NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases

NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases

Multiple Choice Questions

Question 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 could be defined as a state of complete physical, mental and social well¬being. When people are healthy, they are more efficient at work. This increases productivity and brings economic prosperity. Health also increases longevity of people and reduces infant and maternal mortality.

Question 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, helminths, etc. could cause diseases in pants and animals including humans. Such disease causing organisms are called pathogens. Most parasites are therefore pathogens as they cause harm to the host by Jiving in (or on) them. The pathogens can enter our body by various means, multiply and interfere with normal vital activities, resulting in morphological and functional damage.

Question 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) :Salmonella typhi is a pathogenic bacterium which causes typhoid fever in human beings. These pathogens generally enter the small intestine through food and water contaminated with them and migrate to other organs through blood. Sustained high fever (39 to 40°C), weakness, stomach pain, constipation, headache and loss of appetite are some of the common symptoms of the disease. Intestinal perforation and death many occur in severe cases. Typhoid fever could be confirmed by Widal test. Widal test is an agglutination test for the presence of antibodies against the Salmonella organisms that cause typhoid fever.

Question 4.
Diseases are broadly grouped into infectious and non-infectious diseases. In the list given below, identify the infectious diseases.
(1) Cancer
(2) Influenza
(3) Allergy
(4) Small pox
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iv)
Answer:
(d) : Diseases can be broadly grouped into infectious and non-infectious. Diseases which are easily transmitted from one person to another, are called infectious diseases. Influenza and small pox are infectious diseases which are spreaded by direct contact, inhalation and droplet infections. Cancer and allergy are non-communicable diseases, as these diseases remain confined to the person who suffer from them. They are not transmitted from infected person to a healthy person.

Question 5.
The sporozoites that cause infection, when a female Anopheles mosquito bites a person, are formed in
(a) liver of the person
(b) RBCs of mosquito
(c) salivary glands of mosquito
(d) intestine of mosquito.
Answer:
(d) : When a female Anopheles mosquito bites an infected person, gametocyte of Plasmodium enter the mosquito’s gut along with blood meal and fuse to form zygote which elongates and becomes motile, called ookinets. Ookinetes bore through the wall of gut and change to oocysts. Inside oocyst, sporozoites are formed. Mature infective stage(sporozoite) are released in the body cavity of mosquito and migrate to salivary gland of mosquito. When these mosquitoes bite a human, the sporozoites are introduced into his/her body, thereby initiating the disease in the host’s body.

Question 6.
The disease chikungunya is transmitted by
(a) houseflies
(b) Aedes mosquitoes
(c) cockroach
(d) female Anopheles.
Answer:
(b) : Chikungunya is caused by chikungunya virus. The virus was first isolated from human patients and Aedes aegypti mosquitoes from Tanzania in 1952. The disease is transmitted by the bite of Aedes aegypti mosquito.

Question 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, sore throat, headache.
(d) High fever, weakness, stomach pain, loss of appetite and constipation.
Answer:
(a) : Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for the disease pneumonia in humans which infects the alveoli (air filled sacs) of the lungs. As a result of the infection, the alveoli get filled with fluid leading to severe problems in respiration. The symptoms of pneumonia include fever, chills, cough and headache.

Question 8.
The genes causing cancer are
(a) structural genes
(b) expressor genes
(c) oncogenes
(d) regulatory genes.
Answer:
(c) : Cancer is one of the most serious medical problems caused by the abnormal cell growth and proliferation due to the loss of regulation. Cancer-causing genes are oncogenes. Oncogene results from mutation of a normal gene. It is capable of both initiation and continuation of malignant transformation of normal cells.

Question 9.
In malignant tumours, the cells proliferate, grow rapidly and move to other parts of the body to form new tumours. This stage of disease is called
(a) metagenesis
(b) metastasis
(c) teratogenesis
(d) mitosis.
Answer:
(b) : There are two major types of tumours with respect to overall form or growth pattern. If the tumour cells remain in place to form a compact mass, the tumour is benign. In contrast, cells from malignant or cancerous tumours can actively spread, throughout the body in a process known as metastasis, often by floating in the blood and establishing secondary tumours.

Question 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 pros-perous
(c) the patient shows behavioural and social maladjustment
(d) he does not take interest in sports.
Answer:
(c) : If a patient shows behavioural and social maladjustment then, he cannot be considered as healthy. Health is the condition in which the organism (and all of its parts) performs its vital functions normally. It is a state of physical, social and mental well-being and not merely the absence of disease.

Question 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 antibodies 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)
(iv) (d) (i) and (iii)
Answer:
(b) : Rheumatoid arthritis is a disorder of the body’s defence mechanisms in which an immune response is elicited against its own tissues, which are thereby damaged or destroyed i.e., autoimmune disease.
Rheumatoid arthritis is the second most common form of arthritis (after osteoarthritis). It typically involves the joints of the fingers,
wrists, feet, and ankles, with later involvement of the hips, knees, shoulders, and neck. It is a disease of the synovial lining of joints; the joints are initially painful, swollen, and stiff and are usually affected symmetrically. As the disease progresses the ligaments supporting the joints are damaged and there is erosion of the bone, leading to deformity of the joints. Tendon sheath can be affected, leading to tendon rupture.

Question 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) : AIDS is a result of an infection by the human immunodeficiency virus (HIV) :
HIV can be transmitted by the following ways.

  • Transfusion of infected blood.
  • Use of contaminated needles and syringes to inject drugs or vaccines.
  • Use of contaminated razors.
  • Use of contaminated needles for boring pinnae.
  • Sexual intercourse with an infected partner without a condom.
  • From infected mother to child through placenta.
  • Artificial insemination.
  • Organ transplantation.

Question 13.
‘Smack’is a drug obtained from the
(a) latex of Papaver somniferum
(b) leaves of Cannabis sativa
(c) flowers of Datura
(d) fruits of Erythroxylum coca.
Answer:
(a) : Heroin commonly called smack is chemically diacetylmorphine which is a white, odourless, bitter crystalline compound. This is obtained by acetylation of morphine, which is extracted from the latex of poppy plant (Papaver somniferum).

Question 14.
The substance produced by a cell in viral infection that can protect other cells from further infection is
(a) serotonin
(b) colostrum
(c) interferon
(d) histamine.
Answer:
(c) : Interferons are glycoproteins released by living cells in response to viral attack, and make the surrounding cells resistant to viral infection by inhibiting multiplication of viral particles.
Interferons are divided into three groups based on the cell of origin, namely leucocyte (alpha interferon), fibroblast (beta interferon) and lymphocyte (gamma interferon). They are cytokine (chemical mediators) barriers.

Question 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) : Cell mediated immune response is responsible for graft rejection. Tissue matching and blood group matching are essential before undertaking any graft/transplant and even after this the patient has to take immuno¬suppressants throughout his/her life as body is able to differentiate ‘self’ from ‘non-self’.

Question 16.
Antibodies present in colostrum which protect the new born from certain diseases is of
(a) IgGtype
(b) IgA type
(c) IgDtype
(d) IgEtype.
Answer:
(b) : IgA antibody is the major antibody present in the colostrum. It protects the infant from inhaled and ingested pathogens. It is the second most abundant class of immunoglobulins, constituting about 13% of serum immunoglobulins.

Question 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 contains a large number of chemical substances including nicotine, an alkaloid. Nicotine stimulates adrenal gland to release adrenaline and nor-adrenaline into blood circulation, both of which raise blood pressure and increase heart rate.

Question 18.
Anti venom against snake poison contains
(a) antigens
(b) antigen-antibody complexes
(c) antibodies
(d) enzymes.
Answer:
(c) : In case of snakebite, the injection which is given to a patient, contains preformed antibodies against the snakevenom. This type of immunisation is called passive immunisation.

Question 19.
Which of the following is not a lymphoid tissue?
(a) Spleen
(b) Tonsils
(c) Pancreas
(d) Thymus
Answer:
(c) : Lymphoid organs are the organs where origin and/or maturation
and proliferation of lymphocytes occur. The primary lymphoid organs are bone marrow and thymus, where immature lymphocytes differentiate into antigen – sensitive lymphocytes. After maturation the lymphocytes migrate to secondary lymphoid organs like spleen, lymph nodes, tonsils, Peyer’s patches of small intestine and mucosa associated lymphoid tissue (MALT). The secondary lymphoid organs provide the sites for interaction of lymphocytes with the antigen, which then proliferate to become effector cells. Pancreas is not a lymphoid tissue.

Question 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) : The thymus is a lobed organ located near the heart and beneath the breastbone. The thymus is quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained, it reduces to a very small size.

Question 21.
Haemozoin is a
(a) precursor of haemoglobin
(b) toxin released from Streptococcus infected cells.
(c) toxin released from Plasmodium infected cells.
(d) toxin released from Haemophilus infected ‘cells.
Answer:
(c) : Plasmodium enters the human body as sporozoites (infectious form) through the bite of infected female Anopheles mosquito. The parasites initially multiply within the liver cells and then attack the red blood cell (RBCs) resulting in their rupture. The rupture of RBCs is associated with release of a toxic substance, haemozoin, which is responsible for the chill and high fever recurring every three to four days.

Question 22.
One of the following is not the causal organism for ringworm.
(a) Microsporum
(b) Trichophyton
(c) Epidermophyton
(d) Macrosporum
Answer:
(d) : Many fungi belonging to the genera Microsporum, Trichophyton and Epidermophyton are responsible for ringworms which is one of the most common infectious diseases in man. Appearance of dry, scaly lesions on various parts of the body such as skin, nails and scalp are the main symptoms of the disease.

Question 23.
A person with sickle cell anaemia is
(a) more prone to malaria
(b) more prone to typhoid
(c) less prone to malaria
(d) less prone to typhoid.
Answer:
(c) : Sickle cell trait protects against malaria. Malarial parasite is unable to penetrate the erythrocyte membrane of sickle shaped erythrocytes, reducing the number of parasites that actually infect the host thus conferring some protection against the disease.

Very Short Answer Type Questions

Question 1.
Certain pathogens are tissue/organ specific. Justify the statement with suitable examples.
Answer:
Some pathogens attack specific organs or tissues, such as :

  1. Vibrio cholerae – It is a bacterium which causes cholera. It attacks digestive tracts and intestine and spreads through contaminated food and drinks.
  2. Streptococcus pneumoniae – It is a bacterium which causes pneumonia. It is a serious disease of lungs characterised by
    accumulation of mucus/fluid in alveoli and bronchioles hence, breathing becomes difficult.

Question 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) The patient must be suffering from Acquired Immune Deficiency Syndrome (AIDS). .
(b) The causative agent is Human Immuno-deficiency Virus (HIV).
(c) Virus affects helper T cells of the body.

Question 3.
Where are B-cells and T-cells formed? How do they differ from each other?
Answer:
B-cells and T-cells are lymphocytes which are responsible for immune response in the body. Both B-cells and T-cells are formed in bone marrow.
Differences between B-cells and T-cells are

B-cells T-cells
(1) Bone marrow is the site of both formation and maturation. They are produced in bone marrow but mature in thymus.
(2) B-cells are responsible for humoral or antibody mediated immunity. T-cells are responsible for cell mediated immunity.

Question 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:
(c) : Microsporum is a fungus and causes ringworm. It attacks the hair and skin.Filariasis is caused by Wuchereria bancrofti, a helmirith, which results in swelling of limbs, scrotal sacs and breasts.

Question 5.
What would happen to immune system, if thymus gland is removed from the body of a person?
Answer:
T-lymphocytes get differentiated in thymus gland and are responsible for cell mediated immunity (CMI). If thymus gland is removed, then CMI will weaken and person will become more susceptible to infectious diseases.

Question 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:
Physiological barriers of innate immunity protect the body against the pathogens which enter the gut along with food. These barriers are as follows:

  1. Saliva contains lysozyme which kills the microorganisms that are not the normal inhabitants of the buccal cavity and come with food and drinks.
  2. Bile, a bitter alkaline secretion of the liver, checks the growth of foreign bacteria on semidigested food, the chyme, in the intestine.
  3. Acidity of gastric juice kills most of the microorganisms entering the body through digestive tract.

Question 7.
Why is mother’s milk considered the most appropriate food for a new born infant?
Answer:
Mother’s milk (colostrum) contains all the necessary nutrients needed by the newborn and has abundant antibodies (IgA) which provide passive immunity and protect the infant for few months after birth.

Question 8.
What are interferons? How do interferons check infection of new cells?
Answer:
Interferons are cytokine barriers of innate immunity. These are proteins secreted by virus infected cells which protect non- infected cells from further viral infections.

Question 9.
In the figure, structure of an antibody molecule is shown. Name the parts A, B and C
NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases 1
Answer:
A – Variable region of heayy chain
B – Constant region of light chain .
C – Disulphide bond

Question 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:
Withdrawal symptoms shown by an addicted person when not provided with drug or alcohol includes anxiety, nausea, muscle twitch, sweating, nervousness and epilepsy etc.

Question 11.
Why is it that during changing weather, one is advised to avoid closed, crowded and airconditioned places like cinema halls etc.?
Answer:
It is advised to avoid overcrowded places during changing weather because chances of getting infections are very high during this time. This is because changing seasons are the time when infectious agents are more prevalent as moist condition favours pathogen growth. Moreover, people are more vulnerable as their body system is busy in adapting to the changing environmental conditions.

Question 12.
The harmful allele of sickle cell anaemia has not been eliminated from human population. Such afflicted people derive some other benefit. Discuss.
Answer:
Sickle cell anaemia is caused due to a recessive allele in haemoglobin. It is lethal
in homozygous (HbsHbs) condition and the person dies in early age but in heterozygous condition (HbAHbs) person is carrier of disease. Despite having harmful effect, the allele for sickle-cell anaemia continues to persist in human population because it has survival value in malaria infested areas like tropical Africa. Heterozygous individuals show resistance to malarial infection, because malarial parasite is unable to penetrate the membrane of sickle shaped erythrocytes.

Question 13.
Lymph nodes are secondary lymphoid organs. Explain the role of lymph nodes in our immune response.
Answer:
The lymph nodes are small solid structures located at different points along the lymphatic system. Lymph nodes serve to 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.

Question 14.
Why is an antibody molecule represented as H2L2?
Answer:
Antibody molecule is made up of four polypeptide chains-2 Heavy (H) and 2 Light (L), therefore it is represented as H:L2.

Question 15.
What does the term ‘memory’ of the immune system mean?
Answer:
Memory response is a feature of acquired immunity. It develops during first encounter between specific foreign agent and body’s immune system. Second encounter with the same pathogen generates quicker and heightened immune response due to activated memory cells.

Question 16.
lf a patient is advised anti retroviral therapy, which infection is he suffering from? Name the causative organism.
Answer:
The patient must be suffering from Required Immune Deficiency Syndrome (AIDS) caused by Human Immunodeficiency Virus (HIV), which is a member of retrovirus group

short Answer Type Questions

Question 1.
Differentiate between active immunity and passive immunity.
Answer:
Differences between active immunity and passive immunity are as follows:

Active immunity Passive immunity
(1) It develops in response to vaccine or infection. It develops when readymade antibodies are injected from outside.
(2) It is long lasting. It lasts for short period only.
(3) Antibodies are produced within the body. Antibodies are injected from outside.
(4) It has no side effects. It may cause reaction.
(5) Immunity is not immediate. A time lapse occurs for its development. Immunity develops immediately

Question 2.
Differentiate between benign tumour and malignant tumour.
Answer:
Difference between benign tumour and malignant tumour are as follows:

Benign tumour Malignant
(1) It is non-cancerous tumour It is cancerous tumour.
(2) It does not show metastasis and remains confined to the affected organ. It shows metastasis and spreads to other organs of the body.
(3) Rate of growth is slow. It shows definite growth. Rate of growth is rapid. It shows indefinite growth.

Question 3.
Do you consider passive smoking is more dangerous than active smoking? Why?
Answer:
Passive smoking is inhalation of air containing tobacco smoke exhaled by an active smoker. Passive smoker is at higher risk as compared to an active smoker, because active smoker might inhale 10% of the smoke, but passive smoker inhales 90% of the smoke.

Question 4.
“Prevention is better than cure”. Comment.
Answer:
Prevention is always better than cure.This is because there are some diseases which have detrimental effects on our body and lives. Certain diseases like cancer and AIDS etc., are fatal. They can be avoided if our health systems focus on prevention of such diseases rather than concentrating on treatment methods. Many people can be saved from contracting diseases. Similar conditions apply for communicable, infectious diseases. If preventive measures are undertaken, their spread, can be checked and controlled easily.

Question 5.
Explain any three preventive measures to control microbial infections.
Answer:
Maintenance of personal and public hygiene is important to control microbial infections. Some of the measures that can be undertaken are as follows:

  1. Eradication of vectors and destroying their breeding sites
  2. Use of mosquito nets and repellents.
  3. Proper disposal of waste.
  4. Periodic cleaning of water reservoirs.
  5. Checking water stagnation and garbage accumulation.
  6. Implementation of vaccination and immunisation programmes for diseases.

Question 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?
NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases 2
Answer:
(a)
(1) – Viral DN A is produced by reverse transcriptase.
(2) – New viral RNA is produced by the infected cell.
(b) HIV is called retrovirus because viral DNA is produced from viral RNA by reverse transcription.
(c) Infected cell survives but T-lymphocytes decrease in number due to replication and release of virus.

Question 7.
“Maintenance of personal and public hygiene is necessary for prevention and control of many infectious diseases”. Justify the statement giving suitable examples.
Answer:
Maintenance of personal and public hygiene is very important for prevention and control of many infectious diseases because diseases spread through contaminated food, water, air and through vectors. Measures for personal hygiene include keeping the body clean, consumption of clean drinking water, food, etc.
Public hygiene includes proper disposal of waste and excreta, periodic cleaning and disinfection of water reservoirs, pools, and tanks and observing standard practices of hygiene in public catering.
In case of air-borne diseases such as pneumonia and common cold, close contact with the infected persons or their belongings should be avoided. Vector borne diseases can be controlled by eliminating 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, etc.

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

Name of the Disease Causative organism Symptoms
(1) Ascariasis Ascaris
(2) — Trichophyton Appearance of dry, scaly lesions on various parts of the body
(3) Typhoid High fever, weakness, headache, stomach pain, constipation
(4) Pneumonia Streptococcus pneumoniae
(5) Rhino viruses Nasal congestion and discharge, sorethroat, cough, headache
(6) Filariasis Inflammation in lower limbs

Answer:
(i) Abdominal pain, indigestion, nausea, vomiting etc.
(ii) Ringworm
(iii) Salmonella typhi
(iv) Difficulty in breathing, fever, cough, headache, chills, lips and finger nails may turn grey to bluish in colour
(v) Common cold
(vi) Wuchereria bancrofti

Question 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.
NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases 3
Answer:
(a) Cannabinoids
(b) Cannabinoids are generally taken by inhalation and oral ingestion.
(c) They affect cardiovascular system of the body.

Question 10.
Give the full form of CT and MRI. How are they different from each other?
Answer:
CT (Computed Tomography) is recon-struction of three dimensional image made by X rays directly on a computer instead of a photographic film. It is radiological invasive technique. It is used to diagnose diseases of brain, spinal cord, chest and abdomen and detection of tumours. MRI (Magnetic Resonance Imaging) is a non invasive, non radiological technique in which the image is reconstructed on computer by detecting MRI signals generated by nuclei of hydrogen atoms in a magnetic field using non-ionising radiations. It is used to obtain multiplanar imaging of soft living tissues like in case of tumours, muscular disorders, cardiovascular disorders, haemorrhage with superior resolution.

Question 11.
Many secondary metabolites of plants have medicinal properties. It is their misuse that creates problems. Justify the statement with an example.
Answer:
Drugs like barbiturates, amphetamines, benzodiazepines, lysergic acid diethylamides (LSD) and other similar drugs, that are normally used as medicines to help patients coping with mental illnesses like depression and insomnia are secondary metabolites of plants. Misuse of plant metabolites can impair one’s .physical, physiological or functional behaviour creating problem for the society. For example, Cocaine is natural alkaloid obtained from leaves of plant Erythroxylum coca, native to South America. Cocaine has vasoconstrictor properties and therefore, is a good local anaesthetic. It is chewed, eaten or sniffed in its powdered form or taken as a drink. It is a powerful CNS stimulant (interferes with the transport of the neurotransmitter dopamine) and induces a sense of well being and pleasure and delays fatigue. Its overdose causes hallucinations, headache, insomnia, convulsions, etc.

Question 12.
Why cannabinoids are banned in sports and games?
Answer:
Cannabinoids may enhance the performance of some athletes. Smoking cannabis may decrease anxiety, fear, depression, tension, etc., and improve self confidence, relaxation, well-being and sleep, etc. These effects may allow the athletes to perform better, especially under pressure and may reduce stress, experienced before and during competition. The increase in ‘risk taking’ associated with cannabinoids may possibly improve training and performance giving athlete a competitive edge. Moreover, cannabinoids increase bronchodilation (which may improve oxygenation of tissue) and have analgesic (pain relieving) properties which could permit athletes to work during injury. Moreover, long term use of cannabinoids causes various hazards to the health of abuser. Therefore, use of cannabinoids by sportsperson is not justified, as it will lead to unfair competition. Hence, their use is banned in sports and games.

Question 13.
What is secondary metabolism?
Answer:
Secondary metabolism consists of metabolic pathways and products of metabolism that seem to have no direct function in growth and development of plants and are not absolutely required for their survival. Secondary metabolism is responsible for many bioactive compounds used medicinally.

Question 14.
Drugs and alcohol give short-term ‘high’ and long-term’damages’. Discuss.
Answer:
Curiosity, need for adventure and excitement, peer pressure, etc., constitute some common causes which motivate youngsters towards drug and alcohol abuse. Short term effects of drugs/alcohols are generally pleasing and relaxing which habituate the abuser for using these again and again. But long term effects of drugs or alcohol are devastating and detrimental. For example, alcohol when taken occasionally in small quantities, acts as sedative, analgesic and anaesthetic. It stimulates the secretion of gastric juices. But, if a person becomes addicted to it, he consumes high concentration alcohol frequently. Heavy drinking causes depression. Suicide attempt is much common in alcoholics than in the rest of society.

Sexual relationship is usually deteriorated because of erectile dysfunction or rejection by the partner.
Caffeine is a mild stimulant and taken as beverages- tea, coffee, cocoa and cola drinks. Caffeine is CNS stimulant and thus increases alertness and thought. It also acts as a cardiac and respiratory stimulant. It is a mild diuretic and also increases basal metabolic rate. Excessive use may cause anxiety, irritability, diarrhoea, irregular heart beat (arrhythmia) and decreased concentration. It also causes indigestion and disturbs pancreatic and renal functions. If taken empty stomach, it can cause stomach ulcers.

Question 15.
Diseases like dysentery, cholera, typhoid etc., are more common in over crowded human settlements. Why?
Answer:
Overcrowded areas are generally unhygienic. This is because public hygeine is not maintained properly. Water bodies in these areas are generally contaminated with disease causing microbes. Water stagnation and garbage accumulation is often not checked.
Diseases like dysentery, cholera, typhoid etc., are water borne infectious diseases and may easily spread in overcrowded human dwellings.

Question 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 leaves, resins and flowers of hemp plants e.g., Cannabis sativa. Examples of cannabinoids are bhang, ganga, charas etc. They interact with receptors present in the brain and affect cardiovascular system of the body.

Question 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:
Children living in metropolitan cities are more prone to allergies/asthma, mainly due to life style and increased level of pollution. Suspended particulate matter released by vehicles add more to the increasing pollution levels. Moreover children living in metropolitan cities are overprotected in their early childhood days and are more sensitive to allergens.

Allergy is hypersensitive response of the body against foreign particles (allergens). Common symptoms are sneezing, watery eyes, running nose, difficulty in breathing, irritation of throat, trachea and skin etc. Asthma related problems worsen due to changing weather conditions and increased pollution levels.

Question 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:
Vaccine is a preparation of antigenic proteins or inactivated/live but weakened germs of a disease which on inoculation into a healthy person provides temporary or permanent, active immunity by inducing antibody formation.

Vaccination is based on the property of ‘memory’ of the immune system. When a vaccine is injected, it generates primary immune response of low intensity and also produces B cells and T cells. When vaccinated person is again attacked by same pathogen, the existing memory T and B cells recognise the antigen quickly and attack the invaders with a massive production of lymphocytes and antibodies. Hepatitis B vaccine is second generation vaccine that has been produced from yeast by recombinant DNA technology.

Question 19.
What is cancer? How is a cancer cell different from the normal cell? How do normal cells attain cancerous nature?
Answer:
Cancer is an abnormal and uncontrolled division of cells, that invade and destroy surrounding tissues. Cancer cells differ from normal cells in the following ways:

  1. Cancer cells multiply in an uncontrolled manner.
  2. They do not exhibit property of contact inhibition.
  3. They show metastasis.
  4. They have indefinite life span.

Cancer causing agents known as carcinogens include physical, chemical and biological agents which activate the proto-oncogenes or cellular oncogenes to oncogenes and cause deviation in growth pattern leading to uncontrolled growth of cells. They alter the behaviour of normal cells and make them cancerous.

Question 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:
Hypersensitive reaction to foreign substances is known as allergy. Substances which cause allergic reactions are called allergens. The common allergens are dust, pollen, feathers, paint. In allergic chemicals called histamine and serotonin are released from mast cells. Allergy can be avoided by reducing an exposure to allergen and by taking drugs like antihistamines, adrenaline and steroids.

Question 21.
For an organ transplant, it is an advantage to have an identical twin. Why?
Answer:
For an organ transplantation, tissue matching(histocompatibility)andblood group compatibility of donor and recipient are very important. If they do not match, organ may be rejected. This is because, immune system recognises the protein in the transplanted tissue or organ as foreign and initiates cellular immunity. Chances of matching of tissue as well as blood group are very high if donor and recipient are identical twins, because of genetic similarity. Transplantation between identical twins is known as isograft.

Question 22.
What are lifestyle diseases? How are they caused? Name any two such diseases.
Answer:
Diseases which occur due to improper changes in the life style are known as lifestyle diseases. These diseases may be caused due to over-eating or crash dieting, lack of exercise, sedentary lifestyle, smoking, alcoholism, drug addiction etc. Hypertension and obesity are common lifestyle diseases.

Question 23.
If there are two pathogenic viruses, one with DNA and other with RNA, which would mutate faster? And why?
Answer:
Pathogenic virus with RNA as genetic material will mutate faster than the one with DNA as genetic material, because

  1. RNA is chemically and structurally unstable. Uracil present in RNA is less stable than thymine in DNA.
  2. RNA is highly reactive, labile and easily degradable.

Long Answer Type Questions

Question 1.
Represent schematically the life cycle of a malarial parasite.
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases 4

Question 2.
Compare the life style of people living in the urban areas with those of rural areas and briefly describe how the lifestyle affects their health.
Answer:
Lifestyle affects human health in various ways which are as follows:
(1) Food habits – People living in urban areas usually prefer fast food like pasta, pizza, burger, noodles etc., as compared to fresh, green leafy vegetables, fruits which are more consumed by villagers, because of this they suffer from obesity, hypertension and some cardiovascular problems.

(2) Residential areas — Due to lack of space in urban areas, residential areas are generally overcrowded. Sometimes there is no facility for proper disposal of garbage, no proper ventilation etc., therefore people become prone to food, water, and air borne diseases, like cholera, tuberculosis, typhoid, pneumonia etc. On the other hand, villages have clean air, human dwellings are not crowded and the environment is free of pollution. But due to lack of personal and public hygiene and malnutrition, villagers frequently suffer from various diseases.
Moreover, medical facilities in rural areas are not up to the mark. People of urban areas are exposed to good education, proper medical facilities etc.

(3) Exercise and sleep — Because of busy lifestyle, people in urban areas do not get enough rest and have no time for exercise. Whereas people in rural areas do a lot of physical work and have enough time to rest as well.

(4) Due to overprotected environment provided during early life, people in urban areas sometimes have low immunity and are susceptible to certain diseases and are affected by allergens. However, people in rural areas are exposed to harsh environmental conditions therefore develop better immunity.

Question 3.
Why do some adolescents start taking drugs? How can this be avoided?
Answer:
Reasons for drug and alcohol abuse among adolescents are :
(1) Curiosity : Adolescents want to have personal experience of smoking cigarette and taking alcohol and drugs.

(2) Adventure and excitement : A child may go in for use of drug, smoking and alcoholic drink for the sake of adventure and excitement.

(3) Family set up : In certain families, use of alcohol, tobacco, sleeping pills and pain killers are common. It induces the youngsters to taste the
same.

(4) Group or peer pressure : Friends and peer groups often motivate some adolescents to take drugs, alcohol as a defiance of authority and feeling of independence.

(5) Progressiveness : There is a false perception that taking of drugs, alcohol or smoking is a sign of progressiveness in society.

Various measures which help to avoid drug abuse are:

(1) Discipline : Good nurturance with consistent discipline but without suffocating strictness reduces the risk of addictions.

(2) Communication : The child must be able to communicate with the parents seeking clarification of all doubts and discussing problems that arise in studies or develop in the class, with friends, siblings and others.

(3) Independent working : Give responsibility to the child for small tasks and allow him/ her to perform independently. Of course, provide guidance where required.

(4) Education and counselling : Stresses, failures, disappointments and problems are part of life. A child has to be trained, educated and counselled to face them as and when they come.

(5) Looking for danger signs : Teachers and parents should always be careful to look for and identify danger signs that can indicate tendency to go in for addiction.

Question 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:
Behavioural changes which can be observed in an alcoholic are as follows:
(1) Reckless behaviour, vandalism and violence.
(2) Drop in academic performance and unexplained absence from school/ college.
(3) Lack of interest in personal hygiene, isolation, depression, fatigue, aggressive-ness.
(4) Change in eating and sleeping habits. Measures to overcome the problem of alcohol addiction are as follows:

  • Seeking professional advice : Highly qualified psychologists, psychiatrists and de-addiction and rehabilitation programmes can help individuals who are suffering from alcohol abuse.
  • Avoid undue peer pressure : Every person has his/her own choice and personality, which should be kept in mind. So he/ she should not be pressed unduly to do beyond his/her capacities, in work condition and other in social get together or activities.
  • Education and counselling : Helps to overcome the problems, like stresses, disappointments and failure in life. One should utilise one’s energy in some beneficial activities like sports, music, reading, yoga and other extra curricular activities.
  • Seeking help from parents and peers : In case of minors, whenever, there is any problem, one should seek help and guidance from parents and peers. Help should be taken from close and trusted friends. This would help young to share their feelings of anxiety and wrong doings.

Question 5.
What are the methods of cancer detection? Describe the common approaches for treatment of cancer. Answer:
Detection and diagnosis of cancer depends upon histological features of malignant structure. Few methods of cancer detection are as follows:
(1) Bone marrow biopsy and abnormal count of WBCs in leukemia.
(2) Biopsy of tissue, direct or through endoscopy. Pap test (cytological staining) is used for detecting cancer of cervix and other parts of genital tract.
(3) Techniques such as radiography (use of X-rays), CT Scan (computed tomography), MRI Scan (magnetic resonance imaging)
are very useful to detect cancers of the internal organs. Mammography is radiographic examination of breasts for possible cancer.
(4) Monoclonal antibodies coupled to appropriate radioisotopes can detect cancer specific antigens and hence cancer.
Treatment of cancer includes following :
(1) Chemotherapy : In chemotherapy a variety of anti-cancer drugs are used that produce more injury to cancer cells than to normal cells. These drugs interfere with the cell division and growth and affect both normal and cancerous cells. Vincristine and vinblastine from Catharanthus roseus (Vinca rosea) are effective in leukaemia control. Taxol is another anti-cancer drug obtained from Taxus baccata. Tetrathiomolybdate is the new anti-cancer drug. It arrests the tumour growth by starving cancer cells of copper.

(2) Radiotherapy : It is used in addition to chemotherapy. The basic priniciple there is to bombard cancer cells with rays that damage or destroy the ability of cancer cell to grow and divide by damaging the DNA within the tumour cells, but produce minimum damage to the surrounding normal tissue.

(3) Surgery : It is removal of the cancerous cells surgically and has only limited usefulness. In breast tumour and uterine tumour, it is most effective but other treatments are also given to kill any cells that may have been left.

(4) Immunotherapy : Immunotherapy is a form of treatment that enhances the body’s ability to recognise cancer cells and destroy them. It can be given intravenously or by subcutaneous injection.

(5) Blood and marrow transplant : High dose chemotherapy or radiation therapy can destroy bone marrow’s ability to make blood cells. A blood or marrow transplant can be used to replace marrow stem cells which produce blood cells.

Question 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:
LSD is a psychedelic drug or hallucinogen which induces behavioural abnormalities, they cause optical or auditory hallucinations, horrible dreams, emotional outburst and severe damage to central nervous system.

Barbiturates are sedatives which depress CNS activity, give feeling of calmness, relaxation and drowsiness but high doses induces deep sleep.

Amphetamines are synthetic drugs and are CNS stimulants. They cause alertness, self-confidence, talkativeness, wakefulness and increased work capacity. High doses cause euphoria, marked excitement and sleeplessness which may lead to mental confusion. Their use may produce after effects like nausea and vomiting. Amphetamine is one of the drugs included in the ‘dope test’ for athletes.

Question 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 programme is a programme that involves vaccination of children in the age group of 0-5 years with polio vaccine. It began in 1995 with the aim to make India a polio-free country. Oral Polio Vaccine (OPV) is a live-attenuated vaccine, produced by the passage of the virus through non-human cells at a sub-physiological temperature, which produces spontaneous mutations in the viral genome. OPV also proved to be superior in administration, eliminating the need for sterile syringes and making the vaccine more suitable for mass vaccination campaigns. OPV also provides long lasting immunity than the Salk vaccine. One dose of OPV produces immunity to all three poliovirus serotypes in approximately 50% of recipients. WHO declared India and the entire South-East Asia region polio free in March, 2014. Last case of polio in India was reported in January, 2011.

Question 8.
What are recombinant DNA vaccines? Give two examples of such vaccines. Discuss their advantages.
Answer:
Recombinant vaccines are vaccines having a gene encoding the protein of pathogen that causes immunogenic reactions in the host so as to produce antibody against the disease. These vaccines are produced from yeast through recombinant DNA technology. Examples are Hepatitis B vaccine and herpes virus vaccine.

Advantages – By rDNA technology the vaccines can be produced on larger scale so providing greater availabilitv for immunisation. These vaccines are highly specific, pure and elicit strong immune response.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases help you. If you have any query regarding.NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution

NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution

Multiple Choice Questions

Question 1.
Which of the following is used as an atmospheric pollution indicator?
(a) Lepidoptera
(b) Lichens
(c) Lycopersicon
(d) Lycopodium
Answer:
(b): Lichens are the association between specific ascomycetes and certain genera of either green algae or cyanobacteria. Lichens can be used as industrial pollution indicators. They never grow in area that are polluted.

Question 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) : The theory of spontaneous generation states that life originated from non-living things in a spontaneous manner. This concept was held by early Greek philosophers like Thales, Anaximander, Xanophanes,Empedocles, Plato, Aristotle, etc. In ancient Egypt, it was believed that the mud of the Nile could give rise to frogs, toads, snakes, mice and even crocodiles when warmed by the sun. Van Helmont (1577-1644) held that human sweat and wheat grains could give rise to organisms.

Question 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. Animals or plants with desirable characteristics are interbred with the aim of altering the genotype and producing a new strain of the organism for a specific purpose. Traditional breeding techniques have been supplemented by more recent methods of genetic engineering, genetic testing and embryo manipulation. Sequencing of genomes of commercially important animals and plant species has enabled the inheritance of desired genes to be monitored directly by molecular methods.

Question 4.
Palaentological evidences for evolution refer to the
(a) development of embryo
(b) homologous organs
(c) fossils
(d) analogous organs.
Answer:
(c) : Fossils are the remains or traces of any organism that lived in the geological past. In general only the hard parts of organisms become fossilised (e.g., bones, teeth, shells and wood) but under certain circumstances the entire organism is preserved. Fossil records such as number and nature of fossils in early rocks, distribution of fossils in successive strata, etc., have helped the scientists conclude that evolution has taken place from simple to complex forms in a gradual manner

Question 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) they have biochemical similarities.
Answer:
(b) : The bones of forelimbs of whale, bat, cheetah and man are similar in structure, because they are homologous organs. The organs which have the same fundamental structure but are different in functions are called homologous organs. These organs follow the same basic plan of organisation during their development. But in the adult condition, these organs are modified to perform different functions as an adaptation to different environments. The homologous structures are a result of divergent evolution.

Question 6.
Analogous organs arise due to
(a) divergent evolution
(b) artificial selection
(c) genetic drift
(d) convergent evolution.
Answer:
(d) : The organs which have similar functions but are different in their structural details and origin are called analogous organs. The analogous structures are the result of convergent evolution. The wings of an insect are analogous to wings of a bird. It is due to the fact that the basic structure of the wings of insects is different from the wings of bird. However, their function is similar.

Question 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) : Hardy-Weinberg principle describes a theoretical situation in which a population is undergoing no evolutionary changes. According to Hardy-Weinberg principle, the balance in the relative numbers of alleles that is maintained within a large population over a period of time assuming that:

  1. there is no genetic recombination
  2. there is no natural selection
  3. there is no gene migration
  4. there is no mutation. In such a stable population, for a gene with two alleles A (dominant) and a (recessive), if the frequency of A is p and a is q, then the frequencies of the three possible genotypes (AA, Aa and aa) can be expressed by the Hardy-Weinberg equation; p2 + 2pq + q2 = 1.

Question 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) : When a bacterial population encounters a particular antibiotic, those sensitive to it die. But some bacteria having favourable mutations become resistant to the antibiotic. Such resistant bacteria survive and multiply quickly. Soon the resistance providing genes become widespread and entire bacterial population becomes resistant.

Question 9.
Evolution of life shows that life forms had a trend of moving from
(a) land to water
(b) dryland to wet land
(c) fresh water to sea water
(d) water to land.
Answer:
(d) : Life’ originated in the ocean persumably about 4200 million years ago in precambrian era. The effect of electrical discharges and UV radiations on the mixture of methane, water, ammonia and hydrogen present in the sea gave rise to organic molecules like coacervates and microspheres. These protocells collected more substances forming cytoplasm and became first living beings. They were prokaryotic and chemoheterotrophic. When supply of existing organic molecules was exhausted, some heterotrophs evolved into autotrophs capable of synthesising food by chemosynthesis or photosynthesis. With + the advent of photoautotrophs, oxygen was added in sufficient amount in the primitive atmosphere. Eukaryotes originated in the ancient ocean, presumably about 1.5 billion years ago. These later on gave rise to complex forms. With the advent of oxygen, ozone layer was formed in the atmosphere which masked the land against the harmful UV radiations of sun and made it inhabitable. In the mean time, some complex plants could grow on land near water i.e., on boundary line and animal started visiting land, for food etc. Complex plants and animals developed jacketed sex organs and shelled eggs, respectively which reduced their dependence on water for completing their life cycles and they moved to land permanently. In this way, life forms moved from water to land.

Question 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 born leading to more chances of survival
(d) the embryo takes a long time to develop.
Answer:
(c)

Question 11.
Fossils are generally found in
(a) sedimentary rocks
(b) igneous rocks
(c) metamorphic rocks
(d) any type of rock.
Answer:
(a) : Sedimentary rocks are formed by gradual settling down or sedimentation of fragments or earth’s material in regions such as lake or sea. The animals or plants are preserved and fossilised when they are buried in the lava of volcano, in the ice, in an oil rich soil, in swamps, in desiccated deserts, in rocks, or under water, etc. Of all the media mentioned above the most common is water. Dead remains of aquatic animals and plants settle down at the bottom. Remains of terrestrial organisms are also brought to seas and big lakes by rivers and streams. Mud and sand settle down continuously at the bottom.

Sedimentation (deposition of layers) of mud and sand occurs. Fine mineral particles may penetrate the dead bodies. Decay and disintegration of organic remains take place to leave only the harder parts, impression, moulds, casts, etc. The sedimented mud and sand harden with time to form rocks.

Question 12.
FortheMN-blood group system, thefrequencies 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:
(a) : According to Hardy-Weinberg principle, the expected frequency of MN- blood group bearing organism is likely to be
p2 + 2pq + q2 = l
49 + .09 + 2x = 1
.58 + 2x = 1
2x = 1 – 59
= 42%

Question 13.
Which type of selection is industrial melanism, observed in moth, Biston betularia?
(a) Stabilising
(b) Directional
(c) Disruptive
(d) Artificial
Answer:
(b) : In directional selection, the population changes towards one particular direction. It means this type of selection
favours small or large-sized individuals and more individuals of that type will be present in next generation. The mean size, of the population changes. Example are evolution of DDT resistant mosquitoes, industrial melanism in peppered moth and evolution of giraffe.

Question 14.
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)

Question 15.
Which of the following is an example for link species?
(a) Lobe fish
(b) Dodo bird
(c) Sea weed
(d) Chimpanzee
Answer:
(a)

Question 16.
Match the scientists listed under Column ‘A’ with ideas listed under Column ‘B’.
NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution 1

Answer:
(a) A-(i); B-(iv); C-(ii); D-(iii)
(b) A-(iv); B-(i); C-(ii); D-(iii)
(c) A-(ii); B-(iv); C-(iii); D-(i)
(d) A-(iv); B-(iii); C-(ii); D-(i) miim (b)

Question 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)

Question 18.
Variations during mutations of meiotic recombinations are
(a) random and directionless
(b) random and directional
(c) random and small
(d) random, small and directional.
Answer:
(a): Hugo de Vries (1901) put forward a theory of evolution, called mutation theory. The theory states that evolution is a jerky process where new varieties and species are formed by mutations (discontinuous variations) that function as raw material of evolution. Mutations appear all of a sudden. They become operational immediately.

Short Answer Type Questions

Question 1.
What were the characteristics of life forms that had been fossilised?
Answer:

  1. Presence of hard parts in the body like bones, teeth, shell etc.
  2. Buried in a medium where no oxidation of substances occur.

Question 2.
Did aquatic life forms get fossilise? If, yes where do we come across such fossils?
Answer:
Yes, aquatic life forms have better chances of getting fossilised, as their dead bodies are buried at the bottom of water in sedimentary rocks. They get exposed when the rocks come to the surface because of upheavals in the earth crust.

Question 3.
What are we referring to when we say ‘simple organisms’or’complex organisms’?
Answer:
Simple organisms are those organisms which show cellular level of body organisation,
i. e., their body is equivalent to a single cell. These organisms do not show ageing and natural death. As soon as the cell grows to maximum size, it divides to form daughter cells.
Complex organisms have higher level of body organisation as compared to unicellular organisms like tissue level, organ level and organ system level.

Question 4.
How do we compute the age of a living tree?
Answer:
We can compute the age of a living tree by counting the number of annual growth rings of tree. This technique is known as dendrochronology. It depends on the fact that trees in the same locality show a characteristic pattern of growth rings resulting from climatic conditions. Thus, it is possible to assign a definite date for each growth ring in living trees, and to use the ring patterns to date fossil trees or specimens of wood (e.g., used for building or objects on archaeological sites) with life spans that overlap those of living trees.

Question 5.
Give an example for convergent evolution and identify the features towards which they are converging.
Answer:
Tendrils in pea and cucurbits show convergent evolution because tendrils in both provide support to the plant (same function) but pea tendrils develop from leaves, whereas cucurbit tendrils are stem tendrils (different origin).

Question 6.
How do we compute the age of a fossil?
Answer:
Age of fossil can be computed by one of the following methods:

  1. Radioactive carbon dating method
  2. Uranium-lead method
  3. Potassium-argon method
  4. Electron spin resonance method.

Question 7.
What is the most important pre-condition for adaptive radiation?
Answer:
Most important pre-condition for adaptive radiation is that it should occur in s^me geographical area and is shown by different members of same ancestor. It is radiated to different habitats and develops different habits.

Question 8.
How do we compute the age of a rock?
Answer:
Age of rock can be determined by quantity of uranium and lead in the rocks and potassium-argon transformations.

Question 9.
When we talk of functional macromolecules (e.g. proteins as enzymes, hormones, receptors, antibodies etc), towards what are they evolving?
Answer:
Functional macromolecules are evolving towards creation of a complex organism. There are various evidences that are common to simple and complex forms of life which indicate common ancestry. E.g., histone protein tends to be well preserved among all eukaryotes with only one or two amino acids different.

Question 10.
In a certain population, the frequency of three genotypes is as follows:
Genotypes: BB Bb bb
Frequency: 22% 62% 16%
What is the likely frequency of B and b alleles?
Answer:
Frequency of allele B = 100% of BB + 50% of Bb
i. e., 22% + 31% = 53%
Frequency of allele b = 100 % of bb + 50% of Bb
i.e., 16% + 31% = 47%

Question 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:
The other two factors are mutation and natural selection.

Question 12.
What is founder effect?
Answer:
Founder effect is the phenomenon occurring when a population is founded by a small sample of the entire species, perhaps just a handful of individuals. Chance dictates that these founder members will be genetically unrepresentative of the species as a whole, and that the genetic make up of the new population will differ markedly from the main species population.

Question 13.
Who among the Dryopithecus and Ramapithecus was more man like?
Answer:
Ramapithecus was more man like in comparison to Dryopithecus, which was more ape-like.

Question 14.
By what Latin name the first hominid was known?
Answer:
As per the latest discoveries, Sahelanthropous tchadensis is considered the earliest hominid.

Question 15.
Among Ramapithecus, Australopithecus and Homo habilis – who probably did not eat meat?
Answer:
Ramapithecus probably did not eat meat. It was more man-like and lived on the tree tops but also walked on the ground. Its jaws and teeth were like those of humans. Its small canines and large molars suggest that Ramapithecus ate hard nuts and seeds like modern man.

Short Answer Type Questions

Question 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¬existing life or otherwise we will never answer the question as to how the first forms of life arose? Comment.
Answer:
If we trace back the origin of first living organism, this fact will be revealed that life originated on early earth through physio- chemical processes in which atoms combined to form molecules which in turn reacted to produce inorganic and organic compounds.

These compounds then interacted to produce macromolecules which organised to form first living cells. Thus, first life form originated through abiogenesis. Once life originated, abiogenesis could not follow and new life forms then further evolved from pre-existing life (i.e., biogenesis).

Louis Pasteur in his experiments, disproved origin of life through abiogenesis and proved that life can arise from pre-existing life only. This statement holds true to its meaning but contradicts the mechanism of origin of first life form. Hence, it could be corrected as “life evolves from pre-existing life”.

Question 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:
Yes, natural disaster can lead to mass extinction of species from time to time, which is always ‘sudden’, ‘abrupt’ and ‘group specific’. For example extinction of dinosaurs is a good example of mass extinction triggered by a massive meteorite/asteroid impact. Asteroid impact was confirmed by the discovery of high concentrations of metal iridium in a thin layer of earth in almost all parts of the world as iridium is rare on earth and abundantly present in meteorites.

This natural disaster lead to the extinction of many contemporary species present on earth which included, non-avian dinosaurs, benthic foraminiferans, certain marine invertebrate groups etc.

Question 3.
Why is nascent oxygen supposed to be toxic to aerobic life forms?
Answer:
Nascent oxygen in highly reactive and can oxidise various biomolecules like DNA, RNA and proteins etc. If present in the cells of aerobic life forms, it can cause mutations and undesirable changes in their metabolic pathways. Therefore it has been considered as major toxic pollutant.

Question 4.
While creation and presence of variation is directionless, natural selection is directional as it is in the context of adaptation. Comment.
Answer:
The creation and presence of variations is directionless in regard that they occur randomly and spontaneously. Variations in organisms develop due to various reasons like genetic recombination, genetic drift, gene migration, mutations etc. All these variations are non-directional. As a result of variations, population becomes heterogeneous. Natural selection favours more adapted members oven less adapted ones and allows them to multiply at a faster rate to increase their population. This provides direction in a specific way towards evolution.

Question 5.
The evolutionary story of moths in England during industrialisation reveals, that’evolution is apparently reversible’ Clarify this statement.
Answer:
Peppered moth (Biston betularia) exists in two forms, white and melanic. Before industrialisation the trees were covered with whitish lichens. White moths could easily escape from their predators as they remained unnoticed in the surroundings. After industrialisation the bark of the trees got covered with black smoke and white moth were selectively picked up by the predators whereas the black ones escaped. Therefore, the population of white moths decreased. This showed that evolution is reversible and in a mixed population, those that can better adapt to the prevalant environmental conditions, survive and increase their population size but no variant is completely wiped out.

Question 6.
Comment on the statement that “evolution and natural selection are end results or consequences of some other processes but themselves are not processes”.
Answer:
All organisms possess enormous fertility. They multiply in geometric ratio, but the limited food supply, space and ecological niche begins intraspecific or interspecific struggle amongst them. For survival during struggle, organisms develop certain variations due to genetic recombinations, mutations etc., and become heterogenous. The organisms which have favourable variations would survive (survival of the fittest). Those who keep changing themselves according to the changing environment are selected by nature (natural selection). Selected organisms transfer their useful variations to next generation. Accumulation of continuous ’”variations generation after generation leads to evolution of new species. This clearly shows that evolution and natural selection are consequences of some other processes.

Question 7.
State and explain any three factors affecting allele frequency in populations.
Answer:
(1) Gene flow – It is also known as gene migration. It refers to the movement of alleles from one population to another as a result of interbreeding between members of the two population. The genes of two populations intermingle through breeding and result causes variations in the offspring.

(2) Genetic recombination – During sexual reproduction, fusion of male and female gametes takes place. These gametes are formed by reduction division or meiosis. Crossing over during meiosis leads to genetic variation as exchange of genetic material between non-sister chromatids of homologous chromosomes takes place. Hence, new association of alleles are formed in the gametes. Random fusion of gametes leads to genetic variations in the offspring.

(3) Mutations – Mutation is a sudden random change in genetic material of a cell. Somatic mutations affect the non- reproductive cells and are therefore restricted to a single organism but germline mutations which occur in reproductive cells (gametes) are transmitted to the offspring of organism. Mutations are caused by various chemicals or forms of radiations.

Question 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:
Gene flow or gene migration refers to movement of alleles from one population to another as a result of interbreeding between members of two populations. Gene flow occurs through geographical barriers over generations. Changing gene frequencies would indicate that evolution is in progress. By studying specific allele frequencies of specific genes human migratory patterns and evolutionary trends in history and pre-history can be traced as gene migration (affected by human migration) contributes to changing gene frequency in an evolving population. Actually scientists are doing so, under Human Genome Project.

Question 9.
How do you express the meaning of words like race, breed, cultivars or variety?
Answer:
Race is a category used in the classification of organisms that consist of a group t>f individuals within a species that are geographically, ecologically, physiologically distinct from other members of the species e.g., negroid, mongoloid, etc. The term is frequently used in the same sense as subspecies. Breed is a group of animals similar in most characters and are related by descent, e.g., breeds of cow – Jersey, Holstein etc. Variety is a category used in the classification of plants and animals below the species level. A variety consists of a group of individuals that differ distinctly from but can interbreed with other varieties of the same species e.g., cabbage, cauliflower. Cultivar is a plant that has been developed and maintained by cultivation as a result of agricultural or horticultural practices. The term is derived from cultivated variety e.g., Pusa sem of flat beans.

Question 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:
(a) In the struggle for existence, the individuals which have more favourable variations will have a competitive advantage over others which have less favourable or unfavourable variations. They are considered fit and thus survive and reproduce.

Question 11.
Enumerate three most characteristic criteria for designating a Mendelian population.
Answer:
Characteristic criteria for designating a Mendelian population are as follows :

  1. Population must be large enough and there should be free flow of genetic materials among individuals through sexual reproduction.
  2. Random mating of organisms takes place.
  3. There should be either nil or negligible migration.

Question 12.
“Migration may enhance or blur the effects of selection” Comment.
Answer:
Migration involves entry or exit of some members of a species into or from an area. It may cause improvement of race if there is entry of individuals with more desirable genes, or may cause deterioration of race if there is entry of individuals with undesirable traits or exit of individuals with desirable traits. Hence, migration may enhance or blur the effects of selection.

Long Answer Type Questions

Question 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 sum of allele frequencies in a population is stable and remains constant from generation to generation i.e., the gene pool (total genes and their alleles in a population) remain constant. The five factors, called evolutionary agents, which influence the allele frequencies in a population are :

  1. Gene flow – When migration of a section of population to another place occurs, gene frequencies change in the original as well as new population. New alleles are added to the new population and these are lost from the old population. Hence, gene flow changes the allele frequency in a population.
  2. Genetic drift – It refers to the elimination of genes of certain traits when a section of population migrates or dies of natural calamity. It affects the gene frequency of a population.
  3. Genetic recombinations – During sexual reproduction, fusion of male and female gametes takes place. The gametes are formed by reduction division or meiosis. Crossing over during meiosis leads to genetic variation as exchange of genetic material between non-sister chromatids of homologous chromosomes takes place and new association of alleles are formed in the gametes. The offsprings formed by fusion of gametes show new combination of traits.
  4. Mutations – The sudden heritable change in genetic material which is directionless is called mutation. It alters the genetic make up of an individual.
  5. Natural selection – It is the phenomenon by which some members of a population having desirable or favourable traits are favoured and survive to produce new offspring.

Question 2.
Explain divergent evolution in detail. What is the driving force behind it?
Answer:
Similarity in basic structural plan and developmental origin and difference in appearance and functioning of those structures due to adaptations of different needs is called divergent evolution. The driving force behind it is adaptation to new habitat and the prevailing environmental conditions there. As the original population increases in size, it spreads out from its centre of origin to exploit new habitats and food resources. In time this results in a number of populations each adapted to this particular habitat. Eventually these populations differ from each other sufficiently to become new species.
A good example of this process is the evolu tion of the Australian marsupials into species adapted as carnivores, herbivores, burrowers, fliers, etc.

Question 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:
Peppered moth, Biston betularia lives in all parts of England. In the population of peppered moth, two variants exist – dark or melanic and light coloured. Before industrialisation, light coloured moths were prevelant because they blended with the lichen covered bark of trees and remained unnoticed by predatory birds. The melanic moths were easily detected and preyed upon by predatory birds. Hence, the population of light coloured moth was much more as compared to dark  or melanic moth. With industrialisation, the pale tree trunks became more and more blackened due to industrial smoke and soot. As a result the light moths stood out in contrast to their background, increasing the possibility of being easily detected and eaten by their predators, such as birds, in much greater number than the dark melanic variety. Decrease in the number of light moths and increase in the number of dark moths was the ultimate result.

Therefore, evolution favoured the melanic moths to reproduce more successfully for their adaptation in the polluted areas of England. If the industries are removed the area will again become free of pollution leading to growth of lichens. As stated earlier, light coloured moth would easily camouflage with a light background and dark variants would be spotted easily by predators and eaten more frequently. Hence, the population of light coloured moth will again increase.

Question 4.
What are the key concepts in the evolution theory of Darwin?
Answer:
Key concepts in the evolution theory of Darwin are as follows:
(1) Overproduction or enormous fertility:
Living beings have an innate ability of producing their own progeny for the continuity of race.

(2) Struggle for existence : According to Darwin, individuals multiply in geometric ratio whereas space and food remain almost constant.

(3) Variations and heredity : The everlasting competition among the organisms compells them to change according to the conditions so that they can utilise the natural resources and can survive successfully. According to Darwin, the variations are gradual (continuous) and those which are helpful in the adaptations of an organism towards its surroundings would be passed on to the next generation, while the others disappear.

(4) Survival of the fittest or Natural selection: During the struggle for existence only those individuals could survive which exhibit such variations that are more beneficial in facing the hardships and rigours of environment or which change to adapt themselves to the changing conditions. This is known as natural selection.

(5) Origin of species : Natural selection results in modification of traits within a lineage which over a period of time can bring about evolution.

Question 5.
Two organisms occupying a particular geographical area (say desert) show similar adaptive strategies. Taking examples, describe the phenomenon.
Answer:
The phenomenon indicated in the question is convergent evolution wherein organisms which are not closely related, evolve similar traits independently as a result of adaptation to similar environment. There is specific type of environment existing in every geographical area. The environmental factors influence all the organisms living in that area. This can be explained by following examples :
(1) Pectoral fins of sharks and flippers of dolphins are analogous organs. Pectoral fins of sharks are not pentadactyle. The flippers of dolphins are pentadactyle. Thus, basic structure of pectoral fins of sharks and flippers of dolphins are different but both are useful in swimming.

(2) Stings of honey bee and scorpion are analogous structures. The sting of honey bee is a modification of its ovipositor (structure that helps in egg laying) while that of scorpion is modified last abdominal segment. Stings of both arthropods perform similar function.

(3) Leaves are plant organs specialised for photosynthesis. However, there are plants where the leaves are modified or reduced in response to hot and dry environment. The function of leaves is then taken over by other organs like stipules (e.gLathyrus aphaca), petiole (e.g., Acacia auriculiformis) and stem branches (e.g, Ruscus, Asparagus). They are analogous amongst them as well as to the leaves.

Question 6.
We are told that evolution is a continuing phenomenon for all living things. Are humans also evolving? Justify your answer.
Answer:
New researches suggest that human beings are also evolving like all other organisms. By studying specific evolutionary trends in various human characters it can be provfed that human beings are evolving :
(1) Lactose tolerance : Historically the gene that regulated human’s ability to digest lactose was shut down as infants are weaned off of their mother’s breast milk. However, adult human in regions of Africa and Northern Europe developed the ability to tolerate lactose in their diets as recent as 5,000 or 6,000 years ago due to mutations.

(2) Wisdom teeth : Our ancestors had much bigger jaws than we do. Today our jaws are much smaller and wisdom teeth may not erupt in some. Estimates say that they will disappear in the coming population.

Question 7.
Had Darwin been aware of Mendel’s work, would he been able to explain the origin of variations. Discuss.
Answer:
Mendel proposed certain laws of heredity which form the basis of inheritance of characters. He is commonly called ‘father of genetics’ because of his contribution to genetics. Darwin’s theories explain the survival of the fittest, but does not explain the mode of arrival of fittest. He did not explain about any particular substance for inheritance of continuous useful variations from one generation to the other. Whereas Mendel stated that inheritance of characters is particulate and every character is of discrete units called factors (now called alleles). If Darwin had been aware of Mendel’s work, he could have been able to explain the origin of variations and their inheritance in the light of changes in the factors of Mendel.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance

NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 5 Principles of Inheritance and Variation

Multiple Choice Questions

Question 1.
In a DNA strand, the nucleotides are linked together by
(a) glycosidic bonds
(b) phosphodiester bonds
(c) peptide bonds
(d) hydrogen bonds.
Answer:
(b) : A nucleotide has three components – nitrogenous base, a pentose sugar (ribose in case of RNA and deoxyribose for DNA) and a phosphate group. There are two types of nitrogenous bases – purines (adenine and guanine) and pyrimidines (cytosine, thymine and uracil). A nitrogenous base is linked to the pentose sugar through a N-glycosidic linkage to form a nucleoside. When a phosphate group is linked to 5′ -OH of a nucleoside through phosphoester linkage, a correspondingnucleotide is formed. The nucleotides are linked through 3′ – 5′ phosphodiester linkage to form a dinucleotide.

Question 2.
A nucleoside differs from a nucleotide. It lacks the
(a) base
(b) sugar
(c) phosphate group
(d) hydroxyl group.
Answer:
(c) : A purine or pyrimidine base joined with a pentose sugar, either ribose or deoxyribose, is a nucleoside. A nucleotide is a nucleoside with one or more phosphate groups attached to the sugar.

Question 3.
Both deoxyribose and ribose belong to a class of sugars called
(a) trioses
(b) hexoses
(c) pentoses  
(d) polysaccharides.
Answer:
(c) : Pentose sugar is a sugar molecule with five carbon ring structure. Both ribose and deoxyribose are pentoses, the latter is formed by deoxygenation of the former at one carbon atom.

Question 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 semi-conservative nature
(c) uniform width throughout DNA
(d) uniform length in all DNA.
Answer:
(c) : In DNA chain, the base pairs on the two strands are complementary and a specific purine pairs with a specific pyrimidine. This makes the two chains uniformly (2nm) thick. A larger sized purine lies opposite to the smaller-sized pyrimidine i.e., A pairs with T and C pairs with G giving uniform width throughout DNA.

Question 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) : DNA is much more organised in eukaryotic chromatin and is associated with a variety of proteins, most prominent of which are histones. Histones are rich in the basic amino acid residues lysines and arginines. Both these amino acid residues carry positive charges in their side chains. Thus, histones are positively charged. Histones are organised to form a unit of eight molecules called as histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.

Question 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)

Question 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) : Sickle cell anaemia is an autosomal hereditary disorder in which the erythrocytes become sickle shaped under oxygen deficiency, such as during strenuous exercise and at high altitudes. The disorder or disease is caused by the formation of an abnormal haemoglobin molecule called haemoglobin-S. Sickle cell trait protects against malaria. Several studies have suggested that, sickle haemoglobin might get in the way of Plasmodium parasite infecting RBCs, reducing the number of parasites that actually infect the host cell and thus confer some protection against the disease.

Question 8.
One of the following is true with respect to AUG:
(a) It codes for methionine only.
(b) It is also an initiation codon.
(c) It codes for methionine in both prokaryotes and eukaryotes.
(d) All of the above
Answer:
(d) AUG has dual functions : it acts as initiation codon and also codes for methionine (both in prokaryotes and eukaryotes, although formylated in prokaryotes).

Question 9.
The first genetic material could be
(a) protein
(b) carbohydrates
(c) DNA
(d) RNA.
Answer:
(d) : The first genetic material could be RNA. Evidences suggest that, metabolism, splicing and translation all have evolved around RNA. The first biocatalysts were also RNAs. Even now, some enzymes are made of RNAs, e.g., ribozyme. Gradually during evolution, genetic storage function of RNA was taken over by DNA as RNA being single stranded was more reactive, hence unstable. DNA being double stranded with complementary strands is not only more stable but also resists changes as it has evdlved a process of repair. For biocatalysis, RNA was replaced by protein enzymes. The latter were more stable, efficient and occur in many forms.

Question 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) : Eukaryotic transcripts possess extra segments called introns or intervening sequences or noncoding sequences. They do not appear in mature or processed RNA because post transcriptional processing of the transcript includes splicing i.e., the’ process of removal of introns and fusion of exons to form functional RNAs.

Question 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.
Answer:
(c) : Chromosome 1 has 2968 genes while Y-chromosome has 231 genes which respectively are the maximum and minimum number of genes present in a chromosome.

Question 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)

Question 13.
DNA isa polymer of nucleotides whicharelinked 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 deoxyribose.
(c) Remove/Replace 2′ OH group with some other group in deoxyribose.
(d) Both ‘b’ and ‘c’.
Answer:
(b) : If 3′ OH group is removed/replaced in deoxyribose, there will be no formation of phosphodiester bonds which is formed between 3′ hydroxyl (OH) group of one nucleotide and 5′ phosphate of the other, and hence polymerisation of nucleotides will be prevented.

Question 14.
Discontinuous synthesis of DNA occurs in one strand, because
(a) DNA molecule being synthesised is very long
(b) DNA dependent DNA polymearse 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) :The DNA-dependent DNA poly­merases catalyse polymerisation only in one direction, that is 5′ —> 3′. Consequently, on one strand (the template with polarity 3′ —> 5′), the replication is continuous in 5′ —» 3′ direction, while on the other (the template with polarity 5′ —» 3′), it is discontinuous. A RNA primer is added the 3′ end of the opened region of 5′ —» 3′ strand and on this primer, a short stretch of DNA is synthesised in 5′ —> 3′ direction. Then again, when further uncoiling occurs then a new primer is added at 3′ end which is elongated, and the process continues. These discontinuously synthesised fragments are later joined by the enzyme DNA ligase.

Question 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:
(d) : There is single DNA-dependent RN Apoly merase that catalyses transcription of all types of RNA in bacteria. RNA polymerase binds to promoter and initiates transcription (initiation). It uses nucleoside triphosphates as substrate and polymerises them in a template dependent fashion following the rule of complementarity. It somehow also facilitates opening of the helix and continues elongation. Only a short stretch of the prepared RNA remains bound to the enzyme.Once the polymerase reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription. The RNA polymerase alone is only capable to catalyse the process of elongation. It associates transiently with initiation factor(o) and termination factor(р)  to initiate and terminate the transcription respectively. Association with these factors alters the specificity of the RNA polymerase to either initiate or terminate. In eukaryotes, several initiation factors and termination factors along with three types of RNA polymerase each polymerising a specific RNA perform these functions.

Question 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 & c) : Regulation of gene expression refers to a very broad term that may occur at various levels. In eukaryotes, the gene expression can be regulated at transcriptional level, post transcriptional processing level, during transport of mRNA from nucleus to the cytoplasm, and at translational level whereas in prokaryotes, control of the rate of transcriptional initiation is the predominant site for control of gene expression.

Question 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 proteins?
(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) : In a transcription unit, the activity of RNA polymerase at a given promoter is regulated by interaction with accessory proteins, which affect its ability to recognise start sites. These regulatory proteins can act both positively (activators) and negatively (repressors).

Question 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 sequencing of chromosome 1 was completed in May 2006. This was the last of the 24 human chromosomes (22 autosomes and X and Y) to be sequenced.

Question 19.
Which of the following are the functions of RNA?
(a) It is a carrier of genetic information from DNA to 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) : mRNA carries genetic information from DNA to ribosomes for synthesising polypeptide chains. fRNA carries amino acids to ribosomes attached to mRNA for translation. rRNA, is a vital component of ribosomes.

Question 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 a single stranded DNA
(c) it is a double stranded linear DNA
(d) no conclusion can be drawn.
Answer:
(b) : It cannot be a double stranded DNA because as per Chargaff’s rule for a dsDNA, the ratios between adenine and thymine, and guanine and cytosine are constant and equal. Hence, it is a ssDNA.

Question 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) rDNA.
Answer:
(c) : DNA synthesis on RNA template occurs in retroviruses or RNA viruses. They have RNA as their genetic material. When they infect a host cell, they first synthesise DNA on RNA template which is known as cDNA or complementary DNA. This cDNA then continues the process of infection.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.

Question 22.
If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of ,5N/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) : In Meselson and Stahl’s experiment, parent DNA isolated from E.coli is grown in heavy 15N medium. It was then put in light ,4N medium.Then, when replication occurs the new strand synthesised would have l4N. The process can be represented as follows:
NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance 1

Question 23.
If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is 5′ – AT G A A T G – 3′, the sequence of bases in its RNA transcript would be
(a) 5′-AUGAAUG-3′
(b) 5′-UACUUAC-3′
(c) 5′-C All UC AU-3′
(d) 5′ – G U A A G U A – 3′.
Answer:
(a) : The sequence of bases in RNA transcript is similar (not complementary) to the coding DNA strand except that in RNA, uracil is present in place of thymine. So, the correct sequence of bases is:
5′ – AUGAAUG – 3′

Question 24.
The RNA polymerase holoenzyme transcribes
(a) the promoter, the structural gene and the terminator region
(b) the promoter and the terminator region
(c) the structural gene and the terminator region
(d) the structural gene only.
Answer:
(d) : Transcription involves three separate processes : initiation, elongation and termination. Initiation begins when the RNA polymerase binds to the promoter which serves only as a target site for binding of the RNA polymerase and is not transcribed. Each gene contains a specific promoter region for guiding the beginning of transcription. This is followed by region of gene (structural gene) that is transcribed and ends with a terminator that stops transcription and is not transcribed.

Question 25.
If the base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of fRNA pairing with it must be
(a) 5′- UAC – 3′
(b) 5′-CAU-3′
(c) 5’-AUG-3′
(d) 5′- GUA – 3′.
Answer:
(b) : The first base of anticodon in 5′ – 3′ direction binds with the third base in codon (reading in 5′ – 3′ direction). Thus, if the base sequence in codon of mRNA is 5′- AUG – 3′, the complementary anticodon will be 3′-UAC – 3′ or 5′- CAl – 3′.

Question 26.
The amino acid attaches to the fRNA at its
(a) 5′-end
(b) 3′-end
(c) Anticodon site
(d) DHU loop.
Answer:
(b) : fRNA has an anticodon loop that has bases complementary to the code, it also has an amino acid acceptor site at which it binds to amino acids. This site lies at the 3′ end opposite the anticodon. fRNAs are specific for each ahrino acid.

Question 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) : The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. It has two subunits, a large subunit and a small subunit. To start translation, mRNA should bind to small ribosomal unit.

Question 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) : The repressor of the operon is synthesised (all-the-time-constitutively) from the / gene. 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. Thus, the lac operon gets switched on.

Very Short Answer Type Questions

Question 1.
What is the function of histones in DNA packaging?
Answer:
Maim Histones are positively charged basic proteins which help in condensation of DNA. Negatively charged DNA gets wrapped around positively charged histone octamer to form nucleosome.

Question 2.
Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
Answer:
Heterochromatin is thicker, densely packed, dark stained part of chromatin. Euchromatin is thinner, loosely packed lightly stained part of chromatin. Transcriptionally active part is euchromatin.

Question 3.
The enzyme DNA polymerase in 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 serves as a dual polymerase which serves two functions. It primary function is to add nucleotides according to the template strand in 5′ —> 3′ direction. It simultaneously proof reads the newly formed double strand, as it passes through the polymerase molecule. If the wrong base is inserted, the bond is unstable and spontaneous melting of the newly formed stretch occurs. The new strand gets exposed to the 3′ exonuclease site of the enzyme which removes the mismatched base and some additional nucleotides from the 3’ end. Then, the polymerase activity is continued again. This reduces the chances of error in DNA replication from about one in a million to about one in a hundred million base pairs.

Question 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:
New DNA strand is always formed in 5′-3′ direction during DNA replication, over DNA template with 3′-5′ direction. When DNA opens during replication, 3′-5′ strand forms a continuous strand called leading strand. The other template strand with 5′-3′ orientation produces a new short strand over the fork every time when it opens. These short segments are called Okazaki fragments. These segments join together with DNA ligases and form lagging strand.

Question 5.
Given below is the sequence of coding strand of DNA in a transcription unit:
3’AATGCAGCTATTAGG-5′ Write the sequence of
(a) its complementary strand
(b) the mRNA
Answer:
(a) Sequence of complementary strand will be : 5′ -TTACGTCGATAATCC3′
(b) Sequence of mRNA will be : 3′ – A A U G C A G C U A U U A G G 5′

Question 6.
What is DNA polymorphism? Why is it important to study it?
Answer:
Variation at genetic level arisen due to mutations, is called polymorphism. Such variations are unique at particular site of DNA. The polymorphism (variations) in DNA sequences is the basis of genetic mapping and DNA fingerprinting.

Question 7.
Based on your understanding of genetic code, explain the formation of any abnormal haemoglobin molecule. What are the known consequences of such a change?
Answer:
Haemoglobin is made of four polypeptide chains, two a-chains-which are 141 amino acids long and two (3-chains which are 146 amino acids long. In case of sickle cell anaemia, a fault occurs at the sixth amino acid in the (5-chain. The, amino acid should be glutamic acid. In HbS however, it is replaced by valine.
NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance 2

Question 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:
Development of organs in an organism is regulated by expression of different sets of genes in a definite sequence in an orderly manner. Any disturbance in co-ordination ‘         and expression of genes will lead to abnormal  i. organ formation.

Question 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:
Recent studies show that there is a region in the active site cleft of the polymerase enzyme that monitors the 2′ and 3′ substituents of the incoming nucleotide and identifies the sugar component. It ensures that only deoxyribonucleotides are picked up during  replication from the nuclear pool.

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

  1. DNA helicase which stimulates the separation of the two strands.
  2. Topoisomerase which changes the degree of supercoiling in DNA by cutting one or both strands, which forms
  3. Primase which forms RNA primer strands on single-stranded DNA templates.

Question 11.
Name any three viruses which have RNA as the genetic material.
Answer:

  1. Tobacco Mosaic Virus
  2. Human Immunodeficiency Virus
  3. Influenza Virus

Short Answer Type Questions


Question 1.
Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as the genetic material.
Answer:
In 1928, Frederick Griffith performed the transformation experiment using Streptococcus pneumoniae. When he injected heat killed, virulent S strain along with non-virulent, live R strain in mice, then the mice died. It showed that something from dead S strain transformed the non-virulent R strain into virulent one. This phenomenon was called transformation by him. Transformation is the phenomenon by which the transforming principle (as named by Griffith), isolated from one type of cell, when introduced into another type, is able to bestow some of the properties of the former to the latter. This discovery started the quest for identification of this transforming principle. Avery, MacLeod and McCarty later 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 (neither proteins nor RNAs) from S bacteria caused R bacteria to become transformed. They discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation.

Question 2.
Who revealed biochemical nature of the transforming principle? How was it done?
Answer:
Refer answer 1.

Question 3.
Discuss the significance of heavy isotope of nitrogen in the Meseison and Stahl’s experiment.
Answer:
15N is not a radioactive isotope. It is a heavy isotope of N and can be separated from 14N by density gradient centrifugation. This helped Meseison and Stahl to prove that DNA replicates semiconservatively. Meseison and Stahl extracted bacterial DNA and centrifuged it in caesium chloride solution. Depending on the mass of the molecule, the DNA would settle out at a particular point in the tube. They first grew Escherichia coli bacteria in a medium containing heavy isotope 15N for several generations.

The 15N bacteria were then transferred to a growth medium containing the normal, lighter isotope of nitrogen, 14N, where they reproduced by cell division. Extracts of DNA from the first generation offspring were shown to have a lower density, since half the DNA was made up of the original strand containing 15N and the other half was made up of the new strand containing 14N. At succeeding generation times, the DNA extracts were found to separate at lower densities indicating that have a lower proportion of ,5N as more 14N had incorporated into the bacterial DNA. This was conclusive evidence for the semi­conservative method of DNA replication.

Question 4.
Define a cistron. Giving examples, differentiate between monocistronic and polycistronic ‘transcription unit.
Answer:
Cistron is a length of DNA that contains the information for coding a specific polypeptide chain or codes for a functional RNA molecule (i.e., mRNA, transfer RNA or ribosomal RNA).

Monocistronic transcription unit is a type of messenger RNA that can encode only one polypeptide per RNA molecule. In eukaryotic cells, virtually all messenger RNAs are monocistronic.

Polycistronic transcription unit is a type of messenger RNA that can encode more than one pqlypeptide separately within the same RNA molecule. Bacterial messenger RNA is generally polycistronic.

Question 5.
Give any six features of the human genome.
Answer:
Features of the human genome are as follows:

  • Human genome has 3.1647 billion nucleotide base pairs.
  • The average gene size is 3000 base pairs. The largest gene is that of Duchenne muscular dystrophy on X-chromosome. It has 2.4 million (2400 kilo) base pairs. (3-globin and insulin genes are less than 10 kilobases.
  • The human genome consists of about 30,000 genes. Previously, it was estimated to contain 80,000 to 100,000 genes.
  • Chromosome 1 has 2968 genes while Y-chromosome has 231 genes. They are the maximum and minimum genes for the human chromosomes.
  • The function of over 50% of discovered genes is unknown.
  • Less than 2% of the genome represents structural genes that code for proteins.

Question 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:
Whole of DNA does not open in one stretch due to very high energy requirement. The point of separation proceeds slowly towards both the directions in DNA strands. In each direction, it gives the appearance of Y-shaped structure called replication fork. Deoxyribonucleoside triphosphates (dNTPs) serve dual purposes. In addition to acting as substrates, they provide energy for polymerisation reaction (the two terminal phosphates in a deoxynucleoside triphosphates are high-energy phosphates, same as in case of ATP).

Question 7.
Retroviruses do not follow central dogma. Comment.
Answer:
Retroviruses follow reverse of central dogma. Temin and Baltimore reported that double stranded RNA of Rous Sarcoma
Virus (RSV) operates a central dogma reverse (inverse flow of information). RNA of these viruses first synthesises DN A through reverse transcription or feminism. DNA then transfers information to RNA which takes part in translation of coded information to form polypeptide. The mechanism is characteristic of retroviruses, e.g., HIV.
NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance 3

Question 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 bases 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:
In the given question, the distance between two consecutive base pairs is 0.44 nm (0.44 x 109m).
Total number of base pair is 2 x 10bp.
Length of DNA double helix is calculated by multiplying the total number of bp with distance between two consecutive bp i.e., 2 x 109 bp x 0.44 x 10-9 m/bp.Therefore, length of DNA double helix is 0.88 m.

Question 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:
Acidic amino acids are negatively charged. DNA is also negatively charged, Therefore, acidic amino acids will not be able to hold DNA over them. This will lead to failure of packaging of DNA material resulting no chromatin formation.

Question 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
Answer:
RNA is labile and prone to degradation 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 form.

Question 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:
32P is a radioactive isotope but 15N is not. 15N is heavier isotope of nitrogen. Even if it can be detected, 15N will not be able to differentiate between protein and genetic material because it will get incorporated both in DNA as well as proteins. Therefore, the experiment will not give any conclusive result.

Question 12.
There is only one possible sequence of amino acids when deduced from a given nucleotide sequence. But multiple nucleotide sequences can be deduced from a single amino acid sequence. Explain this phenomena.
Answer:
There are 64 triplet codons and only 20 amino acids. Some amino acids are coded by more than one codons. Except tryptophan (UGG) and methionine (AUG) which have single codon each, all other amino acids involved in protein synthesis have more than one codon. In degenerate codons, mostly the first two nitrogen bases are similar while the third one is different. Thus as a codon codes only for a specific amino acid we can deduce a single amino acid sequence from a nucleotide sequence. But, as an amino acid can be coded by multiple codons thus many nucleotide sequences can be coded from a given amino acid sequence.

Question 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:
Sometimes single base mutation may not cause formation of new amino acid. It occurs usually when change’ occurs in third base of triplet codon, e.g. G G A can be replaced by GGU, GGC, and GGG without affecting the incorporation of amino acid glycine. Such mutations are called silent mutations and do not cause any change in the phenotype of the organism.

Question 14.
A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomenon?
Answer:
The lac operon synthesises the permease enzyme which is responsible for transporting lactose into the cell. If permease will not be present, then the lactose cannot enter the cell and cannot act as inducer for lac operon. Hence, a low level of expression of lac operon is always maintained in the cell, so that whenever lactose is present, it can enter the cell and induce lac operon.

Question 15.
How has the sequencing of human genome opened new windows for treatment of various genetic disorders. Discuss amongst your classmates.
Answer:
Human genome sequencing has been very useful in treatment of genetic disorders, such as:

  • It has been found that more than 1200 genes are responsible for common human cardiovascular diseases, endocrine diseases (like diabetes), neurological disorders (like Alzheimer’s disease), cancer and many more. Thus, the medicines that specifically target these genes can be made.
  • Efforts are in progress to determine genes that will change cancerous cells into normal.
  • All the genes or transcripts in a particular t tissue, organ or tumor can be analysed toknow the cause or effect produced in it.

Question 16.
The total number of genes in humans is far less (< 25,000) than the previous estimate (up to 1,40,000 gene). Comment.
Answer:
At the start of genome sequencing project, scientists estimated that human genome has about 1,40,000 genes. The number gradually fell down to 30,000 and finally has been found to be between 20,000 to 25,000. The estimates were so high because scientists have found E.coli (single celled) to have about 4300 genes and Caenorhabditis elegans (about 1000 cell) to have about 19000 genes. Thus, asper the assumption that the more complex an organism the more is number of genes, the human gene number has been estimated to be more than one lakh. But, number of genes has no relationship with complexity of organism. Besides, it also has been found that genome size and gene number is unrelated because only 2% of the genome actually consists of structural genes, rest 98% is non-coding DNA.

Question 17.
Now, sequencing of total genome is getting 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 of genome sequencing are as follows:

  1. It will provide information about genetic abnormalities.
  2. There will be possibility of transfer of certain desired genes to the progeny.
  3. Proneness to certain diseases will make a person avoid certain factors causing those diseases.
  4. Information about timing of normal appearance of degenerative diseases and how to postpone them can be gathered.
  5. Metobolic defects can be taken care of. Disadvantages of genome sequencing are as follows:
  6. Knowing about disease much in advance may lead the person under depression.
  7. People will become keen to improve their genetic constitution.

Question 18.
Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage?
Answer:
Bacteriophages do not have repetitive sequences like VNTR in their genome. Bacterial genome is small in size and with all coding sequences. Therefore DNA finger­printing cannot be done in bacteriophage.

Question 19.
During in vitro synthesis of DNA, a researcher used 2′, 3′ – dideoxycytidine triphosphate as raw nucleotide in place of 2′ -deoxycytidine. What would be the consequence?
Answer:
2′-3′ dideoxycytidine triphosphate cannot form an ester bond necessary for chainformation of DNA, where as 2′ deoxycytidine phosphate can develop such bond. Chain formation of DNA or polymerisation of DNA will stop due to presence of 2′-3′- dideoxycytidine.

Question 20.
What background information did Watson and Crick have made available for developing a model of DNA? What was their contribution?
Answer:
The information already available with them for developing a model of DNA was –

  1. Erwin Chargaff’s generalisations about DNA structure e., the purine and pyrimidines are always in equal amounts A + G = T + C, and the amount of adenine is always equal to that of thymine and the amount of guanine is always equal to that of cytosine A = T and G = C.
  2. X Ray diffraction pictures of crystalline DNA produced by Maurice Wilkins and Rosalind Franklin.

Contributions of Watson and Crick to this background data are:

  1. DNAis a double helical structure with two chains running in antiparallel direction.
  2. Complementary base pairing rule
  3. Semi conservative replication
  4. Mutations occur due to tautomeric changes in nitrogen bases.

Question 21.
What are the functions of (i) methylated guanosine cap, (ii) poly-A “tail” in a mature RNA?
Answer:
Methylated guanosine cap helps in attachment of mRNA to smaller subunit of ribosome during initiation of translation process.
Poly-A-tail provides longevity to mRNA’s life. Tail length and longevity of mRNA are positively correlated.

Question 22.
Do you think that the alternative 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:
Alternative splicing is a controlled process which causes different mRNAs to be produced from a single gene by including some exons in one mRNA under some conditions, while including other exons under other conditions. This greatly increases the number of proteins that can be encoded by < 2$,000 functional genes of the genome. This alternative splicing of exons is sex-specific, tissue-specific and even developmental stage- specific. By alternative splicing of exons, a single gene may encode for several isoproteins and/or proteins of similar class.

Question 23.
Comment on the utility of variability in number of tandem repeats during DNA fingerprinting.
Answer:
Number and type of tandem repeats are specific for every individual. An individual gets them from two parents, but are unique for the individual. Therefore, they can be used to identify the individuals, and their relatives by comparing the repeats using DNA finger printing process.

Long Answer Type Questions

Question 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:
Hershey and Chase experiment is based on the fact that phosphorus is present in DNA but not in the protein, and similarly sulphur is present in proteins but not in DNA. They incorporated radioactive isotope of phosphorus (32P) into phage DNA and that of sulphur (35S) into proteins of a separate phage culture. These phage types were used independently to infect the bacterium Escherichia coli. After some time, this mixture was agitated in a blender to separate the empty phage capsids from the surface of bacterial cells and the two were separated by centrifugation. Hershey and Chase showed that when 32P was used, all radioactivity was associated with bacterial cells and if followed, appeared in the progeny phage.
However, when 35S was used, all radioactive material was limited to phage ‘ghosts’ (empty viral protein coats). These results indicated that the DNA of the bacteriophage and not the protein enters the host, where viral replication takes place. Therefore, DNA is the genetic material of T2 bacteriophage. It directs protein coat synthesis and allows replication to occur. If both DNA and proteins contained phosphorus and sulphur, they could not differentiate or separate protein and DNA and could not have concluded anything.

Question 2.
During 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:

  • It should be able to generate its replica (replication).
  • It should chemically and structurally be stable.
  • It should provide the scope for slow changes (mutation) that are required for evolution.
  • It should be able to express itself in the form of ‘Mendelian Characters’.

Though RNA is known to be the genetic material in some viruses and early cells, it is not a very suitable genetic material because 2′ – OH group present in every nucleotide of RNA is a reactive group. It means RNA is highly reactive, labile and easily degradable. RNA functions as an enzyme and is, therefore, reactive and unstable.Uracil present in RNA is less stable as compared to thymine (methyl uracil) of DNA.

Being unstable, RNA mutates at a much faster rate, that is why RNA viruses have shorter life span and mutate and evolve very fast. Such ‘rapid changes are harmful to higher forms of life.

DNA is the genetic material of most of the organisms because

  • DNA is chemically less reactive and structurally more stable as its nucleotides are not exposed except when they have to express their effect or to be replicated.
  • They are comparatively more stable than RNA. Heat which killed bacteria in Griffith’s experiment did not destroy their DNAs.
  • Presence of thymine in DNA instead of uracil, provides stability to DNA.
  • Hydrogen bonding between purines and pyrimidines and their stacking make DNA more stable for storage of genetic information.
  • DNA is capable of undergoing slow mutations required of genetic material.
  • It has a power of repairing.
    Since DNA is more stable while RNA is more reactive, both the types of nucleic acids have been retained in genetic expression. DNA which is stable enough not to change with different stages of life cycle, age or with change in metabolism of the organism, is retained as better genetic material for the storage of genetic information. It expresses genetic information by protein synthesis through RNA which is more reactive, exposed for quicker action of protein synthesising machinery and thus is better for the transmission of genetic information.

Question 3.
Give an account of post transcriptional modifications of a eukaryotic mRNA.
Answer:
Transcription in eukaryotes occurs within the nucleus. The primary mRNA transcript is longer and localised in the nucleus, where it is also called heterogenous nuclear RNA (/mRNA) or pre-mRNA.
The mRNA is processed from the primary RNA transcript in a process called maturation. Initially, at the 5′ end, a cap (consisting of 7-methyl guanosine or 7 mG) and a tail of poly A at the 3′ end are added. The cap is a chemically modified molecule of guanosine triphosphate (GTP). The primary mRNAs are made up of two types of segments; non¬coding introns and the coding exons. The introns are removed by a process called RNA splicing. Of a pair of small nuclear ribonucleoprotein (snRNPs), one binds to 5′ splice site and the other to 3′ splice site. A spliceosome forms because of interaction between snRNPs and other proteins. This spliceosome uses energy of ATP to cut the RNA and to release the introns. The enzyme ligase joins two adjacent exons to produce mature mRNA.
NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance 4

Question 4.
Discuss the process of translation in detail.
Answer:
The process of decoding of the message from mRNA to protein with the help of fRNA, ribosome and enzyme is called translation (protein synthesis). Protein synthesis occurs over ribosomes. The ribosomes are formed of two subunits. The rosette group formed by ribosomes is called polyribosome. In a polyribosome, ribosomes are held together by strand of /mRNA.
The steps involved in polypeptide synthesis are:
(1) Activation of amino acids :
In the presence of ATP, an amino acid combines with its specific aminoacyl-fRNA synthetase.

(2) Charging of fRNA : The complex formed in the above step reacts with tRNA specific for the amino acid to form aminoacyl- tRNA complex.

(3) Initiation : The small subunit of ribosome (with the mRNA) attaches to the large subunit in such a way that the initiation codon (AUG) comes on the P-site of ribosome.

(4) Elongation : Amino acids carried by the tRNA are added one by one at A site of ribosome in the sequence of the codons and become joined together to form

(5) Termination : When the termination codon present on the /mRNA is reached, the polypeptide synthesis stops and the polypeptide is released.

Question 5.
Define an operon. Giving an example, explain an inducible operon.
Answer:
Operon is a functionally integrated genetic unit for the control of gene expression in bacteria, as proposed in the Jacob-Monod hypothesis. Typically, it comprises a closely linked group of structural genes, coding for protein, and adjacent loci controlling their expression-an operator site and a promoter site. Inducible operons are those which normally remain switched off but can be induced to function under certain conditions. Lac operon is an inducible operon. The ; gene of the lac operon codes for a repressor molecule that inhibits synthesis of the structural genes. Lac operon gets switched ‘on’ in the presence of lactose. The repressor molecule coded by i gene is inactivated by interaction with the inducer (lactose). This allows RN Apolymerase access to the promoter, and transcription proceeds. The operon gets switched ‘off’ in the absence of lactose. The repressor molecule binds again with the operator region of the operon and prevents RNA polymerase from transcribing the operon.
NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance 5

Question 6.
There is a paternity dispute for a child’. Which technique can solve the problem. Discuss the principle involved.
Answer:
The paternity dispute of a child can be solved by DNA fingerprinting.
The procedure in DNA-fingerprinting includes the following:

  1. Extraction – DNA is extracted from the cells in a high-speed, refrigerated centrifuge.
  2. Amplification – Many copies of the extracted DNA are made by polymerase chain reaction.
  3. Restriction digestion – DNA is cut into fragments with restriction enzymes into precise reproducible sequences.
  4. Separation of DNA sequences/restriction fragments – The cut DNA fragments are introduced and passed through electrophoresis containing agarose polymer gel, the separated fragments can be visualised by staining them with a dye that shows fluorescence under ultraviolet radiation.
  5. Southern blotting – The separated DNA sequences are transferred on a nitrocellulose or nylon membrane.
  6. Hybridisation – The nylon membrane is immersed in a bath and radioactive probes are added, these probes target a specific nucleotide sequences that is complementary to them.
  7. Autoradiography – The nylon membrane is pressed on an X-ray film and dark bands develop at the probe sites.

After hybridisation with the radio labelled Variable Number of Tandem Repeats (VNTR) probe and autoradiography, bands of various sizes are formed. The bands form a characteristic pattern which varies from individual to individual. From the patterns developed by the samples A and B it can be confirmed to whether they belong to one individual or two different individuals. If the banding patterns are similar they belong to the same individual but if banding patterns As. are dissimilar then A and B are from different individuals.

Question 7.
Give an account of the methods used in sequencing the human genome.
Answer:
The methods involved two major approaches :
(1) One approach called as Expressed Sequence Tags (ESTs), focussed on identifying all the genes that are expressed as RNA.
(2) Second approach called as sequence annotation, was to simply sequence the whole set of genome, that included all the coding and non-coding sequences and later assigning functions to different regions in the sequences. The steps involved are as follows :

  • The total DNA from the cell is isolated and converted into random fragments of relatively smaller sizes.
  • These fragments are then cloned in suitable hosts using specialised vectors. The commonly used hosts are bacteria and yeast and the vectors are bacterial artificial chromosomes (BAC) and yeast artificial chromosomes (YAC).
  • The fragments are then sequenced using automated DNA sequencers, which work on the principle developed by Frederick Sanger.
  • The sequences were then arranged on the basis of certain overlapping regions present in them. This requires the generation of overlapping fragments for sequencing.
  • Specialised computer based programmes were developed for alignment of the sequences.
  • These sequences were annotated and assigned to the respective chromosomes.
  • The next task was to assign the genetic and physical maps on the genome, this was generated using the information on polymorphism of restriction endonuclease recognition sites and certain repetitive DNA, sequences, called microsatellites.

Question 8.
List the various markers that are used in DNA fingerprinting.
Answer:
Various markers used in DNA fingerprinting are as follows:

  1. Restriction fragment length polymor­phism (RFLPs)
  2. Variable number of tendem repeats (VNTRs)
  3. Short tendem repeats
  4. Microsatellite DNA
  5. Single nucleotide polymorphism (SNPs)

Question 9.
Replication was allowed to take place in the presence of radioactive deoxynucleotide precursors in 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?
NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance 6
NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance 7
Answer:
(a) Two peaks will be obtained, one of high molecular weight strands that are long, continuous strands which were synthesised on leading strand. The other peak is obtained of low molecular weight strands which is due to short Okazaki fragments synthesised on lagging strand because mutated DNA ligase will not join the Okazaki fragments.

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NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction

NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction

Multiple Choice Questions

Question 1.
Choose the incorrect statement from the following. 
(a) In birds and mammals internal fertilisation takes place.
(b) Colostrum contains antibodies and nutrients.
(c) Polyspermy in mammals is prevented by the chemical changes in the egg surface.
(d) In the human female implantation occurs almost seven days after fertilisation.
Answer:
(c) : Polyspermy is the entry of several sperms into the egg during fertilisation. It occurs in animals with yolky eggs (e.g., birds). In humans, although only one sperm nucleus actually fuses with the egg nucleus. Due to acrosomal reaction, plasma membrane of the sperm fuses with the plasma membrane of the secondary oocyte, so that the sperm contents enter the oocyte. Binding of the sperm to the secondary oocyte induces depolarisation of the oocyte plasma membrane. Depolarisation  prevents polyspermy.

Question 2.
Identify the correct statement from the following.
(a) High levels of estrogen triggers the ovulatory surge.
(b) Oogonial cells start to proliferate and give rise to functional ova in regular cycles from puberty onwards.
(c) Sperms released from seminiferous tubules are highly motile.
(d) Progesterone level is high during the post ovulatory phase of menstrual cycle.
Answer:
(d)

Question 3.
Spot the odd one out from the following structures with reference to the male reproductive system.
(a) Rete testis
(b) Epididymis
(c) Vasa efferentia
(d) Isthmus
Answer:
(d) : Females have two oviducts (or Fallopian tubes). Each oviduct has three regions, infundibulum, ampulla and isthmus. Isthmus is the narrow part closest to uterus, while infundibulum is the first part of oviduct that is encountered by released ova.

Question 4.
Seminal plasma, the fluid part of semen, is contributed by
(1) seminal vesicle
(2) prostate
(3) urethra
(4) bulbourethral gland
(a) (i) and (ii)
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) : Seminal plasma is the fluid part of semen and is contributed by seminal vesicles, prostate gland and bulbourethral glands. Seminal vesicles contribute fructose, citric acid and other nutrients as well as fibrinogen and prostaglandins. Secretions from prostate gland contain calcium ions, phosphate ion etc. and are alkaline in nature. Bulbourethral glands secrete alkaline mucus which is important for the lubrication of penis.

Question 5.
Spermiation is the process of the release of sperms from
(a) seminiferous tubules

(b) vas deferens
(c) epididymis
(d) prostate gland.
Answer:
(a) : Spermiation is the release of sperm from seminiferous tubules. From here, they will be transported to vasa efferentia and then to epididymis where maturation of sperm occurs.

Question 6.
Mature Graafian follicle is generally present in the ovary of a healthy human female around
(a) 5-8 day of menstrual cycle
(b) 11-17 day of menstrual cycle
(c) 18-23 day of menstrual cycle
(d) 24-28 day of menstrual cycle.
Answer:
(b) : Starting from puberty, monthly changes in ovaries and uterus also starts. The layer of cells surrounding primary oocyte are called granulosa cells and together, they are  called primary follicle. The growth of the follicles is under the influence of pituitary hormones FSH and LH. Further growth is also stimulated by estrogen. The mature Graafian follicle is present in the ovary around 11-17 day of menstrual cycle and on rupture of which, ovum is released (because of LH surge, 1-2 days before ovulation).

Question 7.
Acrosomal reaction of the sperm occurs due to
(a) its contact with zona pellucida of the ova
(b) reactions within the uterine environment of the female
(c) reactions within the epididymal environ­ment of the male
(d) androgens produced in the uterus.
Answer:
(a) : When head of sperm binds to zona pellucida (ZP) of ovum the acrosome release its contents by exocytosis. The content inside acrosome includes various hydrolytic enzymes like hyaluronidase, corona penetrating enzymes etc. This is called acrosomal reaction. It helps the sperm to reach the plasma membrane of ovum, by dissolving corona radiata and zona pellucida.

Question 8.
Which one of the following is not a male accessory gland?
(a) Seminal vesicle
(b) Ampulla
(c) Prostate
(d) Bulbourethral gland
Answer:
(b) : Ampulla is the part of oviduct between the infundibulum and isthmus. It  is at tfre ampullary-isthmus junction, where fertilisation take place.

Question 9.
The immature male germ cell undergoes division to produce sperms by the process of spermatogenesis. Choose the correct one with reference to above.
(a) Spermatogonia have 46 chromosomes and always undergo meiotic cell division.
(b) Primary spermatocytes divide by mitotic cell division.
(c) Secondary spermatocytes have 23 chromo­somes and undergo second meiotic division.
(d) Spermatozoa are transformed into spermatids.
Answer:
(c) : Spermatogonia are diploid cells on the inside wall of seminiferous tubules that multiply by mitotic divisions. Some of the spermatogonia called primary spermatocyte undergo meiosis-I to give rise to secondary spermatocytes (haploid). Each secondary spermatocyte undergoes meiosis-II to give rise to two haploid spermatids which are transformed to spermatozoa by spermiogenesis.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 1
Question 10.
Match between the following representing parts of the. sperm and their functions and choose the correct option.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 2

(a) A-{ii), B-(iv), C-(i), D-(iii)
(b) A-(iv), B-(iii), C-(i), D-(ii)
(c) A-(iv), B-(i), C-(ii), D-(iii)
(d) A-(ii), B-(i), C-(iii), D-(iv)
Answer:
(b)

Question 11.
Which among the following has 23 chromo­somes?
(a) Spermatogonia
(b) Zygote
(c) Secondary oocyte .
(d) Oogonia
Answer:
(c) : Spermatogonia are the cells on the inside wall of seminiferous tubules and consist of 46 chromosomes. Oogonia are also diploid cells and formed in the foetal ovary. Zygote is also diploid and fertilised ovum formed by fusion of male and female gametes. Secondary oocyte has 23 chromosomes and is formed by meiosis-I of primary oocyte.

Question 12.
Match the following and choose the correct option.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 3
(a) A-(ii), B-(i), C-(iii), D-(iv)
(b) A-(iii), B-(iv), C-(ii), D-(i)
(c) A-(iii), B-(i), C-(ii), D-(iv)
(d) A-(ii), B-(iv), C-(iii), D-(i)
Answer:
(b)

Question 13.
Which of the following hormones is not secreted by human placenta?
(a) hCG
(b) Estrogens
(c) Progesterone
(d) LH
Answer:
(d) : LH is the hormone that is not secreted by human placenta. It is secreted by anteribr pituitary. Initially during pregnancy, hCG performs the function of LH of maintaining corpus luteum for the secretion of progesterone. Later (around 16th week) in pregnancy, when placenta itself starts secreting estrogen and progesterone, corpus luteum regresses.

Question 14.
The vas deferens receives duct from the seminal vesicle and opens into urethra as
(a) epididymis
(b) ejaculatory duct
(c) efferent ductule
(d) Ureter
Answer:
(b)

Question 15.
Urethral meatus refers to the
(a) urinogenital duct
(b) opening of vas deferens into urethra
(c) external opening of the urinogenital duct
(d) muscles surrounding the urinogenial duct.
Answer:
(c) : Urethral meatus refers to the external opening of urinogenital duct, through which in males, urine and semen both exits the body.

Question 16.
Morula is a developmental stage
(a) between the zygote and blastocyst
(b) between the blastocyst and gastrula
(c) after the implantation
(d)between implantation and parturition.
Answer:
(a) : After fertilisation, zygote is formed. Zygote undergoes, mitotic divisions, called cleavage, to form 2, 4, 8 and 16 daughter cells called blastomeres. These divisions start when zygote is moving towards uterus. It is called morula at 8-16 cell stage and it continues to divide to form blastocyst. Blastocyst has two type of cells : trophoblast cells and inner cell mass. Around 7 day after fertilisation, implantation through trophoblast cells occurs.

Question 17.
The membranous cover of the ovum at ovulation is
(a) corona radiata

(b) zona radiata
(c) zona pellucida
(d) Chorion
Answer:
(a) : The outermost membranous cover of the ovum at ovulation is corona radiata. It is formed by follicular cells. Inner to corona radiata is zona pel lucid a, which is made up of three different glycoproteins secreted by the ovum itself.

Question 18.
Identify the odd one from the following.
(a) Labia minora
(b) Fimbriae
(c) Infundibulum
(d) Isthmus
Answer:
(a) : Labia minora are paired folds of tissues under labia majora which in turn surrounds the vaginal opening. Fimbriae, infundibulum and isthmus, along with ampulla are parts of oviduct (or Fallopian tube).

Very Short Answer Type Questions


Question 1.
Given below are the events in human reproduction. Write them in correct sequential order. Insemination, gametogenesis, fertilisation, parturition, gestation, implantation
Answer:
Gametogenesis, insemination, fertilisa­tion, implantation, gestation, parturition.

Question 2.
The path of sperm transport is given below. Provide the missing steps in blank boxes.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 4
Answer:
Seminiferous tubules —>Rete testis —> Vasa efferentia —> Epididymis —> Vas deferens -> Urethra

Question 3.
What is the role of cervix in the human female reproductive system?
Answer:
Role of cervix in human female reproductive system:

  1. Cervical canal e., the cavity of cervix and vagina together form birth canal to facilitate parturition.
  2. It regulates the passage of sperms into the uterus.

Question 4.
Why are menstrual cycles absent during pregnancy?
Answer:
Levels of progesterone and estrogen are high during pregnancy. Their high levels suppress the release of gonadotropin (FSH) responsible for transformation of primary follicles into Graafian follicles, and for ovulation. And, no ovulation means, no menstrual cycle.

Question 5.
Female reproductive organs and associated functions are given below in column A and B. Fill the blank boxes.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 5
Answer:
a – Fertilisation, b – Uterus

Question 6.
From where the parturition signals arise- mother or foetus? Mention the main hormone involved in parturition.
Answer:
Parturition is induced by complex neuroendocrine mechanism. Signals for parturition originate from fully developed foetus and placenta which induce mild uterine contractions called foetal ejection reflex. Oxytocin from maternal pituitary induces strong uterine contractions of myometrium, which leads to expulsion of baby (parturition) through birth canal.

Question 7.
What is the significance of epididymis in male fertility?
Answer:
Epididymis helps in storage, nutrition and physiological maturation of sperms. It also aids motility to sperms.

Question 8.
Give the names and functions of the hormones involved in the process of spermatogenesis. Write the names of the endocrine glands from where they are released.
Answer:
The name of hormones, its endocrine glands and functions of the hormones involved in the process of spermatogenesis are given in the following table:

Hormone Endocrine gland Function
GnRH (Gonado­ tropin releasing hormone) Released from hypothalamus Stimulates the anterior pituitary to release FSH and LH.
Follicle stimulating hormone (FSH) Released from anterior pituitary Stimulates Sertoli cells to secrete certain factors which help in spermatogenesis.
Luteinising hormone (LH) Released from anterior pituitary Stimulates Leydig’s cells of testes to secrete androgens (testosterone) which regulate

Question 9.
The mother germ cells are transformed into a mature follicle through series of steps. Provide the missing steps in the blank boxes.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 6
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 7

Question 10.
During reproduction, the chromosome number (2n) reduces to half (n) in the gametes and again the original number (2n) is restored in the offspring. What are the processes through which these events take place?
Answer:
Chromosome number is reduced to half (n) during the process of gametogenesis, and it is again restored to (2n) as a result of fertilisation.

Question 11.
What is the difference between a primary oocyte and a secondary oocyte?
Answer:
Primary oocyte is a diploid structure surrounded by single layer of follicular  through mitosis and differentiation. Secondary oocyte is a hapoid structure and is surrounded by a few layer of granular cells and theca. It is formed from primary oocyte after it undergoes first meiotic division.

Question 12.
What is the significance of ampullary-isthmic junction in the female reproductive tract?
Answer:
Ampullary-isthmic junction of oviduct is a site for fertilisation.

Question 13.
How does zona pellucida of ovum help in preventing polyspermy?
Answer:
During fertilisation, a sperm comes in contact with zona pellucida layer of ovum and induces changes in the membrane known as cortical reaction. This blocks the entry of additional sperms (polyspermy) and maintains monospermy.

Question 14.
Mention the importance of LH surge during menstrual cycle.
Answer:
Rapid secretion of LH leading to its maximum level during mid cycle is known as LH surge. It induces rupture of Graafian follicle and release of ovum (ovulation).

Question 15.
Which type of cell division forms spermatids from the secondary spermatocytes?
Answer:
Secondary spermatocytes undergo meiosis II to form spermatids. Secondary spermatocytes and spermatid both are haploid and this meiosis II is known as equational division.

Short Answer Type Questions

Question 1.
A human female experiences two major changes, menarche and menopause during her life. Mention the significance of both the events.
Answer:
The first menstrual cycle in females is known as menarche. It indicates the attainment of sexual maturity and puberty stage in females. It signifies the maturation and readiness of the female reproductive system for child bearing. It occurs usually at the age of 10-14 years.Menopause is the period of end of cyclic change in females. It indicates the end of reproductive cycle or fertile period in females. It occurs between 45-55 years of age.

Question 2.
(a) How many spermatozoa are formed from one secondary spermatocyte?
(b) Where does the first cleavage division of zygote take place?
Answer:
(a) A secondary spermatocyte gives rise to two spermatids, which get  transformed into two spermatozoa.
(b) First cleavage division of zygote takes place in Fallopian tubes of females.

Question 3.
Corpus luteum in pregnancy has a long life. However, if fertilisation does not take place, it remains active only for 10-12 days. Explain.
Answer:
Zygote formed after fertilisation gets implanted in the inner lining of uterus, called endometrium. Neural signals are conducted to hypothalamus to sustain LH secretion. This helps in maintaining corpus luteum and continue the secretion of progesterone during gestation period. But when fertilisation does not take place, neural signals are not sent to hypothalamus. As a result corpus luteum starts degenerating and can stay alive only for 10-12 days.

Question 4.
What is foetal ejection reflex? Explain how it leads to parturition?
Answer:
Foetal ejection reflex are the uterine contractions induced by fully developed foetus and placenta which signals for parturi­tion.
This stimulates the posterior pituitary of the mother to release oxytocin. Oxytocin causes stronger uterine contractions of smooth muscles of myometrium called labour pains, which further stimulate more secretion of oxytocin. The stimulatory reflex  continues resulting in stronger and stronger contractions and leading to expulsion of the baby (parturition) through birth canal.

Question 5.
Except endocrine function, what are the other functions of placenta?
Answer:
Functions of placenta other than endocrine functions are:

  1. Nutritive organ – Food materials pass from the mother’s blood into the foetal blood through the placenta.
  2. Digestive organ – The trophoblast of the placenta digests (breaks down) proteins before passing them into the foetal blood.
  3. Respiratory organ – Oxygen diffuses from the maternal blood into the foetal blood through the placenta. Carbon dioxide diffuses from the foetal blood into the maternal blood also through the placenta for elimination by the mother’s lungs. Foetal haemoglobin has a greater affinity for oxygen than adult haemoglobin.
  4. Excretory organ – Nitrogenous wastes, such as urea, pass from the foetal blood into the maternal blood via placenta for elimination by mother’s kidneys.
  5. Storage organ – The placenta stores glycogen for the foetus before liver is formed.
  6. Barrier – Placenta serves as an efficient barrier and allows those materials to pass into the foetal blood that are necessary.

Question 6.
Why doctors recommend breast feeding during initial period of infant growth?
Answer:
Human milk consists of water, minerals, fats, proteins and sugar necessary for development of the child. The milk produced’ during initial days of lactation is called colostrum. It is rich in proteins (lactalbumin and lactoprotein) and various other nutrients. It also contains certain antibodies (IgA), which provide passive immunity to the baby.

Question 7.
What are the events that take place in the ovary and uterus during follicular phase of the menstrual cycle.
Answer:
(1) Changes inside the ovary during follicular phase :-

  • Primary follicle gets transformed into Graafian follicle.
  • Increase in level of estrogen

(2) Changes in uterus during follicular phase :-

  • Proliferation of endometrium.
  • Endometrium becomes highly vascular and glandular.

Question 8.
Given is a flow chart showing ovarian changes during menstrual cycle. Fill in the spaces giving the name of the hormones responsible for events shown.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 8
Answer:
a – FSH and LH
b – LH
c – Progesterone

Question 9.
Give a schematic labelled diagram to represent oogenesis (without descriptions).
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 9

Question 10.
What are the changes in the oogonia during the transition of a primary follicle to Graafian follicle?
Answer:
Oogonia divide by mitosis forming primary oocyte which then gets surrounded by a layer of granulosa cells to form primary follicle. The primary follicles are surrounded by more layers of granulosa cells called secondary follicles.The secondary follicle soon changes into a tertiary follicle which is characterised by a fluid filled cavity called follicular antrum. The tertiary follicle is further converted into mature follicle or Graafian follicle. The formation of Graafian follicle through various stages is called folliculogenesis.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 10

Long Answer Type Questions

Question 1.
What role does pituitary gonadotropins play during follicular and ovulatory phases of menstrual cycle? Explain the shifts in steroidal secretions.
Answer:
The menstrual cycle consists of four phases: menstrual phase, follicular phase (proliferative phase), ovulatory phase and secretory phase. Hormonal control during follicular phase:

  • FSH from pituitary transforms the primary follicle into Graafian follicle.
  • FSH also stimulates Graafian follicular cells to secrete estrogen.
  • Estrogen causes proliferation of endo­metrium of the uterine wall.

Hormonal control during ovulatory phase :

  • Pituitary glands are stimulated by estrogen to release LH.
  • LH surge (maximum level of LH) causes ovulation from the ovary on 14th day of menstruation cycle. LH also induces transformation of Graafian follicle into corpus luteum, inside the ovary. LH stimulates corpus luteum to secrete progesterone to help implantation,placentation, and maintenance of pregnancy.
  • In menstrual phase, there is reduction of progesterone and estrogen. Gonado­tropin releasing hormone (GnRH) stimulates the release of FSH and LH. FSH stimulates the ovarian follicles to produce estrogens during proliferative phase. LH stimulates the ovulation in ovulatory phase.
  • LH develops corpus luteum which causes increased production of progesterone in secretory phase.NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 11

Question 2.
Meiotic division during oogenesis is different from that in spermatogenesis. Explain how and why?
Answer:
Primary spermatocyte divides by meiosis I to form two secondary spermato­cytes. Primary oocyte undergoes meiosis I to form one secondary oocyte and one polar body.
Secondary spermatocyte divides by meiosis II to produce two spermatids. Secondary oocyte divides by meiosis II to form one ovum and one polar body.
A spermatocyte forms four spermatozoa. An oocyte forms only one egg or ovum. Unequal cell divisions during oogenesis, makes the ovum much larger than the other three polar bodies. Ovum has more cytoplasm and more organelles, it has a better chances of surviving. The male makes million of tiny sperms while, the female makes one egg per month because the sperms have to search for the eggs and only a few succeed in this task.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 12

Question 3.
The zygote passes through several developmental stages till implantation. Describe each stage briefly with suitable diagrams.
Answer:

  1. The zygote moves from isthmus to uterus and undergo mitotic divisions called cleavage divisions.
  2. It forms 2, 4, 6, 8,16 daughter cells, called  blastomeres.
  3. The embryo with 8-16 blastomeres is called morula.
  4. The morula continues to get transformed into blastocyst.
  5. The blastomeres in the blastocyst are arranged into an outer layer called trophoblast, and an inner group of cells attached to trophoblast called inner cell mass.
  6. The trophoblast layer gets attached to the endometrium and inner cell mass gets differentiated as the embryo.
  7. After attachment uterine cells divide rapidly and cover the blastocyst.
  8. As a result the blastocyst becomes embedded in the endometrium of the uterus.
  9. This is known as implantation, and it leads to pregnancy.
    NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 13

Question 4.
Draw a neat diagram of thefemale reproductive system and label the parts associated with the following (a) production of gamete, (b) site of fertilisation(c) site of implantation and, (d) birth canal.
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 14

Question 5.
With a suitable diagram, describe the organisation of mammary gland.
Answer:
Mammary glands are paired structures consisting of glandular tissue, the fibrous tissue and variable amount of fat. They are compound tubulo-alveolar modified sweat glands. The glandular tissue of breast consists of about 15-20 lobes of milk glands. Each lobe is formed of many lobules containing cluster of cells called alveoli. The cells of alveoli secrete milk, which is stored in the lumen (cavities) of alveoli. The alveoli open into mammary  tubules, the tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla, which is connected to lactiferous duct, through which milk is sucked out. Mammary glands are functional in females and vestigial in males. Breasts in females are small sized upto puberty. The size increases after puberty under stimulation of estrogen. It further increases during pregnancy and after child birth under the stimulation of prolactin and progesterone.
NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction 15

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NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants

NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants

Multiple Choice Questions

Question 1.
Among the terms listed below, those that are not technically correct names for a floral whorl are
(a) androecium
(b) corolla|
(c) (i) and (iv)
(d) (ii) and (iv)
Answer:
(c) : The technically correct terms for the floral whorls are (from outermost to innermost) calyx, corolla, androecium and gynoecium. They are made up of sepals, petals, stamens and carpels, respectively.

Question 2.
Embryo sac is to ovule as________ is to an anther.
(a) stamen
(b) filament
(c) pollen grain   
(d) androecium
Answer:
(c) : A typical carpel consist of ovary which can have many ovules. Each ovule has an embryo sac, which in turn has a single egg cell. Similarly, in majority of angiosperms each stamen consists of a bilobed anther, which in turn has two pollen sacs in each lobe, consisting of pollen grains.

Question 3.
In a typical complete, bisexual and hypogynous flower the arrangement of floral whorls on the thalamus from the outermost to the innermost is
(a) calyx, corolla, androecium and gynoecium
(b) calyx, corolla, gynoecium and androecium|
(c) gynoecium, androecium, corolla and calyx
(d) androecium, gynoecium, corolla and calyx.
Answer:
(a)

Question 4.
A dicotyledonous plant bears flowers but never produces fruits and seeds. The most probable cause for the above situation is
(a) plant is dioecious and bears only pistillate flowers
(b) plant is dioecious and bears both pistillate and staminate flowers
(c) plant is monoecious
(d) plant is dioecious and bears only staminate flowers.
Answer:
(d) : Fruits can develop from a single ovary of a single flower (simple fruit) or from several free carpels of a single flower (aggregate) or from whole inflorescence (multiple). In total, fruits develop from ovaries. This is why, a dioecious plant (unisexual) bearing only staminate (male) flowers will not produce fruits, whereas monoecious plants (bisexual) or dioecious plants bearing only pistillate (female) flowers or pistillate and staminate both can bear fruits (pollination).

Question 5.
The outermost and innermost wall layers of microsporangium in an anther are respectively
(a) endothecium and tapetum
(b) epidermis and endodermis
(c) epidermis and middle layer
(d) epidermis and tapetum.
Answer:
(d) : The wall layers of a micro­sporangium from outermost to innermost are : epidermis, endothecium, middle layers and tapetum. The first three layers generally provide protection and help in dehiscence of anther. Tapetum performs nutritive function for pollen grains.

Question 6.
During microsporogenesis, meiosis occurs in
(a) endothecium
(b) microspore mother cells
(c) microspore tetrads
(d) pollen grains.
Answer:
(b) : Microsporogenesis is the formation of microspores in the form of tetrads, which later separate and are called pollen grains. Microspore mother cell (2n) undergoes meiosis for the formation of haploid pollen grains formed first in the form of spore tetrads.

Question 7.
From among the sets of terms given below, identify those that are associated with the gynoecium.
(a) Stigma, ovule, embryo sac, placenta

(b) Thalamus, pistil, style, ovule
(c) Ovule, ovary, embryo sac, tapetum
(d) Ovule, stamen, ovary, embryo sac
Answer:
(a) : Stigma,is a part of pistil on which pollen grain lands. Each ovary may have many ovules, which have embryo sacs in them. Placenta is a tissue inside ovary to which ovules are attached. Thalamus is the terminal part of the axis of flower which bears all floral appendages. Tapetum is the innermost layer of microsporangium while stamen is component of androecium.

Question 8.
Starting from the innermost part, the correct sequence of parts in an ovule are
(a) egg, nucellus, embryo sac, integument
(b) egg, embryo sac, nucellus, integument
(c) embryo sac, nucellus, integument, egg
(d) egg, integument, embryo sac, rtucellus.
Answer:
(b) : Egg cell is inside the embryo sac in the ovule. The embryo sac is further enclosed by the parenchymatous tissue, nucellus, which later provides nutrition to developing embryo. Nucellus is ultimately surrounded by integuments.

Question 9.
From the statements given below choose the option that are true for a typical female gametophyte of a flowering plant.
(1) It is 8-nudeate and 7-celled at maturity.
(2) It is free-nuclear during the development.
(3) It is situated inside the integument but outside the nucellus.
(4) It has an egg apparatus situated at the chalazal end.
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)   
(d) (ii) and (iv)
Answer:
(c) : Female gametophyte or embryo ‘* sac is present in the nucellus of an ovule. It is formed by free nuclear mitotic divisions of megaspore which forms 8 nucleate structure (4 nuclei each at micropylar and chalazal end). One nucleus from each side moves to the middle to form polar nuclei which later on fuse to form secondary nucleus. The remaining three nuclei at each end get surrounded by wall to form cells. Hence, female gametophyte is 8-nucleate and 7-celled at maturity because of presence of secondary nucleus. Egg apparatus is situated at the micropylar end whereas 3 antipodal s are situated at chalazal end.

Question 10.
Autogamy can occur in a chasmogamous flower if
(a) pollen matures before maturity of ovule
(b) ovules mature before maturity of pollen
(c) both pollen and ovules mature simultaneously
(d) both anther and stigma are of equal lengths.
Answer:
(c) : Autogamy is pollination within a flower, chasmogamous flowers are those in which anthers and stigma are exposed. For autogamy, in such a flower to take place, pollen and ovule should mature simultaneously and anther and stigma should lie close to each other.

Question 11.
Choose the correct statement from the following.
(a) Cleistogamous flowers always exhibit autogamy.

(b) Chasmogamous flowers always exhibit geitonogamy.
(c) Cleistogamous flowers exhibit both autogamy and geitonogamy.
(d) Chasmogamous flowers never exhibit autogamy.
Answer:
(a) : Autogamy is pollination within the same flower. Geitonogamy is pollination between different flowers of same plant. Xenogamy is pollination between flowers of different plants of same species. Cleistogamous flowers (that do not open at all) always exhibit autogamy, where as chasmogamous flowers (with exposed anthers and stigma) can exhibit autogamy, geitonogamy or xenogamy.

Question 12.
A particular species of plant produces light, non-sticky pollen in large numbers and its stigmas are long and feathery. These modifications facilitate pollination by
(a) insects
(b) water
(c) wind
(d) animals
Answer:
(c) : Light, non-sticky pollens produced in large numbers are generally traits of wind pollinated (anemophilous) plants. Insect and animal pollinated plants have sticky pollens. Long and feathery stigma is also characteristic of anemophilous plants. Maize, Cannabis and many grasses are some of the examples of this category.

Question 13.
From among the situations given below, choose the one that prevents both autogamy and geitonogamy.
(a) Monoecious plant bearing unisexual flowers.
(b) Dioecious plant bearing only male or female flowers.
(c) Monoecious plant with bisexual flowers.
(d) Dioecious plant with bisexual flowers.
Answer:
(b) : Monoecious plant (bisexual) bearing either bisexual or unisexual flowers can exhibit both autogamy as well as geitonogamy. Dioecious (unisexual) plants bearing only male or female flowers will not show autogamy or geitonogamy hence, only xenogamy is possible.

Question 14.
In a fertilised embryo sac, the haploid, diploid and triploid structures are
(a) synergid, zygote and primary endosperm nucleus

(b) synergid, antipodal and polar nuclei
(c) antipodal, synergid and primary endo­sperm nucleus
(d) synergid, polar nuclei and zygote.
Answer:
(a) : Double fertilisation is the fusion of two male gametes brought by a pollen tube with two different cells of the same female gametophyte in order to produce two different structures. It is found only in angiosperms where it was first discovered by Nawaschin in 1898 in Fritillaria and Lilium. Out of the two male gametes one fuses with egg or oosphere to perform generative fertilisation. Generative fertilisation is also called syngamy or true fertilisation. It gives rise to a diploid zygote or oospore. The second male gamete fuses with two haploid polar nuclei or diploid secondary nucleus of the central cell to form a triploid primary endosperm nucleus (PEN). This is called as vegetative fertilisation (or triple fusion).

Question 15.
In an embryo sac, the cells that degenerate after fertilisation are
(a) synergids and primary endosperm cell
(b) synergids and antipodals
(c) antipodals and primary endosperm cell
(d) egg and antipodals.
Answer:
(b)

Question 16.
While planning for an artificial hybridisation programme involving dioecious plants, which of the following steps would not be relevant?
(a) Bagging of female flower
(b) Dusting of pollen on stigma
(c) Emasculation
(d) Collection of pollen
Answer:
(c) : Artificial hybridisation is human performed crossing of two different plants having complementary good traits in order to obtain an overall superior variety. Artificial hybridisation has been used by plant breeders for crop improvement programme. Two precautionary measures in artificial hybridisation are emasculation and bagging. Emasculation is removal of stamens from the floral buds of female parent so that chances of self pollination are eliminated. In case of dioecious (unisexual) plants, emasculation is not required.

Question 17.
In the embryos of a typical dicot and a grass, true homologous structures are
(a) coleorhiza and coleoptile
(b) coleoptile and scutellum
(c) cotyledons and scutellum
(d) hypocotyl and radicle.
Answer:
(c) : During the development of dicot embryo, initially the dicot embryo is globular and undifferentiated. Early embryo with radial symmetry is called proembryo. It is transformed into embryo with the development of radicle, plumule and cotyledons. Two cotyledons differentiate from the sides with a faint plumule in the centre. At this time the embryo becomes heart-shaped. Part of embryo axis between the plumule and cotyledonary node is epicotyl (above the level of cotyledons) while the part between radicle and cotyledonary node is called hypocotyl (below the level of cotyledons). The single cotyledon of monocotyledonous seed (e.g. maize grain) is called scutellum. It occupies the major portion of the embryo regions of grain.

Question 18.
The phenomenon observed in some plants wherein parts of the sexual apparatus is used for forming embryos without fertilisation is called
(a) parthenocarpy
(b) apomixis
(c) vegetative propagation
(d) sexual reproduction.
Answer:
(b) : Apomixis is the term given to any phenomenon that leads to formation of embryo wherein parts of the sexual apparatus are used, but without fertilisation. Fertilisation is also absent in vegetative propagation, but parts of sexual apparatus are not involved. An example of apomixis is Citrus.

Question 19.
In a flower, if the megaspore mother cell forms megaspores without undergoing meiosis and if one of the megaspores develops into an embryo sac, its nuclei would be 
(a) haploid
(b) diploid
(c) a few haploid and a few diploid
(d) with varying ploidy.
Answer:
(b)

Question 20.
The phenomenon wherein, the ovary develops into a fruit without fertilisation is called
(a) parthenocarpy

(b) apomixis
(c) asexual reproduction
(d) sexual reproduction.
Answer:
(a) : Fertilised ovary is technically called fruit. But if ovary develops into fruit, without fertilisation, it is called parthenocarpic fruit. Such fruits are generally seedless. Some common examples found in nature are : Citrus, banana, etc. Parthenocarpy can also be artificially induced by the application of certain plant hormones, specially, auxin and gibberellins.

Very Short Answer Type Questions

Question 1.
Name the component cells of the ‘egg apparatus’ in an embryo sac.
Answer:
Egg apparatus consists of two synergids and one egg cell.

Question 2.
Name the part of gynoecium that determines the compatible nature of pollen grain.
Answer:
Stigma is that part of gynoecium which determines the compatible nature of pollen grain.

Question 3.
Name the common function that cotyledons and nucellus perform.
Answer:
Common function of nucellus and cotyledons is to provide nourishment.

Question 4.
Complete the following flow chart:
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 1
Answer:
2nd PUC Basic Maths Question Bank Chapter 3 Probability 16

Question 5.
Indicate the stages where meiosis and mitosis occur (1, 2 or 3) in the flow chart.
Megaspore mother cell ——-1—— > Megaspores –2— > Embryo sac —-3—- > Egg
Answer:
1 = Meiosis 2 = Mitosis 3 = Mitosis
In the diagram given below, show the path of a pollen tube from the pollen on the stigma into the embryo sac. Name the components of egg apparatus

Question 6.
In the diagram given below, show the path of a pollen tube from the pollen on the stigma into the embryo sac. Name the components of egg apparatus.
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 3
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 4

Egg apparatus consists of two synergids and one egg cell.

Question 7.
Name the parts of pistil which develop into fruit and seeds.
Answer:
Fruit develops from ovary and seeds develop from ovules of a pistil.

Question 8.
In case of polyembryony, if an embryo develops from the synergid and another from the nucellus, which is haploid and which is diploid?
Answer:
In case of polyembryony, if an embryo develops from synergid, it will be haploid and if it develops from nucellus, it will be diploid.

Question 9.
Can an unfertilised, apomictic embryo sac give rise to a diploid embryo? If yes, then how?
Answer:
Apomictic embryo sac develops without the involvement of meiosis and syngamy. Here, a diploid embryo is formed from diploid egg cell or from some other diploid cell of embryo sac.

Question 10.
Which are the three cells found in a pollen grain when it is shed at the three celled stage?
Answer:
The three cells found in a pollen grain when it is shed at the three celled stage are two male gametes and one vegetative cell.

Question 11.
What is self-incompatibility?
Answer:
If a pistil carrying functional female gametes fails to set seeds following pollination with viable and fertile pollen, capable of bringing about fertilisation in another pistil, the two are said to be incompatible, and the phenomenon is known as sexual incompatibility. Sexual incompatibility may be interspecific (between individuals of different species) or intraspecific (between individuals of the same species). The latter is also called self-incompatibility.

Question 12.
Name the type of pollination in self­ incompatible plants.
Answer:
Self incompatible plants do not undergo self pollination. They can undergo only cross pollination.

Question 13.
Draw the diagram of a mature embryo sac and show its 8-nudeate, 7-celled nature. Show the following parts: antipodals, synergids, egg, central cell, polar nuclei.
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 6

Question 14.
Which is the triploid tissue in a fertilised ovule? How is the triploid condition achieved?
Answer:
Endosperm is the triploid tissue in a fertilised ovule which is formed by the division of Primary Endosperm Nucleus (PEN). PEN is formed by the fusion of one male gamete (haploid) with secondary nucleus (diploid) hence, it is triploid.

Question 15.
Are pollination and fertilisation necessary in apomixis? Give reasons.
Answer:
In apomixis, there is no need of pollination and fertilisation. Embryo can develop directly from the nucellus or synergid or egg cell.

Question 16.
Identify the type of carpel with the help of diagrams given below:
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 7

(a) Polycarpellary syncarpous
(b) Polycarpellary apocarpous

Question 17.
How is, pollination carried out in water plants?
Answer:
Plants which occur below the water level undergo epihydrophily or hypohydrophily,i.e pollination takes place by the agency of water. Aquatic plants which are emerged or present at the water surface can undergo entomophily, anemophily or epihydrophily.

Question 18.
What is the function of the two male gametes produced by each pollen grain in angiosperms?
Answer:
The function of the two male gametes produced by each pollen grain in angiosperms are as follows:

  1. One of the male gamete fuses with egg cell to produce zygote (2n).
  2. Second male gamete fuses with two polar nuclei/diploid secondary nucleus to form primary endosperm nucleus (3n)

Short Answer Type Questions

Question 1.
List three strategies that a bisexual chasmogamous.flower can evolve to prevent self pollination (autogamy).
Answer:
Strategies that a bisexual chasmogamous flower can evolve to prevent self pollination are as follows:

  1. Pollen release and stigma receptivity is not synchronised. Either the pollen is released before stigma is receptive or stigma becomes receptive much before pollen release.
  2. Anther and stigma are placed at different positions, so that pollen cannot come in contact with stigma.
  3. When the pollen of the flower reach the stigma of the same flower, pollen grains
  4. do not germinate and the phenomenon is called self-incompatibility.

Question 2.
Given below are the events that are observed in an artificial hybridisation programme. Arrange them in the correct sequential order in which they are followed in the hybridisation programme.
(a) Re-bagging
(b) Selection of parents
(c) Bagging
(d) Dusting the pollen on stigma
(e) Emasculation
(f) Collection of pollen from male parent.
Answer:
(b) —> (e) —> (c) —> (f) —> (d) —> (a)

Question 3.
Vivipary automatically limits the number of offsprings in a litter. How?
Answer:
Viviparity is a form of reproduction in animals in which the development of embryos takes place within the mother’s/ female parent’s body. The embryo obtains its nourishment directly from mother via placenta or by other means and subsequently mother gives birth to the full term young one. It is common in most mammals. Viviparity limits the number of offspring in a litter due to the following reasons:

  1. As the number of eggs released during ovulation are limited during oestrous or menstrual cycles so the number of eggs fertilised during reproductive cycle of female are also limited.
  2. As the entire period of development called gestation is passed within the mother’s/female parent’s body; it restricts the number of embryo that can develop together at one time.
  3. During gestation no ova or egg are released.

Question 4.
Does self incompatibility impose any restrictions on autogamy? Give reasons and suggest the method of pollination in such plants.
Answer:
Self incompatibility is the phenomenon in which self pollens fail to germinate on stigma of pistil. It is a gene physiological process which is controlled by single gene S. If pollen and pistil will have the S alleles in common, pollens will not be functional on that pistil. As self pollens will have common S alleles with pistil (of the same flower), self pollination (autogamy) cannot take place in self incompatible plants. Cross pollens on the other hand, will not have common S alleles with pistil and hence, cross pollination can easily take place in such plants.

Question 5.
In the given diagram, write the names of parts shown with lines.
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 8

Answer:

Question 6.
What is polyembryony and how can it be commercially exploited?
Answer:
The phenomenon of having more than one embryo is called polyembryony, e.g., onion, groundnut.
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 10

Polyembryony is practically important because genetically uniform parental type seedlings are obtained from nucellar embryos. Nucellar embryos are superior to those obtained by vegetative propagation because nucellar embryo seedlings are disease free and maintain their superiority for long time.

Question 7.
Are parthenocarpy and apomixis different phenomena? Discuss their benefits.
Answer:
Parthenocarpy is production and development of seedless fruits. Apomixis is a type of sexual reproduction which does not involve meiosis and syngamy and produces seeds without fertilisation.
Benefits of parthenocarpy are as follows:
(1) It produces fruits which do not contain irritant seeds.
(2) Processing of fruits by food industry requires the removal of seeds which is quite difficult. Therefore, seedless fruits are preferred by food industry.
(3) Fruits can be developed inside green houses where pollinators are not available.
Benefits of apomixis are as follows:

  • Production of hybrid seeds is costly and hence the cost of hybrid seeds becomes too expensive for the farmers. If these hybrids are made into apomicts, there is no segregation of characters in the hybrid progeny. Then the farmers can keep on using the hybrid seeds to raise new crop year after year and do not have to buy hybrid seeds every year.
  • Adventive embryos are better clones than cuttings.
  • Embryos formed through apomixis are generally free from infections.

Question 8.
Why does the zygote begin to divide only after the division of primary endosperm cell (PEC)?
Answer:
Endosperm development precedes embryo development. The primary endo¬sperm cell divides repeatedly and forms the triploid endosperm tissue. The cells of this tissue are filled with reserve food materials and are used for the nutrition of the developing embryo at the micropylar end of the embryo sac. This is an adaptation to provide assured nutrition to the developing embryo.

Question 9.
The generative cell of a two-celled pollen divides in the pollen tube but not in a three- celled pollen. Give reasons.
Answer:
In some plants pollen grains are shed at 2 celled stage whereas in others pollen grains are shed at 3 celled stage. The pollens which are shed at 2 celled stage contain vegetative cell and a generative cell. At the time of pollen germination, generative cell divides to form two male gametes in the pollen tube.
On the other hand, in case of pollens which are shed at 3 celled stage, the generative cell has already divided to form two male gametes before pollination or pollen germination, i.e., formation of pollen tube.

Question 10.
In the ,figure given below, label the following parts: .
Male gametes, egg cell, polar nuclei, synergid and pollen tube.
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 11
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 12

Long Answer Type Questions

Question 1.
Starting with the zygote, draw the diagrams of the different stages of embryo development in a dicot.
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 13

Question 2.
What are the possible types of pollinations in chasmogamous flowers. Give reasons.
Answer:
Chasmogamous flowers are the flowers with stamens and stigma exposed .They may undergo self pollination or cross pollination.
(1) Self pollination in chasmogamous flowers : Bisexual flowers where stigma and the stamens both mature almost at the same time can undergo autogamy or geitonogamy.

  • Autogamy – Transfer of pollen grains from anther of a flower to the stigma of the same flower.
  • Geitonogamy – When pollen grains from one flower are deposited on the stigma of another flower borne on the same plant.

(2) Cross pollination in chasmogamous flowers : Transfer of pollen grains from anther of a flower to the stigma of different flower growing on different plant of same species. Cross pollination is carried by various abiotic agents like wind and water or by biotic agents like insects, birds and animals etc.

Question 3.
With a neat, labelled diagram, describe the parts of a mature angiosperm embryo sac. Mention the role of synergids.
Answer:
Neat and labelled diagram of mature angiospermic embryo sac is as follows :
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 14
Ovule is an integumented megasporangium that encloses an embryo sac. Common type of ovule is anatropous. Parts of mature angiospermic ovule are:
(1) Funicle : It is the stalk of the ovule. It is attached to placenta by funicle. In anatropous ovules the funicle is fused with the body of the ovule lengthwise to form raphe. Place of union of funicle and the body of ovule is called hilum.

(2) Integuments : They are one or two cuticularised coverings of the ovule. The  place of origin of integuments is called chalaza. A pore’occurring on one side of ovule where integuments are absent is known as micropyle.

(3)  Nucellus : It is parenchymatous tissue  contained in the ovule.

(4) Embryo sac : It is female gametophyte which is covered by a thin membrane. Embryo sac has seven cells. Three cells form egg apparatus towards micropylar end. There are two synergids and one egg or oosphere in the egg apparatus. Three cells on the opposite side are called antipodal cells. The seventh cell of the embryo sac is the largest cell called central cell. Central cell has two polar nuclei which may fuse to form a diploid secondary nucleus. Synergids are short­lived (one of them degenerates long before fertilisation and second after entry of pollen tube into embryo sac).
These synergids help :

  • In growth of pollen tube towards egg by secreting chemotropically active substances.
  • In nutrition of embryo sac by absorption and transport of food from nucellus through their filiform apparatus.

Filiform apparatus in the form of finger like projections from cell wall is present in upper part of each synergid. The filiform apparatus is useful for the absorption and transportation of

materials from the nucellus to the embryo sac. Hook like structures help in easy penetration of pollen tube and liberation of male gamete from the pollen tube.

Question 4.
Draw the diagram of a microsporangium and label its wall layers. Write briefly on the role of the endothecium.
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants 15

Question 5.
Endothecium protects the sporogenous tissue and helps in dehiscence of pollen grains.
Embryo sacs of some apomictic species appear normal but contain diploid cells. Suggest a suitable explanation for the condition.
Answer:
Embryo sacs of some apomictic species appear normal but contain diploid cells, because embryo sacs either develop from diploid nucellar cells or from diploid megaspore mother cells without undergoing meiosis. It leads to formation of apomictic diploid embryos.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants help you. If you have any query regarding NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms

NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms

Multiple Choice Questions


Question 1.
A few statements describing certain features of reproduction are given below.

(i) Gametic fusion takes place.
(ii) Transfer of genetic material takes place.
(iii) Reduction division takes place.
(iv) Progeny have some resemblance with parents.
Select the options that are true for both asexual and sexual reproduction from the options given below.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (i) and (iii)
Answer:
(c) : Reproduction is a biological process in which-an organism produces young ones (offspring) similar to itself. In reproduction, offsprings have some resemblance with parents. Both sexual and asexual reproduction involve transfer of genetic material.

Question 2.
The term ‘clone’ cannot be applied to offspring formed by sexual reproduction because
(a) offspring do not possess exact copies of parental DNA

(b) DNA of only one parent is copied and passed on to the offspring
(c) offspring are formed at different times
(d) DNA of parent and offspring are completely different.
Answer:
(a) : In sexual reproduction, there is fusion of male gametes and female gametes, the offspring produced are not identical to their parents. This genetic recombination leads to variations, which play an important role in evolution.

Question 3.
A sexual method of reproduction by binary fission is common to which of the following?
(i) Some eukaryotes
(ii) All eukaryotes
(iii) Some prokaryotes
(iv) All prokaryotes
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (iii) and (iv)
Answer:
(c)

Question 4.
A few statements with regard to sexual reproduction are given below.
(i) Sexual reproduction does not always require two individuals.
(ii) Sexual reproduction generally involves gametic fusion.
(iii) Meiosis never occurs during sexual reproduction.
(iv) External fertilisation is a rule during sexual reproduction.
Choose the correct statements from the options below.
(a) (i) and (iii)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (i) and (iv)
Answer:
(b) : Meiosis is required for the production of haploid gametes during sexual reproduction. External fertilisation is not a rule during sexual reproduction, it can occur internally also.

Question 5.
A multicellular, filamentous alga exhibits a type of sexual life cycle in which the meiotic division occurs after the formation of zygote. The adult filament of this alga has
(a) haploid vegetative cells  and   diploid gametangia
(b) diploid vegetative cells  and   diploid gametangia
(c) diploid vegetative cells  and  haploid gametangia
(d) haploid vegetative cells and  haploid gametangia.
Answer:
(d)

Question 6.
The male gametes of rice plant have 12 chromosomes in their nucleus. The chromo­some number in the female gamete, zygote and the cells of the seedling will be, respectively
(a) 12,24,12
(b) 24,12,12
(c) 12,24,24  
(d) 24,12,24.
Answer:
(c) : Chromosome number in male gamete of rice plant is n = 12 therefore chromosome number in female gamete would also be 12. Zygote is diploid as it is the product of fertilisation and the cells of the seedling would be meiocytes and other diploid cells. Hence, the chromosome number in both zygote and cells of seedling will be 2n = 24.

Question 7.
Given below are a few statements related to external fertilisation. Choose the correct statements.
(i) The male and female gametes are formed and released simultaneously.
(ii) Only a few gametes are released into the medium.
(iii) Water is the medium in a majority of organisms exhibiting external fertilisa­tion.
(iv) Offspring formed as a result of external fertilisation have better chance of survival than those formed inside an organism.
(a) (iii) and (iv)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (i) and (iv)
Answer:
(b) : A large number of gametes are released into the medium to increase the chances of fertilisation. The chances of survival of offsprings from external fertilisation are lesser than those of internal fertilisation as they face more risk from predators.

Question 8.
The statements given below describe certain features that are observed in the pistil of flowers.
(i) Pistil may have many carpels.
(ii) Each carpel may have more than one ovule.
(iii) Each carpel has only one ovule.
(iv) Pistil have only one carpel.
Choose the statements that are true from the options below.
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer:
(a) : Gynoecium is the female part of the flower, a unit of which is called pistil. A pistil may have one or more than one carpels (monocarpellary, bicarpellary etc). Each carpel may have more than one ovules.

Question 9.
Which of the following situations correctly describe the similarity between an angiosperm egg and a human egg?
(i) Eggs of both are formed only once in a lifetime.
(ii) Both the angiosperm egg and human egg are stationary.
(iii) Both the angiosperm egg and human egg are motile transported.
(iv) Syngamy in both results in the formation of zygote.
Choose the correct answer from the options given below.
(a) (ii) and (iv)
(b) (iv) only
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) : Syngamy is the complete and permanent fusion of male and female gametes to form a zygote.

Question 10.
Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because
(a) godes are shorter than internodes
(b) nodes have meristematic cells
(c) nodes are located near the soil
(d) nodes have non-photosynthetic cells.
Answer:
(b) : Vegetative propagation is asexual reproduction from ‘various parts in plants. In plants, such as, sugarcane and ginger, appearance of vegetative propagules from nodes is because of presence of meristematic cells in them. Meristematic cells have the ability to divide to form new cells which can differentiate and give rise to permanent tissues.

Question 11.
Which of the following statements, support the view that elaborate sexual reproductive process appeared much later in the organic evolution?
(i) Lower groups of organisms have simpler body design.
(ii) Asexual reproduction is common in lower groups.
(iii) Asexual reproduction is common in higher groups of organisms.
(iv) The high incidence of sexual reproduction in angiosperms and vertebrates.
Choose the correct answer from the options given below.
(a) (i), (ii) and (iii)
(b) (i), (iii) and (iv)
(c) (i), (ii) and (iv) 
(d) (ii), (iii) and (iv)
Answer:
(c)

Question 12.
Offspring formed by sexual reproduction exhibit more variation than those formed by asexual reproduction because
(a) sexual reproduction is a lengthy process
(b) gametes of parents have qualitatively different genetic composition
(c) genetic material comes from parents of two different species.
(d) greater amount of DNA is involved in sexual reproduction.
Answer:
(b) : Sexual reproduction occurs in almost all types of animals and mostly in higher plants. It is usually biparental. Daughter organisms genetically differ from the parents. Since there are variations, so it contributes to evolution of the species.

Question 13.
Choose the correct statement from amongst the following.
(a) Dioecious organisms are seen only in animals.
(b) Dioecious organisms are seen only in plants.
(c) Dioecious organisms are seen in both plants and animals.
(d) Dioecious organisms are seen only in vertebrates.
Answer:
(c) : Dioecious organisms are those in which male and female sex organs are present in different organisms, where as monoecious organisms are those in which male and female sex organs are present in the same organism. Monoecious organisms are also called as hermaphrodite. Dioecious organisms are seen in both plants and animals. Papaya, date palm and most of the animals are dioecious.

Question 14.
There is no natural death in single celled organisms like Amoeba and bacteria because
(a) they cannot reproduce sexually
(b) they reproduce by binary fission
(c) parental body is distributed among the offspring
(d) they are microscopic.
Answer:
(c) : These are no natural death in single celled organisms like Amoeba and bacteria. It is so, because of asexual reproduction, the body of parent is divided into daughter cells. So, in effect, there is no practical death in Amoeba and bacteria.

Question 15.
There are various types of reproduction. The type of reproduction adopted by an organism depends on
(a) the habitat and morphology of the organism
(b) morphology of the organism
(c) morphology and physiology of the organism
(d) the organism’s habitat, physiology and genetic makeup.
Answer:
(d) : There are various types of reproduction, both asexual (fission, budding, etc.) and sexual (internal and external). The type of reproduction, an organism undergoes depends ultimately on its genetic makeup which influences its physiology. Habitat also influences the type of reproduction, that organism undergoes.

Question 16.
Identify the incorrect statement.
(a) In asexual reproduction, the offspring produced are morphologically and genetically identical to the parent.
(b) Zoospores are sexual reproductive structures.
(c) In asexual reproduction, a single parent produces offspring with or without the formation of gametes.
(d) Conidia are asexual structures in
Answer:
(b) : Spores formation is also a type of asexual reproduction. Zoospores, conidia, oidia, etc. are all asexually reproducing structures. There is generally no gamete formation in asexual reproduction and the offsprings produced are called clones.

Question 17.
Which of the following is a post-fertilisation event in flowering plants?
(a) Transfer of pollen grains
(b) Embryo development
(c) Formation of flower
(d) Formation of pollen grains
Answer:
(b) : Events in sexual reproduction after the fertilisation are called post-fertilisation events. After fertilisation, a diploid zygote is formed in all sexually reproducing organisms. The process of development of embryo from the zygote is called embryogenesis.

Question 18.
The number of chromosomes in the shoot tip cells of a maize plant is 20. The number of chromosomes in the microspore mother cells of the same plant shall be
(a) 20 
(b) 10
(c) 40
(d) 15.
Answer:
(a) : Shoot tip cells and microspore mother cells both are diploid in maize plant. If number of chromosomes in shoot tip cell (2n) = 20, then number of chromosomes in microspore mother cell will be (2n) = 20.

Very Short Answer Type Questions

Question 1.
Mention two inherent characteristics of Amoeba and yeast that enable them to reproduce asexually.
Answer:
The two inherent characteristics for asexual reproduction in Amoeba and yeast are :

  1. Gametes are not formed in Amoeba and yeast.
  2. Uniparental condition (i.e., single parent) is involved in reproduction of both Amoeba and yeast.

Question 2.
Why do we refer to offspring formed by asexual method of reproduction as clones?
Answer:
Offsprings produced by asexual reproduction are morphologically and genetically similar to their parents. Hence, they are known as clones.

Question 3.
Although potato tuber is an underground part, it is considered as a stem. Give two reasons.
Answer:
Reasons for considering potato as a stem are:

  1. Differentiation into nodes and inter­nodes.
  2. The nodes bear buds which can grow to from leaf shoots or plantlets.

Question 4.
Between an annual and a perennial plant, which one has a shorter juvenile phase? Give one reason.
Answer:
Annual plant have shorter juvenile phase as they complete their life cycle in single season i.e., a few weeks to a few months, e.g., wheat, maize, pea, gram.

Question 5.
Rearrange the following events of sexual reproduction in the sequence in which they occur in a flowering plant : embryogenesis, fertilisation, gametogenesis, pollination.
Answer:
The correct sequence of events of sexual reproduction in flowering plants are as follows : gametogenesis, pollination, fertilisation, embryogenesis.

Question 6.
The probability of fruit set in a self-pollinated bisexual flower of a plant is far greater than a dioecious plant. Explain.
Answer:
Presence of male and female repro­ductive organs on same plant provides more chances of self pollination and therefore increases chances of fruit set. However in dioecious plants agents of pollination are needed which may therefore decrease the chances of a fruit set.

Question 7.
Is the presence of large number of chromosomes in an organism a hindrance to sexual reproduction? Justify your answer by giving suitable reasons.
Answer:
The presence of a large number of chromosomes in an organism is not a hindrance to sexual reproduction.

Question 8.
Is there a relationship between the size of an organism and its life span? Give two examples in support of your answer.
Answer:
There is no direct correlation between the life span of organisms and their sizes.

  1. The crow and parrot are nearly of the same size, but life span of crow and parrot are 15 years and 140 years respectively.
  2. The mango and the peepal trees are nearly of the same size but the life span of mango and peepal are 100 years and 1000 years respectively.

Question 9.
In the figure given below the plant bears two different types of flowers marked ‘A’ and ‘B’. Identify the types of flowers and state the type of pollination that will occur in them.
NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms 1

Answer:
The plant shown in figure has two types of flowers:
Flower A –
chasmogamous type with stigma and anthers exposed. Such type of flowers may undergo self or cross pollination.
Flower B – Cleistogamous type – closed type of flowers, where stigma and anthers are not exposed. Such type of flowers undergo only self pollination.

Question 10.
Give reasons as to why cell division cannot be a type of reproduction in multicellular organisms.
Answer:
In multicellular organisms cell division does not divide the whole body into daughter cells as in unicellular organisms. But multicellular organisms have a well developed reproductive organs which help in reproduction.

Question 11.
In the figure given below, mark the ovule and pericarp.
NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms 2
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms 3

Question 12.
Why do gametes produced in large numbers in organisms exhibit external fertilisation?
Answer:
In external fertilisation, there are great chances that the sperm and the eggs released by the organisms can be affected by desiccation, predators, etc. So, to make up for the high fatality rate of the gametes, the organisms produces a lot of gametes.

Question 13.
Which of the followings are monoecious and dioecious organisms.
(a) Earthworm____________
(b) Chara ______________
(c) Marchantia____________
(d) Cockroach ___________
Answer:
(a) Earthworm – Monoecious animal
(b) Chara – Monoecious plant
(c) Marchantia- Dioecious plant
(d) Cockroach – Dioecious plant

Question 14.
Match the organisms given in Column ‘A’ with the vegetative propagules given in Column ‘B’

Column A Column B
i Bryophyllum a. Offset
ii Agave b. Eyes
iii. Potato c. Leaf buds
iv. Water hyacinth d. Bulbils
  1. Rrunphyllum. – (c) Leaf buds
  2. Agave –  (d)   Bulbils
  3. Potato –  (b)   Eyes
  4. Water hyacinth –  (a) Offset

Question 15.
What do the following parts of a flower develop into after fertilisation?
(a) Ovary__________
(b) Ovules__________
Answer:
(a) Ovary – fruit
(b) Ovules-seed

Short Answer Type Questions

Question 1.
In haploid organisms that undergo sexual reproduction, name the stage in the life cycle when meiosis Give reasons for your answer.
Answer:
In zygospore (formed by zygote) meiosis occurs. Because after meiosis it can form meiospores which can develop into haloid organisms.

Question 2.
The number of taxa exhibiting asexual reproduction is drastically reduced in higher plants (angiosperms) and higher animals (vertebrates) as compared with lower groups of plants and animals. Analyse the possible reasons for this situation.
Answer:
Both angiosperms and vertebrates have more complex structural organisation. They have evolved very efficient mechanism of sexual reproduction.

  1. The offspring produced due to sexual reproduction adapt better to the changing environmental conditions.
  2. Genetic recombination, interaction, etc. during sexual reproduction provide vigour and vitality to the offspring.

Question 3.
Honeybees produce their young ones only by sexual reproduction. Inspite of this, in a colony of bees we find both haploid and diploid  individuals Name the haploid and diploid individuals in the colony and analyse the reasons behind their formation.
Answer:
Honeybees produce their young ones by sexual reproduction but parthenogenesis also occurs alongwith sexual reproduction. Fertilised eggs and parthenogenetically developed eggs give rise to different castes. In honey bees, fertilised eggs (zygotes) which is diploid give rise to queens and workers (both are females) and unfertilised eggs (ova) which is haploid develop into drones (males).

Question 4.
With which type of reproduction do we associate the reduction division? Analyse the reasons for it.
Answer:
We can associate the reduction division (meiosis) with sexual reproduction. Sexual reproduction is the process of development of new organisms through the formation and fusion of male and female gametes. It involves meiosis or reduction division i.e., halving the parental chromosomes inside the male and female gamete and their subsequent fusion resulting in the restoration of the original parental number of chromosomes, causing an increased genetic diversity.

Question 5.
Is it possibletoconsider vegetative propagation observed in certain plants like Bryophyllum, water hyacinth, ginger etc., as a type of asexual reproduction? Give two/ three reasons.
Answer:
Yes, it is possible to consider vegetative propagation as a type of asexual reproduction because of the following reasons:

  1. Production of plantlet occurs by a single parent plant without the formation and fusion of gametes.
  2. Plantlets receives all genes from their one parent plant.
  3. Vegetative reproduction involves only mitotic cell division.
  4. There is no gamete formation.

Question 6.
Fertilisation is not an obligatory event for fruit production in certains plants’ Explain the statement.
Answer:
No, fertilisation is not an obligatory event for fruit production in certain plants, as fruits can also develop from unfertilised ovary also. Such fruits are called parthenocarpic  fruits. The formation of this type of fruit takes place without prior’fertilisation of the flower by pollen. The resulting fruits are seedless and therefore do not contribute to the reproduction of the plant, examples are bananas and pineapples. Plant growth substances may have a role in this phenomenon, which can be induced by auxins in the commercial production of tomatoes and other fruits.

Question 7.
In a developing embryo, analyse the consequences if cell divisions are not followed by cell differentiation.
Answer:
Cell division increase the number of cells in the developing embryo while cell differentiation helps the groups of cells to undergo certain modifications and form specialised tissues and organs.

Initially all the cells of the developing embryo are alike with same genetic information. Later however, due to the phenomenon of differentiation depending on their location and internal cellular mechanism, different cells of the embryo develop differently forming embryonal axis, plumule and radicle. However, if cell differentiation does not occur the embryo will remain just a mass of undifferentiated cells or callus. There would not any plumule, radicle, cotyledons or embryo axis. A new plant will not be formed from such an embryo.

Question 8.
List the changes observed in an angiosperm flower subsequent to pollination and fertilisation.
Answer:
In an angiosperm flower, the changes that occur subsequent to pollination and fertilisation are called post-fertilisation changes. Pollen grain germinates over the stigma and forms a pollen tube carrying two gametes. Pollen tube reaches ovary and enters an ovule through one of its synergids. Fertilisation produces a diploid zygote and triploid primary endosperm cell. Zygote produces embryo and primary endosperm cell forms endosperm. Transformation takes place and ovule transforms into seed, ovary transforms into fruit and ovary wall into pericarp. The petals, stamens and style wither away.

Question 9.
Suggest a possible explanation why the seeds in a pea pod are arranged in a row, whereas those in tomato are scattered in the juicy pulp.
Answer:
Arrangement of seeds inside a fruit depends upon the type of placentation and the growth of placental axis. In pea ovary, the ovules are attached to the ventral suture i.e., marginal placentation and carpel is mono- carpellary. So, seeds are arranged in a row. In tomatoes, the gynoecium is tricarpellary with axile placentation. The placentae grow and become pulpy during fruit formation. As a result seeds get scattered in the pulpy mass in tomato.

Question 10.
Draw the sketches of a zoospore and a conidium. Mention two dissimilarties between them and atleast one feature common to both structures.
Answer:
NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms 4

Dissimilarities : Zoospores are motile and flagellated whereas conidia are nonmotile. Zoospore are formed inside a zoosporangium and conidia are formed (exogenously) outside on a conidiophore.
Similarity : Both are reproductive structures for asexual reproduction.

Question 11.
Justify the statement’Vegetative reproduction is also a type of asexual reproduction’.
Answer:
Refer answer 5.

Long Answer Type Questions

Question 1.
Enumerate the differences between asexual and sexual reproduction. Describe the types of asexual reproduction exhibited by unicellular organisms.
Answer:

A sexual reproduction Sexual reproduction
(1) Asexual reproduction involves the participation of single individual parent. Sexual reproduction involves participation of two separate parents.
(2) It generally occurs without the involvement of sex organs. It usually involves the sex organs.
(3) It does not involve meiosis or reduction division. It involves meiosis which occurs at the time of gamete formation.
(4) Asexual reproduction does not involve sexual fusion or fusion of two gametes. Zygotes are not formed. The sexual reproduction requires fertilisation to take place between two opposite gametes leading to the production of a zygote.
(5) Since asexual reproduction does not involve meiosis and fusion of gametes, the offsprings are genetically similar to parents and they do not show variation. The individuals produced as a result of meiosis and gametic fusion exhibit genetic variation and differ from either of the two parents.
(6) It is a very quick method of multiplication and, therefore, used by plant breeders for cloning. It is very slow method of multiplication of individuals.

Asexual reproduction occurs usually in some organisms such as monerans, protists, in plants and in some animals.
Following are the methods of asexual reproduction in unicellular organisms :

(1) Binary fission : In this type of asexual reproduction, the parent organism divides into two halves, each half forms an independent daughter organism e.g.,

(2) Multiple fission : In this process, the parent body divides into many similar daughter individuals. Multiple fission occurs in Amoeba during unfavourable condition.

(3) Bidding : In budding, a daughter individual is formed from a small part or bud, arising from the parent body g., yeast.
Spore formation or sporulation : Spores are minute, single celled, thin or thick walled propagules. In this type of asexual reproduction dispersive structures called spores are released from parent body that germinate under favourable conditions to form new individuals. Motile spores called zoospores are formed in unicellular alga like

Question 2.
Do all the gametes formed from a parent organism have the same genetic composition (identical DNA copies of the parental genome)? Analyse the situation with the background of gametogenesis and provide or give suitable explanation.
Answer:
No, All the gametes formed from a parent organism don’t have the same genetic combination. Formation of two types of gametes-male and female, inside the gametangia, is called gametogenesis. The reproductive units in sexual reproduction are specialised cells called gametes. The gametes are generally of two kinds : male and female. The gametes of all the organisms are usually haploid cells, i.e., possess single set (or n number) of chromosomes. The gametes are usually formed by meiotic divisions. Therefore, they are haploid, i.e., have halved or reduced (n) number of chromosomes. During this processes, random segregation of chromosomes (independent assortment) and exchange of genetic material between homologous chromosomes (crossing over) result in new combinations of genes in the gametes, and this reshuffling increases genetic diversity. The coming together of two unique sets of chromosomes in the zygote forms the genetic basis of variation within the species.

Question 3.
Although sexual reproduction is a long drawn, energy-intensive complex form of reproduction, many groups of organisms in Kingdom Animalia and Plantae prefer this mode of reproduction. Give atleast three reasons for this.
Answer:
Reasons for preference of sexual reproduction in higher groups of organisms are as follows:

  1. During sexual reproduction, there is fusion of gametes, hence genetic recombination takes place causing variations.
  2. The offspring produced due to sexual reproduction adapt better to the changing environmental conditions.
  3. Genetic recombination, interaction, etc. during sexual reproduction provide vigour and vitality to the offspring.
  4. Variation being a major factor of natural selection, therefore, it plays an important role in evolution.

Question 4.
Differentiate between
(a) oestrous and menstrual cycles;
(b) ovipary and vivipary. Cite an example for each type.
Answer:
(a) Differences between menstrual and oestrous cycles are as follows:

Menstrual cycle Oestrous cycle
(1) This cycle consists of menstrual phase, proliferative phase and the secretory phase. It consists of a short period of oestrous or heat (e.g., 12-24 hours in cow) followed by anoestrous or passive period
(2) Blood flows in the last few days of this cycle. Blood does not flow in this cycle.
(3) The broken endometrium is passed out during menstruation The broken endometrium is reabsorbed.
(4) Sex urge is not increased during menstruation. Sex urge is increased during oestrous period.
(5) Female does not permit copulation during menstrual phase of the cycle. Female permits copulation only during oestrous period.
(6) It occurs in primates (monkeys, apes and human beings) only. It occurs in nonprimates such as cows, dogs, etc

 (b) Differences between ovipary and vivipary are as follows:

Ovipary Vivipary
(1) The process in which females lay fertilised/ unfertilised eggs. The process in which females give birth to young ones.
(2) The development of zygote takes place outside the female’s body. The development of zygote takes place inside the female’s body.
(3) Females lay eggs in a safe place in the environment but the chances of survival are less. Females deliver young ones and the chances of survival are more.
(5) Example. All birds, most of reptiles and egg­laying mammals. Example. Mammals except egg-laying mammals

Question 5.
Rose plants produce large, attractive bisexual flowers but they seldom produce fruits. On the other hand a tomato plant produces plenty of fruits though they have small flowers. Analyse the reasons for failure of fruit formation in rose.
Answer:
Both rose and tomato plants are selected by human beings for different characteristics, the rose for its flower and tomato for its fruit. Rose plant is vegetatively propagated and does not produce seeds. Fruit formation fails in rose plant due to following reasons:

  1. Viable pollen may not be produced, resulting in failure of fertilisation.
  2. Functional egg may not be produced.
  3. The ovule produced may be defective or non functional.
  4. There may be pollen-pistil incom­patibility.
  5. The plant may be a hybrid with abnormal segregation of chromosomes resulting in non-viable gametes.
  6. The plant may be a hybrid and sterile.
  7. There may be internal barriers for growth of pollen tube and fertilisation

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