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NCERT Exemplar Class 12 Biology Chapter 8 Human Health and Diseases are part of NCERT Exemplar Class 12 Biology. Here we have given NCERT Exemplar Class 12 Biology Chapter 8 Human Health and Diseases. https://www.cbselabs.com/ncert-exemplar-problems-class-12-biology-human-health-diseases/

NCERT Exemplar Class 12 Biology Chapter 8 Human Health and Diseases

Multiple Choice Questions
Single Correct Answer Type

1. The term ‘Health’ is defined in many ways. The most accurate definition of the health would be
(a) Health is the state of body and mind in a balanced condition
(b) Health is the reflection of a smiling face
(c) Health is a state of complete physical, mental and social well-being
(d) Health is the symbol of economic prosperity.
Answer. (c) Health is not just the absence of disease. It is a state of complete physical, mental, social and psychological well-being.

2. The organisms which cause diseases in plants and animals are called
(a) Pathogens (b) Vectors (c) Insects (d) Worms
Answer. (a) A wide range of organisms belonging to bacteria, viruses, fungi protozoans, helminthes, etc., could cause diseases in man. Such disease causing organisms are called pathogens.

3. The chemical test that is used for diagnosis of typhoid is
(a) ELISA-Test (b) ESR-Test
(c) PCR-Test (d) Widal-Test
Answer. (d) The chemical test that is used for diagnosis of typhoid is Widal-Test.

4. Diseases are broadly grouped into infectious and non-infectious diseases. In the list given below, identify the infectious diseases.
i. Cancer ii. Influenza
iii. Allergy iv. Small pox
(a) i and ii (b) ii and iii –
(c) iii and iv (d) ii and iv
Answer.

5. The sporozoites that cause infection, when a female Anopheles mosquito bites a person being are formed in
(a) Liver of person (b) RBCs of mosquito .
(c) Salivary glands of mosquito (d) Intestine of mosquito
Answer. (d) The sporozoites that cause infection, when a female Anopheles mosquito bites a person being are formed in intestine of mosquito.

6. The disease chikunguniya is transmitted by
(a) House flies (b) Aedes mosquitoes
(c) Cockroach (d) Female Anopheles
Answer. (b) Dengue and chikunguniya are transmitted by Aedes mosquitoes.

7. Many diseases can’be diagnosed by observing the symptoms in the patient. Which group of symptoms are indicative of pneumonia?
(a) Difficulty in respiration, fever, chills, cough, headache
(b) Constipation, abdominal pain, cramps, blood clots
(c) Nasal congestion and discharge, cough, sorethroat, headache
(d) High fever, weakness, stomach pain, loss of appetite and constipation.
Answer. (a)

• Difficulty in respiration, fever, chills, cough, headache: Pneumonia
• Constipation, abdominal pain, cramps, blood clots: Amoebiasis
• Nasal congestion and discharge, cough, sorethroat, headache: Common cold
• High fever, weakness, stomach pain, loss of appetite and constipation: Typhoid

8. The genes causing cancer are
(a) Structural genes (b) Expressor genes
(c) Oncogenes ’ (d) Regulatory genes
Answer. (c) Oncogenes are the cancer causing genes.

9. In malignant tumors, the cells proliferate, grow rapidly and move to other . parts of the body to form new tumors. This stage of disease is called
(a) Metagenesis (b) Metastasis
(c) Teratogenesis (d) Mitosis
Answer. (b) In malignant tumors, the cells proliferate, grow rapidly and move to other parts of the body to form new tumors. This stage of disease is called metastasis.

10. When an apparently healthy person is diagnosed as unhealthy by a psychiatrist, the reason could be that
(a) The patient was not efficient at his work
(b) The patient was not economically prosperous
(c) The patient shows behavioural and social maladjustment
(d) He does not take interest in sports
Answer. (c) When an apparently healthy person- is diagnosed as unhealthy by a psychiatrist, the reason could be that the patient shows behavioural and social maladjustment.

11. Which of the following are the reason(s) for Rheumatoid arthritis? Choose the correct option.
i. The ability to differentiate pathogens or foreign molecules from self cells increases
ii. Body attacks self cells
iii. More afitibodies are produced in the body
iv. The ability to differentiate pathogens or foreign molecules from self cells is lost
(a) i and ii (b) ii and iv (c) iii and iv (d) i and iii
Answer. (b) Rheumatoid arthritis is an autoimmune diseases in which

• Body attacks self cells
• The ability to differentiate pathogens or foreign molecules from self cells is lost

12. AIDS is caused by HIV. Among the following, which one is not a mode of transmission of HIV?
(a) Transfusion of contaminated blood
(b) Sharing the infected needles
(c) Shaking hands with infected persons
(d) Sexual contact with infected persons
Answer. (c) Mode of transmission of HIV:

• Transfusion of contaminated blood
• Sharing the infected needles
• Sexual contact with infected persons

13. ‘Smack’ is a drug obtained fipm the
(a) Latex of Papaver somniferum (b) Leaves of Cannabis sativa
(c) Flowers of Datura (d) Fruits of Erythroxylum coca
Answer. (a)

• ‘ Smack’ is a drug obtained from the latex of Papaver somniferum.
• Smack are brown sugar, the common name of Heroin.
• Heroin is obtained by acetylation of morphine.

14. The substance produced by a cell in viral infection that can protect other cells from further infection is N
(a) Serotonin (b) Colostrum
(c) Interferon (d) Histamine
Answer. (c) The substance produced by a cell in viral infection that can protect other cells from further infection is interferon.

15. Transplantation-of tissues/organs to save certain patients often fails due to rejection of such tissues/organs by the patient. Which type of immune response is responsible for such rejections?
(a) Auto-immune response (b) Humoral immune response
(c) Physiological immune response (d) Cell-mediated,immune response
Answer. (d) Transplantation of tissues/organs to-save certain patients often fails due to rejection of such tissues/organs by the patient. Cell-mediated immune response is responsible for such rejections.

16. Antibodies present in colostrum which protect the new bom from certain diseases is of
(a) Ig G type (b) Ig A type (c) Ig D type (d) Ig E type
Answer. (b) Antibodies present in colostrum which protect the new bom from certain diseases is of Ig A type.

17. Tobacco consumption is known to stimulate secretion of adrenaline and nor¬adrenaline. The component causing this could be
(a) Nicotine (b) Tannic acid (c) Curaimin (d) Catechin
Answer. (a) Tobacco consumption is known to stimulate secretion of adrenaline and nor-adrenaline. The component causing this could be nicotine. Morphine, cocaine, codeine and nicotine are all alkaloids.

18. Anti-venom against’snake poison contains
(a) Antigens (b) Antigen-antibody complexes
(c) Antibodies (d) Enzymes
Answer. (c) Anti-venom against snake poison contains antibodies.

19. Which of the following is not a lymphoid tissue?
(a) Spleen (b) Tonsils (c) Pancreas (d) Thymus
Answer. (c) Spleen, Tonsils and Thymus are lymphoid tissue while pancreas is mixed gland.

20. Which of the following glands is large sized at birth but reduces in size with ageing?
(a) Pineal (b) Pituitary (c) Thymus (d) Thyroid
Answer. (c) Thymus glands is large sized at birth but reduces in size with ageing.

21. Haemozoin is a
(a) Precursor of hemoglobin
(b) Toxin released from Streptococcus infected cells
(c) Toxin released from Plasmodium infected cells
(d) Toxin released from Haemophilus infected cells ’
Answer. (c) Haemozoin is a toxin released from Plasmodium infected cells.

22. One Of the following is not the causal organism for ringworm.
(a) Microsporum (b) Trichophyton
(c) Epidermophyton (d) Macrosporum
Answer. (d) Microsporum, Trichophyton and Epidermophytonare the causal organism for ringworm while Macrosporum is a brown alga.

23. A person with sickle cell anemia is
(a) More prone to malaria (b) More prone to typhoid
(c) Less prone to malaria (d) Less prone to typhoid.
Answer. (c) A person with sickle cell anemia is less prone to malaria.

Very Short Answer Type Questions
1. Certain pathogens are tissue/organ specific. Justify the statement with suitable examples.
Answer. Salmonella typhi causes typhoid and infects small intestine while intestinal endoparasite causes amoebic dysentery and infects large intestine.

2.The immune system of a person is suppressed. In the ELISA test, he was found positive to a pathogen.
a. Name the disease the patient is suffering from.
b. What is the causative organism?
c. Which cells of body are affected by the pathogen?
Answer. a. Acquired Immuno Deficiency Syndrome (AIDS)
b. Human Immuno deficiency virus (HIV)
c. Helper T-lymphocytes (T_H or T₄)

3. Where are B-cells and T-cells formed? How do they differ from each other?
Answer. Both B-cells and T-cells are formed in bone marrow. B-cells matures in bone marrow while T-cells matures in thymus. B-cells provides humoral immunity and T-cells provides cell mediated immunity (CMI).

4. Given below are the pairs of pathogens and the diseases caused by them. Which out of these is not a matching pair and why?
(a) Virus – common cold (b) Salmonella – typhoid
(c) Microsporum – filariasis (d) Plasmodium – malaria
Answer. Pair is mismatched. Microsporum causes ringworm disease.

5. What would happen to immune system, if thymus gland is removed from the body of a person?
Answer. Thymus is the primary lymphoid organ. In thymus gland, immature lymphocytes differentiate into antigen-sensitive lymphocytes. If thymus gland is removed from the body of a person, his immune system becomes weak. As a result the person’s body becomes prone to infectious diseases,

6. Many microbial pathogens enter the gut of humans along with food. What are the preventive barriers to protect the body from such pathogens? What type of immunity do you observe in this case?
Answer. (i) The mucus coating of the epithelium lining of the gut helps in trapping microbes entering the body.
(ii) Saliva in the mouth and hydrochloric acid in gastric juice secreted by stomach prevent microbial growth. This type of immunity is innate immunity.

7. Why is mother’s milk considered the most appropriate food for a new born infant?
Answer. Mother’s milk is considered as the most appropriate food for a new born infant because the yellowish fluid colostrum secreted by mother during the initial days of lactation has abundant antibodies (IgA) to protect the infant.

8. What are interferons? How do interferons check infection of new cells?
Answer. Interferons are natural proteins produced by the cells of immune system in
response to foreign agents such as viruses, tumor cells and parasites and protect non-infected cells from further infection. Interferons inhibit the viral replication within host cells, activate natural killer cells and macrophages, increases antigen presentation to lymphocytes, and induce the resistance of host cells to viral infection. When the antigen is presented to matching T-cells’ and B-cells, these cells multiply and remove the foreign substance.

9. In the figure, structure of an antibody molecule is shown. Name the parts A, B and C. Show A, B and C in the diagram.

Answer. A:—Constant region of heavy chain,
B—Constant region of light chain,
C—Variable region of light and heavy chain

10. If a regular dose of drug or alcohol is not provided to an addicted person, he shows some withdrawal symptoms. List any four such withdrawal symptoms.
Answer. The withdrawal symptoms are:
a. Anxiety b. Shakiness c. Nausea d. Sweating

11. Why is it that during changing weather, one is advised to avoid closed, crowded and air-conditioned places like cinema halls etc.?
Answer. During changing weather, one is advised to avoid closed, crowded and air- conditioned placed like cinema halls, etc., because during this period the infectious agents are more humerous and prevalent to which we are more vulnerable.

12. The harmful allele of sickle cell anemia has not been eliminated from human population. Such afflicted people derive some other benefits. Discuss.
Answer. The harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population because SCA is a harmful condition which is also a potential saviour from malaria.
Those with the benign sickle trait possess a resistance to malarial infection. The pathogen that causes the disease spends part of its cycle in the red blood cells and triggers an abnormal drop in oxygen levels in the cell. In carriers, this drop is sufficient to trigger the full sickle-cell reaction, which leads to infected cells being rapidly removed from circulation and strongly limiting the infection’s progress. These individuals have a great resistance to infection and have a greater chance of surviving outbreaks. This resistance to infection is the main reason the SCA allele and SCA disease still exist. It is found in greatest frequency in populations where malaria was and is still often a serious problem.

13. Lymph nodes are secondary lymphoid organs. Explain the role of lymph nodes in our immune response.
Answer. Lymph nodes are small solid structures located at different points along the lymphatic system. Lymph nodes trap the microorganisms or other antigens, which happen to get into the lymph and tissue fluid. Antigens trapped in the lymph nodes are responsible for the activation of lymphocytes present there and cause the immune response.

14. Why is an antibody molecule represented as H₂ L₂ ?
Answer. Each antibody molecule is made of the two heavy chains (H₂ ) and two light chains (L₂ ), hence represented as H₂ L₂ .

15. What does the term ‘memory’ of the Immune system mean?
Answer. When body encounters a pathogen for the first time produce a response called 1° response. 1° response is of low intensity. When body encounters the same pathogen subsequently then body elicits 2° response. 2° response is highly intensified. This is due to the fact that our body have memory of the first encounter.

16. If a patient is advised Anti Retroviral Therapy, which infection is he suffering from? Name the causative organism.
Answer. The patient is suffering from AIDS. The causative organism for AIDS is HIV (Human Immuno deficiency Virus).

Short Answer Type Questions
1. Differentiate between active immunity and passive immunity.
Answer.

2. Differentiate between benign tumor and malignant tumor.
Answer.

• Benign tumors normally remain confined to their original location and do not spread to other parts of the body and cause little damage.
• Malignant tumors are a mass of proliferating cells called neoplastic or tumor cells. Neoplastic cells grow very rapidly, invading and damaging the surrounding normal tissues. As these cells actively divide and grow they also starve the normal cells by competing for vital nutrients.
• Cells sloughed from such tumors reach distant sites through blood, and wherever they get lodged in the body, they start a new tumor there. This property called metastasis is the most feared property of malignant tumors.

3. Do you consider passive smoking is more dangerous than active smoking? Why?
Answer. Yes, passive smoking is also dangerous as the active smoking because the person is exposed to the same harmful effects of smoking like emphysema, bronchitis, lung cancer, urinary bladder cancer or even peptic ulcer.

4. “Prevention is better than cure”. Comment.
Answer. Prevention is better than cure is true as in same cases the disease is non- curable like AIDS and Hepatitis-B, and in some cases’ the treatment causes financial problems in the family.

5. Explain any three preventive measures to control microbial infections.
Answer. (i) Maintenance of personal and public hygiene is very important for
prevention and control of many infectious diseases. Measures for personal hygiene include keeping the body clean; consumption of clean drinking water, food, vegetables, fruits, etc. Public hygiene includes proper disposal of waste and excreta; periodic cleaning and disinfection of water reservoirs, pools, cesspools and tanks and observing standard practices of hygiene in public catering.
(ii) In cases of air-borne diseases such as pneumonia and common cold, in addition to the above measures, close contact with the infected persons or their belongings should be avoided.
(iii) For diseases such as malaria and filariasis that are transmitted through insect vectors, the most important measure is to control or eliminate the vectors and their breeding places. This can be achieved by avoiding stagnation of water in and around residential areas, regular cleaning of household coolers, use of mosquito nets, introducing fishes like Gambusia in ponds that feed on mosquito larvae, spraying of insecticides in ditches, drainage areas and swamps, etc. In addition, doors and windows should be provided with wire mesh to prevent the entry of mosquitoes.

6. In the given flow diagram, the*replication of retrovirus in a host is shown. Observe and answer the following questions.
(a) Fill in (1) and (2).
(b) Why is the virus called retrovirus?
(c) Can the infected cell survive while viruses are being replicated and released?

Answer. (a) 1. Viral DNA is produced by reverse transcriptase.
2. New viral RNA is produced by infected cell.
(b) HIV is called retrovirus because it forms DNA from RNA by reverse transcription.
(c) Yes, infected cell can survive while viruses are being replicated and released.

7. “Maintenance of personal and public hygiene is necessary for prevention and control of many infectious diseases”. Justify’ the statement giving suitable examples.
Answer. Measures for personal hygiene include keeping the body clean; consumption of clean drinking water, food, vegetables, fruits, etc. Public hygiene includes proper disposal of waste and excreta; periodic cleaning and disinfection of water reservoirs, pools, cesspools and tanks, and observing standard practices of hygiene in public catering. These measures are particularly essential where the infectious agents are transmitted through food and water such as typhoid, amoebiasis and ascariasis.

8. The following table shows certain diseases, their causative organisms and symptoms. Fill the gaps.

Answer.

9. The outline structure of a drug is given below.

(a) Which group of drugs does this represent?
(b) What are the modes of consumption of these drugs?
(c) Name the organ of the body which is affected by consumption of these drugs.
Answer. (a) Cannabinoids
(b) Generally taken by inhalation and oral ingestion
(c) Affect the cardiovascular system of the body

10. Give the full form of CT and MRI. How are they different from each other? Where are they used?
Answer. CT (computed tomography) and MRI (magnetic resonance imaging) are very useful to detect cancers of the internal organs. Computed tomography uses X-rays to generate a three-dimensional image of the internals of an object. MRI uses strong magnetic fields and non-ionising radiations to accurately detect pathological and physiological changes in the living tissue.

11. Many secondary metabolites of plants have medicinal properties. It is their misuse that creates problems. Justify the statement with an example.
Answer. Several plants, fruits and seeds having hallucinogenic properties and have been used for hundreds of years in folk-medicine, religious ceremonies and rituals all over the globe. When these are taken for a purpose other than medicinal use or in amounts/frequency that impairs one’s physical, physiological or psychological functions, it constitutes drug abuse.

12. Why cannabinoids are banned in sports and games?
Answer. As these days cannabinoids are being abused by some sports persons to increase their performance, that is why cannabinoids are banned in sports and games.

13. What is secondary metabolism?
Answer. Secondary metabolism is a term for pathways and small molecule products of metabolism that are not absolutely required for the survival of the organism. Examples of the products include antibiotics and pigments.

14. Drugs and alcohol give short-term ‘high’ and long-term ‘damages’. Discuss.
Answer. Curiosity, need for adventure and excitement, and experimentation, constitute
common causes, which motivate youngsters towards drug and alcohol use.
A child’s natural curiosity motivates him/her to experiment. This is complicated further by effects that might be perceived as benefits, of alcohol or drug use. Thus, the first use of drugs or alcohol may be out of curiosity or experimentation, but later the child starts using these to escape facing problems. Of late, stress, from pressures to excel in academics or examinations, has played a significant role in persuading the youngsters to try alcohol and drugs. The perception among youth that it is ‘cool’ or progressive to smoke, use drugs or alcohol, is also in a way a major cause for youth to start these habits. Television, movies, newspapers, internet also help to promote this perception. Other factors that have been seen to be associated with drug and alcohol abuse among adolescents are unstable or unsupportive family structures and peer pressure.

15. Diseases like dysentery, cholera, typhoid etc., are more common in overcrowded human settlements. Why?
Answer. Diseases like dysentery, cholera, typhoid etc., are more common in overcrowded human settlements because these are infectious diseases that can transmitted from one person to another. In overcrowded settlements there is more chances of transmission of disease from one person to other.

16. From which plant cannabinoids are obtained? Name any two cannabinoids. Which part of the body is effected by consuming these substances?
Answer. Cannabinoids are obtained from the inflorescence of the plant Cannabis sativa. Marijuana, hashish, charas, ganja are some of the cannabinoids. These chemicals interact with cannabinoid receptors of the body, mainly present in the brain. Cardiovascular system is affected adversely.

17. In the metropolitan cities of India, many children are suffering from allergy/ asthma. What are the main causes of this problem? Give some symptoms of allergic reactions.
Answer. Allergy is the exaggerated response of the immune system of certain antigens present in the environment. In metropolitan cities life style is responsible for lowering of immunity and sensitivity to allergens. More polluted environment increases the chances of allergy in children. Some symptoms of allergic reactions are sneezing, watery eyes, running nose and difficulty in breathing.

18. What is the basic principle of vaccination? How do vaccines prevent microbial infections? Name the organism from which hepatitis B vaccine is produced.
Answer. The principle of vaccination is based on the property of ’memory’ of the immune system. In vaccination, a preparation of antigenic proteins of pathogens or inactivated/live but weakened pathogens is introduced into the body. The antigens generate the primary immune response by producing antibodies. The vaccines also generate the memory B-cells and T-cells. When the vaccinated person is attacked by the same pathogens, the existing memory B-cells or T-cells recognise the antigen quickly and overwhelm the invaders with massive production of lymphocytes and antibodies. Hepatitis B vaccine is produced from yeast.

19. What is cancer? How is a cancer cell different from the normal cell? How do normal cells attain cancerous nature?
Answer. An abnormal and uncontrolled division of cells is termed as Cancer. The cancerous cells are different from the normal cells in the following ways.

In our body, the growth and differentiation of cells is highly controlled and regulated. The normal cells show a property called contact inhibition.
The surrounding cells inhibits uncontrolled growth and division of cells. The normal cells lose this property and become cancerous cells giving rise to masses of cells called tumors. Transformation of normal cells into cancerous cells is induced by some physical, chemical and biological agents (carcinogens).

20. A person shows strong unusual hypersensitive reactions when exposed to certain substances present in the air. Identify the condition. Name the cells responsible for such reactions. What precaution should be taken to avoid such reactions?
Answer. Allergy. Mast Cells are responsible for such reactions. To avoid such reactions following precautions must be taken:
(i) The use of drugs like antihistamine, adrenalin and steroids quickly reduce the symptoms of allergy.
(ii) Avoid contact with substances to which a person is hypersensitive.

21. For an organ transplant, it is an advantage to have an identical twin. Why?
Answer. Very often, when some human organs like heart, eye, liver, kidney fail to
function satisfactorily, transplantation is the only remedy to enable the patient to live a normal life. Then a search begins—to find a suitable donor.
Grafts from just any source—an animal, another primate, or any human beings cannot be made since the grafts would be rejected sooner or later. Tissue matching, blood group matching are essential before undertaking any graft/transplant and even after this the patient has to take immuno-suppresants all his/her life. The body is able to differentiate ‘self ’ and ‘nonself’ and the cell-mediated immune response is responsible for the graft rejection. In an identical twin there is no chance of rejection of transplanted organ, so it is advantageous.

22. What are lifestyle diseases? How. are they caused? Name any two such diseases.
Answer. Lifestyle diseases are defined as diseases linked with the way people live their life. This is commonly caused by alcohol, drug and smoking abuse as well as lack of physical activity and unhealthy eating. Diseases that impact on our lifestyle are heart disease, stroke and obesity.

23. If there are two pathogenic viruses, one with DNA and other with RNA, which would mutate faster? And why?
Answer. Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutate at a faster rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.

Long Answer Type Questions
1. Represent schematically the life cycle of a malarial parasite.
Answer.

2. Compare the life style of people living in the urban areas with those of rural areas and briefly describe how the life style affects their health.
Answer. Urban areas
The social environment: Urban environments are more likely to see higher rates of crime and violencfc. ‘
The physical environment: In densely populated urban areas, there is often a lack of facilities and outdoor areas for exercise. In addition, air quality is often lower in urban environments which can contribute to chronic diseases such as asthma. In the developing world, urban dwellers often live in large slums which lack basic sanitation and utilities such as water and electricity. Access to health and social service: Persons of lower socioeconomic status are more likely to live in urban areas and are more likely to lack health insurance. The high prevalence of individuals without health insurance or citizenship creates a greater burden on available systems.
Rural areas
The social environment: Rural dwellers have significantly poorer health status than urban elders. Also, rural residents smoke more, exercise less, have less nutritional diets.
The physical environment: Rural women especially less educated women, are more sedentary than urban women. While poor air quality and crime rates are likely to be less of an issue in rural areas, insufficiencies in the built environment make it difficult for rural residents to exercise and maintain healthy habits.
Access to health and social seryice: Evidence indicates that rural residents have limited access to health care. Some rural areas have a higher proportion of uninsured and individually insured residents than urban areas.

3. Why do some adolescents start taking drugs? How can this be avoided?
Answer. The reasons why adolescents and youngsters take to consumption of drugs are:
(i) Curiosity of child motivates him/her to experiment.
(ii) Need for adventure and excitement.
(iii) Peer group pressure
(iv) Desire to do more physical and mental work.
(v) To overcome frustration and depression, due to failure in examinations or in other activities.
(vi) Unstable or unsupportive family structures.
The following measures can be taken to avoid taking drugs:
(i) Avoid undue pressure on child to perform beyond his/her capability in studies, sports ox any other activities.
(ii) Education and counselling are very important to face problem of stress and failure in life.
(iii) Seeking help from parents, elders and peers. This would help the young to share their feelings and concern.
(iv) Looking for danger signs and taking appropriate measures to treat them.
(v) Seeking professional and medical help for de-addiction and rehabilitation.

4. In your locality, if a person is addicted to alcohol, what kind of behavioural changes do you observe in that person? Suggest measures to overcome the problem.
Answer. The immediate adverse effects of drugs and alcohol abuse are manifested in the form of reckless behaviour, vandalism and violence. Excessive doses of drugs may lead to coma and death due to respiratory failure, heart failure or cerebral hemorrhage. A Combination of drugs or their intake along with alcohol generally results in overdosing and even deaths. The most common warning signs of drug and alcohol abuse among youth include drop in academic performance, unexplained absence from school/college, lack of interest in personal hygiene, withdrawal, isolation, depression,, fatigue, aggressive and rebellious behaviour, deteriorating relationships with family and friends, loss of interest in hobbies, change in sleeping and eating habits, fluctuations in w’eight, appetite, etc. There may even be some far-reaching implications of drug/alcohol abuse. If an abuser is unable to get money to buy drugs/alcohol he/she may turn to stealing. The adverse effects are just not restricted to the person who is using drugs or alcohol. At times, a drug/ alcohol addict becomes the cause of mental and financial distress to his/her entire family and friends.
The age-old adage of‘prevention is better than cure’ holds true here also. It is also true that habits such as smoking, taking drug or alcohol are more likely to be taken up at a young age, more during adolescence. Hence, it is best to identify the situations that may push an adolescent towards use of drugs or alcohol, and to take remedial measures well in time. In this regard, the parents and the teachers have a special responsibility. Parenting that combines with high levels of nurturance and consistent discipline, has been associated with lowered risk of substance (alcohol/drugs/tobacco) abuse. Some of the measures mentioned here would be particularly useful for prevention and control of alcohol and drugs abuse among adolescents.

5. What are the methods of cancer detection? Describe the common approaches for treatment of cancer.
Answer. Cancer detection and diagnosis: Early detection of cancers, is essential as it allows the disease to be treated successfully in many cases. Cancer detection is based on biopsy and histopathological studies of the tissue and blood and bone marrow tests for increased cell counts in the case of leukemias. In biopsy, a piece of the suspected tissue cut into thin sections is stained and examined under microscope (histopathological studies) by a pathologist. Techniques like radiography (use of X-rays), CT (computed tomography) and MRI (magnetic resonance imaging) are very useful to detect cancers of the internal organs. Computed tomography uses X-rays to generate a three-dimensional image of the internals of an object. MRI uses strong magnetic fields and non-ionising radiations to accurately detect pathological and physiological changes in the living tissue.

• Antibodies against cancer-specific antigens are also used for detection of certain cancers. Techniques of molecular biology can be applied to detect genes in individuals with inherited susceptibility to certain cancers. Identification of such genes, which predispose an individual to certain cancers, may be very helpful in prevention of cancers. Such individuals may be advised to avoid exposure to particular carcinogens to which they are susceptible (e.g., tobacco smoke in case of lung cancer).
• Treatment of cancer: The common approaches for treatment of cancer are surgery, radiation therapy and immunotherapy. In radiotherapy, tumor cells are irradiated lethally, taking proper care of the normal tissues surrounding the-tumor mass. Several chemotherapeutic drugs are used to kill cancerous cells. Some of these are specific for particular tumors. Majority of drugs have side effects like hair loss; anemia, etc. Most cancers are treated by combination of surgery, radiotherapy and chemotherapy. Tumor cells have been shown to avoid detection and destruction by immune system. Therefore, the patients are given substances called biological response modifiers such as a-interferon which activates their immune system and helps in destroying the tumor.

6. Drugs like LSD, barbiturates, amphetamines, etc., are used as medicines to help patients with mental illness. However, excessive doses and abusive usage are harmful. Enumerate the major adverse effects of such drugs in humans.
Answer. Drugs like barbiturates, amphetamines, benzodiazepines, and other similar drugs, that are normally used as medicines to help patients cope with mental illnesses like depression and insomnia, are often abused. Morphine is a very effective sedative and painkiller, and is very useful in patients who have undergone surgery. Several plants, fruits and seeds having hallucinogenic properties have been used for hundreds of years in folk-medicine, religious
ceremonies and rituals all over the globe. When these are taken for a purpose other than medicinal use or in amounts/frequency that impairs one’s physical, physiological or psychological functions, it constitutes drug abuse. ,

7. What is Pulse Polio-Programme of Government of India? What is OPV? Why is it that India is yet to eradicate Polio?
Answer. Pulse Polio is an immunisation campaign established by the government of India to eliminate poliomyelitis (polio) in India by vaccinating all children under the age of five years against the polio virus. The project fights poliomyelitis through a large-scale pulse vaccination programme and monitoring for polio cases.

•  In 1995, following the Global Polio Eradication Initiative of the World Health Organization (1988), India launched Pulse Polio immunisation program with Universal Immunization Program which aimed at 100% coverage.
• The last reported cases of wild polio in India were in West Bengal and Gujarat on 13 January 2011. On 27 March 2014, the World Health Organization (WHO) declared India a polio free country, since no cases of wild polio had been reported in for three years.
•  Polio vaccines are the vaccines used to prevent poliomyelitis (polio).One type uses inactivated poliovirus and is given by injection (IPV), while the other type uses weakened poliovirus and is given by mouth (OPV). The World Health Organization recommends all children be vaccinated against polio. The two vaccines have eliminated polio from most of the world. The oral polio vaccine was developed by Albert Sabin and came into commercial use in 1961. _

8. What are recombinant DNA vaccines? Give two examples of such vaccines. Discuss their advantages.
Answer. A recombinant vaccine is a vaccine produced through recombinant DNA technology. This involves inserting the DNA encoding an antigen that stimulates an immune response into bacterial or mammalian cells.
Recombinant DNA technology has allowed the production of antigenic polypeptides of pathogen in bacteria or yeast. Vaccines produced using jthis approach allow large scale production and hence greater availability for immunisation, e.g., hepatitis B vaccine (Recombivax HB) produced from yeast. As of June 2015 one human DNA vaccine had been approved for human use, the single-dose Japanese encephalitis vaccine called IMOJEV, released in 2010 in Australia.
Advantages of recombinant DNA vaccines:
1. No risk for infection
2. Ease of development and production
3. Stability for storage and shipping
4. Cost-effectiveness
5. Expression and purification of recombinant proteins
6. Long-term persistence of immunogen
7. In vivo expression ensures protein more closely resembles normal eukaryotic structure, with accompanying post-translational modifications.

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NCERT Exemplar Class 12 Biology Chaper 7 Evolution

Multiple Choice Questions
Single Correct Answer Type

1. Which of the following is used as an atmospheric pollution indicator?
(a) Lepidoptera (b) Lichens
(c) Lycopersicon (d) Lycopodium
Answer. (b) Lichens are sensitive to SO₂ pollution. Lichens are very good pollution indicators, they do not grow in polluted areas.

2. The theory of spontaneous generation stated that:
(a) Life arose from living forms only
(b) Life can arise from both living and non-living
(c) Life can arise from non-living things only.
(d) Life arises spontaneously, neither from living nor from the non-living.
Answer. (c)

• For a long time it was also believed that life came out of decaying and rotting matter like straw, mud, etc.
• According to theory of abiogenesis, life originates from non-living.

3. Animal husbandry and plant breeding programmes are the examples of
(a) Reverse evolution (b) Artificial selection
(c) Mutation (d) Natural selection
Answer. (b) Animal husbandry and plant breeding programmes are the examples of artificial selection.

4. Paleontological evidences for evolution refer to the
(a) Development of embryo (b) Homologous organs
(c) Fossils (d) Analogous organs
Answer. (c) Palaeritological evidences for evolution refer to the fossils. Fossils provide direct and solid evidence in favour of organic evolution through ages. Fossils are studied for knowing about extinct organisms. Birbal Sahni Institute of Palaeobotany is situated at Lucknow. Birbal Sahni is called ‘Father of Indian Palaeobotany’. Age of fossils is determined by Uranium-Lead method (U²³⁸) or Potassium-Argon method or Radioactive Carbon dating (C¹⁴). ESR (electron-spin-resonance method) is the most accurate method of dating of fossils.

5. The bones of forelimbs of whale, bat, cheetah and man are similar in structure, because
(a) One organism has given rise to another
(b) They share a common ancestor
(c) They perform the same function
(d) The have biochemical similarities
Answer. (b) Organs having same origin but different functions are called homologous organs. For example, whales, bats, Cheetah and human (all mammals) share similarities in the pattern of bones of forelimbs. Homology indicates common ancestry.

6. Analogous organs arise due to
(a) Divergent evolution (b) Artificial selection
(c) Genetic drift (d) Convergent evolution
Answer. (d) Organs having different origin but similar function are called analogous. Wings of butterfly and of birds look alike. They are not anatomically similar structures though they perform similar functions. Hence, analogous structures are a result of convergent evolution (different structures evolving for the same function and hence having similarity).

7. (p + q)²= p² + 2pq + q² = 1 represents an equation used in
(a) Population genetics (b) Mendelian genetics
(c) Biometrics (d) Molecular genetics
Answer. (a) (p + q)²= p²+ 2pq + q² = 1 represents an equation used in population genetics.
Above equation is known as Hardy-Weinberg principle in which
p²—Dominant homozygous
2pq — Heterozygous
q²— Recessive homozygous

8. Appearance of antibiotic-resistant bacteria is an example of
(a) Adaptive radiation
(b) Transduction
(c) Pre-existing variation in the population
(d) Divergent evolution
Answer. (c) Appearance of antibiotic-resistant bacteria is an example of pre-existing variation in the population.

9. Evolution of life shows that life forms had a trend of moving from
(a) Land to water (b) Dryland to wet land
(c) Freshwater to sea water (d) Water to land
Answer. (d) Evolution of life shows that life forms had a trend of moving from water to land.

10. Viviparity is considered to be more evolved because
(a) The young ones are left on their own
(b) The young ones are protected by a thick shell
(c) The young ones are protected inside the mother’s body and are looked after they are bom leading to more chances of survival
(d) The embryo takes a long time to develop
Answer. (c) Viviparity is considered to be more evolved because the young ones are protected inside the mother’s body and are looked after they are bom leading to more chances of survival.

11. Fossils are generally found in
(a) Sedimentary rocks (b) Igneous rocks
(c) Metamorphic rocks (d) Any type of rock
Answer. (a) Fossils are generally found in sedimentary rocks.

12. For the MN-blood group system, the frequencies of M and N alleles are 0.7 and 0.3, respectively. The expected frequency of MN-blood group bearing organisms is likely to be (a) 42% (b) 49% (c) 9% (d) 58%
Answer.

13. The most accepted line of descent in human evolution is
(a) Australopithecus —> Ramapithecus —> Homo sapiens —> Homo habilis
(b) Homo erectus —>Homo habilis —> Homo sapiens
(c) Ramapithecus —> Homo habilis —>Homo erectus —>Homo sapiens
(d) Australopithecus —> Ramapithecus —> Homo erectus —>Homo habilis —> Homo sapiens
Answer. (c) Ramapithecus —> Homo habilis —> Homo erectus —>Homo sapiens

14. Which of the following is an example for link species?
(a) Lobe fish (b) Dodo bird
(c) Seaweed (d) Chimpanzee
Answer. (d) Chimpanzee is an example for link species.

15. Which type of selection is industrial melanism observed in moth, Bistonbitularia?
(a) Stabilising (b) Directional
(c) Disruptive (d) Artificial
Answer. (b) Directional type of selection is industrial melanism observed in moth, Bistonbitularia. (See figure on next page)

16. Match the scientists listed under column ‘A’ with ideas listed column ‘B’

(a) i—M; ii—P; iii—N; iv—O (b) i—P; ii—M; iii—N; iv—O
(c) i—N; ii—P; iii—O; iv—M (d) i—p; ii—O; iii—N; iv—M
Answer. (b)

17. In 1953 S. L. Miller created primitive earth conditions in the laboratory and gave experimental evidence for origin of first form of life from pre-existing non-living organic molecules. The primitive earth conditions created include
(a) Low temperature, volcanic storms, atmosphere rich in oxygen
(b) Low temperature, volcanic storms, reducing atmosphere
(c) High temperature, volcanic storms, non-reducing atmosphere
(d) High temperature, volcanic storms, reducing atmosphere containing CH₄ , NH₃ etc.
Answer. (d) In 1953 S. L. Miller created primitive earth conditions in the laboratory and gave experimental evidence for origin of first form of life from pre-existing non-living organic molecules. The primitive earth conditions created include high temperature, volcanic storms, reducing atmosphere containing CH₄ , NH₃ etc.

18. Variations during mutations of meiotic recombinations are
(a) Random and direction less (b) Random and directional
(c) Random and small (d) Random, small and directional.
Answer. (a) Variations during mutations of meiotic recombinations are random and direction less.

Very Short Answer Type Questions
1. What were the characteristics of life forms that had been fossilised?
Answer. Fossils are remains of hard parts (like bones, teeth, etc.) of life-forms found in rocks.

2. Did aquatic life forms get fossilised? If, yes where do we come across such fossils?
Answer. Yes, aquatic life forms get fossilised in the sediments of the water bodies. Later, sediments form the part of sedimentary rocks in which fossils, are deposited.

3. What are we referring to? When we say ‘simple organisms’ or ‘complex organisms’.
Answer. When we say simple or complex organisms we are talking in terms of evolutionary history of an Organism. A ‘simple organism’ is considered to be primitive and has simple thallus organisation; The level of complexity of metabolism is also low. On the other hand a ‘complex organism’ refers to a more evolved form forming higher levels of structural and functional complexities. They are believed to have arisen from simple organisms.

4. How do we compute the age of a living tree?
Answer. Age of the living tree is calculated by counting the number of annual rings or by the radioactive carbon dating,

5. Give an example for convergent evolution and identify the features towards which they are converging.
Answer. Presence of wings in birds and butterfly is an example of convergent evolution. They are adapted for flying (volant mode).

6. How do we compute the age of a fossil?
Answer. To compute the age of a fossil, we use radiocarbon dating.

7. What is the most important pre-condition for adaptive radiation?
Answer. Pre-existing variations are the most important pre-condition for adaptive radiation.

8. How do we compute the age of.a rock?
Answer. Age of rock is computed by radiocarbon dating, potassium-argon dating, uranium-lead dating and rubidium-strontium dating method.

9. When we talk of functional macro-molecules (e.g., proteins as enzymes, hormones, receptors, antibodies etc.), towards what are they evolving?
Answer. Similarities in proteins and genes performing a given function among diverse organisms give clues to common ancestry. These biochemical similarities point to the same shared ancestry as structural similarities among diverse organisms. Trypsin (ancient enzyme) is present from protozoa to mammals.

10. In a certain population, the frequency of three genotypes is as follows:
Genotypes: BB, Bbbb
Frequency: 22%, 62%, 16%
What is the likely frequency of B and b alleles?
Answer. Frequency of B allele = all of BB + 1/2 of Bb = 22 + 31 = 53%
Frequency of b allele = all of bb + 1/2 of Bb = 16 + 31 = 47%

11. Among the five factors that are known to affect Hardy-Weinberg equilibrium, ‘ three factors are gene flow, genetic drift and genetic recombination. What are the other two factors?
Answer. Natural selection and mutation.

12. What is founder effect?
Answer. Sometimes the change in allele frequency is so different in the new sample of population that they become a different species. Small group of population called founders left their habitat and goes into new habitat. In new habitat this population (founders) shows different genotype frequency from that of the original population and leads to variation. The original drifted population becomes founders and the*effect is called founder effect.

13. Who among the Dryopithecus and Ramapithecus was more man-like?
Answer. Ramopithecus was more man-like wille Dryopithecus was more a ape-like

14. By what Latin name the first hominid was known?
Answer. Homo habilis

15. Among Ramapithecus, Australopithecus and Homo habilis, who probably did not eat meat?
Answer. Homo habilis

Short Answer Type Questions
1. Louis Pasteur’s experiments, if you recall, proved that life can arise from only pre-existing life. Can we correct this as life evolves from pre-existent life or otherwise we will never answer the question as to how the first forms of life arose? Comment.
Answer. We can correct the statement of Louis Pasteur’s because Oparin of Russia and Haldane of England proposed that the first form of life could have come
from pre-existing non-living organic molecules (e.g., RNA, protein, etc.) and that formation of life was preceded by chemical evolution, i.e., formation of diverse organic molecules from inorganic constituents.

2. The scientists believe that evolution is gradual. But extinction, part of evolutionary story, are ‘sudden’ and ‘abrupt’ and also group-specific. Comment whether a natural disaster can be the cause for extinction of species.
Answer. Natural disaster like earth quake can be the cause for extinction of species. During the long period since the origin and diversification of life on earth there were five episodes of mass extinction of species.

3. Why is nascent oxygen supported to be toxic to aerobic life forms?
Answer. Nascent oxygen is highly reactive. It can react readily with different kinds of molecules, including DNA, proteins present in the cells of aerobic life forms. This may lead to mutations and undesirable metabolic changes.

4. While creation and presence of variation is direction less, natural selection is directional as it is in the context of adaptation. Comment.
Answer. Creation and variation occur in a sexually reproducing population as a result of crossing over during meiosis and random fusion of gametes. It is however the organisms that are selected over a period of time which are determined by the environmental conditions. In other words, the environment provides the direction with respect to adaptations so that the organisms are more and more fit in terms of survival.

5. The evolutionary story of moths in England during industrialisation reveals, that ‘evolution is apparently reversible’. Clarify this statement.
Answer. In a collection of moths made in 1850s, i.e., before industrialisation set in, it was observed that there were more white-winged peppered moths (Biston betularia) on trees than dark-winged or melanised moths (Biston carbonaria). However, in the collection carried out from the same area, but after industrialisation, i.e., in 1920, there were more dark-winged moths in the same area, i.e., the proportion was reversed.

• Before industrialisation set in, thick growth of almost white-coloured lichen covered the trees—in that background the white winged moth survived but the dark-coloured moth were picked out by predators. Lichens can be used as industrial pollution indicators. They will not grow in areas that are polluted. During post-industrialisation period, the tree trunks became dark due to industrial smoke and soots.
• Under this condition the white-winged moth did not survive due to predators, dark-winged or melanised moth survived. Hence, moths that were able to camouflage themselves, i.e., hide in the background, survived. This understanding is supported by the fact that in areas where industrialisation did not occur, e.g., in rural areas, the count of melanic moths was low. This showed that in a mixed population, those that can better-adapt, survive and increase in population size.

6. Comment on the statement that “evolution and natural selection are end result or consequence of some other processes but themselves are not processes”.
Answer. The world we see, inanimate .and animate, is only the success stories of evolution. When we describe the story of this world we describe evolution as a process. On the other hand when we describe the story of life on earth, we treat evolution as a consequence of a process called natural selection. We are still not very clear whether to regard evolution and natural selection as processes or end result of unknown processes.

7. State and explain any three factors affecting allele frequency in populations.
Answer. (i) Gene migration or gene flow: When migration of a section of population to another place and population occurs, gene frequencies change in the original as well as in the new population. New genes/alleles are added to the new population and these are lost from the old population. There would be a gene flow if this gene migrationhappens multiple times.
(ii) Genetic drift: If the same change occurs by chance, it is called genetic drift. Sometimes the change in allele frequency is so different in the new sample of population that they become a different species. The original drifted population becomes founders and the effect is called founder effect.
(iii) Mutation: Microbial experiments show that pre-existing advantageous mutations when selected will result in observation of new phenotypes. Over few generations, this would result in speciation. Natural selection is a process in which heritable variations enabling better survival are enabled to reproduce and leave greater number of progeny.

8. Gene flow occurs through generations. Gene flow can occur across language barriers in humans. If we have a technique of measuring specific allele frequencies in different population of the world, can we not predict human migratory patterns in pre-history and history? Do you agree or disagree? Provide explanation to your answer.
Answer. Yes, I agree. Gene flow occurs through generations. By studying specific allele frequencies, we can predict the human migratory patterns in prehistory and history. Studies have used specific genes/chromosomes/mitochondrial DNA to trace the evolutionary history and migratory patterns of humans. (The project is known as the Human Genographics Project).

9. How do you express the meaning of words like race, breed, cultivars or variety?
Answer.

•  Race is a group of people who share similar and distinct physical characteristics.
• A group of animals related by descent and similar in most characters like general appearance, features, size, configuration, etc., are said to belong to a breed.
• A cultivar is a plant or grouping of plants selected for desirable characteristics that can be maintained by propagation.
• A taxonomic category that ranks below species, its members differing from others of the same species in minor but heritable characteristics is called variety.

10. When we say “survival of the fittest”, does it mean that
a. those which are fit only survive, or
b. those that survive are called fit? Comment.
Answer. Those individuals which survive and reproduce in their respective environment are called fit.

11. Enumerate three most characteristic criteria for designating a Mendelian population.
Answer. Population must be sufficiently large with potentialities for free flow of genetic material among individuals (through sexual reproduction). Migration should either be nil or negligible.

12. “Migration may enhance or blur the effects of selection”. Comment.
Answer. Migration may cause enrichment of the gene pool of such alleles that are being selected for, or blur the effects of selection through replenishment of alleles that were selected against by nature.

Long Answer Type Questions
1. Name the law that states that the sum of allelic frequencies in a population remains constant. What are the five factors that influence these values?
Answer. Hardy-Weinberg principle states that the sum of allelic frequencies in a population remains constant. Five factors are known to affect Hardy-Weinberg equilibrium. These are gene migration or gene flow, genetic drift, mutation, genetic recombination and natural selection.

2. Explain divergent evolution in detail. What is the driving force behind it?
Answer. Whales, bats, Cheetah and human (all mammals) share similarities in the
pattern of bones of forelimbs. Though these forelimbs perform different functions in these animals, they have similar anatomical structure. All of them have humerus, radius, ulna, carpals, metacarpals and phalanges in their forelimbs.
Hence, in these animals, the same structure developed along different directions due to adaptations to different needs. This is divergent evolution and these structures are homologous. Homology indicates common ancestry. Driving force behind the divergent evolution is adaptation in different environments.

3. You have studied the story of Pepper moths in England. Had the industries been removed, what impact could it have on the moth population? Discuss.
Answer. In the population of Peppermoth, two variants were already existing in the population, the black and the grey. In the absence of industrialisation the grey moths were prevalent because they blended very well with the lichen and moss covered trees camouflage and the predators cannot spot them. The black ones were easily spotted and killed by predators and therefore were fewer in numbers. With industrialisation the stems got covered with black
soot. This provided better camouflage tb the black variant and their number increased. If the industries had been removed the population of black moths would have declined because as stated earlier they would have been spotted better by predators and therefore be eaten more frequently.

4. What are the key concepts in the evolution theory of Darwin?
Answer. Branching descent and natural selection are the two key concepts of Darwinian Theory of Evolution.

• The novelty and brilliant insight of Darwin was this: He asserted that variations, which are heritable and which make resource utilisation better for few (adapted to habitat better) will enable only those to reproduce and leave more progeny. Hence for a period of time, over many generations, survivors will leave more progeny and there would be a change in population characteristic and hence new forms appear to arise.
• The fitness, according to Darwin, refers ultimately and only to reproductive fitness. Hence, those who are better fit in an environment, leave more progeny than others. These, therefore, will survive more and hence are selected by nature. He called it natural selection and implied it as a mechanism of evolution.

5. Two organisms occupying a particular geographical area (say desert) show similar adaptive strategies. Taking examples, describe the phenomenon.
Answer. One can say that it is the similar habitat that has resulted in selection of similar adaptive features in different groups of organisms but toward the same function. Spins of Opuntia and cactus are modifications of leaves to prevent the loss of water in desert.

6. We are told that evolution is a continuing phenomenon for all living things. Are humans also evolving? Justify your answer.
Answer. Yes, human is also evolving. Fossils give the evidences that evolution is a continuous phenomenon. Our ancestors like Ramapithecus, Australopithecus, Homo habilis, Homo erectus, Neanderthal man and Cro-magnon man continuously evolving and modem man (Homo sapiens sapiens) arises from certain modifications.

7. Had Darwin been aware of Mendel’s work, would he been able to explain the origin of variations. Discuss.
Answer. Yes, Darwin has been aware of Mendel’s work. Even though Mendel had talked of inheritable ‘factors’ influencing phenotype, Darwin either ignored these observations or kept silence.
Darwin would have been able to explain the origin of variations. He asserted that variations, which are heritable and which make resource utilisation better for few (adapted to habitat better) will enable only those to reproduce and leave more progeny. Hence for a period of time, over many generations, survivors will leave more progeny and there would be a change in population characteristic and hence new forms appear to arise.

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NCERT Exemplar Class 12 Biology Chapter 6 Molecular Basis of Inheritance

Multiple Choice Questions
Single Correct Answer Type

NCERT Exemplar Class 12 Biology Chapter 6

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) In a DNA strand the nucleotides are linked together by Phosphodiester bonds.

2. A nucleoside differs from a nucleotide. It lacks the
(a) Base (b) Sugar
(c) Phosphate group t (d) Hydroxyl group
Answer. (b) Nucleoside = Base + sugar
Nucleotide = Base + sugar + phosphate group

3. Both deoxyribose and ribose belong to a class of sugars called
(a) Trioses (b) Hexoses
(c) Pentoses (d) Polysaccharides
Answer. (c) Both deoxyribose and ribose belong to a pentoses class of sugars.

4. The fact that a purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix
(a) The antiparallel nature (b) The semiconservative nature
(c) Uniform width throughout DNA (d) Uniform length in all DNA
Answer. (c) A purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix, uniform width throughout DNA.

Molecular Basis Of Inheritance NCERT Exemplar Class 12

5. The net electric charge on DNA and histones is
(a) Both positive
(b) Both negative
(c) Negative and positive, respectively
(d) Zero
Answer. (c) The net electric charge on DNA and.histones is negative and positive, respectively.

6. The promoter site and the terminator site for transcription are located at
(a) 3′ (downstream) end and 5′ (upstream) end, respectively of the transcription unit
(b) 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit
(c) The 5′ (upstream) end
(d) The 3′ (downstream) end
Answer. (b) The promoter site and the terminator site for transcription are located at 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit.

7. Which of the following statements is the most appropriate for sickle cell anaemia?
(a) It cannot be treated with iron supplements
(b) It is a molecular disease
(c) It confers resistance to acquiring malaria
(d) All of the above
Answer. (d) Sick/e cell anaemia: It is a molecular disease, autosomal recessive disorder and an example of pleiotropy. It cannot be treated with iron supplements and it confers resistance to acquiring malaria.

NCERT Exemplar Molecular Basis Of Inheritance Class 12

8. One of the following true with respect to AUG is
(a) It codes for methionine only 
(b) It is also an initiation codon
(c) It codes for methionine in both prokdryotes and eukaryotes
(d) All of the above
Answer. (d) AUG—It codes for methionine only in both prokaryotes and eukaryotes, it is also an initiation codon.

9. The first genetic material could be
(a) Protein (b) Carbohydrates (c) DNA (d) RNA
Answer. (d) The first genetic material could be RNA but ideal genetic material is DNA.

10. With regard to mature mRNA in eukaryotes,
(a) Exons and introns do not appear in the mature RNA
(b) Exons appear but introns do not appear in the mature RNA
(c) Introns appear but exons do not appear in the mature RNA
(d) Both exons and introns appear in the mature RNA
Answer. (b) In eukaryotes exons appear but introns do not appear in the mature RNA.

Molecular Basis Of Inheritance NCERT Exemplar Solutions Class 12

11. The human chromosome with the highest and least number of genes in them are respectively
(a) Chromosome 21 and Y (b) Chromosome 1 and X
(c) Chromosome 1 and Y (d) Chromosome X and Y
66 NCERT Exemplar Biology: Class XU
Answer. (c)

• Chromosome one have 2968 genes (highest)
• Chromosome Y have 231 genes (lowest)

12. Who amongst the following scientists had no contribution in the development of the double helix model for the structure of DNA?
(a) Rosalind Franklin (b) Maurice Wilkins
(c) Erwin Chargaff (d) Meselson and Stahl
Answer. (d) Meselson and Stahl give the experimental proof of semiconservative DNA replication. He had no contribution to the development of the double helix model for the. structure of DNA.

NCERT Exemplar Class 12 Biology Solutions Chapter 6

13. DNA is a polymer of nucleotides which are linked to each other by 3′-5′ phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?
(a) Replace purine with pyrimidines
(b) Remove/Replace 3′ OH group in deoxy ribose
(c) Remove/Replace 2′ OH group with some other group in deoxy ribose
(d) Both‘b’and‘c’.
Answer. (b) DNA is a polymer of nucleotides which are linked to each other by 3′-5′ phosphodiester bond. Remove/replace 3’OH group in deoxyribose, to prevent polymerisation of nucleotides. .

14. 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) Discontinuous synthesis of DNA occurs in one strand, because DNA dependent DNA polymerase catalyses polymerisation only in one direction (5′ —> 3′).

15. Which of the following steps in transcription is catalysed by RNA polymerase?
(a) Initiation (b) Elongation
(c) Termination (d) All of the above
Answer. (b) Elongation step in transcription is catalysed by RNA polymerase.

Molecular Basis Of Inheritance Exemplar Class 12

16. Control of gene expression takes place at the level of
(a) DNA-replication (b) Transcription
(c) Translation (d) None of the above
Answer. (b) Control of gene expression takes place at the level of transcription.

17. Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. Which of the following statements is correct about regulatory protein?
(a) They only increase expression
(b) They only decrease expression
(c) They interact with RNA polymerase but do not affect the expression
(d) They can act both as activators and as repressors
Answer. (d) Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. They can act both as activators and as repressors.

18. Which was the last human chromosome to be completely sequenced?
(a) Chromosome 1 (b) Chromosome 11
(c) Chromosome 21 (d) Chromosome x
Answer. (a) The Human genome project was completed in 2003. The sequence of chromosome 1 was completed only in May 2006.

19. Which of the following are the functions of RNA?
(a) It is a carrier of genetic information from DNAto ribosomes synthesising polypeptides.
(b) It carries amino acids to ribosomes.
(c) It is a constituent component of ribosomes.
(d) All of the above.
Answer. (d) Functions of mRNA:

• Carrier of genetic information from DNA to ribosomes synthesising polypeptides.
• Carries amino acids to ribosomes.
• Constituent component of ribosomes. .

Molecular Basis Of Inheritance Class 12 NCERT Exemplar

20. While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were:
Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17% Considering the Chargaff’s rule it can be concluded that
(a) It is a double stranded circular DNA
(b) It is single stranded DNA
(c) It is a double stranded linear DNA
(d) No conclusion can be drawn. ‘
Answer. (b) In the DNA of an organism a total number of5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff’s rule it can be concluded that it is a single stranded DNA.

21. In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called
(a) A-DNA (b) B-DNA (c) c DNA (d) r DNA
Answer. (c) In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called c-DNA (Complementary DNA).

22. If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of ₁₅N/₁₅N: ₁₅N/₁₄N: ₁₄N/₁₄N containing DNA in the fourth generation would be
(a) 1 : 1 : 0 (b) 1 : 4 : 0 (c) 0:1:3 (d) 0:1:7
Answer. (d)

•  If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of 15_N/15_N : 15_N/14_N : 14_N/14_N containing DNA in the fourth generation would be 0 : 1 : 7.
• After third generation (60 min.) bacteria contains 25% hybrid (15_N14_N) in 1 :3 ratio.
• After 4th generation (80 min.) bacteria contains 12.5% hybrid and 87.5% light DNA in 1 : 7 ratio.

23. If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is 5′- ATGAATG-3′, The sequence of bases in its RNA transcript would be
(a) 5′ – AU G AAU G – 3′ (b) 5′- U AC U U AC – 3′
(c) 5′-CAUUCAU-3′ (d) 5;-GUAAGUA-3′
Answer. (a)
Coding strand: 5′-AT G A AT G – 3′
Template strand: 3′ – TACTTAC-5″
RNA transcript: 5′-AUGAAUG-3′

24. The RNA polymerase holoenzyme transcribes
(a) The promoter, structural gene and the terminator region
(b) The promoter and the terminator region
(c) The structural gene and the terminator regions
(d) The structural gene only
Answer. (d) The RNA polymerase holoenzyme transcribes the structural gene only.

25. If the base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of tRNA pairing with it must be
(a) 5′ -UAC – 3′ (b) 5′-CAU-3′
(c) 5′-AUG-3′ (d) 5′ – GUA – 3′
Answer. (b) Base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of tRNA pairing with it must be cqpiplementary, i.e., 3′ – UAC – 5′

26. The amino acid attaches to the tRNA at its
(a) 5′-end (b) 3′-end
(c) Anti codon site – (d) DHU loop
Answer. (b) The amino acid attaches to the tRNA at its 3′-end.

27. To initiate translation, the mRNA first binds to
(a) The smaller ribosomal sub-unit (b) The larger ribosomal sub-unit (c) The whole ribosome (d) No such specificity exists
Answer. (a) To initiate translation, the mRNA first binds to the smaller ribosomal sub-unit.

28. In E.coli, the lac operon gets switched on when
(a) Lactose is present and it binds to the repressor
(b) Repressor binds to operator
(c) RNA polymerase binds to the operator
(d) Lactose is present and it binds to RNA polymerase
Answer. (a) In E.coli, the lac operon gets switched on when lactose is present and it binds to the repressor.

Very Short Answer Type Questions
1. What is the function of histones in DNA packaging?
Answer. Histone are the basis protein that are rich in arginine and lysine. Histones form octamer on which DNA is wrapped and form nucleosome that means it helps in packaging of DNA in eukaryotes.

2. Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
Answer.

3. The enzyme DNA polymerase in E.coli is a DNA dependent polymerase and also has the ability to proofread the DNA strand being synthesised. Explain. Discuss the dual polymerase.
Answer. DNA polymerase helps in the replication of DNA molecules. DNA polymerase also helps in the proofreading, if any error/mutation occurred during replication.

4. What is the cause of discontinuous synthesis of DNA on one of the parental strands of DNA? What happens to these short stretches of synthesised DNA?
Answer. The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5′ —> 3′. This creates some additional complications at the replicating fork. Consequently, on leading strand (the template with polarity 3′ —> 5′), the replication is continuous, while on the lagging strain (the template with polarity 5′ —> 3′), it is discontinuous. The discontinuously synthesised fragments are later joined by the enzyme DNA ligase.

5. Given below is the sequence of coding strand of DNA in a transcription unit. 3’AATGCAGCTATTAGG’ Write the sequence of
(a) its complementary strand
(b) the mRNA
Answer. (a) Complementary strand:
5′ – TTACGTCGATAA T C C – 3′
(b) The mRNA
5′ – AAU GCAGCUAUUAGG-3′

6. What is DNA polymorphism? Why is it important to study it?
Answer. Polymorphism (variation at genetic level) arises due to mutations. Allelic sequence variation has traditionally been described as a DNA polymorphism if more than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01. In simple terms, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.
Polymorphism become very useful identification tool in forensic applications. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basis of paternity testing, in case of disputes.

7. Based on your understanding of genetic code, explain the formation of any abnormal hemoglobin molecule. What are the known consequences of such a change?
Answer. The defect is caused by the substitution (trans version) of Glutamic acid (Glu) by Valine (Val) at the sixth position of the betaglobin chain of the haemoglobin molecule. The substitution of amino acid in the glob in protein results due to the single base substitution at the sixth codon of the betaglobin gene from GAG to GUG. The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure.
Sickle-shaped red blood cells that obstruct capillaries and restrict blood flow to an organ resulting in ischaemia, pain, necrosis, and often organ damage.

8. Sometimes cattle or even human beings give birth to their young ones that are having extremely different sets of organs like limbs/position of eye(s) etc. Comment.
Answer. There is a disturbance in co-ordinated regulation of expression of sets of genes associated with organ development.

9. In a nucleus, the number of ribonucleoside triphosphates is 10 times the number of deoxy ribonucleoside triphosphates, but only deoxy ribonucleotides are added during the DNA replication. Suggest a mechanism.
Answer. DNA polymerase is highly specific to recognise only deoxy ribonucleoside triphosphates. Therefore it cannot hold RNA nucleotides.

10. Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them.
Answer. (i) Helicase—opens the helix
(ii) Topoisomerases—removes the super coiling of DNA
(iii) Primase—synthesises RNA primer
(iv) Telomerase—to synthesises the DNA of telomeric end of chromosomes

11. Name any three viruses which have RNA as the genetic material.
Answer. (i) TMV (Tobacco Mosaic virus)
(ii) HIV (Human Immuno Deficiency virus)
(iii) QB bacteriophage

Short Answer Type Questions
1. Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as the genetic material.
Answer. Transforming Principle

• In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumonia), witnessed a miraculous transformation, in the bacteria. During the course of his . experiment, a living organism (bacteria) had changed in physical form.
• When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a culture plate, some produce smooth shiny colonies (S) while others produce rough colonies (R).
• This is because the S strain bacteria have a mucous (polysaccharide) coat, while R strain does not. Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia. .
  S strain -»Inject into mice -» Mice die R strain —»Inject into mice —> Mice live
• Griffith was able to kill bacteria by heating them. He observed that heat- killed S strain bacteria injected into mice did not kill them.
  S strain (heat killed) —> Inject into mice —> Mice live
• When he injected a mixture of heat-killed S and live R bacteria, the mice died. Moreover, he recovered living S bacteria from the dead mice.
  S strain (heat killed ) + —> Inject into mice —> Mice die R strain (live)
• He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria. Some ‘transforming principle’, transferred from the heat-killed S strain, had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material. However, the biochemical nature of genetic material was not defined from his experiments.

2. Who revealed biochemical nature of the transforming principle? How was it done?
Answer. Prior to the work of Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44), the genetic material was thought to be a protein. They worked th determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment. They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed.
They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material, but not all biologists were convinced.

3. Discuss the significance of heavy isotope of nitrogen in the Meselson and Stahl’s experiment.
Answer. Matthew Meselson and Franklin Stahl performed the following experiment in 1958. They grew E. coli in a medium containing 15NH4C1 (15N is the heavy isotope of nitrogen) as the only nitrogen source for many generations. The result was that l5N was incorporated into newly synthesised DNA (as well as other nitrogen containing compounds).
This heavy DNA molecule could be .distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient 15N is not a radioactive isotope, and it can be separated from 14N only based on densities).

4. Define a cistron. Giving examples differentiate between monocistronic and polycistronic transcription unit.
Answer. Portion of DNA having information for an entire polypeptide or trait is called cistron. However by defining a cistron as a segment of DNA coding for a polypeptide, the structural gene in a transcription unit jcould be said as monocistronic (mostly in eukaryotes) or polycistronic (mostly in bacteria or prokaryotes). In eukaryotes, the monocistronic structural genes have interrupted coding sequences—the genes in eukaryotes are split.
The coding sequences or expressed sequences are defined as exons. Exons are said to be those sequence that appear in mature or processed RNA. The exons are interrupted by introns. Introns or intervening sequences do not appear in mature or processed RNA.

5. Give any six features of the human genome.
Answer. Some of the salient Observations drawn from human genome project are as follows:

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.
3. The total number of genes is estimated at 30,000-much lower than previous estimates of 80,000 to 1,40,000 genes. Almost all (99.9 per cent) nucleotide bases are exactly the same in all people.
4. The functions are unknown for over 50 per cent of the discovered genes.
5. Less than 2 per cent of the genome codes for proteins.
6. Repeated sequences make up a very large portion of the human genome.

6. During DNA replication, why is it that the entire molecule does not open in one go? Explain replication fork. What are the two functions that the monomers (dNTPs) play?
Answer. For long DNA molecules, since the two strands of DNA cannot be separated in its entire length (due to very high energy requirement), the replication occur within a small opening of the DNA helix, referred to as replication fork.

7. Retroviruses do not follow central Dogma. Comment.
Answer. Genetic material of retrovirus is RNA. At the time of synthesis of protein, RNA is ‘reverse transcribed’ to its complementary DNA first, which is opposite to the central dogma. Hence, retrovirus are not known to follow central dogma.

8. In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base increases from 0.34 nm to 0.44 nm. Calculate the length of DNA double helix (which has 2 x 10⁹ bp) in the presence of saturating amount of this compound.
Answer. 2 x 10⁹ x 0.44 x 10^-9/bp = 0.88 m.

9. What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?
Answer. If histone proteins were rich in acidic amino acids instead of basic amino acids then they may not have any role in DNA packaging in eukaryotes as DNA is also negatively charged molecule. The packaging of DNA around the nucleosome would not happen. Consequently, the chromatin fibre would not be formed.

10. Recall the experiments done by Frederick Griffith, Avery, MacLeod and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the R-strain into virulent strain? Explain.
Answer. RNA is more labile and prone to degradation (owing to the presence of 2′ OH group in its ribose). Hence heat-killed S-strain may not have retained its ability to transform the R-strain into virulent form if RNA was its genetic material.

11. You are repeating the Hershey-Chase experiment and are provided with two isotopes: ³²P and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer. Use of ¹⁵N will be inappropriate because method of detection of ³⁵P and ¹⁵N is different (³²P being a radioactive isotope while ¹⁵N is not radioactive but is the heavier isotope of Nitrogen). Even if ¹⁵N was radioactive then its presence would have been detected both inside the cell (¹⁵N incorporated as nitrogenous base in DNA) as well as in the supernatant because ¹⁵N would also get incorporated in amino group of amino acids in proteins). Hence the use of ¹⁵N would not give any conclusive results.

12. There is only one possible Sequence of amino acids when deduced from a given nucleotides. But multiple nucleotides sequence can be deduced from a single amino acid sequence. Explain this phenomena.
Answer. Some amino acids are coded by more than one codon (known as degeneracy of codons), hence on deducing a nucleotide sequence from an amino acid sequence, multiple nucleotide sequence will be obtained.
For example, lie has three codous: AUU, AUC AUA hence a depeptide. Met-Ile can have the following nucleotide sequence:
(i) AUG-AUU
(ii) AUG-AUC
(iii) AUG-AUA – And if, we deduce amino acid sequence the above nucleotide sequences, all the three will code for Met-IIe

13. A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do you think the statement is correct? Defend your answer.
Answer. The statement is correct. Because of degeneracy of codons, mutations at third base of codon, usually does not result into any change in phenotype. This is called silent mutations.

14. A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomena?
Answer. In the complete absence of expression of lac operon, permease will not be synthesised which is essential for transport of lactose from medium into the cells. And if lactose cannot be transported into the cell, then it cannot act as inducers, hence cannot relieve the lac operon from its repressed state.

15. How has the sequencing of human genome opened new windows for treatment of various genetic disorders. Discuss amongst your classmates.
Answer. Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and someday prevent the thousands of disorders that affect human beings.

• Besides providing clues to understanding human biology, learning about non-human organisms DNA sequences can lead to an understanding of their natural capabilities that can be applied toward solving challenges in health care, agriculture, energy production, environmental remediation. Deriving meaningful knowledge from the DNA sequences will define research through the coming decades leading to our understanding of biological systems. .
• This enormous task will require the expertise and creativity of tens of thousands of scientists from varied disciplines in both the public and private sectors worldwide. One of the greatest impacts of having the HG sequence may well be enabling a radically new approach to biological research. In the past, researchers studied one or a few genes at a time. With whole-genome sequences and new high-throughput technologies, we can approach questions systematically and on a much broader scale.
• They can study all the genes in a genome, for example, all the transcripts in a particular tissue or organ or tumor, or how tens of thousands of genes and proteins work together in interconnected networks to orchestrate the chemistry of life.

16. The total number of genes in humansds far less (< 25,000) than the previous estimate (upto 1,40,000 gene). Comment.
Answer. Repeated sequences make up very large portion of the human genome. Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution. Less than 2 per cent of the genome codes for proteins.

17. Now, sequencing of total genomes getting is less expensive day by the day. Soon it may be affordable for a common man to get his genome sequenced. What in your opinion could be the advantage and disadvantage of this development?
Answer. Advantages:
1.Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and some day prevent the thousands of disorders that affect human beings.
2. Besides providing clues to understanding human biology, learning about non-human organisms DNA sequences can lead to an understanding of their natural capabilities that can be applied toward solving challenges in health care, agriculture, energy production, environmental remediation.
3. Deriving meaningful knowledge from the DNA sequences will define research through the coming decades leading to our understanding of biological systems.
4. They can study all the genes in a genome, for example, all the transcripts in a particular tissue or organ or tumor, or how tens of thousands of genes and proteins work together in interconnected networks to orchestrate the chemistry of life.
Disadvantages:
1. It creates the problem of patenting of genes.
2. Persons becomes more careless.

18. Would it be appropriate to use DNA probes such as VNTR in DNA finger printing of a bacteriophage?
Answer. Bacteriophage does not have repetitive sequences such as VNTRs in its genome as its genome is very small and have all the coding sequence. DNA finger printing is not done for phages.

19. During in vitro synthesis of DNA, a researcher used 2′, 3′ – dideoxy cytidine triphosphate as raw nucleotide in place of 2′-deoxy cytidine. What would be the consequence?
Answer. Further polymerisation would not occur, as the 3′-OH on sugar is not there to add a new nucleotide for forming ester bond.

20. What background information did Watson and Crick have made available for developing a model of DNA What was their contribution?
Answer. Watson and Crick had the following information which helped them to develop a model of DNA.
(i) Chargaffs’ law suggesting A = T and C = G.
(ii) Wilkins and Rosalind Franklin’s work on DNA crystal’s X-ray diffraction studies about DNA’s physical structure.
(iii) Watson and crick proposed
a. How complementary bases may pair
b. Semi conservative replication and ‘
c. Mutation through tautomerism

21. What are the functions of (i) methylated guanosine cap, (ii) poly-A “tail” in a mature on RNA?
Answer. Methylated Guanine cap helps in binding of mRNA to smaller ribosomal sub-unit during initiation of translation. Poly-A tail provides longevity to mRNA’s life. Tail length and longevity of mRNA are positively correlated.

22. Do you think that the alternate splicing of exons may enable a structural gene to code for several isoproteins from one and the same gene? If yes, how? If not, why so?
Answer. Functional mRN A of structural genes need not always include all of its exons. This alternate splicing of exons is sex-specific, tissue-specific and even developmental stage-specific. By such alternate splicing of exons, a single gene may encode for several isoproteins and/or proteins of similar class. In absence of such a kind of splicing, there should have been new genes for every protein/isoprotein. Such an extravagancy has been avoided in natural phenomena by way of alternative splicing.

23. Comment on the utility of variability in number of tandem repeats during DNA finger printing.
Answer. Tandemness in repeats provides many copies of the sequence for fingerprinting, and variability in nitrogen base sequence in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

Long Answer Type Questions
1. Give an account of Hershey and Chase experiment. What did it conclusively prove? If both DNA and proteins contained phosphorus and sulphur do you think the result would have been the same?
Answer. The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952). They worked with viruses that infect bacteria called bacteriophages. The bacteriophage attaches to the bacteria and its genetic material then enters the bacterial cell. The bacterial cell treats the viral genetic material as if it was its own and subsequently manufacture^ more virus particles.

• Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria. They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not
• Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur. Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender.
• Radioactive phages were allowed to attach to E. Coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge. Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
• Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.
  

2. During the course of evolution why DNA was chosen over RNA as genetic material? Give reasons by first discussing the desired criteria in a molecule that can act as genetic material and in the light of biochemical differences between DNA and RNA.
Answer. A molecule that can act as a genetic material must fulfill the following criteria:
(i) It should be able to generate its replica (Replication):
• Because of rule of base pairing and complementarity, both the nucleic acids (DNA and RNA) have the ability to direct their duplications. The other molecules in the living system, such as proteins fail to fulfill first criteria itself.
(ii) It should chemically and structurally be stable.
• The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of the properties of genetic material was very evident in Griffith’s ‘transforming principle’ itself that heat, which killed the. bacteria,- at least did not destroy some of the properties of genetic material. This now can easily be explained in light of the DNA that the two strands being complementary if separated by heating come together, when appropriate conditions are provided.
• Further, 2′-OR group present at every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable. RNA is also now known to be catalytic, hence reactive. Therefore, DNA chemically is less reactive and structurally more stable when compared to RNA. Therefore, among the two nucleic acids, the DNA is a better genetic material. In fact, the presence of thymine at the place of uracil also confers additional stability to DNA.
(iii) It should provide the scope for slow changes (mutation) that are required
for evolution.
• Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutate at a faster rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.
(iv) It should be able to express itself in the form of ‘Mendelian Characters’.
• RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA, however, is dependent on RNA for synthesis of proteins. The protein synthesising machinery has evolved around RNA. The above discussion indicate that both RNA and DNA can function as genetic material, but DNA being more stable is preferred for storage of genetic information. For the transmission of genetic information, RNA is better.
• RNA being a catalyst was reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable.

3. Give an account of post transcriptional modifications of an eukaryotic mRNA. “
Answer. The primary transcripts (hn-RNA) contain both the exons and the introns and are non-functional. Hence, it is subjected to a process called splicing where the introns are removed and exons are joined in a defined order. Intron is the portion of gene which is transcribed but not translated. In prokaryotes hnRNA is absent so splicing in not required. hnRNA undergoes additional processing called as capping and tailing.
In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5′-end of hnRNA. In tailing, adenylate residues (200-300) are added at 3′-end in a template independent manner. It is the fully processed hnRNA, . now called mRNA, that is transported out of the nucleus for translation.

4. Discuss the process of translation in detail.
Answer. Translation refers to the process of polymerisation of amino acids to form a polypeptide. Ribosome is the site of protein synthesis. The amino acids are joined by a bond which is known as a peptide bond. Formation of a peptide bond requires energy. Therefore, in the first phase itself amino acids are activated in the presence of ATP and linked to their cognate tRNA—a process commonly called as activajjon or charging of tRNA or aminoacylation of tRNA to be more specific.

•  If two such charged tRNAs are brought close enough, the formation of peptide bond between them would be favoured energetically. The presence of a catalyst would enhance the rate of peptide bond formation. The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits; a large subunit and a small subunit.
• When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins. There are two sites in the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme- ribozyme) for the formation of peptide bond or polyribonucleotides.
• A translational unit in mRNA is the sequence of RNA that is flanked by the start codon (AUG) and the stop codon and codes for a polypeptide. An mRNA also has some additional sequences that are not translated and are referred as untranslated regions (UTR). The UTRs are present at both 5′- end (before start codon) and at 3′-end (after stop codon). They are required for efficient translation process.
• For initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA. The ribosome proceeds to the elongation phase of protein synthesis. During this stage, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along the mRNA.
• Amino acids are added one by one, translated into polypeptide sequences dictated by DNA and represented by mRNA. At the end, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.
  

5. Define an operon. Giving an example, explain an Inducible operon.
Answer. Lac Operon

• The elucidation of the lac operon was also a result of a close association between a geneticist, Francois Jacob and a biochemist, Jacque Monod in 1961. They were the first to elucidate a transcriptionally regulated system. In lac operon (here lac refers to lactose), a polycistronic structural gene is regulated by a common promoter and regulatory genes.
• Such arrangement is very common in bacteria and is referred to as operon. To name few such examples, lac operon, trp operon, ara operon, his operon, val operon, etc. Lac operon is a type of inducible operon. In inducible operon, presence of a chemical switch on the operon. The lac operon consists of one regulatory gene (the i gene – here the term i does not refer to inducer, rather it Is derived from the word inhibitor) and three structural genes (z, y and a).
• The z gene codes for beta-galactosidase (p-gal), which is primarily responsible for the hydrolysis of the disaccharide, lactose into its monomeric units, galactose and glucose. The y-gene codes for permease, which increases permeability of the cell to p-galactosides. The a-gene encodes a transacetylase.
• Hence, all the three gene products in lac operon are required for metabolism of lactose. In most other operons as well, the genes present in the operon are needed together to function in the same or related metabolic pathway. Operator gene is switched off in the presence of a repressor. RNA polymerase binds with the promoter gene. Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. Hence, it is termed as inducer.
• In the absence of a preferred carbon source such as glucose,; if lactose is provided in the grcfwth medium of the bacteria, the lactose is transported into the cells through the action of permease (Remember, a very low level of expression of lac operon has to be present in the cell all the time, otherwise lactose cannot enter the cells). The lactose then induces the operon in the following manner.
  
• The repressor of the operon is synthesised (all-the-time constitutively) from the i gene.
  Constitutive genes or housekeeping genes: These genes are constantly expressing themselves because their product is required by cell all the time. E.g.: Genes for ATPalse and glycolysis. The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon.
• In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase access to the promoter and transcription proceeds. Essentially, regulation of lac operon can also be visualised as regulation of enzyme synthesis by its substrate.

6. ‘There is a paternity dispute for a child’. Which technique can solve the problem? Discuss the principle involved.
Answer. DNA finger printing is used to solve the paternity dispute. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.
• The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A : T rich or G : C rich), length of segment, and number of repetitive units, the satellite DNA
• is classified into many categories, such as micro-satellites, mini-satellites etc. These sequences normally do not code for any proteins, but they form a large portion of human genome.
• These sequence show high degree of polymorphism and form the basis of DNA fingerprinting. Since DNA from every tissue (such as blood, hair- follicle, skin, bone, saliva, sperm etc.), from an individual show the same degree of polymorphism, they become very useful identification tool in forensic applications. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basis of paternity testing, in case of disputes. *
• The technique of DNA fingerprinting was initially developed by Alec
Jeffreys. Lalji Singh is called father of Indian DNA fingerprinting or DNA profiling or DNA typing. He used a satellite DNA as probe that shows very high degree of polymorphism. It was called as Variable Number of Tandem Repeats (VNTR).
• The technique, as used earlier, involved Southern blot hybridisation using radiolabelled VNTR as a probe. It included
(i) Isolation of DNA,
(ii) Digestion of DNA by restriction endonucleases,
(iii) Separation of DNA fragments by electrophoresis,
(iv) Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,
(v) Hybridisation using labelled VNTR probe, and
(vi) Detectionof hybridised DNA fragments by autoradiography.

7. Give an account of the methods used in sequencing the human genome.
Answer. Methodologies:
• The methods involved two major approaches. One approach focused on identifying all the genes that are expressed as RNA (referred to as Expressed Sequence Tags (ESTs). The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions (a term referred to as Sequence Annotation).
• For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes (recall DNA is a very long polymer, and there are technical limitations in sequencing very long pieces of DNA) and cloned in suitable host using specialised vectors. The cloning resulted into amplification of each piece of DNA fragment so that it subsequently could be sequenced with ease.
• The commonly used hosts were bacteria and yeast, and the vectors were called as BAC (bacterial artificial chromosomes), and YAC (yeast artificial chromosomes). The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger. Sanger is also credited for developing method for determination of amino acid’sequences in proteins. These sequences were then arranged based on some overlapping regions present in them.
• This required generation of overlapping fragments for sequencing. Alignment of these sequences was humanly not possible. Therefore, specialised computer based programs were developed. These sequences were subsequently annotated and were assigned to each chromosome. Another challenging task was assigning the genetic and physical maps on the genome. This was’-generated using information on polymorphism of restriction endonuclease recognition sites, and some repetitive DNA sequences known as microsatellites.

8. List the various markers that are used in DNA finger printing.
Answer. Different DNA marker systems such as Restriction Fragment Length Polymorphisms (RFLPs), Random Amplified Polymorphic DNAs (RAPDs), Amplified Fragment Length Polymorphisms (AFLPs), Simple Sequence Repeats (SSRs) which also called as microsatellites, Single Nucleotide Polymorphisms (SNPs) and others have been developed.

9. Replication was allowed to take place in the presence of radioactive deoxynucleotides precursors in E. coli that was a mutant for DNA ligase. Newly synthesised radioactive DNA was purified and strands were separated by denaturation. These were centrifuged using density gradient centrifugation. Which of the following would be a correct result?

Answer. (d)

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NCERT Exemplar Class 12 Biology Chapter 5 Principles of Inheritance and Variation are part of NCERT Exemplar Class 12 Biology. Here we have given NCERT Exemplar Class 12 Biology Chapter 5 Principles of Inheritance and Variation. https://www.cbselabs.com/ncert-exemplar-problems-class-12-biology-principles-inheritance-variation/

NCERT Exemplar Class 12 Biology Chapter 5 Principles of Inheritance and Variation

Multiple Choice Questions
Single Correct Answer Type
NCERT Exemplar Class 12 Biology Chapter 5

1. All genes located on the same chromosome
(a) Form different groups depending upon their relative distance
(b) Form one linkage group
(c) Will not form any linkage groups
(d) Form interactive groups that affect the phenotype.
Answer. (b) Linkage groups: A linkage group is a group of linked gene (on the same chromosome) and corresponds to the genome of organism, like human has 23 linkage groups, pea and Ngurospora has 7 linkage groups, Drosophila has 4 linkage groups.

2. Conditions of a karyotype 2n ± 1 and 2n ± 2 are called
(a) Aneuploidy (b) Polyploidy
(c) Allopolyploidy (d) Monosomy
Answer. (a) Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. Aneuploidy arises due to non-disjunction of homologous chromosome.
Aneuploidy is of four types:
1. Monosomy = 2n – 1
2. Nullisomy = 2n – 2
3. Trisomy = 2n + 1
4. Tetrasomy = 2n + 2

NCERT Exemplar Principles Of Inheritance And Variation Class 12

3. Distance between the genes and percentage of recombination shows
(a) a direct relationship (b) an inverse relationship
(c) a parallel relationship (d) no relationship
Answer. (a) Distance between the genes and percentage of recombination shows
direct relationship.

4. If a genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny, the disease is
(a) Autosomal dominant (b) Autosomal recessive
(c) Sex-linked dominant (d) Sex-linked recessive
Answer. (d) Sex-linked recessive disease: A genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny.
E.g.: 1. Haemophilia/Bleeder’s disease
2. Colour-blindness
3. Congenital night blindness
4. DMD (Duchenne’s Muscular Dystrophy)
5. G-6-P dehydrogenase deficiency

Principles Of Inheritance And Variation NCERT Exemplar Class 12

5. In sickle cell anaemia glutamic acid is replaced by valine. Which one of the following are triplets codes for valine? (a) GGG (b) AAG (c) GAA (d) GUG
Answer. (d) The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the betaglobin gene from GAG (Glutamic acid) to GUG (Valine).

6. Person having genotype IA IB would show the blood group as AB. This is because of
(a) Pleiotropy (c) Segregation
Answer. (b)

7. ZZ/ZW type of sex determination is seen in
(a) Platypus (b) Snails
(c) Cockroach (d) Peacock
Answer. (d)

NCERT Exemplar Class 12 Biology Solutions Chapter 5

8. A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?
(a) TT and Tt (b) Tt and Tt ’
(c) TT and TT (d) Tt and tt
Answer. (b)

9. In a dihybrid cross, if you get 9 : 3 : 3 : 1 ratio, it denotes that
(a) The alleles of two genes are interacting with each other
(b) It is a multigenic inheritance
(c) It is a case of multiple allelism
(d) The alleles of two’genes are segregating independently
Answer. (d) Based upon such observations on dihybrid crosses (crosses between plants differing in two traits) Mendel proposed a second set of generalisations that we call Mendel’s Law of Independent Assortment. The law states that “When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters”.

Principles Of Inheritance And Variation Class 12 NCERT Exemplar

10. Which of the following will not result in variations among siblings?
(a) Independent assortment of genes
(b) Crossing over
(c) Linkage
(d) Mutation
Answer. (c)
• Linkage is not result in variations among siblings.
• A sibling is one of two or more individuals having one or both parents in common. The emotional bond between siblings is often complicated and is influenced by factors such as parental treatment, birth order, personality, and personal experiences outside the family. Identical twins share 100% of their DNA. Full siblings are first-degree relatives and, on average, share 50% of their genes out of those that vary among humans. Half-siblings are second-degree relatives and have, on average, a 25% overlap in their human genetic variation.

11. Mendel’s Law of independent assortment holds good for genes situated on the
(a) Non-homologous chromosomes (b) Homologous chromosomes
(c) Extra nuclear genetic element (d) Same chromosome
Answer. (b) Mendel’s Law of independent assortment holds good for genes situated on the homologous chromosomes.

Principles Of Inheritance And Variation NCERT Exemplar Solutions Class 12

12. Occasionally, a single gene, may express more than one effect. The phenomenon is called
(a) Multiple allelism (b) Mosaicism
(c) Pleiotropy (d) Polygeny
Answer. (c) Occasionally, a single gene may express more than one effect. The phenomenon is called pleiotropy.

13. In a certain taxon of insects, some have 17 chromosomes and the others have 18
chromosomes. The 17 and 18 chromosome-bearing organisms are
(a) Males and females, respectively (b) Females and males, respectively (c) All males (d) All females
Answer. (a) The 17 and 18 chromosome-bearing organisms are males and females respectively because it comes under XO-type of sex determination.
Males have only 1 chromosome besides the autosomes whereas females have a pair of X chromosomes besides the autosomes.

14. The inheritance pattern of a gene over generations among humans is studied by the pedigree analysis. Character studied in the pedigree analysis is I equivalent to
(a) Quantitative trait . (b) Mendelian trait
(c) Polygenic trait (d) Maternal trait
Answer. (b) Character studied in the pedigree analysis is equivalent to Mendelian trait.

15. It is said that Mendel proposed that the factor controlling any character is discrete and independent. His proposition was based on the
(a) Results of F₃ generation of a cross
(b) Observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending.
(c) Self pollination of F₁  off springs
(d) Cross pollination of F₁  generations with recessive parental
Answer. (b) Mendel proposed that the factor Controlling any character is discrete and independent. His proposition was based on the observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending.

16. Two genes ‘A’ and ‘B’ are linked. In a dihybrid cross involving these two genes, the F1 heterozygote is crossed with homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation?
(a) 1 : 1 : 1: 1 (b) 9 : 3 : 3 : 1 (c) 3:1 (d) 1:1
Answer. (d) Two genes ‘A’ and ‘B’ are linked. In a dihybrid cross involving these two genes, the Fj heterozygote is crossed with homozygous recessive parental type (aa bb). The ratio of offspring in the next generation will be 1 :1.

17. In the F₂ generation of a Mendelian dihybrid cross, the number of phenotypes and genotypes are
(a) Phenotypes—4; genotypes—16 (b) Phenotypes—9; genotypes—4 (c) Phenotypes—4; genotypes—8 (d) Phenotypes—4; genotypes—9
Answer. (d) In the F₂ generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are 4, 9 respectively.

18. Mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group respectively. What would be the genotype of both mother and father?
(a) Mother is homozygous for ‘A’ blood group and father is heterozygous for ‘B’
(b) Mother is heterozygous for ‘A’ blood group and father is homozygous for ‘B’
(c) Both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively
(d) Both mother and father are homozygous for ‘A’ and ‘B’ blood group, respectively.
Answer. (c) See Answer 2 (Short Answer Type Questions)
Genotype of mother — I^A i
Genotype of father — I^B i
Both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively.

Very Short Answer Type Questions
1. What is the cross between the progeny of F₁  and the homozygous recessive parent called? How is it useful?
Answer. When a progeny of F₁  is crossed with the homozygous recessive parent, it is called test cross. Such a cross is useful to determine the genotype of an unknown, i.e., whether it is heterozygous, or homozygous dominant for the trait.

2. Do you think Mendel’s laws of inheritance would have been different if the
characters that he choose were located on the same chromosome?
Answer. If Mendel choose characters that were located on the same chromosome then Mendel would not find the law of independent assortment.

3. Enlist the steps of controlled cross pollination. Would emasculation be needed in a cucurbit plant? Give reasons for your answer.
Answer. Steps of controlled cross pollination: Selection of parents —> Emasculation —> Bagging —>Collection of pollen from male parent —>Dusting the pollen on stigma —>Re-bagging
Emasculation could not be needed in a cucurbit plant because it has unisexual flowers.

4. A person has to perform crosses for the purpose of studying inheritance of a few traits/characters. What should be the criteria for selecting the organisms?
Answer. Criteria for selecting the organism are:
(i) Shorter life span
(ii) To produce large number of progeny-
(iii) Clear differentiation of sex/traits
(iv) Many type of hereditary variations

5. The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes. Draw your conclusion on the basis of the pedigree.

Answer. The trait is auto some linked and recessive in nature. Both the parents are carrier (i.e., heterozygous) hence among offspring few show the trait irrespective of sex. The other off springs are either normal or carrier.

6. In order to obtain the. F₁  generation Mendel pollinated a pure-breeding tall plant with a pure breeding dwarf plant. But for getting the F₂  generation, he simply self-pollinated the tall F₁  plants. Why?
Answer. Genotype of 50% of the offspring would resemble one parent and the rest the other parent. All the F₁ , off springs of the cross are heterozygous so allowing self-pollination is sufficient to raise F₂  offspring. Also Mendel intended to understand the inheritance of the selected trait over generations.

7. “Genes contain the information that is required to express a particular trait.” Explain.
Answer. The genes present in an organism show a particular trait by way of forming certain product. This is facilitated by the process of transcription and translation (according to central dogma of genetics).

8. How are alleles of particular gene differ from each other? Explain its significance.
Answer. Alleles of a particular gene differ from each other on the basis of certain changes (i.e., mutations) in the genetic material (segment of DNA or RNA). Different alleles of a gene increases the variability or variation among the organisms.

9. In a monohybrid cross of plants with red and white flowered.plants, Mendel got only red flowered plants. On self-pollinating these F₁  plants got both red and white flowered plants in 3:1 ratio. Explain the basis of using RR and rr symbols to represent the geno type of plants of parental generation.
Answer. On crossing red and white flower only red colour flower appeared in the F₁  generation. But the white colour flower again appear in the F₂  generation
which is raised out of the F₁  individuals only. Mendel reasoned that there is a factor of each and every character. Accordingly, there has to be one factor (R) for red flower and other one factor (r) for white flower. In case, an organism possess only one copy of the gene then the possibility of reappearance of white flower in the F₂  generation of the given cross is not there. Also the ratio (3 : 1 of red and white) indicates that each organism must possess two copies of a particular gene.

10. For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.
Answer.

11. A, B, D are three independently assorting genes with their recessive alleles a, b, d, respectively. A cross was made between individuals of Aa bb DD genotype with aa bb dd. Find out the type of genotypes of the off spring produced.
Answer. The given cross is Aa bb dd X aa bb dd. Accordingly the type of offspring produced would .

12. In our society a woman is o|ten blamed for not bearing male child. Do you think it is right? Justify.
Answer. 50 per cent of sperms carry the X chromosome while the other 50 percent carry the Y. After fusion of the male and female gametes the zygote would carry either XX or YY depending on whether the sperm carrying X or Y fertilised the ovum. Sex of the baby is determined by the father and not by the mother.
It is unfortunate that in our society women are blamed for producing female children and have been ostracised and ill-treated because of this false notion.

13. Discuss the genetic basis of wrinkled phenotype of pea seed.
Answer. Wrinkled phenotype of pea seed is due to small grain size produced by double recessive allele (bb).

14. Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?
Answer. A diploid organism has only two alleles of a character, although it has more than two alleles. Multiple alleles can be found only when population studies are made.

15. How does a mutagen induce mutation? Explain with example.
Answer. A mutagen is a physical or chemical agent that changes the genetic material, usually DNA. Different mutagens act on the DNA differently. Powerful mutagens may result in chromosomal instability, causing chromosomal breakages and rearrangement of the chromosomes such as translocation, deletion, and inversion.
Ionizing radiations such as X-rays, gamma rays and alpha particles may cause DNA breakage and other damages. Ultraviolet radiations with wavelength above 260 nm are absorbed strongly by bases, producing pyrimidine dimers. Radioactive decay, such as 14C in DNA which decays into nitrogen.

Short Answer Type Questions
1. In a Mendelian monohybrid cross, the F₂  generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.
Answer. In a monohybrid cross, starting with parents which homozygous dominant and homozygous recessive, F₁ , would be heterozygous for the trait and would express the dominant allele. But in case of incomplete dominance, a monohybrid cross shows the result as follows.

Phenotypic ratio —> Red : Pink : White :: 1 : 2 : 1
Genotypic ratio —> RR : Rr: rr : .1:2:1
Here the genotypic and phenotypic ratios are the same. So, we can conclude that when genotypic and phenotypic ratios are the same, the alleles show incomplete dominance.

2. Can a child have blood group O if his parents have blood group ‘A’ and ‘B’? Explain.
Answer. If parents are heterozygous then child can have O blood group but if parents are homozygous then child cannot have O blood group.

3. What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?
Answer. Down’s syndrome is a human genetic disorder caused due to trisomy of chromosome no. 21. Such individuals are aneuploid and have 47 chromosomes (2n + 1). The symptoms include mental retardation, growth abnormalities, constantly open mouth, dwarfness etc. The reason for the disorder is the non-disjunction (failure to separate) of homologous chromosome of pair 21 during meiotic division in the ovum. The chances of having a child with Down’s syndrome increase with the age of the mother (+40) because ova are present in females. Since their birth and therefore older cells are more prone to chromosomal non-disjunction because of various physico-chemical exposures during the mother’s life-time.

4. How was it concluded that genes are located on chromosomes?
Answer. Morgan confirmed Mendelian laws of inheritance and the hypothesis that genes are located on chromosomes. Morgan had discovered that eye colour in Drosophila expressed a sex-linked trait. All first-generation offspring of a mutant white-eyed male and a normal red-eyed female would have red eyes because every chromosome pair would contain at least one copy of the X chromosome with the dominant trait.

5. A plant with red flowers was crossed with another plant with yellow flowers. If Fj showed all flowers orange in colour, explain the inheritance.
Answer. This is due to the incomplete dominance as the F, do not resemble either of the two parents and is in between the two.

6. What are the characteristic features of a true-breeding line?
Answer. A true-breeding line for a trait is one that, has undergone continuous self¬pollination or brother-sister mating, showing a stability in the inheritance of the trait for several generations.

7. In peas, tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. When a tall plant bearing red flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny in numbers mentioned against them: Tall, Red =138
Tall, White = 132
Dwarf, Red =136
Dwarf, White =128
Mention the genotypes of the two parents and of the four offspring types.
Answer. The result shows that the four types of offspring are in a ratio of 1:1:1:1. Such a result is observed in a test-cross progeny of a dihybrid cross.
The cross can be represented as:
Tall & Red (Tt Rr) x Dwarf & White (ttrr)

8. Why is the frequency of red-green colour blindness is many times higher in males than that in the females?
Answer. For becoming colourblind, the female must have the allele for it in both her X-chromosomes; but males develop colourblindness when their sole X-chromosome has the allele for it.

9. 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. Gene for colour blindness is X-chromosome linked, and sons receive their sole from their mother, not from their father. Male-to-male inheritances is not possible for X-linked traits in humans. In the given case the mother of the child must be a carrier (heterozygous) for colour blindness gene.

10. Discuss why Drosophila has been used extensively for genetical studies.
Answer. Morgan worked with the tiny fruit files, Drosophila melanogaster, which were found very suitable for such studies, as:

1. They could be grown on simple synthetic medium (ripe banana) in the laboratory.
2. They complete their life cycle in about two weeks.
3.  A single mating could produce a large number of progeny flies.
4. There was a clear differentiation of the sexes—the male and female flies are easily distinguishable.
5.  It has many types of hereditary, variations that can be seen with low power microscopes.

11. How do genes and chromosomes share: similarity from the point of view of genetical studies?
Answer. Similarities between Chromosomes and Genes:

1. Both occur in pairs.
2. Both segregate at the time of gamete formation such that only one of each pair is transmitted to a gamete.
3. Independent pairs segregate independently of each other in both.

12. What is recombination? Discuss the applications of recombination from the point of view of genetic engineering.
Answer. The formation of new combinations of genes, either by crossing over or independent assortment is called recombination. Alfred Sturtevant 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. Today genetic maps are extensively used as a starting point in the sequencing of whole genomes as was done in the case of the Human Genome Sequencing Project.

13. What is artificial selection? Do you think it affects the process of natural selection? How?
Answer. A process in the breeding of organisms by which the scientist chooses to only those forms having certain desirable characteristics is called artificial selection.

• Natural selection is the differential survival and reproduction of individuals due to differences in phenotype. In natural selection, the difference in reproductive success based on a particular trait is driven by natural processes like predators, weather conditions, and environmental constraints.
•  Artificial selection is imposed by humans on other organisms. It affects the process of natural selection through change in environmental conditions.

14. With the help of an example, differentiate between incomplete dominance and co-dominance.
Answer.

15. It is said, that the harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population. Why?
Answer. The harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population because SCA is a harmful condition which is also a potential saviour from malaria.
Those with the benign sickle trait possess a resistance to malarial infection. The pathogen that causes the disease spends part of its cycle in the red blood cells and triggers an abnormal drop in oxygen levels in the cell. In carriers, this drop is sufficient to trigger the full sickle-cell reaction, which leads to infected cells being rapidly removed from circulation and strongly limiting the infection’s progress. These individuals have a great resistance to infection and have a greater chance of surviving outbreaks. This resistance to infection is the main reason the SCA allele and SCA disease still exist. It is found in greatest frequency in populations where malaria was and often still is a serious problem.

Long Answer Type Questions
1. In a plant tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents work out a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?
Answer.

Yes, the ratio will deviate if the two genes are interacting with each other. In this condition the genes do not independently assort with each other. So, ratio will be changed from 9:3:3: 1.

2. (a) In humans, males are heterogametic and females are homogametic. Explain.
Are there any examples where males are homogametic and females heterogametic?
(b) Also describe as to who determines the sex of an unborn child?
Mention whether temperature has a role in sex determination.
Answer. (a) The term homogametic and heterogametic refer to the organism depending upon whether all the gametes contain one type of sex chromosome (Homo = same) or two different types of sex chromosomes (Hetero = different). Humans show XX/XY type of sex determination, i.e. Females contain two copies of X chromosome and males contain one X and one Y chromosome. Therefore, ova produced by females contain the same sex chromosome, i.e. X. On the other hand the sperms contain two different types of chromosomes,
i. e. 50% sperms have X and 50% have Y chromosome open from half the autosomes (Meiosis). Therefore, the sperms are different with respect to the composition of sex chromosome. In case of humans, females are considered to be homogametic while males are heterogametic. Yes, there are examples where males are homogametic and females are heterogametic. In some birds the mode of sex determination is denoted by ZZ (males) and ZW (females), (b) 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. In some animals like crocodiles, lower temperature favour hatching of female offsprings and higher temperatures lead to hatching of male off springs.

3. A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart.
Answer.

All daughters are normal visioned; 50% of sons are likely to be colour blind.

4, Discuss in detail the contributions of Morgan and Sturtervant in the area of genetics.
Answer. Experimental verification of the chromosomal theory of inheritance by Thomas Hunt Morgan (Father of experimental genetics) and his colleagues, led to discovering the basis .for the variation that sexual reproduction produced. Morgan worked with the tiny fruit files, Drosophila melanogaster, which were found very suitable for such studies.

• Morgan carried out several dihybrid crosses in Drosophila to study genes that were sex-linked. The crosses were similar to the dihybrid crosses carried out by Mendel in peas. For example, Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F, progeny.
• He observed that the two genes did not segregate independently of each other and the F2 ratio deviated very significantly from the. 9 : 3 : 3 : 1 ratio (expected when the two genes are independent). Morgan and his group knew that the genes were located on the X chromosome and saw quickly 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.
• Morgan attributed this due to the physical association or linkage 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.
• His student Alfred Sturtevant 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. Today genetic maps are extensively used as a starting point in the sequencing of whole genomes as was done in the case of the Human Genome Sequencing Project.

5. Define aneuploidy. How is it different from polyploidy? Describe the individuals having the following chromosomal abnormalities.
a. Trisomy of 21st chromosome
b. XXY
c. XO
Answer. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. – Difference between aneuploidy and polyploidy
1. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism and,this phenomenon is known as polyploidy.
2. Polyploidy occurs due to altering set of chromosome number such as 2n, 3n, 5n, whereas aneuploidy occurs due to altering particular chromosome or part of a chromosome such as 2n + 1 (trisomic) and 2n – 1 (monosomic).
3. Aneuploidy can be seen in human as genetic disorders; for example, Tuner syndrome, Klinefelter syndrome and Down syndrome, whereas polyploidy is common in plants.
(i) Down’s Syndrome (Mongolism)
• The cause of this genetic disorder-is the presence of an additional copy of the chromosome number 21 (trisomy of 21) due to non-disjunction of chromosomes during sperm or ova formation.
• The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease.
• Physical, psychomotor and mental development is retarded.
(ii) Klinefelter’s Syndrome
• This genetic disorder is also caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.
• Such an individual has overall masculine development, however, the feminine development (development of breast, i.e., Gynaecomastia) is also expressed. Such individuals are sterile male.
(iii) Turner’s Syndrome
Such a disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with XO. Such females are sterile as ovaries are rudimentary besides other features including lack of other secondary sexual characters.

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NCERT Exemplar Class 12 Biology Chapter 4 Reproductive Health are part of NCERT Exemplar Class 12 Biology. Here we have given NCERT Exemplar Class 12 Biology Chapter 4 Reproductive Health. https://www.cbselabs.com/ncert-exemplar-problems-class-12-biology-reproductive-health/

NCERT Exemplar Class 12 Biology Chapter 4 Reproductive Health

Multiple Choice Questions
Single Correct Answer Type
Reproductive Health NCERT Exemplar Class 12

1. The method of directly injecting a sperm into ovum is assisted by reproductive technology called
(a) GIFT (b) ZIFT (c) ICSI (d) ET
Answer. (c) Intra cytoplasmic sperm injection (ICSI) is a method in which embryo is formed in the laboratory in which a sperm is directly injected into the ovum.

2. Increased IMR and decreased MMR in a population will
(a) Cause rapid increase in growth rate
(b) Result in decline in grcfwth rate
(c) Not cause significant change in growth rate
(d) Result in an explosive population/exp
Answer. (b) Increased IMR and decreased MMR in a population will result in decline in growth rate.

3. Intensely lactating mothers do not generally conceive due to the
(a) Suppression of gonadotropins
(b) Hypersecretion of gonadotropins
(c) Suppression of gametic transport
(d) Suppression of fertilisation
Answer. will not lead to conception.

NCERT Exemplar Class 12 Biology Chapter 4

4. Sterilisation techniques are generally fool proof methods of contraception with least side effects. Yet, this is the last option for the couples because
i. It is almost irreversible.
ii. Of the misconception that it will- reduce sexual urge/drive
iii. It is a surgical procedure
iv. Of lack of sufficient facilities in many parts of the country
Choose the correct option:
(a) i and iii (b) ii and iii
(c) ii and iv (d) i, ii, iii and iv
Answer. (d) Sterilisation techniques are generally fool proof methods of contraception with least side effects. Yet, this is the last option for the couples because of (i),
(ii) , (iii) and (iv).

5. A national level approach to build up a reproductively healthy society was taken up in our country in (a) 1950s (b) 1960s (c) 1980s (d) 1990s
Answer. (a) A national level approach to build up a reproductively healthy society was taken up in our country in 1950s.

Reproductive Health NCERT Exemplar Solutions Class 12

6. Emergency contraceptives are effective if used within
(a) 72 hrs of coitus (b) 72 hrs of ovulation
(c) 72 hrs of menstruation (d) 72 hrs of implantation
Answer. (a) Combination or IUDs within 72 hours of coitus have been found to be very effective as emergency contraceptives as they could be used to avoid possible pregnancy due to rape or casual unprotected intercourse.

7. Choose the right one among the statements given below:
(a) IUDs are generally inserted by the user herself
(b) IUDs increase phagocytosis reaction in the uterus
(c) IUDs suppress gametogenesis
(d) IUDs once inserted need not be replaced
Answer. (b)

• Hormone- releasing IUDs: Progestasert, LNG-20
• IUDs increases phagocytes of sperm within the uterus

8. Following statements are given regarding MTP. Choose the correct options given below:
i. MTPs are generally advised during first trimester.
ii. MTPs are used as a contraceptive method.
iii. MTPs are always surgical.
iv. MTPs require the assistance of qualified medical personnel.
(a) i and iii (b) ii and iii (c) i and iv (d) i and ii
Answer. (c)

• MTPs are considered relatively safe during the first trimester (up to 12 weeks of pregnancy).
• 2nd trimester abortions are much more risky.
•  MTPs require the assistance of qualified medical personnel.

NCERT Exemplar Reproductive Health Class 12

9. From the sexually transmitted diseases mentioned below, identify the one which does not specifically affect the sex organs.
(a) Syphilis (b) AIDS (c) Gonorrhea (d) Genital warts
Answer. (b) Diseases or infections whieh are transmitted through sexual intercourse called sexually transmitted diseases (STD) or VD (Venereal diseases) or RTI (Reproductive tract infections).
Examples of STDs:

1. HIV (AIDS)
2. Hepatitis-B
3. Genital herpes
4. Chlamydiasis ,
5. Gonorrhoea
6. Genital warts
7.  Syphilis
8. Trichomoniasis

10. Condoms are one of the most popular contraceptives because of the following reasons:
(a) These are effective barriers for insemination
(b) They do not interfere with coital act
(c) These help in reducing the risk of STDs
(d) All of the above
Answer. (d) Condoms are barriers made of thin rubber/latex sheath. Condoms are used to cover the penis in male or vagina and cervix in the female, just before coitus so that the ejaculated semen would not enter into the female reproductive tract. This can prevent conception. Use of condoms has increased in recent years due to its additional benefit of protecting the user from contracting STDs and AIDS. Both male and female condoms are disposable and gives privacy to the user. ‘Nirodh’ is a popular brand of condom for the male. Condom is the most widely used contraceptive by males in India as it is cheap and easily available. It is sirrtple and effective and has no side effects.

Reproductive Health Class 12 NCERT Exemplar

11. Choose the correct statement regarding the ZIFT procedure:
(a) Ova collected from a female donor are transferred to the fallopian tube to facilitate zygote formation.
(b) Zygote is collected from a female donor and transferred to the fallopian tube
(c) Zygote is collected from a female donor and transferred to the uterus
(d) Ova collected from a female donor and transferred to the uterus
Answer. (b)

•  ZIFT (Zygote Intra Fallopian Transfer): The zygote or early embryos (up to 8 blastomeres) could then be transferred into the fallopian tube.
• Transfer of an ovum collected from a donor into the fallopian tube (GIFT: Gamete Intra Fallopian Transfer) of another female who cannot produce one but can provide suitable environment for fertilization and further development.

12. The correct surgical procedure as a contraceptive method is
(a) Ovariectomy (b) Hysterectomy
(c) Vasectomy (d) Castration
Answer. (c) Vasectomy and tubectomy are surgical procedure as a contraceptive method.

NCERT Exemplar Class 12 Biology Solutions Chapter 4

13. Diaphragms are contraceptive devices used by the females. Choose the correct option from the statements given below:
i. They are introduced into the uterus
ii. They are placed to cover the cervical region
iii. They act as physical barriers for sperm entry
iv. They act as spermicidal agents
(a) i and ii (b) i and iii (c) ii and iii (d) iii and iv
Answer. (c) Diaphragms, Cervical caps and Vaults: These are also barrier made of rubber that are inserted into the female reproductive tract to cover the cervix during coitus. They prevent conception by blocking the entry of sperms through the cervix. They are reusable. Spermicidal creams, jellies and foams are usually used along with these barrier to increase their contraceptive efficiency.

Very Short Answer Type Questions
1. Reproductive health refers only to healthy reproductive functions. Comment.
Answer. Reproductive health refers to a total well-being in all aspects of reproduction,
i.e., physical, behavioural, psychological, social and physiological.

2. Comment on the Reproductive and Child Health Care programme of the government to improve the reproductive health of the people.
Answer. Creating public awareness regarding reproduction related aspects and providing facilities to build up a healthy society with added emphasis on the health of mother and child are the basic aims of the RCH programmes.

3. The present population growth rate “in India is alarming. Suggest ways to check it.
Answer. The most important step to overcome this problem is to motivate smaller families by using various contraceptive methods.

• Government of India advertises in the media as well as in posters, showing a happy couple with two children with a slogan “Hum Do Hamare Do” (We two, our two).
• Many couples, mostly the young, urban working ones have even adopted a one child norm.
• Statutory raising of marriageable age of the female to 18 years and that of males to 21 years and incentives given to couples with small families are two of the other measure taken to tackle this problem.

4. STDs can be considered as self-invitedrdiseases. Comment.
Answer. STDs can be considered as self-invited diseases caused due to the conscious behaviour of person. STDs occur in those individuals that have unprotected sex with multiple or unknown partners.

5. Suggest the reproduction-related aspects in which counselling should be provided at the school level.
Answer.

• Introduction of sex education in schools should also be encouraged to provide right information to the young so as to discourage children from believing in myths and having misconceptions about sex-related aspects.
• Proper information about reproductive organs, adolescence and related changes, safe and hygienic sexual practices, sexually transmitted diseases (STD), AIDS, etc., would help people, especially those in the adolescent age group to lead a reproductively healthy life.

6. Mention the primary-aim of the “Assisted Reproductive Technology” (ART) programme.
Answer. The primary aim of the “Assisted Reproductive Technology” (ART) programme is the couples could be assisted to have children through certain special techniques.

7. What is the significance of progesterone-estrogen combination as a contraceptive measure?
Answer. Oral administration of small doses of progesterone-estrogen combination is a contraceptive measure which inhibit ovulation and implantation as well as alter the quality of cervical mucus to prevent/retard the entry of sperms.

8. Strict conditions are to be followed in medical termination of pregnancy (MTP) procedures. Mention two reasons.
Answer. Government of India legalised MTP in 1971 with some strict conditions to avoid its misuse.
Such restrictions are all the more important to check indiscriminate and illegal female foeticides which reported to be high in India.

9. Males in whom testes fail to descend to the scrotum are generally infertile. Why?
Answer. If the testes fail to descend to the scrotum, gametogenesis could be inhibited, the process of spermatogenesis require a marginally lesser ambient temperature than that in the abdominal cavity.

10. Mention two advantages of lactational amenorrhea as a contraceptive method.
Answer. (i) There is no ovulation and therefore the menstrual cycle do not take place.
(ii) As no medicines or devices are used in these methods, side effects are almost nil. .

Short Answer Type Questions
1. Suggest some important steps that you would recommend to be taken to improve the reproductive health standards in India.
Answer.

• Creating awareness among people about various reproduction related aspects and providing facilities and support for building up a reproductively healthy society are the major tasks under these programmes. With the help of audio-visual and the print-media, governmental and non-governmental agencies have taken various steps to create awareness among the people about reproduction-related aspects.
• Parents, other close relatives, teachers and friends, also have a major role in the dissemination of the above information. Educating people, especially fertile couples and those in marriageable age group, about available birth control options, care of pregnant mothers, post-natal care of the mother and child, importance of breast feeding, equal opportunities for the male and the female child, etc., would address the importance of bringing up socially conscious healthy families of desired size.

2. The procedure of GIFT involves the transfer of female gamete to the fallopian tube. Can gametes be transferred to the uterus to achieve the same result? Explain.
Answer. The uterine environment is not congenial 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.

3. Copper ions-releasing IUDs are more efficient than non-medicated methods. Why?
Answer. Unlike non-medicated IUEs, the copper releasing IUDs releases Cu ions that suppress sperm motility and fertilising capacity of sperms.

4. What are the probable factors that contributed to population explosion in India?
Answer. Probable reasons for population explosion are:

1.  A rapid decline in death rate.
2. Decline in MMR (Maternal mortality rate).
3. Decline in IMR (Infant mortality rate).
4. An increase in number of people in reproducible age.

5. Briefly explain IVF and ET. What are the conditions in which these methods are advised? 
Answer. IVF and ET refer to In Vitro Fertilisation and Embryo Transfer. Gametes from the male and female are collected hygienically and induced to fuse in the laboratory set up under simulated conditions. The zygote formed is collected and is introduced into the uterine region of a host dr surrogate mother at an appropriate time (secretory phase). Early embryos (up to 8 cell) are generally transferred to the fallopian tube whereas embryos with more than 8 cells are transferred to the uterus.

6. What are the advantages of natural methods of contraception over artificial methods?
Ans. Advantages of natural methods of contraception over artificial methods:

1. As no medicines or devices are used in these methods, side effects are almost nil.
2.  There is no surgical intervention, so natural method is reversible in nature.

7. What are the conditions in which medical termination of pregnancy is advised?
Answer.

• MTP is used to get rid of unwanted pregnancies either due to casual unprotected intercourse or failure of the contraceptive used during coitus or rapes.
• MTPs are also essential in certain cases where continuation of the pregnancy could be harmful or even fatal either to the mother or to the foetus of both.

8. Comment on the essential features required for an ideal contraceptive.
Answer. An ideal contraceptive should be:

1. User-friendly
2. Easily available
3. Effective
4. Reversible
5. No or least side-effects.
6. Should not interfere with the sexual drive, desire or sexual act of the user

9. All reproductive tract infections RTJs are STDs, but all STDs are not RTIs. Justify with example.
Answer. The common STDs are gonorrhea, syphilis, genital herpes, chlamydiasis, hepatitis-B, AIDs etc. 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 are transmitted through sexual contact and are also infections of the reproductive tract.

Long Answer Type Questions
1. What are the Assisted Reproductive Techniques practised to help infertile couples? Describe any three techniques.
Answer. Inability to conceive or produce children even after 2 years of unprotected sexual cohabitation is called infertility. In India, often the female is blamed ‘ for the couple being childless, but more often than not the problem lies in male partner.
Specialised health care units (infertility, clinics) could help in diagnosis and corrective treatment of some of these disorders and enable these couples to have children. However, where such corrections are not possible, the couples could be assisted to have children through certain special techniques commonly known as ART (Assisted Reproductive Technologies).
1. Test Tube Baby Programme
In vitro fertilisation (IVF) followed by embryo transfer (ET) is a method to treat infertility and commonly known as the ‘Test tube baby’ programme. IVF-Fertilisation outside the body in almost similar conditions as that in the body. In this method ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under simulated conditions in the laboratory.

• ZIFT (Zygote Intra Fallopian Transfer): The zygote or early embryos (up to 8 blastomeres) could then be transferred into the fallopian tube.
• IUT (Intra Uterine Transfer): Embryos with more than 8 blastomeres could be transferred into the uterus, to complete its further development.
• Embryos formed by in vivo fertilization (fusion of gametes within the female) also could be used for such transfer to assist those females who cannot conceive.

2. GIFT
Transfer of an ovum collected from a donor into the fallopian tube (GIFT; Gamete Intra Fallopian Transfer) of another female who cannot produce one but can provide suitable environment for fertilization and further development.
3. AI Technique

• Infertility cases either due to inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates, could be corrected by artificial insemination (AI) technique.
• In this technique, the semen collected either from the husband or a healthy donor is artificially introduced, either into the vagina or into the uterus (IUI-Intra Uterine Insemination) of the female.

2. Discuss the mode of action and advantages/disadvantages of hormonal contraceptives.
Answer.

• Oral administration of small doses of either progestagens or progestogen- estrogen combinations is another contraceptive method used by the females.
• Most important component of oral contraceptive is progesterone. They are used in the form of tablets and hence are popularly called the pills. Pills have to be taken daily for a period of 21 days starting preferably within the first 5 days of menstrual cycle. After a gap of 7 days (during which menstruation occurs) it has to be repeated in the same pattern.
•  Oral contraceptive pills inhibit ovulation and implantation as well as alter (change) the quality of cervical mucus to prevent or retard entry of
  sperms. Progesterone present in OCP- is meant for checking ovulation. Pills are very effective with lesser side effects and are well accepted by the females. Combination or IUDs within 72 hours of coitus have been found to be very effective as emergency contraceptives as they could be used to avoid possible pregnancy due to rape or casual unprotected intercourse.

3. STDs are a threat to reproductive health. Describe any two such diseases and suggest preventive measures.
Answer. Diseases or infections which are transmitted through sexual intercourse called sexually transmitted diseases (STD) or VD (Venereal diseases) or RTI
(Reproductive tract infections).
Examples of STDs:
(i) HIV (AIDS)
(ii) Hepatitis-B
(iii) Genital herpes
(iv) Chlamydiasis
(v) Gonorrhoea
(vi) Genital warts
(vii) Syphilis
(viii) Trichomoniasis
AIDS
• AIDS is caused by HIV virus (Human Immuno deficiency virus or Human T-cell leukemia virus). There is always a time-lag between the infection and appearance of AIDS symptoms (incubation period). Incubation period may vary from a few month to many years (usually 5-10 years).
• ARC (AIDS Related Complex) is a mild or initial form of AIDS which develop after a few mon|h of infection. In AIDS patient, a reduction of 10% weight indicates ARC. After infection, HIV enters in macrophages where RNA of virus replicates to form viral DNA with the help of enzyme reverse transcriptase. Viral DNA incorporated into host cell DNA and directs the infected cell to produce new virus particles. The macrophages continue to produce virus therefore macrophages are called HIV Factory.
• Simultaneously HIV enters into helper T-lymphocytes (TH) or T4 replicates and produce progeny viruses. The progeny viruses released in the blood attack other helper T-lymphocytes. This is repeated leading to a progressive decrease in the number of helper T-lymphocytes in the body of infected person. During this period person suffers from: (i) Bouts of fever, (ii) Diarrhoea, (iii) Weight loss
• Due to the decrease in TH cells, person start suffering from infections that could have been otherwise overcome such as those due to bacteria especially Mycobacterium, virus, fungi and even parasites like Toxoplasma. The patient becomes so immuno-deficient that he/she is unable to protect himself/herself against these infections. Diagnostic test for AIDS is ELISA (Enzyme Linked Immuno Sorbent Assay).Werstem blot is used as confirmatory or supplemental test for AIDS. Routine test of AIDS is PCR.
Hepatitis-B
Hepatitis-B viras is horizontally transmitted by blood transfusions, contaminated needles, body fluids like semen, saliva, sweat, tear and breast milk. Hepatitis-B is also transmitted vertically from infected mother to foetus through placenta. For prevention and control hepatitis-B vaccine is now available. Hepatitis-B vaccine was developed by Blumberg, for which he was awarded Nobel Prize in 1976.

4. Do you justify the statutory ban on amniocentesis in our country? Give reasons.
Answer.

• Intentional or voluntary termination of pregnancy before full term is called MTP or induced abortion. Nearly 45 to 50 millions MTPs are performed in a year all over the world which accounts to l/5th (20%) of the total number of conceived pregnancies in a year. MTP has a significant role in decreasing the population though it is not meant for this purpose.
•  Government of India legalized MTP in 1971 with some strict conditions
  to avoid its misuse. Such restrictions are more important to check indiscriminate and illegal female foeticides which reported to be high in India. ‘ .
• MTP is used to get rid of unwanted pregnancies either due to casual unprotected intercourse or failure of the contraceptive used during coitus or rapes. MTPs are also essential in certain cases where continuation of the pregnancy could be harmful or even fatal either to the mother or to the foetus or both. MTPs are considered relatively safe during the first trimester (up to 12 weeks of pregnancy). 2nd trimester abortions are much more risky. Another dangerous trend is the misuse of amniocentesis to determine the sex of unborn child.
• Amniocentesis is a foetal sex determination test based on the chromosomal pattern in the amniotic fluid surrounding the developing embryo. Amniocentesis is employed for determining hereditary abnormality in embryo. Statutory ban on amniocentesis for sex-determination to legally Check: (i) Increasing female foeticides, (ii) Massive child immunisation.

5. Enumerate and describe any five reasons for introducing sex education to school-going children.
Answer. Proper information about reproductive organs-physiology and its functioning; discourage myths and misconceptions about sex-related aspects; knowledge about safe and hygienic sexual practices; adolescence and related changes, prevention of STDs, AlDs etc.

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