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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.

Odisha State Board CHSE Odisha Class 12 Biology Important Questions Chapter 3 Human Reproduction Important Questions and Answers.

CHSE Odisha 12th Class Biology Important Questions Chapter 3 Human Reproduction

Human Reproduction Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
Which organ’s outer covering is tunica albuginea?
(a) Testis
(b) Ovary
(c) Kidney
(d) Brain
Answer:
(a) Testis

Question 2.
Testicular descent to the inguinal region is effected by
(a) Testosterone
(b) LH
(c) AMH
(d) FSH
Answer:
(c) AMH

Question 3.
The temperature of scrotum is maintained by
(a) perspiration
(b) evaporation
(c) counter-current heat exchange
(d) All of the above
Answer:
(d) All of the above

Question 4.
Prostate gland in males lies
(a) inferior to bladder
(b) posterior to bladder
(c) anterior to bladder
(d) for away from bladder
Answer:
(a) inferior to bladder

Question 5.
Graafian follicle is observed in the ovary of
(a) rohu
(b) Amphioxus
(c) salamander
(d) human
Answer:
(d) human

Question 6.
If fertilisation does not occur. Graafian follicle is converted to which endocrine gland?
(a) Corpus luteum
(b) Corpus albicans
(c) Corpus spongiosum
(d) Corpora cavernosa
Answer:
(b) Corpus albicans

Question 7.
Glands of skene in females are
(a) greater vestibular
(b) paraurethral
(c) Bartholin
(d) homologous to male’s Cowper’s gland
Answer:
(b) paraurethral

Question 8.
Seminal plasma is rich is
(a) fructose
(b) calcium
(c) Both (a) and (b)
(d) None of the above
Answer:
(c) Both (a) and (b)

Question 9.
Corpus luteum secrete
(a) oestrogen
(b) progesterone
(c) relaxin
(d) All of the above
Answer:
(d) All of the above

Question 10.
Spermatids are
(a) motile
(b) immotile
(c) non-dividing
(d) dead
Answer:
(b) immotile

Question 11.
The part of sperm that gets embedded in seminiferous tubules after spermiogenesis is
(a) tail
(b) middle piece
(c) neck
(d) head
Answer:
(d) head

Question 12.
Egg mother cells are formed by
(a) meiosis
(b) mitosis
(c) Both (a) and (b)
(d) None of these
Answer:
(b) mitosis

Question 13.
Human egg is
(a) centrolecithal
(b) telolecithal
(c) alecithal
(d) macrolecithal
Answer:
(c) alecithal

Question 14.
From which part of spermatid is acrosome formed?
(a) Nucleus
(b) Mitochondria
(c) Golgi bodies
(d) Ribosome
Answer:
(c) Golgi bodies

Question 15.
Which type / types of cell division occur (s) in cells of testis at different phases of spermatogenesis?
(a) Only meiotic
(b) Only mitotic
(c) Both (a) and (b)
(d) Amitotic
Answer:
(c) Both (a) and (b)

Question 16.
During which stage of oogenesis the number of chromosomes is reduced to half?
(a) Formation of second polar body
(b) Meiosis-II
(c) Division of secondary oocyte
(d) Formation of first polar body
Answer:
(d) Formation of first polar body

Question 17.
On which day of normal menstrual cycle, ovulation occurs?
(a) 10 th
(b) 13 th
(c) 14 th
(d) 15 th
Answer:
(c) 14th

Question 18.
Generally, what is the site-of fertilisation in humans? (2022)
(a) Ovary
(b) Uterus
(c) Vagina
(d) Fallopian tube
Answer:
(d) Fallopian tube

Question 19.
During fertilisation through which path the male pronucleus moves to meet the female pronucleus?
(a) Penetration path
(b) Copulation path
(c) Migration path
(d)
Answer:
(b) Copulation path

Question 20.
Name the cells formed by the division of zygote,
(a) Blastula
(b) Blastomeres
(c) Blastocoel
(d) None of these
Answer:
(b) Blastomeres

Question 21.
What is the correct sequence of embryo development?
(a) Gamete → Zygote → Morula → Blastula → Gastrula
(b) Gamete → Zygote → Blastula → Morula → Gastrula
(c) Gamete → Neurula → Gastrula
(d) Gamete → Neurula → Morula
Answer:
(a) Gamete → Zygote → Morula → Blastula → Gastrula

Question 22.
Human embryo is protected by which cavity?
(a) Am’niotic cavity
(b) Pleural cavity
(c) Peritoneal cavity
(d) Neural cavity
Answer:
(a) Amniotic cavity

Question 23.
What is the term used to represent the birth of a human foetus?
(a) Ovulation
(b) Abortion
(c) Conception
(d) Parturition
Answer:
(d) Parturition

Question 24.
Cleavage pattern is influenced by …………
(a) cytoplasm
(b) yolk
(c) nucleus
(d) centrosome
Answer:
(b) yolk

Question 25.
Embryo at 16-celled stage and without cavity is called …………..
(a) morula
(b) blastomere
(c) blastula
(d) gastrula
Answer:
(a) morula

Question 26.
What do you call the embryo at 16 celled stage?
(a) Blastocyst
(b) Morula
(c) Blastula
(d) Gastrula
Answer:
(a) Morual

Question 27.
LH surge is seen during
(a) menstrual phase
(b) follicular phase
(c) ovulatory phase
(d) luteal phase
Answer:
(c) ovulatory phase

Question 28.
Slow block to polyspermy is induced by
(a) vitelline membrane
(b) corona radiata
(c) zona pellucida
(d) All of the above
Answer:
(b) corona radiata

Question 29.
Pineal gland is derived from
(a) ectoderm
(b) mesoderm
(c) endoderm
(d) Both (a) and (c)
Answer:
(a) ectoderm

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

Question 1.
Secretion of prostate contains fructose.
Answer:
Secretion of seminal vesicle contains fructose.

Question 2.
The primordial follicles are situated in the medulla of ovary.
Answer:
The primordial follicles are situated in the cortex of ovary.

Question 3.
Corpus albicans is an endocrine structure.
Answer:
Corpus luteum

Question 4.
Efferent ductules in males are 15 in number.
Answer:
12

Question 5.
Process by which sperms are released from seminiferous tubules is called spermiogenesis.
Answer:
Process by which sperms are released from seminiferous tubules is called spermiation.

Question 6.
Degenerated follicles are called Graafian follicles.
Answer:
Degenerated follicles are called atretic follicles.

Question 7.
Mitochondria in tail of sperms provide it motility.
Answer:
Microtubules

Question 8.
Corona radiata is the primary egg membrane. Fill in the blanks
Answer:
Zona pellucida

Question 9.
Three primary germ layers are formed during morula stage of embryonic development.
Answer:
gastrula

Question 10.
Intestine is derived from mesoderm.
Answer:
endoderm

Question 11.
The yellow coloured milk secreted by mother just after childbirth is called neonatal milk.
Answer:
colostrum

Question 12.
Fertilisation occurs in uterus.
Answer:
Fallopian tube

Question 13.
Gonads are derived from endoderm.
Answer:
Mesoderm

Fill in the blanks

Question 1.
Humans reproduce …………….. . (asexually/sexually)
Answer:
sexually

Question 2.
Humans are ………… (oviparous, viviparous, ovoviviparous)
Answer:
viviparous

Question 3.
Fertilisation is ………… in humans. (external/internal)
Answer:
internal

Question 4.
A thin ……….. separates the granulosa cells from theca interna.
Answer:
basement membrane

Question 5.
Rupture of Graafian follicle releases …….
Answer:
secondary oocyte

Question 6.
Each spermatogonium contains …………. chromosomes.
Answer:
46

Question 7.
Oogonia project into stroma of ovary as a chord called ………… .
Answer:
egg tube or Pfluger

Question 8.
FSH and LH are secreted by ……………. gland.
Answer:
pituitary

Question 9.
The sperm’s head gets embedded into ………. after spermiogenesis.
Answer:
Sertoli cells

Question 10.
Primary oocyte gets arrested at …………. stage of meiosis-I.
Answer:
diplotene

Question 11.
The fusion of male and female pronuclei is called …………. .
Answer:
fertilisation

Question 12.
During fertilisation, the sperm’s acrosome releases ……………..
Answer:
hyaluronidase

Question 13.
The covering of egg is called …………… membrane.
Answer:
vitelline

Question 14.
The primitive gut that forms during gastrulation is called ……………. .
Answer:
archenteron

Question 15.
The hormone …………. stimulates the secretion of milk.
Answer:
oxytocin

Question 16.
Zygote is ………….
Answer:
diploid

Question 17.
The process of release of ovum from a mature follicle is called ………….. .
Answer:
ovulation

Question 18.
Ovulation is induced by a hormone ………….. .
Answer:
LH

Question 19.
Zygote divides to form ………… which is implanted in uterus.
Answer:
blastocyst

Express the following in one or two word(s)

Question 1.
Give the name of structures by which testes are suspended in the scrotum.
Answer:
Spermatic cords

Question 2.
Name the cells that nourish the germ cells in the testes. Where are these cells located in the testes?
Answer:
Sertoli cells. They are found on the inner lining of seminiferous tubule.

Question 3.
Name the type of cells found in the secondary follicle.
Answer:
Cuboidal-shaped granulosa cells.

Question 4.
Name the structure formed after the degeneration of corpus luteum.
Answer:
Corpus albicans.

Question 5.
Which hormone is secreted by Leydig cells?
Answer:
Testosterone

Question 6.
Somatic chromosome number is 40.
What shall be the chromosome number in the cells of seminiferous tubules?
Answer:
40 chromosomes

Question 7.
From which part of spermatid is acrosome formed?
Answer:
Head

Question 8.
How many spermatids are produced from a single primary spermatocyte?
Answer:
4

Question 9.
State the stage at which oogonia reach their maximum number.
Answer:
Before birth at five months of foetal life.

Question 10.
When are polar bodies formed in female ovary?
Answer:
During the production of secondary oocyte and ovum.

Question 11.
How many sperms will be produced from 10 primary spermatocytes?
Answer:
40

Question 12.
Name the process of release of spermatozoa from Sertoli cells into cavity of seminiferous tubule.
Answer
Spermiation

Question 13.
Generally, what is the site of.fertilisation in human being?
(Vagina, Uterus, Fallopian tube and Ovary)
Answer:
Fallopian tube

Question 14.
During fertilisation through which path the male pronucleus moves to meet the female pronucleus?
(Penetration path, Copulation path, Migration path and Zygotic path)
Answer:
Migration path

Question 15.
In which stage of development are the three germinal layers formed ?
Answer:
Gastrula

Question 16.
Which germ layer forms the nervous system?
Answer:
Ectoderm

Question 17.
From which germ layer does the coelom arise in vertebrates?
Answer:
Mesoderm

Question 18.
When does ovulation occur?
Answer:
Day 14 of menstrual cycle

Question 19.
What do you call the layer of cells forming the outer wall of the blastocyst?
Answer:
Trophoblast

Question 20.
Name the thin clear coat surrounding mammalian egg.
Answer:
Plasma membrane

Short Answer Type Questions

Question 1.
Write a note on human male reproductive system.
Answer:
Human Male reproductive system It consists of various glands, ducts and supporting structures. The structures present in this system are either paired or unpaired.
Testes are primary sex organs, posterior to the penis within the scrotum, produce spermatozoa and testosterone.

Seminal vesicles are club-shaped glands posterior to prostate, secrete alkaline fluid containing nutrients, fructose and prostaglandins.Scrotum is the pouch of skin, posterior to the penis, encloses and protects testes.

Penis is the unpaired organ, anterior to the scrotum and attached to pubis, conveys urine to outside of body.
Prostate gland and Cowper s gland secrete fluids which neutralise acidic environment of vagina and lubricate urethra and penis, respectively.

Question 2.
Draw a labelled diagram of male reproductive system.
Answer:

Question 3.
Why are the human testes located outside the abdominal cavity? Name the pouch in which they are present.
Or
Why are testes extra-abdominal in human males?
Answer:
Human testes are responsible for the synthesis of male gametes (sperms) and they need slightly lower temperature than the normal body temperature for this function. Thus, testes are located outside the abdominal cavity in sac-like structure called scrotum. Its temperature is about 2-3° lower than the normal body temperature.

Question 4.
Describe the structure of a seminiferous tubule.
Or Write a note on seminiferous tubules.
Answer:
Each testicular lobule contains 1-3 highly coiled seminiferous tubules. Each tubule is lined on its inside by two types of cells.

The cuboidal cells are called male germ (spermatogenic) cells and few large pyramidal (supporting cells-Sertoli or Nurse cells).
The former gets divided to form sperms while the latter provides nutrition to the germ cells.

Question 5.
Write the location and function of the following in human testes
(i) Sertoli cells
(ii) Leydig cells
Answer:
(i) Sertoli cells are found in the seminiferous tubules and they help to support and nourish the male gametes.
(ii) Leydig cells are found at interstitial region between seminiferous tubules and secrete androgens.

Question 6.
What is the significance of epididymis in male fertility?
Answer:
The epididymis helps sperm in attaining maturity, acquiring increased motility and fertilising capacity. It also stores sperms for a short period before they enter the vasa deferens.

Question 7.
How the spermatozoa are moved through the vasa efferentia and the epididymis?
Answer:
Spermatozoa or sperm formed in the seminiferous tubules reaches the vasa efferentia through rete testis. Several vasa efferentia open into the epididymis and carry sperm outside the body.

Question 8.
What are the major components of the seminal plasma?
Answer:
Seminal fluid is a cumulative secretion of accessory sex glands of male that is seminal vesicles, prostate gland and Cowper’s glands. It is rich in fructose, calcium and certain enzymes.

Question 9.
What are the major functions of male accessory ducts and glands?
Answer:
The male accessory ducts are rete testis, vasa efferentia, epididymis and vas deferens. These ducts store and transport the sperms from the testes to the outside through the urethra. The male accessory glands are paired seminal vesicles, prostate gland and paired bulbourethral glands.

The secretion of these glands form seminal plasma, which is rich in fructose, calcium and certain enzymes. The secretion of bulbourethral glands lubricate the penis.

Question 10.
How would a test confirm sexual intercourse by analysing vaginal swab?
Answer:
In this test, vagina is analysed for the presence of fructose, which is not produced anywhere in the human body . except semen (contributed by seminal vesicles). Therefore, its presence in the vagina indicates coitus/sexual intercourse.

Question 11.
Enlist the disorders of male reproductive system.
Answer:

Disorder	Description
Cryptorchidism	Failure of lestes to descend into the scrotum.
Benign prostatic hypertrophy	Enlargement of the prostate, often occurs in old age. Causes difficulty in urinating, dribbling and nocturia.
Htydrocoele	Accumulation of fluids around a testicle.
Erectile dysfunction (impotance)	Inability to develop or maintain an erection of the penis during sexual activity. Causes are drugs, ageing, diabetes, etc.
Inguinal hernia	Protrusion of the abdominal cavity contents through the inguinal canal.

Question 12.
Draw a labelled diagram of female reproductive system.
Answer:

Question 13.
Write a note on human ovary.
Answer:
Ovaries
Ovaries are the primary sex organs of female that is situated in the pelvic cavity.
They produce female gametes (ovum) and several steroid hormones (ovarian hormones). The ovaries are almond-shaped, lying in the lower part of the abdomen and held to the broad ligament by double folds of peritoneum called mesovarium. Each ovary is about 2-4 cm in length and is connected by an ovarian ligament to the uterus and by a suspensory ligament to the lateral pelvic wall. Each ovary is covered by a thin epithelium layer called germinal epithelium.

Question 14.
What is the difference between a primary oocyte and a secondary oocyte?
Answer:
Primary oocyte is a diploid cell formed in foetal ovary when the gamete mother cell, oogonia is arrested at prophase-I of meiosis. Secondary oocyte is the haploid cell formed from primary oocyte that completes its first meiotic division, during puberty and produces the female gamete ova (n).

Question 15.
What are the changes in the oogonia during the transition of a primary follicle to Graafian follicle?
Answer:

1. The germinal epithelial cells divide repeatedly until many diploid oogonia are formed.
2. The oogonia grow to form primary oocytes. Each primary oocyte then gets surrounded by a layer of granulosa cells and is called the primary follicle.
3. These get surrounded by more layers of granulosa cells and called secondary follicles.
4. The latter soon transforms into a tertiary follicle, which is characterised by a fluid-filled cavity called antrum.
5. The primary oocyte within the tertiary follicle undergoes meiotic division to become a secondary oocyte and a first polar body (haploid).
6. The tertiary follicle further changes into the mature follicle or Graafian follicle that ruptures to release the secondary oocyte (ovum) from the ovary by the process called ovulation.

Question 16.
Name and explain the role of the inner and middle walls of the human uterus.
Answer:
Uteus has a thick, highly vascular wall composed of three layers of tissues

1. Outer perimetrium A thin single-layered membranous covering of uterus wall that is composed of squamous epitbelial cells.
2. Middle myometrium A thick layer of smooth muscle fibres, which contracts strongly during the delivery of the baby.
3. Inner endometrium A glandular layer with many blood vessels. It consists of simple cuboidal or columnar epithelium. It undergoes cyclic changes during the menstrual cycle.

Question 17.
The diagram shows the side view of the female reproductive system

(i) Label the parts A to D.
(ii) In which region are sperms released during intercourse?
Answer:
(i) A-Uterus, B-Fallopian tube, C-Ovary, D-Vagina.
(ii) During intercourse, the penis reaches into the vaginal area and thus, the sperms are released in vagina.

Question 18.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Answer:
It is the formation of spermatozoa (sperms) in the testes, which originate from the cuboidal cells called Primordial Germ Cells (PGCs). The process of spermatogenesis starts at puberty.
It is completed in two steps

• Formation of spermatids (spermatocytogenesis)
• Transformation of spermatids to sperms (spermiogenesis)

Question 19.
Define spermiogenesis and spermiation.
Or Write a note on spermiogenesis.
Answer:
Spermiogenesis:
It is the formation of spermatozoa (sperms) in the testes, which originate from the cuboidal cells called Primordial Germ Cells (PGCs). The process of spermatogenesis starts at puberty.
It is completed in two steps

• Formation of spermatids (spermatocytogenesis)
• Transformation of spermatids to sperms (spermiogenesis)

Spermiogenesis:
The spermatids (immotile) transform into motile spermatozoa by the process called spermiogenesis. It is also called differentiation phase during which sperm heads get embedded in the Sertoli cells and are finally released from the seminiferous tubules by the process of spermiation.

Question 19.
Draw and label the various parts of human sperm.
Answer:

Question 20.
Spermatogenesis in human males is a hormone regulated process. Justify.
Answer:
Spermatogenesis initiates due to a significant increase in the secretion of Gonadotropin Releasing Hormone (GnRH) by the hypothalamus at the age of puberty. The increased level of GnRH acts on the anterior pituitary and stimulates the secretion of two gonadotropins, i.e. Luteinizing Hormone (LH) and Follicle Stimulating Hormone (FSH). LH acts on the Leydig cells thus, stimulating the synthesis and secretion of androgens. This in turn stimulate the process of spermatogenesis. FSH acts on Sertoli cells and stimulates the secretion of some factors like inhibins, activins and transferrins which help in spermiogenesis.
Note Androgens are testicular hormones which function to maintain the accessory ducts and glands in males.

Question 21.
State the function of the following
(i) Acrosome
(ii) Sperm tail
Answer:
Their functions are
(i) Acrosome It contains hydrolytic enzymes that are required to dissolve the membranes of the ovum for fertilisation.
(ii) Sperm tail It helps in the sperm movement in the female genital tract for fertilisation.

Question 22.
What is the number of chromosomes in the following cells of a human male?
(i) Spermatogonial cells
(ii) Spermatids
(iii) Primary spermatocytes
(iv) Sertoli cells
Answer:
(i) 46
(ii) 23
(iii) 46
(iv) 46

Question 23.
Identify A, B, C and D with reference to gametogenesis in humans in the flow chart given below.

Answer:
A – GnRH,
B – FSH,
C – Spermatogenesis,
D – Spermiogenesis

Question 24.
Write the steps in the formation of an ovum from an oogonium in human.
Answer:
This process involves following phases
(i) Multiplication phase Cells of germinal epithelium undergo mitotic division to produce undifferentiated germ cells called oogonia or egg mother cells within each foetal ovary.

1. The oogonia further divides by mitosis and projects as a chord into the stroma of the ovary. This chord is known as egg tube or pfluger, which later become a round mass called egg nest.
2. In the egg nest, one cell grows, enters into prophase-I of meiosis and gets arrested at diplotene stage temporarily. This cell becomes the primary oocyte.
3. Other oogonia in the nest surround this primary ooctye to form primary follicle. These follicles protects and nourish the primary oocyte.
4. About 2 million of primary follicles exist in females before birth. A large number of these follicles degenerate during the phase from birth to puberty by the process called follicular atresia. Therefore, at puberty only 60000- 80000 primary follicles are left in each ovary.

(ii) Growth phase

1. The primary follicles get surrounded by more layers of granulosa cells and a new theca to form secondary follicles.
2. This follicle then gets converted into a tertiary follicle that possesses a fluid-filled cavity called antrum surrounded by theca interna and theca externa.

(iii) Maturation phase The primary oocyte within this tertiary follicle grows in size. The fully grown primary oocyte completes its first meiotic division producing two daughter nuclei in which larger haploid cell is called secondary oocyte and the tiny one is called first polar body or polocyte. The secondary oocyte contains bulk of the nutrient rich cytoplasm of primary oocyte and becomes tertiary follicle.

Question 25.
Why are polar bodies formed during oogenesis, but not in spermatogenesis?
Answer:
Polar bodies are formed during maturation phase of oogenesis. Two small polar bodies are formed during I and II meiotic divisions. Due to unequal cytokinesis, the cytoplasmic content of polar bodies is negligible as compared to ovum. They help to get rid of haploid set of chromosomes during oogenesis. But, during spermatogenesis equal cytokinesis occurs thus, polar bodies are not formed in this.

Question 26.
How many eggs do you think were released by the ovary of a female dog, which gave birth to six puppies?
Answer:
The ovaries of a female dog must have released 6 eggs, each of which was fertilised.
So, six zygotes were formed, that developed into six puppies.

Question 27.
Identify the statements as True/False. Correct each false statement to make it true.
(i) Oogenesis takes place in corpus luteum.
(ii) Spermatozoa get nutrition from Sertoli cells.
Answer:
(i) False-Oogenesis takes place in ovaries.
(ii) True.

Question 28.
What is menstrual cycle? Which hormones regulate menstrual cycle?
Answer:
The hormones that regulate menstrual cycle along with their functions are

1. Pituitary gonadotropins (LH, FSH and Oestrogen) These help in the proliferation of Graafian follicle and endometrium of uterus.
   1. FSH Stimulates the growth of ovarian follicle and secretion of oestrogen.
   2. Oestrogen Increased level of oestrogen gives positive feedback and stimulates more LH production.
   3. LH Stimulates ovulation.
2. Progesterone Essential for the maintenance of endometrium.

The hormones involved in the process are

• FSH, LH stimulate growth of follicle and maturation of ovum.
• Oestrogen is involved in endometrial repair and growth.
• Oestrogen and progesterone together prepare endometrium and other parts of the body for pregnancy.

Question 29.
Write the source and effect of the high concentration of LH on a mature Graafian follicle.
Answer:
The changes in the ovary during follicular phase stimulates anterior pituitary to release more LH. In the middle of cycle, i.e. during ovulatory phase, LH attains a peak level and it is known as LH surge. High level of LH ruptures the Graafian follicles and thereby releases the ovum.

Question 30.
Draw a labelled diagram of a Graafian follicle.
Answer:

Question 31.
Describe, how the changing levels of FSH, LH and progesterone during menstrual cycle induce changes in the ovary and the uterus in human female.
Answer:
Both LH and FSH attain a peak level in the middle of cycle (about 14th day).

1. Rapid secretion of LH leading to its maximum level during the midcycle is called LH surge. It induces rupture of Graafian follicle and thereby, the release of ovum, i.e. ovulation.
2. The ovulation (ovulatory phase) is followed by the luteal phase.

Question 32.
Given alongside is a flowchart showing ovarian change during menstrual cycle. Fill in the spaces with the hormonal factor(s) responsible for the events shown.

Answer:
A – FSH and oestrogen, B – LH and C – Progesterone.

Question 33.
Corpus luteum in pregnancy has a long life. However, if fertilisation does not take place, it remains active only for 10-12 days. Why?
Answer:
During pregnancy, there is no menstruation. The corpus luteum secretes large amount of progesterone in case the fertilisation occurs.
This is to maintain the corpus luteum as long as the embryo remains there. In the absence of fertilisation, the corpus luteum cannot be maintained for more than 10-12 days and hence, it degenerates.

Question 34.
Write the function of following
(i) Corpus luteum
(ii) Endometrium
Answer:
(i) Corpus luteum It secretes progesterone required for the maintenance of endometrium.
(ii) Endometrium It is the innermost layer of uterus.
It provides nutrition for the development of foetus. Implantation of blastocyst occurs in endometrium.

Question 35.
Write short note on fertilisation.
Or What is fertilisation? In man, where does it take place?
Answer:
Fertilisation:
The process of fusion of a sperm with an ovum is called fertilisation. It involves discharge of semen by the penis into the female’s vagina close to the cervix during coitus is called insemination.
The motile sperms move through the cervix, enter the uterus and reach the junction of the isthmus and ampulla (ampullary-isthmus junction) of the Fallopian tube.

The ovum released from the ovary also reaches the ampullary-isthmic junction, where the two gametic nuclei fuse. This process is called amphimixis or syngamy. Fertilisation can only occur if the ovum and sperms are transported simultaneously to this junction. This explains why all copulations do not lead to fertilisation and pregnancy.

Question 36.
In our society the women are often blamed for giving birth to daughters. Can you explain why this is not correct?
Answer:
Women are blamed for giving birth to daughters. This is wrong because sex of the baby is determined by the sperm that can have either X or Y-chromosomes.
Women have only one type of chromosome (X) in all the ova.

If the sperm having X-chromosome fertilises the ovum (X), the resulting zygote (XX) will become a female.
If the sperm having Y-chromosome fertilises the ovum (X), the resulting zygote (XY) will become a male.

Question 37.
A sperm has just fertilised a human egg in the Fallopian tube. Trace the events that the fertilised egg will undergo upto the implantation of the blastocyst in the uterus.
Answer:
Fertilisation
The process of fusion of a sperm with an ovum is called fertilisation. It involves discharge of semen by the penis into the female’s vagina close to the cervix during coitus is called insemination.
The motile sperms move through the cervix, enter the uterus and reach the junction of the isthmus and ampulla (ampullary-isthmus junction) of the Fallopian tube.

The ovum released from the ovary also reaches the ampullary-isthmic junction, where the two gametic nuclei
fuse. This process is called amphimixis or syngamy. Fertilisation can only occur if the ovum and sperms are transported simultaneously to this junction. This explains why all copulations do not lead to fertilisation and pregnancy.
In humans, fertilisation occurs in following steps

(i) The sperm comes in contact with ovum. The sperm secrete a chemical called anti-fertilizin which react with the chemical called fertilizin secreted by the ovum. This reaction is called agglutination reaction. The fertilizin and anti-fertilizin are complementary only when they belong to same species. This reaction helps to prevent the progress of multiple sperms towards the egg.

(ii) The sperm undergoes capacitadon and acrosomal reacdon. The latter reaction involves the release of sperm lysins which include-hyaluronidase, Corona Penetrating Enzyme (CPE) and Zonalysins (acrosin). These enzymes help in the digestion of primary, secondary egg membranes so that the sperm can penetrate the egg.

(iii) While penetrating the egg, contact between sperm and zona pellucida depolarise the membrane (fast block) so as to block the entry of additional sperms. Similarly, the contact between sperm and corona radiata induces cortical reaction (slow block and results in the formation of a fertilisation membrane.

(iv) Simultaneously, secondary oocyte restart cell cycle by the breakdown of MPF to APC (Anaphase Promoting Complex) due to the entry of sperm.

(v) After the completion of second meiotic division, ovum is formed. Its nucleus condenses to form female pronucleus. Simultaneously, male pronucleus is formed by the separation of sperm head from middle piece.

(vi) The fusion of male and female pronuclei results in the formation of zygote (2n – 46).

Ovum surrounded by sperms

Question 38.
Name the embryonic stage that gets implanted in the uterine wall of the human female. Also draw its diagram.
Answer:
The blastocyst gets embedded in the endometrium of the uterus.
This is called implantation and it leads to pregnancy.

Question 39.
When and where do chorionic villi appear in humans? State their function.
Answer:
Chorionic villi appear just after implantation. They appear as finger-like projections on the trophoblast and become interdigitated with the surrounding uterine tissues and form the placenta.

Question 40.
A woman passes out hCG in the urine during the pregnancy. Why?
Answer:
The presence of human Chorionic Gonadotropin (hCG) is the basis of pregnancy test. Placenta formed during pregnancy releases hCG, which comes in urine of pregnant woman.

Question 41.
Mention the name of hormones and their source organs produced only during pregnancy.
Answer:
The hormones are
(i) human Chorionic Gonadotropin (hCG) It is secreted by the placenta.
(ii) Relaxin It is secreted by the ovary.

Question 42.
Placenta acts as an endocrine gland. Explain.
Or
Write the function of placenta.
Or
Write a note on placenta.
Answer:
The placenta secretes hormones like human gonadotropins, human placental lactogen, oestrogen and progesterone that are necessary to maintain pregnancy. Due to the secretion of these hormones, it is called an endocrine gland.

Question 43.
When and how does placenta develop in human female?
Answer:
Soon after implantation, finger-like projections appear on k the trophoblast. These are known as chorionic villi and surrounded by the uterine tissue and maternal blood.

The chorionic villi and the uterine tissue become inter-digitated with each other. Together, they form a structural and functional unit between foetus and maternal body called placenta. In humans the placenta is of haemochorial metadiscoidal and deciduous type. After the sixth week of pregnancy, placenta take over the function of corpus luteum and produce oestrogen and progesterone.

Question 44.
Where is morula formed in humans? Explain the process of its development from zygote.
Answer:
The development of embryo from fertilised ovum involves the following events. The mitotic division starts as the zygote (fertilised ovum) moves through the isthmus of the oviduct towards the uterus, forming 2, 4, 8, 16 daughter cells called blastomeres. This process is called cleavage. During holoblastic cleavage in humans, the young embryo is slowly moving down the Fallopian tube towards the uterus and following events take place

1. The embryo with 8-16 blastomeres is called morula, But, it is not larger than a zygote.
2. The morula continues to divide and transforms into blastocyst as it moves further into the uterus. The cavity of the blastocyst is called blastocoel. This process of transformation is called blastulation.
3. The blastomeres in the blastocyst are arranged into an outer layer called trophoblast and the inner group of cells attached to trophoblast called the inner cell mass.
4. The trophoblast layer then gets attached to the endometrium and the inner cell mass differentiates into the embryo. After attachment, the uterine cells divide rapidly and cover the blastocyst.
5. As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is called implantation and it leads to pregnancy (i.e. the time from conception to birth).
   

Question 45.
State the function of yolk.
Answer:
The function of yolk are

• It provides nutrition to the developing embryo.
• Cleavage, i.e. the division of cells in the early embryo mostly depends on the amount of yolk in the egg.
• The cleavage can be holoblastic (total cleavage) or meroblastic (partial cleavage).

Question 46.
What is the effect of yolk on cleavage?
Answer:

1. The zyogte undergoes mitotic divisions called cleavage and develops into an embryo.
2. The pattern of cleavage is determined by the amount of yolk and its distribution in the egg.
3. If the egg is having moderate to sparse amounts of yolk then the egg divides completely and this type of cleavage is called holoblastic cleavage.
4. If the egg contains a large amount of yolk then the egg cannot divide completely as the cleavage furrow cannot penetrate the yolk. This type of cleavage is called meroblastic cleavage.

Question 47.
Draw the following diagrams related to human reproduction and label them.
(i) The zygote after first cleavage division
(ii) Morula stage
(iii) Blastocyst stage (sectional view)
Answer:
(i) The zygote after first cleavage division

(ii) Morula stage

(iii) Blastocyst stage (sectional view)

Question 48.
Study the figure given below and answer the questions that follows.

(i) Name the stage of human embryo the figure represents.
(ii) Identify ‘A’ in the figure and mention its function,
(iii) Mention the fate of the inner cell mass after implantation in the uterus.
Answer:
(i) Blastocyst.
(ii) ‘A’ is trophoblast. The trophoblast layer gets attached to endometrium and later forms chorionic villi.
(iii) The inner cell mass gets differentiated into an embryo.

Question 49.
What is parturition? Which hormones are involved in induction of parturition?
Or
Why parturition is called a neuroendocrine mechanism? Explain.
Or
What are gestation and parturition?
Answer:
Parturition
The time period during which the embryo remains in the uterus, i.e. after the last day of menstruation to the day of parturition is called gestation. The gestation period of humans is about 9 months.
Vigorous contraction of the uterus at the end of pregnancy causes expulsion/delivery of the foetus. This process of delivery of the foetus (childbirth) is called parturition. It is induced by a complex neuroendocrine mechanism.
The signals of parturition originate from the fully developed foetus and the placenta, which induce mild uterine contractions. It is called foetal-ejection reflex.

1. This triggers the release of oxytocin from the maternal pituitary. It promotes contractions of uterine muscles, which in turn stimulate further secretion of oxytocin.
2. The stimulatory reflex between the uterine contraction and oxytocin secretion continues resulting in stronger contractions. This leads to the expulsion of the baby out of the uterus through the birth canal.
3. Soon after the birth of baby, placenta is also expelled out of the uterus. Relaxin hormone relaxes the pelvic ligaments of mother to prepare for the childbirth.

Question 50.
What is lactation? Mention its significance.
Answer:
Lactation:
The mammary glands in females undergo differentiation during pregnancy. The production and release of milk by the mammary glands of female after birth of a young one is called lactation. The first milk, which comes out from the mother’s mammary glands just after childbirth is called colostrum. It is rich in proteins and energy along with the Antibodies (IgA) that provide passive ‘ immunity to the newborn. Breast feeding during the initial period of infant growth is highly recommended by the doctors for bringing up a healthy baby.

Differentiate between the following (for complete chapter)

Question 1.
Graafian follicle and Corpus luteum.
Answer:
Differences between Graafian follicle and corpus luteum are as follows

Graafian follicle	Corpus luteum
It consists of an oocyte, zona pellucida, cellular membranous granulosa surrounded by the theca interna and the externa.	It consists of luteum cells, fibrin and blood clot.
It contains follicular antrum (follicular cavity) filled with follicular fluid.	It contains blood clot.
It is formed by the germinal epithelium of the ovary.	It is formed after the release of secondary oocyte from the Graafian follicle.
Its granular cells secrete oestrogens.	It secretes progesterone.

Question 2.
Corpus luteum and Corpus albicans.
Answer:
Differences between corpus luteum and corpus albicans are as follows

Corpus luteum	Corpus albicans
Graafian follicle changes into corpus luteum after ovulation in ovary.	Corpus luteum degenerates and forms corpus albicans if there is no pregnancy.
It is made of luteal cells.	It is made of scar tissues.
Secretes progesterone.	Do not secrete any hormone.

Question 3.
Menarche and Menopause.
Answer:
Differences between menarche and menopause are as follows

Menarche	Menopause
It is the first menstrual period in human female.	It is the end of menstrual period in human female.
It occurs at around 11-16 years of age.	It occurs at around 45-50 years of age.
It marks the beginning of reproductive phase of a female.	It marks the end of reproductive phase of a female
There is elevated level of oestrogen.	There is decline in the level of oestrogen.

Question 4.
Endometrium and Myometrium.
Answer:
Differences between endometrium and myometrium are as follows

Endometrium	Myometrium
It is inner glandular layer of the wall of the uterus.	It is thick muscular middle layer of the wall of the uterus.
It undergoes cyclic changes during menstrual cycle. Implantation of blastocyst takes place on endometrium.	It is involved in the uterine movements.

Question 5.
Spermatocytes and Oocytes.
Answer:
Differences between spermatocytes and oocytes are as follows

Spermatocytes	Oocytes
Primary spermatocytes are formed from the spermatogonia in the seminiferous tubules of testes by mitosis.	Primary oocytes are formed from the oogonia in the ovary of the foetus.
Each primary spermatocyte undergoes meiosis-i and forms the two haploid secondary spermatocytes.	Each primary oocyte undergoes meiosis-l and forms haploid secondary oocytes and haploid first polar body.
Each secondary spermatocyte undergoes meiosis-ll and forms two haploid spermatids.	The secondary oocyte undergoes meiosis-ll and forms one ovum and one second polar body.
Each primary spermatocyte forms four haploid spermatids.	Each primary oocyte forms one ovum and three polar bodies.

Question 6.
Cleavage and Typical mitosis.
Answer:
Differences between cleavage and typical mitosis are as follows

Cleavage	Typical mitosis
It occurs in zygote or parthenogenetic egg,	It occurs in most of body cells.
Interphase is short.	Interphase is of long duration.
Growth does not occur.	Growth occurs during interphase.
Oxygen consumption is high as it is very rapid process.	Oxygen consumption is low as it is slow process.
Size of blastomeres decreases.	Size of daughter cells remains same after growth.
DNA synthesis is faster.	DNA synthesis is slower.
Nuclear cytoplasmic ratio increases.	Nuclear cytoplasmic ratio remains same.

Long Answer Type Questions

Question 1.
Give an account of the human male reproductive system.
Answer:
Human Reproductive System
Human beings show prominent sexual dimorphism in which male and female differ from each other by a number of primary and secondary sex characters. Primary sex characters are those which are present since birth in both males and females, e.g. external genitalia.
The secondary sex characters appear after puberty and include a number of traits like the presence of thick facial hairs, low pitch voice, etc., in males and well-developed mammary glands, high pitch voice, etc., in females.
On the basis of their functions, reproductive organs can be divided into the following types

1. Primary sex organs These produce gametes and sex hormones, e.g. ovary in females and testes in males.
2. Secondary sex organs These do not produce gametes or sex hormones, but help in morphological discrimination among two sexes, e.g. breast in females, penis in males, etc.
3. Accessory organs These do not form gametes, but are essential for reproduction, e.g. prostate gland in males, Fallopian tube in females,

Male Reproductive System:
The male reproductive system is located in the pelvis region. It consists of scrotum a pair of testes, excurrent ducts and accessory glands.

The male reproductive system : (a) Anterior view (b) Sagittal view

Scrotum:
It is a pouch-like structure of pigmented skin arising from the lower abdominal wall and hanging between the leg. It is divided internally into right and left scrotal sacs by a muscular partition called septum scrota. The testis originates in the abdominal cavity, but later during the seventh month of development, it descend into the respective scrotal sacs through the passages called inguinal canals.

Testicular descent to inguinal region during gestation period is affected by Antimullerin Hormone (AMH) secreted by Sertoli cells. 90% of babies have completely descended testes, when they are born. In 10% of new-borns, testes fail to descend and retained in the abdominal cavity. This condition is called cryptochidism. It can be treated by gonadotropins or surgery.

In cold weather, the testes are elevated by the contraction of a band of muscle, known as cremasteric muscle to get the warmth of the trunk. This effect is known as cremasteric reflex.

The same effect occurs when the thigh of a man is stroked. In the baby, this stimulus causes the testes to ascend up into the abdominal cavity through the inguinal canal.

Functions of Scrotum:
Scrotum keeps the temperature of testes about 2-2.5°C lower than the internal body temperature, which is essential for the production of sperms, i.e. spermatogenesis. This temperature is maintained by perspiration and evaporation from the scrotal surface. These processes occur due to the presence of pampiniform plexus that surround testiculae arteries which ascend from the testis. Venous blood returning from testis through pampiniform plexus is cooler than the blood in testicular arteries. The countercurrent heat exchange mechanism between the arterial and venous blood cools down the arterial blood.

Testes: Microscopic Anatomy:
Testes are the primary sex organs in men. These are the male gonads, i.e. the site where the male gametes or sperms are made. A pair of testis is situated outside the abdominal cavity within the scrotum. Each testis is oval in shape and measures about 4-5 cm in length and 2-3 cm in width. These are suspended in the scrotum by spermatic cords.

Microscopic Anatomy:
The outermost covering of the testis is formed by a dense connective tissue called tunica albuginea. It thickens and extends inwardly into each testis as mediastinum testis which further forms thin connective tissue septum. The latter divides each testis into 200-300 compartments called testicular lobules. Each lobule contains 1-3 highly convoluted seminiferous tubules that is lined by stratified cuboidal epithelium.

Seminiferous Tubule:
It is the structural and functional unit of testis in which sperms are produced. Each seminiferous tubule is present in the mass of loose connective tissue that contain fibroblast, epithelial cells, nerves, blood and lymphatic vessels, etc. Inside these tubules, two types of highly specialised cells are present namely Sertoli cells (nurse cells) and spermatogenic cells.

Transverse section through human testis : (a) A single seminiferous tubule with cells of Leydig in the interstitial tissue, (b) A magnified part of the seminiferous tubule showing the blood-testis barrier, Sertoli cells and its relationship to the differentiating spermatogenic cells and (c) A Sertoli cell with spermatogenic cells in different stages of development

1. Spermatogenic cells These are majority of dividing cells or male germ cells (cuboidal cells) which produce spermatogonia by mitotic division. The spermatogonia further grows into primary spermatocytes and undergo meiosis producing haploid cells, first secondary spermatocytes and then spermatids.

2. Sertoli or Sustentacular cells They are tall, pyramidal, non-dividing somatic cells. They serve the supporting and nourishing function for the spermatogenic cells in different stages of their differentiation. The basal lamina of the germinal epithelium, muscle-like myoid cells at the base of the basal lamina and tight junctions between adjacent Sertoli cells constitute a blood testis barrier which performs the following functions

Prevents many macromolecules from moving into the tubular lumen.
Prevents the blood borne noxious chemical agents from entering into the tubule.
Prevents the passage of antigenic agents from the tubule into the blood, which are likely to generate an autoimmune response.
The region outside the seminiferous tubules called interstitial spaces contain small blood vessels and masses of cells called interstitial cells or Leydig cells. These cells synthesise and secrete the testicular hormones called androgens. Leydig cells are endocrine in function and thus, regulate and maintain male sex characteristics. Other immunologically competent cells are also present in this region.

Accessory or Excurrent Duct System:
These ducts store and transport the sperms along with other glandular secretions from the testis to the outside through urethra. The male’s sex accessory ducts include-tubuli recti, rete testis, vasa efferentia (intratesticular ducts), epididymis, vas deferens and urethra (extratesticular ducts). The ends of seminiferous tubules converge and join to form short straight tubules called tubuli recti.

Duct system in human testis

The tubuli recti open into a network of wider, irregular tubules called rete testis, that further open into 12 short efferent ductules or ductuli efferentes.

The latter open into the epididymis and carry sperms outside the testis. Epididymis is a long tube, which is greatly coiled and tightly packed to form an elongated flattened body. It is located along the posterior surface of each testis.

Sperms undergo physiological maturation, acquiring increased motility and fertilising capacity (i.e. capacitation) in epididymis. After that, they pass down to the tail of epididymis, where they stay for a very short period before entering the two vas deferens or ductus deferens.

The vas deferens ascends to the abdomen and lopps over urinary bladder. It receives a duct from seminal vesicle to form ejaculatory duct. The ejaculatory duct enters the prostate gland and opens into the prostatic urethra.

The prostatic urethra extends through the penis to its external opening called urethral meatus or penile urethra. The urethra carries urine from bladder as well . as spermatozoa and secretions from the Cowper’s and prostate glands.

Accessory Glands:
The accessory glands or the secondary glands in males are described below
1. Prostate gland It is a single gland that lies inferior to the bladder. It receives two ejaculatory ducts that joins to form prostatic urethra. The latter receives ductules from prostate. It secretes a milky, slightly alkaline fluid, that contains lipids, enzymes, citric acid, etc.

It is released during ejaculation and helps to neutralise the acidic medium of vagina, making sperms more active to swim. Prostatic fluid accounts for nearly 20-30% of semen volume.

2. Seminal vesicles This paired gland is present posterior to the bladder above prostate. Its ducts empties in the ampulla of ductus deferens.
They secrete mucus and a watery alkaline fluid that contains fructose (acts as an energy source for the sperms). Prostaglandins (stimulate uterine contractions for sperm movement) and a clotting protein form a temporary clot after ejaculation. Later, fibrolysins dissolves the coagulate or clot so that it assume a liquid form. Calcium and certain other enzymes are also a part of seminal plasma.

Fructose, produced by seminal vesicles, is not present elsewhere in the body. Therefore, during forensic test for rape, its presence in females genital tract confirms sexual intercourse. The seminal vesicles and prostrate degenerates or atrophy after the removal of testes because they are androgen dependent glands.

3. Bulbourethral or Cowper’s gland These are situated beneath the bladder and behind the urethra. There ductules discharge into the prostatic urethra. They secrete mucus and an alkaline fluid into the urethra. The mucus helps in the lubrication of penis and neutralises any urinary acids in urethra

Semen
The secretions of accessory sex glands (seminal plasma) and mucus are added to the sperms to form seminal fluid or semen. The main constituent of the semen is live spermatozoa present in an alkaline viscous medium. It is maintained at a pH range of 7.2-7.8. Another main constituent is fructose contributed by the seminal vesicles. It is the chief energy source for the spermatozoa.

The semen is forcibly expelled through the urethra by the process known as ejaculation. The volume of ejaculate is 1.5-5.0 mL. The normal count of sperm is 40-250 million/mL. The secretions have many functions

• Provide medium for transport of sperms.
• Nourish and activate sperms to keep them viable and motile.

Question 2.
Study the given figure.

(i) Pick out and name the cells that undergo spermiogenesis.
(ii) Name ‘A’ and ‘B ’ cells. What is the difference between them with reference to the number of chromosome?
(iii) Pick out and name the motile cells.
(iv) What is ‘F’ cell? Mention its function.
(v) Name the structure of which the given diagram is a part.
Answer:
(i) A-Spermatids, undergo spermiogenesis.
(ii) A-Spermatids, B-Spermatogonium.
Spermatogonium (B) are diploid and have 46-chromosomes in humans while, spermatids (A) are haploid and have 23-chromosomes in case of human male.
(iii) E-Spermatozoa, are motile cells.
(iv) F-Sertoli cell, it provides nutrition to the germ cells.
(v) Seminiferous tubule.

Question 3.
Give an account of the human female reproductive system.
Answer:
Female Reproductive System:
It consists of a pair of ovaries, glands, a duct system and many supporting structures. Its components are found in both paired and unpaired condition.

The female reproductive system consists of the following organs

Ovaries:
Ovaries are the primary sex organs of female that is situated in the pelvic cavity.
They produce female gametes (ovum) and several steroid hormones (ovarian hormones). The ovaries are almond-shaped, lying in the lower part of the abdomen and held to the broad ligament by double folds of peritoneum called mesovarium. Each ovary is about 2-4 cm in length and is connected by an ovarian ligament to the uterus and by a suspensory ligament to the lateral pelvic wall. Each ovary is covered by a thin epithelium layer called germinal epithelium.

Next to this layer, dense irregular connective tissue mass called tunica albuginea is present. Inner to this, cortex is present which is followed by highly vascularised connective tissue called cortex or stroma ovarian stroma, which contains connective tissues, blood vessels and mature follicles. The stroma is divided into two parts, i.e. a peripheral cortex and an inner medulla or stroma. There is no distinct boundary between the cortex and medulla.

Microscopic Anatomy of Ovary:
The production of eggs in females begins before birth, i.e. during the embryonic development stage, but is completed only after fertilisation. It takes place in ovaries. Cells of germinal epithelium undergo mitotic or equational division during multiplication phase, producing undifferentiated germ cells called oogonia or egg mother cells within each foetal ovary.

These cells enter into the phase of maturation and start two meiotic divisions to enter prophase-I of meiotic division as primary oocytes. They get temporarily arrested at diplotene state and remains as such until the onset of puberty.

Primordial and primary follicles in the cortex of the ovary

Each primary oocyte then gets surrounded by a single layer of squamous follicle cells and is called the primordial follicle. About 2 million of these follicles exist in ovarian cortex of females before birth. A large number of these follicles degenerate during the phase from birth to puberty. Therefore, at puberty only 60000-80000 primary follicles are left in each ovary.

The primordial follicles get surrounded by more layers of cuboidal granulosa cells and a new theca. This process is stimulated by the pituitary gonadotropins-FSH and LH. Later the squamous follicular cells of primordial follicle changes to cuboidal or low columnar cells. This follicle is called primary follicle.

The follicular cells of primary follicle divide by mitosis and forms the new cuboidal-shaped granulosa cells. At this stage follicle is called secondary follicle.

The innermost layer of granulosa cells is corona radiata that surrounds the oocyte. Between the oocyte and eorona radiata, a glycoprotein layer called zona pellucida appears. The stromal cells that surround granulosa cells differentiate to form thecal cells. These cells differentiate as an outer theca externa and inner theca interna.

A thin basement membrane separates granulosa cells from the theca interna. This follicular stage is known as mature or Graafian follicle stage.

Structure of a Graafian (mature) follicle : A part(inset) is magnified to show the elaborate structure of the egg with the associated follicle cells

Development of Ovarian Follicle:

1. The primary follicle grow in size and a fluid called liquor folliculi or follicular fluid accumulates in a cavity called antrum of granulosa cells. These follicles with an antrum are called antral or secondary follicles.
2. The synthesis of liquor folliculi continues, antrum grows in volume and further segregation of granulosa cells occur. Due to this, few cells surround the oocytes and some other are displaced to periphery.
3. Theca externa cells form multiple layers around theca interna cells which leads to the formation of mature or Graafian follicle. At this stage, the first meiosis of primary oocyte gets completed to form a secondary oocyte.
4. The Graafian follicle, ruptures to release the secondary oocyte from the ovary.
5. The secondary ooctye is captured by the fimbriae of Fallopian tube which transport it to the uterus.
6. The remaining portion of Graafian follicle gets filled with blood to form corpus haemorrhagicum.
7. The theca and granulosa cells proliferate and become glandular to form theca lutein and granulosa lutein cell, respectively. These lutein cells are the source of oestrogen and progesterone. This post ovulatory follicle is called corpus luteum (endocrine structure) which secrete estradiol, progesterone and relaxin (peptide hormone).

Fate of Corpus Luteum:

1. If fertilisation occurs Corpus luteum persist and no mensuration occurs, zygote undergoes embryonic development and implants in the endometrium of uterus.
2. If fertilisation does not occur Corpus luteum degenerates to form corpus albicans and menstruation occurs.
   

Not all ovarian follicles reach maturity. Only one dominant follicle gets matured and ovulated. Rest of them degenerates by the process called follicular atresia and the resultant follicles are known as atretic follicles which are replaced by connective tissue.

Fallopian Tube (Uterine Ducts):
These are two small accessory tubes of 10-12 cm length, lying on either sides of uterus near the kidney. These tubes carry the egg from the ovary to the uterus and also provides the appropriate environment for its fertilisation.

The Fallopian tubes or oviducts show four regions, i.e. infundibulum, ampulla, isthmus and interstitial region,

1. Infundibulum is broad and funnel-shaped with its edges bearing motile, finger-like projections called fimbriae. It helps in the collection of ovum after ovulation.
   
   Female reproductive system
2. Ampulla is the long, wide, thin-walled part of the Fallopian tube next to the infundibulum where fertilisation takes place.
3. Isthmus is a short narrow lumen that follows ampulla and joins the uterus.
4. Interstitial region passes through the uterine wall and communicates with the uterine cavity.

Uterus (Womb):
It is about 7.5 cm long, 5 cm wide, like an inverted pear in shape. It is supported by ligaments attached to the pelvic wall. It lies between the urinary bladder and rectum.

Uterus has a thick, highly vascular wall, composed of three layers of tissues

1. Outer perimetrium A thin single-layered membranous covering of uterus wall that is composed of squamous epithelial cells.
2. Middle myometrium A thick layer of smooth muscle fibres, which contracts strongly during the delivery of the baby.
3. Inner endometrium A glandular layer with many blood vessels. It consists of simple cuboidal or columnar epithelium. It undergoes cyclic changes during the menstrual cycle.

Blood Supply to Uterus:
The arteries that supply blood to uterus gets divided into arcuate arteries and spreads in the myometrium. The arcuate arteries further gets divided into parts so as to supply blood to endometrium.
These include

• Straight arteries These are short, supply blood to basalis layer.
• Spiral arteries These are long and coiled, supply blood to functionalis layer. Decreased supply to this layer during menstruation causes it to degenerate and cast off.

The endometrium descends into the lamina propria and form numerous uterine glands. Functionally, the layer is divided into two layers

• Luminal stratum functionalis During menstruation, this layer cast off along with blood vessels and uterine glands.
• Stratum basalis It helps to form a new functionalis layer.

Functions of Utreus:
Uterus receives the ovum from the Fallopian tube, forms placenta for the development of foetus and also expels the young one at the time of birth.

Cervix:
It is the narrow entrance of the uterus into the vagina. It has strongest sphincter in its wall and is normally blocked by a plug of mucus. The cavity of cervix is called cervical canal which along with vagina forms the birth canal.

Vagina:
It is about 10 cm long and acts as a receptacle for the penis during copulation, for allowing menstrual flow and for serving as a birth canal during parturition. It opens to the exterior by an aperture called vaginal orifice situated posterior to the opening of urethra.

Accessory Glands:
1. A pair of greater vestibular or Bartholin’s gland occurs on each side of the vaginal orifice. They are small rounded bodies.
These glands correspond-to the Cowper’s gland of the male and secrete a clear, viscid fluid under sexual excitement. This fluid serves as a lubricant during copulation.

2. Numerous glands of skene are present on either sides of urethral orifice. They are homologous to male’s prostate and secrete mucus. These are also called lesser vestibular or paraurethral glands.

External Genitalia (Vulva):
The external genitalia of females consists of the following parts
1. Mons pubis It is the anteriormost structure of vulva and is covered with skin and pubic hairs.

2. Two longitudinal folds of tissue called labia majora form the boundary of the vulva. It also covers two additional folds of tissue called the labia minora.
Both labia majora and labia minora protect the vaginal and urethral openings.

3. Clitoris It is the small erectile organ, which lies at the upper junction of the labia minora above the urethral opening. It is a homologous structure to glans penis of males.

Frontal view of vagina with its associated structures

4. Hymen The vaginal orifice is normally covered by a membrane, called hymen. It is a thin mucous membrane, which covers the vaginal opening either partly or completely. It is often torn during the first coitus (intercourse), but may be present in some women even after coitus. Hymen may also get stretched or torn by normal activities such as horseback riding and therefore, the presence or absence is not an accurate indicator of a woman’s virginity or sexual experience.

Question 4.
Illustrate in detail the process of spermatogenesis in human.
Or Describe the process of spermatogenesis.
Answer:
Spermatogenesis:
It is the formation of spermatozoa (sperms) in the testes, which originate from the cuboidal cells called Primordial Germ Cells (PGCs). The process of spermatogenesis starts at puberty.
It is completed in two steps

• Formation of spermatids (spermatocytogenesis)
• Transformation of spermatids to sperms (spermiogenesis)

Spermatocytogenesis
It includes the following stages

1. Multiplication phase The undifferentiated germ cells present on the inside wall of seminiferous tubules of the testes are called spermatogonia (sing : spermatogonium) or sperm mother cells. They increase in number by repeated mitotic divisions so that the newly formed spermatogonium possesses the same number of chromosomes.
Each spermatogonium is diploid and contains 46-chromosomes.

2. Growth phase Some of the spermatogonia grow by obtaining nourishment from Sertoli cells and differentiate to primary spermatocytes.

3. Maturation phase or Formation of spermatids Each of these primary spermatocytes undergoes first meiotic division (reductional) and produces two secondary spermatocytes (23-chromosomes) each which further undergoes second meiotic division (equational) to produce four haploid spermatids.

Spermiogenesis:
The spermatids (immotile) transform into motile spermatozoa by the process called spermiogenesis. It is also called differentiation phase during which sperm heads get embedded in the Sertoli cells and are finally released from the seminiferous tubules by the process of spermiation.

Hormones Essential for Spermatogenesis
Spermatogenesis initiates due to a significant increase in the secretion of Gonadotropin Releasing Hormone (GnRH) by the hypothalamus at the age of puberty. The increased level of GnRH acts on the anterior pituitary and stimulates the secretion of two gonadotropins, i.e. Luteinizing Hormone (LH) and Follicle Stimulating Hormone (FSH). LH acts on the Leydig cells thus, stimulating the synthesis and secretion of androgens. This in turn stimulate the process of spermatogenesis. FSH acts on Sertoli cells and stimulates the secretion of some factors like inhibins, activins and transferrins which help in spermiogenesis.

Question 5.
(i) When and where does spermatogenesis occur in a human male?
(ii) Draw the diagram of a mature human male gamete. Label the following parts: acrosome, nucleus, middle piece and tail.
(iii) Mention the functions of acrosome and middle piece.
Answer:
(i) Spermatogenesis occurs at the age of puberty in testes.

(ii)

(iii) Head It is flat, oval and composed of a large haploid nucleus and a small anterior cap-like structure, called acrosome. The acrosome lies at the tip of the nucleus and is formed from the Golgi complex. It contains hydrolytic enzymes and is used to contact and penetrate the egg (ovum) during fertilisation.

(ii) Middle piece It is cylindrical and known as the , powerhouse of the sperm. It possesses many mitochondria to produce energy for the movement of the tail that facilitates sperm motility essential for fertilisation.

Question 6.
What is oogenesis? Give a brief account of oogenesis.
Or Write a note on oogenesis.
Or Describe the process of oogenesis.
Answer:
Oogenesis:
It is the process of formation of a mature female gamete called egg or ova. The production of eggs in females begins before birth, i.e. during the embryonic development stage, but is completed only after fertilisation. It takes place in ovaries.
This process involves following phases

(i) Multiplication phase Cells of germinal epithelium undergo mitotic division to produce undifferentiated germ cells called oogonia or egg mother cells within each foetal ovary.

1. The oogonia further divides by mitosis and projects as a chord into the stroma of the ovary. This chord is known as egg tube or pfluger, which later become a round mass called egg nest.
2. In the egg nest, one cell grows, enters into prophase-I of meiosis and gets arrested at diplotene stage temporarily. This cell becomes the primary oocyte.
3. Other oogonia in the nest surround this primary ooctye to form primary follicle. These follicles protects and nourish the primary oocyte.
4. About 2 million of primary follicles exist in females before birth. A large number of these follicles degenerate during the phase from birth to puberty by the process called follicular atresia. Therefore, at puberty only 60000- 80000 primary follicles are left in each ovary.

(ii) Growth phase

1. The primary follicles get surrounded by more layers of granulosa cells and a new theca to form secondary follicles.
   This follicle then gets converted into a tertiary follicle that possesses a fluid-filled cavity called antrum surrounded by theca interna and theca externa.

(iii) Maturation phase The primary oocyte within this tertiary follicle grows in size. The fully grown primary oocyte completes its first meiotic division producing two daughter nuclei in which larger haploid cell is called secondary oocyte and the tiny one is called first polar body or polocyte. The secondary oocyte contains bulk of the nutrient rich cytoplasm of primary oocyte and becomes tertiary follicle.

1. Tertiary follicle changes into mature Graafian follicle. Zona pellucida that is formed by secondary oocyte, surrounds it.
2. Graafian follicle ruptures to release secondary oocyte. This process is called ovulation.
3. The secondary oocyte completes its maturation in Fallopian tube by second maturation division (unequal division), which is triggered by sperm penetration (egg activation) and leads to the formation of ootid and a second polar body. First polar body usually degenerates during this division.

The ootid undergoes few changes and gets transformed into mature ovum or egg, which is ready to be fertilised.

Odisha State Board CHSE Odisha Class 12 Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants Important Questions and Answers.

CHSE Odisha 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants

Sexual Reproduction in Flowering Plants Class 12 Important Questions CHSE Odisha

Very Short Answer Type Questions

Choose the correct option

Question 1.
The cushion of parenchymatous cells that joins ovary and ovule is known as
(a) nucellus
(b) placenta
(c) hilum
(d) funiculus
Answer:
(b) placenta

Question 2.
The narrow pore at one end of the ovule is called as
(a) funiculus
(b) chalaza
(c) micropyle
(d) hilum
Answer:
(c) micropyle

Question 3.
Megagametogenesis is the process of formation of embryo sac from
(a) pollen grain
(b) microspore
(c) ovule
(d) megaspore
Answer:
(d) megaspore

Question 4.
Antipodal cells are three in number and occur towards
(a) chalazal pole
(b) micropylar pole
(c) Both (a) and (b)
(d) None of the above
Answer:
(a) chalazal pole

Question 5.
Filiform apparatus is a characteristic feature of
(a) egg
(b) synergid
(c) zygote
(d) suspensor
Answer:
(b) synergid

Question 6.
Of the three cells in the micropylar end of the embryo sac, the central one is the egg cell and the other two are called ……..
(a) synergids
(b) antipodals
(c) generative cells
(d) vegetative cells
Answer:
(a) synergids

Question 7.
Wind pollination is common in
(a) lilies
(b) grasses
(c) orchids
(d) legumes
Ans.
(b) grasses

Question 8.
Nocturnal flowers like Nyctanthes attract insects by their …….. .
(a) colour
(b) nectar
(c) scent
(d) edible sap
Answer:
(c) scent

Question 9.
Adaptation of some floral parts, which acts as barriers to self-pollination is called …….. .
(a) dichogamy
(b) herkogamy
(c) homogamy
(d) cleistogamy
Answer:
(c) homogamy

Question 10.
In …….. pollen tube enters through micropyle into the ovule.
Or Entry of pollen tube through micropyle during fertilisation is called …….. .
(a) porogamy
(b) chalazogamy
(c) mesogamy
(d) herkogamy
Answer:
(a) porogamy

Question 11.
Fusion of the male gamete with …….. in the embryo sac of angiosperms forms the primary endosperm nucleus.
(a) antipodals
(b) synergids
(c) polar nuclei
(d) egg cells
Answer:
(c) polar nuclei

Question 12.
Individual part or segment of calyx is called
(a) sepal
(b) petal
(c) tepal
(d) corolla
Answer:
(a) sepal

Question 13.
Chalazal pole is present
(a) opposite to micropyle
(b) at the origin of integuments
(c) opposite to nucellus
(d) near the embryo sac
Answer:
(a) opposite to micropyle

Question 14.
In angiosperms, triple fusion is required for the formation of
(a) embryo
(b) endosperm
(c) seed coat
(d) fruit wall
Answer:
(b) endosperm

Question 15.
Zygote divides by an asymmetric mitotic division to form two cells. Out of these, the cells towards chalazal side is known as
(a) apical cell
(b) basal cell
(c) Both (a) and (b)
(d) None of these
Answer:
(a) apical cell

Question 16.
Basal cell divides to produce
(a) haustorium
(b) suspensor
(c) hypobasal cell
(d) epibasal cell
Answwer:
(b) suspensor

Question 17.
The part of embryonal axis above the level of cotyledons is called
(a) epicotyl
(b) hypocotyl
(c) Both (a) and (b)
(d) None of these
Answer:
(a) epicotyl

Question 18.
Root cap enclosed in an undifferentiated sheath is called
(a) epicotyl
(b) coleorhiza
(c) coleoptile
(d) scutellum
Answer:
(b) coleorhiza

Question 19.
Which type of endosperm is found in Asphodelusl
(a) Helobial
(b) Cellular
(c) Nuclear
(d) Both (a) and (b)
Answer:
(a) Helobial

Question 20.
True polyembryony occurs in
(a) Citrus
(b) Mangifera
(c) Opuntia
(d) All of these
Answer:
(d) All of these

Question 21.
Adventive embryony in Citrus is due to
(a) nucellus
(b) integuments
(c) zygotic embryo
(d) fertilised egg
Answer:
(a) nucellus

Question 22.
Formation of gametophyte directly from sporophyte without meiosis is
(a) apospory
(b) apogamy
(c) parthenogenesis
(d) amphimixis
Answer:
(a) apospory

Question 23.
Albuminous seed
(a) has no endosperm
(b) has thick cotyledons
(c) have food storage in cotyledons
(d) Both (a) and (c)
Answer:
(c) have food storage in cotyledons

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

Question 1.
Plants with male and female reproductive structures present on the same plant is called dioecious.
Answer:
monoecious

Question 2.
Stamen helps in production of megaspores, fruits and seeds.
Answer:
Gynoecium

Question 3.
Potential pollen mother cell gives rise to megaspores.
Answer:
microspore tetrad

Question 4.
The inner sterile tissue that provides nourishment to the developing microspores in microsporogenesis is called endothecium.
Answer:
tapetum

Question 5.
The mode of arrangement of ovule along the placenta in the cavity of the ovary is known as style.
Answer:
placentation

Question 6.
In protandrous flower, carpels mature earlier than stamens.
Answer:
protogynous

Question 7.
An ovule is a differentiated megasporangium.
Answer:
integumented

Question 8.
Megaspore mother cell is found near the region of nucellus.
Answer:
nucellus

Question 9.
The tegmen marks the point of attachment to the stalk.
Answer:
The hilum marks the point of atta’chment to the stalk.

Question 10.
Perisperm is remnents of embryo.
Answer:
Perisperm is remnents of nucellus.

Question 12.
Apomixis is the development of seeds with fertilisation.
Answer:
Apomixis is the development of seeds without fertilisation.

Question 13.
Thalamus contributes in the fruit formation in apple.
Answer:
True

Fill in the blanks

Question 1.
The stalk of ovule by which it is connected to placenta is called …………. .
Answer:
funiculus

Question 2.
A mass of parenchyma cells, surrounded by integuments and encloses embryo sac is called ………. .
Answer:
nucellus

Question 3.
………. is formed of a chemical called sporopollenin.
Answer:
Exine

Question 4.
A flower is said to ………. when either of the two
sexes is missing.
Ans. imperfect

Question 5.
Aquatic plants like water hyacinth and water lily are pollinated by ……….. .
Answer:
entomophily (insects)

Question 6.
Intine is made up of ………….. .
Answer:
cellulose and pectin

Question 7.
In a zygote, the terminal cell situated towards chalazal pole is called ………….. .
Answer:
apical cell or embryonal cell.

Question 8.
The position of plumule in monocot embryo is ………… .
Answer:
lateral

Question 9.
The part of pistil which develops into fruits is …………. .
Answer:
ovary

Question 10.
The thick swollen embryonal leaf filled with reserve food is called …………… .
Answer:
cotyledon

Express in one or two word(s)

Question 1.
Name the pollination preferred by snails.
Answer:
Malacophily

Question 2.
The flowers which are bisexual and never open.
Answer:
Cleistogamous flowers.

Question 3.
Stigma of a flower matures earlier than the anther.
Answer:
Protogyny.

Question 4.
The individual members of a corolla.
Answer:
Petals.

Question 5.
Give the name of the type of ovule in which the hilum, chalaza and the micropyle lie in the same longitudinal axis.
Answer:
Orthotropous

Question 6.
Name the type of pollination as a result of which genetically different types of pollen grains of same species land on the stigma.
Answer:
Xenogamy

Question 7.
A type of endosperm, which is an intermediate between cellular and nuclear type.
Answer:
Helobial endosperm.

Question 8.
The portion of the embryonal axis above the level of attachment of scutellum.
Answer:
Epicotyl

Question 9.
An embryo sac directly produced from a nucellar cell.
Answer:
Aposporous embryo sac.

Question 10.
Embryonal axis above the level of cotyledons.
Answer:
Plumule

Question 11.
Embryonal axis below the level of cotyledons.
Answer:
Radicle

Short Answer Type Questions

Question 1.
What is microsporogenesis? Where does it occur in angiosperms? What is its significance?
Answer:
Microsporogenesis:
Each cell of the sporogenous tissue is a potential Pollen Mother Cell (PMC) or microspore mother cell and can give rise to microspore tetrad. This process of formation of microspore from a pollen mother cell through the process of meiosis is called microsporogenesis.

As the anthers mature and dehydrate, the microspores dissociate from one another and form tetrad and develop into pollen grains. Inside each microsporangium, several thousands of microspores or pollen grains are formed that are released with the dehiscence of anther. In general, dehiscence of anther occurs through the rupture of anther lobe walls which causes release of several thousands of pollen grains at a time.

Question 2.
What is triple fusion? Where and how does it take place? Give the name of nuclei involved in triple fusion.
Answer:
The diploid zygote finally develops into’the embryo. The second male gamete fuses with the two polar nuclei or secondary nucleus in the central cell to form the triploid Primary Endosperm Nucleus (PEN). The process is called triple fusion as three haploid nuclei are involved in the fusion.

After triple fusion, the central cell becomes the Primary Endosperm Cell (PEC) which gives rise to the endosperm, while the zygote develops into the embryo. As both the fusions, syngamy and triple * fusion, occur in an embryo sac, the phenomenon is termed as double fertilisation.

Question 3.
Name all the haploid cells present in an unfertilised mature embryo sac of a flowering plant. Write the total number of cells in it.
Answer:
An unfertilised embryo sac of angiosperm is composed of 8-nuclei and 7 cells. Among 8 nuclei, 6 are enclosed by cell walls and organised into cell, while the remaining 2 nuclei (called polar nuclei) are situated above the egg apparatus.
Out of 6 cells, 3 ate grouped at micropylar end and constitute the egg apparatus, made up of 2 synergids and 1 egg cell. The other 3 are located at chalazal end and are called antipodals.

Question 4.
Explain the role of tapetum in the formation of pollen grain wall.
Answer:
Tapetum is the innermost layer of the microsporangium. It produces the exine layer of the pollen grains, which is composed of the sporopollenin, the most resistant fatty substance.
During microsporogenesis, the cells of tapetum produce various enzymes, hormones, amino acids and other nutritious materials required for the development of pollen grains.

Question 5.
How does the pollen grow through the style? Explain briefly.
Answer:
Growth of a Pollen Tube:
The pollen grains reach the receptive stigma of the carpel by the act of pollination. Pollen grains after getting attached to the stigma absorb water and swell. After compatible pollination, the pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores. The contents of the pollen grain then move into this tube.

The pollen tube grows through the tissues of stigma and style by secreting the enzymes that digest them. In some plants that shed pollen at two-celled condition, the generative cell divides and forms the two male gametes during the growth of pollen tube in the stigma.
In plants which shed pollen in the three-celled condition, pollen carries the two male gametes from the beginning.

Question 6.
Make a list of any three outbreeding devices that flowering plants have developed and explain how they help to encourage cross-pollination.
Answer:
Adaptations (Contrivances) for Cross-Pollination or Outbreeding Devices
Continued self-pollination leads to chances of inbreeding depression. Thus, flowering plants have developed many devices to discourage self-pollination and to encourage cross-pollination. These include
1. Dichogamy In some plant species, receptivity of stigma and pollen release is not synchronised, i.e. often the pollen is released before the stigma becomes receptive (protandry) or stigma becomes receptive before the release of pollen (protogyny). This condition is called dichogamy.

2. Heterostyly In some other species, the anther and stigma are placed at different positions, so that the pollen cannot come in contact with the stigma of same flower. This condition is called heterostyly.
Both the above mentioned methods will prevent autogamy.

3. Self-incompatibility or Self-sterility is the third device to prevent inbreeding. It is a genetic phenomenon of preventing the pollen from fertilising ovules by the same flower by inhibiting pollen germination or pollen tube growth in the pistil. Self-incompatibility may be due to genotype of sporophyte known as sporophytic incompatibility, whereas if it is due to genotype of pollen, it is known as gametophytic incompatibility.

4. Dicliny or Unisexuality effectively prevents self-pollination. It is the presence of unisexual flowers in plants that prevents autogamy but not geitonogamy, e.g. castor, maize, etc.

5. Herkogamy is seen in orchids where male or female sex organs themselves prove as a barrier to prevent self-pollination by some structural abnormalities.

6. Dioecy Both autogamy and geitonogamy is prevented in several species like papaya, where male and female flowers are present on different plants, i.e. each plant is either male or female (dioecy).

Question 7.
What is significance of pollination?
Answer:

1. Pollination is essential for fertilisation and production of seeds and fruits, which is necessary for continuity of plant life.
2. Cross-pollination results in the production of plants with variations having new combination of characters of two different plants of the same species.

Question 8.
What is hydrophily? Name any hydrophilous plant and give its three important characters which help in pollination.
Answer:
It is the transfer of pollen grains from anthers to the stigma through the agency of water.
Vallisneria is a hydrophilous plant. It is a dioecious plant in which male flowers and female flowers are borne on separate plants. The important characteristics are pollination are .

• Flowers are small, colourless, nectarlless.
• Calyx and corolla and other floral parts are the unwettable. Mature male flowers abscise and float on water.
• Stigmas are well-exposed, long and sticky.
• Pollen grains and stigmas are unwettable.

After pollination female flowers are pulled inside the water.

Question 9.
Geitonogamous flowering plants are genetically autogamous but functionally cross-pollinated. Justify.
Answer:
Geitonogamous flowering plants have open bisexual flowers in which pollen grains from the anther are transferred to the stigma of another flower of same plant (genetically similar to autogamy) involving different

• pollination agent like the cross-pollination (functionally similar to cross-pollination).

Question 10.
Explain the function of each of the following
(i) Coleorhiza
(ii) Germ pores
Answer:
(i) Coleorhiza is a cap-like structure present over the radicle in case of monocot seed, e.g. maize grain. During germination, it grows to form a small tubular structure, which protects the enclosed growing radicle, in early stages of development.
(ii) Germ pores are the areas on the wall of pollen grains where the exine is thin or absent. The areas may have thickened intine or deposition of callose. The pollen tube comes out through one of the germ pores.

Question 11.
Where is sporopollenin present in plants? State its significance with reference to its chemical nature.
Answer:
Sporopollenin is a chemical substance which forms the hard outer layer, the exine of the pollen grains. It is one of the most resistant organic material. It can withstand high temperatures, strong acids and alkali.
No enzyme can degrade sporopollenin. Pollen grains are preserved as fossils because of the presence of sporopollenin.

Question 12.
What is pollen kit? Write a note on pollen viability.
Answer:
Pollen kit is a yellow, oily, sticky layer present on the surface of pollen grains of entomophilous plants. It protects the pollen from atmospheric harmful radiations. Pollen viability is the period for which pollen grains retain the ability to germinate and produce pollen tubes. It varies from few minutes (rice, wheat) to several months (Rosaceae, Leguminoseae).
It however depends upon temperature and humidity. Pollen can be stored for years in liquid nitrogen in pollen banks for future plant breeding programmes.

Question 13.
Write the mode of pollination in Vallisneria and water lily. Explain the mechanism of pollination in Vallisneria.
Answer:
Mode of pollination in (i) Vallisneria- epihydrophily (ii) water lily-entomophily.
In Vallisneria, a dioecious plant male flower abscises from the submerged spadix and rises to surface of water.
The female plant produces long stalked solitary flowers which are also brought to the surface of water. They have large sticky trifid stigmas.
The male flowers come in contact with stigma of female flowers. The anther bursts and pollination is performed.
After pollination, female flowers are pulled inside water by coiling of stalk.

Question 14.
Trace the development of microsporophyte in the anther to a mature pollen grain.
Answer:
Microsporophyte or microspore mother cells / pollen mother cells are polygonal and closely packed in microsporangia initially. Pollen grains develop from them in pollen sacs of anthers. As anther enlarges, pollen mother cells lose cytoplasmic connections and become loosely arranged.

A few microsporocytes become non-functional and are absorbed by developing microspores. Each microsporocyte becomes rounded, undergoes meiosis and forms four haploid cells microspores or pollen grains. They are arranged in tetrahedral manner.
Later on from each tetrad, microspores separate and develop into a mature pollen grain.

Question 15.
What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer.
Answer:
Chasmogamous flowers are the bisexual flowers that are open with exposed anther and stigma.
Cross-pollination cannot occur in cleistogamous flowers because they remain in bud form throughout their life and form seeds via self-pollination.

Question 16.
Why do some plants have both chasmogamous and cleistogamous flowers?
Answer:
Viola (common pansy) and Commelina can produce both chasmogamous and cleistogamous flowers on the same plant. These flowers produce seeds by outcrossing and selfing respectively and the mixed breeding system is considered a successful reproductive strategy.

Question 17.
Why is geitonogamy also referred to as genetical autogamy?
Answer:
Geitonogamy is functionally cross-pollination, involving a pollinating agent. But genetically, it is similar to autogamy since, the pollen grains come from the same plant. So, it is also referred to as genetical autogamy.

Question 18.
Some plants have a mechanism of shedding of pollen before maturation of stigma. Why?
Answer:
The strategy of shedding of pollen before maturation of stigma is to prevent self-fertilisation and reduce inbreeding depression. Such mechanisms facilitate cross-pollination and produce genetically better yielding crops.

Question 19.
What is self-incompatibility? Why does self-pollination not lead to seed formation in self-incompatible species?
Answer:
Self-incompatibility is a genetic mechanism to prevent self-pollination from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.
Due to the inhibition of self-fertilisation, self-pollination does not lead to seed formation.

Question 20.
Mention two strategies evolved to prevent self-pollination in flowers.
Or
Explain any two devices by which autogamy is prevented in flowering plants.
Answer:
(1) Dichogamy In some plant species, receptivity of stigma and pollen release is not synchronised, i.e. often the pollcn is released before the stigma becomes receptive (protandry) or stigma becomes receptive before the release of pollen (protogyny). This condition is called dichogamy.

(2) Self-incompatibility or Self-sterility is the third device to prevent inbreeding. ft is a genetic phenomenon of preventing the pollen from fertilising ovules by che same flower by inhibiting pollen germination or pollen tube growth in the pistil Self-incompatibility may be duc co genotype of sporophyte known as sporophytic incompatibility, whereas if it is due to genotype of pollen, t is known as gametophytic incompatibility.

Question 21.
Write short note on out breeding devices.
Answer:
1. Dichogamy In some plant species, receptivity of stigma and pollen release is not synchronised, i.e. often the pollen is released before the stigma becomes receptive (protandry) or stigma becomes receptive before the release of pollen (protogyny). This condition is called dichogamy.

2. Heterostyly In some other species, the anther and stigma are placed at different positions, so that the pollen cannot come in contact with the stigma of same flower. This condition is called heterostyly.
Both the above mentioned methods will prevent autogamy.

3. Self-incompatibility or Self-sterility is the third device to prevent inbreeding. It is a genetic phenomenon of preventing the pollen from fertilising ovules by the same flower by inhibiting pollen germination or pollen tube growth in the pistil. Self-incompatibility may be due to genotype of sporophyte known as sporophytic incompatibility, whereas if it is due to genotype of pollen, it is known as gametophytic incompatibility.

4. Dicliny or Unisexuality effectively prevents self-pollination. It is the presence of unisexual flowers in plants that prevents autogamy but not geitonogamy, e.g. castor, maize, etc.

5. Herkogamy is seen in orchids where male or female sex organs themselves prove as a barrier to prevent self-pollination by some structural abnormalities.

6. Dioecy Both autogamy and geitonogamy is prevented in several species like papaya, where male and female flowers are present on different plants, i.e. each plant is either male or female (dioecy).

Question 22.
“Not all hydrophytes are pollinated by water.”  Justify by giving two examples.
Answer:
All hydrophytes are not pollinated by water.
Those hydrophytes whose flowers emerge above the surface of water are pollinated by insects or wind, e.g. water hyacinth and water lily are pollinated by insects.

Question 23.
Write short note on endosperm.
Answer:
Endosperm development precedes embryo development. The process takes place by following steps
Primary endosperm cell divides repeatedly and forms a triploid endosperm tissue. The cells of this tissue are filled with reserve food materials and are used for the nutrition of the developing embryo.

Types of Endosperm:
On the basis of its development, endosperm is classified into three main groups
(i) Cellular endosperm It is found in 25% of the angiosperm families particularly in dicotyledons. In this type, the primary endosperm mother cell divides by mitotic divisions and each division is followed by a cell wall formation, e.g. Adoxa, Verbascum, Cetranthus, Scutellaria, Datura, Myostis arvensis, Impatiens, Magnolia, etc.

(ii) Nuclear endosperm It is a more common type of endosperm and occurs in about 56% of the angiosperm families. In this type, PEM cell divides by free-nuclear divisions, which are not followed by the cell wall formation.

Thus, free nuclei remain in the cytoplasm of embryo sac. Later on, cell wall formation takes place from periphery to the centre (centripetal) leading to the formation of a cellular type of embryo sac, e.g. Acer, Arachis, Gravillea, Lomatia, Phoenix, Calotropis, Primula, Citrus, Malva, Acalypha, etc.

In coconut, Primary Endosperm Nucleus (PEN) undergoes a number of free nuclear divisions. Hence, the embryo sac gets filled with a clear fluid, this fluid is known as coconut water.

Question 24.
Why do you think that the zygote is dormant for sometime in a fertilised ovule?
Answer:
Zygote is dormant for some time because the outer conditions like high temperature, humidity, improper light, etc., are not favourable for germination.

Question 25.
In angiosperms, zygote is diploid while primary endosperm cell is triploid. Explain.
Answer:
Zygote is a product of fusion of haploid male gamete and haploid female gamete, i.e. egg cell. Thus, it is diploid.

Whereas, primary endosperm cell is a product of fusion of secondary nucleus (2n) and a haploid male gamete so, it is triploid.

Question 26.
Why do the integuments of an ovule harden and the water content gets highly reduced as the seed matures?
Answer:
Integuments of ovule harden to form tough protective seed coats. The micropyle remains as a small pore in the seed coat. This facilitates the entry of oxygen and water into the seed during germination.

Question 27.
Strawberry is sweet and eaten raw just like any other fruit. Why do botanists call it a false fruit?
Answer:
In most of the plants, by the time fruit develops from ‘ the ovary, other floral parts degenerate and fall off.
However, in case of strawberry, thalamus also contributes in fruit formation. So, it is called false fruit.

Question 28.
Are pollination and fertilisation necessary in apomixis?
Answer:
No, pollination and fertilisation are not necessary in apomixis. Apomixis is a form of asexual reproduction that mimics sexual reproduction, it is the formation of seeds without fertilisation.

Question 29.
Give reasons why hybrid seeds are to be produced year after year.
Answer:
If the seeds are collected from the hybrid variety and are sown the plants in the progeny will segregate and would not maintain the hybrid characters. So, hybrid seeds are to be produced year after year.

Question 30.
Starting with the zygote, draw the diagrams of the different stages of embryo development in a dicot.
Answer:
The lowermost cell of the suspensor is called hypophysis (h). The hypophysis cell divides to form 8-cells which are arranged in 2 times. The lower tier gives rise w root cap and epidermis whereas che other tier forms initials which finally develop into root corta.

Successive stages of the development of proembryo in Capsella bursa-pastons

Question 31.
(i) Describe the endosperm development in coconut.
(ii) Why is tender coconut considered a healthy source of nutrition?
Answer:
(i) In coconut, endosperm formation is nuclear type.
The primary endosperm nucleus undergoes nuclear division without cell wall formation.
(ii) Soft coconut is an endosperm. It is rich in nutrients like fats, proteins, carbohydrates, minerals, vitamins, etc. Hence, it is considered as a healthy source of nutrition.

Question 32.
Double fertilisation is reported in plants of both, castor and groundnut. However, the. mature seeds of groundnut are non-albuminous and castor are albuminous. Explain the post-fertilisation events that are responsible for it.
Answer:
Double fertilisation is reported in both castor and groundnut but their mature seeds are different in terms of endosperm. The primary endosperm nucleus formed after fertilisation divides mitotically without cytokinesis ‘ to initiate the formation of endosperm. At this stage, the endosperm is called free nuclear endosperm.

Then, cell wall formation occurs and the endosperm becomes cellular type. The number of free nuclei formed before cellularisation varies gready. Endosperm ‘ may be completely utilised by the developing embryo before the maturation of seeds as in groundnut. Such seeds are called non-albuminous or non-endospermic seeds. When a portion of endosperm remains in seeds and is used up during seed germination, such seeds are called albuminous or endospermic seeds, e.g. castor.

Question 33.
Write short note on parthenocarpy.
Answer:
Parthenocarpy
The formation of fruits without fertilisation is termed as parthcnocarpy. The term parthenocarpy was coined by Noll (1902). The parthenocarpic development of fruit may require the pollination stimulus (stimulative parthenocarpy) or it may occur in unpollinated flowers (vegetative parthenocarpy).

Seedless fruits should not be considered synonymous to parthenocarpic fruits because in a seedless fruit the ovules may have been fertilised and later aborted, e.g. Vizir vinifrra.
There may be parthenocarpic fruits with seeds in them. Auxin treatments arc known o produce seeded parthenocarpic fruits in Citrus, grapes, etc.

Question 34.
What is apomixis? Comment on its significance. How can it be commercially used?
Or Write short note on apomixis.
Answer:
Apomixis is a type of asexual reproduction that mimics sexual reproduction, i.e. in which seeds are produced without fertilisation. Here, the embryos can develop directly from the nucellus, integuments, synergid or egg of a diploid embryo sac. The plants produced by apomixis are genetically similar to the parent plant. They maintain and exhibit all the good characteristics of the parent plant without any risk of segregation of hybrid characters as in case of sexually reproducing plants. The information gained from genetics of apomixis can be commercially used in transfer of apomictic genes into hybrid varieties. The production of hybrid seeds by sexual reproduction every year, is too expensive for farmers.

Long Answer Type Questions

Question 1.
Give an account of the development of male gametophyte in angiosperms.
Answer:
Development of a Male Gametophyte

Microspore is the first cell of male gametophyte. It involves formation of microsporangium and development of male gametophyte at pre-pollination and post-pollination events. The structures are stages which lead to development of male gametophyte are as follows

Stamen:
It is the male reproductive unit of angiosperm.
It consists of two parts
(i) The long and slender stalk called the filament.
(ii) The terminal generally bilobed structure called the anther.
The anther and filament are connected by a connective.

Stamen (a) Ventral view, (b) Dorsal view, (c) TS of anther (enlarged)

Structure of an Anther:
A typical angiosperm anther is bilobed structure with each lobe having two theca (dithecous) and separated by a longitudinal groove running lengthwise.
In a cross-section, the anther is a four-sided (tetragonal) structure consisting of four microsporangia located at the corners, two in each lobe. Later, the microsporangia develop and become pollen sacs, which are packed with the pollen grains.

Formation of Microsporangium (Pollen Sacs):
A typical microsporangium is surrounded by four wall layers, i.e. the epidermis, endothecium, middle layer and the tapetum. The outer three wall layers are protective in function and help in dehiscence of anther to release pollen grains. Tapetum (innermost layer) nourishes the developing microspores or pollen grains and the cells of tapetum possess dense cytoplasm and generally have more than one nucleus. When the anther is young, a group of compactly arranged homogenous cells called the sporogenous tissue occupies the centre of each microsporangium.

Transverse section of a mature anther

A young anther possesses a homogenous mass of hypodermal cells bounded by epidermis. After some time, this homogenous mass appears like a tetra-angular mass. Inner to this epidermis, some cells at each angle contain a prominent nucleus and abundant protoplasm, they are . called archesporial cells.

Development of microsporangium (a-e) successive stages of the development of microsporangium, (f) A mature pollen sac in a

These cells undergo periclinal division (parallel to the outerwall of the epidermal cells) to form outer parietal cells and inner sporogenous cells. The parietal cells divide by anticlinal and periclinal divisions to form 2-5 layered microsporangial wall. The sporogenous cells either directly function as microspore mother cells or go through some mitotic divisions and then junction as Microspore Mother Cells (MMCs).

Microsporogenesis:
Each cell of the sporogenous tissue is a potential Pollen Mother Cell (PMC) or microspore mother cell and can give rise to microspore tetrad. This process of formation of microspore from a pollen mother cell through the process of meiosis is called microsporogenesis.

As the anthers mature and dehydrate, the microspores dissociate from one another and form tetrad and develop into pollen grains. Inside each microsporangium, several thousands of microspores or pollen grains are formed that are released with the dehiscence of anther. In general, dehiscence of anther occurs through the rupture of anther lobe walls which causes release of several thousands of pollen grains at a time.

The Microspore or Pollen Grain:
It is a haploid, uninucleate and minute spore produced in large numbers by meiosis in the microspore mother cell. They vary in their size, shape, colour, design, etc., from species to species.

Structure of pollen grain

Pollen grains are generally spherical measuring about 25-50 micrometer in diameter. It has a two layered wall (also called sporoderm), outer hard layer exine is made up of sporopollenin. It is the most resistant organic material as it can withstand high temperature and strong acids and alkali.

No known enzyme can degrade it. Thus, pollens are well preserved as fossils. Pollen grains have prominent distal aperture for germination called germ pore, where sporopollenin is absent. The inner intine layer is thin and chiefly composed of cellulose and pectin.

In insect pollinated pollen grains, the exine is covered by a yellowish viscous and sticky substance called pollenkitt which emits smell. The definite function of pollenkitt is not known but it is believed that it helps in attracting insects and protects the pollen from ultraviolet radiation.
Study of pollen grains is known as palynology. The pollen grain on further development forms a male gametophyte.

Formation of Male Gametophyte:
(i) Pre-pollination development Development of male gametophyte starts in pollen grains while still present in the microsporangium or pollen sact The nucleus of pollen grain grows in size, moves to one side near the wall and divides mitotically to form a vegetative cell and a generative-cell. A layer of callose develops around the generative cell. Later on, the callose dissolves and the naked generative cell comes to lie freely in the cytoplasm of the tube cell. The cytoplasm of generative cell does not ‘ contain much of stored food. The shape of generative cell is elongated which helps in its passage through pollen tube.

The vegetative cell contains stored food and also contains protein granules. The microspore containing the two cells, i.e. vegetative and generative cell, is shed at this stage from microsporangia. Pollen grains are generally shed at this 2-celled stage in over 60% of angiosperms. In the remaining species, the generative cell divides mitotically and gives rise to 2 male gametes before pollen grains are shed.

(ii) Post-pollination development On reaching the stigma, pollen grain absorbs water and nutrients from the stigmatic secretion through its germ pores and the tube (vegetative) cell enlarges. The intine of the pollen grain protrudes out through one of the germ pores and a pollen tube is formed. The pollen tube pierces the stigmatic surface and moves down through the style of the pistil.

Now in the generative cell, the nucleus divides mitotically to form two male nuclei which become surrounded by a thin cytoplasmic sheath and appear as distinct non-motile male gametes. The nucleus of the generative cell, migrates to pollen tube. As the pollen tube elongates, the distal part becomes highly vacuolated and becomes separated from the anterior part containing the three nuclei, by formation of callose plug.

(a-b) Mature pollen grain, (c) Pollen germination

Question 2.
Gives the structure of a typical pollen grain and its pre and post-pollination changes.
Answer:
Formation of Male Gametophyte:
(i) Pre-pollination development Development of male gametophyte starts in pollen grains while still present in the microsporangium or pollen sact The nucleus of pollen grain grows in size, moves to one side near the wall and divides mitotically to form a vegetative cell and a generative-cell. A layer of callose develops around the generative cell. Later on, the callose dissolves and the naked generative cell comes to lie freely in the cytoplasm of the tube cell. The cytoplasm of generative cell does not ‘ contain much of stored food. The shape of generative cell is elongated which helps in its passage through pollen tube.

The vegetative cell contains stored food and also contains protein granules. The microspore containing the two cells, i.e. vegetative and generative cell, is shed at this stage from microsporangia. Pollen grains are generally shed at this 2-celled stage in over 60% of angiosperms. In the remaining species, the generative cell divides mitotically and gives rise to 2 male gametes before pollen grains are shed.

(ii) Post-pollination development On reaching the stigma, pollen grain absorbs water and nutrients from the stigmatic secretion through its germ pores and the tube (vegetative) cell enlarges. The intine of the pollen grain protrudes out through one of the germ pores and a pollen tube is formed. The pollen tube pierces the stigmatic surface and moves down through the style of the pistil.

Now in the generative cell, the nucleus divides mitotically to form two male nuclei which become surrounded by a thin cytoplasmic sheath and appear as distinct non-motile male gametes. The nucleus of the generative cell, migrates to pollen tube. As the pollen tube elongates, the distal part becomes highly vacuolated and becomes separated from the anterior part containing the three nuclei, by formation of callose plug.

(a-b) Mature pollen grain, (c) Pollen germination

Question 3.
Give an account of the development of female gametophyte in angiosperms.
Answer:
Female Gametophyte (Embryo Sac)
In general, the development of embryo sac is monosporic, e.g. in Polygonum. In this type of development, only one megaspore situated towards chalazal end remains functional, while the remaining three megaspores gradually degenerate and finally disappear. Following are the different stages in development of female gametophyte
1. The functional haploid megaspore is the first cell of female gametophyte of angiosperm.
2. It enlarges in size to form the female gametophyte, also called embryo sac.
3. Its nucleus undergoes mitotic division to form 2-nuclei that move to opposite poles forming 2-nucIeate embryo sac.
4. The 2-nucleate embryo sac undergoes two more sequential mitotic divisions giving rise to the 4-nucleate stage and later 8-nucleate stage of embryo sac. This stage comprises of a micropylar end and a chalazal end with four nuclei at each end.
5. Six of the eight nuclei are surrounded by cell walls and get organised into cells. Three cells present towards the micropylar end grouped together, constitute the egg apparatus, i.e. two synergids and one egg cell.
6. Three cells of the chalazal end are called the antipodals. The large central cell is formed by the fusion of 2-polar nuclei. Thus, a typical angiospermic embryo sac or female gametophyte at maturity consists of 8-nuclei and 7-cells.
The egg cell combines with a male gamete to form zygote which becomes the embryo. The pollen tube makes its way through the synergids releasing the male gametes.

One male gamete fuses with female gamete(egg) called syngamy. The two polar nuclei which have fused to form secondary nucleus combine with second male gamete and form primary endosperm cell. This primary endosperm cell develops into endosperm which provides nutrition to the developing embryo. The antipodal cells degenerate.

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 17 Environmental Issues Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 17 Question Answer Environmental Issues

Environmental Issues Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Multiple choice questions

Question 1.
Which gas leaked from Union Carbide’s Pesticide Plant in December 1984 is responsible for Bhopal Gas Tragedy?
(a) Methyl salicylate
(b) Methyl isocyanate
(c) Ammonia
(d) Hydrogen sulphide
Answer:
(b) Methyl isocyanate

Question 2.
Minamata disease is caused by the consumption of fish contaminated with
(a) lead
(b) copper
(c) zinc
(d) mercury
Answer:
(d) mercury

Question 3.
The toxic metal used as an anti-knocking agent in petrol for automobiles is
(a) chelated copper
(b) tetraethyl lead
(c) iron sulphide
(d) lead chloride
Answer:
(b) tetraethyl lead

Question 4.
Bharat Stage Emission standards came into force from the year
(a) 1998
(b) 2000
(c) 2008
(d) 2010
Answer:
(b) 2000

Question 5.
Bone and tooth decay disease is caused by drinking water contaminated with
(a) fluoride
(b) borate
(c) silicate
(d) aluminium
Answer:
(a) fluoride

Express in one or two word(s)

Question 1.
Removal of toxic substances from water by using living organisms.
Answer:
Bioremediation

Question 2.
Toxic compound formed by the reaction of carbon monoxide with haemoglobin in blood.
Answer:
Carboxyhaemoglobin

Question 3.
Enrichment of water bodies with excess amount of nutrients as a result of runoff from surrounding land leading to overgrowth of plants and algae.
Answer:
Eutrophication

Question 4.
A kind of air pollutant named for the mixture of smoke and fog in the air.
Answer:
Smog

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

Question 1.
Fifth June of each year is usually observed as Word Food security Day.
Answer:
Environment

Question 2.
The process of nutrient enrichment in water bodies is called as biomagnification.
Answer:
Eutrophication

Question 3.
Particulate matter formed by the combination of gas and water vapour is called as smog.
Answer:
Statement is correct

Question 4.
Chipko Movement was organised for the protection of water bodies.
Answer:
Forest

Fill in the blanks

Question 1.
The Environment Protection Act was enacted in
the year
Answer:
1986

Question 2.
The common refrigerant responsible for the , depletion of ozone layer of the atmosphere is ……………. .
Answer:
chlorofluorocarbon

Question 3.
Carbon monoxide binds with haemoglobin forming …………….. .
Answer:
carboxyhaemoglobin

Question 4.
Depletion of ozone layer is speeded up by the …………. atom present in CFC.
Answer:
chlorine

Short Answer Type Questions

Write within three valid points

Question 1.
Aerosol.
Answer:

• Aerosol is a colloid of fine solid particles or liquid droplets in air or another gas.
• Aerosols can be natural or anthropogenic. Examples of natural aerosols are fog and geyser steam.
  Examples of artificial aerosols are haze, dust, particulate, air pollutant and smoke.
• These are emitted through jet and supersonic aeroplanes. It causes depletion of ozone layer.

Question 2.
Greenhouse effect.
Answer:
Greenhouse Effect
The term, ‘greenhouse effect’ has been derived from a phenomenon, which occurs inside a greenhouse. In a greenhouse, the glass panel lets the light in, but does not allow heat to escape. This results in warming up of the greenhouse.

The greenhouse effect is a naturally occurring phenomenon that is responsible for heating of earth’s surface and its atmosphere. Without greenhouse effect, the average temperature at surface of earth would have been chilly, i.e. approximately -18°C rather than the present average of 15°C.

Question 3.
Eutrophication.
Answer:
Eutrophication The process of eutrophication occurs in young lake where water is cold and clear to support life. It is the accelerated ageing of lakes due to the sewage, agricultural and industrial wastes.
The water body gets enriched with excess of nutrients such as nitrates and phosphorous promoting the overgrowth of microbes and algae. The algae release toxins in water and gradually cause deficiency of dissolved oxygen in water.

Question 4.
Acid rain.
Answer:

1. The term acid rain was given by Robert August. It is rainfall with a pH of less than 5.65.
2. Acid rain is a mixture of H₂SO₄ and HN₃ and the ratio of the two may vary depending on the relative quantities of oxides of sulphur and nitrogen emitted on an average 60-70% of acidity is ascribed to H₂SO₄ and 30-40 % to HNO₃.
3. It damages foliage and growing points of plant. Causes leaching of essential minerals of soil like Ca, Mg, NO₃^– and SO₄^-2

Question 5.
Photochemical smog.
Answer:
Photochemical Smog:
Smog refers to a combination of smoke and fog formed during winter. Its formation takes place when water vapour surrounds the smoke, dust or soot particles resulting in the formation of secondary particles. These particles remain suspended in the air.

The vehicular emissions consisting of oxides of nitrogen, sulphur and hydrocarbons undergo a series of photochemical reactions forming many photochemical oxidants. This process takes place during warmer sunny days photochemical oxidants react with troposphere ozone resulting in the formation of a brownish hozy fume which is known as photochemical smog.

Question 6.
Global warming.
Answer:
Global Warming
The gradual and continuous increase in average temperature of surface of the earth has resulted in global warming.
Climate Earth temperature has increased by 0.6°C during past century, most of it in last three decades. This increased temperature caused changes in precipitation patterns.

Differentiate between the following

Question 1.
Aerosol and Smog.
Answer:
Differences between aerosol and smog are as follows

Aerosol	Smog
These are present in the vapour form. Refrigerators and air conditioners use aerosol as refrigerant.	It is an opaque or dark fog having condensed water vapours, dust, smoke and gases.
It causes depletion of ozone layer.	It causes silvering / glazing and necrosis in plants, allergies and asthma/bronchitis in humans.

Question 2.
Renewable resources and Non-renewable resources.
Answer:
Differences between renewable resources and non-renewable resources are as follows

Renewable resources	Non-renewable resources
These resources have an ability to renew themselves in a given period of time.	These resources connot be renewed after exhaustion.
These are the energy resources which cannot be exhausted.	They are the energy resources which can be exhausted one day.
It has low carbon emission and hence environment friendly.	It causes high carbon emission and hence not environment friendly.
Solar energy, wind energy, tidal energy etc., are the examples of renewable resources.	Goal, petroleum, natural gases are the examples of non-renewable resources.

Question 3.
Bioremediation and Eutrophication.
Answer:
Differences between bioremediation and eutrophication are as follows

Bioremediation	Eutrophication
It is a waste management technique that involves the use of biological organisms to neutralize pollutants from a contaminated site.	It is the enrichment of a water body with nutrients.
Microorganisms used to perform the function of bioremediation are known as bioremediators.	This process induces growth of plants and algae and due to the biomass load may result oxygen depletion of the water body
It is of two types, i.e., in-situ and ex situ.	It is also of two types, i.e. accelerated and natural eutrophication.
It reduces pollution.	It occurs as a result of water pollution.

Question 4.
Primary pollutants and Secondary pollutants.
Answer:
Differences between primary pollutants and secondary pollutants are as follows

Primary pollutants	Secondary pollutants
These pollutants enter the environment directly from the source.	These pollutants are produced by the interaction of primary pollutants with other constituents.
They are less harmful.	They are more harmful.
They are of various categories such as particulate matter, aerosols and gases, which remain in their original form.	They generally undergo wide range of photochemical reactions and get modified.
e.g. CO, CO2, arsenic.	e.g. ozone, sulphuric and nitric acids.

Long Answer Type Questions

Question 1.
Give an account of secondary air pollutants.
Answer:
Secondary Pollutants:
These are not directly emitted but are formed when primary pollutants react in atmosphere, e.g. ozone, sulphuric acid, nitric acid, Peroxyacetyl Nitrate (PAN), etc.

1. Tropospheric Ozone:
Under the influence of UV-radiation, the nitrogen dioxide released in atmosphere undergoes dissociation. This results in the formation of nitric oxide (NO) and nascent oxygen (O). This nascent oxygen undergoes reaction with molecule oxygen and forms ozone in the troposphere. Ozone traps heat causing greenhouse effect and also causes formation of photochemical smog.

Chemical reactions in the troposphere generating ozone.

2. Peroxyacetylnitrate (PAN):
The emission from vehicles contains a lot of primary pollutants.
The precursors of PAN are methyl glyoxyl, acetaldehyde and several byproducts of oxidation of aromatic compounds. The sunlight undergoes reaction with non-methane hydrocarbons and nitrogen oxides resulting in the formation of PAN. PAN is one of the important component of photochemical smog.

3. Photochemical Smog:
Smog refers to a combination of smoke and fog formed during winter. Its formation takes place when water vapour surrounds the smoke, dust or soot particles resulting in the formation of secondary particles. These particles remain suspended in the air.

The vehicular emissions consisting of oxides of nitrogen, sulphur and hydrocarbons undergo a series of photochemical reactions forming many photochemical oxidants. This process takes place during warmer sunny days photochemical oxidants react with troposphere ozone resulting in the formation of a brownish hozy fume which is known as photochemical smog. This photochemical smog causes damage to vegetation, rubber goods and irritation in eyes and lungs.

4. Acid Rain:
Primary pollutants like oxides of nitrogen, sulphur dioxide and chlorine are released in atmosphere from the fossij fuel burning, vehicular exhaust, forest fire, etc. These primary pollutants react with water vapour present in atpiosphere resulting in the formation of acids such as nitrip acid and sulphuric acid. Their acids fall on the surface of earth in the form of rain known as acid rain. The acid present in rain has harmful effects on living organisms. Acid rain causes deterioration of historical monuments. One of the examples of deterioration of monuments include Taj Mahal in Agra.

Question 2.
How can the industrial and vehicular emissions be controlled? Describe.
Answer:
Control of Industrial Emission
Industrial emission can be controlled by two practices either by confining the pollutants of gaseous nature at the source or by diluting them in the atmosphere. The first practice involves two methods.

It can be done by modifying the process of formation of pollutants so that their formation does not occur beyond the permissible level. The second method involves reducing the concentration of pollutants before they are released into the environment. These practises takes place via the following steps

(a) Combustion It is performed when the pollutants are of organic nature. It comprises of flame combustion and catalytic combustion. It is involved in the conversion of pollutants to water vapour and less harmful carbon dioxide. Catalytic combustion makes use of catalytic converters and flame combustion uses incinerator.
(b) Absorption A scrubber is used to remove or modify emitted gas. It contains a liquid absorbent through which emitted gas is passed.
(c) Adsorption In this process, the gas is passed through a porous solid adsorbent, like activated carbon silica gel and lime stone. The pollutants are held at the interface of the adsorbent.

The working of a scrubber can be described as follows Scrubber is used to remove harmful gases like SO₂ from the industrial exhausts. The exhaust is passed through a spray of lime or water. Water dissolves the gases and lime reacts with SO₂ to form a precipitate of calcium sulphate and sulphide.

Scrubber

Control of Vehicular Emission:
Catalytic converters are fitted into automobiles (major cause of air pollution in metro cities) for reducing emissions of poisonous gases like CO and NO₂. They are made with expensive metals like- platinum-palladium and rhodium as catalysts. As the exhaust passes through catalytic converter, following changes occur
(i) Unburnt hydrocarbons get burnt completely into CO₂ and H₂O.
(ii) Carbon monoxide and nitric oxide are converted into CO₂ and N₂ gases, respectively.
Motor vehicles equipped with catalytic converter should use unleaded petrol, as lead present in petrol inactivates the catalyst.

Question 3.
Write the causes of ground water pollution and state how this can be controlled.
Answer:
Ground water pollution occurs when pollutants are released to the ground and make their way down into groundwater contaminants found in groundwater cover a broad range of physical, inorganic chemical, organic chemical, bacteriological and radioactive material.

Ground water pollutants Arsenic and fluoride have been the major pollutant of ground water. The metalloid arsenic can occur naturally in groundwater. Arsenic in groundwater can also be present where mining operations or mine waste dumps that will leach arsenic. Pathogens contained in faeces can lead to ground water pollution. Viruses and protozoans commonly found in polluted groundwater. Ground water that is contaminated with pathogens can lead to fatal faecal-oral transmission of diseases.

Nitrate is the most common chemical contaminant in the world’s groundwater and aquifers. Nitrate levels above 10 mg/L in ground water can cause Blue baby syndrome.

Organic compounds are a dangerous contaminant of groundwater. They are generally introduced to the environment through careless industrial practices.

Causes of Ground Water Pollution % Natural causes The natural arsenic pollution occurs because aquifer sediments contain organic matter that generates anaerobic conditions in the aquifer. These conditions generate arsenic.
Sanitation system Ground water pollution with pathogens and nitrate can also occur from the liquids infiltrating into the ground from on site sanitation system such as pit latrines and septic tanks.

Fertilisers and pesticides Nitrate can also enter the groundwater via excessive use of fertilisers, including manure spreading. High application rates of nitrogen- containing fertilisers combined with the high water- solubility of nitrate leads to increased runoff into surface water well as leaching into groundwater.

Commercial and industrial leaks A wide variety of both inorganic and organic pollutants have been found in aquifers underlying commercial and industrial activities.

Prevention of Ground Water Pollution:

• Landfills should be properly designed, maintained and operated, located away from sensitive groundwater areas.
• Underground storage tanks should be able to meet regulatory compliance policies on their installation and maintenance.

Deep groundwater should be regularly tested and inspected.

• Fertilisers should be used in minimum amount

Question 4.
Write about the different classes of solid wastes.
Answer:
Classes of Solid Wastes:
The various classes are

1. Domestic wastes These include wastes from homes, offices, schools, etc. It generally consists of paper, leather, textile, rubber, glass, waste food materials, etc.

2. Industrial wastes These include wastes like scraps, toxic heavy metals, flyash (oxides of iron, silica and aluminium), etc., generated by industries.

3. Biomedical wastes These include disinfectants and other harmful chemicals generated by the hospitals.

4. Electronic wastes (e-wastes) comprise the damaged electronic goods and irreparable computers. It contains harmful chemicals like copper, zinc, aluminium, etc.

5. Defunct ships Old defunct ships are broken down in developing countries like India, Bangladesh and Pakistan because of cheap labour and demand for scrap metal. These ships however, possess a number of toxic materials like asbestos, lead, mercury and polychlorinated biphenyls. The people involved in ship breaking are exposed to these toxic materials and thus suffer from various diseases. The coastal areas where ship breaking is undertaken also become polluted.

6. Agricultural wastes Solid organic wastes from agricultural practice during the growing and harvesting seasons are dumped on the soil, which decompose and are washed away into nearby water bodies. These bring about eutrophication of water bodies, which affects the local biotic potential.

7. Radioactive waste Radioactive waste is generated from nuclear power plants, nuclear weapon manufacturing facilities, cancer treating hospitals and research laboratories using radioisotopes in investigations. This waste is to be disposed off safely by observing the standard guideline because if it remains for a very long period and continues to emit ionising radiation, it will be extremely hazardous to health of all forms of life.

8. Construction waste Due to demolition or construction of buildings a large amount of waste material in different forms is produced in urbon areas.

9. Extraction and processings industry waste mining and quarrying operations also produce solid wastes. Food processing industries produce a large amount of organic waste.

10. Plastic It is non-biodegradable and also increase the volume of municipal waste. Plastic has an adverse effect on animals and birds who consume it.
Burning of these results in the generation of toxic fumes which add to air pollution.

11. Waste from natural disasters Disasters like flood, earthquake, volcanic eruption and cyclone generate a lot of ash, slag, dust, smoke and organic silt.

Question 5.
What are greenhouse gases? Write about their effects on the environment.
Answer:
Greenhouse Gases
These are the gases which trap the heat causing greenhouse effect. The carbon dioxide is the most prominent one. The various greenhouse gases are
(i) Carbon Dioxide:
It is most common and abundant greenhouse gas. Its rise has been due to the large scale deforestation, change in land use and large scale combustion of fossil fuels.
Burning of petrol and diesel contributes 36%, coal contributes 35% and natural gas contributes 20% of the carbon dioxide. The carbon dioxide level has increased from 315 ppm (parts per million) in 1958 to 355 ppm in 1992 4nd then to 389 ppm in 2010. The countries which are major contributors of carbon dioxide are USA, Russia, European countries and China.

(ii) Methane
Its concentration was 700 ppb in pre-industrial times and 1750 ppb in 2000. Methane is produced by incomplete biomass combustion and incomplete decomposition mostly by anaerobic methanogens.

(iii) Nitrous Oxide
It is produced by combustion of nitrogen rich fuels, livestock wastes, breakdown of nitrogen fertilisers in soil, nitrate contaminated water, etc.

(iv) Chlorofluorocarbons(CFCs)
It is used as a common refrigerant and aerosol propellant. Bromine atom from halon used in fire extinguishers has a similar effect. These two are potent greenhouse gases.

Relative contribution of various greenhouse gases to total global warming

(v) Water Vapour:
It has the capacity to trap heat radiating from the surface of the earth. Rain along with the presence of sunshine causes increase in temperature. Such climate is referred to as hot and humid.

(vi) Fluorinated Gases
These include perfluorocarbons and sulphur hexafluoride. They are industrial byproducts. They are also used as substitutes for CFCs.

(vii) Tropospheric Ozone
Nitrogen dioxide dissociates into nitric oxide and nascent oxygen in presence of UV-radiation. The nascent oxygen produced reacts with molecular oxygen forming ozone which acts as a greenhouse gas.

Question 6.
Write the causes and consequences of global warming.
Answer:
Global Warming
The gradual and continuous increase in average temperature of surface of the earth has resulted in global warming.

Effects of Global Warming:
It has been estimated by computer application studies that there may be a rise of 3°C by the year 2100 on an average.
The major effects of global warming include
1. Climate Earth temperature has increased by 0.6°C during past century, most of it in last three decades. This increased temperature caused changes in precipitation patterns.
2. Glaciers and ice caps Scientists have proposed that this rise in temperature causes deleterious changes in the environment, resulting in odd climatic changes. Thus, leading to melting of the polar ice caps and Himalayan snow caps.
3. Animals and humans The new warmer temperature conditions lead to eruption of diseases in animals and thousands of species will become extinct in a very short period of time. People from coastal areas will start migrating due to climate change.
4. Ocean and coasts The increase in temperature causes melting of polar ice caps and glaciers. This will result in the rise of ocean water level. The increased level of ocean water will cause the submerging of many island nations and coastal cities. The high temperature will cause accelerated vanishing of coral reefs.
5. Water and agriculture The increased temperature will cause decreased productivity in agricultural practice.

Reducing Greenhouse Gases
Due to the harmful effets of global warming, there was a need to control it. For this following steps were taken
1. World Meterological Organisation and United Nations Environment Programme (UNEP) jointly set up the Intergovernmental Panel on Climate Change (IPCC) in 1988.
2. At Earth Summit (1992), there was created the United Nations Framework Convention on Climate Change (UNFCC) which came into force in 1994.
3. Kyoto Protocol (1997) was signed by 160 countries to reduce the emission of CO 2 NO, CH4 by 5% and also to reduce CFCs emission.
4. Copenhagen conference was held in 2009 under UNFCC. However, it failed as there was no unanimity in agreement among the participating countries.

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 16 Biodiversity and its Conservation Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 16 Question Answer Biodiversity and its Conservation

Biodiversity and its Conservation Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Multiple choice questions

Question 1.
Genetic diversity refers to
(a) variation in the genetic material
(b) variation in the populations
(c) variation in the number of species
(d) variation in the animal distribution
Answer:
(a) variation in the genetic material

Question 2.
Species diversity means
(a) number of species
(b) relative abundance of species
(c) pecies composition
(d) genetic diversities
Answer:
(b) relative abundance of species

Question 3.
The Forest Conservation Act was enacted in
(a) 1972
(b) 1952
(c) 1980
(d) 1991
Answer:
(c) 1980

Question 4.
Conservation of wild animals and plants in sanctuaries and national parks is
(a) ex situ conservation
(b) in vivo conservation
(c) in vitro conservation
(d) in situ conservation
Answer:
(d) in situ conservation

Question 5.
Corbett national park is situated in
(a) Uttarakhand
(b) Jharkhand
(c) Uttar Pradesh
(d) Himachal Pradesh
Answer:
(a) Uttarakhand

Question 6.
Following mass extinctions, recovery to the same level of biodiversity has taken
(a) hundreds of years
(b) millions of years
(c) thousand of years
(d) billions of years
Answer:
(b) millions of years

Express in one or two words

Question 1.
A species originated in one place and found no where else.
Answer:
Endemic species

Question 2.
Organism whose no living representative is seen.
Answer:
Extinct

Question 3.
Biogeographic region with high endemism and habitat destruction.
Answer:
Biodiversity hotspot.

Question 4.
Conservation of biodiversity in its natural site.
Answer:
In situ conservation.

Question 5.
Diversity of all life forms in the earth.
Answer:
Biodiversity.

Correct the statement by changing the underlined word(s)

Question 1.
Hybrid plants of a species are the source of disease resistant genes.
Answer:
Disease resistant

Question 2.
Bhitarkanika is a hotspot.
Answer:
national park of India.

Question 3.
Hotspots are characterised by low endemism and habitat destruction.
Answer:
high

Question 4.
Botanical gardens are meant for in situ conservation of biodiversity.
Answer:
ex situ

Question 5.
WWF has enlisted endangered species in Red Data Book.
Answer:
IUCN

Fill in the blanks

Question 1.
The term ‘biodiversity’ was coined by ……….. .
Answer:
Thomas E. Lovejoy

Question 2.
The three levels of biodiversity are ………… diversity, species diversity and ……….. diversity.
Answer:
genetic, ecological

Question 3.
There are ……….. numbers of megabiodiversity countries in the world.
Answer:
17

Question 4.
India had ………….. numbers of biodiversity hotspots.
Answer:
three

Question 5.
The Wildlife Protection Act was enacted in …………… .
Answer:
1972

Question 6.
The UN conference of human environment was held in ……………. in 1972.
Answer:
Stockholm

Question 7.
The expanded form of IUCN is …………….. .
Answer:
International Union of Conservation of Nature and Natural Resources.

Question 8.
The first national park of India is …………. national park.
Answer:
Hailey (Jim Corbett)

Question 9.
Odisha has ……….. numbers of national park.
Answer:
two

Question 10.
There are ………… numbers of wildlife sanctuaries in Odisha.
Answer:
19

Question 11.
India has …………. numbers of biosphere reserves.
Answer:
18

Question 12.
The concept of biosphere reserve made a beginning under ……….. programme instituted by a UN body, namely ……….. .
Answer:
Man and Biosphere (MAB), UNESCO

Short Answer Type Questions

Write short note on each of the following

Question 1.
Ecological diversity
Answer:
Ecological Diversity (Diversity among Communities):
It explains about the variety of ecosystems present in the biosphere. The community composition, i.e., assemblage of several interacting populations in a given space at a particular time is affected directly by the environment. Thus, it is the diversity at the level of communities and ecosystems of a region.

Question 2.
Wildlife Protection Act (1972).
Answer:
Wildlife Protection Act (1972) It was enacted to provide protection to wild flora and fauna and other natural resources. This act offers protection based on two approaches that are
(a) Species based approach for specific endangered species which are protected by special projects such as Project Tiger.
(b) Habitat based approach which is conservation of endangered wild flora and fauna in National Parks and Wildlife Sanctuaries.

Question 3.
In situ conservation.
Answer:
It involves the to protection of plants, animals and microorganisms within their natural ecosystems. The i in situ conservation is the most effective way of protecting the species and improving the quality of the habitat they live in. The in situ approach is preferable because of the fact that not much diversity can be conserved outside the centres of diversity.

Biodiversity at all its levels can be conserved in situ by comprehensive system of protected areas such as the national parks, wildlife sanctuaries, natural reserves, natural monuments, cultural landscapes, biosphere reserves, wetlands, etc. So far, in situ practice is considered most effective method of protecting and propagating the species and improving the quality of their habitats.

Question 4.
Ex situ conservation
Answer:
Ex situ (Off-site) Conservation:
It refers to conservation of biological diversity outside the boundaries of their natural habitats by perpetuating sample population in genetic resource centres, e.g. zoos, botanical gardens, culture collections, etc., or in the form of gene pools and gametes storage for fish, germplasm banks for seeds, pollen, semen, ova, cell, etc. Zoos also help in captive breeding of organisms which are endangered, whereas botanical gardens have seed gene banks, tissue culture labs and other technologies for storing and growing germplasm.

Question 5.
Biosphere reserve
Answer:
These are large tracts of protected land used for preserving the genetic diversity of an ecosystem by preserving wildlife. The creation of biosphere reserves was initiated in 1975 under the Man and Biosphere (MAB) programme of UNESCO. It is category V protected area designated by IUCN.

Each biosphere reserve integrates human activities and has following zonation

• Core zone Strictly protected.
• Buffer zone Sustainable and recreation activities allowed.
• Transition zone Anthropogenic activities like research and sustainable development allowed.
• Zone of human encroachment Normal anthropogenic activities allowed.

Differentiate between the following

Question 1.
In situ and Ex situ conservations.
Answer:
Differences between in situ conservation and ex situ conservation are as follows

In situ conservation	Ex situ conservation
It is the conservation of species in their natural habitats.	It is the conservation of species outside their natural habitats.
The endangered species are protected from predators.	The endangered species are protected from all adverse factors.
Augmentation of depleted resources is done.	Animals or plants are kept under human supervision and provided with all the essential necessities for survival.
e.g. national parks, sanctuaries and biosphere reserves.	e.g. zoos, botanical gardens, cultural collections, etc.

Question 2.
Genetic diversity and Species diversity.
Answer:
Differences between genetic diversity and species diversity are as follows

Genetic diversity	Species diversity
It is related to the number and type of genes and their alleles found in organisms.	It is related to the number, type and distribution of species found in given area.
It is the trait of the species.	It is the trait of the community.
It influences adaptability and distribution of a species in diverse habitat.	it influences biotic interaction and stability of the community.

Question 3.
National park and Sanctuary.
Answer:
Differences between national park and sanctuary are as follows

National park	Sanctuary
It is meant for protection of flora and fauna of the area.	It is meant for protection of one or more group of wild animals.
Cultivation of land, grazing and forestry are not allowed.	Cultivation of land, grazing and forestry are allowed.
Private ownership is not permitted.	Private ownership is permitted.
Boundary is well demarcated.	Boundary is not well demarcated.
e.g. Corbett National Park (Uttarakhand)	e.g. Bird Sanctuary Chilika (Odisha).

Question 4.
Extinct species and Endangered species
Answer:
Differences between extinct species and endangered species is as follows

Extinct species	Endangered species
A taxon is extinct when there is no reasonable doubt that the last individual has died, e.g. Indian cheetah.	A species is endangered when it is facing a very high risk of extinction in the wild in the near future, e.g. giant panda, polar bear.

Long Answer Type Questions

Question 1.
What is meant by biodiversity? Write the causes of loss of biodiversity.
Answer:
The term ‘biodiversity’ was first used by Thomas E. Lovejoy (1980) to refer the number of species of a region. It is the degree of variation of life occurring at different levels like genetic, organismal and ecological. These levels forms a hierarchy of biodiversity

Loss of Biodiversity:
The loss of biodiversity is a global crisis. Extinction of species is a natural phenomenon aided by the physical changes in the environment. However, the accelerated rates of species extinctions, that the world is facing now are largely due to human activities.
Till now, five episodes of mass extinction of species have occured in the history of biological evolution. The sixth episode of extinction of species however, is credited to human activities, which otherwise would not have occurred.

According to IUCN estimates, 12259 species have become extinct since the time of origin of life on the earth. The major cause of the biodiversity losses are called drivers, which belong to two classes namely, direct and indirect.

Direct Drivers Factors
The directly influence the ecosystem processes which bring about the mass extinction of species.
The various processes associated with direct drivers includes

1. Habitat destruction and fragmentation Conversion of forest land for agriculture, development projects, mining operation, etc., leads to the destruction of the natural habitats of the organisms.
Indiscriminate agricultural practices involving use of chemical fertilisers and pesticides are potent factors for the destruction of habitats. When a large population fragments into smaller ones there is more inbreeding and inbreeding pressure leading to population decline.

2. Overexploitation of natural resources Humans are dependent on nature for food and shelter, but when ‘need’ turns to ‘greed’, it leads to overexploitation.

To meet the need for increased housing, the natural habitats of animals and plants are being destroyed. This results in habitat loss and extinction of species. It has caused extinction of many species in last 500 years. In addition, indiscriminate hunting of wild animals has made their status in Red book as endangered or critically endangered.

3. Introduction of alien invasive species When alien species are intentionally or unintentionally introduced in a particular area, they might turn invasive and cause decline or extinction of endogenous species, e.g., Eichhornia is known as the ‘Terror of Bengal’. It was introduced as ornamental plants but it became wild in India because of invasiveness.
Similarly, Lantana and Parthenium were important due to their ornamental and food values, respectively but, they become wide spread due to favourable environmental conditions.

4. Climate change The global climate is changing, due to the anthropogenic activities like greenhouse gases and it has led to global warming. This is causing melting of glaciers, polar ice caps, etc. This may submerge low lying coastal habitats and also plants and animals are unable to adapt themselves to this change which is causing their elimination.

5. Environmental pollution It is another major factor for species extinction. Pollution may reduce and eliminate populations of sensitive species. Environmental pollution is most commonly caused by accumulation of non-biodegradeble wastes like plastics. Agricultural chemicals like pesticides enters the food chain and get deposited in the body of higher organisms. This effects the population of fish eating birds and falcons by disturbing their reproductive process.

Question 2.
How can the biodiversity be conserved? Add a note on importance of biodiversity.
Answer:
Biodiversity is directly or indirectly involved in maintaining the well-being of human society. Thus, scientists have classified the values of diversity as ecosystem goods, i.e., natural products harvested from ecosystems and directly used by humans and ecosystem services, involving different ecosystem processes which indirectly benefit human life.

Biodiversity Conservation:

The protection, uplift and scientific management of biodiversity at its optimum level for present and future generations is known as biodiversity conservation.
The International Union for Conservation of Nature and Natural Resources (IUCN) or World Conservation Union, World Wide Fund for Nature (WWF), Food and Agricultural Organisation (FAO) and United Nations Educational Scientific and Cultural_Organisation (UNESCO) formed the world conservation strategy in 1980 for the conservation and sustainable use of biological resources. For this, two major types of conservation strategies were framed. These are

1. In situ (On-site) Conservation:
It involves the to protection of plants, animals and microorganisms within their natural ecosystems. The i in situ conservation is the most effective way of protecting the species and improving the quality of the habitat they live in. The in situ approach is preferable because of the fact that not much diversity can be conserved outside the centres of diversity.

Biodiversity at all its levels can be conserved in situ by comprehensive system of protected areas such as the national parks, wildlife sanctuaries, natural reserves, natural monuments, cultural landscapes, biosphere reserves, wetlands, etc. So far, in situ practice is considered most effective method of protecting and propagating the species and improving the quality of their habitats.

2. Ex situ (Off-site) Conservation:
It refers to conservation of biological diversity outside the boundaries of their natural habitats by perpetuating sample population in genetic resource centres, e.g. zoos, botanical gardens, culture collections, etc., or in the form of gene pools and gametes storage for fish, germplasm banks for seeds, pollen, semen, ova, cell, etc. Zoos also help in captive breeding of organisms which are endangered, whereas botanical gardens have seed gene banks, tissue culture labs and other technologies for storing and growing germplasm.

Biodiversity Preservation Methods and Sites

A protected area, as defined by IUCN, is an area (either land or sea) especially dedicated for the protection and maintenance of biological diversity through legal and ‘ other effective ways. IUCN has classified protected area into six different types. Some of these protected areas are, discussed below

National Parks:
India’s first National Park (IUCN category-II protected area) was Hailey National Park, now known as Jim Corbett National Park, established in 1935. According to National Wildlife Database, there were 103 National parks in India as in April, 2015. A national park is an area maintained by government and dedicated to conserve the environment, natural and historical objects and the wildlife therein. Operations such as plantation, cultivation, grazing forestry are not allowed in national parks. Private ownership rights and habitat manipulation are also prohibited. IUCN (1975) has adopted following keypoints to define a national park.

1. A national park is a relatively large area reserved for the betterment of the wildlife. The habitats of native plant and animals becomes the site of scientific, educational and recreative interests along with maintenance of its aesthetic values.
   Since, human intervention is nil or limited, the operating ecosystem remains unaltered and conserved.
2. It is also defined as an area where the highest authority take measures to prevent exploitation and enforce conservation measures.
3. A place where visitors are permitted to enter only in special conditions like inspirational, cultural and recreative purposes.
   

Hotspots of Biodiversity:
The concept of hotspot was given by Norman Myers in 1988. Hotspots are areas that are extremely rich in species diversity, have high endemism and are under constant threat. There are 34 hotspots which cover less than 2% of earth land area of the world. In these sites nearly 75% population of world’s most threatened mammals, birds and amphibians, approximately 50% plants and 42% land vertebrates are conserved (Conservation International, 2003).

According to Myers, the key criteria for an area to be assigned as biodiversity hotspot are

• It must contain atleast 0.5% or 1,500 of the world’s 3,00,000 species of vascular plants as endemics.
• It should have lost atleast 70% of its primary vegetation.

Tropical forests appear in 15 hotspots, Mediterranean – type zones in 5, 9 hotspots are mainly or completely made up of islands and 16 hotspots are in the tropics. About 20 % of the human population lives in the hotspot regions.

Tropical Andes hotspot has 20,000 endemic plants and 1567 vertebrates and it is at the top of the list. Four regions of India that fulfills the criteria of hotspots are The Western Ghats and Sri Lanka, The Eastern Himalayas, Indo-Burma (North-Eastern India South of Brahmaputra river)and Sundarland (Nicobar Islands). These sites are also known as Gade of speciation.

The Western Ghats are a chain of hills that lies parallel to the Western coast of peninsular India. These regions have moist deciduous forest and rain forest and have high species diversity and high levels of endemism. Nearly 77% of the amphibians and 62% of the reptile species found here are found nowhere else. Over 6000 vascular plants of over 2500 genera are found in this hotspot, of which over 3000 are endemic.

Much of the world’s species like black pepper and cardamom have their origins in the Western Ghats. It also harbors over 450 bird species, about 140 mammalian species, 260 reptiles and 175 amphibians. Over 60% of the reptiles and amphibians are completely endemic to this hotspot.

The Eastern Himalayan hotspot has approximately 163 globally threatened species including the one-horned rhinoceros, the Wild Asian Water buffalo and in all 45 mammals, 50 birds, 17 reptiles, 12 amphibians, 3 invertebrate and 36 plant species.
Thus, hotspots are the most precious sites for biodiversity conservation and should be protected from exploitation.

Wildlife Sanctuary:
It is IUCN category IV of protected area. A wildlife sanctuary can be established by a gazette notification from the State Forest Department, where protection is provided to vulnerable, endangered and critically endangered wild animals life. Operations such as procuring timber and minor forest products and private ownership are allowed provided they do not cause any adverse effects on the animals. Till 2015, there were 520 wildlife sanctuaries in India, covering 122867.34 km2 (3.74%) of land in India. The state of Odisha has 19 wildlife sanctuaries, which are listed below

Biosphere Reserve:
These are large tracts of protected land used for preserving the genetic diversity of an ecosystem by preserving wildlife. The creation of biosphere reserves was initiated in 1975 under the Man and Biosphere (MAB) programme of UNESCO. It is category V protected area designated by IUCN. Today, India has 18 biosphere reserves, some of which are also included under National Parks and Odisha has a share of one in Simlipal.

Each biosphere reserve integrates human activities and has following zonation

1. Core zone Strictly protected.
2. Buffer zone Sustainable and recreation activities allowed.
3. Transition zone Anthropogenic activities like research and sustainable development allowed.
4. Zone of human encroachment Normal anthropogenic activities allowed.

Out of 18 biosphere reserves, 8 of them are a part of world network of Biosphere Reserves based on MAB Programme of UNESCO.

The objectives of this programme are

• Conserving representative samples of ecosystems.
• In situ conservation of genetic diversity.
• Facilitating basic and applied research in ecology and environmental biology on site.
• Create opportunities for environmental education k and training.
• Promoting and creating awareness about sustainable management of living resources.
• Promoting international cooperation.

Therefore, the biosphere reserves have three junctions, that includes

• Conservation of ecosystems and genetic variations.
• Promotion of sustainable economic and human development.
• They serve as examples of education and training local, regional, national and international issues of sustainable development.

A protected area to be declared as a biosphere reserve should have the following essential features

• Abundant genetic diversity should be present.
• It should be unique in itself.
• The area should be legally protected for long term.
• Appropriate size for effective maintenance of natural populations so that there is no genetic drift.
• Sufficient natural resource available for ecological research, education and training. It should be a natural home for the endangered species of plants and animals.

Question 3.
Give an account of biodiversity and its conservation measures.
Or
Give an account of the concept of biodiversity.
Answer:
The term ‘biodiversity’ was first used by Thomas E. Lovejoy (1980) to refer the number of species of a region. It is the degree of variation of life occurring at different levels like genetic, organismal and ecological. These levels forms a hierarchy of biodiversity. The integration of several sciences such as ecology and genetics to sustain biological diversity at all its levels is called conservation biology.

Levels of Biodiversity

In 1986, Norse and Me Manus explained the three levels of biodiversity. These include

1 Genetic Diversity:
It involves variations in genetic composition among the individuals of a species. This variations could be in the nucleotides, genes, entire genome or chromosomes. This type of diversity arises due to genetic recombination during sexual reproduction and mutation.

Variations in the genes of a species increases with the increase in size and environmental parameters of the habitat. Genetic diversity is useful as it helps an individual to adapt to changing environmental condition, natural selection and is essential for healthy breeding. It also helps in speciation or evolution of new species.

2. Organismal or Species Diversity (Diversity among Species):
It is the variety in the number and richness of a species of a region. Sometimes, a species remains confined to a particular area and is found only in that area. Such species are said to be endemic, e.g., Indian giant squirrel is endemic to Panchmarhi hills in Madhya Pradesh.

The IUCN (International Union for Conservation of Nature and Natural Resources) recognises three types of species diversity, i.e.

• Alpha (α) diversity It refers to the variety of species within a community. It is also referred to as species richness, i.e., the number of species per unit area.
• Beta (β) diversity It refers to the diversity of species among communities.
• Gamma (γ) diversity It refers to the diversity of species across a wide geographical range.

The important features of species diversity to the ecosystem are as follows

• Increased biodiversity provides resistance to the ecosystem against natural disasters.
• Ecosystem with more species shows more yields and greater productivity with variation of biomass.
• Community with more species generally tends to be more stable than those with less species.

3. Ecological Diversity (Diversity among Communities):
It explains about the variety of ecosystems present in the biosphere. The community composition, i.e., assemblage of several interacting populations in a given space at a particular time is affected directly by the environment. Thus, it is the diversity at the level of communities and ecosystems of a region.

Patterns of Biodiversity

Biodiversity is not uniform throughout the world. It varies with the changes in latitude and altitude. For many groups of animals and plants, there are specific patterns in diversity based on the favourable environmental conditions.
The pattern of biodiversity among different regions is discussed below

Latitudinal Gradients:
Species diversity decreases as we move away from the equator towards the poles. It means biodiversity is more at lower latitude (equator) than the higher latitude (poles). The biodiversity gradient is steep in Northern hemisphere than the Southern hemisphere.

Biodiversity in Tropics
Tropics (latitudinal range of 23.5°N to 23.5°S) harbour more species than temperate or polar areas, e.g., Colombia located near the equator, has 1,400 species of birds.

New York located at 41°N has 105 species of birds, while Greenland at 71°N has only 56 species of birds. India, with most of its area in tropical latitude has more than 1200 species of birds. A forest of equal area in tropical region (like equator) has 10 times more species of vascular plants than in temperate region (like Mid-West of USA). Amazonian rainforest in South America has the greatest biodiversity on the earth with more than 40,000 species of plants, 3,000 of fishes, 1,300 of birds, 427 of mammals and amphibians, 378 of reptiles and more than 1,25000 of invertebrates.

Reasons for Greater Biodiversity in Tropics
The factors making tropical rainforests rich in biodiversity are

1. Tropical latitudes have remained undisturbed for million of years allowing speciation.
2. Tropical environments are relatively constant throught the year which promotes niche specialisation and greater diversity.
3. High productivity leads to greater diversity. It is also found that species diversity increases with area (species area curves), how it peaks in areas with intermediate productivity or intermediate rates of disturbance. The more variable the habitat, the greater the species diversity within it. This pattern was offered as one of the reasons why there are more species in bigger area as more area covers a greater variety of habitat.

Importance of Biodiversity

Biodiversity is directly or indirectly involved in maintaining the well-being of human society. Thus, scientists have classified the values of diversity as ecosystem goods, i.e., natural products harvested from ecosystems and directly used by humans and ecosystem services, involving different ecosystem processes which indirectly benefit human life.

Direct Value

1. Food It includes all the plant and animal products used as food by humans, e.g., cereals, pulses, vegetables, fruits, milk, beverages, etc.
2. Clothing It includes natural fabric made out of cotton, jute and natural silk (harvested from silk moth).
3. Shelter It includes raw material obtained from ecosystem for making houses, e.g., wood, etc.
4. Medicines Large number of substances with therapeutic properties are obtained from variety of plant species and animals, e.g.
   1. Quinine Antimalarial drug, obtained from the bark of Cinchona plant.
   2. Anticoagulants Antihemorrhagic drugs, extracted from blood sucking animals.
   3. Snake venom and toxins Drugs for neural and muscular disorders.
   4. Penicillin, tetracyclins and streptomycins Antibiotics extracted from microorganisms.
   5. Biocidal compounds Used in manufacturing antibiotics, extracted from beetles, millipedes, snails and ants.
5. Industrial products A variety of industrial products are directly made out of biological resources, e.g. timber, fuel, dyes, oil, etc., from plants and leather, etc., from animals skin.

Indirect Value:

1. Biological control It is the use of microorganisms for the manufacture of antibiotics, oral contraceptives, etc., and management of pests, increasing soil fertility, cleaning oil spill (super bug), treatment of sewage and solid waste, recovery of metals (bioleaching), monitoring pollution generating biofuel, etc.

2. Environmental modulation. It includes some animals and plants which influence and modulate the environment directly and indirectly. Such animals and plants are known as ecosystem engineers. One of these are keystone species whose extinction reduces abundance of other species in the community.

A well known example of ecosystem engineer is beaver (a mammal). It influence plant and animal communities and the entire biodiversity of watershed area by
(a) Creating dams using logs in river channels
(b) Modifying nutrient cycling
(c) Influencing decomposition dynamics.

3. Ecosystem functions and services Biodiversity plays a major role in many ecosystem services and functions such as replenishing oxygen through photosynthesis, pollination through bees, regulation of global climate, retention of rainwater in aquifers and reservoir, control of floods, etc.
Nature always key a check on these activities to maintain a state of equilibrium (homeostasis), which further helps in the sustainable development of resources. However, overexploitation of biolgical resources by humans leads to destabilisation of ecosystem balance.
For example,

• Decreased flora of an area leads to CO₂ increase in the atmosphere which causes temperature elevation of that area.
• Increased carnivore population decreases the herbivore population by predation which then increase the vegetation.
• Declined population of microflora prevents the recycling between complex organic matter and simple inorganic matter.

4. Ecotourism The diverse biological resource of a country motivates people from around the world to undertake recreational activities like tours to enjoy the diverse wildlife and charismatic landscape. In return the host country earns a large sum of foreign exchange as revenue.

Loss of Biodiversity

The loss of biodiversity is a global crisis. Extinction of species is a natural phenomenon aided by the physical changes in the environment. However, the accelerated rates of species extinctions, that the world is facing now are largely due to human activities.

Till now, five episodes of mass extinction of species have occured in the history of biological evolution. The sixth episode of extinction of species however, is credited to human activities, which otherwise would not have occurred.

According to IUCN estimates, 12259 species have become extinct since the time of origin of life on the earth. The major cause of the biodiversity losses are called drivers, which belong to two classes namely, direct and indirect.

Direct Drivers Factors

The directly influence the ecosystem processes which bring about the mass extinction of species.
The various processes associated with direct drivers includes

1. Habitat destruction and fragmentation Conversion of forest land for agriculture, development projects, mining operation, etc., leads to the destruction of the natural habitats of the organisms.
Indiscriminate agricultural practices involving use of chemical fertilisers and pesticides are potent factors
for the destruction of habitats. When a large population fragments into smaller ones there is more inbreeding and inbreeding pressure leading to population decline.

2. Overexploitation of natural resources Humans are dependent on nature for food and shelter, but when ‘need’ turns to ‘greed’, it leads to overexploitation.
To meet the need for increased housing, the natural habitats of animals and plants are being destroyed. This results in habitat loss and extinction of species. It has caused extinction of many species in last 500 years. In addition, indiscriminate hunting of wild animals has made their status in Red book as endangered or critically endangered.

3. Introduction of alien invasive species When alien species are intentionally or unintentionally introduced in a particular area, they might turn invasive and cause decline or extinction of endogenous species, e.g., Eichhornia is known as the ‘Terror of Bengal’. It was introduced as ornamental plants but it became wild in India because of invasiveness.

Similarly, Lantana and Parthenium were important due to their ornamental and food values, respectively but, they become wide spread due to favourable environmental conditions.

4. Climate change The global climate is changing, due to the anthropogenic activities like greenhouse gases and it has led to global warming. This is causing melting of glaciers, polar ice caps, etc. This may submerge low lying coastal habitats and also plants and animals are unable to adapt themselves to this change which is causing their elimination.

5. Environmental pollution It is another major factor for species extinction. Pollution may reduce and eliminate populations of sensitive species. Environmental pollution is most commonly caused by accumulation of non-biodegradeble wastes like plastics. Agricultural chemicals like pesticides enters the food chain and get deposited in the body of higher organisms. This effects the population of fish eating birds and falcons by disturbing their reproductive process.

Indirect Drivers
These influences or changes one or more the direct driver. They include

1. Population growth Rapid increase in human population causes loss of biodiversity becatise population explosion results in rapid growth of exploitation of natural resources such as water, food and minerals. If this trend of population growth continue, the resources will be depleted faster and most species will face the risk of extinction.
2. Income and lifestyle Today people’s income has increased tremendously which has led to more luxurious lifestyle of people. For leading this, they tend to use more and more of natural resources which causes pollution, degradation of environment and biodiversity loss.

Extinction of Species

In the history of earth many species have disappeared and new ones got evolved over million of years. The major threat to biodiversity is extinction of species Extinction is the total elimination or dying out of species (fossilisation) form the earth. As, we already know, the extinction of species is a natural process which accelerates due to human activities.

There are generally three types of extinction
1. Natural extinction It is a slow process of replacement of existing species with the better adapted species due to alternate evolution, changes in environmental condition, predation and diseases. Extinction of species occurs due to combination of genetic and demographic factors.

2. Mass extinction It occur due to catastrophes, which struck the earth several times. A mass extinction occurred about 225 million years ago in Permian period when 90% of shallow marine invertebrates disappeared.
Another mass extinction occurred between cretaceous and tertiary period over 60 million years ago when dinosaurs and a number of other organisms disappeared.

3.  Anthropogenic extinction These are extinction of organisms due to human activities like hunting, overexploitation and habitat destruction, e.g. dodo (Raphus cucullatus), Tasmanian wolf, etc. Anthropogenic extinction is causing a sixth extinction of species. It is 100-1000 times more faster than the rate of natural extinctions.

IUCN and Red List
IUCN is International Union of Conservation of Nature and Natural Resources which is now called World Conservation Union (WCU). It has its headquarters at Morges, Switzerland. IUCN maintains a Red Data Book (RDB) or Red List which is a catalogue of taxa facing risk of extinction. Red Data Book was initiated in 1963.
The purpose of red list is to

(i) Provide awareness to the degree of threat to biodiversity.
(ii) Provide global index about already declined of biodiversity.
(iii) Identification and documentation of species at high risk of extinction.
(iv) Preparing conservation priorities and help in conservation plan.
(v) Information about international agreements like conservation on biological diversity and CITES (Convention on International Trade in Endangered Species) of Wild Fauna and Flora.

The IUCN Red List 2004 has recorded a total loss of 784 species in the last 500 years. These include 733 animals (mostly vertebrates and molluscs), 110 plants and one red alga. The extinction of dodo in Mauritius, quagga in Africa were notable extinctions in the recent years.

The species that became extinct in 2003 was the plant Nesiota elliptica, St. Helena Olive (a small tree in Saint Helena Island) in the South Atlantic Ocean. The IUCN red list of threatened species founded in 1964, is the worlds most comprehensive inventory of global conservation status of biological species.The IUCN Red List has listed 132 species of plants and animals from India as ‘Critically Endangered’.

Red List assign categories to each species. These are as follows

1. Extinct A taxon is extinct when there is no reasonable doubt that its last individual has died, e.g. dodo, Indian cheetah.
2. Extinct in the Wild (EW) A number of domesticated animals and plants have become extinct in the wild. A taxon is extinct in the wild when it is known to survive only under cultivation.
3. Critically Endangered (CR) A taxon is critically , endangered when it is facing an extremely high risk of extinction in the wild in immediate future (925 animals and 1014 plants), e.g, One horned rhinoceros.
4. Endangered (EN) A taxon is endangered when it is not critically endangered, but facing a very high risk of extinction in the wild in near future, e.g, Giant panda and polar bear.
5. Vulnerable (VU) A taxon is vulnerable when it is not critically endangered or endangered, but it is facing a high risk of extinction in the wild in the medium term future, e.g. sparrow.
6. Threatened Species is the one which is liable to become extinct if not allowed to realise its full biotic potential by providing protection from the exotic species, e.g. in black buck.
7. Low Risk (LR) A taxon is at low risk when evaluated, it does not qualify for any of the categories like critically endangered, endangered or vulnerable.
8. Data Deficient (DD) A taxon is data deficient when there is inadequate information to make a direct or indirect assessment of its risk of extinction based on its distribution or population status.
9. Not Evaluated (NE) A taxon is under the category of not evaluated, when it has not yet been assessed against above criteria.

Biodiversity is directly or indirectly involved in maintaining the well-being of human society. Thus, scientists have classified the values of diversity as ecosystem goods, i.e., natural products harvested from ecosystems and directly used by humans and ecosystem services, involving different ecosystem processes which indirectly benefit human life.

Biodiversity Conservation:

The protection, uplift and scientific management of biodiversity at its optimum level for present and future generations is known as biodiversity conservation.
The International Union for Conservation of Nature and Natural Resources (IUCN) or World Conservation Union, World Wide Fund for Nature (WWF), Food and Agricultural Organisation (FAO) and United Nations Educational Scientific and Cultural_Organisation (UNESCO) formed the world conservation strategy in 1980 for the conservation and sustainable use of biological resources. For this, two major types of conservation strategies were framed. These are

1. In situ (On-site) Conservation:
It involves the to protection of plants, animals and microorganisms within their natural ecosystems. The i in situ conservation is the most effective way of protecting the species and improving the quality of the habitat they live in. The in situ approach is preferable because of the fact that not much diversity can be conserved outside the centres of diversity.

Biodiversity at all its levels can be conserved in situ by comprehensive system of protected areas such as the national parks, wildlife sanctuaries, natural reserves, natural monuments, cultural landscapes, biosphere reserves, wetlands, etc. So far, in situ practice is considered most effective method of protecting and propagating the species and improving the quality of their habitats.

2. Ex situ (Off-site) Conservation:
It refers to conservation of biological diversity outside the boundaries of their natural habitats by perpetuating sample population in genetic resource centres, e.g. zoos, botanical gardens, culture collections, etc., or in the form of gene pools and gametes storage for fish, germplasm banks for seeds, pollen, semen, ova, cell, etc. Zoos also help in captive breeding of organisms which are endangered, whereas botanical gardens have seed gene banks, tissue culture labs and other technologies for storing and growing germplasm.

Biodiversity Preservation Methods and Sites

A protected area, as defined by IUCN, is an area (either land or sea) especially dedicated for the protection and maintenance of biological diversity through legal and ‘ other effective ways. IUCN has classified protected area into six different types. Some of these protected areas are, discussed below

National Parks:
India’s first National Park (IUCN category-II protected area) was Hailey National Park, now known as Jim Corbett National Park, established in 1935. According to National Wildlife Database, there were 103 National parks in India as in April, 2015. A national park is an area maintained by government and dedicated to conserve the environment, natural and historical objects and the wildlife therein. Operations such as plantation, cultivation, grazing forestry are not allowed in national parks. Private ownership rights and habitat manipulation are also prohibited. IUCN (1975) has adopted following keypoints to define a national park.

1. A national park is a relatively large area reserved for the betterment of the wildlife. The habitats of native plant and animals becomes the site of scientific, educational and recreative interests along with maintenance of its aesthetic values.
   Since, human intervention is nil or limited, the operating ecosystem remains unaltered and conserved.
2. It is also defined as an area where the highest authority take measures to prevent exploitation and enforce conservation measures.
3. A place where visitors are permitted to enter only in special conditions like inspirational, cultural and recreative purposes.
   

Hotspots of Biodiversity:
The concept of hotspot was given by Norman Myers in 1988. Hotspots are areas that are extremely rich in species diversity, have high endemism and are under constant threat. There are 34 hotspots which cover less than 2% of earth land area of the world. In these sites nearly 75% population of world’s most threatened mammals, birds and amphibians, approximately 50% plants and 42% land vertebrates are conserved (Conservation International, 2003).

According to Myers, the key criteria for an area to be assigned as biodiversity hotspot are

• It must contain atleast 0.5% or 1,500 of the world’s 3,00,000 species of vascular plants as endemics.
• It should have lost atleast 70% of its primary vegetation.

Tropical forests appear in 15 hotspots, Mediterranean – type zones in 5, 9 hotspots are mainly or completely made up of islands and 16 hotspots are in the tropics. About 20 % of the human population lives in the hotspot regions.

Tropical Andes hotspot has 20,000 endemic plants and 1567 vertebrates and it is at the top of the list. Four regions of India that fulfills the criteria of hotspots are The Western Ghats and Sri Lanka, The Eastern Himalayas, Indo-Burma (North-Eastern India South of Brahmaputra river)and Sundarland (Nicobar Islands). These sites are also known as Gade of speciation.

The Western Ghats are a chain of hills that lies parallel to the Western coast of peninsular India. These regions have moist deciduous forest and rain forest and have high species diversity and high levels of endemism. Nearly 77% of the amphibians and 62% of the reptile species found here are found nowhere else. Over 6000 vascular plants of over 2500 genera are found in this hotspot, of which over 3000 are endemic.

Much of the world’s species like black pepper and cardamom have their origins in the Western Ghats. It also harbors over 450 bird species, about 140 mammalian species, 260 reptiles and 175 amphibians. Over 60% of the reptiles and amphibians are completely endemic to this hotspot.

The Eastern Himalayan hotspot has approximately 163 globally threatened species including the one-horned rhinoceros, the Wild Asian Water buffalo and in all 45 mammals, 50 birds, 17 reptiles, 12 amphibians, 3 invertebrate and 36 plant species.
Thus, hotspots are the most precious sites for biodiversity conservation and should be protected from exploitation.

Wildlife Sanctuary:
It is IUCN category IV of protected area. A wildlife sanctuary can be established by a gazette notification from the State Forest Department, where protection is provided to vulnerable, endangered and critically endangered wild animals life. Operations such as procuring timber and minor forest products and private ownership are allowed provided they do not cause any adverse effects on the animals. Till 2015, there were 520 wildlife sanctuaries in India, covering 122867.34 km2 (3.74%) of land in India. The state of Odisha has 19 wildlife sanctuaries, which are listed below

Biosphere Reserve:
These are large tracts of protected land used for preserving the genetic diversity of an ecosystem by preserving wildlife. The creation of biosphere reserves was initiated in 1975 under the Man and Biosphere (MAB) programme of UNESCO. It is category V protected area designated by IUCN. Today, India has 18 biosphere reserves, some of which are also included under National Parks and Odisha has a share of one in Simlipal.

Each biosphere reserve integrates human activities and has following zonation

1. Core zone Strictly protected.
2. Buffer zone Sustainable and recreation activities allowed.
3. Transition zone Anthropogenic activities like research and sustainable development allowed.
4. Zone of human encroachment Normal anthropogenic activities allowed.

Out of 18 biosphere reserves, 8 of them are a part of world network of Biosphere Reserves based on MAB Programme of UNESCO.

The objectives of this programme are

• Conserving representative samples of ecosystems.
• In situ conservation of genetic diversity.
• Facilitating basic and applied research in ecology and environmental biology on site.
• Create opportunities for environmental education k and training.
• Promoting and creating awareness about sustainable management of living resources.
• Promoting international cooperation.

Therefore, the biosphere reserves have three junctions, that includes

• Conservation of ecosystems and genetic variations.
• Promotion of sustainable economic and human development.
• They serve as examples of education and training local, regional, national and international issues of sustainable development.

A protected area to be declared as a biosphere reserve should have the following essential features

• Abundant genetic diversity should be present.
• It should be unique in itself.
• The area should be legally protected for long term.
• Appropriate size for effective maintenance of natural populations so that there is no genetic drift.
• Sufficient natural resource available for ecological research, education and training. It should be a natural home for the endangered species of plants and animals.

Odisha State Board CHSE Odisha Class 12 Biology Solutions Chapter 5 Heredity and Variation Textbook Questions and Answers.

CHSE Odisha 12th Class Biology Chapter 4 Question Answer Heredity and Variation

Heredity and Variation Class 12 Questions and Answers CHSE Odisha

Very Short Answer Type Questions

Multiple choice questions

Question 1.
The experimental plant material used by Mendel was
(a) cowpea
(b) garden pea
(c) wild pea
(d) sweet pea
Answer:
(b) garden pea

Question 2.
Which of the following characters is not among the seven characters considered by Mendel for his hybridisation experiments?
(a) Seed colour
(b) Pod shape
(c) Flower position
(d) Flower shape
Answer:
(d) Flower shape

Question 3.
Which law Mendel would not have proposed, if the phenomenon of linkage was known to him?
(a) Law of unit character
(b) Law of dominance
(c) Law of segregation
(d) Law of independent assortment
Answer:
(d) Law of independent assortment

Question 4.
The number of genotypes produced in F2-generation in Mendel’s monohybrid cross was
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 5.
In which of the crosses, half of the offsprings show dominant phenotype?
(a) Tt × Tt
(b) TT × tt
(c) Tt × tt
(d) TT × TT
Answer:
(c) Tt × tt

Question 6.
Two allelic genes are located on the
(a) same chromosome
(b) two homologous chromosomes
(c) two non-homologous chromosomes
(d) any two different chromosomes
Answer:
(b) two homologous chromosomes

Question 7.
Red (RR) Antirrhinum is crossed with white (rr) one. The F₁-hybrid is pink. This is an example of
(a) complete dominance
(b) codominance
(c) incomplete dominance
(d) complete recessive
Answer:
(c) incomplete dominance

Question 8.
In a dihybrid cross in F₂-generation, the parental types are far greater in number than the recombinants. This is due to
(a) linkage
(b) incomplete dominance
(c) multiple allelism
(d) complete dominance
Answer:
(a) linkage

Express in one or two word(s)

1. A pair of Mendelian factors (genes) that appear at a particular location on a particular chromosome and control the same characteristic.
Answer:
Alleles

2. Phenomenon where in the heterozygous condition an intermediate phenotype is observed.
Answer:
Incomplete dominance

3. The phenomenon of a single gene contributing to multiple phenotypic traits.
Answer:
Pleiotropy .

4. Genes which move together and do not show independent assortment.
Answer:
Linked gene

5. A cross between the F₁-hybrids with any one of the homozygous parents.
Answer:
Back cross

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

1. The process of transmission of characters through generations is known as variation.
Answer:
inheritance

2. In Mendel’s monohybrid cross, the dwarf phenotype is always homozygous.
Answer:
Correct statement

3. In Mendel’s dihybrid cross in F₂-generation, nine phenotypes are produced.
Answer:
four

4. The phenomenon of linkage disproved the principle of independent assortment.
Answer:
Correct statement

5. In a test cross, always dominant parent is used.
Answer:
recessive

6. The distance between genes in a constructed gene map is expressed as Mendel unit.
Answer:
Morgan

Fill in the blanks

1. Monohybrid cross in Regeneration yields ____ number of phenotypes.
Answer:
two

2. Monohybrid cross in Regeneration yields ____ number of genotypes.
Answer:
three

3. The name of scientist often coined with linkage is ____ .
Answer:
TH Morgan

4. Genotype of a plant showing the dominant phenotype can be determined by ____ cross.
Answer:
test

5. In a cross between AaBB and aaBB, the genotypic ratio in Ft-generation will be ____ .
Answer:
1 : 1

Short Answer Type Questions

Write notes on the following

Question 1.
Law of independent assortment
Answer:
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 of9 + 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 2.
Multiple alleles
Answer:
Multiple allelism and Inheritance of Blood Groups:
Each gene has alternative forms or allelomorphs. For example, the genes for rail and dwarf characters of pea plant arc ailcics or allelomorphs. Here, former is called normal or wild type and Iatcr as mutant type.
Sometimes, there may no be any aiternative 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.

Question 3.
Chromosomal basis of inheritance
Answer:
It was proposed independently by Walter Sutton and Theodore Boyen in 1902. They united the knowledge of chromosomal segregation with Mendelian principles and called it chromosornal theory of inheritance.
According to this theory

• All hereditary characters must be with sperms and egg cells as they provide bridge from one generation to the other.
• The hereditary factors must be carried by the nuclear material.
• Chromosomes are also found in pairs like the Mendelian alleles.
• The two alleles of a gene pair are located on homologous sites on the homologous chromosomes.

Question 4.
Codominance
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.

Question 5.
Incomplete dominance
Answer:
Incomplete Dominance:
It is a phenomenon in which phenotype of the F₁-hybrid offsprings does not resemble any of the parent, but is an intermediate between the expression of two alleles in their homozygous state. Carl Correns was the one who reported incomplete dominance in plant Mirabilis jalapa. He showed the petal colour inheritance in this plant. Here, the phenotypic ratio deviates from Mendel’s monohybrid ratio but the parental characters reappear in F₂-generation.

Question 6.
Law of segregation
Answer:
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 7.
Linkage
Answer:
The genes of a particular chromosome show the tendency to inherit together. This phenomenon of genic inheritance in which genes of a particular chromosome show their tendency to inherit together, i.e tendency to retain their parental combination even in the offsprings is known as linkage.

Question 8.
Recombination
Answer:
They attributed this due to physical association of the two genes and coined the term ‘linkage’ to describe this physical association of genes on a chromosome and the term ‘recombination’ to describe the generation of non-parental gene combinations. Morgan performed a test cross by crossing heterozygous grey-bodied and long-winged with homozygous recessive black-bodied and vestigial-winged fly

Question 9.
Test cross
Answer:
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.

Question 10.
Back cross
Answer:
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.

Differentiate between the following

Question 1.
Homozygous and Heterozygous
Answer:
Differences between homozygous and heterozygous are as follows

Homozygous	Heterozygous
It is a condition when both alleles of a gene are similar.	It is a condition when both alleles of a gene are dissimilar.
The genotype is expressed as TT or tt.	The genotype expressed as Tt.
They are true breeding to purelines.	They are not true breeding.
The gametes produced by them are similar in genotype.	The gametes produced by them are of two types, one with dominant allele and other with recessive allele.

Question 2.
Genotype and Phenotype.
Differences between phenotype and genotype are as follows
Answer:

Phenotype	Genotype
It refers to observable traits or characters.	It refers to the genetic constitution of an individual.
It results from expression of genes.	It constitutes single gene pair or sum total of all the genes.
The phenotypic ratio of Mendel’s monohybrid cross is 3 : 1.	The genotypic ratio of Mendel’s monohybrid cross is 1:2:1.
It may change with age and environment.	It remains the same throughout the life of an individual.

Question 3.
Dominant genes and Recessive genes.
Answer:
Differences between dominant genes and recessive genes are as follows

Dominant genes	Recessive genes
When an allele expresses itself in the presence of its recessive allele, it is called dominant trait.	It can only express in the absence of its dominant allele and remain masked in its presence.
Dominant allele forms a complete functional enzyme due to which complete polyeptide is formed to express.	Recessive allele forms incomplete polypeptide enzyme due to which non-functionai polypetide is formed and fails to express completely.

Question 4.
Back cross and Test cross.
Answer:
Differences between back cross and test cross are as follows

Back cross	Test cross
It is a cross involving F1-progeny and either of the parents.	It is a cross involving
It is used by scientists to improve a breed or variety of plant or animal.	F1-individual and its recessive parent.

Question 5.
Qualitative inheritance and Quantitative inheritance.
Answer:
Differences between qualitative inheritance and quantitative inheritance are as follows

Qualitative inheritance	Quantitative inheritance
It deals with the inheritance of qualitative characters.	It deals with the inheritance of quantitative characters.
Each character is controlled by one pair of contrasting alleles.	Each character is controlled by more than one pair of non-allelic genes (Polygenes).
Each character has two distinct expressions, i.e. exhibits two distinct phenotypes.	Each character has an intergrading range of phenotypes.
The degree of expression remains the same whether the character is controlled by one or both the dominant genes.	The degree of expression depends on the number of the dominant genes.
Phenotypic expression is not affected by the environment.	Phenotypic expression is influenced by environmental factors.
Monogenic inheritance exhibits discontinuous pattern of inheritance.	Polygenic inheritance represents continuous pattern of inheritance.
F1-individuals resemble the dominant parent.	F1-individuals exhibit intermediate expression between the two parents.
F2-individuals exhibit 3:1 ratio. Intermediate expressions are not found.	In F2-generation, individuals with intermediate genotype and phenotype are maximum.
Examples of monogenic or qualitative inheritance are yellow or green coat color or round or wrinkled seed character in pea seeds.	Examples of polygenic or quantitative inheritance are height, weight, intelligence and skin color in human beings, milk yield in cattle and egg production in poultry.

Long Answer Type Questions

Question 1.
Give an account of Mendel’s monohybrid cross. What inference did Mendel draw from this experiment?
Answer:
Monohybrid Cross
The study of inheritance of a single pair of alleles or factors of a trait at a time (monohybrid cross) is called one gene inheritance. When a cross is made between pure tall and pure dwarf plant (for purity, the pureline is taken into consideration) in F₁-generation, all plants will be tall.

When F₁ -plants are self-pollinated, then in F₂-generation both tall and dwarf plants are found in approximate ratio of 3 : 1.

The dwarf plants of F₂ on self-pollination, produce dwarf plants generation after generation, while among tall plants, only 1 /3rd show this character generation after generation (pureline) and rest 2/3rd produce tall and dwarf in 3 : 1 ratio again (F₃ -generation).

Explanation Mendel’s monohybrid cross explained that in each main pair of alternative character one is expressed and other is masked.
The character which is expressed in F₁-generation is called dominant and the one which is not expressed is called recessive.

The monohybrid cross between tail and dwarf

In F₂-generation, the genotypic ratio is 1 : 2 : 1 and phenotypic ratio is 3 : 1. Mendel came to the conclusion that progeny possessing similar factors is called homozygous and the one which is hybrid is called heterozygous.

Mendel used english letters to record his observations of breeding experiments. He assigned capital letters for dominant characters and small letters for recessive characters which tabulated in the given below table

Characters	Dominant	Recessive
Seed shape	Round (R)	Wrinkled (r)
Seed colour	Yellow (Y)	Green (y)
Pod shape	Full (F)	Constricted (f)
Pod colour	Green (G)	Yellow (g)
Flower/Pod position	Axial (A)	Terminal (a)
Seed coat colour/Flower colour	Red/Violet (R/V)	White (r/v)
Plant height	Tall (T)	Dwarf (t)

Based on his observations on monohybrid crosses, Mendel proposed two general rules in order to consolidate his understanding of inheritance in monohybrid crosses.

Based on the Mendel’s observations, the German scientist Carl Correns formulated certain principles of heredity. These now known as Mendel’s laws of inheritance or the principles or laws of inheritance.
These are
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 2.
State and explain Mendel’s laws of inheritance.
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 allele or the hybrids have Tt.

8. Since, these hybrids contain alleles which express contrasting traits, the plants are heterozygous.

Question 3.
What do you mean by back cross and test cross? Explain test cross through an example.
Answer:
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 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.

Question 4.
Describe Mendel’s dihybrid cross.
Answer:
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 F₁-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

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 linkage and recombination.
Answer:
Linkage, Crossing Over and Recombination:
According to ‘chromosomal theory of inheritance’, the chromosomes are vehicles of inheritance. Hence, the number of genes per individual for exceed the number of chromosomal pairs, i.e. each chromosome bears many genes. These genes are arranged in linear fashion over the chromosome and cannot show independent assortment.

In other words, we can say that the genes of a particular chromosome show the tendency to inherit together. This phenomenon of genic inheritance in which genes of a particular chromosome show their tendency to inherit together, i.e tendency to retain their parental combination even in the offsprings is known as linkage.

Morgan and his group observed in Drosophila that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type.

They attributed this due to physical association of the two genes and coined the term ‘linkage’ to describe this physical association of genes on a chromosome and the term ‘recombination’ to describe the generation of non-parental gene combinations. Morgan performed a test cross by crossing heterozygous grey-bodied and long-winged with homozygous recessive black-bodied and vestigial-winged fly. They obtained the following results

Phenotype	Per cent of occurrence
Grey body long wing	41.5
Black body vestigial wing	41.5
Grey body vestigial wing	8.5
Black body long wing	8.5

This result was not in accordance with Mendel’s law of inheritance. Now suppose in order to explain, we assume the alphabets G and g for grey and black body colours and L and 1 for long and vestigial wings, respectively.

Thus, linkage is a phenomenon of genic inheritance in which genes of a particular chromosome show their tendency to inherit together.
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 figure


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.

Linkage Groups:
All the genes linked together in a single chromosome constitute a linkage group. The number of linkage group in an organism is equal to their haploid number of chromosomes. This hypothesis was proved by TH Morgan by his experiments on Drosophila.

Morgan and his group hybridised yellow-bodied and white-eyed females with brown-bodied and red-eyed males (wild type) and intercrossed their F₁-progeny (cross A). It was observed that the two genes did not segregate independently of each other and the F₂-ratios deviated significantly from 9:3:3 :1 ratio.

In F₂-generation, parental combinations were 98.7% and the recombinants were 1.3%. In another cross (cross-B), between white-bodied female fly with miniature wing and a male fly with yellow body and normal wing, parental combinations were 62.8% and recombinants were 37.2% in F₂-generation. Thus, it was proved from the crosses that the linkage between genes for yellow body and white eyes is stronger than the linkage between the white body and miniature wing.

Chromosome Maps or Linkage Maps:
Alfred Sturtevant (Morgan’s student) used the frequency of recombination between gene pairs on the same chromosome as a measure of the .distance between genes and ‘mapped’ their position on the chromosome. Genetic maps are now used as a starting point in the sequencing of whole genomes as done in case of human genome sequencing project.

The frequency of recombination Cross Over Value (COV) is calculated by using the formula

A linkage or genetic chromosome map is a linear graphic representation of the sequence and relative distances of the various genes present in a chromosome. 1% crossing over between two linked genes is known as 1 map unit or Morgan (after TH Morgan, who is considered as ‘Father of Experimental Genetics’).