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

CHSE Odisha 12th Class Biology Important Questions Chapter 14 Organisms and Environment

Organisms and Environment Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
Plants and animals living in a particular area constitute
(a) flora and fauna
(b) community
(c) ecosystem
(d) ecology
Ans.
(b) community

Question 2.
Biome is defined as
(a) sum of ecosystem in a geographical area
(b) sum of ecosystem of the whole earth
(c) biotic potential of a population
(d) biotic component of a ecosystem
Ans.
(a) sum of ecosystem in a geographical area

Question 3.
The adaptation of aquatic plants which roots are poorly developed is seen in
(a) Eichhornia
(b) Nymphaea
(c) Dunaliella
(d) Vallisneria
Ans.
(d) Vallisneria

Question 4.
Plenty of aerenchyma are found in
(a) hydrophytes
(b) mesophytes
(c) xerophytes
(d) halophytes
Ans.
(a) hydrophytes

Question 5.
The plant that does not belong to the ecological group, represented by the other plants is
(a) Pistia
(b) Casuarina
(c) Jussiaea
(d) Hydrilla
Ans.
(b) Casuarina

Question 6.
Sunken stomata are seen in
(a) hydrophytes
(b) xerophytes
(c) parasites
(d) symbionts
Ans.
(b) xerophytes

Question 1.
The organisms and their environment in a particular area
(a) bioregion
(b) biosphere
(c) ecosystem
(d) biome
Answer:
(c) ecosystem

Question 2.
Community is a group of independent and interacting population of
(a) different species
(b) same species
(c) same species in a specific area
(d) different species in a specific area
Answer:

Question 3.
Group of two or more than two plant species is called as
(a) plant community
(b) animal ecosystem
(c) plant ecosystem
(d) ecological niche
Answer:
(a) plant community

Question 4.
Which of the following is an animal of benthic zone?
(a) Frog
(b) Chemosynthetic bacteria
(c) Rat
(d) Human being
Answer:
(b) Chemosynthetic bacteria

Question 5.
Example of submerged hydrophyte is
(a) Hydrilla
(b) Lemna
(c) Nelumbium
(d) Eichhornia
Answer:
(a) Hydrilla

Question 6.
An association of two organism living together and benefitting each other is called ……….
(a) mutualism
(b) saprophytism
(c) parasitism
(d) commensalism
Answer:
(a) mutualism

Question 7.
A high density of tiger population is an area can result is
(a) predation
(b) interspecific competition
(c) intraspecific competition
(d) protocooperation
Answer:
(c) intraspecific competition

Question 8.
Which of the following shows detrimental effects on species
(a) mutualism
(b) predation
(c) parasitism
(d) competition
Answer:
(d) competition

Question 9.
Plasmodium is an example of
(a) predator
(b) endoparasite
(c) prey
(d) ectoparasite
Answer:
(b) ectoparasite

Question 10.
Monarch butterfly is not eaten by predators because of
(a) rough skin
(b) bitter taste
(c) foul smell
(d) colouration
Answer:
(b) bitter taste

Question 11.
The association of animals where both partners are benefitted is
(a) commensalism
(b) amensalism
(c) mutualism
(d) parasitism
Answer:
(c) mutualism

Question 12.
There are two optimal ways of exploitation one way is parasitism. Which is the other one?
(a) Antibiosis
(b) Competition
(c) Predation
(d) Commensatism
Answer:
(c) Predation

Question 13.
The most important factor which determined the increase in human population in India during 20th century is
(a) natality
(b) mortality
(c) immigration
(d) emigration
Answer:
(c) immigration

Question 14.
Population density is represented by
(a) N/S
(b) N/t
(c) t/S
(d) DNn/Dt
Answer:
(d) DNn/Dt

Question 15.
Natality increases the
(a) population density
(b) population size
(c) number of organisms in the population
(d) All of the above
Answer:
(d) All of the above

Fill in the blanks

Question 1.
Some xerophytes have multiple epidermis like …………. .
Answer:
Nerium

Question 2.
Leaves are large, broad and thin in …………. plants.
Answer:
mesophytic

Question 3.
Fish, amphibians and reptiles are ……………. .
Answer:
stenothermal animals

Question 4.
Root caps are present in ………….. .
Answer:
mesophytes

Question 5.
………. is the mechanism evolved by competing species for co-existence.
Answer:
Resource partitioning

Question 6.
Fig and wasp show ralationship.
Answer:
mutualistic

Correct the statements, if required by changing the underlined words

Question 1.
Plants which grow in bright light are called sciophytes.
Answer:
heliophytes

Question 2.
The organism which can tolerate a wide range of temperature are called stenothermal organisms.
Answer:
eurythermai

Question 3.
Plants growing in moist habitat are known as xerophytes.
Answer:
mesophytes

Question 4.
Plants growing in dry land are called mesophytes.
Answer:
xerophytes

Question 5.
Lichens represent an intimate mutualistic relationship between fungus and algae.
Answer:
It is correct

Question 6.
Mediterranean orchid ophrys employs sexual compatibility to get pollinated by bee.
Answer:
sexual deceit

Express in one or two words

Question 1.
The type of habitat in which plants are adapted to live in water scarcity.
Answer:
Xeric habitat.

Question 2.
The type of organisms who change their osmotic concentration according to the environment.
Answer:
Osmoconformers.

Question 3.
The factors which are related to soil in a habitat.
Answer:
Edaphic factors.

Question 4.
Plants growing or adapted to live in the shade.
Answer:
Sciophytes

Question 5.
Plants that grow best in direct sunlight.
Answer:
Heliophytes

Question 6.
The mechanism in which one animal kills other and eat it.
Answer:
Predation

Question 7.
The mechanism in which one species depend on the other for food and shelter.
Answer:
Parasitism

Short Answer Type Questions

Question 1.
Write different features of mesophytes.
Answer:
These are plants which grow in moist habitats and need well-aerated soils. These show the following characteristics

1. Root system is well-developed. They are branched, with root caps and root hairs.
2. Stems are aerial and freely branched.
3. Leaves are large, broad and thin.
4. Cuticle in all aerial parts is moderately developed.
5. The stomata are dorsiventral in dicotyledons and isobilateral position in monocot leaves.
6. The photosynthetic are in leaf, i.e., mesophyll tissue in differentiated into palisade, parenchyma and spongy parenchyma.
7. Water and food conductive tissue (vascular tissue) and mechanical tissue (collenchyma and sclerenchyma are well-developed.)

Question 2.
Write a note on hydrophytes.
Answer:
Hydrophytes
The plants growing in abundance of water or wet place are called hydrophytes. These plants may be partially or wholly submerged in water.
The aquatic habit at provides the following to the plant to grow

1. Availabity of nutrients in the water.
2. Plant growth matrix.
3. Approximate constant temperature (with least variation).
4. Availability of light.
5. Movement of water (waves strong or weak).

Based on the relation of plants to water and air.

Question 3.
Write short note on hydrophytic adaptations in roots of plants.
Answer:
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.

• Roots may be entirely absent, e.g. Wolffia, Salvinia or poorly developed, e.g. Hydrilla.
• Roots are well-developed with distinct root caps, e.g. Ranunculus (emergent hydrophytes), aerenchyma present.
• 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 geotrophic.

Question 4.
What are free-floating and rooted hydrophytes?
Answer:
Free-floating hydrophytes These plants are absolutely float freely on the water surface and are not linked to the soil or substratum, e.g. Duck weed (Lemna and Wolffia), water hyacinth (Eichhomia crassipes), water ferns (Azolla and Salvinia).

Question 5.
What is habitat? Name some factors which define habitat of an organism.
Answer:
Habitat is a natural abode or a locality where a plant or animal grows or in other words, where a species lives.
The factors which define habitat can be climatic temperature, humidity, edaphic, i.e. related to soil, topographic or physical.

Question 6.
List various adaptations shown by epidermis in xerophytic plants.
Answer:
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.

Question 7.
Does light factor affect the distribution of organisms? Write a brief note giving suitable examples of plants.
Answer:
Plants require sunlight for photosynthesis. Therefore, light is an important factor that affects the distribution of plants, e.g.

1. Many species of small plants (herbs and shrubs) growing in forests are adapted to photosynthesis optimally under very low light conditions so, they are seen distributed in shady areas under tall, canopied trees.
2. Many plants in the shade will grow vertically to gain access to light. These plants will appear to have smaller leafs than others of the same species of the , same age found in conditions with better sunlight.
3. Large sized trees will be present in areas that get abundant sunlight.

Question 8.
List any four characteristics that are employed in human population census.
Answer:
A population has the following characteristics that are employed in human population census

• Natality and mortality
• Sex-ratio
• Population density
• Age distribution

Question 9.
Describe the mutual relationship between the fig tree and wasp and comment on the phenomenon that operates in their relationship.
Answer:
The relationship between fig tree and wasp shows mutualism. The wasp while searching for sites to lay its eggs, pollinates the fig’s inflorescence.
On the other hand, the fig not only provides shelter (fruit) for oviposition, but also allows wasp’s larva to feed on its seeds.

Question 10.
What is mutualism? Mention any two examples where the organisms involved are commercially (2018) exploited in agriculture.
Answer:
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

Question 11.
Name important defence mechanisms in plants against herbivory.
Answer:
The herbivores are predators of plants and nearly 25% insects are phytophagous (feeding on plants). So, plants show morphological as well as chemical defence against herbivores such as

1. Thorns of rose and Acacia as well as cactus.
2. Certain plants produce chemicals, such as Opium, quinine, caffeine, nicotine, to protect them against being grazed by the animals.
3. Calotropis produces highly poisonous cardiac glycosides. So, the cattle and goats do not eat this plant.

Question 12.
What is predation? Explain with the help of suitable examples why is it required in a community with rich biodiversity.
Answer:
Predation is an interaction where one organisms (predator) kills and eats the other weaker organisms called prey.
Predation is a natural way of transferring the energy fixed by plants, to higher trophic levels.
Examples-snake eating a frog, tiger killing and eating a deer. Predators keep prey population under control which otherwise could achieve very high population densities and cause instability in ecosystem.
They also help in maintaining a species diversity in a community by reducing. The intensity of competition among competing prey species.

Question 13.
Name the interaction in each of the following:
(i) Cuckoo lays her eggs in the crow’s nest.
(ii) Orchid grows on a mango tree.
(iii) Ticks live on the skin of dogs.
Answer:
(i) Brood parasitism.
(ii) Commensalism, orchid is an epiphyte.
(iii) Parasitism, ticks are ectoparasites.

Question 14.
Explain the S-shaped pattern of population growth. How is J-shaped pattern different from it and why?
Answer:
S-shaped pattern of population growth form shows an initial gradual increase, followed by an exponential increase and then a gradual decline to a near constant level. It is different from J-shaped curve because J-shaped pattern shows exponential population growth and its abrupt crash after attaining the peak value. A-When resources are not limiting the growth, plot is exponential. B-When resources are limiting the growth, plot is logistic, K is the carrying capacity.

Question 15.
Explain diagrammatically the age structure of expanding, stable and declining population.
Answer:
The pyramids can be of three difference types as follows
1. 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.

2. 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.

3. 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.

Differentiate between the following (for complete chapter)

Question 1.
Population and Community.
Answer:
Differences between population and community are as follows

Population	Community
It is a grouping of individuals of a single species found in an area.	It is grouping of individuals of different species found in an area.
All the individuals of a population are morphologically and behaviourly similar.	Different members of a community are morphologically and behaviourly dissimilar.
Individuals of a population interbreed freely.	Interbreeding is absent amongst different members of a community.

Question 2.
Mesophytes and Hydrophytes.
Answer:
Differences between mesophytes and hydrophytes are as follows

Mesophytes	Hydrophytes
The plants grown on terrestrial habitate (land).	These are found in aquatic habitate (water).
The root septum is very cell developed in there plants.	Root system is not very well-developed or absent.
Leaves are broad, large and thin, mucilage covering is absent.	Leaves are, thin ribbon-shaped and covered with mucilage.

Question 3.
Mutualism and Commensalism.
Differences between mutualism and commensalism are as follows

Mutualism	Commensalism
In this two species are involved, both derive benefit from each other.	Two species involved but only one ge{ benefit other remain unharmed.
Example-see-anemone and hermit-crab.	Example-Sucker fish and shark.

Question 4.
Parasitism and Predation.
Answer:
Differences between parasitism and predation are as follows

Parasitism	Predation
It is host specific.	Predators have choice of prey.
Parasites are smaller is size.	Predators are large in size.
These have high reproductive potential.	These have low reproductive potential.

Question 5.
Ectoparasites and Endoparasites.
Answer:
Differences between ectoparasites and endoparasites are as follows

Ectoparasites	Endoparasites
Ectoparasites live on the surface of the host.	Endoparasites live in the body of the host.
They can be temporary, intermittent or permanent.	They are generally permanent parasites.
They can be hemiparasites or holoparasites.	They are usually holoparasites.
Respiration is aerobic.	Respiration is ofter anaerobic.
Specialisation has lead to loss of fewer strutures, e.g. wings in fleas, bedbugs and lice.	Specialisation has led the loss of several structures, e.g, digestive organs in Taenia.

Question 6.
Immigration and Emigration.
Answer:
Differences between immigration and emigration are as follows

Immigration	Emigration
It is permanent inward movement of some individuals into a local population.	It is a permanent outward movement of some individuals from a local population.
Size of gene pool and local population in increases.	Size of gene pool and local population decreases.
It is caused by availability of better living conditions.	It is caused by occurrence of deficiencies and calamities.

Question 7.
Camouflage and Mimicry.
Answer:
Differences between camouflage and mimicry are as follows

Camouflage	Mimicry
It is the ability of animals to blend with the background.	It is resemblance of on species of animals with another species.
Camouflage allows the animals to remain unnoticec from a distance.	Mimicry hides the true 1 identity of the animal species.
It is advantageous to both prey as well as predator.	It is advantageous to mimics against predation.

Question 8.
Intraspecific competition and Interspecific competition.
Answer:
Differences between intraspecific competition and interspecific competition are as follows

Intraspecific competition	Interspecific competition
It is competition among individuals of the same species.	The competition is amongst the members of different species.
The competition is for all the requirements.	The competition is for one or a few requirements.
The competing individuals have similar type of adaptation.	The competing individuals have different types of adaptations.
It is more severe due to similar needs and adaptations.	It is less severe as the similar needs are a few and the adaptations are different.

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

CHSE Odisha 12th Class Biology Important Questions Chapter 13 Applications of Biotechnology

Applications of Biotechnology Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
A biotech company that released first ever genetically manipulated flower into the market is
(a) Cryobank
(b) Eli Lilly
(c) Florigene
(d) Genentech
Ans.
(c) Florigene

Question 2.
When an abnormal gene is replaced by normal gene, what do you call it? (2023)
(a) Gene mutation
(b) Gene donning
(c) Gene therapy
(d) Gene ligation
Ans.
(c) Gene therapy

Question 3.
Cry II Ab and Cry I Ab produce toxins that control
(a) cotton bollworms and corn borer, respectively
(b) corn borer and cotton bollworms, respectively
(c) tobacco budworms and nematodes, respectively
(d) nematodes and tobacco budworms, respectively
Ans.
(a) cotton bollworms and corn borer, respectively

Question 4.
First genetically modified plant commercially released in India is
(a) golden rice
(b) slow ripening tomatoes
(c) Bt-bringal
(d) Bt-cotton
Ans.
(d) Bt-cotton

Question 5.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because
(a) bacteria are resistant to the toxin
(b) toxin is immature
(c) toxin is inactive
(d) bacteria enclose toxin in a special sac
Answer:
(c) toxin is inactive

Question 6.
Basic principle of developing transgenic animals is to introduce the gene of interest into the nucleus of
(a) somatic cell
(b) vegetative cell
(c) germ cell
(d) body cell
Answer:
(c) germ cell

Question 7.
Which among the following pharmaceutical products is harvested by using transgenic animals as bioreactors?
(a) Urokinase
(b) Insulin
(c) Lactoferrin
(d) All of these
Answer:
(d) All of these

Question 8.
The superbug can be used in
(a) oil spills
(b) water pollution
(c) eutrophication
(d) air pollution
Answer:
(a) oil spills

Question 9.
Biopatents are usually awarded for the discovery of
(a) new cell lines
(b) new DNA sequences
(c) GM strains
(d) All of these
Answer:
(d) All of these

Question 10.
Exploitation of patent biological resources of a country by another country is known as?
(a) biopatent
(b) biopiracy
(c) biowar
(d) All of these
Answer:
(b) biopiracy

Question 11.
Patents are given for
(a) discoveries
(b) inventions
(c) biopiracy
(d) gene therapy
Answer:
(b) inventions

Questions 12.
Which one of the following known as ‘Superbug’?
(a) Pseudomonas putida
(b) E. coli
(c) Aspergillus niger
(d) Acetobacter aceti
Answer:
(a) Pseudomonas putida

Questions 13.
US Patent on turmeric was challanged by
(a) CSIR
(b) EPO
(c) FSSAI
(d) FDI
Answer:
(a) CSIR

Questions 14.
Biopiracy is
(a) the use of biological patent
(b) thefts of plants and animals
(c) the use of bioresources of a country without proper authorisation
(d) stealing of biological resources
Answer:
(c) the use of bioresources of a country without proper authorisation

Correct the statements, if required, by changing the underlined words

Question 1.
An amorphous mass of parenchyma cells developed by tissue culture is called embryo.
Answer:
Callus

Question 2.
Petunia is ice minus strain which when sprayed on crops prevents frost formation.
Answer:
Pseudomonas

Question 3.
The natural source of vitamin-E is α-tocopherol.
Answer:
γ-tocopherol

Question 4.
The first GMO was created by Watson.
Answer:
Herbert Boyer and Stanley Cohen.

Question 5.
Soil bacterium Nitrosomonas syringae promotes ice nucleation in plants.
Answer:
Pseudomonas

Question 6.
The C-peptide is added during the maturation of pro-insulin to insulin.
Answer:
deleted

Question 7.
ADA treatment uses monocytes.
Answer:
lymphocytes

Question 8.
Antitrypsin is an agent that dissolves blood clot.
Answer:
Tissue Plasminogen Activator (TPA).

Question 9.
The first transgenic cow was Lilly.
Answer:
Rosie

Question 10.
A Pacific transgenic whale was generated by a growth hormone transgene.
Answer:
salmon

Question 11.
Transgenic mouse is smaller than the normal mouse.
Answer:
larger

Question 12.
Human protein α-2-trypin is used to treat emphysema.
Answer:
α-1-antitrypsin

Express in one or two word(s)

Question 1.
Define callus.
Answer:
It is undifferentiated mass of totipotent cells in the culture media.

Question 2.
Name one plant used to create novel transgenic plants.
Answer:
Petunia

Question 3.
Name the drugs isolated from Catharanthus roseus for cancer treatment.
Answer:
Vincristine and vinblastine.

Question 4.
Which microorganism is used as cloning host cell to produce humulin?
Answer:
E. coli

Question 5.
State the number of polypeptides found in mature human insulin.
Answer:
Two

Question 6.
What do you mean by the term transgene?
Answer:
oreign gene that is incorporated in an orgainsm to bring about desirable changes.

Question 7.
Where the LDL receptors are present?
Answer:
On the surface of hepatocytes

Question 8.
What is the name of the scientist who coined a sheep named Dolly?
Answer:
Keith Campbel and Ian Wilmut.

Question 9.
Name the institute that came up with a cloned sheep, named Dolly.
Answer:
Ian Wilmut of Roslin Institute in Scottland.

Question 10.
Which department of the Goverment of India is the nodal centre for Indian biosafety network?
Answer:
Department of biotechnology.

Question 11.
Which bacterium species is involved in Diamond vs Chakraborty case?
Answer:
Pseudomonas

Question 12.
What are transgenic animals?
Answer:
Genetically modified organism

Question 13.
Name the first transgenic cow that produced human protein enriched milk.
Answer:
Rosie

Question 14.
Name a transgenic animal being used in testing the safety of polio vaccine.
Answer:
Mouse

Question 15.
Mention the name of two common diseases that can be treated by medicines that contain biological products of transgenic animals.
Answer:
Cystic fibrosis and rheumatoid arthritis

Question 16.
What is patent?
Answer:
It is an open latter, a set of legal right, privilege and authority granted by a sovereign state to a person or institution for an invention for a limited period of time.

Fill in the blanks

Question 1.
…………… is a mammalian protein that have been successfully expressed in plants.
Answer:
Enkephalin.

Question 2.
Genetically engineered rice rich in vitamin-A is known as ……………… .
Answer:
golden rice

Question 3.
The recombinant human insulin is known as …………. .
Answer:
humulin

Question 4.
The full form of ELISA is ………… .
Answer:
Enzyme Linked Immunosorbant Assay.

Question 5.
Primarily, insulin is synthesised as ……………. .
Answer:
single polypeptide.

Question 6.
SCID is caused due to failure of synthesis of enzyme ………….. .
Answer:
adenosine deaminase.

Question 7.
…………. is the precursor of vitamin-A.
Answer:
ß-carotene

Question 8.
………… infects the roots of tobacco plants.
Answer:
Agrobacterium

Question 9.
The transgenic mouse is called as …………….. .
Answer:
Super mouse

Question 10.
…………. is a transgenic sheep.
Answer
Dolly

Question 11.
…………. is an infant nutrition formula that have been harvested using transgenic animals as bioreactors.
Answer:
Lactoferrin.

Question 12.
In 1990 ……….. the transgenic ewe was born in Scottland.
Answer:
Tracy

Question 13.
Full from of TPA is ……………. .
Answer:
Tissue Plasminogen Activator.

Question 14.
Neem patent case was first awarded in favour of ……………. .
Answer:
USA.

Short Answer Type Questions

Question 1.
Find out from the internet what is golden rice.
Answer:
Golden rice is a genetically modified rice with high levels of ß-carotene and other carotenoids. This rice is modified in order to enhance the quantity of vitamin-A in it. It is called golden due to the gold-like colour it gets from ß-carotene.

Question 2.
Can a disease be detected before its symptoms appear? Explain the principle involved.
Answer:
When the symptoms of the disease are not visible and the pathogen concentration is very low, then detection by conventional diagnostic tests is very difficult. However, detection at the above stated stage is made possible by molecular diagnostic techniques like the amplification of their nucleic acid by Polymerase Chain Reaction (PCR). The principle involved here is that a single DNA molecule can be copied endlessly in a test tube, using primers, DNA polymerase enzyme and free nucleotides and appropriate conditions.

Question 3.
How is DNA recombinant technology helpful in detecting the presence of mutated genes in the cancer patients?
Answer:
Molecular diagnosis in DNA recombinant technology uses a single-stranded DNA or RNA tagged with a radioactive molecules. It is allowed to hybridise to its complementary DNA in a clone of cells followed by detection using autoradiography. The clone having the mutated gene will not appear on the photographic film, because the probe used will not be complementary to the mutated gene. In this way, presence of mutated genes can be detected.

Question 4.
Why is the functional insulin produced, considered better than the ones used earlier by diabetic patients?
Answer:
Insulin prepared by rDNA technology does not produce sensitive allergic reactions and complications to the foreign protein which occurred in the case of earlier extracted insulin from the pancreas of slaughtered cattle or pigs. Thus, it is considered better than the earlier used insulin.

Question 5.
How is a mature, functional insulin hormone different from its pro-hormone form?
Answer:
Mature functional insulin is obtained by the processing of pro-hormone which contains extra peptide called C-peptide. This C-peptide is removed during the maturation of pro-insulin to insulin.

Question 6.
Refer to the diagram given below and answer the questions that follows

(i) The diagram shown above is insulin or proinsulin? Justify.
(ii) How is mature insulin synthesised?
Answer:
(i) The diagram is proinsulin as it contains C-peptide.
(ii) Mature insulin is synthesised by the removal of extra stretch called C-peptide.

Question 7.
How did an American Company, Eli Lilly use the knowledge of rDNA technology to produce human insulin?
Or
How did Eli Lilly synthesise the human insulin? Mention one difference between this insulin and the one produced by the human pancreas.
Or
What is humulin?
Answer:
The production cost was high due to its complex extraction and purification processes. Additionally, the purified insulin was contaminated by many pathogenic viruses. These problems have been overcome by the use of recombinant DNA technology. Insulin that is produced by recombinant DNA technology is known as recombinant human insulin.

Question 8.
Recombinant DNA technology is of great
importance in the field of medicine. With the help of a flow chart, show how this technology has been used in preparing genetically engineered human insulins.
Answer:
Insulin production by using recombinant DNA technology is shown in flow chart below

Question 9.
What is gene therapy? Name the first clinical case in which it was used.
Or
What is gene therapy? Illustrate using the example of Adenosine Deaminase (ADA) deficiency.
Answer:
It is a method of treatment which allows correction of a biochemical (like phenylketonuria) or a genetic defect*that has been diagnosed in a child or embryo. The defective mutant alleles of the gene are replaced by the normal gene insertion to take over the function of and compensate for the non-functional gene. Gene therapy is widely used to treat

• Biochemical disorder, e.g. alkaptonuria, phenylketonuria, albinism, etc.
• Chromosomal and gene disorders, e.g. Down’s syndrome, Turner’s, syndrome, Klinefelter’s syndrome, fragile X-syndrome, cri-du chat, Huntington’s disease, Tay-Sach’s syndrome, etc.

Question 10.
Write a note on genetically modified organism.
Answer:
The plants, bacteria, fungi and animals whose genes have been altered by manipulation are called Genetically Modified Organisms (GMO).
These are also called transgenic organisms, as they contain and express one or more foreign genes called transgenes. Herbert Boyer and Stanley Cohen developed the first GMO in 1973.

They transferred the Kanamycin antibiotic resistance gene of a bacterium into the another Kanamycin sensitive bacterium.
The genes which are being transferred are called transgenes.
The recepient bacterium later acquired Kanamycin resistance. Following many such discoveries, GMOs were developed. Novel plants and animals were created by genetic manipulation for human welfare.

Question 11.
Write a short note on transgenic 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 12.
Write a short note with 2-3 important points on Bacillus thuringiensis.
Or
Why do the toxic insecticide proteins secreted by Bacillus thuringiensis kill insects?
Answer:
Bacillus thuringiensis (Bt) It is a soil-borne, Gram-positive bacterium. It is used to create transgenic plants having resistance to different pests. The genes having insecticidal properties in the bacterium are isolated and incorporated into plants by using advanced biotechnological methods to create Bt plants. During sporulation, many Bt strains produce crystal proteins (proteinaceous inclusions) called S-endotoxins, that have insecticidal action.

When consumed by insect, these toxins bind to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis, leading to the death of an insect, e.g. 5r-cotton, fir-tomato, soybean, coffee, etc.

Question 13.
Why does Bt toxin cannot kill the bacterium that produces it, but kills the insect that ingests it?
Answer:
Bt toxin is produced by a soil bacterium called Bacillus thuringiensis. This toxin does not kill the bacterium which produces it, because in them, it is present in an inactive and crystalline form. It becomes active and toxic only when it is consumed by insects such as lepidopterans, etc. due to the alkaline pH of the gut.

Question 14.
Bt cotton is resistant to pests, such as lepidopterans, dipterans and coleopterans. Is Bt cotton resistant to other pests as well?
Answer:
Bt cotton is made resistant only to certain specific taxa of pests. It is quite likely that in future some other pests may infest this Bt cotton. It is similar to immunisation against smallpox which does not provide immunity against other pathogens like those that cause cholera, typhoid, etc.

Question 15.
Why certain cotton plants are called Bt cotton?
Answer:
Cotton plants are called Bt cotton because they bear specific Bt toxin genes which were isolated from Bacillus thuringiensis and incorporated into certain cotton plants. This helps the host plants in developing resistance against and various pests like bollworms, etc.

Question 16.
Differentiate the terms ‘Cry’ and ‘cry’.
Answer:
‘Cry’ refers to protein symbol and its first letter is always capital. It is written in Roman letters, ‘cry’ refers to the gene which is usually written in small letters and is invariably in italics.

Question 17.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Or
Name the source and type of cry genes used for incorporation into crops by biotechnologists. Explain, how have these genes brought beneficial changes in the genetically modified crops.
Answer:
The proteins encoded by the gene named cry are called Cry proteins. Organism that produces Cry proteins is Bacillus thuringiensis.
The cry genes are incorporated in several crop plants, which then develop resistance to a specific targeted pest, e.g. cry IAc and cry IIAb control the cotton bollworms and cry LAb controls corn borer.

Question 18.
Expand GMO. How is it different from a hybrid?
Answer:
GMO stands for Genetically Modified Organism.
It differs from a hybrid because in a hybrid, cross is done between total genomes of two species or strains, whereas in a GMO, foreign genes from entirely dilferent species are introduced in the organism and are usually maintained as extrachromosomal entity or are integrated into the genome of the organism.

Question 19.
Describe any three potential applications of genetically modified plants.
Answer:
Potential applications of genetically modified plants are

• Nutritional enhancement, e.g vitamin-A enriched rice.
• Stress tolerance crops are more tolerant to abiotic stresses such as cold, drought, etc.
• Creation of tailor made plants by using GM plants to supply alternative resources to industries in the form of starches, biofuels, etc.

Question 20.
What is meant by transgenic animals?
Answer:
Animals that have had their DNA manipulated to possess and express an extra (foreign) gene are known as transgenic animals, e.g. transgenic rats, rabbits, pigs, sheep, cows and fish. Over 95% of all the existing transgenic animals are mice. The gene that is being transferred is called transgene.

Question 21.
Comment on how transgenic animals have proved to be beneficial in
(i) Production of biological products?
(ii) Chemical safety testing?
Answer:
(i) The transgenic animals have been proved to be beneficial in the production of biological products like human protein α-1 antitrypsin (by coding genes from that protein only), in the treatment of emphysema and production of human protein (α-lactalbumin) enriched milk by transgenic cow, i.e. Rosie. This milk was more nutritionally balanced for human beings than natural cow’s milk.
(ii) Transgenic animals are studied for testing toxicity of drugs and other chemicals as they carry genes that make them more sensitive to toxic substances.

Question 22.
What is the utility of transgenic animals?
Or
With respect to understanding diseases, discuss the importance of transgenic animal models.
Answer:
Transgenic animals are important in the following fields
(i) They are being used in basic science research to elucidate the role of genes in the development of diseases like cancer, cystic fibrosis, rheumatoid arthritis and Alzheimer’s disease.
(ii) They are valuable tools in the drug development process itself.
(iii) Milk producing transgenics can produce medicines or human proteins (insulin, growth hormone, etc.) in large quantities.
(iv) Transgenics can be a source of transplant organs as well.

Question 23.
(i) Which animals are being used for testing the safety of vaccines? Name the vaccine for which trials are going on.
(ii) Name the first transgenic cow. Why is it important?
Answer:
(i) Transgenic mice are used for testing the safety of vaccines.
The trials are going on for polio vaccine.
(ii) Rosie was the first transgenic cow. It is important because it produces human protein enriched milk, even better than a natural cow’s milk.

Question 24.
(i) Explain alpha lactalbumin. Where is it produced in human body?
(ii) In what manner biotechnology has helped in production of more nutritionally balanced milk?
Answer:
(i) Alpha lactalbumin is a human milk protein which helps to increase the production of lactose in the body. It is produced in human milk.
(ii) Biotechnology has lead to production of transgenic cow, Rosie that produced around 2.4 g/L human protein enriched milk. This milk contained the human alpha lactalbumin and was nutritionally more balanced than a natural cow’s milk.

Question 25.
While creating genetically modified organisms, genetic barriers are not respected. How can this be dangerous in the long run?
Answer:
Genetic modification of organisms can have unpredictable results when such organisms are introduced into the ecosystem. Because the real effects of gene manipulation are visible only when such organisms interact with other components and organisms of the ecosystem.

Question 26.
Biopiracy should be prevented. State why and how?
Answer:
Biopiracy should he prevented because
(i) The countries and people concerned are not given adequate compensatory payment.
(ii) The countries/people also lose their right to grow and use breeding experiments to improve the other varieties of the same species.
It may be prevented by implementing specific laws that takes into consideration all the biopatents and biopiracy related issues.

Question 27.
State the initiative taken by the Indian Parliament against biopiracy.
Answer:
The Indian Parliament has recently passed the second amendment to the Indian Patents Bill that takes action against biopiracy. In India, the Patent Act was enacted in 1970 to protect their resources and traditional knowledge from being exploited by other countries.
This act has undergone many amendments in 1999, 2002, 2005 and 2006. The Indian biosafety network is headed by the Department of Biotechnology.

Question 28.
Write a self-explanatory note on biopatent.
Or What is patent?
Answer:
When an individual develops a new product or process innovation using his intellect, the innovation becomes his own. Various rules at national and international lands protects the misuse of this innovation, and also safeguards the rights of the innovater.
The new inventions can be safe protected by patents, design trademark, trade secrets and copy rights, etc.

It is a set of exclusive legal rights granted by a government to the inventors or their assignee for a limited period of time to prevent others from commercial use of their invention. When patent is granted for biological entities and for products derived from them, they are called biopatents. Primarily USA, Japan and members of European Union are awarding biopatents.

Question 29.
Name a set of principles that may be used to regulate human activities in relation to the biological world. Why are they important?
Answer:
A set of principles that may be used to regulate human activities in relation to the biological world are called bioethics.
These are important because the genetic modification of an organism can have unpredictable results when such organisms are introduced into the ecosystem.

Question 30.
What is meant by biopiracy?
Answer:
It refers to the use of bioresources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment. The majority of industrialised nations are financially rich but poor in biodiversity and traditional knowledge, in comparison to developing and underdeveloped countries.

Another cause of biopiracy is bioprospecting which means a thorough survey of a source material to expand the knowledge and applications in biotechnology. During the course of bioprospecting, scientists may transfer any biological resource which they may consider as novel.

Differentiate between the following (for complete chapter)

Question 1.
Herbicide resistant plants and Frost resistant plants.
Answer:
Differences between herbicide resistant plants and frost resistant plants are as follows

Herbicide resistant plants	Frost resistant plants
Generated by transferring bacterial herbicide resistant gene into plant cells.	Generated by deleting a gene that promotes ice nucleation.
e.g. Petunia contains glyphosate resistant gene which is being isolated to generate glyphosate resistant plants.	e.g. Pseudomonas syringae contains gene promoting ice nucleation. It is deleted by genetic engineering to produce ice minus strain.

Question 2.
Humulin and Wosulin.
Answer:
Differences between humulin and wosulin are as follows

Humulin	Wosulin
Manufactured by Eli Lilly corporation, USA.	Manufactured by wokhardt limited India.
First recombinant drug approved by FDA for human use.	General drug to treat diabetes.

Question 3.
Ex vivo gene therapy and In vivo gene therapy.
Answer:
Differences between Ex vivo gene therapy and In vivo gene therapy are as follows.

Ex vivo gene therapy	In vivo gene therapy
The cells are removed from the patient and genetic material is inserted in them in vitro, prior to transplantation of modified cells.	The genetic material is transferred directly into cells within a patient.
This approach is applicable to tissues that can be removed from the body and returned later and survive for longer period of time, e.g. hematopoietic cells.	It is only possible in tissues where the individual cells cannot be cultured in vitro in sufficient numbers or where cultured cells cannot be efficiently reimplanted, e.g. brain cells.

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

CHSE Odisha 12th Class Biology Important Questions Chapter 12 Principles and Processes of Biotechnology

Principles and Processes of Biotechnology Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
GAATTC is the recognition site for which of the following restriction endonucleases?
(a) Hind III
(b) Eco RI
(c) Bam I
(d) Hae III
Answer:
(b) Eco RI

Question 2.
Given below is a sample of a portion of DNA strand giving the base sequence on the opposite strands. What is so special shown in it?
Img 1
(a) Replication completed
(b) Deletion mutation
(c) Start codon at the 5′ end
(d) Palindromic sequence
Answer:
(d) Palindromic sequence

Question 3.
Agarose extracted from sea weeds is used in
(a) spectrophotometer
(b) tissue culture
(c) PCR
(d) gel electrophoresis
Answer:
(d) gel electrophoresis

Question 4.
The blotting of RNA is called
(a) Northern blot
(b) Southern blot
(c) Western blot
(d) Eastern blot
Answer:
(a) Northern blot

Question 5.
Eco RI cleaves the DNA strands to produce
(a) blunt ends
(b) sticky ends
(c) satellite ends
(d) ori replication end
Answer:
(b) sticky ends

Question 6.
DNA fragments generated by the restriction endonucleases in a chemical reaction can be separated by
(a) electrophoresis
(b) restriction mapping
(c) centrifugation
(d) polymerase chain reaction
Answer:
(a) electrophoresis

Question 7.
Commonly used vectors for human genome sequencing are
(a) T-DNA
(b) BAC and YAC
(c) expression vector
(d) T/A cloning vectors
Answer:
(b) BAC and YAC

Question 8.
Which procedure is followed for amplification of DNA?
(a) Electrophoresis
(b) Autoradiography
(c) Polymerase chain reaction
(d) Southern blotting
Answer:
(c) Polymerase chain reaction

Question 9.
The Polymerase Chain Reaction (PCR) is a technique that is used for
(a) in vivo replication of specific DNA sequence using thermostable DNA polymerase
(b) in vitro synthesis of mRNA
(c) in vitro replication of specific DNA sequence using thermostable DNA polymerase
(d) in-vivo synthesis of mRNA
Answer:
(c) In-vitro replication of specific DNA sequence using thermostable DNA polymerase.

Question 10.
The figure below shows three steps (A, B, C) of Polymerase Chain Reaction (PCR). Select the option giving correct identification together with what it represents?
Img 5
(a) B – denaturation at a temperature of about 98°C separating the two DNA strands
(b) A – denaturation at a temperature of about 50°C
(c) C – extension in the presence of heat stable DNA polymerase
(d) A – annealing with two sets of primers
Answer:
(a) B – denaturation at a temperature of about 98°C separating the two DNA strands.

Question 11.
In recombinant DNA technique, the term vector refers to
(a) plasmids that can transfer foreign DNA into a living cell
(b) cosmids that can cut DNA at specific base sequence
(c) plasmids that can join different DNA fragments
(d) cosmids that can degrade harmful proteins
Answer:
(a) plasmids that can transfer foreign DNA into a living cell

Question 12.
The rDNA molecule is introduced into the cell of bacterium with the help of.
(a) restriction endonuclease
(b) DNA ligase
(c) electroporation
(d) None of the above
Answer:
(a) restriction endonuclease

Question 13.
The bacterial source of Hpa I is
(a) Haemophilus influenzae
(b) Bacillus amyloliquefaciens
(c) Providentcia stuarth
(d) Haemophilus parainfluenzae
Answer:
(d) Haemophilus parainfluenzae

Question 14.
Klenow fragment does not possess
(a) 5′ → 3′ exonuclease
(b) 3′ → 5′ exonuclease
(c) polymerase
(d) All of the above
Answer:
(a) 5′ → 3′ exonuclease

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

Question 1.
Restriction enzymes are used to cut single-stranded DNA.
Answer:
double-stranded DNA.

Question 2.
‘CO’ part in Eco RI stands for coenzyme.
Answer:
coli.

Question 4.
The first isolated restriction endonuclease was Hind III.
Answer:
Eco RI

Question 5.
DNA ligase forms phosphodiester bonds to ligate the DNA fragments.
Answer:
It is correct

Question 6.
Plasmids present in bacterial cells are linear double helical DNA molecules.
Answer:
circular

Question 7.
Taq polymerase is used between annealing and denaturation during PCR.
Answer:
extension.

Question 8.
A hybrid of plasmid and phage is YAC.
Answer:
BAC

Question 9.
Bacteria phage are autonomously replicating circular DNA.
Answer:
Bacteriophage

Fill in the blanks

Question 1.
DNA polymerase can be obtained from ……….. .
Answer:
Thermus aquaticus

Question 2.
The usual source of restriction endonuclease used in gene cloning is ………….. .
Answer:
bacteria .

Question 3.
Other than E. coli ………….. bacteria is used in recombinant DNA technology.
Answer:
Salmonella typhimurium.

Question 4.
First letter of restriction enzymes represents …………. .
Answer:
genus

Question 5.
Alkaline phosphatase is ………. an enzyme.
Answer:
DNA ligase

Question 6.
The vector for T-DNA is ………….. .
Answer:
Agrobacterium tumefaciens.

Question 7.
In biolistic method ………….. particles coated with foreign DNA are bombarded into target cells.
Answer:
gold or tungesten.

Question 8.
……………. is most commonly used process of foreign DNA injection in animal cells.
Answer:
Microinjection

Question 9.
…………. helps in selecting transformants and eliminating non-transformants.
Answer:
Conventional method

Express in one or two word(s)

Question 1.
Molecular scissors used in recombinant technology are known as
Answer:
restriction enzymes.

Question 2.
Which enzyme helps in joining DNA fragments?
Answer:
DNA ligase.

Question 3.
Name the process of transfer of protein molecules onto a membrane.
Answer:
Western blotting

Question 4.
Single-stranded fragments are transferred onto nitrocellulose filter paper by which process?
Answer:
Northern blotting

Question 5.
Name the bacterium that yields thermostable DNA polymerases.
Answer:
Thermus aquaticus

Question 6.
What is particle gun?
Answer:
It is a technique of bombarding microparticles of gold or tungsten with foreign DNA into target cell with high velocity.

Question 7.
Name one chemical that helps foreign DNA to enter host cell.
Answer:
Transformation

Short Answer Type Questions

Question 1.
Write a short note on genetic engineering.
Answer:
Genetic engineering is a modification, of chemical nature of genetic material (DNA/RNA) and their introduction into another organism to change the phenotypic characters of that organism.
This involves recombinant nucleic acid (DNA or RNA) techniques to form new combinations of heritable genetic material followed by the incorporation of material indirecdy through vectors or directly through microinjection and other techniques.

Question 2.
What are restriction enzymes? Mention their functions in recombinant DNA technology.
Answer:
The enzymes used for cutting the DNA during recombinant DNA technology are called restriction enzymes.
Restriction enzymes function as chemical knives or molecular scissors in genetic engineering. It recognises specific nucleotide sequence and makes cuts.

Question 3.
(i) Mention the difference in the mode of action of exonuclease and endonuclease.
(ii) How does restriction endonuclease function?
Answer:
(i) Exonucleases cleave base pairs of DNA at their terminal ends (either 5′ or 3 0 while, the endonucleases cleave DNA at any point within DNA segment at specific position except terminal ends.

(ii) Restriction endonuclease Eco RI cuts the DNA strands a little away from the palindromic sequences, but between the same two bases on the two strands
Img 2

Question 4.
How are ‘sticky ends’ formed on a DNA strand? Why are they so called?
Or
A recombinant DNA is formed when sticky ends of the vector DNA and the foreign DNA join. Explain how sticky ends are formed and get joined?
Answer:
Restriction enzymes cut the strands of the DNA, a little away from the centre of the palindromic sites, but between the same two bases on opposite strands. This leaves sticky single-stranded position at the ends. These overhanging stretches are called ‘sticky ends’. These are named so, because they form hydrogen bonds with their complementary cut counterparts.

Question 5.
What does Hind and ‘III’ refer to in the enzyme Hind III?
Answer:

1. The first letter ‘H indicates the genus of the organism, from which the enzyme was isolated, i.e. H-genus, Haemophilus.
2. The roman number (III) denotes the sequence in which the restriction endonuclease from that particular genus, species and strain of bacteria have been isolated, i. e., third restriction endonuclease to be isolated from this species.

Question 6.
Collect five examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base pair rules.
Answer:
Img 3

Question 7.
Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer:
Enzymes should not have more than one site of action.
This is because vectors have very few recognition sites for commonly used restriction enzymes.
If it will have more than one restriction sites it will generate various fragments. This will complicate the process of genetic engineering.

Question 8.
Draw agarose gel electrophoresis apparatus. Description is not required.
Answer:
Img 4
(a) Agarose gel apparatus, (b) Fluorescent bands on the agarose gel slab containing resolved DNA fragments with differing molecular weights.

Question 9.
A mixture of fragmented DNA was electrophoresed in an agarose gel. After staining the gel with ethidium bromide, no DNA bands were observed. What could be the reason?
Answer:
DNA bands may not be observed when a mixture of fragmented DNA was electrophoresed, due to following reasons

1. Concentration of agarose in the gel was not proper, as greater the concentration of agarose gel used, the greater will be separation of small DNA fragments, whereas smaller the concentration of agarose, higher will be the resolution of bands.
2. If the concentration of salt in the buffer was not proper.
3. If DNA sample is contaminated with RNA or any other impurity or if the concentration of DNA is too low.

Question 10.
Name and describe the technique that helps in separating the DNA fragments by the use of restriction endonuclease.
Or
(i) Name the technique used for the separation of DNA fragments.
(ii) Write the type of matrix used in this technique.
(iii) How is the prepared DNA visualised and extracted for use in recombinant technology?
Answer:
(i) DNA fragments can be separated by the technique called gel electrophoresis.
(ii) The most commonly used matrix is agarose gel, which is a natural polymer extracted from sea weeds.
(iii) A compound called Ethidium Bromide (EtBr) stains DNA, which on exposure with ultraviolet radiationals gives orange light. Hence, DNA fragments appear as orange bands.

Question 11.
A plasmid DNA and a linear DNA (both of the same size) have one site for a restriction endonuclease. When cut and separated on agarose gel electrophoresis, plasmid shows one DNA band while, linear DNA shows two fragments. Explain.
Answer:
This is because plasmid is a circular DNA molecule.
When cut with enzyme, it becomes linear, but does not get fragmented. Whereas, a linear DNA molecule gets cut into two fragments. Hence, a single DNA band is observed for plasmid while, two DNA bands are observed for linear DNA in agarose gel.

Question 12.
Write the role of ‘ori’ and ‘restriction site’ in a cloning vector pBR322.
Answer:
Ori is the site where replication starts. This site is responsible for controlling the copy number of a vector. Restriction site is the site of ligation of alien/foreign DNA in the vector, in one of the two antibiotic resistance sites or coding sequence of β- galactosidase.

Question 13.
Write a short note on plasmids.
Answer:
These are small and double-stranded extrachromosomal DNA molecules having an origin of replication. They occur naturally in bacteria. Naturally occurring plasmids are not suitable for the cloning of genes. They are genetically engineered and made suitable for cloning.

Such plasmids are called cloning plasmids. A suitable cloning plasmid should preferably have the following properties

• It should be non-conjugative.
• It should have a relaxed replication and a high copy number.
• It should have an origin of replication.
• It should have unique restriction sites.
• It should have antibiotic marker gene for selection.

The most common cloning plasmid is designated as pBR322, where p stands for the word plasmid and B and R signify the names of its engineers (B. Boliver and R. Rodriguez) 322 is a numerical designation that has a relevance to these workers, who worked out the plasmid.

Question 14.
You have chosen a plasmid as vector for cloning your gene. However, this vector plasmid lacks a selectable marker. How would it affect your experiment?
Answer:
In a gene cloning experiment, first a recombinant DNA molecule is constructed, where the gene of interest is ligated to the vector and introduced inside the host cell (transformation). Since, not all the cells gets transformed with the recombinant/plasmid DNA, in the absence of selectable marker, it will be difficult to distinguish between transformants and non-transformants. Because the role of selectable marker is in the selection of transformants.

Question 15.
Draw a schematic sketch of pBR322 plasmid and label the following in it.
(i) Any two restriction sites
(ii) Ori and rop genes
(iii) An antibiotic resistant gene
Answer:
The labelled diagram of pBR322 plasmid is shown in the figure with
(i) Eco RI and Bam HI as restriction enzymes
(ii) Ori and rop genes
(iii) amp^R (an antibiotic resistant gene)
Img 6
E. coli cloning vector pBR322

Question 16.
(i) How are recombinant vectors created?
(ii) For creating one recombinant vector only one type of restriction endonuclease is required. Give reason.
Answer:
(i) The vector DNA is cut at a particular restriction site using a restriction enzyme (to cut the desired DNA segment). The alien DNA is then linked with the plasmid DNA using an enzyme called ligase to form the recombinant vector.

(ii) Since, a restriction enzyme recognises and cuts the DNA at a particular sequence is called recognition site, the same restriction enzyme is used for cutting the DNA segment from both the vector and the other source to achieve as ends.

Question 17.
How bacterial cells are made competent to take up DNA?
Or
Describe the role of CaCl₂ in the prepration of competent cell.
Answer:
Since, DNA is a hydrophilic molecule, it cannot pass through the cell membranes. In order to force bacteria to take up the plasmid, the bacterial cells must first be made competent to take up foreign DNA. This is done by treating them with a specific concentration of a divalent cation such as calcium (Ca²⁺), which increases the efficiency with which DNA enters the bacterium through the pores in its cell wall.

Recombinant DNA can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock), and then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

Question 18.
PCR is a useful tool for early diagnosis of an infectious disease. Comment.
Answer:
PCR is a very sensitive technique, which enables the specific amplification of desired DNA from a limited amount of DNA containing sample. Hence, it can detect the presence of an infectious organism in the patient at an early stage of infection (even before the infectious organism has multiplied to large number).

Question 19.
(i) Explain how to find whether an E coli bacterium has transformed or not, when a recombinant DNA bearing ampicillin-resistance gene is transferred into it
(ii) What does the ampicillin-resistant gene act as, in the above case?
Answer:
(i) The recombinant/transformant can be selected out from the non-recombinants/ non-transformants by plating the transformants on ampicillin-containing medium. The transformants will grow in it, while the non-transformants will not grow.
(ii) It acts as a selectable marker.

Question 20.
Name the source of the DNA polymerase used in PCR technique. Mention, why it is used ?
Or
Give the name of the organism from where the thermostable DNA polymerase is isolated. State its role in genetic engineering.
Answer:
Bacterium Thermus aquaticus is a source of enzyme Taq polymerase. As, it is a thermostable enzyme and works at high temperature, it is used to amplify DNA in vitro by PCR. The amplified fragment of desired DNA can be used to ligate with the vector for further cloning.

Question 21.
Rearrange the following in the correct sequence to accomplish an important biotechnological reaction
(a) In vitro synthesis of copies of DNA of interest
(b) Chemically synthesised oligonucleotides
(c) Enzyme DNA polymerase
(d) Complementary region of DNA
(e) Genomic DNA template
(f) Nucleotides provided
(g) Primers
(h) Thermostable DNA-polymers (From Thermus aquaticus)
(i) Denaturation of dsDNA
Answer:

(i) Denaturation of dsDNA
↓
(e) Genomic DNA template
↓
(g) Primers
↓
(b) Chemically synthesised oligonucleotides
↓
(d) Complementary region of DNA
↓
(f) Nucleotides provided
↓
(c) Enzyme DNA polymerase
↓
(h) Thermostable DNA polymerase (from Thermus aquaticus )
↓
(a) In vitro synthesis of copies of DNA of interest

Question 22.
While doing a PCR, denaturation step is missed. What will be its effect on the process?
Answer:
If denaturation of double-stranded DNA does not take place, then primers will not be able to anneal to the template. Amplification will not occur and no extension will take place.

Question 23.
A schematic representation of PCR up to the extension stage is given below. Give answers of the following questions.
Img 7
(i) Name the process A
(ii) Identify B.
(iii) Identify C and mention its importance in PCR.
Answer:
(i) A – Denaturation process
(ii) B – Primers
(iii) C – Taq DNA polymerase.
Taq polymerase is a thermostable enzyme, which remains active during the high temperature and induces denaturation of DNA.

Differentiate between the following (for complete chapter)

Question 1.
Exonucleases and Endonucleases.
Answer:
Differences between exonucleases and endonucleases are as follows

Exonucleases	Endonucleases
Remove nucleotide from the outer ends of the DNA.	Make cuts at specific positions within the DNA.
It makes cut at only one of the two strands of DNA.	It makes cuts at both strands of DNA simultaneously.
e.g., snake venom and spleen phosphodiesterase.	e.g., deoxyribonuclease 1 and II.

Question 2.
Plasmid and Chromosomal DNA.
Answer:
Differences between plasmid and chromosomal DNA are as follows

Plasmid DNA	Chromosomal DNA
Extrachromosomal circular DNA.	Generally linear.
Not associated with histone proteins.	Associated with histone proteins.
Contains very few genes, but may not be necessary for the cell.	It Consists of complete genome vital for the cellular functions.
Replicates independently.	Replicates with genome.

Question 3.
Electroporation and Microinjection.
Answer:
Differences between electroporation and microinjection are as follows

Electroporation	Microinjection
In this method, electric field is applied to cells in order to increase the permeability of the cell membrane, allowing DNA to be introduced into the cell.	This is a vectorless method of direct delivery of recombinant DNA into plant and animal cells.
In this, brief pulses of high voltage electric currents are passed through the medium.	The foreign DNA is delivered into the nucleus with the help of microinjection or micropipette.
This procedure is mostly used for transforming plant protoplasts.	This method is particularly used for mammalian cells.

Question 4.
Type-I endonuclease and Type II endonuclease.
Answer:
Differences between type-I endonuclease and type-II endonuclease are as follows

Type-I endonuclease	Type-II
They consist of 3 different subunits.	They are structurally simple.
Require ATP, Mg2+, S-adenosyl-methionine for restriction.	Required Mg2+; for restriction.
Recognise specific sequences, but do not cut these sites.	Recognise specific sites and cut them.

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

CHSE Odisha 12th Class Biology Important Questions Chapter 11 Microbes in Human Welfare

Microbes in Human Welfare Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
The bacterium that converts milk into curd is ………. .
(a) Lactobacillus
(b) Azotobacter
(c) Rkizobium
(d) Clostridium
Answer:
(a) Lactobacillus

Question 2.
Fermentation of milk sugar, ……… is done by Lactobacillus.
(a) glucose
(b) fructose
(c) sucrose
(d) lactose
Answer:
(d) lactose

Question 3.
The antibiotic chloramphenicol can be obtained from the bacterium ……. .
(a) Streptomyces griseus
(b) S. aureofaciens
(c) S. venezuelae
(d) S. noursei
Answer:
(c) Streptomyces venezuelae

Question 4.
Acetobacter is involved in the production of
(a) citric acid
(b) acetic acid
(c) gluconic acid
(d) fumaric acid
Answer:
(b) acetic acid

Question 5.
Yeast and Acetobacter are both involved in the production of ……… from carbohydrates.
(a) penicillin
(b) citrate
(c) methane
(d) vinegar
Answer:(d) vinegar

Question 6.
Lipase enzyme is produced by the activity of
(a) Trichoderma viride
(b) Rhizopus sp.
(c) Aspergillus sp.
(d) None of these
Answer:
(b) Rhizopus sp.

Question 7.
The secondary treatment of sewage includes
(a) biological treatment
(b) chemical treatment
(c) filtration
(d) All of the above
Answer:
(a) biological treatment

Question 8.
In primary treatment of sewage, non-biodegradable particles are removed through a process of …………. .
(a) sedimentation
(b) sterilisation
(c) chemical treatment
(d) biological treatment
Answer:
(a) sedimentation

Question 9.
In secondary treatment of sewage in open bioreactors, the microorganisms grow and multiply to form ………. .
(a) manure
(b) compost
(c) sludge
(d) sediment
Answer:
(c) sludge

Question 10.
Milk enzyme that coagulates during curd formation is
(a) protease
(b) casein
(c) rennin
(d) pectinase
Answer:
(c) rennin

Question 11.
Name the bacterium, which produces Swiss cheese.
(a) Saccharomyces cerevisiae
(b) Propionibacterium shermanii
(c) Penicillium roqueforti
(d) Lactobacillus
Answer:
(b) Propionibacterium shermanii

Question 12.
Nystatin is obtained from
(a) Penicillium sp.
(b) Streptomyces griseus
(c) Streptomyces noursei
(d) Aspergillus sp.
Answer:
(c) Streptomyces noursei

Question 13.
The raw material used for the production of wine is
(a) grapes
(b) barley grain
(c) vinegar
(d) None of these
Answer:
(a) grapes

Question 14.
During commercial production of vinegar, secondary fermentation occurs under
(a) anaerobic conditions
(b) aerobic conditions
(c) Both (a) and (b)
(d) unfavourable conditions
Answer:
(b) aerobic conditions

Question 15.
The bacterium which acts on bio-waste to produce biogas is
(a) Methanobacterium
(b) Trichoderma
(c) Bacillus thuringiensis
(d) Azotobacter
Answer:
(a) Methanobacterium

Question 16.
The principal component of biogas is ……….. .
(a) hydrogen
(b) carbon dioxide
(c) ozone
(d) methane
Answer:
(d) methane

Question 17.
Which one of the following microbes is used as biocontrol agent?
(a) Papilloma virus
(b) Baculovirus
(c) Herpes virus
(d) Pox virus
Answer:(b) Baculovirus

Question 18.
Biopesticides are preferred over chemical pesticides due to their
(a) high pest specificity
(b) biodegradability
(c) non-biodegradability
(d) Both (a) and (b)
Answer:
(d) high pest specificity and biodegrdability

Question 19.
Which one of the following fungi makes symbiotic association with the plants?
(a) Glomus
(b) Nostoc
(c) BGA
(d) Rhizobium
Answer:
(a) Glomus

Question 20.
Which one of the following organisms has been commercialised as blood cholesterol lowering agent?
(a) Trichoderma polysporum
(b) Monascus purpureus
(c) Saccharomyces cerevisiae
(d) Aspergillus niger
Answer:
(b) Monascus purpureus

Question 21.
Which one of the following is mainly produced by the activity of anaerobic bacteria on sewage?
(a) Mustard gas
(b) Biogas
(c) Laughing gas
(d) Propane
Answer:
(b) Biogas

Question 22.
Secondary sewage treatment is mainly a
(a) chemical process
(b) biological process
(c) physical process
(d) mechanical process
Answer:
(b) biological process

Question 23.
Organisms called methanogens are most abundant in a
(a) polluted stream
(b) hot spring
(c) sulphur rock
(d) rumen of these
Answer:
(d) rumen of these

Question 24.
A biofertiliser is
(a) Rhizobium
(b) Azotobacter
(c) Nostoc
(d) All of these
Answer:
(d) All of these

Fill in the blanks

Question 1.
Roquefort cheese is ……….. by growing a specific ………… on it.
Answer:
ripened, fungus

Question 2.
The enzyme ……… is used in detergent formulations to remove oily stains from the laundry.
Answer:
lipase

Question 3.
The ……… from secondary treatment of sewage is generally released into ………. water bodies.
Answer:
effluent, natural

Question 4.
During sewage treatement, aerobic microbes are converted into mesh-like structure called ……………. .
Answer:
floes

Question 5.
Species of ………. is used for production of roquefort cheese.
Answer:
Penicillium

Question 6.
………. is an antibiotic that was used to treat soldiers during World War II.
Answer:
Penicillin

Question 7.
………. enzyme produced by Aspergillus niger is used for clarifying bottled juices.
Answer:
Pectinase

Question 8.
…………. is used as a blood cholesterol lowering agent.
Answer:
Statins

Question 9.
During the secondary treatment of primary effluents, the BOD level ……….. .
Answer:
decreases

Question 10.
During sludge digestion, bacteria produce gases like …………. .
Answer:
CH₄,H₂S, CO₂

Question 11.
The technology of biogas production was developed in India by ……….. .
Answer:
Khadi and Village Industries Commision (KVIC)/Indian Agricultural Research Institute (IARI).

Question 12.
The use of biological methods for controlling plant diseases and pests is called ……….. .
Answer:
biocontrol

Question 13.
…………. are used as biocontrol agents to get rid of moquitoes.
Answer:
Dragonflies

Question 14.
Baculovirus is a rod-shaped ………….. DNA virus.
Answer:
double-stranded

Question 15.
Rhizobium is a ………. bacteria that serves as biofertiliser.
Answer:
symbiotic

Question 16.
The first antibiotic was discovered by ………… .
Answer:
Alexander Fleming

Question 17.
……….. are used in biological control of aphids.
Answer:
Ladybirds

Question 18.
Azospirillum and Azotobacter are ………… bacteria.
Answer:
free-living

Question 19.
Plants with mycorrhizal association show tolerance to …………. and ……….. .
Answer:
salinity, drought

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

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

Question 2.
The first antibiotic to be extracted from bacterial culture was nystatin.
Answer:
streptomycin

Question 3.
In ethyl alcohol production, the unicellular fungus, Penicillium is used.
Answer:
Saccharomyces cerevisiae

Question 4.
Acetic acid is produced by Lactobacillus sp.
Answer:
Acetobacter

Question 5.
In the secondary treatment of sewage, unwanted coarse, non-biodegradable particles are removed.
Answer:
primary treatment

Question 6.
Biogas production technology was developed in India mainly by NBRI and CDRI.
Answer:
KVIC, IARI

Question 7.
Methanogens are present in the liver of cattle.
Answer:
rumen

Question 8.
The major component of biogas is carbon dioxide.
Answer:
methane

Question 9.
Nostoc is present in root nodule of leguminous plants and fixes atmospheric nitrogen.
Answer:
Rhizobium

Question 10.
Biopesticides are non-biodegradable compounds.
Answer:
biodegradable

Question 11.
In paddy fields, blue-green algae fix phosphorus to enrich the soil fertility.
Answer:
nitrogen

Question 12.
Bacillus thuringiensis is a Gram-negative bacteria.
Answer:
positive

Question 13.
Baculoviruses are single-stranded RNA viruses.
Answer:
double-stranded DNA

Question 14.
Mycorrhizae provide cobalt to the roots of plants.
Answer:
phosphorus

Question 15.
Microorganisms belonging to the genus-Penicillium are used as agents of biological control.
Answer:
Nucleopolyhedrovirus

Express in one or two word (s)

Question 1.
The drink produced by fermentation of sap of palm tree.
Answer:
Toddy

Question 2.
The source of antibiotic streptomycin.
Answer:
Streptomyces griseus

Question 3.
Name the molecules which are used as blood cholesterol lowering agents?
Answer:
Statins

Question 4.
An enzyme used for the clarifying bottled fruit juice.
Answer:
Pectinase

Question 5.
Which organic acid gets converted into methane in the final step of biogas production?
Answer:
Acetate

Question 6.
Name a bacterium used as biopesticide.
Answer:
Bacillus thuringiensis.

Question 7.
Give an example of free-living fungi present in root ecosystem.
Answer:
Trichoderma

Question 8.
Name any two major components of biogas.
Answer:
CH₄ and CO₂

Question 9.
Name any one biocontrol agent.
Answer:
Baculovirus

Question 10.
A cheese with large holes produced by Propionibacterium shermanii.
Answer:
Swiss cheese

Question 11.
Name any two cyanobacteria used as biofertilisers
Answer:
Nostoc and Anabaena

Short Answer Type Questions

Question 1.
Write a note on fermentation by microbes and its two applications.
Or
Write a short note on fermentation.
Answer:
Fermentation is the process of conversion of carbohydrates to alcohol and CO₂ by some microorganisms in the absence of O₂. Microbes via fermentation are utilised for the synthesis of a number of products valuable for human beings.

Some of the applications of microbes are

• Production of bread using baker’s yeast.
• Wine, beer and other alcoholic drinks are produced by fermentation.

Question 2.
Mention a product of human welfare obtained with the help of each one of the following microbes.
(i) LAB
(ii) Saccharomyces cerevisiae
(iii) Propionibacterium shermanii
(iv) Aspergillus niger
Answer:

Microbe	Product of human welfare
(i) LAB	Curd
(ii) Saccharomyces cerevisiae	Bread, cakes, wines, beer
(iii) Propionibacterium shermanii	Swiss cheese
(iv) Aspergillus niger	Citric acid

Question 3.
How are fermented beverages prepared?
Or
Write a note on alcoholic beverages.
Answer:
Fermented beverages include wine, beer, whisky, brandy and rum which are obtained by fermenting malted cereals and fruit juices with Saccharomyces cerevisiae or ‘ brewer’s yeast to produce ethanol.

The production of variety of alcoholic drinks depends on the type of raw material used and the’type of processing, e.g. wine and beer are produced without distillation. Whisky, brandy and rum are produced by the distillation of the fermented broth.

Question 4.
How streptomycin is obtained?
Answer:
Streptomycin is an antibiotic that is obtained from Streptomyces griseus. The bacteria are cultured on medium containing glucose, soyameal and mineral salts. The pH of the medium is maintained at 7.4 -7.5. The fermentation is carried out under submerged condition for 5-7 days. It is done at 25-30° C.

Question 5.
Write a short note on alcoholic fermentation.
Answer:
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

Question 6.
What are anaerobic sludge digesters?
Answer:
Sludge is the remaining part of organic matter after secondary treatment of sewage. In sludge digesters, other kinds of bacteria, which grow anaerobically, digest the bacteria and the fungi present in the sludge. During this process, bacteria produce a mixture of gases, such as methane, hydrogen sulphide and carbon dioxide which form biogas. It can be used as a source of energy.

Question 7.
Write a short note on secondary treatment of sewage.
Answer:
Secondary treatment of sewage is as follows
(i) This treatment is also known as biological treatment because it involves the use of microbes or biota for the treatment of sewage. The effluent from primary treatment is passed into large aeration tanks, where it is constantly mechanically agitated and air is pumped into it.

(ii) This air helps in the growth of useful aerobic, microbes into floes (masses of bacteria associated with fungal filament to form mesh like structures). The growth of microbes consumes major part of the organic matter, converting it into microbial biomass and releasing lot of minerals. This significantly reduces the BOD (Biochemical Oxyen Demand) of the water.

Question 8.
Write a short note on biogas.
Ans.
It is a complex mixture of gases, like CH₄, CO₂ and H₂, but its major content is methane gas. It is produced by the microbial activity during anaerobic digestion of biomass. Biogas is used as fuel. The type of gas produced by microbes during their growth and metabolism depends upon the microbes and the organic substrates they utilise. Certain bacteria, which grow anaerobically on cellulosic material, produce large amount of methane gas along with CO₂ and H₂ by the process called methanogenesis.

These bacteria are called methanogens and one such common bacterium is Methanobacterium. Methanogens produce large amount of biogas which contains about 50-70% CH₄, 30-40% CO₂, 1-5% H₂ and 0.01% oxygen.

Question 9.
Write a short note on biocontrol agents.
Answer:
Biocontrol agents involve the use of biological methods for controlling plant diseases and pests. Bacteria, fungi and viruses can act as biocontrol agents.
These microbes reduce the target species population through many ecological mechanisms, including pathogenicity, competition, production of chemicals and other interactions.

Question 10.
Write a short note on Bacillus thuringiensis (Bt).
Answer:
Bacillus thuringiensis is a soil-borne, Gram-positive bacterium. It is used to create a transgenic crop plant having resistance to several diseases. The genes encoding for insecticidal properties in the bacterium are isolated and incorporated into crop plants by using advanced biotechnological methods.

The bacterial spores can also be sprayed on the crops. They are ingested by insects and create pores in their gut wall which leads to their death. Thus Br-toxins work as biocontrol agent.

Question 11.
Why is Rhizobium categorised as a symbiotic bacterium? How does it act as a biofertiliser?
Answer:
Rhizobium lives in the root nodules of leguminous plants and fixes the atmospheric nitrogen by converting them into nitrogenous compounds.
These can be utilised by the plants as nutrients. Since, bacteria make the nitrogen available to plants in utilisable form, therefore, it is known as a biofertiliser.

Question 12.
How do plants benefit from having mycorrhizal symbiotic association?
Answer:

• The fungus absorbs phosphorus from the soil and passes it to the plant.
• Plants with mycorrhizal association show resistance to root borne pathogens.
• They show increased tolerance to salinity and drought.

Question 13.
Write short note on biofertiliser.
Answer:
Nutrients in optimum amount are important for the healthy growth, development and maximum productivity of plants.
Many nutrients required for the same include nitrogen and phosphorus which are the constituents of proteins, nucleic acids, coenzymes and some lipids like important biomolecules.
Phosphorus is present as insoluble phosphate in the soil sediments. On the other hand, nitrogen (80%) is present in atmosphere.

However, plants cannot use this atmospheric nitrogen and insoluble phosphates rather they either depend on microbes or chemical fertilisers to meet this basic necessity but, the continuous use of chemical fertilisers causes soil sickness, environmental pollution, etc. Thus, there is a pressure to shift to organic farming. Biofertilisers enrich the nutrient quality of soil by enhancing the availability of nutrients to the crops.

Differentiate between the following (for complete chapter)

Question 1.
Primary and Secondary sewage treatment.
Answer:
Differences between primary and secondary sewage treatment are as follows

Primary sewage treatment	Secondary sewage treatment
It is a physical process.	It is a biological process.
It involves the removal of grit and large pieces of organic matter.	It involves the digestion of organic matter by microbes.
It is carried out by the sedimentation and filtration process.	It is carried out by aerobic and anaerobic biological units.
It is relatively simple and less time consuming process.	It is relatively complex and takes a long time for its completion.

Question 2.
Primary sludge and Activated sludge.
Answer:
Differences between primary sludge and activated sludge are as follows

Primary sludge	Activated sludge
It is sludge formed during primary sewage treatment.	It is sludge formed during secondary sewage treatment.
It does not possess floes of decomposer microbes.	It possesses floes decomposer microbes.
It does not require aeration.	Formation of activated sludge requires aeration.
Little decomposition occurs during the formation of primary sludge.	A lot of decomposition occurs during the formation of activated sludge.

Question 3.
Swiss cheese and Roquefort cheese.
Answer:
Differences between Swiss cheese and Roquefort cheese are as follows

Swiss cheese	Roquefort cheese
They possess large holes.	They have relatively small holes.
These are made by Propionibacterium shermanii	These are ripened by Penicillium roquefort.

Question 4.
Methanogenic bacteria and Symbiotic bacteria.
Answer:
Differences between methanogenic bacteria and symbiotic bacteria are as follows

Methanogenic bacteria	Symbiotic bacteria
There bacteria are involved in energy production.	They are used as biofertilisers.
They live in the rumen of cattle.	They live in the root nodules of leguminous plants.
They produce methane gas by converting cellulose components into simple compounds, e.g. Methanobacterium.	They fix atmospheric nitrogen by converting it into nitrogenous compounds, e.g. Rhizobium.

Question 5.
Biopesticides and Biofertilisers.
Differences between biopesticides and biofertilisers are as follows

Biopesticides	Biofertilisers
They inhibit the growth of pests.	They help in the growth of crop plants.
They increase the yield by killing insects and pests, e.g, Bacillus thuringiensis.	They increase the yield by providing nutrients to the plants, e.g. Nostoc.

Long AnswerType Question

Question 1.
Discuss how waste water treatment can be done.
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.

They need to be treated before their disposal. Conventional methods of waste treatment include cell pits, septic tanks, sewage farms, gravel beds percolating filters and activated sludge process with anaerobic digestion. These methods are less effective and sometimes non-productive.

The domestic sewage contains a large amount of degradable organic compounds. These can be degraded into simpler compounds with the help of microbes. The adequate supply of nutrients, oxygen and other essential compounds helps in the growth of microbes. It consequently enhances the rate of chemical degradation of sewage.

Therefore, modern method includes treatment of sewage before its disposal in local water bodies. Sewage is treated in Sewage Treatment Plants (STPs) in order to make it less polluting. The treatment of waste water is done by the heterotrophic microbes, that are naturally present in the sewage.

This treatment is carried out in the following three stages

Primary Treatment:
It is also known as physical treatment because it basically involves the physical removal of small and large, floating and suspended, non-biodegradable solids from sewage through filtration and sedimentation. Initially, floating debris is removed by sequential filtration. Then, the grit (soil and small pebbles) are removed by sedimentation process in settling tanks. Aluminium or iron sulphate is added in certain places for flocculation.

All solids that settle down during this step form the primary sludge. It traps lots of microbes and debris. The supernatant forms the effluent which is then taken from the primary settling tank for secondary treatment.

Secondary Treatment:
This treatment is also known as biological treatment because it involves the use of microbes or microbiota for the treatment of sewage. The effluent from primary treatment is passed into large aeration tanks, where it Is constantly, mechanically agitated and air is pumped into it.

This air helps in the growth of useful aerobic microbes” into floes (masses of bacteria associated with fungal filament to form mesh-like structures). While growing, these microbes consume major part of the organic matter, converting it into microbial biomass and releasing lot of minerals.
This significantly reduces the Biochemical Oxygen Demand (BOD) of water.

When the BOD of effluent is reduced significantly, it is then passed into a settling tank, where the bacterial ‘floes’ are allowed to sediment. This sediment is called activated sludge. A small part of the activated sludge is then pumped back into the aeration tank to serve as the inoculum. The remaining part of the sludge is pumped back into large tanks called anaerobic sludge digesters, in which other anaerobic bacteria (methanogens) are also present.

They digest the organic mass as well as aerobic microbes (bacteria and fungi of the sludge). During the digestion, mixture of gases like methane (CH₄), hydrogen sulphide (H₂S), carbon dioxide (CO₂), etc., are produced. These gases form biogas that is. used as a source of energy because it is inflammable.
The effluents from secondary treatment plant are released into natural water bodies like rivers and streams.

Tertiary Treatment:
It is an optional process including chemical precipitation. Its effectiveness depends on the number of microbes coming in contact with the pollutant organic molecules. Therefore, the secondary and tertiary treatments are performed in a constantly stirred open bioreactor that are supplied with nutrients.

Other Methods:
Sometimes a percolating or trickling fibre bioreactor is also used. In this method, a stone gravel or plastic sheet is used on which microbes have been immobilised. The sewage is allowed to flow on it. Deep shaft fermentation system is another method of waste water treatment.

This system contains a hole (deep shaft) in the ground. It is divided to allow the cycling and mixing of waste water, air and microbes. In some countries where sunlight hours are high, the algae-bacterial bioreactors are used. The biomass is used in the production of biogas and animal feed.

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

CHSE Odisha 12th Class Biology Important Questions Chapter 10 Improvement in Food Production

Improvement in Food Production Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
Inbreeding increases the frequency of
(a) recessive homozygotes
(b) dominant homozygotes
(c) heterozygotes
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 2.
Heterosis is
(a) hybrid inviability
(b) hybrid sterility
(c) hybrid vigour
(d) hybrid incompatibility
Answer:
(c) hybrid vigour

Question 3.
If you want to develop hybrid seeds within a bisexual flower which of the following parts need to be removed from the same flower?
(a) Stigma
(b) Ovary
(c) Anther
(d) Oviduct
Answer:
(c) Anther

Question 4.
Emasculation is one of the major steps during ………… .
(a) hybridisation
(b) pureline selection
(c) micropropagation
(d) crop improvement
Answer:
(a) hybridisation

Question 5.
Nutritional quality of crop plants like rice, maize, etc. is improved through ………… .
(a) biomagnification
(b) bioremediation
(c) biofortification
(d) bioaccumulation
Answer:
(c) biofortification

Question 6.
The capacity of a plant cell to give rise to a new plant is called …………. .
(a) cotipotency
(b) reproduction
(c) budding
(d) regeneration
Answer:
(a) totipotency

Question 7.
Single Cell Protein (SCP) is
(a) protein obtained from a clone of cells
(b) protein obtained from unicellular organisms
(c) biomass obtained from microorganisms
(d) proteins obtained from biomass of microorganisms
Answer:
(c) biomass obtained from microorganisms

Question 8.
Somatic hybridisation can be used for
(a) gene transfer
(b) transfer of cytoplasm
(c) formation of allopolyploids
(d) All of the above
Answer:
(d) All of the above

Question 9.
The emasculation is required for
(a) purelines
(b) cross pollination
(c) natural hybridisation
(d) selective hybridisation
Answer:
(d) selective hybridisation

Question 10.
An example of improved wheat
(a) Jaya
(b) Pusa Swarnim
(c) Sonalika
(d) Pusa Shubra
Answer:
(c) Sonalika

Question 11.
The organic nutrient of a plant tissue culture medium are
(a) sucrose
(b) glucose
(c) fructose
(d) All of these
Answer:
(d) All of these

Question 12.
Embryo like structure produced in vitro culture is called
(a) embryoid
(b) zygote
(c) somatic hybrid
(d) cybrid
Answer:
(a) embryoid

Question 13.
Which one of the following statement regarding pomato is correct ?
(a) It is a product of somatic hybridisation
(b) It is a product of gene manipulation
(c) product of sexual hybridisation
(d) product of cloning
Answer:
(a) It is a product of somatic hybridisation

Question 14.
An indigenous breed of cattle developed through cross breeding is
(a) Red Dane
(b) Karan Swiss
(c) Jersey (d) Rathi
Ans.
(b) Karan Swiss

Question 15.
The cross-breed of cattle is
(a) Ongole
(b) Sunandini
(c) Tharparkar
(d) Kangayam
Ans.
(b) Sunandini

Question 16.
In India, the milk yield of cattle is low due to
(a) inferior breeds
(b) inadequate food
(c) Both (a) and (b)
(d) use of medicines
Answer:
(c) Both (a) and (b)

Question 17.
Which of the following is a disease of cattle?
(a) Ranikhet disease
(b) Coryza
(c) Marek’s disease
(d) All of these
Answer:
(d) All of these

Question 18.
Fowlpox is caused by
(a) ectoparasites
(b) endoparasites
(c) bacteria
(d) virus
Answer:
(d) virus

Question 19.
Which of the following species of bee is commercially cultivated?
(a) Apis dorsata
(b) Apis mellifera
(c) Apis florea
(d) Apis indica
Answer:
(b) Apis mellifera

Question 20.
Which of the following species of honeybee is reared in artificial hives?
(a) Apis indica
(b) Apis dorsata
(c) Apis florea
(d) None of these
Answer:
(a) Apis indica

Question 21.
The term used for birds raised under domesticaton for economic purpose is called
(a) fisheries
(b) poultry
(c) apiculture
(d) aquaculture
Answer:
(b) poultry

Question 22.
Which of the following breed of buffalo has maximum milk fat percentage in its milk?
(a) Nagpuri
(b) Bhadawari
(c) Mehsana
(d) Murrah
Answer:
(d) Murrah

Question 23.
High milk yielding varieties of cows are obtained by
(a) superovulation
(b) artificial insemination
(c) use of surrogate mothers
(d) All of the above
Answer:
(d) All of the above

Question 24.
Artificial insemination is better than natural insemination in cattle because
(a) semen of good bulls can be provided everywhere
(b) there is no likelihood of contagious diseases
(c) it is economical
(d) All of the above
Answer:
(a) semen of good bulls can be provided everywhere

Question 25.
Which of the following about breeding is incorrect?
(a) By inbreeding purelines cannot be evolved
(b) Continued inbreeding, especially closed inbreeding reduces fertility and productivity
(c) Cross breeding allows desirable qualities of different breeds to be confined
(d) Inbreeding exposes harmful recessive genes that are eliminated by selection
Answer:
(a) By inbreeding purelines cannot be evolved

Question 26.
Which one of the following methods of selection is mainly useful for cross pollinated crops?
(a) Mass selection
(b) Pureline selection
(c) Progeny selection
(d) Clonal selection
Answer:
(b) Pureline selection

Question 27.
Which one of the following breeds of buffaloes is in most demand?
(a) Surti
(b) Jaffrabadi
(c) Murrah
(d) Bhadawari
Answer:
(c) Murrah

Question 28.
Heterosis cannot be maintained in sexually reproducing plants as it disappears on
(a) outbreeding
(b) inbreeding
(c) cross-breeding
(d) None of these
Answer:
(c) cross-breeding

Question 29.
Bagging is done to
(a) achieve desired pollination
(b) prevent contamination of unwanted pollen
(c) avoid self-pollination
(d) avoid cross-pollination
Answer:
(b) prevent contamination of unwanted pollen

Question 30.
A nutritional disease, which is found in poultry birds is
(a) rickets
(b) Ranikhet
(c) fowl cholera
(d) aspergillosis
Answer:
(a) rickets

Question 31.
In honeybees, the drones develop from
(a) fertilised egg
(b) unfertilised egg
(c) schizogony
(d) asexual reproduction
Answer:
(a) fertilised egg

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

Question 1.
The genetically superior individuals are called clones.
Answer:
hybrids

Question 2.
Selection is mixing out plants with desirable characters in a population.
Answer:
sorting out

Question 3.
Hybridisation is a cross between genetically similar organisms.
Answer:
genetically dissimilar

Question 4.
Hybridisation is one of the best methods adopted for crop improvement.
Answer:
Correct, no change

Question 5.
An amorphous mass of parenchyma cells developed by tissue culture is called embryo.
Or
In tissue culture, amorphous mass of thin-walled parenchymatous cells developing from proliferating cells is called explant.
Answer:
Callus

Question 6.
Naked plant cell without cell wall is called plasmalemma.
Answer:
Protoplast

Question 7.
Tissue culture technique was first attempted by Hanning.
Answer:
Haberlandt

Question 8.
Mutational variations are the ones produced during tissue culture.
Answer:
Somaclonal

Question 9.
NDRI is situated in Lucknow.
Answer:
Karnal

Question 10.
The giant honeybee, yielding maximum honey is Apis mellifera.
Answer:
Apis dorsata

Question 11.
In honeybee, the process of development of male bee without fertilisation is termed as swarming.
Answer:
Parthenogenesis

Question 12.
The branch of agriculture which deals with feeding shelter, health and breeding of domestic animals is called animal husbandry.
Answer:
It is correct

Question 13.
Animal food is generally rich in micronutrients.
Answer:
vitamins and roughage

Question 14.
Removal of stamens before release of pollen grains from the plants is called anthesis.
Answer:
emasculation

Question 15.
Mass selection is the simplest method of plant breeding applied mainly in case of self-pollinated crops.
Answer:
cross-pollinated

Express in one or two word(s)

Question 1.
Name the hormones used in tissue culture.
Answer:
2,4-D(auxin) and cytokinin

Question 2.
What is the use of polyethylene glycol in somatic hybridisation?
Answer:
Promotes protoplast fusion

Question 3.
Name an alga used as single cell protein.
Answer:
Chlorella

Question 4.
Which organs are develop during organogenesis?
Answer:
Root and shoot

Question 5.
The birds reared for meat purpose.
Answer:
Broilers.

Question 6.
Breeding between unrelated individuals.
Answer:
Outbreeding.

Question 7.
The important monosaccharide present in – honey.
Answer:
Levulose.

Question 8.
Name the strategy used to increase the homozygosity in cattles for desired traits.
Answer:
Inbreeding

Question 9.
Name an exotic breed of cow.
Answer:
Jersey

Question 10.
List any two economically important products for human obtained from Apis indica.
Answer:
Honey and beeswax

Question 11.
Name any two poultry diseases.
Answer:
Fowl pox and Ranikhet disease

Fill in the blanks

Question 1.
During hybridisation, emasculated buds need to be ……….. .
Answer:
bagged

Question 2.
Technique for production of disease-free plants is …………… .
Answer:
tissue culture

Question 3.
Crosses between the plants of the same variety are called ………. .
Answer:
inter-varietal

Question 4.
The most convenient way of removal of stamens is ……… .
Answer:
emasculation

Question 5.
Continued close inbreeding reduces fertility and productivity which is called ……….. .
Answer:
inbreeding depression

Question 6.
Milk yielding cattle breeds are known as ………. breeds.
Answer:
milch

Question 7.
Exotic breed of poultry bird is ……….. .
Ans. playmouth rock

Question 8.
The juvenile bees are reared in ……….. chamber of the honeycomb.
Answer:
brood

Question 9.
In bees, dance is meant for ………… .
Answer:
direction and distance

Question 10.
The loss of vigour due to continuous inbreeding is known as …………… .
Answer:
inbreding depression

Question 11.
The characteristic flight of the queen bee during fertilisation is known as ……………… .
Answer:
nuptial flight

Question 12.
Culturing of honeybee on commercial basis is known as …………. .
Answer:
apiculture

Short Answer Type Questions

Question 1.
Write note on hybridisation with 2-3 important points.
Answer:
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

• To produce variations in progeny which are useful.
  It is achieved by recombination of characters.
• To make the use of hybrid vigour which is the superiority of progeny over its parents.
• To develop high yielding varieties which are also resistant to diseases.

Question 2.
Write a short note on 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.
(i) Mention two ways of inducing artificial mutation in a crop field.
(ii) List two steps that help in introducing the desired mutation into the crop.
Answer:
(i) Artificial mutation can be induced in a crop field by

• Using chemicals (like aniline).
• Radiations (like gamma-radiations).

(ii) Two steps that help in introducing the desired mutation into the crops are

• Screening the plant for resistance.
• Selecting the desirable plants for multiplication and for breeding.

Question 4.
How has mutation breeding helped in improving the production of mung bean crop?
Answer:
Mutation breeding is a phenomenon by which genetic variation is achieved through changes in base sequences within genes.
This creates a new character or trait absent in parental generation. It is the process of breeding by artificially inducing mutations using chemicals or radiations, e.g. in moong bean, resistance to yellow mosaic virus and powdery mildew were introduced by this method.

Question 5.
Discuss the importance of testing of new plant varieties in a geographically vast country like India.
Answer:
Before the new plants are generated through the plant breeding programmes, they need to be evaluated for their yield and other agronomic traits of quality, disease resistance, etc. The testing is done on the farmer’s field for atleast three growing seasons, at different locations in the country representing all the agroclimatic zones, where the crop is usually grown.

The material is evaluated in comparison to the best available local crop cultivar known as a check or reference cultivator.

Question 6.
What is meant by biofortification? Explain.
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

• Protein, oil content and quality.
• Vitamin content.
• Micronutrients and mineral content.

Question 7.
Why are biofortified maize and wheat considered nutritionally improved?
Answer:
Biofortified maize and wheat are considered nutritionally improved, because of following reasons

• Maize hybrids have twice the amount of amino acid, lysine and tryptophan as compared to existing maize hybrids.
• Atlas-66 has been used as a donor for developing wheat varieties with improved protein content.

Question 8.
Enumerate four objectives for improving the nutritional quality of different crops for the health benefits of the human population by the process of ‘biofortification’.
Answer:
Four objectives for improving nutritional quality of crops

• Protein content and quality.
• Oil content and quality.
• Vitamin content.
• Micronutrient and mineral content.

Question 9.
Write a note on single cell protein.
Answer:
Microbial biomass passess about 45-55% protein, but in certain bacteria, the protein content may be upto almost 80%. The term ‘single cell protein’ refers to microbial biomass which acts as a source of mixed proteins. They are extracted from pure or mixed culture of organisms or cells.

SCP act as a supplement or alternative source of protein that is not supplied by the traditional or conventional agriculture production. Mirobes are being grown commercially as a sources of SCP. They are

• Yeast Saccharomyces cerevisiae
• Filamentous fungi Eusarum graminearum
• Bacteria Methylophilus methybtrophus
• Cyanobacteria Spirulina
• Algae Chlorella

Question 10.
How can healthy potato plants be obtained from a desired potato variety which is viral infected? Explain.
Answer:
Healthy potato plants can be obtained from a desired potato variety which is viral infected with the help of tissue culture.
The apical and axillary meristems of the infected plant remain virus free. Hence, they are removed and grown in vitro to obtain healthy potato plants. This is one of the application of tissue culture.

Question 11.
Write a note on micropropagation with 2-3 important points.
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 12.
(i) Why are the plants raised through micropropagation termed as somaclones?
(ii) Mention two advantages of this technique.
Answer:
(i) The plants produced by micropropagation are genetically identical to the original plant from which they are grown, so they are called somaclones.

(ii) Advantages of this technique are

• More number of plants can be produced in a short time
• Disease free plants can be developed from diseased plants. Also seedless plants can be multiplied.

Question 13.
Why are plants obtained by protoplast culture called somatic hybrids?
Answer:
Plants obtained by protoplast culture are called somatic hybrids because they are formed by the fusion of isolated protoplasts from two different varieties of plants, each having a desirable character, to obtain a hybrid protoplast which can be further grown to form a plant.

Question 14.
Name the steps represented in the following process.

Answer:
A – Collection of germplasm.
B – Cross hybridisation among selected plants.
C – Selection and testing of superior recombinants.

Question 15.
Explain in brief the role of animal husbandry in human welfare.
Answer:
Domesticated animals of any breed or population of animal, which are intentionally kept in an agricultural setting for the benefit of human beings are referred to as ‘lifestock’. The agricultural practice of breeding and raising livestock is called animal husbandry.

It deals with care and breeding of animals like buffaloes, cows, pigs, horses, cattle, sheep, camels, goats, etc. by humans for various products like milk, meat, wool, etc. Poultry farming and fisheries are also considered as the

part of animal husbandry. According to a survey, about 70% of world’s livestock population is in India and China. Despite this huge production, their contribution to the world farm produce is only 25%, i.e the productivity per unit is very low.

Question 16.
What is meant by the term breed? What are the objectives of animal breeding?
Answer:
A breed is a group of animals related by descent and similar in most character like general appearance, features, size configuration, etc. Red Dane, Jersey, Brown Swiss are the examples of foreign breeds of cow.

Objectives of animal breeding are

• To increase the yield of animals.
• To produce disease resistant varieties of animals.
• To improve the desirable qualities of the animal produce such as milk, etc., having high protein content, etc.

Question 17.
What is the difference between breed and species? Give an example for each category.
Answer:
Breed is a specific group of animals or plants having homogeneous appearance, behaviour and other characteristics that distinguishes it from other animals or plants of the same species, e.g. Red Dane, Jersey, Brown Swiss are some common breeds of cow.

Species is one of the basic units of biological classification and a taxonomic rank. It can be defined as the largest group of organisms capable of interbreeding and producing fertile offspring, e.g. lion, cow, dog.

Question 18.
Enumerate any six essentials of good, effective ‘dairy farm management practices?
Answer:
Dairying is the management of animals for milk and its products for human consumption. Dairy farming integrated with agricultural farming has been the base of Indian economy since long time. It mainly deals with the processes and systems to improve quality and quantity of milk. Milk yield mainly depends on the quality of breeds. The dairy farm management includes the following main processes

1. Selection of good breeds with high yielding potential (under the climatic conditions of the area) and resistance to various diseases.
2. Cattle should be housed-well, have sufficient water and should be kept in diseased-free conditions.
3. They should be fed in a scientific manner with an emphasis on quality and quantity of fodder.
4. Regular inspection and keeping proper records of all the activities of dairy is also important.
5. Regular visits of a veterinary doctor is necessary.
6. Stringent cleanliness and hygiene of both the cattle and the handler are very important during milking, storage and transport of milk and its products.

Question 19.
Enlist the steps involved in inbreeding of cattle. Suggest two disadvantages of this practice.
Answer:
The mating of closely related animals within the same breed for 4-6 generations is called inbreeding. The strategies for inbreeding are

1. Superior males and females of the same breed are identified and then mated.
2. The progeny obtained from such type of matings are evaluated and superior males and females among them are identified for further mating.
3. In case of cattle, more milk per lactation is the criteria for superior female for cow and buffalo. Whereas a superior male is the one who give rise to superior progeny.

When the inbreeding is repeated it is called upgrading. Inbreeding is advantageous in the introduction of beneficial genes without changing the original genetic composition. But, inbreeding has disadvantages also. Inbreeding may lead to the expression of harmful effects of the deleterious gene. This results in inbreeding depression as it brings harmful recessive genes together. It cause decrease fertility and hybrid vigour of, in cattle.

Question 20.
State the disadvantage of inbreeding among cattle. How it can be overcome?
Or
How does inbreeding depression set in? Mention the procedure you would suggest to reverse this.
Answer:
Inbreeding depression usually reduces fertility and even productivity. It sets in due to continued inbreeding especially close inbreeding.
When inbreeding depression takes place or happens, selected animals of the breeding population should be mated with the unrelated superior animals of the same breed. This will help to restore the fertility and yield.

Question 21.
Why interspecific crosses are rare in nature and intergeneric crosses almost unknown?
Answer:
In interspecific crosses, male and female animals of two different related species are mated. The resultant progeny may combine desirable features of both the parents are infertile. Thus are rare in nature.
The same applies to intergeneric crosses. It is the crossing of two different animals/plants of different genus. It is almost unknown in nature as the gametes show species specificity.

Question 22.
What is artificial insemination? Mention two ways in which it is useful in breeding of dairy animals.
Answer:
Artificial insemination is a method of improvement in animals by using controlled breeding methods. During this process, the semen collected from the selected superior quality male parent is injected into the reproductive tract of selected female parent by the breeder. Breeding of dairy animals by the artificial insemination methods is useful in following two main ways

1. The bull of superior quality can be made to inseminate with many cows so, as to produce a large number of offsprings with desired trait.
2. The offsprings produced by artificial insemination have higher yield as compared to the normal offspring.

Question 23.
What is meant by transgenic animals?
Answer:
Refer to text on page no. 280.

Question 24.
What is the utility of transgenic animals?
Answer:
Transgenic animals are used to obtain various products like CC-antitrypsin, haemoglobin, iron binding protein, lacteferrin, etc. In Japan, gynogenesis is being used to improve fish size.

Question 25.
What is apiculture? How is it important in our lives?
Answer:
Refer to text on page no. 280.

Question 26.
Briefly describe swarming in honeybees.
Answer:
During early summer, when the beehive becomes loaded with honey and overcrowded by bees, the queen leave the hive with some drones and workers to establish a new colony at some other new place. This process is called as swarming.

Question 27.
Why are beehives kept in a crop field during flowering period ? Name any two crop fields where this is practised.
Answer:
Bees, while collecting nectar from flowers, transfer the pollen grains. Beehives are kept in a crop field during flowering period to enhance the pollination of the crop, which increases the crop yield. Also, bees can easily collect huge amounts of nectar from .the flowers of the crop in a close reach without much foraging.
This in turn increases the production of honey, e.g. this technique is practised in apple and mustard fields.

Question 28.
Social life of honeybees.
Or
Write a note on queen bee.
Answer:
Honeybees are social and polymorphic insects. They are of three main types

1. Queen It is fertile female. It is diploid as it develops from a fertilised egg. It feeds on royal jelly. It has well-developed ovary for laying eggs.
2. Drone It is a fertile male. It is haploid as it develops from unfertilised egg by the process of parthenogenesis. It copulates with queen.
3. Worker It is a sterile female. It is diploid. They are workers of beehives.

Worker bees have some specific structure such as long proboscis to collect nectar pollen basket to collect pollen, wax secreting glands in abdomen to secrete wax and powerful sting for defence.

Question 29.
What is beeswax?
Answer:
Beeswax is a secretory product of hypodermal glands of workers bees abdomen. It is used in the industries for the manufacture of cosmetics, polishes, paints ointments and lubricants.

Differentiate between the following (for complete chapter)

Question 1.
Hybrids and Cybrids.
Answer:
Differences between hybrids and cybrids are as follows

Hybrids	Cybrids
When nuclear and cytoplasmic fusion occur, the product is known as hybrid.	When only cytoplasmic fusion occurs which may be followed by the loss of any one of the nucleus, it is known as cybrid.

Question 2.
Inbreeding and Outbreeding.
Answer:
Differences between inbreeding and outbreeding are as follows

Inbreeding	Outbreeding
Mating of closely related animals of the same breed is called inbreeding.	Mating of unrelated animals belonging to different breeds or different species is called outbreeding.
It Increases homozygosity by which the prepotency of inbreed line increases.	It increases heterozygosity which results in hybrid vigour.
It reduces genetic variance within lines.	It increases genetic variance within lines.

Question 3.
Explant and Callus.
Answer:
Differences between explant and callus are as follows

Explant	Callus
These are small pieces of plant part of tissues that are aseptically cut and used to propagate a plant in vitro.	It is an unorganised and undifferentiated mass of proliferative cells produced when explants are grown on artificial culture medium in vitro.
Explants are readily available from parent plants.	Callus is obtained from culturing of explant within 2-3 weeks.

Question 4.
Callus culture and Suspension culture.
Answer:
Differences between callus culture and suspension culture are as follows

Callus culture	Suspension culture
In this culture, cell division in explant forms a callus. Callus is an irregular unorganised and undifferentiated mass of actively dividing cells.	It consists of single cells and small groups of cells suspended in a liquid medium.
The culture is maintained in agar medium	The culture is maintained in liquid medium
The medium contains growth regulators the auxin such as 2,4-D and cytokinin like BAP.	The medium contains growth regulator auxin such as 2, 4-D only.
Callus is obtained within 2-3 weeks.	Suspension cultrue grow much’faster than callus culture.
It does not need to be agitated.	It must be constantly agitated at 100-250 rpm (revolutions per minute).

Question 5.
Somatic embryogenesis and Somatic hybridisation.
Answer:
Differences between somatic embryogenesis and somatic hybridisation are as follows

Somatic Embryogenesis	Somatic hybridisation
A Somatic Embryo (SE) is an embryo derived from a somatic cell, other than zygote.	Somatic hybrids are formed through the fusion of protoplasts of two plants belonging to different varieties, species and even genera.
Somatic embryos are obtained usually on culture of the somatic cells in vitro, this process is called somatic embryogenesis	The process through which somatic hybrids are formed is known as somatic hybridisation
It is induced by a relatively high concentration of auxin, like 2, 4-D.	The fused protoplasts are either cultured in liquid or semi-liquid agar medium plates.

Question 6.
Broilers and Layers.
Answer:
Differences between broilers and layers are as follows

Broilers	Layers
These are comprised of both male and females.	These are females only.
Broilers are reared for meat production.	Layers are reared for egg production.
These attain a body weight of 2.2-2.4 kg within 6 weeks, this is due to faster growth.	Layers attain a body weight of approximately 1.5-1.8 kg at sexual maturity (20-2 weeks).

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

CHSE Odisha 12th Class Biology Important Questions Chapter 9 Health and Diseases

Health and Diseases Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

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

Question 2.
The malignant malaria is caused by
(a) Plasmodium vivax
(b) Plasmodium malariae
(c) Plasmodium ovale
(d) Plasmodium falciparum
Answer:
(d) Plasmodium falciparum

Question 3.
Which is the infective stage of Plasmodium falciparum?
(a) Sporozoite
(b) Trophozoite
(c) Cryptozoite
(d) Merozoite
Answer:
(a) Sporozoite

Question 4
………… causes common cold.
(a) Virus
(b) Bacteria
(c) Protozoa
(d) Fungus
Answer:
(a) Virus

Question 5.
The confirmative test for typhoid is ………….. .
(a) Widal
(b) Mantoux
(c) CBC
(d) PCR
Ans.
(a) Widal

Question 6.
Through which vector is Wuchereria bancrofti transmitted?
(a) Anopheles
(b) Aedes
(c) Culex
(d) Tes-tse fly
Answer:
(c) Culex

Question 7.
The disease filariasis is caused by
(a) Treponema pallidum
(b) Neisseria gonorrhoeae
(c) Mycobacterium leprae
(d) Wuchereria bancrofti
Answer:
(d) Wuchereria bancrofti

Question 8.
Which parasitic causes amoebic dysentery?
(a) Escherichia coli
(b) Amoeba proteus
(c) Entamoeba histolytica
(d) Plasmodium vivax
Answer:
(c) Entamoeba histolytica

Question 9.
Which of the following is not a congenital disease?
(a) Haemophilia
(b) Down’s syndrome
(c) Cold
(d) Colour blindness
Answer:
(c) Cold

Question 10.
Common symptoms of typhoid are
(a) high fever and weakness
(b) stomach pain and constipation
(c) headache and loss of appetite
(d) All of the above
Answer:
(d) All of the above

Question 11.
Female Anopheles is a vector of
(a) fllariasis
(b) malaria
(c) typhoid
(d) AIDS
Answer:
(b) malaria

Question 12.
Which of the following is a pair of bacterial diseases?
(a) Typhoid and pneumonia
(b) Malaria and AIDS
(c) Ringworm and AIDS
(d) Cold and malaria
Answer:
(a) Typhoid and pneumonia

Question 13.
During allergic reactions which of the following is secreted?
(a) Allergens
(b) Histamines
(c) Immunoglobulins
(d) Pyrogens
Answer:
(b) Histamines

Question 14.
Cancer is
(a) non-malignant tumour
(b) controlled division of cells
(c) unrestrained division of cells
(d) microbial infection
Answer:
(c) unrestrained division of cells

Question 15.
The spread of cancerous cells to distant sites is termed as
(a) metamorphosis
(b) metagenesis
(c) metastasis
(d) metachrosis
Answer:
(c) metastasis

Question 16.
Blood cancer is called
(a) leukaemia
(b) haemophilia
(c) thrombosis
(d) haemolysis
Answer:
(a) leukaemia

Question 17.
Immunodeficiency syndrome could develop due to
(a) defective thymus
(b) defective liver
(c) weak immune system
(d) AIDS virus
Answer
(d) AIDS virus

Question 18.
Cirrhosis of liver is caused by the chronic intake of
(a) alcohol
(b) tobacco (chewing)
(c) cocaine
(d) opium
Ans.
(a) alcohol

Question 19.
Caffeine is a stimulant present in
(a) coffee
(b) tea
(c) cold drinks
(d) All of these
Ans.
(d) All of these

Question 20.
Cannabis sativa is the source of
(a) opium
(b) LSD
(c) marijuana
(d) cocaine
Ans.
(c) marijuana

Question 21.
The drug which is used for reducing pain is
(a) opium
(b) hashish
(c) bhang
(d) marijuana
Ans.
(a) opium

Question 22.
A tranquilliser is drug which
(a) relieves pain
(b) gives soothing effect to mind
(c) induces sleep
(d) has stimulating effect
Answer:
(a) relieves pain

Question 23.
The drug changes the person’s thought is
(a) cocaine
(b) barbiturate
(c) hallucinogens
(d) insulin
Answer:
(c) hallucinogens

Question 24.
Which one of the following is not correctly matched?
(a) Glossina palpalis – Sleeping sickness
(b) Culex – Filariasis
(c) Aedes – Yellow fever
(d) Anopheles – Leishmaniasis
Answer:
(d) Anopheles – Leishmaniasis

Question 25.
Stomach cleans pathogen by
(a) HCl
(b) gastric hormones
(c) Both (a) and (b)
(d) None of these
Answer:
(b) gastric hormones

Question 26.
Ability of body to fight against disease is called
(a) susceptibility
(b) immunity
(c) vulnerability
(d) irritability
Answer:
(b) immunity

Question 27.
Antigen binding site is present on antibody between
(a) two heavy chains
(b) two light chains
(c) one heavy and one light chain
(d) normal chains
Answer:
(b) two light chains

Question 28.
Which of the following is an opiate narcotic?
(a) Morphine
(b) LSD
(c) Amphetamines
(d) Basbiturates
Answer:
(a) Morphine

Question 29.
Which part of the brain is involved in loss of control when a person drinks alcohol?
(a) Cerebrum
(b) Medulla oblongata
(c) Cerebellum
(d) Pons Varolii
Ans.
(c) Cerebellum

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

Question 1.
The diseases which occur due to change in chromosomal structure are called degenerative diseases.
Answer:
congenital diseases

Question 2.
Plasmodicum vivax causes cerebral malaria.
Answer:
Plasmodium falciparum

Question 3.
The toxin released due to rupture of RBCs in malaria is haemoglobin.
Answer:
haemozoin

Question 4.
The other name for filariasis is amoebic dysentery.
Answer:
elephantiasis

Question 5.
Ringworm is a viral disease.
Answer:
fungal disease

Question 6.
Typhoid is diagnosed by Mantoux test.
Answer:
Widal test

Question 7.
Antibody-mediated immunity helps the body to differentiate between self and non-self cells during organ transplantation.
Answer:
Cell-mediated immunity

Question 8.
Cellular changes in body as a result of any wound or injury is called immunisation.
Answer:
inflammation.

Question 9.
Cellular barriers forms first line of defence.
Answer:
Physical barriers.

Question 10.
Viral oncogenes on activation lead to tumour formation.
Answer:
Cellular oncogenes

Question 11.
Which test is conducted to identify HIV ?
Answer:
ELISA

Question 12.
HIV is treated using a combination of madicines called antibacterial therapy.
Answer:
antiretroviral.

Question 13.
The ookinete penetrates through the stomach wall and encysts as an gamete.
Answer:
oocyst

Question 14.
Active immunisation against tetanus and diphtheria is achieved by antibiotics.
Answer:
exotoxins or toxoides

Question 15.
Caffeine is used to make tobacco.
Answer:
Nicotiana tabacum is used to make tobacco.

Question 16.
The mood altering drugs are tranquilisers.
Answer:
The mood altering drugs are psychotropic drugs.

Question 17.
Benzodiazepines is used as a stimulant.
Answer:
Cocaine is used as a stimulant.

Question 18.
Antibody production is assisted by monocytes.
Answer:
B-cells

Question 19.
Cell-mediated immunity is mainly function of paratope.
Answer:
T-cells

Express in one or two word(s)

Question 1.
What is called protein pathogen that does not contain nucleic acid?
Answer:
Prions.

Question 2.
The nature of spread of communicable disease.
Answer:
Epidemiology

Question 3.
Name the test specifically employed to determine the presence of disease causing Salmonella typhi.
Answer:
WIDAL Test

Question 4.
Causative organism of malignant malaria.
Answer:
Plasmodium falciparum

Question 5.
Ringworms belong to the fungal genus.
Answer:
Microsporum

Question 6.
The disease transmitted through contact.
Answer:
Measles

Question 7.
An oral dose of drug given for amoebiasis treatment.
Answer:
Metronidazole

Question 8.
Colostrum is rich in which type of antibody(20l9>
Answer:
IgA

Question 9.
The type of immunisation performed for treating snake bite.
Answer:
Passive immunisation.

Question 10.
The type of immunity responsible for graft rejection.
Answer:
Cell-mediated immunity.

Question 11.
The genetic material of HIV?
(Only DNA, only RNA, Both DNA and RNA, Nucleoproteins)
Answer:
Only single-stranded RNA.

Question 12.
The tissue affected in sarcoma.
Answer:
Bone, muscle and lymphnode

Question 14.
The test conducted to detect HIV infection.
Answer:
ELISA, PCR test

Question 15.
What is the source of LSD?
Answer:
Claviceps purpurea

Question 16.
Give one psychological disorder that occurs due to drug addiction.
Answer:
Epilepsy

Question 17.
Which type of drug is used as tranquliser?
Answer:
Benzodiazepines

Question 18.
Name the drug obtained from coca plant.
Answer:
Cocaine

Fill in the blanks

Question 1.
The asexual cycle of Plasmodium in its primary host …………. .
Answer:
humans

Question 2.
The infective stage of malaria parasite is ………….. .
Answer:
sporozoite

Question 3.
Filariasis is caused by ………… .
Answer:
Wuchereria bancrofii

Question 4.
A parasite …………. causes abdominal pain, constipation, cramps, faeces with excess mucus and blood clots.
Answer:
Entamoeba histolytica

Question 5.
A disease causing agent is called …………. .
Answer:
pathogen

Question 6.
The vaccine used against typhoid fever is …………. .
Answer:
Vi antigen

Question 7.
…………. is acquired through vaccines which generate antibodies when introduced in body.
Answer:
Artificial active immunity

Question 8.
IgE are the antibodies involved in ………….. .
Answer:
allergic reaction

Question 9.
Cancer of muscle is named as …………… .
Answer:
sarcoma

Question 10.
………………. tests are conducted to know number of cell counts during cancer.
Answer:
Blood and bone marrow

Question 11.
HIV virus belongs to a group of …………… .
Answer:
retrovirus

Question 12.
Anti Tetanus Serum (ATS) administration generates ………….. immunity in the body.
Answer:
artificial passive

Question 13.
AIDS virus has ……….. RNA.
Answer:
single-stranded

Question 14.
…………. is the most common skin problem that occur during adolescence.
Answer:
Acne

Question 15.
The strong sense of fear in reference to a particular situation or thing is called …………… .
Answer:
phobia

Question 16.
LSD is a natural drug.
Answer:
psychedlic/hallucinogens

Question 17.
Cocaine is obtained from the plant
Answer:
Erythroxylon coca

Question 18.
…………. are antisleep drugs.
Answer:
Amphetamines

Question 19.
Tobacco is obtained from ……………. .
Answer:
Nicotiana tabacum leaves

Question 20.
Pathogen causing ascariasis is ………….. .
Answer:
Ascaris lumbricoides

Question 21.
Phenomenon of rejection of self cells is called ………….. .
Answer:
Autoimmunity

Short Answer Type Questions

Question 1.
Define health and disease.
Answer:

• Health is defined as a state of complete physical, mental and social well-being.
• Disease is a state when functioning of one or more organs or systems of the body is adversely affected.

Question 3.
What are the two basic groups of diseases? Give one example of each group.
Answer:
Two basic groups of diseases are

• Infectious diseases are those which has the ability of transmitting from one person to another, e.g. AIDS, common cold, etc.
• Non-infectious diseases which does not have the ability of transmitting from one person to another, e.g. cancer, diabetes, etc.

Question 4.
Write a short note on pathogens.
Answer:
Organisms that cause infectious diseases are known as pathogens. These can harm any living individual or organism by living on them or living in them.
These disrupt the normal physiology of organisms either plants or animals and express certain symptoms. These enter in our body through some source like air, water, food, soil, etc.
The major classes of pathogens are viruses, bacteria, fungi, prions and parasitic.

Question 5.
Name two bacterial diseases of human.
Answer:
Two bacterial diseases of human are

• Typhoid It is caused by bacterium (Salmonella typhi) which enters the body through food.
• Pneumonia It is caused by bacterium (Streptococcus pneumoniae) which infect alveoli of the lungs.

Question 6.
State the symptoms of typhoid.
Answer:
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.

Question 7.
Answer the following
(i) Name the stage of Plasmodium that gains entry into the human body.
(ii) Explain the cause of periodic recurrence of chill and high fever during malaria attack in human.
Answer:
(i) Sporozoite stage.
(ii) When the Plasmodium enters the RBCs, it causes the rupture of red blood cells. The rupture of RBCs, is associated with the release of chemical haemozoin which causes frequent chills and high fever.

Question 8.
Name the toxin responsible for the appearance of symptoms of malaria in human. Why do these symptoms occur periodically?
Answer:
The toxin responsible for symptoms of malaria is haemozoin. It is released when RBCs get ruptured due to erythrocytic schizogony of Plasmodium that takes place every 48 hours. This is the reason behind periodic occurrence of symptoms.

Question 9.
What is the causative organism of amoebiasis? State the symptoms of the disease.
Answer:

• Amoebiasis is caused by an intestinal endoparasite, Entamoeba histolytica.
• Symptoms of amoebiasis are abdominal pain constipation, cramps, faeces with excess mucous and blood clots.

Question 10.
List the symptoms of ascariasis. How does a healthy person acquire this infection?
Answer:
Symptoms
Most of the patients during light infection do not show any symptoms of Ascaris. However, patients with moderate to heavy infections show some symptoms depending upon the infected organ

1. If intestine is infected Ascaris eggs reach to small intestine through contaminated food or water. They hatch and develop from larva to adult worms in small intestine and remain there till they die. In mild ascariasis, symptoms are mild abdominal pain, nausea and vomitting or diarrhoea with blood stool.
In severe infection, large number of worms are present in a person which may cause severe abdominal pain, fatigue, vomitting and weight loss.

2. If lungs are infected After ingestion of Ascaris eggs, these hatch into larvae in small intestine. These larvae migrate into lungs via blood or lymph. At this stage, symptoms are similar to asthma or pneumonia with cough, breathlessness and wheezing. After 6-10 days in lungs, larvae travel into throat where these are coughed up and swallowed.

Question 11.
(i) What is immunity?
(ii) Give an account of cell-mediated immunity.
Answer:
(i) Immunity is the capactiy of an organism to resist or defend itself from the development of a disease. It has two main types, i.e. innate immunity and acquired immunity.

(ii) Immune response is the specific reactivity induced in a host by an antigentic stimulus. The immune response is of following types
• Humoral Antibody Mediated Immunity (AMI)
• Cell-Mediated Immunity (CMI)

Cell-Mediated Immunity:
It is the responsibility of a sub-group of T-cell, called T-cytotoxic cells. An activated T-cytotoxic cell is specific to a target cell which has been infected and kills the target cell by a variety of mechanisms.

It prevents the completion of life cycle of the pathogen since it depends on an intact host cell. Cell mediated immunity is also invovled in killing of cancer cells. T-cells attack the following
(a) Cells that have become infected by a microorganism most commonly a virus.
(b) Transplanted organs and tissues.
(c) Cancer causing cells.
The whole cell is involved in the attack, so this type of immunityis described as cell-mediated immunity.
T-cells do not relase antibodies.

Question 12.
Write a short note on innate immunity.
Or What is 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 Cells
These are non-phagocytic granular lymphocytes which are present in spleen, lymph nodes and bone marrow.

Question 13.
What are interferons?
Ans.
Interferons are antiviral glycoproteins released by the virus infected host cells. They do not inactivate to kill the virus but enter the neighbouring uninfected host cells to prevent viral multiplication in them.
Interferons are host specific as they are produced by one host and will not work in another host.

Q 14.
Mention the origin and importance of T-cells.
Ans.
T-cells or T-lymphocytes are originates in bone marrow in immature forms. In thymus, these cells are transformed into mature T-!ymphocytes.

There are four types of T-cells, i.e. helper T-cells, suppressor T-cells, memory T-cells and cytotoxic T-cells. Helper T-cells stimulate B-cells to produce antibodies and cytotoxic T-killer cells attack foreign cells.
Suppressor T-cells suppress the functions of cytotoxic and helper T-cells and memory cells recognise the original invading antigen to create a more intense immune response.

Question 15.
What are immunoglobins?
Or What is antibody?
Answer:
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 16.
Why is mother’s milk considered the most appropriate food for a newborn infant?
Answer:
Mother’s milk is considered most appropriate for a newborn infant as it provides immunity in the initial period of its life. The yellowish fluid colostrum secreted by mother during the initial days ofdactation has abundant antibodies (IgA) to protect the infant from primary infections like cold, flu, etc.

Question 17.
Write a short note on vaccination.
Answer:
Vaccination refers to the process of injecting a biological or chemical agent that enhances the immunity of a person against the specific disease. It generally involves injection of dead organisms into . host organism.
Vaccination is the best preventive measure against any disease. A number of specific vaccines are given against many disease, e.g. polio vaccine, diphtheria vaccine, etc.

Question 18.
Why are tumour cells dangerous?
Answer:
Tumour cells are dangerous due to the following reasons

• These grow rapidly and also damage normal cells.
• Cells from tumours can get sloughed off and may reach other parts of the body via blood and start a new tumour, i.e. metastasis.
• Tumour cells compete for nutrients and may starve the other cells.

Question 19.
Write a short note on types and causes of cancer.
Or Write short note on 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.

Question 20.
What is HIV?
Answer:
The HIV name was given in 1986 to AIDS virus. The two more strains of HIV, namely HIV 1 and 2 have been discovered. HIV viruses have originated from non-primates from West-central Africa and transferred to human. HIV 1 is more pathogenic and distributed world-wide, while HIV 2 is less prevalent and less pathogenic. It is distributed in West Africa only.

Question 21.
How is the AIDS provirus formed?
Answer:
HIV first attaches to the host cell surface by adsorption. It then enters CD 4 cells and the HIV RNA changes to DNA (provirus). This HIV provirus gets inserted into DNA of CD \ cell and them begins to replicate using the host’s cellular.

Question 22.
How AIDS can be prevented?
Answer:
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.

Question 23.
Write short note on drug abuse.
Answer:
Drug Abuse and Addiction:
Drugs are chemicals which are used in the treatment of a disease under the supervision of a physician. But prolonged unnecessary use of a drug makes a person dependent and an addict to that drug.

Question 24.
From which plant cocaine is obtained? Why sports persons are often known to abuse this drug?
Answer:
Cocaine is derived from the leaves and young branches of South American plant Erythroxylum coca or coca plant.
Sports persons abuse this drug because it is a strong stimulant and acts on central nervous system, producing a sense of increased energy. It is highly addictive.

Question 25.
Write the source and effects of following drugs.

1. Morphine
2. Cocaine
3. Marijuana

Answer:

1. Morphine It is obtained from latex of Papaver somniferum.
   It has a sedative effects that slows down the body function.
2. Cocaine is obtained from plant Erythroxylon coca. It affects central nervous system and produces increased sense of happiness.
3. Marijuana It is obtained from Cannabis sativa and affects cardiovascular system of the body.

Differentiate between the following (for complete chapter)

Question 1.
Congenital diseases and Acquired diseases.
Answer:
Differences between congenital diseases and acquired diseases are as follows

Congenital diseases	Acquired diseases
Diseases present from birth	Disease occurs only after birth.
These are occur due to gene or chromosomal mutations.	These are non-heritable but are caused due to some causative agents like bacteria, fungus, virus.
e.g. Down syndrome, haemophilic, etc.	e.g. leprosy, typhoid, malaria, etc.

Question 2.
Amoebiasis and Filariasis.
Answer:
Differences between amoebiasis and filariasis are as follows

Amoebiasis	Filariasis
Entamoeba histolytica is the causative organism of amoebiasis.	Filariasis is caused by filarial nematodes, Wuchereia bancrofti and W. malayi.
Houseflies are mechanical carriers and transmit the parasite from faeces of the infected person to food and contaminate them.	Filarial infectin is caused by different species of mosquitoes, e.g. Culex, Aedes, etc.
Symptoms include constipation, abdominal pain, cramps, etc.	Symptoms include oedema, swelling of lower exteremities and deformation of genital organs.

Question 3.
Primary immune response and Secondary immune response.
Answer:
Differences between primary immune response and secondary immune response are as follows

Primary immune response	Secondary immune response
It occurs as a result of the first contact of the individual with an antigen.	This immune response occurs during second and subsequent contacts with the same antigen.
Its response is feeble to moderate.	Its response is strong.
It takes a longer time to establish immunity.	It is rapid.
It declines rapidly.	It lasts longer, sometimes lifelong.
Receptors for the antigen develops during response.	Receptors are already present.

Question 4.
Active immunity and Passive immunity.
Answer:
Differences between active immunity and passive immunity are as follows

Active immunity	Passive immunity
Develops due to contact with pathogen or its antigen.	Develops when readymade antibodies are injected into the body.
Slow but long lasting.	Fast but lasts for few days.
No or few side effects.	May cause side effects.

Question 5.
Antibody-mediated immune system and Cell- mediated immune system.
Answer:
Differences between antibody mediated immune system and cell-mediated immune system are as follows

Antibody-mediated immune system	Cell-mediated immune system
It is mediated by B-lymphocytes.	It consists of T-lymphocytes.
It operates through formation of antibodies.	It operates directly through T-cells.
It acts on pathogens that invade body fluids.	It operates against those pathogens which invade body cells.
It hardly has any effect against cancers and transplants.	It operates against cancer cells and transplants.

Question 6.
B-lymphocytes and T-lymphocytes.
Answer:
Differences between B-lymphocytes and T-lymphocytes are as follows

B-iymphocytes	T-lymphocytes
B-cells form humoral or Antibody Mediated Immune System (AMIS)	T-cells form Cell-Mediated Immune System (CMIS),
They defend against viruses and bacteria that enter the blood and lymph.	They defend against pathogens including protists and fungi that enter the cells.
They form plasma cells and memory cells by the division.	They form killer, helper and suppressor cells by the division of lymphoblasts.

Question 7.
Antigens and Antibodies.
Answer:
Differences between antigens and antibodies are as follows

Antigens (Immunogens)	Antibodies (Immunoglobins)
They are usually foreign materials such as a protein or polysaccharide molecules.	These are protein molecules.
These trigger the formation of antibodies.	These are synthesised in the body to combat foreign materials.
These may occur on the surface of microbes or as free molecules.	These may occur on the surface of plasma cells and in body fluids.
Antigens bind to macrophages to reach helper T-cells to initiate immune response.	These directly join antigens to destroy them.
They produce diseases or allergic reactions.	These are protective and immobilise or lyse antigenic molecules.

Question 8.
IgG and IgM.
Answer:
Differences between IgG and IgM are as follows

IgG	IgM
It is the most abundant Immunoglobulin in blood.	It is third abundant immunoglobulin in blood.
IgM is replaced by IgG and becomes the principal antibody.	It is first antibody synthesised by the newborn.
It is monomer.	It is pentamer.

Question 9.
Cancer cells and Normal cells.
Answer:
Differences between cancer cells and normal cells are as follows

Cancer cells	Normal cells
The lifespan is not definite.	These cells have a definite lifespan.
These ceils divide in an unregulated and uncontrolled manner.	These cells divide in a regulated manner.
These cells do not have contact inhibition.	The cells show contact inhibition.
These cells do not undergo differentiation.	These cells undergo- differentiation.
These cells do not remain adhered and have lost cell to cell contact.	These cells remain adhered, i.e. have cell to to cell contact.

Question 10.
Stimulants and Hallucinogens.
Answer:
Differences between stimulants and hallucinogens are as follows

Stimulants	Hallucinogens
These increase the activity of CNS.	These damage the CNS performance.
These induce alterness, more wakefulness and excitement	These change thought of a person, feeling and perceptions, cause hallucinations.
e.g. caffeine, cocaine, amphitamines.	e.g. charas, ganja, bhang and marijuana.

Odisha State Board CHSE Odisha Class 12 Biology Important Questions Chapter 7 Molecular Basis of Inheritance Important Questions and Answers.

CHSE Odisha 12th Class Biology Important Questions Chapter 7 Molecular Basis of Inheritance

Molecular Basis of Inheritance Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
In a DNA strand, the nucleotides are linked together by
(a) glycosidic bonds
(b) phosphodiester bonds
(c) peptide bonds
(d) hydrogen bonds
Answer:
(b) phosphodiester bonds

Question 2.
In DNA double helix, thymine is paired with …………… .
(a) guanine
(b) uracil
(c) cytosine
(d) adenine
Answer:
(d) adenine

Question 3.
Semiconservative mode of replication of DNA was proved by
(a) Hershey and Chase
(b) Griffith
(c) Watson and Crick
(d) Meselson and Stahl
Answer:
(d) Meselson and Stahl

Question 4.
Discontinuous synthesis of DNA occurs in one strand, because
(a) DNA molecule being synthesised is very long
(b) DNA dependent DNA polymerase catalyses polymerisation only in one direction (5′ → 3′)
(c) it is a more efficient process
(d) DNA ligase has to have a role
Answer:
(b) DNA dependent DNA polymerase catalyses polymerisation only in one direction (5′ → 3′)

Question 5.
The enzyme not associated with DNA replication is
(a) polymerase
(b) helicase
(c) topoisomerase
(d) transcriptase
Answer:
(d) transcriptase

Question 6.
Which is the enzyme used for joining the fragments of DNA?
(a) Ligase
(b) Polymerase
(c) Endonuclease
(d) Transferase
Answer:
(a) Ligase

Question 7.
To form a continuous DNA molecule, the enzyme ………… joins Okazaki fragments.
(a) primase
(b) polymerase
(c) helicase
(d) ligase
Answer:
(d) ligase

Question 8.
DNA ligase is commonly known as ……….. .
(a) molecular scissor
(b) molecular marker
(c) molecular probe
(d) milecular glue
Answer:
(d) molecular glue

Question 9.
In eukaryotic cells, the RNA transcribed from DNA is called ………….. .
(a) rRNA
(b) cistron
(c) cDNA
(d) heterogenous mRNA
Answer:
(d) heterogenous mRNA

Question 10.
At 5′ end of a polynucleotide chain
(a) H-bond is present
(b) -OH group is attached
(c) PO^–₄ group is attached
(d) pentose sugar is attached
Answer:
(c) PO^–₄ group is attached

Question 11.
In which one of the following, double-stranded RNA is present?
(a) Bacteria
(b) Chloroplast
(c) Mitochondria
(d) Reovirus
Answer:
(d) Reovirus

Question 12.
DNA replication is
(a) semiconservative, directional and continuous
(b) semiconservative, bidirectional
(c) semiconservative and semidiscontinuous
(d) only semiconservative
Answer:
(c) semiconservative and semidiscontinuous

Question 13.
A double-stranded RNA segment has 120 adenine and 120 cytosine bases. The total number of nucleotides present in the segment is
(a) 120
(b) 240
(c) 60
(d) 480
Answer:
(b) 240

Question 14.
Termination codon which stops, further addition of amino acids to the polypeptide chain is
(a) AAU
(b) GUG
(c) AUG
(d) UAG
Answer:
(d) UAG

Question 15.
Which one is not a non-sense codon?
(a) UAA
(b) UGA
(c) UCA
(d) UAG
Answer:
(c) UCA

Question 16.
A phenomenon where the third base of tRNA at its 5′ end can pair with a non-complementary base ofmRNAis called
(a) universality
(b) colinearity
(c) degenerency
(d) wobbling
Answer:
(d) wobbling

Question 17.
Translation is the synthesis of
(a) DNA from a mRNA template
(b) protein from a mRNA template
(c) RNA from a mRNA template
(d) RNA from a DNA template
Answer:
(b) protein from a OTRNA template

Question 18.
The peptide bonds are present between
(a) nucleic acids
(b) organic acids
(c) fatty acids
(d) amino acids
Answer:
(d) amino acids

Question 19.
Gene which is responsible for the synthesis of a polypeptide chain is called
(a) operator gene
(b) regulatory gene
(c) promoter gene
(d) structural gene
Answer:
(d) structural gene

Question 20.
Repressor protein is produced by
(a) regulator gene
(b) operator gene
(c) structural gene
(d) terminator gene
Answer:
(a) regulator gene

Question 21.
In split genes, the coding sequences are called
(a) cistrons
(b) operons
(c) exons
(d) introns
Answer:
(c) exons

Question 22.
The non-sense codons
(a) have no role in biological systems
(b) act as terminators during protein synthesis
(c) are of little value in transcription
(d) have a poor role in transcription
Answer:
(b) act as terminators during protein synthesis

Question 23.
If a cell is treated with a chemical that blocks nucleic acid synthesis, which of the following processes is the most likely one to be affected first?
(a) DNA replication
(b) tRNA synthesis
(c) mRNA synthesis
(d) Protein synthesis
Answer:
(a) DNA replication

Question 24.
Aminoacyl synthetase takes part in
(a) attachment of mRNA to 30S ribosome
(b) transfer of activated amino acids to tRNA
(c) activation of amino acid
(d) hydrolysis of ATP to AMP
Answer:
(c) activation of amino acid

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

Question 1.
Watson and Griffith proposed the double helical structure of DNA.
Answer:
Watson and Crick.

Question 2.
The helical turns are left-handed in Z-DNA.
Answer:
It is correct.

Question 3.
Okazaki fragments are formed on both leading and lagging strand.
Answer:
Okazaki fragments are formed only on lagging strand.

Question 4.
Cytosine is common for both DNA and RNA.
Answer:
It is correct.

Question 5.
RNA does not have guanine as nitrogenous base.
Answer:
Thymine

Question 6.
The complementary base of adenine in RNA molecule is thymine.
Answer:
Uracil

Question 7.
The process of formation of RNA from DNA is translation.
Answer:
Transcription

Question 8.
DNA polymerase-I is mainly responsible for synthesis of new strand during DNA replication.
Answer:
DNA Polymerase

Question 9.
The genetic information from DNA transferred to ribosomes through ribosomal RNA.
Answer:
messenger

Question 10.
The initiation codon AUG normally codes for formylated cystine.
Answer:
methionine

Question 11.
CCC is the initiation codon.
Answer:
AUG

Question 12.
The split genes are needed constantly for cellular activity.
Answer:
housekeeping

Question 13.
The lac operon consists of four regulatory genes only.
Answer:
three

Question 14.
A regulated unit of genetic material for prokaryotic gene expression is called operon.
Answer:
It is correct.

Question 15.
64 codons code for all the 20 essential amino acids.
Answer:
61 codons

Question 16.
Prokaryotic /nRNA is monocistronic.
Answer:
polycistronic

Question 17.
The structural genes are regulated as a unit by a single regulator in operon.
Answer:
promoter

Question 18.
Galactose is an inducer molecule.
Answer:
Lactose

Question 19.
tRNA carries the codes for amino acid sequence.
Answer:
It is correct.

Fill in the blanks

Question 1.
Frederick Griffith discovered the phenomenon called …………… .
Answer:
transformation

Question 2.
The two strands of polynucleotides forming DNA are ……….. and antiparallel.
Answer:
complementary

Question 3.
A ……………. is located towards 3’ end of the coding strand.
Answer:
OH

Question 4.
The correspondence between triplets in DNA (or RNA) and amino acids in protein is known as ……….. .
Answer:
genetic code

Question 5.
……… is a short sequence of DNA where the repressor binds, preventing RNA polymerase from attaching to the ……….. .
Answer:
Operator, promoter

Question 6.
DNA fingerprinting works on the principle of …………. in DNA sequences.
Answer:
polymorphism

Question 7.
RNA can give rise to DNA through the enzyme ……………. .
Answer:
reverse travscriptase

Question 8.
The movement of a ribosome from 5′-3′ end of mRNA to recognise all codons during protein synthesis is called ………… .
Answer:
elongation

Express in one or two word(s)

Question 1.
The enzyme which joins Okazaki fragments to form a continuous DNA molecule.
Answer:
Ligase

Question 2.
The organism on which Meselson and Stahl (1958) provided strong evidence for semiconservative mode of DNA replication?
Answer:
E. coli

Question 3.
The strand which is transcribed into mRNA (RNA transcript).
Answer:
Template strand

Question 4.
The scientist who formulated central dogma of molecular biology in 1958?
Answer:
Crick

Question 5.
The first X-ray diffraction pattern of DNA was given by which scientist?
Answer:
Wilkins

Question 6.
The scientist who suggested that the genetic code should be made of a combination of three nucleotides.
Answer:
George Gamow

Question 7.
The codon which acts as initiation codon and also codes for amino acid methionine.
Answer:
AUG

Question 8.
All terminator codons begin with nucleotide of which base?
Answer:
U

Question 9.
The scientist who proposed the operon concept?
Answer:
Jacob and Monod

Short Answer Type Questions

Question 1.
Write a short note on nitrogenous bases.
Answer:
Nitrogenous bases are heterocyclic compounds in which the rings contain both nitrogen and carbon atoms.
There are two types of nitrogenous bases

• Purines (with double rings) adenine and guanine
• Pyrimidines (with single ring) cytosine, uracil and thymine.

Out of the pyrimidines, cytosine is common for both DNA and RNA while thymine is present only in DNA. Uracil is present in RNA in place of thymine.

Question 2.
If a double-stranded DNA has 20% of cytosine, calculate the percentage of adenine in the DNA.
Answer:
Given, cytosine = 20%
∴ Percentage of guanine = 20%
Now according to Chargaff’s rule,
A + T = 100 – (G + C)
⇒ A + T =100 – 40
∴ Percentage of thymine = Percentage of adenine
= \(\frac{60%}{2}\) = 30%

Question 3.
The base sequence in one of the strands of DNA is TAGCATGAT.
(i) Give the base sequence of the complementary strand.
(ii) How are these base pairs held together in a DNA molecule?
(iii) Explain the base complementarity rule. Give the name of the scientist who framed this rule?
Answer:
(i) ATCGTACTA
(ii) Base pairs are held together by weak hydrogen bonds, adenine pairs with thymine by two H-bonds and guanine pairs with cytosine forming three H-bonds.
(iii) According to base complementarity rule formed by Erwin Chargaff for a double-stranded DNA, the ratios between adenine-thymine and guanine-cytosine are constant and equal to one.

Question 4.
A DNA segment has a total of 1000 nucleotides, out of which 240 of them are adenine containing nucleotides. How many pyrimidine bases this DNA segment possesses?
Answer:
According to ChargafFs rule, ratio of purines to pyrimidines is equal, i.e. A + G = C + T
Since, the number of adenine (A) is equal to the number of thymine (T) and A = 240 (given)
Therefore, T = 240
Also, the number of guanine (G) is equal to cytosine (C)
Thus, G+C = 1000 – (A + T)
G + C = 1000-480 = 520
Hence, G = 260, C = 260
The number of pyrimidine bases, i.e.
C + T = 240 + 260 = 500

Question 5.
It is established that RNA is the first genetic material. Explain giving reasons.
Answer:
RNA is the first genetic material because

• It is capable of both storing genetic information and catalysing chemical reactions.
• Essential life processes such as metabolism, translation, splicing, etc., have evolved around RNA.
• It can directly code for protein synthesis and hence, can easily express the character.

Question 6.
Write short note on RNA.
Answer:
RNA is a genetic material in some viruses.
RNA differs from DNA in having uracil in place of thymine and most RNAs are single-stranded.
It is of three main types, i.e. messenger RNA (mRNA), transfer RNA (tRNA) and ribosomal RNA (rRNA).
mRNA associates with ribosomes for protein synthesis, tRNA is helical and transfers specific amino acids from the cytoplasm to site of protein synthesis while rRNA is part of ribosomes.

Question 7.
Write a short note on tRNA.
Answer:

• tRNA is the smallest form of RNA and functions as in transfering amino acids from cytoplasm to the ribosomes at the time of protein synthesis.
• tRNA has a secondary structure like clover leaf. But its three dimensional structure depicts it as an inverted L-shaped molecule.
• tRNA has five arms or loops, i.e. anticodon loop, amino acid acceptor end, T-loop, D-loop and variable loop.
• tRNAs are specific for specific amino acid.

Question 8.
Explain the two factors responsible for conferring stability to double helix structure of DNA.
Answer:
The factors responsible for stability of double helix structure of DNA are as follows

• Stacking of one base pair over other.
• H-bond between nitrogenous bases.

Question 9.
Which property of DNA double helix led Watson and Crick to hypothesise semiconservative mode of DNA replication? Explain.
Answer:
In the double helical structure of DNA, the two strands of DNA have complementary base pairing and run in opposite direction. This property of DNA double helix led Watson and Crick to hypothesise the semiconservative mode of DNA replication.

Question 10.
Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them.
Answer:
The enzymes involved in DNA replication other than DNA polymerase and ligase are listed below with their functions

• Helicase – Opens the helix
• Topoisomerases – Removes the tension caused due to unwinding
•  DNA ligase – Joins the cut DNA strands

Question 11.
State the dual role of deoxyribonucleoside triphosphates during DNA replication.
Answer:

• The deoxyribonucleoside triphosphates are the building blocks for the DNA strand (polynucleotide chain) is as substrate.
• These also serve as energy source in the form of ATP and GTP from two terminal phosphates.

Question 12.

Why do you see two different types of replicating strands in the given DNA replication fork? Explain. Name of these strands.
Answer:
Both the parent strands function as template strands.
On the template strand with 3′ → 5′ polarity, the new strand is synthesised as a continuous strand.
The DNA polymerase can carry out polymerisation of the nucleotides only in 5′ → 3′ direction. This is called continuous synthesis and the strand is called leading strand.
On the other template strand with 5′ → 3′ polarity, the new strand is synthesised from the point of replication fork, also in 5′ → 3′ direction. But, in short fragments, they are later joined by DNA ligases to form a strand called lagging strand.

Question 13.
Write a short note on centrol 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.

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.

Question 14.
Briefly describe transcription in prokaryotes.
Answer:
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 / σ^H under heat shock, σ⁵⁴ / σ^N under nitrogen starvation and σ²⁸ for chemotaxis.

A typical bacterial transcription unit

Question 15.
Describe the initiation process of transcription in bacteria.
Answer:
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.

Question 16.
Explain (in one or two lines) the function of the following
(i) Introns
(ii) Exons
Answer:
The primary transcript contains both the exons and introns and these are non-functional.
Such genes are called split genes/interrupted genes. Therefore, they undergo a process called splicing to remove the introns and to join the exons in a proper order to allow translation to take place.

Expression of an interrupted/split gene and RNA splicing

The presence of introns is reminiscent of antiquity and the process of splicing represents the dominance of RNA world. The bnRNA undergoes two additional processes, i. e. post-transcriptional modifications.

Question 17.
State the difference between the structural genes in a transcription unit of prokaryotes and eukaryotes.
Answer:
Prokaryotic structural genes are found in continuity without any non-coding region, while eukaryotic structural genes are divided into exons (coding DNA) and introns (non-coding DNA). Exons appear in mature RNA. Introns are spliced out during splicing.

Question 18.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:

• DNA-dependent DNA polymerase uses DNA template to catalyse the polymerisation of deoxynucleotides.
• DNA-dependent RNA polymerase catalyses transcription of all types of RNAs in bacteria.
• In eukaryotes, there are three types of DNA-dependent RNA polymerase
  • RNA polymerase-I transcribes rRNAs.
  • RNA polymerase-II transcribes precursor of mRNA.
  • RNA polymerase-III transcribes tRNA, srRNA and snRNAs.

Question 19.
Why hnRNAis required to undergo splicing?
Answer:
hnKNA is required to undergo splicing because of the presence of introns (the non-coding sequences) in it. These need to be removed and the exons (the coding sequences) have to be joined in a specific sequence for translation to take place.

Question 20.
Write a note on genetic code.
Answer:
Francis Crick conducted an experiment in Viral DNA in 1961 and concluded that genetic code is triplet and with any punctuation it is read continuously. Once the triplet nature of codon was established, different scientists then tried to establish codons for 20 different amino acids found in proteins.

1. Marshall Nirenberg In 1961, he used a synthetic twRNA of uracil only. He found that the translated polypeptide was composed of amino acid-phenylalanine only Thus, he concluded the UUU codon codes for phenylalanine.
2. Nirenberg and Philip Leder In- 1964, they found 47 out of 64 possible codons by employing the technique of triplet binding.
3. Har Gobind Khorana He worked out remaining 17 codons by employing artificial bzRNA.
   Note In 1968, Nirenberg and Khorana shared Nobel Prize with RW Holley (gave details of fRNA structure).

On the basis of above discoveries, the spellings of the genetic code were put together in a checkerboard, as given below

Question 21.
Write short note on peptide bonds.
Answer:
Amino acids are joined together in proteins by peptide bonds. A peptide bond is a covalent bond formed between two molecules when the carboxyl group of one molecule reacts with the amine group of the other molecule. This releases a water molecule. CONH is called a peptide link.

Question 22.
Unambiguous, universal and degenerate are some of the terms used for the genetic code. Explain the salient features of each of them.
Answer:
Unambiguous code means that one codon codes for only one amino acid, i.e. AUG codes only for methionine. Genetic code is universal, as particular codon codes for the same amino acid in all organisms. It is degenerate because some amino acids are coded by more than one codon, e.g. UUU and UUC, both code for phenylalanine.

Question 23.
Write short note on aminoacylation in translation.
Ans.
Amino acids in the cytoplasm are inactive. They cannot take part directly in protein synthesis or translation. The formation of peptide bond requires energy. In the presence of ATP, amino acids become activated by binding with aminoacyl fRNA synthase enzyme, i.e. aminoacylation.

Question 24.
Write short note on operon.
Or
Write a note on operon concept.
Answer:
An operon is a unit of prokaryotic gene expression which includes sequentially regulated structural genes and control elements recognised by the regulatory gene product. F Jacob and J Monod gave the operon concept and were the first ones to describe a transcriptionally regulated system.

Question 25.
What is DNA fingerprinting? Mention its applications.
Or Write a note on DNA fingerprinting.
Answer:
DNA Fingerprinting:
The technique of DNA fingerprinting or DNA typing or DNA profiling was developed and established by British geneticist Dr. Alec Jeffreys based on the fact that like every individual organism is unique in its fingerprints. The DNA pattern also differs in every individual.

Fingerprints can be altered by surgery but there is no known procedure available to alter the DNA design of an individual. For obtaining the DNA fingerprints of an individual, highly polymorphic genes that occur in multiple forms in different individuals are selected.

Applications of DNA Fingerprinting:
This technique can he applied in various fields such as

• Used as a tool in forensic investigations.
• To settle paternity disputes.
• To study evolution by determining the genetic diversities among population.

Question 26.
(i) Explain DNA polymorphism as the basis of genetic mapping of human genome.
(ii) State the role of VNTR in DNA fingerprinting.
Answer:
(i) Polymorphism is inherited from parents to children. So, it is useful for the identification (forensic application) and paternity testing. It arises due to mutations and also plays an important role in evolution and speciation. These mutations in the non-coding sequences have piled up with time and form the basis of DNA polymorphism. It is the basis of genetic mapping of human genome as well as DNA fingerprinting.

(ii) Variable Number of Tandem Repeats (VNTRs) belong to a class of satellite DNA called as minisatellite. VNTR are used as probes in DNA fingerprinting.

Long Answer Type Questions

Question 1.
Describe the structure of DNA with a neat and labelled diagram.
Or Describe the structure of DNA molecule as per the model proposed by Watson and Crick.
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

• One end with a free phosphate moiety at 5′ end of deoxyribose sugar. This is marked as 5′ end of polynucleotide chain.
• 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.

A Ploynucleotide chain

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
1. DNA is made up of two polynucleotide chains. The backbone is constituted by sugar phosphate, while the nitrogenous bases project inwards.
2. 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.
3. 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.
4. 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).
5. Percentage calculation of bases is done by A + T = 100 – (G + C).
6. 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.

Structure of DNA : (a) Watson and Crick model of double helix, (b) Double-stranded polynucleotide chain sequence showing hydrogen bonds

Question 2.
Give an account of Griffith’s experiment on transformation.
Answer:
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:

1. 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.
2. The mice that were infected with the R-strain (R-II) did not develop pneumonia and they lived.
3. 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.
4. 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

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

• R-II + DNA extract of S-III + no enzyme = R-II colonies + S-III colonies
• R-II + DNA extract of S-III + Ribonuclease = R-II colonies + S-III colonies
• R-II + DNA extract of S-III + Protease = R-II colonies + S-III colonies
• R-II + DNA extract of S-III + Deoxyribonuclease = Only R-II colonies

Question 3.
State the aim and describe Meselson and Stahl’s experiment.
Answer:
Meselson and Stahl in 1958, aimed to prove that DNA replicates in a semiconservative fashion. The semiconservative DNA replication suggests that, after the completion of replication, each DNA molecule will have one parental and one newly synthesised strand.

Meselson and Stahl’s experiment
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 ¹⁵NH₄Cl (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 ¹⁴N 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.

Meselson-Stahl experiment to demonstrate semiconservative replication

Similar experiments on a eukaryote, ‘ Vicia faba’ (faba beans) 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 4.
Describe the process 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.

Watson and Crick model for semiconservative DIMA replication

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:
Meselson and Stahl in 1958, aimed to prove that DNA replicates in a semiconservative fashion. The semiconservative DNA replication suggests that, after the completion of replication, each DNA molecule will have one parental and one newly synthesised strand.

Meselson and Stahl’s experiment
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 ¹⁵NH₄Cl (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 ¹⁴N 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.

Meselson-Stahl experiment to demonstrate semiconservative replication

Similar experiments on a eukaryote, ‘ Vicia faba’ (faba beans) 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.

Pre-Requisites of DNA Replication:
DNA replication is a complex process that requires many enzymes and protein factors. This process is very fast and accurate. It is seen in both prokaryotes and eukaryotes and involves some basic steps that are listed below
(i) Two parental strands unwind and get separate.
(ii) One of the parental strand acts as a template for the synthesis of new strand. Hence, a new strand is formed by the winding of one old and one new strand.

Major components involved in the process of DNA replication are as follows
I. On (Origin of Replication)

1. It is the specific site on DNA where replication starts and proceeds in one or both directions. This site is known as origin of replication (Ori).
2. Ori specifying DNA segments can be isolated from E. coli, Coli phages, plasmids, yeasts and eukaryotic viruses.
3. Ori in E. coli is called Ori C. It is a DNA sequence of about 245 base pairs that is rich in A-T bases. Hence, the two strands easily get separated at the origin.
4. Ori in Yeast is called Autonomous Replication . Sequence (ARS) which is 150 base pair long. It acts as the binding site for Origin Recognition Complex (ORC).
5. During replication, Ori is recognised by replication initiator complex and the process of replication starts. It proceeds along the replication forks.
6. Each Ori has two termini. A replicon is one Ori (or origin) with its two unique termini. In prokaryotes (E.coli), the entire circular DNA acts as a single replicon. In eukaryotes, DNA is larger and hence, they have several Ori per DNA.
7. In bidirectional replication, the two strands separate at the origin or Ori. It results in the formation of a replication eye and both ends move along the replication, e.g. θ-replication (Theta) in prokaryotes.
   
8. In unidirectional replication, one of the two ends of the replication eye moves along the replication fork while the other end remains stationary, e.g. replication of mitochondrial DNA (mt DNA) in vertebrates.

II. DNA Polymerase

1. It is the main enzyme which uses a DNA template to catalyse the polymerisation of deoxynucleotides. The average rate of polymerisation is 2000 bp (base pairs) per second approximately.
2. DNA polymerase adds deoxyribonucleotides to the 3′-OH end of polynucleotide by the removal of pyrophosphate from nucleoside triphosphate.
3. Polynucleotide (n) + d NTP → Polynucleotide (n + 1) + PPi

Types of DNA Polymerases

1. In prokaryotes, there are three types of DNA polymerases, i.e. DNA polymerase-I, II and III, whereas in eukaryotes, five different DNA polymerases have been indentified, i.e. DNA polymerases α, ß, γ, δ and ε.
2. DNA polymerase was first isolated from Exoli by Arthur Kornberg in Washington University in 1956. It was first called Kornberg enzyme but later its name was changed to DNA polymerase-I due to the discoveries of other polymerases.
3. DNA polymerase-I and II Involved in DNA repair and proofreading in prokaryotes.
4. DNA polymerase-III It has exonuclease activity. It can remove nucleotides from 3′ end of DNA strand, i.e., 3’→5′ exonuclease. It also helps in proofreading so that wrong nucleotide added at 3′ end can be removed.
5. Exonuclease activity of different DNA polymerases is listed below
   

Working of DNA Polymerase
DNA polymerase (Pol) requires a template to synthesise a new strand in 5′-3′ direction. For this, they add a primer to the template strand. The new nucleotides are then added to the 3′-OH end of primer so that the synthesis of DNA proceeds in 5′-3′ direction.
Note A prime is small DNA or RNA strand that is to template strand through hydrogen bonds.

III. Other Enzymes
Besides DNA polymerases, other enzymes involved in the process of DNA replication are as follows

1. Helicase It unwinds the DNA strand, i.e. separates the two strands from one point, for the formation of a
   replication fork.
2. Topoisomerase (DNA gyrase) The unwinding of DNA creates a tension in the DNA strands, which get released by the enzyme topoisomerase.
3. DNA Ligase It facilitates the joining of DNA strands together by catalysing the formation of phosphodiester bond. It plays a role in repairing single-strand breaks in duplex DNA.

Mechanism of DNA Replication:
All the enzymes and protein factors involved in DNA replication constitutes a replicase system or replisome. . The process of replication proceeds in the following steps

1. An initiator protein recognises the Origin of replication (Ori) and binds to it.
2. DNA helicase enzyme breaks the hydrogen bonds between nitrogenous bases and unwinds the ds DNA.
3. To prevent rewinding and attack by single stranded nuclease, Single-Stranded Binding proteins (SSB proteins) bind to the separated strands. SSB also help to keep these ssDNA in extended position.
4. The combined action of helicase and SSB proteins results in the formation of V-shaped replication fork at the origin.
5. As the replication fork moves, DNA unwinds and a positive super coil is formed in the unreplicated portion of DNA, i.e. in front of the fork.
6. The super coil is like a knot that hinders the fork movement. It is removed by the enzyme topoisomerase (gyrase) in E.coli and topoisomerase-I in eukaryotes.

Question 5.
How do mRNA, tRNA and ribosomes help in the process of translation?
Answer:
(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 tRNA 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.

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.

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 tRNA and another for specific amino acid molecule.

Question 6.
Describe the process of translation in prokaryotes.
Or Describe the initiation step of translation in prokaryotes.
Answer:
Translation requires a machinery which consists of ribosome, wzRNA, rRNAs, aminoacyl rRNA synthetase (enzyme that helps in combining amino acid to particular rRNA) and amino acids.

Initiator tRNA
It is a specific rRNA for the process of initiation and there are no rRNAs for stop codons.

Ribosome
It occurs in cytoplasm and responsible for protein synthesis. It consists of structural RNAs and around 80 different proteins. Ribosome exists as two subunits in its inactive stage
(i) Small subunit When the small subunit encounters an mRNA, translation of mRNA to protein begins.
(ii) Large subunit It consists of two sites where amino acids can bind to and be close to each other for the formation of a peptide bond. Ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme ribozyme) for peptide bond formation.

Translational Unit:
It is the sequence of RNA flanked by the start codon (AUG) and the stop codon in mRNA. It codes for a polypeptide that has to be produced.

Untranslated Regions (UTR):
These are some additional sequences in an wRNA that are not translated. They are present at both the ends, i.e. at 5′ end (before start codon) and at 3′ end (after stop codon). They improve the efficiency of translation process.

Mechanism of Translation:
The main steps in translation include
(i) Binding of mRNA to ribosome
(ii) Activation of amino acids (aminoacylation of tRNA).
(iii) Transfer of activated amino acids to tRNA.
(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 tRNA 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.

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.

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 tRNA 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
(a) tRNAf^met for initiation carrying formyl methionine and
(b) tRNA^met for carrying normal methionine to growing polypeptide.

The initiation of polypeptide synthesis requires the following components
mRNA, 30S subunit of ribosome, formylmethionyl-tRNA (f^met-tRNAf^met), 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-tRNAf^met (the specific tRNA aminoacylated to formyl methionine) binds to the AUG codon at the P-site. The tRNAfmct is the only tRNA that binds to its codon present on the P-site. All other fRNA 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-tRNAf^met 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-rRNA fmet bound to initiation codon at P site is known as initiation complex.

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.

Binding of the second aminoacyl tRNA to the A site of ribosome

Formation of a peptide bond

(vi) Translocation of Polypeptide:
The peptidyl tRNA 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.

(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.
Note In eukaryotes, only one release factor is known. It iseRF₁.

Odisha State Board CHSE Odisha Class 12 Biology Important Questions Chapter 6 Sex Determination Important Questions and Answers.

CHSE Odisha 12th Class Biology Important Questions Chapter 6 Sex Determination

Sex Determination Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
Non-homologous segment of Y-chromosome carries
(a) dominant genes
(b) recessive genes
(c) holandric genes
(d) None of the above
Answer:
(c) holandric genes

Question 2.
Which type of sex determination is found in humans?
(a) XX – XY
(b) Z\V – ZZ
(c) XX – XO
(d) ZZ – ZO
Answer:
(a) XX – XY

Question 3.
In XO-type of sex-determination
(a) females produce two different types of gametes
(b) males produce two different types of gametes
(c) females produce gametes with Y-chromosome
(d) males produce single type of gametes
Answer:
(b) males produce two different types of gametes

Question 4.
Which of the following types of sex-determination is found in grasshopper?
(a) XX female and XY male
(b) ZW female and ZZ male
(c) XX female and XO male
(d) XX male and XO female
Answer:
(c) XX female and XO male

Question 5.
Sex chromosomes of a female bird are represented by
(a) XO
(b) XX
(c) ZW
(d) ZZ
Answer:
(c) ZW

Question 6.
ZZ/ZW type of sex-determination is seen in
(a) snails
(b) peacock
(c) platypus
(d) cockroach
Answer:
(b) peacock

Question 7.
In gynandromorph
(a) all cells have XX genotype
(b) all cells have XY genotype
(c) all cells with XXY genotype
(d) some cells of the body contain XX and some cells with XY genotype
Answer:
(d) some cells of the body contain XX and some cells with XY genotype

Question 8.
In which chromosome is the gene for haemophilia located?
(a) X-chromosome
(b) Y-chromosome
(c) Autosome
(d) Both (a) and (b)
Answer:
(a) X-chromosome

Question 9.
A colourblind person cannot distinguish
(a) all colours
(b) green
(c) red
(d) red and green
Answer:
(d) red and green

Question 10.
Which chromosome-linked genes do cause the genetic metabolic Phenylketonuria (PKU)?
(a) Somatic dominant gene
(b) Somatic recessive gene
(c) Y-linked gene
(d) X-linked gene
Answer:
(b) Somatic recessive gene

Question 11.
Down’s syndrome is an example of
(a) triploidy
(b) polyteny
(c) polyploidy
(d) aneuploidy
Answer:
(d) aneuploidy

Question 12.
What is the diploid chromosome number in a person suffering from Down syndrome?
(a) 45
(b) 46
(c) 47
(d) 48
Answer:
(c) 47

Question 13.
Which is the genotype of Turner’s syndrome?
(a) XO
(b) XXY
(c) XYY
(d) XXX
Answer:
(a) XO

Question 14.
Number of Barr bodies present in Turner’s syndrome is
(a) 0
(b) 1
(c) 2
(d) Both (b) and (c)
Answer:
(a) 0

Question 15.
In which of the following diseases, the man has an extra X-chromosome?
(a) Bleeder’s disease
(b) Turner’s syndrome
(c) Klinefelter’s syndrome
(d) Down’s syndrome
Answer:
(c) Klinefelter’s syndrome

Question 16.
A colour blind person cannot distinguish colour/colours.
(a) all
(b) red
(c) green
(d) red and green
Answer:
(d) red and green

Question 17.
The extra inactive X-chromosome in karyotype of Klinefelter syndrome is called
(a) Barr body
(b) barr chromosome
(c) dosage body
(d) None of these
Answer:
(a) Barr body

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

Question 1.
Heterogametic individual produces similar type of gametes.
Answer:
Homogametic individual produces similar type of gametes.

Question 2.
D. melanogaster with 2A +XX chromosome complement is female.
Answer:
Correct statement.

Question 3.
Gynandromorphs die due to failure of segregation.
Answer:
Correct statement.

Question 4.
Mary F lyon discovered X-chromosome in male bug and described it as X-bodv.
Answer:
Barr body

Question 5.
The genotype of a carrier haemophila is XhXh.
Answer:
XXh

Question 6.
Cystic fibrosis is a sex-linked recessive disease, which is transmitted from an unaffected carrier female to some of the male offsprings.
Answer:
Haemophilia

Question 7.
Down syndrome is an inherited blood disorder, in which the body makes an abnormal form of haemoglobin.
Answer:
Thalassemia

Question 8.
Klinefelter syndrome is an abnormal condition caused by the presence of an extra Y chromosome.
Answer:
X

Fill in the blanks

Question 1.
In humans, males are heterogametic, whereas females are
Answer:
homogametic.

Question 2.
According to genic balance theory, the sex index of 1.0 is a ………….. .
Answer:
Female

Question 3.
The unfertilised egg of honeybee develops into …………….. .
Answer:
drones

Question 4.
In grasshopper, female is ……………. and the male is …………. .
Answer:
XX and XO

Question 5.
…….. is also known as bleeder’s disease.
Answer:
Haemophilia

Question 6.
Down’s syndrome is due to ………….. of chromosome 21.
Answer:
trisomy

Question 7.
Turner’s syndrome is caused due to of one of the X-chromosome.
Answer:
the absence

Question 8.
………….. is an inherited disorder which results in the failure to distinguish red and green colours.
Answer:
Colour Blindness

Express in one or two word(s)

Question 1.
The sex of the child developed from 44A+XX zygote.
Answer:
Female

Question 2.
At high temperature, what sex of turtle is produced?
Answer:
Female

Question 3.
Name the environmental factor that determines the sex in Bonellia.
Answer:
Temperature

Question 4.
Name any one autosomal recessive disease.
Answer:
Thalassemia.

Question 5.
Name the scientist who discovered Down’s syndrome.
Answer:
Langdon Down.

Question 6.
A heritable disorder linked to genes on the non-sex chromosomes.
Answer:
Down’s syndrone

Question 7.
A heritable disease caused by the presence of one defective allele.
Answer:
Thalassemia

Question 8.
Chromosomes fail to sort properly during meiosis.
Answer:
Down’s syndrome

Short Answer Type Questions

Question 1.
What is Barr body?
Answer:
A Barr body is a small darkly stained mass of X-chromosome, which in inactive and are found only in the female cells. Out of the two X-chromosomes in feamales only one is functional and the other remain as Barr body.

Question 2.
How the sex is determined in humans?
Or Write a short note on sex-determination in human.
Answer:
The male and female individuals normally differ in their chromosomal constituents. There are two types of chromosomes, i.e.

1. Sex chromosomes or Allosomes The chromosomes responsible for sex determination, e.g. X and Y-chromosomes.
2. 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

Mechanism of sex-determination

Question 3.
What is male heterozygosity?
Answer:
Male heterozygosity is a type of mechanism of sex-determination in organisms. XX and XY type of sex-determination shows the phenomenon of male heterogamety or heterozygosity because in both these cases, males produce two different types of gametes such as • Either with or without X-chromosome.

• Some gametes with X-chromosome and some with Y-chromosome.

Question 4.
Describe sex-determination in grasshoppers.
Answer:
Grasshoppers have XX-XO method of sex-determination. In this, female has XX and produces homogametic eggs, while male has only one chromosome and produces two types of sperms, e.g. gymnosperms (with X) and angiosperms (without X).

Question 5.
(i) why grasshopper and Drosophila show male heterogamety? Explain.
(ii) Explain female heterogamety with the help of examples.
Answer:
(i) Male heterogamety is shown by male grasshopper and Drosophila, as they both produce two types of gametes having 50% X-chromosomes and other with 50% Y-chromosomes.

(ii) In this case, the total number of chromosomes are same in both males and females. But two different types of gametes having different sex chromosomes are produced by females.

1. ZZ-ZW Mechanism
This mechanism of sex-determination is seen in birds, fowls and fishes. Females have one Z and one W-chromosome (i.e. heterogametic) along with autosomes whereas males have a pair of Z-chromosomes (i.e. homogametic). Thus, the sex of an organism is determined by the type of ovum that is fertilised to produce an offspring.

Determination of sex (ZZ-ZW) in fowl and birds

2. ZZ-ZO
In this mechanism of sex-determination, the female is heterogametic (ZO) and male is homogametic (ZZ). It occurs in lepidoptera, e.g. certain butterflies and moths.

Determination of heterogametic and homogametic female and male

Question 6.
Write the types of sex-determination mechanisms of the following crosses. Give an example of each type.
(i) Female XX with male XO
(ii) Female ZW with male ZZ
Answer:
(i) The type of sex-determination mechanism shown in female XX with male XO is male heterogamety, e.g. grasshoppeer.
(ii) The type of sex-determination mechanism shown in female ZW with male ZZ is female heterogamety, e.g. birds.

Question 7.
Write short note on sex-determination in Bonellia viridis.
Answer:
In Bonellia viridis (worm), the environment determines the sex differentiation. In these, when the young ones are reared alone they develop into females, but when the newly hatched eggs are reared in close proximity to an adult female (i.e. attached to female proboscis) they become male.
This is due to the hormones released by female proboscis which induces larvae to differentiate into males.

Question 8.
What is criss-cross inheritance?
Answer:
Criss-cross inheritance is defined as the inheritance of sex-linked characters transmitted from father to daughter, who pass it on to the grandsons. The trait is expressed only in males in alternate generations, e.g. red-green colours blindness, haemophilia, etc.

Question 9.
What is sex-linked inheritance? Discuss the inheritance of haemophilia in man.
Answer:
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


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

Question 10.
Haemophilia is a sex-linked inheritance condition in humans where a simple cut causes non-stop bleeding. Study the pedigree chart showing the inheritance of haemophilia in a family. Give reasons, which explain that haemophilia is (i) sex-linked and (ii) caused by X-linked gene.

Answer:
(i) Haemophilia is sex-linked because

• It is transmitted from an unaffected carrier female to some of the male offsprings.
• Female rarely becomes haemophilic as her mother has to be atleast a carrier and father should be haemophilic.

(ii) Gene for haemophilia is present on X-chromosome because the heterozygous female for haemophilia may transmit the disease to sons.

Question 11.
What is sex-linked inheritance? Discuss this taking colour blindness as an example.
Answer:
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.

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 (X^CY) 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.

Question 12.
Write short note on inheritance of colour blindness in man.
Answer:
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.

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 (X^CY) 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.

Question 13.
If a father and son are both defective in red-green colour vision, is it likely that the son inherited the trait from his father. Comment.
Answer:
No, defective gene for red-green colour vision cannot be inherited from father to his son. Gene for colour blindness is X-chromosome linked and sons receive their sole X-chromosome from their mother, not from their father. Male to male inheritance is not possible for X-linked traits in humans.
In the given case, the mother of the son must be a carrier (heterozygous) for colour blindness gene, thus transmitting the gene to her son.

Question 14.
A colourblind child is born to a normal couple. Work out a cross to show how it is possible. Mention the sex of this child.
Answer:
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 (X^CY) 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.

Question 15.
Write the symptoms of Down’s syndrome.
Answer:
The symptoms of Down’s syndrome are

1. Broad forehead
2. Short and broad neck
3. Short and stubby fingers
4. Partially open mouth, furrowed tongue
5. Mental retardation.

Question 16.
How is the child affected, if it has grown from the zygote formed by an XX-egg fertilised by a Y-carrying sperm? What do you call this abnormality?
Answer:

• The zygote will be XXY. It means the zygote is male with feminine characters.
• This abnormality is called Klinefelter’s syndrome.

Question 17.
Name a disorder, give the karyotype and write the symptoms, where a human male suffers as a result of an additional X-chromosome.
Answer:
The disorder is Klinefelter’s syndrome. It is a chromosomal disorder, which occurs in males. The presence of an additional copy of X-chromosome results in karyotype 44 + XXY.

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 of X 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.

Long Answer Type Questions

Question 1.
Explain the chromosomal basis of sex-determination in animals.
Or Give an account of chromosomal theory of sex-determination.
Or Explain the chromosomal theory of sex-determination in animals.
Or Discuss the chromosomal theory of sex-determination in animal’s.
Or Describe the chromosomal basis of sex-determination in human, honeybee and birds.
Or Discuss sex-determination in birds and honey bees.
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.

1. Sex chromosomes or Allosomes The chromosomes responsible for sex determination, e.g. X and Y-chromosomes.
2. 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

Mechanism of sex-determination

Sex-Determination involving Heterogametic Males:
It is the mechanism in which male produces two different types of gametes. The two conditions that can occur in males are

• Only one X chromosome containing gamete is present.
• Some gametes with X-chromosome and some with Y-chromosome.

(i) XX-XY Type or Lygaeus Mechanism:
This mechanism was first studied by Wilson and Stevens in the milkweed bug called Lygaeus turcicus.
It is present in certain insects like Drosophila melanogaster and mammals including human.
In males, an X-chromosome is present but its other part is very small called as Y-chromosome, whereas, females have a pair of only X-chromosome, i.e. XX.

Both males and females bear same number of chromosomes. The males have autosomes plus XY-chromosomes and females have autosomes plus XX-chromosomes. The males produce two types of gametes containing X or Y sex chromosome (heterogametic) and females produce only one type of gametes with an X-chromosome (homogametic).

(a) Determination of sex in Drosophila
(b) Pattern of sex chromosomal inheritance in human

Thus, in this type of sex-determination, the presence of Y-chromosomes determines the maleness.

(ii) XX-XO Mechanism
In this pattern, the female has two X-chromosomes (called XX), while male has only one X-chromosome (called XO). The Y-chromosome is completely absent here and it is denoted by the letter O. Thus, the presence of unpaired X-chromosome determines the masculine sex. The female just like the previous method produces, only one type of eggs and male produces two types of sperms, i.e. 50% with one X-chromosome and 50% without any sex chromosome. The sex of offspring depends upon the type of sperm, which fertilises the egg.

Determination of sex in grasshopper (XX-XO type)

Eggs fertilised by sperms having an X-chromosome become females and those fertilised by sperms that do not have X-chromosome become males, e.g. grasshopper and bugs.

Sex-Determination involving Heterogametic Female:
In this case, the total number of chromosomes are same in both males and females. But two different types of gametes having different sex chromosomes are produced by females.

(i) ZZ-ZW Mechanism
This mechanism of sex-determination is seen in birds, fowls and fishes. Females have one Z and one W-chromosome (i.e. heterogametic) along with autosomes whereas males have a pair of Z-chromosomes (i.e. homogametic). Thus, the sex of an organism is determined by the type of ovum that is fertilised to produce an offspring.

Determination of sex (ZZ-ZW) in fowl and birds

(ii) ZZ-ZO
In this mechanism of sex-determination, the female is heterogametic (ZO) and male is homogametic (ZZ). It occurs in lepidoptera, e.g. certain butterflies and moths.

Determination of heterogametic and homogametic female and male

Question 2.
(i) Also describe as to, who determines the sex of an unborn child?
(ii) Mention whether temperature has a role in sex-determination.
Answer:
(i) As a rule, the heterogametic organism determines the sex of the unborn child. In case of humans, since males are heterogametic it is the father and not the mother who decides the sex of the child.
(ii) In some animals like crocodiles, lower temperature favours the hatching of female offsprings and higher temperature leads to hatching of male offsprings.
This pattern is reversed in Bonellia worm where females are produces in high temperature and males are produced in lower temperature.

Question 3.
Describe various environmental factors that help in sex-determination.
Answer:
Environmental Factors in Sex-Determination:
In some lower animals, sex-determination is non-genetic and depends on the factors present in the external environment.
Different environmental factors responsible for sex-determination are given below

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.

Sex-determination in Bonellia

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.

Question 4.
What is sex-linked inheritance? Discuss how sex-linked gene inheritance occurs in human, giving two examples.
Or What is sex-linked inheritance? Discuss the mechanism with reference to haemophilia.
Answer:
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. X^CX^C. 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


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 (X^CY) 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

Question 5.
(i) Why are thalassemia and haemophilia categorised as Mendelian disorders? Write the symptoms of these diseases. Explain their pattern of inheritance in humans.
(ii) Write the genotypes of the normal parents producing a haemophilic son.
Answer:
(i) Thalassemia and haemophilia are categorised as
Mendelian disorders because these disorders are due to alteration in a single gene. Also, they are transmitted to offsprings through Mendelian principles of inheritance. Symptoms and pattern of inheritance are given below

(a) Thalassemia It is an autosomal linked recessive blood disorder characterised by defect in a, (3 and 8 chain resulting in abnormal Hb molecule. Symptom Anaemia.
Inheritance Two mutant alleles (one from each parent) must be inherited for an individual to be affected, i.e. homozygous. Heterozygous are carriers and may pass the mutant allele to children.

(b) Haemophilia It is a sex-linked recessive disorder whose gene is located on X-chromosome. Symptom Prolonged clotting time and internal bleeding, even in a minor injury.
Inheritance The gene is present on X-chromosome, so it is inherited by males as they have a single X-chromosome. Affected males are said to be hemizygous. Females have two X-chromosomes, thus possibility of them being affected is rare as the mother of such female has to be atleast carrier and father should be haemophilic.

(ii) Genotypes of the normal parents producing a haemophilic son are X CX (carrier mother) and XY (father).

Carrier Haemophilic Normal Normal daughter son daughter son

Question 6.
Answer the following questions related to Down’s syndrome.

1. When was Down’s syndrome first described?
2. What is its frequency in human and does it also occur in other animals?
3. How does Down’s syndrome arise?
4. Relate it with chromosome dysfunctions.

Answer:

1. It was first described in 1866 by J Langdon Down, so it is called Down’s syndrome.
2. About 1 in every 750 children exhibits Down’s syndrome throughout the world. It is also seen in chimpanzees and other related primates.
3. In humans, it occurs as a result of non-disjunction of chromosome 21 during egg formation. The cause of this primary non-disjunction is not known.
4. The defect is associated with a particular small portion of chromosome 21. When this particular chromosome segment is present in three copies, instead of two, due to translocation Down’s syndrome occurs.

Odisha State Board CHSE Odisha Class 12 Biology Important Questions Chapter 5 Heredity and Variation Important Questions and Answers.

CHSE Odisha 12th Class Biology Important Questions Chapter 5 Heredity and Variation

Heredity and Variation Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
The process of physical removal of anthers is called …………
(a) emasculation
(b) mass selection
(c) introduction
(d) mutation
Answer:
(a) emasculation

Question 2.
Which term did Mendel use to denote something in germ cells responsible for transmission of characters?
(a) Chromosome
(b) Element
(c) Factor
(d) Gene
Answer:
(c) Factor

Question 3.
The character that is expressed in the F₁ is called ……….
(a) recessive character
(b) dominant character
(c) codominant
(d) None of these
Answer:
(b) dominant character

Question 4.
An individual who has two different alleles of a gene is called ……………..
(a) allelopathic
(b) homozygous
(c) heterozygous
(d) codominant
Answer:
(c) heterogyzous

Question 5.
A cross of F₁ with the recessive parent is known as
(a) back cross
(b) hybrid cross
(c) test cross
(d) double cross
Answer:
(c) test cross

Question 6.
The crose between F₁ hybrid and the double recessive parent is called
(a) test cross
(b) double cross
(c) reciprocal cross
(d) complementary cross
Answer:
(c) test cross

Question 7.
If tallness (TT) is dominant and dwarfness (tt) is recessive, then a cross between ……….. will yield offspring of which 50% are dwarf.
(a) TT × tt
(b) Tt × tt
(c) tt × tt
(d) Tt × Tt
Answer:
(b) Tt × tt

Question 8.
The work of Mendel was published in 1866 before the Natural Science Society of Brunn in a paper named
(a) hybridisation on pea plant
(b) inheritance pattern in pea plant
(c) experiments in plant hybridisation
(d) Mendelian experiments on pea plant
Answer:
(c) experiments in plant hybridisation

Question 9.
The scientist not associated with the rediscovery of Mendel’s work is
(a) Hugo de Vries
(b) Carl Correns
(c) Erich von Tschermak
(d) William Bateson
Answer:
(d) William Bateson

Question 10.
Which one is not the reason for the success of Mendel?
(a) Made statistial analysis of the offsprings
(b) Kept accurate records
(c) Select pea plant
(d) Only did self-pollination in plants
Answer:
(d) Only did self-pollination in plants

Question 11.
Which one is a heterozygous condition?
(a) RR
(b) rr
(c) Rr
(d) RRrr
Answer:
(c) Rr

Question 12.
Phenotype is
(a) the genetic make up of an individual
(b) the same for parent and offspring
(c) the account of physiological activities
(d) the appearance of an individual
Answer:
(d) the appearance of an individual

Question 13.
To determine the heterozygosity of a cross, one has to perform
(a) back cross
(b) test cross
(c) reciprocal cross
(d) All of the above
Answer:
(b) test cross

Question 14.
The genetic ratio of 9 :3: 3:1 is due to
(a) segregation of characters
(b) crossing over of characters
(c) independent assortment of genes
(d) homologous pairing between chromosomes
Answer:
(c) independent assortment of genes

Question 15.
Organisms phenotypically similar but genotypically different are said to be
(a) heterozygous
(b) monozygous
(c) multizygous
(d) homozygous
Answer:
(a) heterozygous

Question 16.
Lack of independent assortment of two genes A and B in fruitfly is due to
(a) repulsion
(b) recombination
(c) linkage
(d) crossing over
Answer:
(c) linkage

Question 17.
The gene which controls many characters is called
(a) codominant gene
(b) polygene
(c) pleiotropic gene
(d) multiple gene
Answer:
(c) pleiotropic gene

Question 18.
A linkage group is explained as
(a) different groups of genes located on the same chromosome
(b) all the linked genes of a chromosome
(c) all genes of a chromosome
(d) None of the above
Answer:
(b) all the linked genes of a chromosome

Question 19.
Crossing over brings about
(a) recombination of genes
(b) no significant change
(c) sturdy offspring
(d) cytoplasmic reorganisation
Answer:
(a) recombination of genes

Question 20.
Different mutations referrable to the same locus of a chromosome gives rise to
(a) multiple alleles
(b) pseudoalleles
(c) polygenes
(d) oncogenes
Answer:
(a) multiple alleles

Question 21.
ABO blood group is an example of
(a) pseudoalleles
(b) isoalleles
(c) multiple alleles
(d) cytoplasmic inheritance
Answer:
(c) multiple alleles

Question 22.
Sutton gave chromosomal theory of inheritance, he united the knowledge of chromosomal segregation with
(a) recombination
(b) crossing over
(c) Both (a) and (b)
(d) Mendelian principle of segregation
Answer:
(d) Mendelian principle of segregation

Question 23.
The phenomenon of single gene contributing to multiple phenotypic traits is called
(a) pleiotropy
(b) codominance
(c) incomplete dominance
(d) polygenic inheritance
Answer:
(a) pleiotropy

Question 24.
Multiple alleles control the inheritance of in man.
(a) phenylketonuria
(b) colour blindness
(c) sickle-cell anaemia
(d) blood groups
Answer:
(d) blood groups

Question 25.
An incomplete dominance is shown by
(a) Pisum sativum
(b) Neurospora
(c) Mirabilis jalapa
(d) Lathyrus odoratus
Answer:
(c) Mirabilis jalapa

Question 26.
Pleiotropy occurs when a gene has
(a) a complementary gene elsewhere
(b) a small effect on one trait
(c) reversible effects on the phenotype, depending on age
(d) many effects on the phenotype
Answer:
(d) many effects on the phenotype

Question 27.
Crossing over occurs at
(a) 2 strand stage
(b) 4 strand stage
(c) Both (a) and (b)
(d) None of these
Answer:
(b) 4 strand stage

Question 28.
The number of linkage group found in Drosophila is
(a) 1
(b) 3
(c) 2
(d) 4
Answer:
(d) 4

Fill in the blanks

Question 1.
The basic unit of heredity is …………… .
Answer:
gene

Question 2.
An organism in which two alleles of a trait are unlike is called …………….. .
Answer:
heterozygous

Question 3.
An allele of T is …………
Answer:
t

Question 4.
Test cross is a cross between and ……….. .
Answer:
Tt, tt

Question 5.
A man with blood group ‘AB’ marries a woman with ‘O’ blood group. The blood group of offsprings will be ……………. .
Answer:
either ‘A’ or ‘B’ blood group

Question 6.
Phenylketonuria is an example of …………….. .
Answer:
pleiotropy

Question 7.
During incomplete dominance, phenotypic ratio is ………….. .
Answer:
1:2:1

Question 8.
Skin colour inheritance in humans is an example of …………..
Answer:
Polygenic inheritance

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

Question 1.
Griffith coined the term ‘gene’ for Mendelian factor.
Answer:
Wilhelm Johannsen.

Question 2.
Inheritance is the degree by which progeny differs from their parents.
Answer:
Variation

Question 3.
Monohybrid cross yields two numbers of genotype.
Answer:
one

Question 4.
The genetic ratio in F₂-generation of Mendel’s monohybrid cross is 9:3 : 3:1.
Answer:
1:2:1

Question 5.
A cross in which parents differ in a single pair of contrasting character is called dihybrid cross.
Answer:
monohybrid cross

Question 6.
Allelomorphs or alleles indicate identical characters of an individual.
Answer:
contrasting

Question 7.
The number of phenotypic classes is same as to the genotype in complete dominance.
Answer:
incomplete

Question 8.
In a cross between red and white flowered plants, F₁-hybrids are pink. This is called quantitative dominance.
Answer:
incomplete

Question 9.
When two or more genes equally express themselves, they are called dominant genes.
Answer:
codominant

Question 10.
Inheritance of skin colour in man is monogenic.
Answer:
polygenic

Question 11.
Inheritance of skin colour in man is monogenic.
Answer:
polygenic

Question 12.
Multiple allelism is the phenomenon in which a single gene regulates several phenotypes.
Answer:
Pleiotropy

Express in one or two word(s)

Question 1.
These express contrasting characters of an individual.
Answer:
Alleles

Question 2.
Sum total of heredity or genetic makeup.
Answer:
Genome

Question 3.
Represent the genetic make up or gene complement of an organism.
Answer:
Genotype.

Question 4.
It is a simple square-shaped diagram which is drawn to show the possible combinations of male and female gametes of F₁-parents.
Answer:
Punnett square.

Question 5.
Genes located in the same chromosome and being inherited together.
Answer:
Linked genes.

Question 6.
Skin colour inheritance in humans is an example of
Answer:
polygenic inheritance

Question 7.
In this phenomenon, the number of phenotypic classes is same as to the genotype.
Answer:
Incomplete dominance

Question 8.
In this phenomenon, both the genes of allelomorphic pair express themselves equally in the F₁-hybrids.
Answer:
Codominance

Short Answer Type Questions

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:
Mendel’s Experimental Material
He selected garden pea plant as a sample due to the following reasons

1. Pea plants are readily available on large scale and has bisexual cleistogamous flowers.
2. Peas are self-pollinated and can be cross-pollinacd also.
3. These are annual plants with short life cycle. So, several generations can be studied within a short period.
4. Pea plants could be easily raised, maintained and handled.
5. Pea plants differ in distinct/contrasting characteristics, which provide many easily detectable contrasting characters.

Question 2.
Write a note on emasculation.
Answer:
Mendel conducted artificial/cross-pollination experiments by performing emasculation (removal of anthers) on several true-breeding pea lines. A true-breeding line refers to one that have undergone continuous self-pollination and show stable trait inheritance and expression for several generations.
Mendel selected 14 true-breeding pea plant varieties as pairs, which were similar except for one character with contrasting traits.

Question 3.
Write short note with 2-3 important points on law of segregation.
Answer:
Principle of Segregation:
This principle states that, though the parents contain two alleles during gamete formation, the factors or alleles of a pair segregate from each other, such that a gamete receives only one of the two factors. Hence, the alleles do not show any blending and both the characters are recovered as such in the F₂-generation though one of these is not seen in the F₁ -generation.

Question 4.
A cross was carried out between two pea plants showing contrasting traits of height of the plants. The result of the cross showed 50% parental characters.
(i) Work out the cross with the help of a Punnett square.
(ii) Name the type of the cross carried out.
Answer:
Following inferences were made by Mendel based on his observations

1. He proposed that some ‘factors’ passed down from parent to offsprings through the gametes over successive generations. Now-a-days, these factors are known as genes. Genes are hence, the units of inheritance. Genes which code for a pair of contrasting traits are known as alleles or allelomorphs, i.e. they are slightly different forms of the same gene.
2. Genes occur in pairs in which, one dominates the other called as the dominant factor or the gene which expresses itself, while the other remains hidden and is called recessive factor.
3. Allele can be similar in case of homozygous (TT or tt) and dissimilar in case of heterozygous (Tt).
4. In a true-breeding tall or dwarf pea variety, the allelic pair of genes for height are identical or homozygous.
5. TT and tt are called genotype (sum total of heredity or genetic make up) of the plant, while the term tall and dwarf are the phenotype.
6. When tall and dwarf plants produce gametes by process of meiosis, the alleles of the parental pair segregate and only one of the alleles gets transmitted to a gamete. Thus, there is only 50% chance of a gamete containing either allele, as the segregation is a random process.
7. During fertilisation, the two alleles, ‘T’ from one parent and V from other parent are united to produce a zygote, that has one ‘T’ and one ‘t’ allele or the hybrids have Tt.
8. Since, these hybrids contain alleles which express contrasting traits, the plants are heterozygous.

Question 5.
Using a Punnett square, work out the
distribution of phenotypic features in the first (F₁) filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Answer:
Homozygous female – WW
Hetrozygous – Ww

In this case, all F₁ -generation is with dominant trait, i.e. 50% with homozygous and 50% with heterozygous nature.

Question 6.
When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seeds (Ttyy), what proportions of phenotype in the offsprings could expected to be?
(i) Tall and green
(ii) Dwarf and green
Answer:

Tall green – 3/8 or 6/16
Dwarf green – 1/8 or 2/16
phenotypic ratio = 3 : 1

Question 7.
How are alleles of particular gene differ from each other? Explain its significance.
Answer:
Alleles are polymorphs that differ in their nucleotide sequence resulting in contrasting phenotypic expression. Alleles are the alternative forms of a same gene, e.g. genes for height have two alleles, one for dwarfness (t) and one for tallness (T). Significance of alleles are
(i) A character may have two or more contrasting phenotypic expressions, thus resulting in variation in a population.
(ii) These are used in the studies of inheritance and in understanding their behaviour.

Question 8.
During a monohybrid cross involving a tall pea plant with a dwarf pea plant, the offspring populations were tall and dwarf in equal ratio. Workout a cross to show how it is possible.
Answer:
In a monohybrid test cross, involving a heterozygous tall plant (Tt) and a pure dwarf plant, the progeny consists of tall and dwarf plants in the ratio of 1:1.
This can be shown as given below

Question 9.
Write a short note on Mendel’s principle of dominance.
Answer:
Principle of Dominance:
It states that when two contrasting alleles for a character come together in an organism, only one is expressed completely and shows visible effect. This allele is called dominant and the other allele of the pair which does not express and remains hidden is called recessive.
For example, in the monohybrid cross when dwarf plant is crossed with tall plant, the Frgeneration are all tall plants. This shows that allele for tallness is dominant.

Question 10.
Define and design a test cross.
Answer:
Back Cross and Test Cross
Back cross is a cross of F₁ -progeny back to one of their parents. In back cross, there can be two possibilities, i.e. F₁ -hybrid to be crossed with homozygous dominant parent or with homozygous recessive parent.

(a), (b): Diagrammatic representation of back cross

A special back cross to the recessive parent is known as test cross. This method was devised by Mendel to determine whether the dominant phenotype is homozygous or heterozygous.
For example, in a monohybrid cross between violet colour flower (W) and white colour flower (w), the F₁-hybrid was a violet colour flower. If all the F₁-progenies are of violet colour, then the dominant flower is homozygous and if the progenies are in 1:1 ratio, then the dominant flower is heterozygous.

Diagrammatic representation of a test cross

Question 11.
A teacher wants his/her students to find the genotype of pea plants bearing purple coloured flowers in their school garden. Name and explain the cross that will make it possible.
Answer:
Test cross is a method devised by Mendel to determine the genotype of plant with dominant phenotype (purple flower in this case).
In a test cross, the unknown dominant genotype is crossed with recessive parent (white, WW in the given case).
(i) If the progeny consists of purple and white flowers in ratio of 1 : 1, the purple flower is a hybrid with – PW genotype. It can be seen from the given cross.

(ii) If the progeny obtained have all purple flowers, both parents are homozygous, i.e. genotype of purple flower; is PP. It can be seen from the cross that follows

Question 12.
When tall pea plants were selfed, some of the offsprings were dwarf. Explain with the help of a Punnett square.
Answer:
Punnett Square
The production of gametes by the parents, the formation of zygotes, the F₁ and F₂-generations can be explained by a diagram called Punnett square. It was developed by British geneticist Reginald C Punnett.
It is a graphical representation to calculate the probability of all possible genotypes of offsprings in a genetic cross, as shown in the figure (5.2).
The 1/4 : 1/2 : 1/4 ratio of TT : Tt : tt is mathematically condensable in the form of the binomial expression (ax+ by)², that has the gametes bearing genes T or t in equal frequency of 1/2. The expression is expanded as
(1/2T + 1/2t)² = (1/2T + 1/2t) x (1/2T + 1/2t)
= \(\frac{1}{4}\) TT + 1/2T t + 1/4tt
Though the genotypic ratios can be calculated using mathematical probability, but it is not possible to know the genotypic composition by looking at the phenotype of a dominant trait.
Therefore, to solve this problem Mendel devised test cross.

A Punnett square used to understand a typical monohybrid cross conducted by Mendel between true breeding tall plant and true-breeding dwarf plant

Question 13.
A pea plant with purple flowers was crossed with white flowers producing 50 plants with only purple flowers. On selfing, these plants produced 482 plants with purple flowers and 162 with white flowers. What genetic mechanism accounts for these results? Explain.
Answer:

Phenotypic ratio is
Purple flowered plants : White flowered plants
3(482) : 1(162)
Mechanism is as follows
(i) Factors segregate from each other that remained together in a hybrid during gamete formation.
(ii) A homozygous parent produces all gametes that are similar while heterozygous parent produces two kinds of gametes in equal ratio.

Question 14.
Write a short note on law of independent assortment.
Answer:
Law of Independent Assortment
It states that when two pairs of traits are combined in a hybrid, segregation of one pair of traits is independent to the other pair of traits. As in the dihybrid cross of Mendel the presence of new combinations, i.e. round-green and wrinkled-yellow suggests that the genes for shape of seed and colour of seed are assorted independently. The results (9:3:3:1), indicate that yellow and green seeds appear in the ratio of 9+3 : 3+1 = 3:1. Similarly, the round and wrinkled seeds appear in the ratio of 9+3 : 3+1 = 3:1.

This indicates that each of the two pairs of alternative characters viz yellow-green cotyledon colour is inherited independent of the round-wrinkled characters of the cotyledons. It means that at the time of gamete formation the factor for yellow colour enters the gametes independent of R or r, i.e. gene Y can be passed on to the gametes either with gene R or r.

Question 15.
With the help of one example, explain the phenomenon of codominance and multiple allelism in human population.
Answer:
Codominance
It is the phenomenon in which two alleles express themselves independently when present together in an organism. In other words, it is the phenomenon in which offspring shows resemblance to both the parents, e.g.
ABO blood grouping in humans.
ABO blood groups are controlled by the gene-I.

The plasma membrane of the red blood cells has sugar polymers that protrude from its surface and the kind of sugar is controlled by the gene. The gene I has three alleles I^A, I^B and i. The alleles I^A and I^B produce a slightly different form of the sugar while, allele i does not produce any sugar. In humans, each person possesses any two of the three I gene alleles. I^A and I^B are completely dominant over i. When I^B and i are present, only I^B expresses (because i does not have any sugar), same is the case with I^A and i.

But when I^A and I^B are present together, they both express their own types of sugars, this is due to codominance. Therefore, the red blood cells have both A and B-types of sugars. Since, there are three different types of alleles, there can be six different combinations. More information about blood type is discussed under multiple alleles.

In case of plants, when a red and white flowered plants are crossed, the progeny bears flowers that have red and white spots. When these plants are self-pollinated,
F₂-generation will have genotypic ratio of 1 : 2 : 1 (Red : Spotted : White).

Multiple Allelism and Inheritance of Blood Groups:
Each gene has alternative forms or allelomorphs. For example, the genes for tall and dwarf characters of pea plant are alleles or allelomorphs. Here, former is called normal or wild type and later as mutant type. Sometimes, there may not be any alternative form such mutation that results in complete elimination of a gene is known as null mutation. Sometimes silent mutation occurs in which mutation does not have any effect of all.

These mutations occur in wild gene in any direction with a possibility of formation of many alternative alleles. Some genes may occur in more than two allelic forms, i.e. a gene can mutate several times to produce several alternative expressions such genes are called multiple alleles.

The multiple alleles can be defined as set of 3, 4 or more allelomorphic genes or allele, which have arises as a result of mutation of the normal gene and which occupy the same locus in the homologous chromosomes.
A mutant gene series is said to be multiple alleles only if they possess following characteristics

(i) Occupy same locus in homologous chromosome pair.
(ii) Since, only rwo chromosomes (homologous pair) of each type present in each diploid cell hence, only two alleles of a gene are found in a cell or in a given individual.
(iii) Gametes normally contain only one chromosome out of the pair. Hence, only one allele of a gene is found in one gamete.
(iv) Due to their presence on same locus, they do not exhibit the phenomenon of crossing over in themselves.
(v) They always regulate same characteristic, but with a variable degree of efficiency.
(vi) Within a multiple series, a wild type is always dominant over normal genes and others may be dominant or intermediate.

Question 16.
How does the gene T control ABO blood groups in humans? Write the effect the gene has on the structure of red blood cells.
Answer:
ABO Blood Group
ABO blood group in humans serves as a very good example of multiple alleles. It shows the phenomenon of both complete dominance and codominance. In humans, group-A, B, AB and O types of blood groups are discovered. A set of three multiple alleles on the chromosome-9 is responsible for the four blood types.

These different types of blood groups are due to the occurrence of different types of antigens present on the surface of RBC. The genes which are responsible for the production of these antigens is being called as I. This gene is present in the form of three alleles which are I^A, I^B and VI i. I^A codes for enzyme that adds galactosamine on the surface of RBC. So, blood group-A individuals have only galactosamine added on the cell surface of RBC.

IB codes for the enzyme that add galactose on the surface of RBC and so blood group-B individuals have only galactose added to the cell surface of RBC.
Individuals belonging to AB blood group possess both the sugars added on the surface of RBC and individuals with blood group-O do not have any sugar added on the surface of RBC. People homozygous for the recessive i allele belong to blood group-O.

The Relationship Between ABO Genotype and Blood type (group)

Both I^A and IB are dominant to i. Thus, you will express blood group-A, if you are either I^A/Aor I^Ai and you will express blood group-B, if you are either I^B / I^B or I^B/i. Heterozygous I^A/I^B individuals express blood group-AB.

Question 17.
Briefly mention the contribution of TH Morgan in genetics.
Answer:
Morgan and his group also found that even when genes were grouped on the same chromosome, some genes were tightly linked, i.e. linkage is stronger between two genes, if the frequency of recombination is low (cross-A). Whereas, the frequency of recombination is higher, if genes are loosely linked, i.e. linkage is weak between two genes (cross-B) as given in below.

Linkage : Results of two dihybrid crosses conducted by Morgan. Cross ‘A’ shows crossing between genes y and w ; Cross ‘B’ shows crossing between genes w and m. Here, dominant wild type alleles are represented with ( + ) sign in superscript

Those traits present on same chromosome, which do not show any production of recombinants are completely linked which is known as complete linkage and it is very rare.

Question 18.
Linkage and crossing over of genes are alternatives of each other. Justify with the help of an example.
Answer:
Linkage is the tendency of certain loci or alleles (genes) to be inherited together while, crossing over is the segregation of genes, e.g. the genes on a chromosome either follow linkage path or cross over, to form the gametes during gametogenesis in human.

Question 19.
Two heterozygous parents are crossed. If the two loci are linked, what would be the distribution of phenotypic features in F₁-generation for a dihybrid cross?
Answer:
If the two loci in parents are completely linked, there would be no segregation. The F₁-generation will exhibit parental characters only. But, if the two loci are incompletely linked, then segregation would occur partly and the F₁-generation will exhibit both parental and recombinant characteristics, but the recombinants will be in a very small proportion.

Question 20.
Explain two situations, when independent assortment of genes occurs, resulting in 50% recombinants.
Answer:
Two situations are
(i) When the genes of different traits are located on the same chromosome and must be distantly located to enhance the recombination frequency.
(ii) When the genes of different traits may be located on different chromosomes.

Question 21.
In a dihybrid cross, white-eyed, yellow-bodied female Drosophila was crossed with red-eyed, brown-bodied male Drosophila, produced in F₂-generation are 1.3% recombinants and 98.7% progeny with parental type combinations. This observation of Morgan deviated from Mendelian F₁-phenotypic .dihybrid ratio. Explain giving reasons, Morgan’s observations.
Answer:
Morgan saw that when the two genes in dihybrid cross were situated on same chromosome, the proportion of parental gene combinations were higher than non-parental type.

This is due to physical association of genes on a chromosome or linkage. In Morgan’s experiment, the genes for eye and body colour show linkage and do not always allow crossing to over during gamete formation. Hence, parental type progeny is in greater ratio than that of recombinants.

Question 22.
How do biologists use cross over frequencies to map genes on chromosomes?
Answer:
The farther the genes are on a chromosome, the more frequently they will cross over. By comparison, genes that are close together on a chromosome, are less likely to be separated. The analysis of how often the traits appear together, helps to establish linkage map, which shows the relative positions of genes on chromosomes.

Question 23.
(i) Explain the phenomenon of multiple allelism and codominance taking ABO blood group system as an example.
(ii) What is the phenotype of the following
(a) I^Ai (b) ii
Answer:
(i)
These mutations occur in wild gene in any direction with a possibility of formation of many alternative alleles. Some genes may occur in more than two allelic forms, i.e. a gene can mutate several times to produce several alternative expressions such genes are called multiple alleles.

It is the phenomenon in which two alleles express themselves independently when present together in an organism. In other words, it is the phenomenon in which offspring shows resemblance to both the parents, e.g.
ABO blood grouping in humans.
ABO blood groups are controlled by the gene-I.

The plasma membrane of the red blood cells has sugar polymers that protrude from its surface and the kind of sugar is controlled by the gene. The gene I has three alleles IA, IB and i. The alleles IAand IB produce a slightly different form of the sugar while, allele i does not produce any sugar. In humans, each person possesses any two of the three I gene alleles. IA and IB are completely dominant over i. When IB and i are present, only IB expresses (because i does not have any sugar), same is the case with IA and i.

But when IA and IB are present together, they both express their own types of sugars, this is due to codominance. Therefore, the red blood cells have both A and B-types of sugars. Since, there are three different types of alleles, there can be six different combinations. More information about blood type is discussed under multiple alleles.

(ii) (a) I^Ai – ‘A’ blood group
(b) ii – ‘O’ blood group

Question 24.
A child has blood group ‘O’. If the father has blood group ‘A’ and mother’s blood group is B, workout the genotypes of the parents and the possible genotypes of the other offsprings. .
Answer:
The child with blood group ‘O’ will have homozygous recessive alleles. Therefore, both the parents should be heterozygous, i.e. the genotype of father will be I^Ai afid the mother will be I^Bi.

The other possible genotypes of offsprings will be ‘A’, ‘B’ and ‘AB’ blood groups.

Question 25.
Explain the following
(i) What is the most common example of pleiotropy in humans?
(ii) How are the pleiotropic genes useful?
Answer:
(i) In humans, the phenylketonuria, a disorder caused by mutation in the gene coding the enzyme phenylalanine hydroxylase serves as the best example. In this disorder, phenylalanine hydroxylase enzyme is deficient and is needed to convert essential amino acid phenylalanine to tyrosine.

Thus, tyrosine is not present which is being required for normal functioning of the body. Phenylalanine gets converted to phenylpyruvate which also creates problems. Tyrosine is required for protein biosynthesis and also as a precursor for neurotransmitters like norepinephrine, for pigment melanin and for hormone thyroxine. Thus, it leads to hair and skin pigmentation and mental retardation.

(ii) The pleiotropic genes provide us valuable information regarding the evolution of different genes and gene families. The pleiotropy reflects the fact that most proteins have multiple roles in distinct cell types.
Thus, any genetic change that alters gene expression or function can potentially have wide ranging effects in a variety of tissues.

Question 26.
State the significance of crossing over.
Answer:
The significance of crossing over are as follows
(i) Crossing over produced recombinations or new gene combinations which form variations which are the raw material in the formation of new species and evolution.
(ii) Useful recombination produced by crossing over are used in producing new varieties of plants and animals in breeding programme.
(iii) It is used for preparing linkage maps or determining location of genes in chromosomes.

Question 27.
What is chiasma?
Answer:
Crossing over takes place in the pachytene stage of meiosis-I, where there is exchange of segments between two non-sister chromatids. It occurs at several points all along their length and the point of contact between these non-sister chromatids of homologous chromosomes is called chiasma.

Question 28.
What is linkage? Mention its significance.
Answer:
The physical association of two genes on a chromosome is called linkage. It is the tendency of genes on a chromosome to remain together and is passed as such in the next generation. It is significant as it brings more parental gene combination types.

Question 29.
Write a short note on crossing over.
Answer:
Crossing over is the combination of genes due to the exchange of genetic material resulting from interchange of corresponding segments or parts of homologous chromosomes. This takes place in the pachytene stage of meiosis-I. It brings out new combination or recombination of genes.

Question 30.
Define multiple allelism.
Answer:
More than two alternative forms (allele) of a gene occupying the same locus on a chromosome in a population are known as multiple alleles. The ABO blood grouping is a good example of multiple alleles. In this case, more than two, i.e. three alleles are present governing the same character. Multiple alleles can be found only when population studies are made.

Question 31.
During his studies on genes in Drosophila that were sex-linked, TH Morgan found F₂-population phenotypic ratio to be deviated from expected 9: 3: 3:1. Explain the conclusion he arrived at.
Answer:
Conclusion of Morgans experiments are
(i) Genes were located on the X-chromosome.
(ii) When the two genes in a dihybrid cross were studied on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type. Morgan stated this association as ‘linkage’ to describe the physical association of genes on a chromosome.
(iii) Morgan and his group also found that even when genes were grouped on the same chromosome, some genes were tightly linked (low recombination) and others were loosely linked (high recombination).

Question 32.
A woman with blood group ‘A’ marries a man with blood group ‘O’. Discuss the possibilities of the inheritance of the blood group in the following starting with ‘Yes’ or ‘No’ for each.
(a) They produce children with blood group ‘A’ only.
(b) They produce children some with ‘O’ blood group and some with ‘A’ blood group.
Answer:
(a) No, it is not necessary as mother could have a genotype I^AI^A or I^Ai. If the genotype is I^A, all the offsprings would have ‘A’ blood group, but in the second case, offsprings can have either ‘A’ or ‘O’ blood group as their father has ‘O’ blood group.
(b) Yes, if the mother is A (genotype I^Ai) and father has ‘O’ (genotype ii) blood group, then the blood group of some children can be ‘O’ and some can be with blood group ‘A’.

Question 33.
Persons of which blood group are universal recipient?
Answer:
ABO blood group in humans serves as a very good example of multiple alleles. It shows the phenomenon of both complete dominance and codominance. In humans, group-A, B, AB and O types of blood groups are discovered. A set of three multiple alleles on the chromosome-9 is responsible for the four blood types.

These different types of blood groups are due to the occurrence of different types of antigens present on the surface of RBC. The genes which are responsible for the production of these antigens is being called as I. This gene is present in the form of three alleles which are I^A, I^B and I⁰/i. I^A codes for enzyme that adds galactosamine on the surface of RBC. So, blood group-A individuals have only galactosamine added on the cell surface of RBC.

Question 34.
Explain ‘ABO blood group’.
Answer:
ABO blood group in humans serves as a very good example of multiple alleles. It shows the phenomenon of both complete dominance and codominance. In humans, group-A, B, AB and O types of blood groups are discovered. A set of three multiple alleles on the chromosome-9 is responsible for the four blood types.

These different types of blood groups are due to the occurrence of different types of antigens present on the surface of RBC. The genes which are responsible for the production of these antigens is being called as I. This gene is present in the form of three alleles which are I^A, I^B and I⁰ / i. I^A codes for enzyme that adds galactosamine on the surface of RBC. So, blood group-A individuals have only galactosamine added on the cell surface of RBC.

I^B codes for the enzyme that add galactose on the surface of RBC and so blood group-B individuals have only galactose added to the cell surface of RBC.

Individuals belonging to AB blood group possess both the sugars added on the surface of RBC and individuals with blood group-O do not have any sugar added on the surface of RBC. People homozygous for the recessive i allele belong to blood group-O.

Question 35.
Inheritance pattern of flower colour in garden pea plant and snapdragon differs. Why is this difference observed? Explain showing the crosses upto F₂-generation.
Answer:
(i) Inheritance pattern of flower colour in garden pea plant

Phenotypic Ratio Purple : White
3 : 1
Genotypic Ratio PP : Pp : pp
1 : 2 : 1
Inheritance of flower colour in garden pea shows true dominance. In F₁ -generation, dominant colour purple is expressed and in F₂-generation both dominant (purple) and recessive (white) colour are expressed in the ratio of 3:1.
(ii) Inheritance pattern of flower colour in snapdragon is given below

Phenotypic Ratio Red : Pink : White
1 : 2 : 1
Genotypic Ratio RR : Rr : rr
1 : 2 : 1
Inheritance in snapdragon flower colour shows incomplete dominance. In this phenomenon, neither of the two alleles are completely dominant over the other and the hybrid is-intermediate between the two. Hence, red is a homozygous dominant, white is a homozygous recessive, while hybrid is an intermediate, i.e. pink.

Question 36.
Explain the genetic basis of blood grouping in human population.
Answer:
‘ABO’ blood grouping in human shows the phenomenon of codominance (alleles are able to express themselves independently when present together) and multiple allelism (more than two alleles govern the same character).

ABO blood group in humans serves as a very good example of multiple alleles. It shows the phenomenon of both complete dominance and codominance. In humans, group-A, B, AB and O types of blood groups are discovered. A set of three multiple alleles on the chromosome-9 is responsible for the four blood types.

These different types of blood groups are due to the occurrence of different types of antigens present on the surface of RBC. The genes which are responsible for the production of these antigens is being called as I. This gene is present in the form of three alleles which are I^A, I^B and I⁰ / i. I^A codes for enzyme that adds galactosamine on the surface of RBC. So, blood group-A individuals have only galactosamine added on the cell surface of RBC.

I^B codes for the enzyme that add galactose on the surface of RBC and so blood group-B individuals have only galactose added to the cell surface of RBC.

Individuals belonging to AB blood group possess both the sugars added on the surface of RBC and individuals with blood group-O do not have any sugar added on the surface of RBC. People homozygous for the recessive i allele belong to blood group-O.

Question 37.
Can a child have blood group ‘O’ if his/her parents have blood group ‘A’ and ‘B’? Explain.
Answer:
The child with blood group ‘O’ will have homozygous recessive alleles. Therefore, both the parents should be heterozygous, i.e. genotype of father will be I^A i or I^B i and of mother will be I^A i or I^B i. A child have blood group ‘O’ in the following two cases
Case I When father is I^A i and mother is I^B

The offsprings will have the above possible blood groups, i.e. AB, A, B and O.
Case II When father is I^B i and mother is I^A i.

Question 38.
How do genes and chromosomes share similarity from the point of view of genetical studies?
Answer:
By 1902, the chromosome movement during meiosis had been worked out. Walter Sutton and Theodore Boveri (1902) noted that the behaviour of chromosomes was parallel to the behaviour of genes and used chromosome movement to explain Mendel’s laws. They studied the behaviour of chromosomes during mitosis (equational division) and during meiosis (reduction division). The chromosomes as well as genes occur in pairs and the two alleles of a gene pair are located on homologous sites of homologous chromosomes.
Chromosomes segregate when germ cells are formed.

Long Answer Type Questions

Question 1.
(i) Explain monohybrid cross taking seed coat colour as a trait in Pisum sativum. Workout the cross upto F₂-generation.
(ii) State the laws of inheritance that can be derived from such a cross.
(iii) How is the phenotypic ratio of F₂-generation different in a dihybrid cross?
Answer:
(i) In a monohybrid cross, when homozygous dominant and homozygous recessive parents are crossed, F₁ -hybrid would be heterozygous for the trait and would express the dominant allele.

(ii) The hybrid is heterozygous containing both alternative alleles (Y and y) but only one trait,
i.e. yellow colour appeared and green colour trait is suppressed in F₁ -generation. This shows that yellow seed colour is dominant over the green seed colour trait. This explains Mendel’s law of dominance.
(iii) Phenotypic ratio in F₂-generation, Yellow seeds : Green seeds (3 : 1) in monohybrid cross and in dihybrid cross 9 : 3 : 3 : 1.

Question 2.
Workout a typical Mendelian dihybrid cross and state the law that he derived from it.
Answer:
Dihybrid Cross:
When two or more than two characters are taken in a cross it is called as polyhybrid cross, e.g. dihybrid cross, trihybrid cross, etc. A dihybrid cross is a cross involving two pairs of contrasting characters. For example, when a cross is made between yellow-round and wrinkled green seeds (both pureline homozygous), plants with only yellow round seeds are seen in F₁-generation but in F₂-generation, four types of combinations are observed.

Two of these combinations are similar to the parental combinations and others are new combinations. These are round green and wrinkled yellow.
The cross can be seen as shown in the figure

Phenotypic Ratio
Round yellow : Round green : Wrinkled yellow : Wrinkled green = 9 : 3 : 3 : 1
Genotypic Ratio 1 :2:2:4:1 :2:1 :2:1
Dihybrid cross between the two parents differed in two pairs of contrasting traits, i.e. seed colour and seed shape

The ratio of four combinations in F₂-generation comes out to be 9 (round, yellow) : 3 (round, green) : 3 (wrinkled, yellow) : 1 (wrinkled, green). This ratio is called phenotypic dihybrid ratio. Phenotypic ratio of dihybrid test cross is 1 : 1 : 1 : 1.

Mendel’s Postulate Based on Dihybrid Cross:
Based on the result obtained from dihybrid crosses or two gene interaction, Mendel proposed the fourth postulate, i.e. law of independent assortment.

Question 3.
You are given tall pea plants with yellow seeds, whose genotypes are unknown. How would you find the genotype of these plants? Explain with the help of cross.
Answer:
The given tall pea plant with yellow seeds need to be crossed with a dwarf plant with green seeds.
(i) The dominant traits are tallness and yellow colour of seeds. The recessive traits are dwarfness and green colour of seeds.
(ii) Cross between tall plant/yellow seeds and dwarf plant/green seeds.
Cross showing heterozygous nature for both traits.

In this cross, the F₁ -generation shows four phenotypes in the ratio of 1 : 1 : 1 : 1. So, the given plant is heterozygous for both the traits.

Question 4.
Explain Mendel’s dihybrid cross with a checker board.
Or Describe Mendel’s dihybrid cross with checker board.
Answer:
Dihybrid Cross:
When two or more than two characters are taken in a cross it is called as polyhybrid cross, e.g. dihybrid cross, trihybrid cross, etc. A dihybrid cross is a cross involving two pairs of contrasting characters. For example, when a cross is made between yellow-round and wrinkled green seeds (both pureline homozygous), plants with only yellow round seeds are seen in F₁-generation but in F₂-generation, four types of combinations are observed.

Two of these combinations are similar to the parental combinations and others are new combinations. These are round green and wrinkled yellow.
The cross can be seen as shown in the figure

Phenotypic Ratio
Round yellow : Round green : Wrinkled yellow : Wrinkled green = 9 : 3 : 3 : 1
Genotypic Ratio 1 :2:2:4:1 :2:1 :2:1
Dihybrid cross between the two parents differed in two pairs of contrasting traits, i.e. seed colour and seed shape

The ratio of four combinations in F₂-generation comes out to be 9 (round, yellow) : 3 (round, green) : 3 (wrinkled, yellow) : 1 (wrinkled, green). This ratio is called phenotypic dihybrid ratio. Phenotypic ratio of dihybrid test cross is 1 : 1 : 1 : 1.

Mendel’s Postulate Based on Dihybrid Cross:
Based on the result obtained from dihybrid crosses or two gene interaction, Mendel proposed the fourth postulate, i.e. law of independent assortment.

Question 5.
Give an account of Mendel’s law of inheritance.
Or State and explain Mendel’s low of dominance.
Answer:
Dihybrid Cross:
When two or more than two characters are taken in a cross it is called as polyhybrid cross, e.g. dihybrid cross, trihybrid cross, etc. A dihybrid cross is a cross involving two pairs of contrasting characters. For example, when a cross is made between yellow-round and wrinkled green seeds (both pureline homozygous), plants with only yellow round seeds are seen in F₁-generation but in F₂-generation, four types of combinations are observed.

Two of these combinations are similar to the parental combinations and others are new combinations. These are round green and wrinkled yellow.
The cross can be seen as shown in the figure

Phenotypic Ratio
Round yellow : Round green : Wrinkled yellow : Wrinkled green = 9 : 3 : 3 : 1
Genotypic Ratio 1 :2:2:4:1 :2:1 :2:1
Dihybrid cross between the two parents differed in two pairs of contrasting traits, i.e. seed colour and seed shape

The ratio of four combinations in F₂-generation comes out to be 9 (round, yellow) : 3 (round, green) : 3 (wrinkled, yellow) : 1 (wrinkled, green). This ratio is called phenotypic dihybrid ratio. Phenotypic ratio of dihybrid test cross is 1 : 1 : 1 : 1.

Mendel’s Postulate Based on Dihybrid Cross:
Based on the result obtained from dihybrid crosses or two gene interaction, Mendel proposed the fourth postulate, i.e. law of independent assortment.

Law of Independent Assortment:
It states that when two pairs of traits are combined in a hybrid, segregation of one pair of traits is independent to the other pair of traits. As in the dihybrid cross of Mendel the presence of new combinations, i.e. round-green and wrinkled-yellow suggests that the genes for shape of seed and colour of seed are assorted independently. The results (9:3:3:1), indicate that yellow and green seeds appear in the ratio of 9+3 : 3+1 = 3:1. Similarly, the round and wrinkled seeds appear in the ratio of 9+3 : 3+1 = 3:1.

This indicates that each of the two pairs of alternative characters viz yellow-green cotyledon colour is inherited independent of the round-wrinkled characters of the cotyledons. It means that at the time of gamete formation the factor for yellow colour enters the gametes independent of R or r, i.e. gene Y can be passed on to the gametes either with gene R or r.

Odisha State Board CHSE Odisha Class 12 Biology Important Questions Chapter 4 Reproductive Health Important Questions and Answers.

CHSE Odisha 12th Class Biology Important Questions Chapter 4 Reproductive Health

Reproductive Health Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
A sexually transmitted disease symptomised by the development of chancre on the genitals is caused by
(a) Hepatitis -B virus
(b) Treponema pallidum
(c) HIV
(d) Neisseria
Answer:
(b) Treponema pallidum

Question 2.
The preventive measures of sexually transmitted diseases include
(a) sex hygiene
(b) avoiding multiple sex partners
(c) use of condom
(d) All of the above
Answer:
(d) All of the above

Question 3.
Which of the following is the component of oral pills?
(a) Progesterone
(b) Oxytocin
(c) Relaxin
(d) FSH
Answer:
(a) Progesterone

Question 4.
Oral contraceptives check ovulation by inhibiting the secretion of
(a) follicle stimulating hormone
(b) luteinizing hormone
(c) Both (a) and (b)
(d) All of the above
Answer:
(c) Both (a) and (b)

Question 5.
Which one is not a terminal birth control method?
(a) Vasectomy
(b) Tubectomy
(c) Hysterectomy
(d) Copper-T
Answer:
(d) Copper-T

Question 6.
What is the surgical method for preventing pregnancy in which vas deferens is incised?
(a) Tubectomy
(b) Vasectomy
(c) Sterilisation
(d) Hysterectomy
Answer:
(b) Vasectomy

Question 7.
AIDS cannot be spread from one person to other through
(a) sexual contacts
(b) blood transfusion
(c) placental contacts
(d) kissing each other
Answer:
(d) kissing each other

Question 8.
One of the legal methods of birth control is
(a) by having coitus at the time of day break
(b) by a premature ejaculation during coitus
(c) abortion by taking an appropriate medicine
(d) by abstaining from coitus from day 10-17 of the menstrual cycle
Answer:
(d) by abstaining from coitus from day 10-17 of the menstrual cycle

Question 9.
What is a safe period?
(a) A week before and after menses
(b) A week before menses
(c) Two weeks after menses
(d) Two weeks before menses
Answer:
(b) A week before menses

Question 10.
The ovulation time during menstrual cycle is marked by
(a) changes in cervical mucous
(b) changes in body temperature
(c) changes in eating habit
(d) changes in behavioural habits
Answer:
(d) changes in behavioural habits

Question 11.
GIFT was first attempted by
(a) Steptoe
(b) Edwards
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Question 12.
ZIFT is a spin-off of
(a) GIFT
(b) ICSI
(c) IVF
(d) IUI
Answer:
(a) GIFT

Question 13.
Blockage of Fallopian tube can be identified using
(a) blood test
(b) hysterosalpingography
(c) ovarian reserve testing
(d) general physical testing
Answer:
(b) hysterosalpingography

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

Question 1.
Oral pills are very popular contraceptives among the rural women.
Answer:
Oral pills are very popular contraceptives among the urban women.

Question 2.
Surgical methods of contraception prevent gamete formation.
Answer:
The surgical methods of contraception prevent gamete transfer from the organs of production to the site of fertilisation.

Question 3.
Vaults are hormone releasing IUDs.
Answer:
LNG-20

Question 4.
The mixing of the sperm and egg is called ICSI.
Answer:
insemination

Question 5.
Follicular atresia is done to remove eggs from woman’s body
Answer:
Follicular aspiration

Question 6.
The first Indian test tube baby was Rahul.
Answer:
Kumari Harsha

Fill in the blanks

Question 1.
A contraceptive device consisting of a small thimble-shaped cup that is placed over the uterus to prevent the entrance of spermatozoa …………… .
Answer:
cervical cap

Question 2.
A surgical procedure performed on males in which the vas deferens are cut and tied is known as …………… .
Answer:
vasectomy

Question 3.
Removal of gonads, often referring to the removal of male testes is called ……………… .
Answer:
castration

Question 4.
Conventional vasectomy is called ……….. surgery.
Answer:
scalpel

Question 5.
Surgical removal of both the testes is called ………….. .
Answer:
vasectomy

Question 6.
The ejaculatory duct obstruction in males is confirmed by ………….. .
Answer:
ultrasound

Question 7.
Fertility treatment with donor eggs is usually done using …………….. .
Answer:
IVF

Question 8.
Success rate of ZIFT is …………. .
Answer:
64.8%

Question 9.
In …………….. egg is fertilised outside the body and then inserted into oviduct.
Answer:
in vitro technique

Express in one or two word(s)

Question 1.
Give the name of STD, which can be transmitted through contaminated blood.
Answer:
AIDS.

Question 2.
Name the IUD that promotes the cervix hostility to sperms.
Answer:
Progestasert

Question 3.
Name the technique other than amniocentesis which is used to determine genetic disorder in foetus.
Answer:
Chorionic villi sampling.

Question 4.
Name the causative agent of AIDS.
Answer:
HIV

Question 5.
Name the organisation that developed Saheli pill.
Answer:
CDRI

Question 6.
Enlist two possible causes of fertility in relation to ovulation.
Answer:
Polycystic ovarian syndrome and hyperprolactinemia.

Question 7.
Give the full form of ZIFT.
Answer:
ZIFT stands for Zygote Intra Fallopian Transfer.

Question 8.
Name the test performed to find out how effective the eggs are after ovulation.
Answer:
Ovarian reserve testing

Short AnswerType Questions

Question 1.
Write note on sexually transmitted diseases.
Answer:
Sexually Transmitted Diseases (STDs) spread from one person to other through intimate contact. STDs can affect male or female of any age and background. These diseases are also known as Venereal Diseases (VDs) or Reproductive Tract Infections (RTIs) or Sexually Transmitted Infections (STIs). STDs are major threat to a healthy society as these are reported to be high among youths of age 15-24 years. Except HIV, hepatitis-B and genital herpes, all other diseases are completely curable.

Question 2.
Why STDs are considered as self-invited diseases?
Answer:
Sexually Transmitted Diseases (STDs) can be considered self-invited diseases because one could be free of these infections, by following the simple principles given below

1. Avoid sex with unknown partners/multiple partners.
2. Always use condoms during coitus.
3. In case of doubt one should go to a qualified doctor for early detection and get complete treatment if diagnosed with disease.

If all the above said precautions are not strictly adopted by people, they are inviting STDs to infect them.

Question 3.
All Reproductive Tract Infections (RTIs) are STDs, but all STDs are not RTIs. Justify with example.
Answer:
Among the common STDs, hepatitis-B and AIDS are not infections of the reproductive organs though their mode of transmission could be through sexual contact also.

All other diseases like gonorrhoea, syphilis, genital herpes, hepatitis-B are transmitted through sexual contact and are also infections of the reproductive tract so, these are STDs as well as RTIs, whereas, AIDS and hepatitis are STDs, but not RTIs.

Question 4.
Write a note on barrier and surgical method of birth control.
Answer:
Barrier methods These methods prevent sperms and ovum, from physically meeting in order to prevent fertilisation. These methods are available for both males and females. These are as follows

1. Condoms These are made of thin rubber or latex sheath used to cover the penis in males and vagina and cervix in females.
2. Cervical caps These are also made of rubber and are inserted into the female reproductive system to cover the cervix during intercourse.

Surgical (sterilisation) methods These are the terminal methods used by male/female partner to prevent any more pregnancies.
These are the permament methods, which block the transport of gametes and thereby contraception. It is available in the form of vasectomy in men and tubectomy in womem.

Question 5.
Removal of gonads cannot be considered as a contraceptive option. Why?
Answer:
Removal of gonads cannot be considered as a contraceptive option because it not only stops the production of gametes, but also stops the secretion of many other-hormones which are required for normal body function. This is also an irreversible method that means once these are removed, then these cannot be replaced and the person will remain infertile for whole life.

Question 6.
Is the use of contraceptives justified? Give reasons.
Answer:
Use of contraceptives is justified due to the following reasons

1. In the absence of contraceptives, the population growth rate will rise at explosive rate and there will be scarcity of even the basic necessities.
2. Contraceptives provide an option for planning the family by spacing the pregnancies and avoiding unwanted pregnancies.
3. Contraceptives also guard against STDs to some extent.

Question 7.
Define spermicides. Also, mention their use and mode of contraceptive action.
Answer:
The chemicals that are used to kill the sperms to prevent physical meeting of sperms and eggs are called spermicides. They are available in the form of creams, jellies and foams and are applied to the uterine lining to kill the sperms.

Question 8.
Why intensely lactating mothers do not generally conceive? Explain.
Answer:
Yes, breastfeeding is one of the natural contraceptive methods. It reduces fertility by affecting the production of certain reproductive hormones. It is known to suppress the production of Gonadotropin Releasing Hormone (GnRH) and Follicle Stimulating Hormone (FSH). The release of these hormones triggers ovulation. Breast feeding also leads to increase level of prolactin, a hormone that inhibits ovulation.
So, even when a woman ovulates, her likelihood of conceiving is low if she is breast feeding.

Question 9.
Mention three advantages of lactational amenorrhea as a contraceptive method.
Answer:
The three advantages of lactational amenorrhea as a contraceptive method are mentioned as below

• If the mother is breast feeding completely, she would not ovulate, so chances of conception would be low.
• It does not require the use of any pill or devices for birth control.
• It does not have any side effects.

Question 10.
How is tubectomy effective in birth control?
Or What is tubectomy?
Answer:
Tubectomy is a surgical method of female fertilisation, where small part of Fallopian tubes is removed or tied up through a small incision in the abdomen or vagina in female. This process is irreversible and do not allow the reunion of both the ends to block the passage of ova through them. Thus, tubectomy is an effective method of birth control.

Question 11.
Classify the following contraceptive measures into different methods of birth control.
(i) Saheli
(ii) Tubectomy
(iii) Vasectomy
(iv) Diaphragm
(v) Cervical caps
Answer:
(i) Oral pills
(ii) Surgical method
(iii) Surgical method
(iv) Barrier method
(v) Barrier method

Question 12.
Write note on amniocentesis.
Or What is amniocentesis? How is it misused?
Answer:
Amniocentesis is a prenatal diagnostic test to detect the chromosomal pattern of the cells in the amniodc fluid that surrounds the developing foetus in the womb.
It is misused to determine the sex of pre-born child that leads to female foeticide of an unborn girl child.

This ban is necessary as this sex determination technique has been misused to kill girl child before birth.
The ban is justified to prevent female foeticide which could lead to change in sex ratio of the population.

Question 13.
If implementation of better techniques and new strategies are required to provide more efficient care and assistance to people, then why is there a statutory ban on amniocentesis? Write the use of this technique and give reason to justify the ban.
Answer:
Amniocentesis is one such technique that helps to determine any chromosomal abnormalities or genetic disorder and sex as well as foetal infections, by using minute amount of amniotic fluid, surrounding the foetus. Though implementation of better techniques and new strategies are required to provide more efficient care and assistance to people, but there is a statutory ban on amniocentesis to legally check female foeticide. This ban is necessary as this sex determination technique has been misused to kill girl child before birth.
The ban is justified to prevent female foeticide which could lead to change in sex ratio of the population.

Question 14.
Describe sexually transmitted diseases.
Answer:
Sexually Transmitted Diseases

• Chlamydia
• Ganital herpes [Caused by Herpes Simplex Virus (HSV)]
• Genital warts
• Gonorrhoea
• Hepatitis-B [caused by Hepatitis-B Virus (HBV)]
• HIV and AIDS
• Pelvic Inflammatory Disease (PID)
• Public lice infection (caused by crabs)
• Syphilis
• Trichomoniasis

Question 15.
What will happen, if gametes are directly transferred to the uterus?
Answer:
The procedure of GIFT involves the transfer of female gamete to the Fallopian tube. Gametes cannot be transferred to the uterus to achieve the same result because the uterine environment is not congenital for the survival of the gamete. If directly transferred to the uterus, they will undergo degeneration or could be phagocytosed and hence, viable zygote would not be formed.

Question 16.
Explain, how surrogacy is helpful in case of an infertile woman?
Or What is surrogate mother?
Answer:
Surrogacy is a practice in which a woman (a surrogate mother) bears a child for a couple unable to produce children, usually because the wife is infertile or unable to carry a pregnancy to full term.
The surrogate mother is impregnated with the use of artificial insemination or through the implantation of the embryo produced by in vitro fertilisation.

Question 17.
Why is ZIFT a boon to childless couples? Explain the procedure of ZIFT.
Or Write a note on ZIFT.
Answer:
Zygote Intra Fallopian Transfer (ZIFT) is a boon to the couple where the female cannot conceive naturally.
In this method, fertilisation is carried out in vitro (in the laboratory conditions) and the zygote nearly embryo so, formed (with upto 8-blastomeres) is transferred into the Fallopian tube.

Question 18.
Enlist five reasons that may cause semen to be abnormal.
Answer:
The following reasons may cause the semen to be abnormal

• Testicular infection, cancer or surgery
• Diseases like-anaemia, diabetes, thyroid malfunctioning
• Overheating of the testicles
• Ejaculation disorders
• Genetic abnormality

Question 19.
How is sperm donation helpful to infertile couples? Explain.
Answer:
Due to change in life style and eating habits, many couples are suffering from infertility, now-a-days, i. e. they are not able to conceive. Infertility may occur in both men and women.
In order to solve the problems related to infertility, the advancement in technologies has come to the rescue. Sperm donation has become boon for the couple, who suffer from infertility and want their own child.
Sperm donation is helpful in following cases in males

• When sperms are blocked by an abnormality in the epididymis and in the testis.
• Sexual dysfunction.
• The absence of sperms in males.

Question 20.
Males in whom testes fail to descend to th*e scrotum are generally infertile. Why?
Answer:
Testes are very sensitive to temperature. If they do not descend into the scrotum prior to adolescence, then they would stop producing sperms and will lead to infertility in males, a condition known as cryptochoridism.

Question 21.
Mention the primary aim of the ‘Assisted Reproductive Technology’ (ART) programme.
Answer:
‘Assisted Reproductive Technology’ (ART) is the collection of certain special techniques. The primary aim of the ART programmes is to assist infertile couples to have children through certain special techniques (like ZIFT, IUT, GIFT, ICSI, AI, etc.) when corrective treatment for infertility problems is not possible.

Question 22.
What is IVF?
Answer:
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.

Question 23.
Shyam Lai and his wife have been advised by the gynaecologist to go for artificial insemination. What may have been the reason for the doctor to give such an advice? Explain.
Answer:
It signifies in the case of male infertility, where the sperm count is low. That’s why the couple has been advised to undergo artificial insemination.

Differentiate between the following (for complete chapter)

Question 1.
In vitro and In vivo fertilisation.
Answer:
Differences between in vitro and in vivo fertilisation are as follows

In vitro fertilisation	In vivo fertilisation
Fertilisation of egg with sperm occurs in lab under controlled conditions.	Fertilisation occurs inside human body.
It is performed by collecting the contents from a woman’s Fallopian tubes or uterus to mix with sperm.	The mixing of ova and sperm for fertilisation occurs in uterus of woman only.
It is an artificial process.	It is a natural process.

Question 2.
Syphilis and Gonorrhoea.
Answer:
Differences between syphilis and gonorrhoea are as follows

Syphilis	Gonorrhoea
It is a STD caused by pathogen Treponema pallidum.	It is a STD caused by bacterium Neisseria gonorrhoeae.
It generally affects external genital organs, penis of male and vagina of female.	It causes infection in urethra in males and Bartholin’s gland in females.
Penicillin may control the disease at primary and secondary stages.	For treatment of disease spectinomycin and tetracycline are given.

Question 3.
GIFT and ZIFT
Answer:
Differences between GIFT and ZIFT are as follows

GIFT	ZIFT
In this technique eggs are removed from woman’s ovaries and placed in one of the Fallopian tubes, along with man’s semen.	In this technique eggs are removed from ovulating woman’s ovaries and in vitro fertilised.
Fertilisation take place in woman’s uterus.	Fertilisation take place outside the uterus.
After fertilisation in uterus resulting zygote then implants and women become pregnant.	After invitro fertilisation, resulting zygote placed in the Fallopian tube by the use of laproscopy.