etc. These limbs have undergone adaptive modifications so as to perform the required funtions to adapt to their environment.

Similar adaptive modification rule also applies to mammals. In figure, a typical pentadactyl limb is seen in a terrestrial mammal.
CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 3
Adaptive radiation in the limb structure of mammals

This pattern has been modified for different functions like running (cursorial), swimming (aquatic), flying (aerial), climbing (arboreal) and burrowing (fussorial). Thus, all mammals have originated from an ancestral terestrial mammal through adaptive modifications of the basic pentadactyl limb plan.

Analogous Organs and Analogy:
In contrast to homologous organs, the analogous organs are different in their basic structure and developmental origin, but appear same and perform similar functions.
This relationship between the structures of different groups of animals due to their similar functions is called analogy or convergent evolution.

Examples of analogy are as follows
1. Wings of an insect a bird, Pterosaur (extinct flying reptile and a bat (flying mammal) show analogy. The wings are modified forelimbs that are adapted for flight.
The internal organisation of vertebrate (reptile, bird and bat) wings is same and they are composed of muscles and bones whereas, the wings of insect do not possess bones and muscles. They are only thin membranous extentions of exoskeleton and are made up of chitin.
CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 4
2. Flippers of dolphin and penguin.
3. Fins of fishes and flippers of whales.
4. Tracheae of an insect and lungs of the vertebrates are adapted for respiration, but are not homologous, as tracheae are ectodermal in origin, whereas the lungs are endodermal in origin.

Adaptive Convergence (Convergent Evolution):
In adaptive convergence, separate lineages show similar morphology under the influence of similar environmental factors. The existence of analogous structures also suggest the occurrence of convergent evolution. It may be explained in terms of the environment acting through the agency of natural selection and favouring those variations which confer increased survival and reproductive potential of the organisms possessing them.
Marsupial mammals in Australia are a good example to study adaptive convergence or convergent evolution.

Vestigial Organs:
The degenerated, rudimentary organs which are non-functional in the possessor, but were functional in their ancestor and in related animals are called vestigial organs. There are more than 90 vestigial organs in the human body. Some examples in human are coccyx (tailbone), nictitating membrane ( semilunar fold or plica semilunaries or 3rd eyelid), caecum, vermiform appendix, canines, wisdom teeth, tonsils, body hair, auricular muscles, mammary glands in males, etc.
CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 5
Vestigial organs are present in some other animals also, e.g. splint bones representing metacarpals of digits II and IV in horse, small bones representing hindlimbs and pelvic girdle in python and boas, wings and feathers in flightless bird kiwi of New Zealand, etc.

In plants, Dandelions and some other asexually reproducing plants retains flower which produce pollen grains necessary for sexual reproduction.

Atavism or Reversion:
It is the sudden reappearance or refunctioning of some ancestral organs, which have either completely disappeared or are present as vestigial organs.

It supports the idea of organic evolution, that living organisms have the ability to develop even lost or non-functional structures. For example, long and dense body hair, ability to move pinna in .some individuals, birth of a human baby with a small tail, presence of additional mammae in some individuals, elongated canine teeth, etc.

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 3.
Give an account of the embryological evidences of organic evolution.
Answer:
Evidences from Embryology:
Similarities and degree of intimacy in the embryonic development of various animals provide the supportive evidences of organic evolution. Some conclusions are derived from the study of comparative embryology, which are as follows

1. Common developmental pattern A common pattern of development is found in all the multicellular organisms. The development of embryo by sexual reproduction starts from diploid zygote or fertilised egg. The zygote undergoes repeated cleavage or cell division to form a solid structure called morula. The morula divides to form blastula, i.e. a single-layered hollow structure, which finally develops into gastrula, i.e. two to three-layered structure.

The animals with two-layered gastrula are termed as diploblastic, e.g. in coelenterates. The animals in which three-layered gastrula is found are known as triploblastic, e.g. frog, lizard, etc. These two or three layers of gastrula are termed as primary germ layers, which give rise to the entire animal. Thus, the similar early embryonic development shows close relationship among all the multicellular organisms.

2. Similarity in early embryos of vertebrates If a comparative study of embryos of vertebrates at same age is done, it is observed that they resemble one another. Such similarities suggest that these animals have common ancestry.

Some similarities in early embryonic stages are as follows
Presence of gill clefts, notochord, tail, rudimentary eyes and ears, etc. in all vertebrates from fishes to mammals.
Notochord is replaced by vertebral column in all adult vertebrates.
Gills are replaced by lungs in adult amphibians, reptiles and mammals.
CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 6

3.Recapitulation in embryos Von Baer stated that during the embryo development, distantly related animals depart more and more than do closely animals. Ernst Haeckel (1905) reinterpreted Baer’s law in the form of recapitulation theory in the light of evolution. The theory of recapitulation or biogenetic law states that ontogeny (development of embryo) recapitulates phylogeny (ancestral sequence).

This means that the life history of an animal reflects its evolutionary history. For example, during the life history, frog’s tadpole larva resembles fishes in habits and structure. It suggests that amphibians have evolved from fish ancestors.
Embryo logical Evidences in Plants

These include

  • Presence of filamentous green algae-like structure, i. e. protonema during development of Funaria (moss).
  • Pteridophytes and primitive -gymnosperms like Cycas and Ginkgo have flagellated sperms and they depend on water for fertilisation. It strengthens the fact that gymnosperms have evolved from pteridophytes.

Question 4.
Describe palaeontological evidences of organic evolution.
Answer:
Evidences from Palaeontology:
Palaeontology is the study of fossils of plants and animals that lived in prehistoric times. Leonardo da Vinci (1452-1519), an Italian painter is known as the ‘Father of Palaeontology’ and Baron Georges Cuvier (1769-1832) is known as ‘The Founder of Modern Palaeontology’.

The study of fossil in different sedimentary layers indicates the geological period in which they existed. It also shows that the life forms varied over time and certain life forms are restricted to certain geological time scale. Hence, new forms of life have evolved at different times in the history of earth. All this is called palaeontological evidence.

Fossils:
These are the material remains (bones, teeth, shells) or traces (physical or chemical) of ancient organisms induding plants and animals. According to Charles Lyell, fossil is any body or traces of body of animal or plant buried and preserved by the natural causes. .

Fossilisation is the process of formation of fossils. Fossils are generally preserved in sedimentary rocks in which multiple layers are present and the lowermost layer gets harden into rock under pressure. These are formed when parts of dead organisms decay with the passage of time and get replaced by inorganic materials. The hard parts of the body (i.e. bone, teeth, shell, etc.), are preserved more readily than soft parts, into rocks. Both animals as well as plants can be fossilised as additional layers get deposited with time.

Fossils are also formed by processes other than petrification, e.g. an organism may get buried intact in ‘ preservatives like resins, snow, oil, tar, volcanic, ash, etc.

Sometimes, the organism or its parts get washed away to water bodies and settle down at the bottom. Gradually they get covered by the layers of mud and sand. Particularly when buried in rapidly hardening mud, they decay completely and the space it occupies, becomes filled with another kind of material forming moulds and casts.

Types of Fossils:
Some general types of fossils are given below
(i) Unaltered It includes animals, plants and humans who got embedded in permafrost of arctic/alpine snow and remain preserved in actual state, e.g. wooly mammoth (25000 years old in Siberia) and insects trapped in the amber of plants.
(ii) Petrifications The fossils in which hard body parts (organic matter) get replaced by mineral matter like silica, pyrites and calcium, etc. In some petrified fossils, even cellular details are found. This process is used to preserve original structures of organisms.
(iii) Moulds They are hardened encasements formed in the outer parts of extinct organic remains which later decayed leaving cavities.
(iv) Casts They are hardened pieces of mineral matter deposited in the cavities of moulds.
(v) Impressions/Imprints They are external features of organisms or their parts that are left due to hardening rocky matter before they completely decay.
CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 7
(a) Imprint of a crustacean, (b) Mould of a bivale, (c) Insect trapped in amber, (d) Petrified softwood, (e) Petrified cone of Araucaria mirabilis (a coniferous tree)

Important Characteristics of Fossils:
The fossil records are direct evidences that support organic evolution due to the following reasons
1. Fossils of different ages are mostly found in the different layers of sedimentary rocks in an ascending order, i.e. from simple to complex forms.

2. The lower layers of rocks of early era contain fossils of simple nature. In upper layers more recent fossils, which are more recent and complex in structure. Fossils are not found in the rocks of the Archaeozoic (first) Era.

3. In the rocks of the second era, i.e. Proterozoic, only few fossils are found. These are simple, soft-bodied organisms, such as marine invertebrates. In the upper strata of rocks, fossils are more in number, belonging to organisms of later ages.

4. The fossils of two consecutive strata are different from each other indicating the occurrence of progressive changes in course of time.

5. Certain mammals, such as horse, elephant, camel and man, have complete fossil records. They clearly explain the gradual evolution of these species.
CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution 8

6. Fossils of some transitional forms (also called connecting links or missing links) explain the emergence of a new species from its ancestor. For example, fossil of Archaeopteryx discovered from the rocks of Jurassic Period at Bavaria, Germany in 1861 is a connecting link between birds and reptiles. As a transitional form, it possessed the characters of both reptiles and birds.

7. Living fossils are the living organisms which are similar in appearance to a recorded fossil of distant ancestors but they usually have no close relatives. Such organisms have undergone very slow changes over a long span of time, e.g Latimeria (a coelacanth fish).

8. The approximate ages of fossils can be determined by different radioactive dating methods.

9. Extinction of species can be explained by the fossil records of that species, e.g. dinosaurs extinct about 66 million years ago.

CHSE Odisha Class 12 Biology Solutions Chapter 8 Evolution

Question 5.
Describe Darwin’s theory of natural selection and origin of species and discuss about the criticisms.
Answer:
Darwinism (Natural Selection Theory):
Charles Robert Darwin was born in 1809. In 1831, he accepted an unpaid post of naturalist on the survey ship, called HMS Beagle. In his voyage, he spent five years in sea charting the East Coast of South America. During a five week stay on the Galapagos Archipelago Islands, he was struck by the similarities shown by the flora and fauna of the islands and mainland. In particular, he was intrigued by the characteristic distribution of species of tortoises and finches.

Darwin observed different types of beaks in the same population of finches. He termed this phenomenon as adaptive radiation which explains that the changes in beak structure were the result of adaptations to the available food to the native finches. Over the years, the ancestral beak evolved into diverse types of beaks. Thus, Darwin realised the importance of competition and adaptation in the evolution of finches.

After his return, Darwin formulated his concept of organic evolution. He was also influenced by a paper published by Robert Malthus (1838) on populations, which states that the population increases in a geometric progression, while the food supply increases more slowly. Therefore, the food supply becomes a limiting factor. In the meantime, another naturalist Alfred Russel Wallace, came to the same conclusions as Darwin regarding natural selection. The content of Wallace’s write-up was similar to Darwin’s thinking.

Darwin and Wallace presented papers on their ideas which were published in the ‘Journal of the Proceedings of The Linnaean Society of London in 1858. Darwin published a book entitled ‘On the Origin of Species by Means of Natural Selection (later changed to ‘Origin of Species’ In its 6th edition in 1872), embodying his observations and conclusions in 1859.

Postulates of Darwinism:
The main postulates, which formed the basis of Darwin’s theory of natural selection are as follows
(i) Prodigality of Reproduction (Overproduction):
All organisms possess enormous fertility. They multiply in a geometric proportion with some organisms producing very large number of species. Despite of this high rate of reproduction of a species, its number remains constant under fairly stable environment. The production of more offsprings by some organisms and fewer by others is termed as differential reproduction.

(ii) Limiting Factors
The resources like food, space, etc., remain limited inspite of rapid multiplication of the individuals of all the species. It helps to check the increased number of animals and plants.

(iii) Struggle for Existence
The limited amount of resources and overproduction of organisms are the main causes of struggle for existence. Various types of struggle help an organism to cope up with unfavourable environmental conditions.
The three types of struggles are as follows
(a) Intraspecific struggle It is the struggle among the individuals of same species for their common requirements like food, shelter, mate, breeding places, etc.
(b) Interspecific struggle It is the struggle between the individuals of different species for their similar requirements like food and space.
(c) Environmental struggle It is the struggle of living forms against the environmental conditions like extreme heat, cold, drought, earthquakes, storms, disease, volcanic eruption, etc.

(iv) Variations and Heredity
All individuals are dissimilar in some of their characters except the identical twins. This dissimilarities are mainly due to the variations. These are the small or large differences among the individuals. Variations allow some individuals to better adjust with their environment.

Variations can be categorised into the following types
1. Somatic variations These variations affect the somatic cells of an organism. They are also called modifications or acquired characters because they are aquired by an individual during its lifetime. These are caused by various environmental factors, use and disuse of organs and conscious efforts, etc.
2. Germinal variations These are inheritable variations recognised by Darwin but he had no idea of inheritance of characters. They are formed mostly in germinal cells. They are further of two types
• Continuous (gradual) variations These are fluctuating variations, which oscillate due to race, variety and species.
• Discontinuous (sudden) variations These appear suddenly and show no ‘spots’ gradation. These variations were termed as ‘spots’ by Darwin and ‘mutation’ by Hugo de Vries. Darwin regarded continuous variations to be more important because the discontinuous variations being mostly harmful would not be selected again.

(v) Survival of the Fittest and Natural Selection:
The organisms, which have inherited favourable variations generally survive. This is termed as ‘survival of the fittest’ (the phrase being originally used by Herbert Spencer). Whereas, the organisms without such variations appear unfit and get eliminated. Nature plays a decisive role in selecting the fit organisms.

Natural selection is based on merit and is without any prejudice or bias. It eliminates the unfit ones and selects those organisms that are most fit to survive in a particular environment and to produce offsprings. Survival alone does not make any sense from evolution point of view.
The fit organisms must reproduce to contribute to the next generation. Lerner (1959) says, ‘Individuals having more offsprings are the fit ones’.

(vi) Origin of New Species (Speciation):
Darwin considered that as a result of struggle for existence, variability (continuous variations) and inheritance, species became better adapted to their environment. These beneficial adaptations are preserved and accumulated in the individuals of species generation after generation. This results into the origin of new species or speciation and the resultant offsprings become visibly distinct from their ancestors.

Criticism to Darwinism:
Darwin’s theory was widely accepted, but Sir Richard Owen and Adam Sedgewick criticised it due to following reasons
(i) Darwin emphasised on inheritance of useful variations,. However, sometimes inheritance of small variations, which are not useful to individuals are also seen.
(ii) He could not explain the presence of vestigial organs and concept of use and disuse of organs.
(iii) Darwinism failed to explain the arrival of the fittest.
(iv) Darwinism failed to differentiate between the somatic and germinal variations and considered all types of variations as heritable.
(v) Darwin’s natural selection theory was based on the mistaken concept of artificial selection. He wrongly believed that changes brought on by domestication of animal were also heritable.
(vi) Darwin failed to recognise the large fluctuating variations (occurring due to mutation). He only believed in the occurrence of small continuous variations.

Darwin proposed ‘theory of pangenesis’ explaining that pangenes or gemmules are transmitted from one generation to next. However, this theory was refuted by Weismann’s germplasm theory.

Question 6.
Discuss about the synthetic theory of organic evolution.
Answer:
Modern Synthetic Theory of Organic Evolution:
The fundamental mechanisim of evolution as explained by Darwin and his contemporaries underwent major modification with the progress in genetics.

Mendel’s laws of inheritance were applied to various theories (like natural selection). The validity of these laws were later verified by Correns, Tschermark and de Vries. It came to light that mutation and genetic recombination were the cause of genetic variation in living organisms.

These variations were used as the raw material on which natural selection acted on, leading to evolution of new species. The process of evolution as a population character not an individual one was later on proposed by GH Hardy and W Weinberg. According to them, a disturbance in the gene pool of a population results in evolution. Among these new developments, the concept of modern synthetic theory (post-Darwinian synthesis) was proposed.

The modern synthetic theory is based on the work of a number of scientist namely-Dobzhansky’s (1937) Julian Huxley (1942), Ernst Mayr (1970), RA Fisher (1958), JBS Haldane, Sewall Wright (1968) and GL Stebbins (1971). Stebbins in his book ‘Process of Organic Evolution discussed the modern synthetic theory.

This theory is a collective explanation of the fundamental mechanism of evolution. Homologous recombination, mutation, natural selection, isolation, genetic drift and migration form the basis of the mechanism of evolution.

Genetic Recombination:
These are the homologous combinations between genes present on different chromosomes (i.e. paternal and maternal) during gametogenesis. It can occur by following ways
(i) Crossing over Mutual exchange of genes between non-sister chromatids of homologus chromosomes during meiosis-I. It forms multiple variations in a population.
(ii) Independent assortment of chromosomes It forms genetically different haploid gametes during meiosis which bring about variations in new generation.
(iii) Random fusion of gametes During sexual reproduction, random fusion of male and female gametes produces a new individual.

Changes in Chromosome Number and Structure:
Chromosomal mutations or aberrrations arise due to change in number and structure of chromosomes. Chromosomal number may change in followinng two ways
(i) Polyploidy (increase in number of chromosome sets)
(ii) Aneuploidy (change in number of one or both chromosomes of a homologous pair). When the change occurs in the chromosomal morphology, it is called chromosomal aberration.

These are of four types
(i) Deletion Loss of a segment of a chromosome.
(ii) Duplication Doubling of a chromosomal segment.
(iii) Inversion Reversal in the order of genes.
(iv) Translocation Mutual exchange of a segment of chromosome between two non-homologous chromosomes.

Gene Mutations:
In 1901, Hugo de Vries carried out experiments on evening primrose plant (Oenothera lamarckiana) and proposed the mutation theory of evolution. This theory states that the evolution occurs by the sudden large differences or mutations in the population. Mutation is the sudden change in appearance or variations in an individual or a population. When mutation affects only a single nucleotide, it is called point mutation.

However, when more than one nucleotide is involved in mutation, it is called gross mutation. These mutations results in drastic changes which can be lethal or insignificant or useful. They lead to the new phenotypes. Though mutations are random and occur at very slow rates, they are sufficient to create considerable genetic variations for speciation to occur.

Natural Selection:
It is the most widely accepted theory for explaining the mechanism of evolution, profounded by Charles Darwin and Alfred Russel Wallace. Natural selection selects the favourable variation and allows such organisms to reproduce. Meanwhile, harmful or non-adaptive variations are discouraged by natural selection and such organisms are eliminated from the population. Therefore, variations act as raw material natural selection which decides the favourable traits to continue through generations.

Isolation:
The mechanism by which a population gets segregated into two or more subtypes by a geographical barrier, such that each subtype influences a different environment, is called isolation.
This segregation can occur due to
(i) Geographical isolation When two related populations occupy geographically or spatially separated areas.
Such species are called as allopatric species. These species become so different to each other that when they are brought together, they fail to reproduce.
(ii) Reproductive isolation The prevention of interbreeding between the population of two different species. Such reproductively isolated species are called as sympatric species.

CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 7 Question Answer Molecular Basis of Inheritance

Molecular Basis of Inheritance Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Fill in the blanks with correct answers from the choices given in the brackets of each bit

Question 1.
In split genes, the coding sequences are ……….. (introns, operons, exons, cistrons)
Answer:
exons

Question 2.
The smallest part of the gene is called …………… (recon, muton, exon, cistron)
Answer:
cistron

Question 3.
The enzyme referred to as Kornberg enzyme is ……….. (DNA polymerase-I, DNA polymerase-II, RNA polymerase, ligase)
Answer:
DNA polymerase-I

Question 4.
The polymerase that has 5′-3′ exonuclease property is known as ………… (DNA pol-I, DNA pol-II, RNA pol, DNA ligase)
Answer:
DNA pol-I

Question 5.
The termination factor that recognises the termination codon UAG is ………….. (only RF1, only RF2, both RF1 and RF2, neither RF1 and RF2)
Answer:
Only RF

Question 6.
The enzyme that removes formyl group from the first amino acid methionine of a newly synthesised polypeptide is ………… (RF3 translocase, deformylase, exoaminopeptidase)
Answer:
deformylase

Question 7.
The word gene was coined by ……….. (Garrod, Johannsen, Meischer, Griffith)
Answer:
Wilhelm Johannsen

Question 8.
In 1869 …………… discovered DNA.(Garrod, Meischer, GrifFith, Wilkins)
Answer:
Friedrich Meischer

Question 9.
The virulent, pneumococcus possessed a …………. coat for its protection. (protein, lipid, phospholipid, polysaccharide)
Answer:
polysaccharide

Question 10.
Complete sequence of amino acids in ………….. was proposed by Sanger. (insulin, haemoglobin, kinetin, polymerase)
Answer:
insulin

Question 11.
RNAs lack ………. as nitrogenous base. (adenine, guanine, cytosine, thymine)
Answer:
thymine

Question 12.
One complet turn of B-DNA contains ………….. number of nitrogenous bases. (10,11,9,12)
Answer:
10 .

Question 13.
The most stable form of RNA is …………. RNA. (messenger, transfer, ribosomal, small nuclear)
Answer:
ribosomal

Question 14.
When a codon codes or more than one amino acid, it is called ………….. code.
(commaless, degenerate, nonsense, universal)
Answer:
degenerate

Question 15.
The start codon is ………… (UAA, UGA, AUG, UGA)
Answer:
AUG

Express in one or two word(s)

Question 1.
If in a double-stranded DNA there is 25% of thymine, then calculate the per cent of guanine.
Answer:
25%

Question 2.
What is the complementary base of adenine in RNA?
Answer:
Uracil

Question 3.
In a double helix if one stand is on 5′ → 3′, what will be arrangement of other strand?
Answer:
3′ → 5′

Question 4.
What are the basic proteins called in eukaryotic DNA?
Answer:
Histones

Question 5.
What is called to amino acids with more than one codon?
Answer:
Degenerate

Question 6.
What type of genes do express continuously ?
Answer:
Housekeeping genes

Question 7.
What type of RNAs do carry amino acids to the site of protein synthesis ?
Answer:
tRNA

Correct the sentences in each bit without changing the underlined words

Question 1.
Watson and Griffith proposed double helical structure of DNA.
Answer:
Watson and Crick proposed the double helical structure of DNA.

Question 2.
A nucleoprotein is building block of all nucleic acids.
Answer:
Nucleotides are the building blocks of all nucleic acids.

Question 3.
The strand of the DNA double helix represent nucleotide phosphate backbone and are antiparallel.
Answer:
The DNA double helix consist of a polynucleotide chain with a backbone formed of sugar and phosphate groups. These chains are antiparellel to each other.

Question 4.
The helical turns are right handed is Z DNA.
Answer:
The helical turns are left-handed in Z-DNA.

Question 5.
Avery, McCarty and MacLeod experimentally proved that the transforming principle is a protein.
Answer:
Avery, McCarty and MacLeod experimentally proved that transforming principle is DNA.

Question 6.
Meischer proposed the transforming principle.
Answer:
Frederick Griffith proposed the transforming principle.

Question 7.
The enzyme ligase is responsible for transcription.
Answer:
The enzyme DNA dependent RNA polymerase is responsible for transcription.

Question 8.
The operator is under the control of a repressor molecule synthesised by structural gene which is not a part of operon.
Answer:
The operator is under the control of a repressor molecule synthesised by regulator gene (i-gene) which is a part of operon.

Question 9.
The example of regulatory gene is genes of respiratory enzymes.
Answer:
The example of constitutive gene is genes of respiratory enzymes.

Question 10.
P-site in prokaryotes only accepts tRNAmet.
Answer:
P-site in prokaryotes only accepts tRNA fmet.

Question 11.
The coding or translatable sequences are introns.
Answer:
The intervening or non-coding sequences are introns.

Question 12.
The structural gens transcribe tRNA and rRNA.
Answer:
The structural gene do not transcribe tRNA and rRNA.

Question 13.
A primer is a small DNA or RNA strand hydrogen bonded to a template.
Answer:
A primer is a short strand of RNA or DNA that serves as a starting point for DNA synthesis.

Question 14.
In DNA replication, as per semiconservative model, two new strands synthesised, form new DNA molecules.
Answer:
In DNA replication as per semiconservative model, two parental strands would separate and act as a template for the synthesis of new complementary strands. Hence, a new strand consist of one parental strand and other new replicated strand.

Fill in the blanks

Question 15.
The enzyme …………. hydrolyses DNA molecules.
Answer:
exonucleases

Question 16.
Clover leaf model of fRNA was proposed by ………………
Answer:
Holey

Question 17.
The segment of DNA that expresses specific character is called …………..
Answer:
gene

Question 18.
The enzyme ………… helps to join nucleotides.
Answer:
ligase

Question 19.
The DNA strand which takes part in transcription is called ………………
Answer:
leading strand

Question 20.
UAG is …………. codon.
Answer:
stop

Question 21.
The gene which becomes active due to the presence of specific substance is called ………….. gene.
Answer:
inducible

Question 22.
To identify criminals DNA ………… is done.
Answer:
fingerprinting

Short Answer Type Questions

Write notes on the following with atleast 2 valid points

Question 1.
Inducible operon
Answer:
Inducible operon An inducible operon is an operon, in which the presence of a key metabolic substance induces transcription of the structural genes. One example of an inducible operon is lac operon and the inducer of this operon is lactose.

Question 2.
Repressible operon
Answer:
Repressible operon A repressible operon is an operon which always transcribes structural genes unless a repressor is present. One example of a repressible operon is trp operon.

Question 3.
Housekeeping genes
Answer:
House keeping genes are involved in basic cell maintenance and therefore are expected to maivlain constant constant expression levels in all cells and conditions. They are also called constituline gene

Question 4.
Adaptor molecules
Answer:
Adapter molecules tRNA acts as adapter molecules in translation. It serves as the physical link between the mRNA and the amino acid sequence of proteins. It carries an amino acid to the ribosome as directed by a three nucleotide sequence in a mRNA.

Question 5.
Split genes
Answer:
Split genes It is an interrupted gene that contains sections of DNA called exons which are expressed as RNA and proteins, interrupted by sections of DNA called introns, which are not expressed.

Question 6.
RNA splicing
Answer:
RNA splicing Splicing is the editing of the nascent precursor of mRNA transcript into a mature mRNA. After splicing, introns are removed and exons are joined together. The process of splicing involves two successive transesterification reaction. The RNA splicing is carried out with the help of a large complex called spliceosome.

Question 7.
Termination of translation
Answer:
Termination:
It is accomplished by the presence of any of the three termination codon on mRNA. Which are recognised by the release (termination) factor, RF1, RF2 and RF3.

RF1 and RF2 They resemble the structure of tRNA. They compete with tRNA to bind the termination codon at A-site of ribosome. This is called molecular mimicry. RF1 recognise UAG and RF2 recognise UGA. UAA is recognised by both of them. RF3 helps to split the peptidyl tRNA body by using GTP.
Note In eukaryotes, only one release factor is known. It iseRF1.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 1

Question 8.
Okazaki fragments
Answer:
Okazaki fragments During DNA replication, the new strand elongation occurs in opposite directions, as the two strands of the DNA are antiparallel.
On the leading strands, a continuous synthesis of new strand occurs while on the other strand (5′ → 3′ strand) discontinuous synthesis occurs. These discontinuous pieces of new DNA strands are called Okazaki fragments which join by DNA ligase to form the continuous strand

Question 9.
Central dogma
Answer:
Central Dogma:
It was proposed by Francis Crick (in 1958). According to the central dogma in molecular biology, the flow of genetic information is unidirectional, i.e.
DNA → RNA → Protein.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 2
Central dogma
But later in 1970, HM Temin reported that the flow of information can be in reverse direction also, i.e. from RNA to DNA in some viruses (e.g. HIV) which is called as reverse transcription.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 3
Modified central dogma

Note: Eukaryotic gene expressions were obtained from yeast,
Arabidopsia thaliana with developed techniques.

Differentiate with atleast 2 valid points

Question 1.
Genes and Chromosomes.
Answer:
Differences between genes and chromosomes are as follows

Genes Chromosomes
Gene is a segment of DNA on the chromosome that codes for a functional protein and RNAs like tRNA, rRNA or ribozymes. Chromosome is the structure formed by the condensation of chromatin during cell division.
Genes basically refers to the DNA fragment that directs the synthesis of a protein. Chromosome consists of long DNA strand wrapped around histone proteins.
Gene contain coding sequence called exons and non-coding sequence called introns on the chromosome that directs synthesis of a protein. Chromosome is a long DNA strand containing bdth coding (genes) and non-coding DNA (junk DNA or spacer DNA) between genes.

Question 2.
DNA and RNA.
Answer:
Differences between DNA and RNA are as follows

DNA RNA
It is mainly confined to the nucleus, but also occurs in mitochondria and chloroplasts in small amount. It mainly occurs in the cytoplasm. A small quantity is found in the nucleus.
It contains deoxyribose. It contains ribose sugar.
It pyrimidines are cytosine and thymine. Its pyrimidines are cytosine and uracil.
It consists of two polynucleotide chains held together by hydrogen bonds and coiled into a double helix. Some viruses (ϕ x 174) have single- stranded DNA. It consists of a single polynucleotide chain. It may fold on itself and get hydrogen bonded and coiled into a pseudohelix. Some viruses (reovirus) have double-stranded RNA.

Question 3.
Purines and Pyrimidines.
Answer:
Differences between purines and pyrimidines are as follows

Purines Pyrimidines
It contains two carbon-nitrogen rings and four nitrogen atoms. It contains one carbon nitrogen ring and two nitrogen atoms.
Purines are adenine and guanine Pyrimidines are cytosine, thymine and uracil.
120.11 g mol molar mass 80.088 g mol<sup>-1</sup> molar mass.

Question 4.
Exons and Introns.
Answer:
Differences between exons and introns are as follows

Exons Introns
These are the nucleotide sequence of genes that are expressed and those are found at either sides of an intron. These are sequences of nucleotides present in the genes between exons.
These are nucleotide sequences which code for proteins. These are nucleotide sequences do not code for proteins.
These are translated regions on mRNA. These are untranslated regions on mRNA.
After RNA splicing, only exons are present on the mature mRNA. These are removed from mRNA during RNA splicing.

Question 5.
B-DNA and Z-DNA.
Answer:
Differences between B-DNA and Z-DNA are as follows

B-DNA Z-DNA
It is with a right-handed type of helix. It is with a left-handed type of helix.
Its helical diameter is 2.37 nm. Its helical diameter is 1.84 nm.
It has 0.34 nm rise per base pair. It has 0.37 nm rise per base pair.
It has 3.4 nm distance per complete turn. It has 4.5 nm distance per complete turn.
Number of base pair per complete turn are 10. Number of base pair per complete turn are 12.

Question 6.
Answer:
Replication and Transcription.
Answer:
Differences between replication and transcription are as follows

Replication Transcription
It occurs in the S-phase of cell cycle. It occurs in the G<sub>1</sub> and G<sub>2</sub> phases of cell cycle.
It is catalysed by DNA polymerase enzyme. It is catalysed by RNA polymerase enzyme.
Deoxyribonucleoside triphosphate (dATP, dGTP, dTTP) serve as raw materials. Ribonucleoside triphosphate (ATP, UTP, GTP, CTP) serve as raw materials.
Replication occurs along the strands of DNA. IT takes place along one strand of DNA.
It involves unwinding and splitting of the entire DNA molecule. It involves unwinding and splitting of only those genes which are to be transcribed.
It involves copying of the entire genome. It involves copying of certain individual genes only.
Two double-stranded DNA molecules are formed from one DNA molecule. A single one-stranded RNA molecule is formed from a segment of one strand.
Serve to conserve the genome for the next generation of cells and individuals. Serve to form DNA copies of individual genes for immediate use in protein synthesis.
It require RNA primer to start replication. No primer is required to start.
It produces normal DNA molecules that do not need It produces primary RNA transcript molecules which need processing to acquire final form and size.

Question 7.
Transcription and Translation.
Answer:
Differences between transcription and translation are as follows

Transcription Translation
It is the formation of RNA from DNA. It is the synthesis of polypeptide over ribosome.
The template is antisense strand of DNA. The template is mRNA.
It occurs inside the nucleus in eukaryotes and cytoplasm in prokaryotes. It occurs in cytoplasm.
The raw materials are four types of ribonucleoside triphosphates – ATP, GTP, CTP and UTP. The raw materials are 20 types of amino acids.
It forms three types of RNAs, i.e. rRNA, tRNA and mRNA All the three types of RNAs take part in translation.
Transcription requires RNA polymerases and some transcription factors. Translation requires initiation, elongation and releasing factors.
Polymerase moves over the template DNA. Ribosome moves over mRNA.
An adapter molecule is not required. Adaptor molecules bring amino acids over the template.
Product often requires splicing. Splicing is absent.

Question 8.
Housekeeping gene and Inducible gene.
Answer:
Differences between housekeeping gene and inducible gene are as follows

Housekeeping gene Inducible gene
These genes are constantly expressing themselves in a cell because their products are required for the normal cellular activities, e.g. genes for glycolysis, ATPase. These genes are switched on in response to the presence of a chemical substance or inducer which is required for the functioning of the product of gene activity, e.g, nitrate for nitrate reductase.

Question 9.
Degenerate codon and Nonsense codon.
Answer:
Differences between degenerate codon and nonsense codon are as follows

Degenerate codon Nonsense codon
Genetic code is degenerate for a particular amino acid, that is more than one codon can code for a single amino acid. A codon for which no normal tRNA molecule exists does not code for any amino acid.
These codon causes translation. The presence of nonsense codon causes termination of translation ending polypeptide synthesis chain.
These are many e.g. phenylalanine has two Codon is UAA and UUC. These are three nonsense codons and are called amber(UAG), ochre(UAA) and opal(UGA).

Long Answer Type Questions

Question 1.
Give the structure of DNA. Add a note on different forms of DNA.
Answer:
Primary Structure of DNA:
Two nucleotides when linked through a 3′-5′ phosphodiester linkage, form a dinucleotide. The phosphodiester linkage is formed when each phosphate group esterifies to the 3′ hydroxy group of a pentose and to the 5′ hydroxyl group of the next pentose.

In a similar fashion, more nucleotides may join to form a polynucleotide chain (fig. structure of DNA). The polymer chain thus, formed has
(i) One end with a free phosphate moiety at 5′ end of deoxyribose sugar. This is marked as 5′ end of polynucleotide chain.
(ii) The other end with a free hydroxyl 3′-OH group marked as 3′ end of the polynucleotide chain.
Thus, the sugar and phosphates form the backbone in a polymer chain and the nitrogenous bases linked to sugar moiety project from this backbone. In RNA, there is an additional -OH group at 2′ position in the ribose of every nucleotide residue.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 4

Secondary Structure of DNA:
Watson and Crick proposed the secondary structure in the form of the famous double helix model in 1953 on the basis of following observations
1. Erwin Chargafif (in 1950) formulated important generalisation on the base and other contents of DNA, called as ChargafFs rule. It states that for a double-stranded DNA, the ratios between adenine (A) and thymine (T) and guanine (G) and cytosine (C) are constant and equal to one.
i.e. \(\frac{A + T}{G + C}\) = 1

2. X-ray diffraction studies by Wilkins in 1952, suggested a helicoidal configuration of DNA.

One of the important features of this model was the complementary base pairing. It means if the sequence of bases in one strand is known, the sequence in other strand can be easily predicted. Also, if each strand from a DNA acts as a template for synthesis of a new strand, the daughter DNA thus produced would be identical to the parental DNA molecule.

Watson and Crick Model of DNA:
Watson and Crick worked out the first correct double helix model of DNA, which explained most of its properties.
The salient features of double helix structure of DNA are as follows
(i) DNA is made up of two polynucleotide chains. The backbone is constituted by sugar phosphate, while the nitrogenous bases project inwards.
(ii) The two chains have anti-parallel polarity, i.e. when one chain has 3′ → 5′ polarity, the other has 5′ → 3′ polarity. Hence, orientation of deoxyribose sugar is opposite in both the strands.
(iii) The two strands are complementary to each other, i.e. purine base of one strand has pyrimidine counterpart on other strand. The complementary bases in two strands are paired through hydrogen bonds (H-bonds) to form base pairs.
(a) Adenine is bonded with thymine of the opposite strand with the help of two hydrogen bonds.
(b) Guanine is bonded with cytosine of the opposite strand with the help of three hydrogen bonds. So, a purine bonds with a pyrimidine always. Thus, maintaining a uniform distance between the two strands of the helix.
(iv) The two polypeptide chains are coiled in a right-handed fashion. Pitch of the helix, i.e. length of DNA in one complete turn = 3.4 nm or
3.4 × 10-9 or 34 Å.
Number of base pairs in each turn = 10. Distance between a base pair in a helix = 0.34 nm. The diameter of DNA molecule is 20 Å (2nm).
(v) Percentage calculation of bases is done by
A + T = 100 – (G + C).
(vi) The plane of one base pair stacks over the other in double helix. This provides the stability to the helical structure, in addition to H-bond.
The length of DNA in E. coli is 1.36 mm, while in humans it is 2.2 m.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 5
Structure of DNA : (a) Watson and Crick model of double helix, (b) Double-stranded polynucleotide chain sequence showing hydrogen bonds

Structural Forms of the Double Helix:
Double helical DNA exists in three structural forms namely the A-form DNA, the B-form DNA (described by Watson and Crick) and the Z-form DNA. The transition among these three forms plays an important role in the regulation of gene expression.
(i) B-DNA It is right-handed helix, contains 10 residues per 360° turn.

  • The planes of bases are perpendicular to the helix axis.
  • It is primarily found in chromosomal DNA.

(ii) A-DNA

  • It is formed by the moderate dehydration of B-DNA.
  • It is right-handed helix, contains 11 base pairs (residues) per 360° turn.
  • The planes of bases are tilted 20° away from the perpendicular to helical axis.
  • It is mainly found in DNA-RNA hybrid or RNA-RNA double-stranded regions.

(iii) Z-DNA It has zig-zag backbone and hence has the name.

  • It is a left-handed helix, twelve base pairs are present per turn.
    CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 6
  • The Z-DNA stretches occurring naturally in DNA have a sequence of alternating purines and pyrimidines (i.e. poly GC regions).
  • Besides these three major forms, the other two right-handed forms are
    • C-DNA with nine base pairs per turn.
    • D-DNA with eight base pairs per turn. There are many forms of DNA molecules in viruses.

Various Forms of DNA Molecules Found in a Variety of Viruses
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 7

Question 2.
Describe the semiconservative model of DNA replication.
Answer:
DNA Replication
In addition to the double helical structure of DNA, Watson and Crick also proposed a scheme for DNA replication. According to this model, the two strands of double helix separate and act as a template for the synthesis of new complementary strands in which the base sequence of one strand determines the sequence on the other strand.

This is called base complementarity and it ensures the accurate replication of DNA. After the completion of replication, each DNA molecule have one parental and one newly synthesised strand.
This scheme for DNA replication was termed as semiconservative DNA replication.

DNA Replication is Semiconservative:
Matthew, Meselson and Franklin Stahl in 1958 conducted an experiment in California Institute of technology with Escherichia coli to prove that DNA replicates semiconservatively as follows

1. They grew many generations of E.coli in a medium containing 15NH5Cl (15N is the heavy isotope of nitrogen) as the only source of nitrogen.
The result was that 15N got incorporated into the newly synthesised DNA after several generations. By centrifugation in a cesium chloride (CsCl) density gradient, this heavy DNA molecule could be distinguished from the normal DNA.

2. The bacterial culture was then washed to make the medium free and the cells were then transferred into a medium containing normal 14NH4Cl.

3. The samples were separated independently on CsCl gradient to measure the densities of DNA.

4. At definite time intervals, as the cells multiplied, samples were taken and the DNA which remained as double-stranded helices was extracted.

5. The DNA obtained from the culture, one generation after the transfer from 15N to 14N medium (i.e. after 20 minutes because E.coli divides in 20 minutes) had a hybrid or intermediate density.
This hybrid density was the result of replication in which DNA double helix had separated and the old strand (N15) had synthesised a new strand (N14).

6. The DNA obtained from the culture after another generation (II generation) was composed of equal amounts of hybrid DNA containing (N15 molecule) and ‘light’ DNA (containing 4 N molecule).

7. With the preeceding growth generations in normal medium, more and more light DNA was present. Hence, the semiconservative mode of DNA replication was confirmed.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 8
Meselson-Stahl experiment to demonstrate semiconservative replication

were conducted by Taylor and colleagues in 1958, involving use of radioactive thymidine, i.e. bromodeoxy-uridine. The replicated chromosomes were then, stained with fluorescent dye and Giemsa. The old and new strand were stained differently which confirm that the DNA in chromosomes also replicates semiconservatively.

Question 3.
Give evidences of DNA as genetic material.
Answer:
DNA as Genetic Material:
The discovery of nuclein by Meischer and the proposition of principal of inheritance by Mendel were almost at the same time, i.e. 1869 and 1866, respectively. But the fact that DNA acts as a genetic material took a long time to be discovered and proven.

By 1926, the quest to determine the mechanism for genetic inheritance had reached the molecular level and gradually the question, what molecule acts as genetic material got answered.

Transforming Principle:
Frederick Griffith in 1928, carried out a series of experiments with Diplococcus pneumoniae (a bacterium that causes pneumonia). He observed that when these bacteria were grown on a culture plate, some of them produced smooth, shiny colonies (S-type), whereas the others produced rough colonies (R-type).

This difference in appearance of colonies (smooth/rough) is due to the presence of mucous (polysaccharide) coat on S-strains (virulent/pathogenic) but not on R-strains (avirulent/non-pathogenic).

Experiment:

  • He first infected two separate groups of mice. The mice that were infected with the S-strain (S-III) died from pneumonia as S-strains are the virulent strains causing pneumonia.
  • The mice that were infected with the R-strain (R-II) did not develop pneumonia and they lived.
  • In the next set of experiments, Griffith killed the bacteria by heating them. The mice that were injected with heat-killed S-strain bacteria did not die and lived.
  • Whereas, on injecting a mixture of heat-killed S-strain and live R-strain bacteria, the mice died. Moreover, living S-bacteria were recovered from the dead mice.
    These steps are summarised below
    CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 9

From all these observations Griffith concluded that the live R-strain bacteria, had been transformed by the heat-killed S-strain bacteria, i.e. some ‘transforming principle’ had transferred from the heat-killed S-strain, which helped the R-strain bacteria to synthesise a smooth polysaccharide coat and thus, become virulent.

This must be due to the transfer of the genetic material. However, he was not able to define the biochemical nature of genetic material from his experiments.

Biochemical Characterisation of Transforming Principle:
Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44) worked in Rockfellar Institute, New Xork, USA to determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment in an in vitro system. Prior to this experiment, the genetic material was thought to be protein.

During this experiment, purified biochemicals (i.e. proteins, DNA, RNA, etc.) from the heat-killed S-III cells were taken, to observe which biochemicals could . transform live R-cells into S-cells.

They discovered that DNA alone from heat-killed S-type bacteria caused the transformation of non-virulent R-type bacteria into S-type virulent bacteria.
They also discovered that protein digesting enzymes (proteases) and RNA digesting enzymes (ribonuclease) did not inhibit this transformation. This proved that the ‘transforming substance’ was neither protein nor RNA.

DNA-digesting enzyme (deoxyribonuclease) caused inhibition of transformation, which suggests that the DNA caused the transformation. This provided the first evidence for DNA as transforming principle or the genetic material.

The steps of this experiment are summarised below:

  1. R-II + DNA extract of S-III + no enzyme = R-II colonies + S-III colonies
  2. R-II + DNA extract of S-III + Ribonuclease = R-II colonies + S-III colonies
  3. R-II + DNA extract of S-III + Protease = R-II colonies + S-III colonies
  4. R-II + DNA extract of S-III + Deoxyribonuclease = Only R-II colonies

Question 4.
Explain the mechanism of translation in prokaryotes.
Answer:
Mechanism of Translation:
The main steps in translation include
(i) Binding of iwRNA to ribosome
(ii) Activation of amino acids (aminoacylation of rRNA).
(iii) Transfer of activated amino acids to rRNA.
(iv) Initiation of polypeptide chain synthesis.
(v) Elongation of polypeptide chain.
(vi) Termination of polypeptide chain formation.

(i) Binding of mRNA to Ribosome
Ribosomes occur in the cytoplasm as separate smaller and larger units.
In prokaryotes, IF-3 binds to 30 S subunit to prevent association of subunits. The incoming tRNA containing specific amino acid binds to the A-site. The prokaryotic mRNA has a leader sequence called shine-Delgarno sequence that is present just prior to initiation codon-AUG.

It is homologous to 3′ end of 16S rRNA (ASD region) in 30S subunit. This complementarity ensures correct binding of 30S subunit on mRNA. The peptidyl mRNA containing elongating polypeptide then binds to P-site.
The bacterial ribosome contains another site, the E-site or Exit-site to which the discharged tRNA or tRNA whose peptidyl has already been transferred binds before its release from ribosome.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 10
A bacterial ribosome with three sites; E, exit site, P, peptidyl site and A, aminoacyl site

(ii) Aminoacylation
Aminoacylation of tRNA The activation of amino acids takes place through their carboxyl groups. The amino acids are activated in the presence of ATP and linked to their cognate tRNA. In the presence of ATP, amino acids become activated by binding with aminoacyl tRNA synthetase enzyme.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 11
The amino acid AMP-enzyme or AA-AMP enzyme complex is called an activated amino acid.
In the second step, this complex associated with the 3′-OH end of tRNA. AMP gets hydrolysed to form an ester bond between amino acid and tRNA and the enzyme is released.
AA – AMP – enzymes + tRNA ——– AA – tRNA + AMP + enzyme.

(iii) Transfer of Activated Amino Acids to tRNA:
The amino acids are attached to the tRNA by high energy bonds. These bonds are formed between the carboxyl group of amino acid and 3′ hydroxy terminal of ribose of terminal adenosine of CCA and of tRNA.
The complete reaction is carried out by enzyme synthetase which has two active sites, i.e. one for rRNA and another for specific amino acid molecule.

(iv) Initiation of Polypeptide Chain Synthesis:
The protein synthesis begins from the amino terminal end of the polypeptide, proceeds by the addition of amino acids through peptide bond formation and ends at the carboxyl terminal end. In prokaryotes, the initiation amino acid is formylated methionine while in eukaryotes it is methionine.
Initiation in Prokaryotes
In prokaryotes, two types of tRNA are present for methionine

  • tRNAfmet for initiation carrying formyl methionine and
  • tRNAmet for carrying normal methionine to growing polypeptide.

The initiation of polypeptide synthesis requires the following components
mRNA, 30S subunit of ribosome, formylmethionyl-tRNA (fmet – tRNAfmet), initiation factors IF-1, IF-2 and IF-3, GTP, 50S ribosomal subunit and Mg+2.

The sequence of events occurring during initiation process are

1. The smaller 30S subunit of ribosome binds to the transcription factor IF-3. It prevents the premature association of two ribosomal subunits.

2. Interaction of SD region of mRNA and ASD region of ribosome helps the mRNA to bind to 30S subunit. It also helps AUG to correctly positioned at the P-site of the ribosome.

3. The fMet-tRNAfmet (the specific tRNA aminoacylated to formyl methionine) binds to the AUG codon at the P-site. The tRNAfmet is the only tRNA that binds to its codon present on the P-site. All other tRNA along with their respective amino acids bind to their codon present at the A-site. Therefore, AUG codon present as initiation codon codes for formylmethionine. When it is present at other position it codes for normal methionine.

4. The initiation factor IF-1, binds to the A-site. It prevents the binding of any other aminoacyl tRNA to the codon at the A-site during initiation.

5. The GTP bound IF-2 (GTP-IF-2) and the initiating f Met-tRNAfmet attaches to the complex of 30S subunit-IF3-IF1-mRNA.

6. 50S subunit then attaches the complex formed in the previous step. The GTP bound to IF-2 is hydrolysed to GDP and Pi. After this step, all the three initiation factors leave ribosome. This complex of 70S ribosome, mRNA and f Met-tRNA fmet bound to initiation codon at P site is known as initiation complex.
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Stepwise formation of initiation complex in prokaryote

(v) Elongation of Polypeptide Chain:
In this step, another charged aminoacyl tRNA complex binds to the A-site of the ribosome, following the hydrolysis of GTP to GDP and Pi. A peptide bond forms between carboxyl group (—COOH) of amino acid at P-site and amino group (—NH3) of amino acid at A-site by the enzyme peptidyl transferase.
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Binding of the second aminoacyl tRNA to the A site of ribosome
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 14
Formation of a peptide bond

(vi) Translocation of Polypeptide:
The peptidyl fRNA bounded to A-site comes to the P-site of ribosome.
The empty tRNA comes to E-site and a new codon occupies the A-site for next aminoacyl tRNA.

  • This is achieved by the movement or translocation of ribosome by a codon in 5′ to 3′ direction of mRNA in the presence of EF-G (translocase) and GTP.
  • tRNA interact with E-site on 50S subunit through it CCA sequence at 3′ end.

The tRNA molecule is then, transferred from A site to P-site and from P-site to E-site by the movement of two subunits of ribosomes.
Finally, the deacylated tRNA is released to cytosol from E-site.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 15

(vii) Termination:
It is accomplished by the presence of any of the three termination codon on mRNA. Which are recognised by the release (termination) factor, RF1, RF2 and RF3.

RF1 and RF2 They resemble the structure of tRNA. They compete with tRNA to bind the termination codon at A-site of ribosome. This is called molecular mimicry. RF1 recognise UAG and RF2 recognise UGA. UAA is recognised by both of them. RF3 helps to split the peptidyl tRNA body by using GTP.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 16

Question 5.
Describe transcription in prokaryotes.
Answer:
Transcription in Prokaryotes:
All three RNAs are needed for synthesis of a protein in a cell. DNA dependent RNA polymerase is the single enzyme that catalyses the transcription of all types of bacterial RNA. But for the expression of different genes, different sigma factors may associate with same core enzymes.

In E.coli, σ70 is used in normal condition σ32 / σH under heat shock, σ54N under nitrogen starvation and σ28 for chemotaxis.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 17
A typical bacterial transcription unit

The transcription process in prokaryotes occurs in following steps
I. Initiation

  1. The holoenzyme binds to the promoter region of transcription unit.
  2. The sigma polypeptide binds loosely to the promoter sequences so as to form a loose, closed, binary complex.
  3. It is followed by the formation of a transcription eye or bubble due to the denaturation of adjacent sequence of DNA, lying next to the complex.
  4. The transcription bubble along with the bounded holoenzyme is called open binary complex.
  5. In 90% of cases, the start point of transcription is a purine.
  6. At the elongation site of enzyme, two nucleotides complementary to the first two nucleotides of template strand binds.
  7. A phosphodiester bond is formed between these two ribonucleotides.
  8. At this stage, the complex is called ternary complex that consists of partly denatured DNA bounded with holoenzyme having a di-ribonucleotide.
  9. The same process continues till a RNA chain of about nine nucleotides is synthesised. The holoenzyme does not move throughout this process.
  10. After the completion of initiation process, sigma factor dissociates from RNA polymerase. This facilitates the promoter clearance so that a new holoenzyme can bind to promoter for second round of transcription.
    CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 18
    Binding of RNA polymerase and initiation of RNA synthesis

II. Elongation

  • The RNA chain grows in 5′-3′ direction by the addition of ribonucleotides to the 3′-end of RNA.
  • The transcription bubble moves in 3′ → 5′ direction of template strand.
  • The movement of holoenzyme along the bubble unwinds (denature) the DNA in growing point and rewinds at the opposite end.
  • In each elongation cycle, the growing site (leading product) of enzyme gets filled with 10 newly added nucleotides. The opposite site (lagging product) contains the previous segment of RNA.
  • About 40 nucleotides are added per 37°C.
  • In some phages, e.g. T3 and T4, RNA pol synthesise RNA at much rapid rate of about 200 nucleotides per 37° C.
    CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 19
    Elongation of RNA chain by the movement of RNA polymerase and transcription bubble in 5′ to 3′ direction

III. Termination
It is achieved by certain termination signals on DNA called terminators. In E.coli, terminators are of two types

(a) Intrinsic Terminators

  • These are rho independent protein factors in which the RNA at 3′-end contains a long stretch of U residues.
  • These residues are hydrogen bonded to the long stretch of A residues of the temple. In the stem of the RNA, a stretch of G-C rich segment is present which results in a hair-pin loop formation in the RNA stem.
  • Due to the weak association between A-U base pairs, the long stretch of termination sequence break and the RNA is released.
  • It occurs due to the formation of hair-pin loop in the stem of RNA before the termination signal slows down transcription and as a result dA-rU bonds break at any one point so as to release RNA from RNA-DNA hybrid.

(b) Extrinsic Terminator

  • These are rho dependent protein factors and are extensively used in E. coli.
    This protein is active as an hexamer (having six identical subunits). Its molecular weight is about 46,000 and it also has ATP hydrolysing activity.
  • To terminate the process of transcription, rho factor binds to the 5′-end of nascent mRNA and moves along the length of mRNA until it reaches the termination point. Due to this, the transcription process slows down, rho breakdown ATP and utilises the energy to denature the RNA-DNA hybrid. Hence, the RNA is released from the bubble.

In prokaryotes (bacteria), mRNA does not require any processing to become active and both transcription and translation take place in same compartment (as there is no separation of nucleus and cytosol in bacteria). Therefore, translation can start much before the mRNA is fully transcribed, i.e. transcription and translation can be coupled.

Question 6.
Give an account of the operon model.
Answer:
The Operon Model:
F Jacob and J Monod gave the operon concept and were the first ones to describe a transcriptionally regulated system. An operon is a unit of prokaryotic gene expression which includes sequentially regulated (structural) genes and control elements recognised by the regulatory gene product. The various components of an operon are

  1. Structural genes These are the regions of DNA which transcribe wRNA for polypeptide synthesis.
  2. Promoter gene This is the sequence of DNA where RNA polymerase binds and initiates transcription.
  3. Operator This is the sequence of DNA found adjacent to promoter where specific repressor protein binds. It is under the control of a repressor.
  4. Regulator It is the gene which codes for the repressor protein binding to the operator and suppresses its activity, so that transcription does not occur. It is also represented as ‘i’ gene. It is synthesised by a regulator gene which is not a part a operon.
  5. Inducer Its main role is to prevent the repressor from binding to the operator. This makes the process of transcription to switch on. An inducer can be any metabolite, hormone, etc.

Loc Operon in E. coli:
The lac operon found in E.coli is an inducible system. It is responsible for the synthesis of enzymes found in lactose (the milk sugar). It has an operator sequence of 26 base pairs and three structural genes. Its first structural gene (SG) is lac Z which is of 3063 base pairs. It is responsible for the synthesis of the emzyme ß-galactosidase.
CHSE Odisha Class 12 Biology Solutions Chapter 7 Molecular Basis of Inheritance 20
The operator is a part of lac Z. The other two genes are lac y (helps in the synthesis of ß-galactoside permease) and lac a (synthesise fbgalactoside transacetylase). ß-galactoside permease is a transmembrane protein that pumps galactose into the cell, ß-galactosidase helps to breakdown lactose to galactose and glucose.

When lactose is available to the bacterium, the active repressor produced by the regulator forms an inactive dimer with lactose (acts as an inducer).

As a result, the inactive dimer cannot bind to the operator and the three contiguous structural genes are transcribed into a polycistronic or polygenic mRNA which is then translated into three proteins (enzymes). In the absence of lactose (the inducer), the product of regulator enzyme activates inhibitor dimer that, binds to the operator and prevents transcription.

CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 6 Question Answer Sex Determination

Sex Determination Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
A cross between Fj-hybrid and a recessive parent gives the ratio of
(a) 3 : 1
(b) 2 : 1
(c) 1 : 1
(d) 4 : 1
Answer:
(c) 1 : 1

Question 2.
A cross of F1 with the recessive parent is known as
(a) back cross
(b) test cross
(c) hybrid cross
(d) double cross
Answer:
(b) test cross

Question 3.
A woman with albinic father marries an albinic man. The proportion of her progeny is
(a) 2 normal : 1 albinic
(b) all normal
(c) all albinic
(d) 1 normal : 1 albinic
Answer:
(d) 1 normal : 1 albinic

Question 4.
Y-chromosome is called
(a) sex chromosome
(b) androsome
(c) autosome
(d) gynosome
Answer:
(a) sex chromosome

Question 5.
Which one is a sex-linked disorder?
(a) Leukemia
(b) Cancer
(c) Night blindness
(d) Colour blindness
Answer:
(d) Colour blindness

CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination

Question 6.
A haemophilic man marries a normal homozygous woman. What is the probability that their son will be haemophilic?
(a) 100%
(b) 75%
(c) 50%
(d) 0%
Answer:
(d) 0%

Question 7.
What is the probability that their daughter will be haemophilic?
(a) 100%
(b) 75%
(c) 50%
(d) 0%
Answer:
(d) 0%

Question 8.
A fruitfly exhibiting both male and female trait is
(a) heterozygous
(b) gynandromorph
(c) hemizygous
(d) gynandev
Answer:
(b) gynandromorph

Question 9.
Genes located on Y-chromosome are
(a) mutant genes
(b) autosomal genes
(c) holandric genes
(d) sex-linked genes
Answer:
(c) holandric genes

Question 10.
A colourblind person cannot distinguish
(a) all colours
(b) red colour
(c) green colour
(d) red and green colours
Answer:
(d) red and green colours

Question 11.
The gene responsible for haemophilia is linked to which chromosome?
(a) X
(b) Y
(c) Both X and Y
(d) Autosome
Answer:
(a) X

CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination

Question 12.
Red-green colour blindness in man is
(a) sex-linked character
(b) sex-limited character
(c) sex influenced character
(d) sexual character
Answer:
(a) sex-linked character

Question 13.
Sex-linked characters are
(a) dominant
(b) recessive
(c) lethal
(d) not inherited
Answer:
(b) recessive

Question 14.
Which gene is present in the Y-chromosome that codes for the protein TDF?
(a) cry
(b) sty
(c) try
(d) tra
Answer:
(b) sty

Question 15.
In birds, which type of chromosomal basis of sex-determination is present?
(a) XX – XY
(b) XX – XO
(c) ZW – ZZ
(d) ZZ – ZO
Answer:
(c) ZW – ZZ

Question 16.
When the ratio of X/A=0.67 in genic balance theory, which type of sex is expressed?
(a) Super female
(b) Intersex
(c) Super male
(d) Triploid female
Answer:
(b) Intersex

Question 17.
Which type of sex-determination is found in Bonellia?
(a) Temperature dependent
(b) Chemotactic
(c) Holandric
(d) Pseudoautosomal
Answer:
(b) Chemotactic

Question 18.
In a person with Turner syndrome, the number of X-chromosome is
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(a) 1

CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination

Question 19.
A Down syndrome will be
(a) 45 + XX
(b) 44 + XY
(c) 44 + XXY
(d) 22 + XY
Answer:
(a) 45 + XX

Question 20.
Number of Barr bodies present in Turner syndrome is
(a) 0
(b) 1
(c) 2
(d) Either (b) or (c)
Answer:
(a) 0

Express in one or two word(s)

Question 1.
Name two sex-linked diseases of human being.
Answer:
Haemophilia and colour blindness.

Question 2.
How Down’s syndrome is caused?
Answer:
Down’s syndrome is caused due to the presence of an extra-chromosome number 21, i.e. 21 trisomy.

Question 3.
In which chromosome is the gene for haemophilia located?
Answer:
X-chromosome

Question 4.
What is the chromosomal formula for Turner’s syndrome?
Answer:
44 + XO

Question 5.
Which sex is usually a carrier?
Answer:
Female sex

Question 6.
Who proposed the genic balance theory?
Answer:
Calvin Bridges

Question 7.
What are holandric genes?
Answer:
Genes located on Y-chromosomes are known as holandric genes

Question 8.
In which chromosome, the factors for haemophilia and colour blindness are found?
Answer:
X-chromosome

Question 9.
What is the other name of Bleeder’s disease?
Answer:
Haemophilia

Question 10.
Which protein is in sry gene of Y-chromosome?
Answer:
TDF (Testis Determining Factor)

Question 11.
What is gynandromorph?
Answer:
Exhibiting both male and female characters are called gynandromorphs.

Question 12.
What is freemartin?
Answer:
Sterile female with many male characteristics.

CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination

Question 13.
What is criss-cross inheritance?
Answer:
The transmission of characters from grandfather to grandson through daughter is called criss-cross inheritance.

Question 14.
Which type of defect is found in thalassemia?
Answer:
The mutation or deletion of the genes controlling the formation of globin chains of haemoglobin result in an abnormal form of haemoglobin.

Question 15.
Who first described Klinefelter’s syndrome?
Answer:
H. F. Klinefelter in 1942.

Short Answer Type Questions

Write brief notes on the following (within 50 words each)

Question 1.
Criss-cross inheritance
Answer:
The transmission of characters from grandfather to grandson through daughter is called criss-cross inheritance.

Question 2.
Holandric gene
Answer:
Genes located on Y-chromosomes are known as holandric genes

Question 3.
Haplo-diploidy mechanism of sex-determination
Answer:
Haplo-Diploidy Mechanism:
In insects-like honeybees wasps, ants, etc., the sex chromosomes are not differentiated and sex is determined on the basis of ploidy of the individual. In honeybees, drones are males and are haploid (n =16), which had developed from unfertilised eggs of females (Arrhenotoky).
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 1

Question 4.
Genic balance theory
Answer:
Genic Balance Mechanism
The investigations on Drosophila by C. Bridges showed that female determiners were located on the X-chromosomes and that of male were on the autosomes. Hence, autosomes also plays an important role in determining sex in Drosophila melanogaster. Genic balance’ theory by Bridges proposed the sex-determination mechanism based on the ration of number of X-chromosomes (X) and sets of autosomes (A).
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 2

CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination

Question 5.
Freemartin
Answer:
It is an infertile female mammal with masculinised behaviour and non-functioning ovaries. Freemartinism is the normal outcome of mixed sex twins in all cattle species, i.e. it occurs when the twins of opposite sex are born also occurs occasionally in other mammals including sheep, goats and pigs.
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 3
In most cattle twins, the blood vessels in the chorions become interconnected, creating a shared circulation for both twins. Mostly the male hormones are produced first. If both foetuses are of the same sex this is of no significance, but if they are of different sex, male hormones pass from the male twin to the female twin.
The male hormones then masculinise the female twin and the result is a freemartin or sterile masculine female.

Question 6.
Gynandromorph
Answer:
Abnormal chromosomal behaviour in insects can result in the formation of gynandromorphs or sexual mosaics in which some parts of the animal exhibit female characters and other parts exhibit male characters.

Some gynandromorphs in Drosophila are bilateral intersexes with male colour pattern, body shape, and sex comb on one half of the body and female characteristics on the other half.
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 4
The failure of segregation (non-disjunction) of X-chromosomes at cleavage leads to formation of gynandromorphs. The chromosome complement of zygote is 2A + 2X. During first cleavage, one of the X-chromosomes is lost in one of the blastomeres.
As a result, one of the blastomeres acquires 2A + 2X complement which forms the female half while the other blastomere with 2A + X complement forms the male half. Thus, half of the body is female while the other half is male.

Question 7.
Single gene effect
Answer:
Single Gene Effect:
In certain organisms-like Drosophila, human, Asparagus and several fishes, a single gene pair is responsible for the determination and expression of sex.
In Drosophila, the sex is expressed by a recessive gene called tra (transformer) present on the third autosome. Males and female members with dominant (tra+) allele are mostly fertile. However, a normal female (i.e. AA + XX), having homozygous recessive tra alleles, develops into sterile male.

As studied earlier, in humans, the Y-chromosome has a sry gene which influences the development of testis in males. Its absence results in development of ovaries in females.
Thus, an XX female with sry gene or an XY male without sry gene ultimately develops into a sterile female.

Question 8.
Sex reversal
Answer:
Sex Reversal:
Artificial removal of gonads of either sex before puberty (castration or ovariectomy) results in the development of secondary sexual characters of the opposite sex. It is observed in fishes, amphibians, birds and some mammals, including humans.
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 5
In the given figure, the hen develops male secondary sex characteristics after the removal of ovaries. Thus, the bird is still female genotypically but phenotypically it becomes male.

Question 9.
Temperature dependent sex-determination
Answer:
Temperature Dependent Sex-Determination:
In some reptiles, the temperature at which the fertilised eggs are incubated prior to hatching plays a major role in determining the sex of the offspring. Surprisingly high temperature during incubation have opposite effect on sex-determination in different species.

In turtles, high incubation temperature (above 30°C) of eggs results in the production of female progeny whereas in the lizard and crocodiles, high incubation temperature results in the production of male offspring. At the lower temperature range between 22.5-27°C, male turtles are produced. This pattern is reversed in lizards and crocodiles.

CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination

Question 10.
Chemotactic sex-determination
Answer:
Chemotactic Sex-Determination:
It is seen in males of the marine worm Bonellia. These are small, degenerate and live within the reproductive tract of the larger female. All organs of male worm’s body are degenerate except those of the reproductive system.
In Bonellia, the larvae of male and female are genetically and cytolosically similar, i.e. it is hermaphrodite. A newly hatched worm if reared from a single cell kept in isolation, it develops into a female. If the larvae are reared with mature females in water, they adhere to the proboscis.

Later they transform into males who eventually migrate into the female reproductive tract, where they become parasitic.
It has been found that a chemotactic substance secreted by the proboscis of a mature female Bonellia induces the differentiation of larva into males.
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 6

Question 11.
Thalassemia
Answer:
It is an autosomal recessive blood disease, which is caused due to reduced synthesis of either the α or ß-chains of haemoglobin, as a result of mutation in the genes of α or ß-chains. It was discovered by Cooley (1925) and is prevalent in Asia, middle-East, Africa and Mediterranian countries. Anaemia is the characteristic symptom of this disease. Depending upon the globin chain affected, thalassemia is classified into following types, i.e.
1. α-Thalassemia is caused by defective α-chain. The OC-globin is controlled by two genes present on chromosome 16, i.e. HBA1 and HBA2. α-thalassemia is of two types-haemoglobin H-disease and Hydrops Foetalis. The later is more severe as all the four globin genes are mutated and the defective alleles kill the foetus resulting in still birth or death soon after delivery.
Haemoglobin H-disease occurs when there are three defective alleles out of four α-globin genes.

2. ß-Thalassemia is caused by decreased synthesis of ß-globin chain. It is further classified into thalassemia major (Cooley’s anaemia), i.e. when both the alleles for ß-globin are defective or absent. It is more severe in comparison to the second type called thalassemia intermedia, i.e. when only one allele is defective in ß-globin.

Symptoms:
The common symptoms include tiredness, pale skin with severe anaemia, enlarged spleen, yellowish skin and dark urine.

Diagnosis Treatment and Prevention:
The disease is diagnosed by blood test and genetic analysis. There are two treatment options, i.e. blood transfusion and bone marrow transplantation. Genetic councelling is not recommended as it makes the persqn concious about the consequences of the disease.

Question 12.
Down’s syndrome
Answer:
Down’s Syndrome (Mongolism):
This syndrome was previously called mongolism because the affected persons were of short stature. The estimated frequence of birth of individual with Down’s syndrome is 1/700.
It was described by J Langdon Down in 1866 but its cause was found by Lejeune in 1959.

Genetic Basis:
It occurs due to chromosomal aberration, known as aneuploidy (trisomy). The individuals suffering from Down’s syndrome posses an extrachromosome number 21. Both the chromosomes of 21 position passes into a single egg due to primary non-disjunction which may occur during meiosis-I or II in maturation phase of gametogenesis. Thus, the egg instead of possessing 23 chromosomes have 24 chromosomes and the offspring has 47 chromosomes (45 + XY in males, 45 +XX in females). It is also seen in chimpanzees and other related primates.

CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination

Question 13.
Turner’s syndrome
Answer:
Turner’s Syndrome:
This condition is characterised by one missing X-chromosome which result in 45 + XO chromosomal complement in affected person. It was first described by H.H. Turner in 1938. The estimated birth frequency of Turner’s syndrome is 1/2500 live female births. It total frequency in human population is 1/5000.

Genetic Basis:
It is a disorder which is caused due to chromosomal aberration, known as aneuploidy (monosomy). Due to absence of one of the X-chromosome, the condition is 45 with XO. Primary non-disjunction in either of meiotic divisions during gametogenesis results in this condition.

Question 14.
Klinefelter’s syndrome
Answer:
Klinefelter’s Syndrome:
HF Klinefelter first described this condition in 1942.
This genetic disorder occurs due to the presence of an additional copy of the X-chromosome. It is also known as trisomy of X-chromosome. Its estimated birth frequency is 1/500 live male births.

Genetic Basis:
The union of an abnormal XX-egg with a normal Y-sperm or a normal X-egg with an abnormal XY-sperms results in the karyotype of 47, XXY in males or 47, XXX in females.

The abnormal eggs and sperms are formed due to the v primary non-disjunction ofX and Y chromosomes during the maturation phase of gametogenesis. Although the usual karyotype of this condition is 47 + XXY but sometimes more complex karyotypes also occurs, e.g. XXXY, XXXXY, XXXXXY, XXXXYY, etc.

Differentiate between two words in the following pairs of words

Question 1.
Phenotype and Genotype.
Answer:
Differences between phenotype and genotype are as

Phenotype Genotype
It refers to observable traits or characters. It refers to the genetic constitution of an individual.
It results from expression of genes. It constitutes single gene pair or sum total of all the genes.
The phenotypic ratio of Mendel’s monohybrid cross is 3 : 1. The genotypic ratio of Mendel’s monohybrid cross is 1 : 2 :1.
It may change with age and environment. It remains the same throughout the life of an individual.

Question 2.
Autosome and Allosome.
Answer:
Differences between autosomes and allosomes are as follows

Autosome Allosomes
They are somatic chromosomes which control the body character  or somatic characters. They are sex chromosomes which determine the sex of an individual.
In humans, out of the total 23 pairs, of chromosomes 22 pairs are autosomes. In humans, the 23rd pair of chromosome is called sex  chromosome.

Question 3.
X-chromosome and Y-chromosome.
Answer:
Differences between X and Y-chromosomes are as follows

X-chromosome Y-chromosome
It is sex chromosome. It is also a sex chromosome.
Females have two X-chromosomes. It is absent in females.
It is larger than Y-chromosome. It is smaller than X-chromosome.
It does not contain sry gene. It carries male determining gene called sry gene.

Question 4.
Supermale and Superfemale.
Answer:
Differences between superfemales and supermales are as follows

Superfemales Supermales
Such individuals have 47(44 + XXX), 48(44 + XXX) chromosomes. Such individuals have 47(44+YYY) chromosomes.
These females have abnormal sexual development and mentally retarded. These males are characterised by abnormal height, mental retardation.

Question 5.
Sex differentiation and Sex reversal.
Answer:
Differences between sex differentiation and sex reversal are as follows

Sex differentiation Sex reversal
It is the process of the differences between males and females from an undifferentiated zygote. It is the phenomenon of development of secondary sexual characters of the opposite sex.
It is induced by specific genes by hormones and by anatomy. It is induced by some chemicals.

Question 6.
Gynandromorph and Freemartin.
Answer:
Differences between gynandromorphs and freemartin are as follows

Gynandromorphs Freemartin
It is the phenomenon in which a part of the body exhibits female characters, while the other part exhibits male characters, e.g. Drosophila. When the,twins of the opposite sex are born, the male is normal but the female is sterile with many male characters. Such sterile females are called freemartin, e.g. cattle.
These develop due to failure of segregation of X-chromosomes at cleavage. These develop due to influence of male hormone.

Question 7.
Down’s syndrome and Turner’s syndrome.
Answer:
Differences between Down’s syndrome and Turner’s syndrome are as follows

Down’s syndrome Turner’s syndrome
It occurs due to the presence of an additional copy of the chromosome number 21. This condition is called trisomy of 21 chromosome. It is a disorder caused due to the absence of one of the X-chromosome, i.e. 45 with XO.
It is an autosomal genetic disorder. It is a sex-linked chromosomal genetic disorder.

Long Answer Type Questions

Question 1.
Discuss the chromosomal theory of sex-determination.
Answer:
Chromosomal Mechanism of Sex-Determination:
The male and female individuals normally differ in their chromosomal constituents. There are two types of chromosomes, i.e.

  • Sex chromosomes or Allosomes The chromosomes responsible for sex determination, e.g. X and Y-chromosomes.
  • Autosomes The chromosomes which determines the somatic characters.

X-chromosome was first discovered by Henking (1891). He named this structure as X-body. Scientists further explained that X-body was a chromosome and called it as X-chromosome. The concept of autosomes and allosomes was proposed by Wilson and Stevens (1902-1905) in chromosomal theory of sex-determination.
There are various types of chromosomal sex-determination mechanism observed in different animals as follows
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 7

Sex Differentiation:
Sex-determination in lower grade animals can be explained successfully by the genic balance and chromosomal theory. However, this is not the case with vertebrates and in some invertebrates, where the embryo develops certain traits of opposite sex along with its own. This indicates that sex of an organism changes under specific conditions. This may happen as a result of hormones secreted from the gonads of such organisms.
Some examples of sex differentiation are given below

Sex Reversal:
Artificial removal of gonads of either sex before puberty (castration or ovariectomy) results in the development of secondary sexual characters of the opposite sex. It is observed in fishes, amphibians, birds and some mammals, including humans. ,
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 5
In the given figure, the hen develops male secondary sex characteristics after the removal of ovaries. Thus, the bird is still female genotypically but phenotypically it becomes male.

Freemartin:
It is an infertile female mammal with masculinised behaviour and non-functioning ovaries. Freemartinism is the normal outcome of mixed sex twins in all cattle species, i.e. it occurs when the twins of opposite sex are born: also occurs occasionally in other mammals including sheep, goats and pigs.
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 3
In most cattle twins, the blood vessels in the chorions become interconnected, creating a shared circulation for both twins. Mostly the male hormones are produced first. If both foetuses are of the same sex this is of no significance, but if they are of different sex, male hormones pass from the male twin to the female twin.
The male hormones then masculinise the female twin and the result is a freemartin or sterile masculine female.

Question 2.
What is genic balance theory and explain its role in sex-determination?
Answer:
Genic Balance Mechanism:
The investigations on Drosophila by C. Bridges showed that female determiners were located on the X-chromosomes and that of male were on the autosomes. Hence, autosomes also plays an important role in determining sex in Drosophila melanogaster. Genic balance’ theory by Bridges proposed the sex-determination mechanism based on the ration of number of X-chromosomes (X) and sets of autosomes (A).

The table given below describes about the phenotypic sex of ‘ D. melanogaster based on X/A values
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 2

Gynandromorph in Drosophila as a Proof of Chromosomal Mechanism of Sex-Determination
Abnormal chromosomal behaviour in insects can result in the formation of gynandromorphs or sexual mosaics in which some parts of the animal exhibit female characters and other parts exhibit male characters.

Some gynandromorphs in Drosophila are bilateral intersexes with male colour pattern, body shape, and sex comb on one half of the body and female characteristics on the other half.
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 4
The failure of segregation (non-disjunction) of X-chromosomes at cleavage leads to formation of gynandromorphs. The chromosome complement of zygote is 2A + 2X. During first cleavage, one of the X-chromosomes is lost in one of the blastomeres.

As a result, one of the blastomeres acquires 2A + 2X complement which forms the female half while the other blastomere with 2A + X complement forms the male half. Thus, half of the body is female while the other half is male.

Question 3.
Explain sex-linked inheritance. Discuss the phenomenon with the example of colour blindness.
Answer:
Sex-Linked Inheritance:
Sex chromosomes contain genes primarily concerned with the determination the sex of the organism. In addition to sex genes, they also contain the genes to control other body characters, thus are called sex-linked genes. The somatic characters whose genes are located on sex chromosomes are known as sex-linked characters. The inheritance of a trait (phenotype) that is determined by a gene located on one of the sex chromosome is called sex-linked inheritance.

Sex-Linked Genes:
The sex-linked genes are of the following types
1. X-linked Genes These are sex-linked genes which lie on X-chromosomes, e.g. genes for colour blindness and haemophilia. These X-linked traits have a unique mode of inheritance as females have two doses of X-linked genes, while males have only one. Thus, males are hemizygous for X-linked traits they possess only half the number of X-chromosomes a female possess. An X-linked gene can be dominant or recessive due to which a female can be a heterozygous carrier of X-linked trait.

2. Y-linked Genes These are sex-linked genes, which are inherited straight from father to son or male to male, e.g. genes for hypertrichosis. Any gene which occurs exclusively on Y-chromosome is said to be holandric and it shows holandric inheritance pattern,

3. Pseudoautosomal Genes Genes located on homologous parts of both X and Y-chromosomes.

Inheritance of Sex-linked Characters:
The alleles for sex-linked traits are recessive to their normal alleles. These alleles express themselves in males, i.e. in heterogametic condition whereas in females, they express themselves only in homozygous condition (XCXC). In case the female is heterozygous for sex-linked gene (XCX), the trait is not expressed in F1-generation, but the female becomes carrier of the allele.

The X-chromosomes of carrier female is distributed equally to her children, while through male, it is distributed in daughters (in F1-generation). The daughter then distributes her X-chromosomes equally to her children during F2-generation (son or daughter). Thus, the X-chromosome does not pass directly from father to son but follows a criss-cross inheritance.

In other words, a male transmits his (X-linked) traits to his grandson through her daughter whereas, a female passes her traits to grand-daughter through her son. This pattern of inheritance where a trait skips a generation or criss-crosses the F1-generation while passing the trait to F2 is known as criss-cross inheritance.

Inheritance of Haemophilia
It is a sex-linked recessive disease, which is transmitted from an unaffected carrier female to some of the male offsprings. Due to this, patient continues to bleed even during a minor injury because of defective blood coagulation and hence, it is also called as bleeders disease. The gene for haemophilia is located on X-chromosome and it is recessive to its normal allele. In this disease, a single protein that is part of cascade of proteins involved in blood clotting is affected.

The possibility of a female becoming a haemophilic is extremely rare because mother of such a female has to be atleast carrier and father should be haemophilic, e.g. females suffer from this disease only in homozygous condition, i.e. XCXC. The haemophilic alleles shows criss-cross inheritance and they follow Mendelian pattern of inheritance. The family pedigree of Queen Victoria (who was a carrier of haemophilia) shows a number of haemophilic individual.

The inheritance is explained below
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 8
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 9
Four crosses explaining the inheritance of haemophilia allele in human, (a) Normal mother and haemophilic father, (b) Haemophilic mother and normal father, (c) Carrier mother and normal father and (d) Carrier mother and haemophilic father

Inheritance of Red-Green Colour Blindness:
It is a sex-linked recessive disorder, which results in defect in either red or green cone of eye. It does not mean the incapability to see any colour at all, infact it leads to the failure in discrimination between red and green colour. The gene for colour blindness is present on X-chromosome.

It is present mostly in males (XCY) because of the presence of only one X-chromosome as compared to two chromosomes in females. A heterozygous female has normal vision, but is a carrier and passes on the disorder to some of her sons. Its inheritance pattern is similar to that of haemophilia.
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 10

Question 4.
Give an account of sex linkage in Drosophila and man.
Ans.
Sex linkage is the phenotypic expression of an allele related to the allosome of the individual. In autosomal chromosomes both sexes have the same probability of existing but since humans have many more genes on the female X-chromosome than on the male Y-chromosome, these are much more common than Y-linked traits. Examples of sex-linked traits in humans are haemophilia and colour blindness (for detail refer to text on page no. 147-148).

Thomas Hunt Morgan discovered sex linkage in fruitfly. It supported the chromosomal theory of heredity.
Morgan proposed that the inheritance of eye colour is related to the sex of the offspring. He found that the gene for eye colour is located on X-chromosome. There is no corresponding allele for this trait on Y-chromosome.

Example During the cross between white-eyed male and red-eyed female, the F1-flies (both male and female) were all red-eyed indicating that white eye colour is recessive to the normal red eye colour.
CHSE Odisha Class 12 Biology Solutions Chapter 6 Sex Determination 11
A cross between red-eyed female and white-eyed male showing sex-linked inheritance in Drosophila

CHSE Odisha Class 12 Biology Solutions Chapter 4 Reproductive Health

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 4 Reproductive Health Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 4 Question Answer Reproductive Health

Reproductive Health Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Choose the correct options

Question 1.
The method of directly injecting a sperm into an ovum is assisted by reproductive technology called
(a) GIFT
(b) ZIFT
(c) ICSI
(d) ET
Answer:
(c) ICSI

Question 2.
Intensely lactating mothers do not generally conceive due to the
(a) suppression of gonadotropins
(b) hypersecretion of gonadotropins
(c) suppression of gametic transport
(d) suppression of fertilisation
Answer:
(a) suppression of gonadotropins

Question 3.
Which is not a spacing method of family planning?
(a) Natural method
(b) Terminal method
(c) Chemical method
(d) Hormonal method
Answer:
(b) Terminal method

Question 4.
Intrauterine Devices (IUDs) are not made up of
(a) plastic
(b) metal
(c) rubber
(d) Both (a) and (b)
Answer:
(c) rubber

CHSE Odisha Class 12 Biology Solutions Chapter 4 Reproductive Health

Question 5.
Creams, jelly and foam tablets are chemical contraceptions of which method of birth control?
(a) IUD
(b) Chemical method
(c) Hormonal method
(d) Natural method
Answer:
(b) Chemical method

Question 6.
Which is not a constituent of chemical method?
(a) Lactic acid
(b) Boric acid
(c) Malic acid
(d) Citric acid
Answer:
(c) Malic acid

Question 7.
Which is not a method of tubectomy?
(a) Conventional transabdominal surgery
(b) Conventional laprotomy
(c) Implants
(d) Milaparatomy
Answer:
(c) Implants

Question 8.
In which type of pill, both oestrogen and progestin are present in nearly the same amount?
(a) Monophasic combined
(b) Multiphasic combined
(c) Mini
(d) Antiprogesteron
Answer:
(b) Multiphasic combined

Question 9.
Which is not a type of Intrauterine Device (IUD)?
(a) Vaginal vault
(b) Loops
(c) Spirals
(d) Ts
Answer:
(a) Vaginal vault

Question 10.
Which is not a common type of Sexually Transmitted Disease (STD)?
(a) Genital warts
(b) Syphilis
(c) Cancer
(d) Gonorrhoea
Answer:
(c) Cancer

Question 11.
Which is a fertility treatment for men?
(a) Intrauterine insemination
(b) Erectile dysfunction
(c) Assisted hatching
(d) In vitro fertilisation
Answer:
(a) Intrauterine insemination

Question 12.
Emergency contraceptives are effective if used within
(a) 72 hours of coitus
(b) 72 hours of ovulation
(c) 72 hours of menstruation
(d) 72 hours of implantation
Answer:
(a) 72 hours of coitus .

Question 13.
The correct surgical procedure as contraceptive method is
(a) ovariectomy
(b) hysterectomy
(c) vasectomy
(d) castration
Answer:
(c) vasectomy

Fill in the blanks

Question 1.
The scientific study of human population is called ……………..
Answer:
demography

Question 2.
In India, the sex ratio of 1:1 is found in ………….
Answer:
Kerala

Question 3.
The common brand provided by family welfare services is …………
Answer:
NIRODH

Question 4.
Fern shield is otherwisely known as female ……………
Answer:
condom

Question 5.
Loops and bows are the type of …………….
Answer:
IUDs

Question 6.
Cooper-T has a local ……….. effect.
Answer:
antifertility.

CHSE Odisha Class 12 Biology Solutions Chapter 4 Reproductive Health

Question 7.
Sponge (Today) is a foam suppository or tablet containing ………….. as spermicides.
Answer:
nonoxynot-9

Question 8.
The example of chemical contraceptive in the form of cream is ……………
Answer:
delfem

Question 9.
An antiprogesterone pill ………… is a single pill treatment for oral contraceptive.
Answer:
mifepristone

Question 10.
The sterilisation procedure in males is called ……….. and in females is called …………. .
Answer:
vasectomy, tubectomy

Question 11.
………. and ……….. are combined injectable contraceptives.
Answer:
Cyclofem, mesigna

Question 12.
Human …………… infection is a known cause of cancer of the cervix.
Answer:
papilloma virus

Question 13.
The method of preserving sperm in frozen condition is called ……………. .
Answer:
sperm banking or semen cryopreservation

Question 14.
The monthly release of eggs is called …………. .
Answer:
ovulation

Question 15.
The ejaculatory duct obstruction in males is confirmed by ………….. .
Answer:
ultrasound test

Question 16.
Fertility treatment with donor eggs is usually done using ………….. .
Answer:
IVF

Express in one or two word(s)

Question 1.
In which state of India, the sex ratio is favourable for females?
Answer:
Kerala

Question 2.
Which device provides protection against sexually transmitted diseases including AIDS?
Answer:
Condom

Question 3.
Which device prevents the entry of sperms into the uterus?
Answer:
Cervical caps

Question 4.
Which toxic substance is released by local antifertility effect of copper-T?
Answer:
Cytokines

CHSE Odisha Class 12 Biology Solutions Chapter 4 Reproductive Health

Question 5.
What activities are caused due to hormone releasing lUDs?
Answer:
Hormone releasing IUDs result in the prevention of maturation of ovum, so that no ovulation occurs in females.

Question 6.
The chemical method of birth control absorbs, what?
Answer:
Sperms

Question 7.
STDs can be considered as self-invited diseases. Comment.
Answer:
STDs can be self-invited if unprotected sexual intercourse is done with an infected person. Therefore, unprotected sex should be avoided.

Question 8.
Mention the primary aim of the ‘Assisted Reproductive Technology’ (ART) programme .
Answer:
The primary aim of ‘Assisted Reproductive Technology’ (ART) is to cure infertility by successful conception and giving birth to healthy babies.

Question 9.
What is the significance of progesterone – oestrogen combination as a contraceptive, method?
Answer:
The combination of oestrogen and progesterone as contraceptive method inhibits the ovulation. Hence, no egg is made available for pregnancy.

Question 10.
Males whose testes fail to descend to the scrotum are generally infertile, why?
Answer:
Scrotum provides optimum temperature for sperm . production. If testes fail to descend to scrotum, sperms are not formed due to high body temperature.

Question 11.
Name the process of bringing eligible couples under family planning measures.
Answer:
Couple protection.

Short Answer Type Questions

Question 1.
Mention the different barrier methods of family planning.
Answer:
The common barrier methods are condoms, diaphragm, fern shield, cervical cap and vault cap. These are mechanical devices which prevent the release of sperms into the vagina and hence, their passage into the uterus.

Question 2.
What are the different types of IUDs?
Answer:
IUDs are of three types- copper releasing, hormone and non-medicated IUDs.

  • Copper releasing IUDs has local antifertility effect by releasing the toxic cytokines.
  • Hormone releasing IUDs suppress endometrial changes which cause an ovulation.
  • Inert IUDs are made up of polyethylene, impregnated with barium sulphate, their exact mechanism is not clear.

Question 3.
Why copper-T are to be replaced every 3-5 years?
Answer:
Copper-T are to be replaced every 3-5 years when copper release slows down due to calcium deposition.

Question 4.
What changes occur in the endometrium due to inert IUD contraception?
Answer:
Inert IUDs histologically and biochemically change the endometrium, which have gametotoxic and spermicidal effects.

CHSE Odisha Class 12 Biology Solutions Chapter 4 Reproductive Health

Question 5.
What are the hormone releasing IUDs?
Answer:
Hormone releasing IUDs are contraceptive devices which release small quantities of hormones to suppress endometrial changes and cervical mucous, cause an ovulation and insufficient luteal activity. Hormone releasing IUDs are of three types

  • Oral contraceptives
  • Non-oral contraceptives
  • Emergency contraceptives

Question 6.
What are the different types of hormonal methods of family planning?
Answer:
Hormonal methods of family planning involve the use of oral contraceptives, hormone releasing IUDs, implants, injectable contraceptives and emergency contraceptives.

Question 7.
What are the morning after pills of oral contraceptive?
Answer:
The drugs used in emergency contraception are called morning after pills. They include noral, norgynon and ovidon. An antiprogesterone pill (mifepristone) is a single pill treatment.

Question 8.
Mention the different natural methods of birth control.
Answer:
Natural methods of birth control include

  • Safe period or Rhythm method Avoiding sexual intercourse upto 48 hours of an ovulation prevents conception.
  • Coitus interruptus Withdrawal of penis before ejaculation.
  • Lactational amenorrhea Period of intense breast feeding prevents ovulation and hence, pregnancy.

Question 9.
What are the different ways of tubectomy?
Answer:
Tubectomy is performed by conventional transabdominal surgery, conventional laparoscopy and milaparotomy.

Question 10.
What are the different diseases, which show no symptoms?
Answer:
STDs like chlamydia, genital herpes and gonorrhoea may be present, but express no symptoms, especially in women.

Question 11.
What are the STDs in women, which show no symptoms?
Answer:
STDs like chlamydia, genital herpes and gonorrhoea may be present, but express no symptoms, especially in women.

CHSE Odisha Class 12 Biology Solutions Chapter 4 Reproductive Health

Question 12.
What is ‘ectopic pregnancy’ ?
Or The pregnancy that occurs outside the uterus is called pregnancy.
Answer:
Ectopic (tubular) pregnancy is a complication of pregnancy in which the embryo attaches outside the uterus.

Question 13.
Name the different types of semen problems.
Answer:
Abnormal semen is responsible for male infertility in more than 75% cases. The following semen problems are possible

  • Low sperm count (oligospermia) Sperm count should be 20 million sperms/mL3 of semen. If the count is under 10 million, there is a low sperm concentration.
  • No sperm When the man ejaculates, there is no sperm in the semen.
  • Low sperm motility The sperm cannot swim as effectively as it should.
  • Abnormal sperm Sometimes the sperms have unusual structures, making these more difficult to swim towards the egg and fertilise it.

Differentiate between two words in the following pairs of words

Question 1.
Vasectomy and Tubectomy
Answer:
Differences between vasectomy and tubectomy are as follows

Vasectomy Tubectomy
It is a sterilisation method of contraception in males. It is a sterilisation method of contraception in females.
In this method, a small part of the vas deferens is removed or tied up through a small incision on the scrotum. In this method, a small part of Fallopian tube is removed or tied up through a small incision in the abdomen/vagina.
Man’s body continues to produce sperms, they are simply reabsorbed back into body and not ejaculated. This keeps egg away from the uterus, site of its fertilisation.

Question 2.
Spacing method and Terminal method
Answer:
Differences between spacing and terminal method are as follows

Spacing method Terminal method
It is a temporary method. It is a permanent method.
Birth of children can be planned later in life. Birth of children cannot be planned after this procedure.
It may use physical barriers, hormones, chemicals, etc., to induce contraception. It is a surgical procedure to block the passage of sperms and ova.
It includes use of condoms, pills, rhythm method, etc. It includes methods like vasectomy and tubectomy.

Question 3.
Chemical and Natural methods of birth control.
Answer:
Differences between chemical and natural method of birth control are as follows

Chemical method of birth control Natural method of birth control
In this, chemicals are used to kill spermatozoa. It works on the principle of avoiding the chances of sperm meeting an ovum.
Spermicides like creams, foams, jellies which contain lactic acid, citric acid, zinc, sulphate, etc. are used. They include various methods like periodic abstinence, coitus interrupts or withdrawal and lactational amenorrhea.
In this, chemicals are introduced into the vagina before coitus. No such chemicals are introduced into the vagina before coitus.

CHSE Odisha Class 12 Biology Solutions Chapter 4 Reproductive Health

Question 4.
Safe period and Unsafe period.
Answer:
Differences between safe and unsafe period are as follows

Safe period Unsafe period
Period of monthly cycle minus ± 2 days of ovulation. Period upto 48 hours after ovulation.
Pregnancy does not occur. Pregnancy chances are higher during this period,
Body temperature is raised around 1°F above average. Body temperature dips below average.
Cervical mucous is thick. Cervical mucous is slippery to allow easy passage of sperm.

Question 5.
Conventional vasectomy and Non-scalpel vasectomy.
Answer:
Differences between conventional and non-scalpel vasectomy are as follows

Conventional vasectomy Non-scalpel vasectomy
Incision is made on scrotum’s skin with the help of scalpel over the area of vas deferens. Instead of scalpel, dissecting forceps and ringed forceps are used.
As vas deferens is exposed and cut, the two ends are separated and tied. The skin is punctured, vas deferens is taken out. It is occulated, followed by ligation of ends.
A gap of 1 -4 cm must be maintained. A gap of 1 -2 cm is maintained.

Long Answer Type Questions

Question 1.
Discuss the mode of action and advantages / disadvantages of hormonal contraceptives.
Answer:
Hormonal Methods
These are hormones possessing contraceptive properties, usually employed on women for suppressing ovulation. Hormonal methods are of three types-oral contraceptives (oral pills), non-oral hormonal contraceptives and emergency contraceptives.
(i) Oral contraceptives These are the preparations of hormones either progestin or progestin-oestrogen combinations in the form of pills (tablets), used by the females.
These oral contraceptives act in four ways
(a) Inhibition of ovulation.
(b) Alternation in the uterine endometrium to make it unsuitable for implantation.
(c) Modification in cervical mucus secretion to impair the ability of transportation of sperm.
(d) Inhibition of motility and secretory activity of Fallopian tubes.

Oral pills are of following types
1. Pill (combined) It contains the female sex hormones oestrogen and progesterone. Oestrogen prevents development of eggs and ovulation by inhibiting secretion of FSH. Progesterone inhibits LH production; acts on cervical mucous to prevent penetration of sperm. They can also prevent the blastocyst implantation, e.g. Mala-D Mala-L. One pill taken orally each day during first 3 weeks of cycle. After 4 weeks, menstruation starts and the pill is taker) again.

2. Saheli (non-steroidal pill) Selectively, it is a non-hormonal pill that contains centchroman. It acts by inhibiting oestrogen activity. It was developed by Central Drug Research Institute (CDRI), Lucknow. It is taken twice a week for 3 months.

3. Mini pill or Progestin Only Pill (POP) It contains progestin, a synthetic progesterone only. Ovulation may occur, but cervical mucous is thickened, preventing entry of sperms. Must be taken within 3 hours of same time every day, but newer generation have the relaxations of 12 hours.

(ii) Non-oral Contraceptives
These are categorised as of following two types ,
1. Injectables hormone injections (Depo-Provera) are progestin derivative preparations.
They are convenient and highly effective with no side effects. These injectables inhibit the release of hormone to prevent ovulation.
For example, Depot-Medroxy Progesterone Acetate (DMPA) with dose of 150 mg every 3 months or 300 mg for 6 months and Norethisternone Enanthate (NET EN) with dose 200 mg every 2 months.
Cyclofem and mesigna are combined injectable – contraceptives that contain progestin and oestrodiol.

2. Implant (Norplant) is a new method of ‘ contraception, which is inserted under the skin inside upper arm or forearm through a small incision.
Though, these act similarly to oral contraceptives by blocking ovulation and thickening the cervical mucous, to prevent sperm transport, their effective periods are longer. One implant is effective for about five years.
Implanon is a single rod-like device (40 mm x 2mm). It is implanted through a wide bored needle. It contains about 60mg of 3-keto desogestrel and remains functional for three years. Norplant is progestin device contains six small permeable capsules (34 mm x 2.4 mm). Each capsule comprises of 36 mg levonorgestrel.
Norplant is effective for 5 years.

3. Emergency contraceptives (Morning-After Pills) These are the most common form of kits which consist of high dose of birth control pills.
These, if taken within 72 hours of coitus have been very effective as emergency contraceptives as they could avoid possible pregnancy due to rape or casual unprotected intercourse. Their side effects are menstrual irregulation, vomiting, etc., e.g. I-pill, pill-72, unwanted-72, mifepristone (antiprogesterone pill), norigynon, ovidon and ovral. Oral pills are taken two tablets at begining and two tablets in later 12 hours give result to prevent implantation.

Question 2.
What are advantages of natural methods of contraceptive over artificial methods?
Answer:
Natural methods of contraceptives are safe and they do not pose any threat in future. But, these methods are not reliable. In contrast, artificial methods leads to many complications like abdominal pain, hair less or unwanted facial hair, infection in reproductive tract (due to IUDs), hormonal imbalance, etc. Natural methods are easy and safe to use, but do not demand any cost and also does not interfere with the sexual drive of humans.

Natural/Traditional Methods:
These methods are based on the principle of avoiding chances of sperms and ovum meeting.
Some of them are as follows
1. Periodic abstinence (or Rhythm method) It is a method in which couples avoid coitus (intercourse) from 10 to 17th days of the menstrual cycle, because ovulation can occur mosdy during this time (it is called the fertile period). Thus, by abstaining from coitus during this period, conception could be prevented. This method is also known as rhythm period or safeperiod.

2. Coitus interruptus or Withdrawal In this method, the male partner withdraws his penis from the vagina just before ejaculation so as to avoid insemination.

3. Lactational amenorrhea (Absence of menstruation) From ancient times, women have extended breast feeding so, as to avoid a new pregnancy.
It refers to the absence of menstruation during the period of intense lactation following parturition. Because ovulation does not occur in this period, the chances of conception are almost nil. This method is reliable for a maximum period of six months following delivery. Side effects are almost nil, but chances of failure of this method are high.

Question 3.
Why are the assisted reproductive techniques practised to help infertile couples? Describe any three techniques.
Answer:
Due to certain issues both men and women suffer from infertility. Some women are also unable to conceive normally. There are several methods that are practiced under Assisted Reproductive Technologies to help such women to conceive.

1. Intrauterine Insemination (IUI):
A fine catheter is inserted through the cervix, into the uterus to place a sperm sample directly into the uterus. This procedure may be done when ovulation occurs.
The woman may be administered a low dose of Luteinizing Hormone (LH) before the practice to initiate ovulation. This method is also applicable in men having low sperm count and severe erectite dysfunction.

2. Donation of Sperms or Eggs:
If partners are unable to produce eggs or sperms then they can receive eggs or sperms from a donor. The egg is fertilised in vitro and transplanted into the uterus of pseudopregnant woman.

3. Gamete Intra Fallopian Transfer (GIFT):
The ovum collected from donor is transferred to the Fallopian tube of another female, who cannot produce ova, but can provide a suitable environment for fertilisation and further development.
This is termed as Gamete Intra Fallopian Transfer (GIFT). This technique was first attempted by Steptoe and Edwards and later led by Lichardo Ascho.

4. Assisted Hatching
It improves the possibilites of the implantation of embryo on the wall of uterus. In this method, an opening or small hole is created on the outer membrane of embryo, i.e. zona pellucida by an expert practitioner. This opening enables the embryo to leave its shell and get implanted into the uterine lining. This method is also useful in unsuccessful IVF [In Vitro Fertilisation) procedures or when the female is unable to produce good quality of eggs, i. e. at the age of above 35 years.

5. Electrical or Vibratory Simulation to Induce Ejaculation:
Men who are unable to ejaculate naturally are provided with the electrical or vibratory simulations.

6. Surgical Sperm Aspiration:
In this method, the semen is removed from the male reproductive tract (vas deferens or epididymis) of a person’ who himself cannot ejaculate during coitus. This procecjure is done through minor surgery.

7. In Vitro Fertilisation (IVF):
It is a technique in which fertilisation occurs outside the female body. It is followed by the embryo transfer in which embryo is placed inside the uterus. This method is also called as test tube baby technique because sperms are placed with unfertilised eggs in petridish for fertilisation. Donated sperms or eggs can be used in this purpose. IVF techinque can also be employed in gestational surrogacy.
In this case, fertilised egg is implanted in the uterus of surrogate mother.

Surrogacy is banned in few countries because it is unethical to build business on women’s reproductive capabilites. The practice of surrogacy is expensive too. Thus, these two obstacels has opened a new area of tourism, i.e. fertility tourism in those countries where surrogacy is comparatively cheaper.

Question 4.
Give an account of Medical Termination of Pregnancy (MTP).
Answer:
Medical Termination of Pregnancy (MTP):
Intentional or voluntary termination (abortion) of pregnancy before full term or before the foetus becomes viable is called Medical Termination of Pregnancy (MTP). Nearly 45-50 million MTPs are performed in a year all over the world which accounts to 1/5th of the total number of conceived pregnancies in a year.

Government of India legalised MTP in 1971 with some strict conditions to avoid its misuse, especially to check indiscriminate and illegal female foeticides, which are reported to be high in India.

Misoprostol (a prostaglandin) combined with mifepristone (antiprogesterone) is an effective combination to terminate pregnancy. Vacuum aspiration and surgical procedure are also used for medical termination of pregnancy.

Question 5.
Discuss the spacing method and intrauterine devices of family planning.
Answer:
Intrauterine Devices (IUDs) or Intrauterine Contraceptive Devices (IUCDs)
They are made up of plastic, metal or combination of both. They are called loops, spirals, rings, bows or shields. These devices contain either copper or progesterone and are inserted by doctors in the uterus through vagina. These are a form of long reversible contraceptive method.

They may be categorised as
1. Copper releasing IUDs They are commonly called copper-Ts having ionised copper. It slowly diffuses at the rate of around 50 mg/day. It has a local antifertility effect by bringing about the release of toxic cytokines.
The device is to replaced every 3-5 years when copper release slows down due to calcium deposition, e.g. Cu-T, Cu-7, Multiload 375 and paragard.
2. Hormone releasing IUDs, e.g. Progestasert, LNG-20, Mirena are some of the well-known hormone releasing IUDs.
3. Non-medicated or Inert IUDs These are made up – of polyethylene, barium sulphate impregnated or stainless steel.

IUDs prevent contraception in the following ways

  • By increasing phagocytosis of sperms within the uterus.
  • By suppressing the sperm motility and fertilising ability of sperms by releasing Cu ions.
  • The hormone releasing IUDs make the uterus unsuitable for implantation.
  • IUDs are also suppress endometrial changes and cervical mucus, cause an ovulation and insufficient luteal activity.
  • IUDs are ideal contraceptives for females who want to maintain space among children and/or delay pregnancy. It is one of most widely accepted methods of contraception in India.

Chemical Methods:
Spermicides available in the form of creams (e.g. delfem) jellies (e.g. perception, volpar) and foam tablets (e.g. aerosol foam, chlorimin-T or contab) are usually used along with the above stated barrier methods to increase their contraceptive efficiency.

Placed in vagina to cover lining of vagina and cervix. Effective for about 1 hour. These are applied at the surface of vagina before intercourse. These contain spermicides (that kill spermatozoa) such as lactic acid, citric acid, boric acid, zinc sulphate and potassium permanganate. These can be used by anyone who is not allergic to these spermicides.

Now-a-days Sponge is a foam suppository or tablet containing nonoxyneet-9 as a spermicide. Fitted over cervix upto 24 hours before intercourse. It should be left in place atleast 6 hours after intercourse.

Hormonal Methods:
These are hormones possessing contraceptive properties, usually employed on women for suppressing ovulation. Hormonal methods are of three types-oral contraceptives (oral pills), non-oral hormonal contraceptives and emergency contraceptives.
(i) Oral contraceptives These are the preparations of hormones either progestin or progestin-oestrogen combinations in the form of pills (tablets), used by the females.

These oral contraceptives act in four ways

  • Inhibition of ovulation.
  • Alternation in the uterine endometrium to make it unsuitable for implantation.
  • Modification in cervical mucus secretion to impair the ability of transportation of sperm.
  • Inhibition of motility and secretory activity of Fallopian tubes.

Oral pills are of following types
(a) Pill (combined) It contains the female sex hormones oestrogen and progesterone. Oestrogen prevents development of eggs and ovulation by inhibiting secretion of FSH. Progesterone inhibits LH production; acts on cervical mucous to prevent penetration of sperm. They can also prevent the blastocyst implantation, e.g. Mala-D Mala-L. One pill taken orally each day during first 3 weeks of cycle. After 4 weeks, menstruation starts and the pill is taker) again.

(b) Saheli (non-steroidal pill) Selectively, it is a non-hormonal pill that contains centchroman. It acts by inhibiting oestrogen activity. It was developed by Central Drug Research Institute (CDRI), Lucknow. It is taken twice a week for 3 months.

(c) Mini pill or Progestin Only Pill (POP) It contains progestin, a synthetic progesterone only. Ovulation may occur, but cervical mucous is thickened, preventing entry of sperms. Must be taken within 3 hours of same time every day, but newer generation have the relaxations of 12 hours.

(ii) Non-oral Contraceptives
These are categorised as of following two types
(a) Injectables hormone injections (Depo-Provera) are progestin derivative preparations.
They are convenient and highly effective with no side effects. These injectables inhibit the release of hormone to prevent ovulation.
For example, Depot-Medroxy Progesterone Acetate (DMPA) with dose of 150 mg every 3 months or 300 mg for 6 months and Norethisternone Enanthate (NET EN) with dose 200 mg every 2 months.
Cyclofem and mesigna are combined injectable-contraceptives that contain progestin and oestrodiol.

(b) Implant (Norplant) is a new method of contraception, which is inserted under the skin inside upper arm or forearm through a small incision.
Though, these act similarly to oral contraceptives by blocking ovulation and thickening the cervical mucous, to prevent sperm transport, their effective periods are longer. One implant is effective for about five years.

Implanon is a single rod-like device (40 mm x 2mm). It is implanted through a wide bored needle. It contains about 60mg of 3-keto desogestrel and remains functional for three years. Norplant is progestin device contains six small permeable capsules (34 mm x 2.4 mm). Each capsule comprises of 36 mg levonorgestrel.
Norplant is effective for 5 years.

(iii) Emergency contraceptives (Morning-After Pills) These are the most common form of kits which consist of high dose of birth control pills.

These, if taken within 72 hours of coitus have been very effective as emergency contraceptives as they could avoid possible pregnancy due to rape or casual unprotected intercourse. Their side effects are menstrual irregulation, vomiting, etc., e.g. I-pill, pill-72, unwanted-72, mifepristone (antiprogesterone pill), norigynon, ovidon and ovral. Oral pills are taken two tablets at begining and two tablets in later 12 hours give result to prevent implantation.

Natural/Traditional Methods:
These methods are based on the principle of avoiding chances of sperms and ovum meeting.
Some of them are as follows
(i) Periodic abstinence (or Rhythm method) It is a method in which couples avoid coitus (intercourse) from 10 to 17th days of the menstrual cycle, because ovulation can occur mosdy during this time (it is called the fertile period). Thus, by abstaining from coitus during this period, conception could be prevented. This method is also known as rhythm period or safeperiod.

(ii) Coitus interruptus or Withdrawal In this method, the male partner withdraws his penis from the vagina just before ejaculation so as to avoid insemination.

(iii) Lactational amenorrhea (Absence of menstruation) From ancient times, women have extended breast feeding so, as to avoid a new pregnancy.

It refers to the absence of menstruation during the period of intense lactation following parturition. Because ovulation does not occur in this period, the chances of conception are almost nil. This method is reliable for a maximum period of six months following delivery. Side effects are almost nil, but chances of failure of this method are high.

Medical Termination of Pregnancy (MTP):
Intentional or voluntary termination (abortion) of pregnancy before full term or before the foetus becomes viable is called Medical Termination of Pregnancy (MTP). Nearly 45-50 million MTPs are performed in a year all over the world which accounts to 1/5th of the total number of conceived pregnancies in a year.

Government of India legalised MTP in 1971 with some strict conditions to avoid its misuse, especially to check indiscriminate and illegal female foeticides, which are reported to be high in India.

Misoprostol (a prostaglandin) combined with mifepristone (antiprogesterone) is an effective combination to terminate pregnancy. Vacuum aspiration and surgical procedure are also used for medical termination of pregnancy.

Question 6.
Give an account of risk factors of infertility.
Answer:
Some Risk Factors of Infertility
These are as follows

  1. Age A woman’s fertility starts to drop when she is above 32 years old or more. A 50 year old man is less fertile than a man in his 20 year (male fertility begins to drop after the age of 40).
  2. Smoking It significantly increases the risk of infertility in both men and women. It may also undermine the effects of fertility treatment. When a pregnant woman smokes she develops a higher risk of miscarriage.
  3. Alcohol consumption A woman’s pregnancy can be seriously affected by alcohol consumption. Alcohol abuse may lower’ male fertility. Moderate alcohol consumption ‘has not been shown to lower fertility in most men, but is thought to lower fertility in men who already have a low sperm count.
  4. Obese or Overweight In developed countries, overweight (obesity) and a sedentary lifestyle are the principal causes of female and male infertility. An overweight man may suffer from lower sperms count or form abnormal sperms.
  5. Eating disorders Women who are in the category of serious underweight may have fertility problems.
  6. Being vegetarian If a person is on a strict vegetarian diet, he/she must make sure that intake of iron, folic acid, zinc and vitamin-B12 should be adequate in a diet, fertility may be affected.
  7. Overexercise A woman who exercises for more than seven hours every week may have ovulation problems.
  8. Sedentary life-style Leading a sedentary life-style may increase the chance of lower fertility in both men and women.
  9. Sexually Transmitted Infections (STls) Chlamydia can damage the Fallopian tubes and influences man’s scrotum. Some other STls may also cause infertility.
  10. Exposure to some chemicals Some pesticides, herbicides, heavy metals (lead) and solvents may cause fertility problems in both men and women.
  11. Mental stress Studies indicate that ovulation and sperms production may be affected by mental- stress. If a partner is stressed, than the frequency of sexual intercourse decreases, resulting in a lower chance of conception.

Question 7.
Describe the various causes of infertility in women.
Answer:
Causes of Infertility in Women
(i) Ovulation disorders Problems with ovulation are the • common causes of infertility in women. Ovulation is the monthly release of an egg. Some women never release eggs, while some do not release eggs during regular cycles.

Ovulation disorders can be due to

  1. Premature ovarian failure The ovaries stop functioning properly before 40 years of age.
  2. PCOS (Polycystic Ovary Syndrome) The ovaries function abnormally. Such women have abnormally high levels of androgens. About 5% to 10% of women in the reproductive age range are affected by this syndrome.
  3. Hyperprolactinemia If the prolactin is high to those women who are not pregnant or not in breast feeding period, then they may effect of an ovulation and infertility.
  4. Poor egg quality Eggs that are damaged or develop genetic abnormality cannot sustain a pregnancy. Older women are at a higher risk.

(ii) Problems in the uterus or Fallopian tubes The egg travels from the ovary to the uterus (womb) through the Fallopian tube, where the fertilisation of the egg takes place. If there is something wrong in the uterus or the Fallopian tubes the woman may not be able to conceive naturally.
This may be due to

  1. Surgery Pelvic surgery may sometimes cause damage to the Fallopian tubes. Cervical surgery may sometimes cause shortening of the cervix. The cervix is the neck of the uterus.
  2. Submucosal fibroids Benign or non-cancerous tumours found in the muscular wall of the uterus, occurr in 30-40% of women in the child bearing age. They may interfere with implantation. They may also block the Fallopian tube preventing sperm from fertilising the egg. Large submucosal uterine fibroids make the uterus cavity bigger, increasing the distance the sperm has to travel.
  3. Endometriosis Cells that are normally found within the lining of the uterus start growing elsewhere in the body.
  4. Previous sterilisation treatment If a woman has her Fallopian tubes blocked, it is possible to reverse the process. But, the chances of becoming fertile again are low.

(iii) Medications Some drugs can affect the fertility of a woman. These include

  1. NSAIDs (Non-Steroidal Anti-Inflammatory Drugs) Women who take aspirin or ibuprofen in a long term may find it harder to conceive.
  2. Chemotherapy Some medications used in chemotherapy can result in ovarian failure. In some cases, this side effect of chemotherapy may be permanent.
  3. Radiotherapy If radiation therapy is aimed near the woman’s reproductive organs, there is a higher risk of fertility problems.
  4. Illegal drugs Some women who take marijuana or cocaine may have fertility problems.

Question 8.
What is Sexually Transmitted Diseases (STDs)? Give an account of the prevention and treatment of STDs.
Answer:
Sexually Transmitted Diseases (STDs) are the diseases that are mainly passed from one person to another during sexual intercourse. They are also known as Venereal Diseases (VDs) or Reproductive Tract Infections (RTls). These can be prevented by avoiding unprotected sexual intercourse.

Prevention and Treatment It is much easier to prevent STDs than treating. The only way to completely prevent STDs is to abstain from all types of sexual contact. If some one is going to have intercourse, the best way to reduce the chance of getting an STD is by using a condom. People who are in the habit of having frequent sexual contract should get regular gynaecological or male genital examinations.

There are two benefits from this. First, these examinations give the doctor an opportunity to teach people about the consequences of contracting STDs and the methods of protecting themselves. Secondly, regular examinations give a better opportunity for an effective treatment, while the infections are in their earliest.

The visiting person needs to disclose all facts about his / her sexual contact to the physician without fear, so that the physician decides on a course of action. Consequently, the physician will prescribe an investigation and then an effective treatment schedule.
If the person feels embarassed to visit a physician, with whom he/she is familiar, he/she may seek the assistance of experts by calling STD hotline operated by some national organisations.

The experts will advice in respect of the STD clinics undertaking the investigation and treatment. In doing so a confidentiality about the identity of the person in question is maintained.

Not all infections in the genitalia are caused by STDs. Sometimes, people may express symptoms that are . similar to those of STDs, although they have never had sex. For example, in girls, a yeast infection is sometimes confused with an STD. Males may worry about bumps on the penis that turn out to be pimples or irritated hair follicles. It is therefore, important to visit a physician to solve many of the above mentioned problems.

Question 9.
How infertility in men and women can be diagnosed by tests?
Answer:
Tests for Females:

  1. General physical examination is cross examination of female patient by the gynaecologist about her medical history, menstrual cycle and sexual habits. The gynaecologist may also examine some other gynaecological habits.
  2. Blood test is analysis of the blood sample to check the female hormone levels and also ovulation period in women.
  3. Hysterosalpingography is radio-opaque fluid which when injected into the woman’s uterus appears on the X-ray film. X-rays are taken to know about blockages in the uterus and Fallopian tubes.This is followed by a surgery to rectify the problem.
  4. Ovarian reserve testing is done to find out of effectiveness of the eggs after ovulation.
  5. Pelvic ultrasound is the sonography of the pelvic region of the female to know about the normal structures of the ovary, uterus and Fallopian tube.
  6. Thyroid function test checks the fertility of women due to thyroid malfunctions. According to National Health Service (UK), between 1.3% and 5.1% of women are suffering thyroid disease. This test is undertaken and used to correct it by medication.

Tests for Males

  1. General physical examination is done by the andrologist (physician specialised in male reproduction) when he knows about the patient medical history, medications and sexual habits. The physician is also examine of his genitals. The testicles are also checked for lumps or deformities, while the shape and structure of penis are examined for any abnormality.
  2. Semen analysis is the analysis of semen, sperm count, motility, colour, quality and infections in a laboratory.
  3. Blood analysis is analysis of testosterone and other male hormone concentrations in blood.
  4. Ultrasound test is used to detect any ejaculatory duct obstruction, retrograde ejaculation or other abnormal functioning.
  5. Chlamydia test is used to dignose chlamydia in man.

Question 10.
Describe the treatment options for infertility in men and women.
Answer:
Treatment for Infertility:
The treatment of infertility depends on many factors
including age, medical issues, personal choice, etc., of patient.

Fertility Treatment for Women:
1. Ovulation disorders It can be treated by drugs including Clomifene, Metformin, Bromocriptine, HMG, FSH, Human chorionic gonadotropins, GnRH, etc.

2. Surgical Procedure for Women
(a) Fallopian tube surgery If the Fallopian tubes are blocked or scarred, surgery may repair them and make them easier for eggs to pass through them.
(b) Laparoscopic surgery A small incision is made in the woman’s abdomen. Laparoscope is inserted through the incision. In endometriosis, laparoscopy removes implants and scar tissue restoring fertility.

Fertility Treatment for Men:

  • Erectile dysfunction or Premature ejaculation Medication and/or some behaviour approaches can help men to improve general sexual problems and fertility.
  • Blockage of the ejaculatory duct In case of blockage sperms are extracted direcdy from the testicles and injected into an egg in vitro in the laboratory condition.
  • Retrograde ejaculation Sperms are directly taken from the bladder for injecting into an egg in vitro.
  • Surgery for epididymal blockage If the epididymis is blocked, it can be surgically repaired and sperms can be ejaculated properly.

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Odisha State Board CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance Questions and Answers.

CHSE Odisha 12th Class Economics Chapter 13 Question Answer Public Finance

Group – A

Short type Questions with Answers
Answer with in Two/Three sentence.

Question 1.
What is Public finance?
Answer:
Public finance studies the income, expenditure of the government & its rational adjustment with each other for the purpose of economic progress. It also includes the revenue & capital expenditures of the government & sources of revenue to be collected.

Question 2.
Balanced budget is the best budget
Answer:
In case of balanced budget, the government receipt becomes equal to government expenditure. Here, there is no surplus nor any deficit in respect of govt, expenditure or income & so it is treated as the best budget.

Question 3.
Deficit budget is beneficial to the UDCs
Answer:
In deficit budget the govt, expenditure is more than the govt, receipts. There is ample scope for developmental activities. Hence it is beneficial to the UDCs.

Question 4.
What is Surplus budget?
Answer:
Surplus budget is one type of unbalanced budget in which revenue receipts for the budget period are greater than the expenditure to be incurred. Hence, the total revenue earned through various sources of the Government has not been exhausted and some amount of income remains unspent or unutilised.

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 5.
Write down two demerits of Deficit budget.
Answer:
Deficit budget is prone to inflation. If it is not properly managed, there will be wide gap between demand and supply, Inflation will come up. Deficit budget creates inflation. So, price of the goods to be exported increases. The volume of export comes down. It will reduce the amount of foreign currency in the country.

II. Answer within Five/Six sentence :

(A) write shot notes on :
1. Public Finance:
Answer:
Public finance deals with the income and expenditure of public authorities and with the adjustment of one to another.” Obviously, public finance also deals with the problems of adjustments of income and expenditure of the Government. The methods of expenditure of public bodies and income of public bodies as well as borrowing by public bodies are known as operations of public finance. According to Bastable “Public finance deals with expenditure and income of public authorities of the State and their mutual relation as also with the financial administration and control”. Thus public finance was concerned with the explanation of revenue and expenditure process of the public authorities.

2. Budget:
Answer:
Budget is a financial statement of income and expenditure of public authorities. It is a reflection of not only taxation and public expenditure policy, but also of a plan for future course of action. It is designed to Secure the normative ideas of allocation, distribution, stabilisations and growth. Budget, thus refers to the financial arrangement covering the income and expenditure of the Government especially for a particular year. It is an aimual plan of income and expenditure of the public authorities which reflects the economic activities to be undertaken by the Government for a particular year.

3. Balanced budget:
Answer:
Budget is viewed as a balanced budget if the Government revenues are equal to Government expenditures. In such budget, there exists no gap between the total income and total expenditure of the Government during a particular time-period. This sort of budget is appreciated for underdeveloped countries which suffer from chronic destability due to considerable amount of deficit. Besides, this budget avoids extravagant expenditure and thus ensures effective allocation and mobilisation of the available resources.

4. Deficit Budget:
Answer:
Deficit budget is that budget in which the government expenditure exceeds government revenue. In this case government receipts for the budget period is inadequate to meet the government expenditure. The gap between the government receipt & government expenditure is met through public debt & issue of new currency notes, Deficit budget is expansionary in nature. Deficit budget controls unemployment, deflation etc.

5. Surplus budget:
Answer:
The surplus budget is that in which anticipated revenue of the government is more than the anticipated expenditure. In this budget, revenue receipts during budget period are greater than the cost payments. The outcome of surplus budget is very discouraging in modem world. It has got contractionary effect. Besides, it checks unproductive expenditure & reduces debt burden. This budget is not favourable for developing countries.

(B) Distinguish Between
6. Public finance & Private finance:

  1. Public finance deals with the income & expenditure of the public authorities i.e. local, state & central government. But the private finance deals with the income & expenditure of the private individual.
  2. Public finance aims at collective welfare of the society whereas private finance aims at individual welfare.
  3. In public finance, government adjusts its income to its expenditure whereas in private finance, the individual adjusts its expenditures to its income.
  4. In public finance, the government can not avoid or postpone its expenditure whereas in private finance, it can be avoided or postponed.

Group – B

Long Type Questions With Answers

Question 1.
What is public finance ? Compare & contrast between Private finance & Public finance.
Answer:
The two terms ‘public’ & ‘finance’ denotes income & income expenditure of the government. Here public implies government which collectively considers the people of the State. Similarly, finance refers to income & expenditure, public finance denotes income & expenditure of the government. According to H. Dalton, public finance is a “subject which is concerned with the income & expenditures of public authorities & with the adjustment of the one with the other”. In simple way, public finance deals with the income of the public authorities (government) i.e. local, state or central. It was a practice among the writers on public finance to compare public with private finance. Such a comparison will enable us to have a clearer picture of the public finance.

(1) It is said that an individual will adjust his expenditure to his income. On the other hand, it is pointed out that the state will adjust its income to expenditure. In the case of the individual, income determines the expenditure. In the case of the state, expenditure determines its income. Ofcourse, this statement has its exception. An individual too tries to increase his income when it is not enough to meet his essential expenditure. On the other hand, the state too curtails its expenditure when it is not possible to raise its income. After all, the sources of income to the state also are not completely elastic.

(2) Both the individual and state attempt to equalise income and expenditure. Both of them try to earn more or at least borrow to meet the additional expenditure but there is one difference. An individual can borrow from another individual or institution, i.e., externally. A state can borrow externally from foreign citizens, institutions or government as well as internally from its own citizens. Further, the government may also inflate, i.e., print more notes to meet its expenditure, This course is entirely out of reach for the individual.

(3) Generally, the state adopts a year as its accounting period and attempts to equalize revenue and expenditure during that period. An individual does not have any such-rigid scheme of balancing income and expenditure in every year. However even with regard to a state. There is no particular sanctity attached to a year. The budget period may will be changed to three or four years.

(4) An individual is supposed to distribute the expenditure among various items so as. to equalize their marginal utilities. In case of the state, this is not considered to be possible since it isnot a person. But even in the case of the state, the statesmen should so distribute the expenditure that marginal utilities to the community on all forms of expenditure must be the same. Ofcourse, neither the individual nor the public authorities will be able to do these calculations very accurately. There can at best, be only an approximately correct distribution following this principle.

(5) The government will be in a much better position to make deliberate changes in its income and expenditure than an individual. An individual has limited resources. The government can similarly change the pattern of expenditure with greater than the individual.

(6) The government spends on certaqin ojects like security, peace and order, national defence etc. These are entirely beyond the scope of private finance.

(7) the individuals can afford to discont future at a higher rate. But the state cann’t do so since it will be in perpetual existence. It has to provide for the future. It has to compensate for the deficiency in the individual’s provision for future.

(8) In private business, ‘special service and special payment’ is the principle. But in the case of the state here is usually no relationship between the service rendered by the state and the payment it demands from individuals.

(9) The individual will have an eye on quick returns and high profits while making investments, the state doesn’t follow this principle. It sometimes undertakes even follow this principle. It sometimes undertakes even singly, unproductive projects which are beneficial to the society in the long run.

(10) The effects of expenditure on the part of the state are different from the effects of expenditures by individuals. Ofcourse the effect of expenditure in creating effective demand for goods is the same in regard to both. However, the variation of the public expenditure has become an important means of anti cyclical policy because of its greater administrative convenience.

(11) In Private business, cost of production can be compared to the price of the product. But in case of the state, it is rather difficult to compare the cost of the service with the value of the service. Both cost and value can n’t be easily estimated in all cases. Defence service provides an example for this.

(12) Generally the financial operations of the individual are keep secret. They will not be open for public qaze. But the financial operations of the state are always public. They have to be approved before hand by the appropriate authority like the parliament. The people have a right to scrutinize the expenditure since they meet the bills through paying taxes.

(13) An individual can ebcome bankrupt when his assets are short of his liabilities. But there is ordinarily, no question of a state going bankrupt. The>liabilities of the state can be as certained, but its assets can’t be so precisely ascertained because they are just the pme as the assets of all the citizen together.

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 2.
What is budget ? Discuss various types of budget.
Answer:
Budget is a financial statement of income and expenditure of public authorities. It is a reflection of not only taxation and public expenditure policy, but also of a plan for future course of action. It is designed to secure the normative ideas of allocation, distribution, stabilisations and growth. Budget, thus refers to the financial arrangement covering the income and expenditure of the Government especially for a particular year. It is an annual plan of income and expenditure of the public authorities which reflects the economic activities to be undertaken by the Government for a particular year.

Purpose : The purpose of a Government budget is varied. There are a number of objectives which the budget seeks to attain simultaneously. The overall purpose is to use the budget as an instrument of economic policy. A budget is such a plan which explicitly mentions the programmes that are to be taken up in the course of the fiscal year. Secondly, different programmes may need different durations for completion. What part of the action can be completed within a fiscal needs to be specified so that implementing agency clearly can know as to which portion of the programme should be completed within a specific year.

Thus budget spells out such year wise responsibility. Thirdly, the implementation of a programme requires availability of necessary funds. The extent of availability depends upon the budgetary sources of revenue. Hence, the programme structure has to be built which can be supported by funds. The budget draws up schemes of revenue mobilisation on the one hand and programmes of the public expenditure on the other. Fourthly, to achieve efficiency in revenue collection, the expenditure on the accounts of the collectorate is specified on the basis of the past trend, present level of cost of collection in the budget. F ifthly, to achieve efficiency in public expenditure physical targets of attainment are specified in the budget.

Sixthly, formulation of future programmes on the basis of past experience is an important purpose of the Government budget. Seventhly, an important purpose of the budget is to allow the legislature and the people to appreciate the overall programme framed in it in the back drop of prevailing economic situations. This is why a good budget is accompanied by an analytical account prevailing economic situation and financial position. Eighthly, investment, consumption and capital formation to assess the trend of growth in the economy. Ninthly, the main purpose of the budget in developed countries is to act as an analytical fiscal weapon. This is done through the manipulation of the budget balance. Lastly, the budget serves the purpose of public accountability of funds to a considerable extent.

Types of budget: As a suitable fiscal device, the budget may be classified into two categories like balanced budget and unbalanced budget. This classification of the budget is made on the basis of the difference between the total revenue (income) and total expenditure of the Government.

Balanced budget: Budget is viewed as a balanced budget if the Government revenues are equal to Government expenditures. In such budget, there exists no gap between the total income and total expenditure of the Government during a particular time-period. This sort of budget is appreciated for underdeveloped countries which suffer from chronic destability due to considerable amount of deficit. Besides, this budget avoids extravagant expenditure and thus ensures effective allocation and mobilisation of the available resources. In spite of these merits, this type of budget is not conductive to economic development which requires huge funds.

Surplus budget: Surplus budget is one type of unbalanced budget in which revenue receipts for the budget period are greater than the expenditure to be incurred. Hence, the total revenue earned through various sources of the Government has not been exhausted and some amount of income remains unspent or unutilised. This budget, though checks extravagant expenditure, is not suitable for economic development It is because the amount earned is not totally utilised. If this budget is adopted economic prosperity cannot be expected. But this budget is a cure during the period of inflation.

Deficit budget: Deficit budget is one in which revenue receipts for the budget period arc less than the expenditures required for.. In this budget, the income Of the Government becomes inadequate to meet expected expenditure. Scope for economic development. Secondly, this budget is a cure during the period of depression. But this budget is found to be inflationary if the deficit becomes, excessive. However this budget has got its overall importance as it results in economic prosperity.

In the advanced countries, a balanced budget is pursued at a time when the economy suffers neither from inflation nor from unemployment or depression so that objective of maintaining full employment with price-stability is achieved. When the economy suffers from inflation, a surplus budget is operated while a deficit budget is purchased when the economy suffers from unemployment. The developing and underdeveloped countries normally suffer from idle resources and to make their proper use, additional expenditures are incurred and hence, they mostly pursue deficit budgets.

Question 3.
Budget in an instrument of economic policy. Justify.
Answer:
Government budget is an important instrument of economic policy in both developed and developing countries. In the developed countries, the economy operates at full employment level and hence, there does not exist unemployed resources. But the economy is subject to trade cycle and therefore, occasionally faces the problems of depression of unemployment and inflation or pressure of excess purchasing power. In the underdeveloped countries, the economy operates at less than full empolyment level and hence, the main problem is how to attain economic growth. In these poor countries, growth process is faced with a number of problems. They are allocational, distributional and stabilisational. Budget serves as an important device to achieve economic development is these countries also. The following are the important ways in which the Government budget can influence the economy of a country.

(i) Revenue raising device : The Government requires enough revenue to discharge its fiscal responsibility. Modem countries have increasingly become welfare states with larger and larger State activities coming under the fold of public sector. Hence, resources have to be found in sufficient quantity. Budget secures this purpose through a financial plan. The receipts side of the budget clearly mentions the sources and the extent of funds for the purpose of financing state activities.

(ii) Incentive to economic activity: Budgetary receipts as well as expenditures ca eatly influence economic activities both in the industrial and agricultural sectors. Through tax concessions and discriminatory taxes, the budget can influence production and productivity in favour of these secto; s which the Government likes to promote. Important industries in the priority list of Government may be granted tax holiday or tax concessions in order to attract promising extrepreneurs to these ventures. Similarly, agricultural activity and production can be increased through budgetary provisions of free or subsidised supply of agricultural inputs and extension services.

(iii) Human capital formation : The most important need for a country’s economic development is human capital formation. The level of human capital formation like education, medical and public health, etc. is very poor in the underdeveloped countries. Unless people are educated and healthy, they cannot be good workers and their productivity cannot be increased. Budgetary provisions can serve this purpose. Since investment on such human capital formation is heavy and subjected to long gestation period, funds will not come from private sector. It is only the Government which can rise the level of general and technical education and of health and productive capacity by providing educational and health facilities through budgetary outlays.

(iv) Building of economic overheads : The main reason of underdevelopment of the poor countries is absence of proper economic infrastructure. Without proper transport and communication training facilities for workers and entrepreneurs, industrial development is not possible. Similarly agricultural production and productivity cannot improve in the absence of proper irrigation facilities, flood control measures, technological improvements with research and development activities etc. These facilities must be provided by the Government. The cost of supplying these services is heavy and cannot be raised directly from the beneficiaries. Therefore, these facilities are supplied free of direct charges through budgetary provision. Thus budget has tremendous influence on the industrial and agricultural development.

(v) Diversion of resources to more useful production :Free market mechanism leads to production of those goods and which give maximum investment is generally concentrated on the production of luxury commodities. It is therefore, necessary to divert resources to the production of more useful goods and services, particularly of the kind of mass consumption ones. This can be done by Government interference thorough the budget. Imposition of heavy tax on harmful and less essential goods and tax exemption or tax concessions granted to more essential goods and services can divert resources to the production of right kind of goods and services. Grant of facilities through budgetary expenditure can also do the same job.

(vi) Proper allocation of resources : Most efficient allocation of resources is given by the equality between marginal cost and price which is possible only under perfect market conditions. Underdeveloped countries seriously suffer from mal-allocation of resources. The general market conditions in private sectors are set by existence of monopoly, monopolistic competition and oligopoly. To correct this mal-allocation, the Government has to interfere either in the form, of production subsidy or supply of goods and services by public authorities so that the gap between average revenue (price) and the marginal cost is reduced as far as possible. This is the reason why the heavy investment public welfare industries which are subjected to decreasing cost conditions are increasingly coming under the fold of public sector.

(vii) Balanced development: Underdevelopment countries suffer from regional imbalance in economic development. Left to the private sector which is motivated by profit maximisation, the industries will be located in the urban and already developed areas. The Government can correct this geographical imbalance by setting up public sector industries will be located in the urban and already developed areas. The Government can correct this geographical imbalance by setting up public sector industries in backward areas. Moreover, the development of agriculture and small scale and village industries can be secured through Government patronage in the terms of supply of infrastructure facilities and various incentives or subsidy measures. This will develop the econoniy of rural areas.

(viii) Income and employment: Since underdeveloped countries, are low income economies, people live in poverty and hence, saving and investment is very low. Income of the people can be increased only through increased productivity and production. Budgetary provisions can go a ’ ng way to achieve this. When agricultural technology is improved though budgetary programmes, the income of the people engaged in agriculture rises. People get gainful employment in the sector. Improvement in small scale industries inthe backward regions will increase employment opportunities in these industries. The budgetary provisions of employment-related tax concessions can influence creation of employment opportunity in the private sector also.

(ix) Saving and investment: In-underdeveloped countries the level of saving and investment is very low. Moreover without increased saving and investment, economic growth cannot be achieved. Due to low level of income, marginal propensity to consume is very high and hence, the mass people cannot save. Public saving is, therefore necessary. Taxation of various types serves this purpose. The saving and investment of private individuals are also influenced by the savings investment related tax concessions and other budgetary subsidy programmes. Capacity and wiHjngness to work, save and invest of the people is increased through various human capital formation measures and creation of employment opportunities. These are all done through budgetary expenditures.

(x) Poverty removal: Poverty removal programme is a part and parcel or the budget in underdeveloped countries. All expenditure measures are designed in such a way that they directly or indirectly influence reduction of poverty inthe economy. Thus when budgetary resources are spent on account of education, whether general or technical and vocational, or on health measures, land
reforms, flood control and irrigation, an important objective is to remove poverty of people. Direct budgetary programmers for poverty removal are those of increasing employment opportunities and creation of community assets like those under I.R.D.P., N.R.E.P. or R.L.E.GP. schemes as in India, employment insurance, social security, consumption subsidy, public distribution system and price support programmes, low-income housing, area development, input supply, agricultural wage restructuring etc.

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 4.
What is balanced budget ? State its merits & demerits.
Answer:
A budget is said to be balanced whose anticipated revenue and expanditure are equal. In the opinion of Prof. Dalton, “A balanced budget is that over a period of time, revenue does not fall short of expenditure.

Balanced budget may be true in accounting sense. It is nothing but a balance sheet approach. But in practice it is not easy to have a balanced budget as it tends to be either surplus or deficit as per the necessity of time. Budget can be balanced only in average, by taking a number of years together. So, balanced budget for a particular year is often theoretical. However, attempts should be taken to maintain balance as far as possible.

Merits of Balanced budget: Balanced Budget was the brain child of the classical economists for its following merits.
(i) Checks Wasteful Expenditure : In case of balanced budget, the Government designs its public expenditure according to its revenue. To assure welfare to the community, expenditure is made through rational planning. There can not be reckless and unproductive public expenditure. So, financial discipline of the Government can be assured.

(ii) Economic Stability : In case of balanced budget, what Government collects from the people and what it spends are the same. So, as believed by the classical economists, national income does not change. It brings room for neither inflation nor deflation. Economy experiences every type of normalcy.

(iii) Controls over Public Debt: In balanced budget, anticipated income and expenditure are equal. So, as there is no excess expenditure, there is no scope for public debt.

(iv) Favourable for Developed Countries : Balanced budget is a boon for developed countries. The countries which are already in the apex of development only need to continue the existing rate of development with stability. It can very well be possible through balanced budget.

(v) Limited Role of Government: In modem times, in most of the countries, Government does not like to shoulder more responsibilities Most of the economic functions are left to the private hands. So, public budget tends to be smaller. It creates room for balanced budget. All these argue in favour of balanced budget.

Demerits of Balanced Budget:- However, balanced budget has got so many demerits. Those are noted below.
(i) Paper Cultivation : A public budget should aim at provision of the maximum welfare for the community. Otherwise, all the activities, involved in preparation of budget will go down the drain. It will only be mere paper cultivation. It happens in case of balanced budget.

(ii) Not Effective in Modern World : In classical economy, Say’s Law of Market is obeyed. It suggests for a normal situation as whatever supplied is automatically demanded. Equilibrium has to prevail everywhere. In support of this, balanced budget is advocated. But in the modem world, normalcy is a dream. There are a number of problems like unemployment, over production, under production, economic instability etc. To tackle these problems, balanced budget can not be the right solution.

(iii) Not Effective for Developing Countries : Every developing country like India requires higher productive public expenditure for optimum utilisation of available resources. It can not be possible if the country follows balanced budget as it suggests for the least public expenditure and the minimum Government involvement.

(iv) Wasteful Expenditure : In the opinion of Prof. Arthur Smithies, “It is quite possible that to have balanced budget during a period of time the Government may lead to wasteful public expenditure. ” Balanced budget suffers from inadequate finance. Some of the Government projects remain incomplete.

(v) Restricts Freedom of Public Authority : Modem state is welfare oriented. To provide welfare, it has to spend on various purposes which require Himalayan public expenditure. But, due to balanced budget, the Government does not have free hand to spend and to assure welfare. All these demerits of balanced budget suggest to follow unbalanced budget as an appropriate fiscal policy.

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 5.
What is deficit budget ? Discuss its merits & demerits.
Answer:
In case of deficit budget, anticipated expenditure is higher than anticipated revenue. Prof. Taylor has rightly mentioned deficit budget as that whose revenue receipts for the bqdget period are less than the cost payments. Deficit budget is met by the Government through public debt, issue of new notes etc.

Deficit budget has expansionary effect. When more is spent for the people than what is taxed, the national income increases in multiple. It is helpful to control deflation and to bring economic stability. It provides free hand to the Government for all round development. So most of the developing countries are following this budgetary policy.

Merits of Deficit Budget: Deficit Budget has the following merits.
(i) Increases National Income : In case of deficit budget, Government’s expenditure is higher than its income. That means, people receive more than what they pay in term of tax. As a result national income increases in multiple. There is every scope for higher economic growth.

(ii) Controls Unemployment: When deficit budget increases national income in multiple, the effective demand goes up. It encourages higher investment and employment. Employment of the economy goes on inerteasing in multiple.

(iii) Controls deflation : Deficit budget has got the capacity to increase national income in multiple. So, deflation which refers to decrease in price due to low income can be cured.

(iv) Not inflationary : Inflation occurs due to higher demand than what is supplied. It can be effectively controlled if production can be simultaneously increased with increase in demand. In case of deficit budget Government enjoys free hand to spend in various productive purpose. It leads to higher increase in national production. It brings mild inflation which is favourable to the economy.

(v) Maximum Utilisation of Resources : Deficit budget creates scope for higher productive expenditure. The available resources can be properly utilised. So, economy can grow up to its full extent. It can bring a new look to the economy.

(vi) Proper Distribution : Through deficit budget the Government enjoys scope to spend for the poor. Because of more effective employment, they can earn more. They can have better standard of living. As a result, the gap between the rich and the poor can be lessened.

(vii) Befitting for Emergency : A country confronts emergencies like war, natural calamities, epidemics etc. To face these emergencies, the Government requires to spend more. It can only be possible if it follows deficit budget. All these advantages of deficit budget present its importance as a fiscal policy. These confirm why deficit budget is a blessing for developing countries.

Demerits of Deficit Budget:  Deficit budget is not an unmixed blessing. It has got the following demerits.
(i) Inflationary : Deficit budget is prone to inflation. If it is not properly managed, there will be wide gap between demand and supply, Inflation will come up.

(ii) Less Export: Deficit budget creates inflation. So, price of the goods to be exported increases. The volume of export comes down. It will reduce the amount of foreign currency in the country.

(iii) Wasteful Expenditure : Deficti budget provides free hand to the Government to spend. There is every possibility of unproductive public expenditure. To be more popular the party in power will spend for its own benefit at the cost of the benefit of the economy. It leads to misutilisation of resources.

(iv) Debt Burden : The Government meets deficit through public debt. In case of deficit budget, all the debts are not spent on productive purposes. Those become burden to the community. The country enters into debt trap. It compels the country to make fresh debt to repay old debts. Despite its demerits, deficit budget has been always considered as an effective tool of growth and development in developing countries.

Group – C

Objective type Questions with Answers
I. Multiple Choice Questions with Answers :

Question 1.
Public finance deal with
(i) income & expenditure of the government
(ii) development of the state
(iii) individual welfare
(iv) all of the above
Answer:
(i) income & expenditure of the government

Question 2.
Which is not the objective of public finance?
(i) welfare of the state
(ii) maximisation of income
(iii) individual welfare
(iv) rational adjustment of income & expenditure
Answer:
(iii) individual welfare

Question 3.
Which is not in the subject matter of public finance?
(i) public expenditure
(ii) it has got general acceptability
(iii) public debt
(iv) all of the above
Answer:
(iv) individual income

Question 4.
Which budget is preferable for UDCs?
(i) currefl depOSit
(ii) fixed deposit
(iii) saving deposit
(iv) all of the above
Answer:
(ii) fixed deposit

Question 5.
What type of dep sit is appreciated by the business man?
(i) fixed deposit
(ii) saving deposit
(iii) deposit made with bonds & securities
(iv) None of these
Answer:
(iii) deposit made with bonds & securities

Question 6.
Public finance deals with the income and expenditure of the:
(i) Central Govt.
(ii) current deposit
(iii) Local Govt.
(iv) fixed deposit
Answer:
(iv) All of the above

Question 7.
Who adjusts expenditure to income :
(i) Govt.
(ii) current deposit
(iii) Both (i) and (ii)
(iv) recurring deposit
Answer:
(ii) Individual

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 8.
Surplus budget is a cure for:
(i) Inflation
(ii) draft
(iii) Both (i) and (ii)
(iv) both (i) and (ii)
Answer:
(i) Inflation

Question 9.
Deficit budget is a cure for :
(i) Inflation
(ii) discounting bills of exchange
(iii) Both (i) and (ii)
(iv) direct loan
Answer:
(ii) Deflation.

Question 10.
Which is applicable to balance budget?
(i) It controls unnecessay expenditure
(ii) Does not creates economy crises
(iii) Does not creates inflation
(iv) All of the above
Answer:
(iv) All of the above

Question 11.
Which is the feature of deficit budget?
(i) Increases level of income and employment
(ii) Removes the unemployment problem
(iii) Eliminates deflationary pressure
(iv) All of the above
Answer:
(iv) All of the above

Question 12.
Which is the liability of the commercial banks?
(i) all types of deposits
(ii) authorised.capital
(iii) borrowing from other banks
(iv) all of the above
Answer:
(iv) all of the above

Question 13.
Which is not the asset of the commercial banks
(i) loans & advances
(ii) cash with RBI
(iii) Reserve funds
(iv) investments
Answer:
(iii) Reserve funds

Question 14.
Which is the most liquid asset of the commercial banks?
(i) cash in hand
(ii) saving deposit
(iii) loans and advances
(iv) investments
Answer:
(i) cash in hand

Question 15.
Which is not a function of central bank?
(i) lender of the last resort
(ii) advisor to the govt.
(iii) advances loan to people
(iv) custodian of foreign exchange
Answer:
(iii) advances loan to people

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 16.
As monopoly of note issue, RBI prints all types of notes except
(i) two rupee notes
(ii) one rupee notes
(iii) five rupee notes
(iv) fifty rupee notes
Answer:
(ii) one rupee notes

Question 17.
Which notes are issued by Ministry of finance?
(i) one rupee notes
(ii) two rupee notes
(iii) five rupee notes
(iv) all of the above
Answer:
(i) one rupee notes

Question 18.
Which bank controls credit?
(i) RBI
(ii) SBI
(iii) Regional rural banks
(iv) all of the above
Answer:
(i) RBI

Question 19.
Which is a method of credit control?
(i) bank rate
(ii) open market operation
(iii) variable cash reserve ratio
(iv) all of the above
Answer:
(iv) all of the above

Question 20.
When Reserve Bank of India increases bank rate, the demand for loan
(i) increases
(ii) decreases
(iii) not affected
(iv) none of the above
Answer:
(ii) decreases

Question 21.
Barteris:
(i) Indirect exchange of goods against goods is called barter
(ii) Direct exchange of goods against goods is called bartar.
(iii) Both (i) and (ii)
(iv) None of the above
Answer:
(ii) Direct exchange of goods against goods is called bartar.

Question 22.
Deiine moeny:
(i) Money is what money does
(ii) Direct exchange of goods against goods in money
(iii) Anything that possesses general acceptability is money
(iv) None of the above
Answer:
(iii) Anything that possesses general acceptability is money

Question 23.
The function of money are:
(i) A medium and a measure
(ii) A standard and a store
(iii) Both (I) and (ii)
(iv) None of the above
Answer:
(iii) Both (1) and (ii)

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 24.
Money serves as a link between:
(i) The present and past
(ii) The past and future
(iii) The present and future
(iv) None of the above
Answer:
(iii) The present and future

Question 25.
Moeny facilities:
(i) Barter transaction
(ii) credit trangaction
(iii) All of the above
(iv) None of the above
Answer:
(ii) credit transaction.

Question 26.
Example of near money is:
(i) Time or fixed deposits
(ii) Bills of exchange and Treasury bill
(iii) Stock and share
(iv) All of the above
Answer:
(I) All of the above

Question 27.
A command Bank has:
(i) Unlimited credit creation power
(ii) Limited credit creation power
(iii) All of the above
(iv) None of the above
Answer:
(ii) Limited credit creation power

Question 28.
Primary function of a commercial bank is:
(i) To finance Internal and External trade
(ii) Creation of moeny
(iii) Acceptance of deposits
(iv) None of the above
Answer:
(iii) Acceptance of deposits

Question 29.
The right-hand side of the balance sheet shows the items under the:
(i) Liabilities
(ii) Assets
(iii) Cash
(iv) None of the above
Answer:
(ii) Assets.

Question 30.
Cash-in-hand is otherwise known as:
(i) Till money
(ii) Cashin-vault
(iii) Both (i) and (ii)
(iv) None of the above
Answer:
(iii) Both (j) and (ii)

Question 31.
Moeny at call and short-notice is a:
(i) Long period loans
(ii) Very short term loans
(iii) Both (i) and (ii)
(iv) None of the above
Answer:
(i) Very short term loans

Question 32.
One rupee not is issued by:
(i) R. B. I
(ii) Commercial Bank
(iii) Govt. of India
(iv) None of the above
Answer:
(iii) Govt. of India

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 33.
Time deposits are withdrawn:
(i) On the demand
(ii) After the expiry of the period
(iii) All of the above
(iv) None of the above
Answer:
(ii) After the expiry of the period

Question 34.
Most liquid asset of a commercial bank is:
(i) Demand deposits
(ii) Investment
(iii) Cash
(iv) None of the above
Answer:
(iii) Cash

Question 35.
Under which principle the central Bank of India issues notes:
(i) Proportional reserve system
(ii) Minimum reserve system
(iii) Maximum reserve system
(iv) None of the above
Answer:
(ii) Minimum reserve system

Question 36.
Quantitative credit control method refers to:
(i) Control the use of credit
(ii) Bring change in the total volume of credit in general
(iii) All of the above
(iv) None of the above
Answer:
(ii) Bring change in the total volume of credit in general

Question 37.
The selective credit control methods adopted by the central Bank to control credit are:
(i) Open market operation
(ii) Regulation of margin-requirements
(iii) Regulation on of consumers credit
(iv) Both (ii) and (iii)
Answer:
(i) Both (ii) and (iii)

Question 38.
The function of central Bank:
(i) Lender of the lust resort
(ii) Clearing agent
(iii) Banker’s Bank
(iv) Ail of the above
Answer:
(iv) All of the above

Question 39.
Central Bank acts as a financial advisor to the:
(i) General public
(ii) Commercial Banks
(iii) Govt.
(iv) None of the above
Answer: (iii) Govt.

II. Fill in the blanks :

Question 1.
_____ adjusts income to expenditure
Answer:
Government

Question 2.
_____ expenditure is compulsory in nature.
Answer:
Public

Question 3.
Government prefers _____ budget.
Answer:
deficit

Question 4.
_____ budget creates inflation. .
Answer:
deficit

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 5.
_____ budget creates deflation.
Answer:
surplus

Question 6.
_____ budget does not create any economic crises.
Answer:
balanced

Question 7.
_____ budget is cure for inflation.
Answer:
surplus

Question 8.
_____ budget is preferred in under developed countries.
Answer:
deficit

Question 9.
_____ budget eliminates unemployement problem.
Answer:
deficit

Question 10.
_____ budget is a cure for deflation.
Answer:
deficit

III. Correct the Sentences :

Question 1.
Government adjusts expenditure to income.
Answer:
Incorrect
Correct: Government adjusts income to expenditure

Question 2.
Private expenditure is compulsory in nature.
Answer:
Incorrect.
Correct: Public expenditure is compulsory in nature.

Question 3.
Government prefers surplus budget.
Answer:
Incorrect.
Correct: Government prefers deficit budget.

Question 4.
Budget has been derived from a Greek word.
Answer:
Incorrect.
Correct: Budget has been derived from the French words.

Question 5.
Budget is always balanced.
Answer:
Incorrect
Correct: Budget may be balanced or unbalanced.

Question 6.
Budget is meant for a financial year.
Answer:
Correct

Question 7.
Deficit budget checks wasteful expenditure.
Answer:
Incorrect.
Correct: Balanced budget checks wasteful expenditure.

Question 8.
Price stability is assured in case of surplus budget.
Answer:
Incorrect.
Correct: Price stability is assured in case of balanced budget.

Question 9.
Balanced budget controls public debt.
Answer:
Correct.

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 10.
Surplus budget increases national income .
Answer:
Incorrect.
Correct: Deficit budget increases national income.

Question 11.
In surplus budget. Government receipts are higher than government expenditure.
Answer:
Correct.

Question 12.
Surplus budget controls unemployment.
Answer:
Incorrect.
Correct: Deficit budget controls unemployment.

Question 13.
Surplus budget leads to maximum utilisation of resources.
Answer:
Incorrect
Correct: Deficit budget leads to maximum utilisation of resources.

Question 14.
Deficit budget is contractionary; but surplus budget is expansionary
Answer:
Incorrect.
Correct: Surplus budget is contractionary but deficit budget is expansionary.

Question 15.
Surplus budget increases debt burden.
Answer:
Incorrect
Correct: Deficit budget increases debt burden.

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 16.
Surplus budget checks unproductive expenditure.
Answer:
Correct

Question 17.
Deficit budget is inflationary.
Answer:
Correct.

Question 18.
In deficit budget price of exported goods increases.
Answer:
Correct

IV. Answer the following questions in one word :

Question 1.
What is public finance?
Answer:
Public finance is a subject that deals with the income and expenditure of the government.

Question 2.
Write a similarity that exists in both public finance & private finance.
Answer:
Both public finance & private finance have same objective i.e. satisfaction of human wants.

Question 3.
What is budget?
Answer:
Budget is the annual financial statement of the anticipated receipts & expenditure of the government for a financial year. ‘

Question 4.
From which word, the term ‘Budget’ is derived?
Answer:
The term budget is derived from the French word “Budgett”, which means small leather bag.

Question 5.
What is consolidated fund?
Answer:
Consolidated fund consists of all revenues & loan received by the government.

Question 6.
What is balanced budget?
Answer:
Balanced budget is that budget in which the anticipated receipts & expenditure of the government are equal.

CHSE Odisha Class 12 Economics Solutions Chapter 13 Public Finance

Question 7.
State one merit of the balanced budget?
Answer:
Balanced budget checks unproductive expenditure.

Question 8.
What is deficit budget?
Answer:
If the estimated revenue falls short of the estimated expenditure in the budget period, it is called deficit budget.

Question 9.
What is surplus budget?
Answer:
In surplus budget, the anticipated government receipts are higher than the government expenditure during the budget period.

Question 10.
What type of budget is framed in UDCS?
Answer:
UDCS frame deficit budget.

Question 11.
Give a demerit of deficit budget?
Answer:
Deficit budget is inflationary in nature.

Question 12.
Which budget leads to maximum utilisation of resources?
Answer:
Deficit budget.