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Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 9 Strategies for Enhancement in Food Production Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production

PSEB 12th Class Biology Guide Strategies for Enhancement in Food Production Textbook Questions and Answers

Question 1.
Explain in brief the role of animal husbandry in human welfare.
Answer:
Animal husbandry deals with the scientific management of livestock. It includes various aspects such as feeding, breeding, and control diseases to raise the population of animal livestock. Animal husbandry usually includes animals such as cattle, pig, sheep, poultry, honeybee, silkworm and fish which are useful for humans in various ways. These animals are managed for the production of commercially important products such as milk, meat, wool, egg, honey, silk, etc. The increase in human population has increased the demand of these products. Hence, it is necessary to improve the management of livestock scientifically.

Question 2.
If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production?
Answer:
Dairy farm management deals with processes which aim at improving the quality and quantity of milk production. Milk production is primarily dependent on choosing improved cattle breeds, provision of proper feed for cattle, maintaining proper shelter facilities, and regular cleaning of cattle.

Choosing improved cattle breeds is an important factor of cattle management. Hybrid cattle breeds are produced for improved productivity. Therefore, it is essential that hybrid cattle breeds should have a combination of various desirable genes such as high milk production and high resistance to diseases. Cattle should also be given healthy and nutritious food consisting of roughage, fibre concentrates, and high levels of proteins and other nutrients.

Cattle should be housed in proper cattle-houses and should be kept in well ventilated roofs to prevent them from harsh weather conditions such as heat, cold, and rain. Regular baths and proper brushing should be ensured to control diseases. Also, time-to-time check ups by a veterinary doctor for symptoms of various diseases should be undertaken.

Question 3.
What is meant by the term ‘breed’? What are the objectives of animal breeding?
Answer:
A breed is a special variety of animals within a species. It is similar in most characters such as general appearance, size, configuration, and features with other members of the same species. Jersey and Brown
Swiss are examples of foreign breeds of cattle. These two varieties of cattle have the ability to produce abundant quantities of milk. This milk is very nutritious with high protein content.
Objectives of animal breeding are as follows:

• To increase thd yield of animals.
• To improve the desirable qualities of the animal produce.
• To produce disease-resistant varieties of animals.

Question 4.
Name the methods employed in animal breeding. According to you which of the methods is best? Why?
Answer:
Animal breeding is the method of mating closely related individuals. There are several methods employed in animal breeding, which can be classified into the following categories:

(A) Natural Methods of Breeding Include Inbreeding and Out-breeding : Breeding between animals of the same breed is known as inbreeding, while breeding between animals of different breeds is known as out-breeding. Out-breeding of animals is of three types:
(a) Out-crossing: In this type of out-breeding, the mating of animals occurs within the same breed. Thus, they have no common ancestors up to the last 4-5 generations.

(b) Cross-breeding: In this type of out-breeding, the mating occurs between different breeds of the same species, thereby producing a hybrid.

(c) Interspecific hybridisation: In this type of out-breeding, the mating occurs between different species.

(B) Artificial Methods of Breeding Include Modern Techniques of Breeding : It involves controlled breeding experiments, which are of two types:
(a) Artificial insemination : It is a process of introducing the semen (collected from the male) into the oviduct or the uterus of the female body by the breeder. This method of breeding helps the breeder overcome certain problems faced in abnormal mating.

(b) Multiple ovulation embryo technology (MOET) : It is a technique for cattle improvement in which super-ovulation is induced by a hormone injection. Then, fertilisation is achieved by artificial insemination and early embryos are collected. Each of these embryos are then transplanted into the surrogate mother for further development of the embryo.

The best method to carry out animal breeding is the artificial method of breeding, which includes artificial insemination and MOET technology. These technologies are scientific in nature. They help overcome problems of normal mating and have a high success rate of crossing between mature males and females. Also, it ensures the production of hybrids with the desired qualities. This method is highly economical as a small amount of semen from the male can be used to inseminate several cattle.

Question 5.
What is apiculture? How is it important in our lives?
Answer:
Apiculture is the practise of bee-keeping for the production of various products such as honey, bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines. It is
useful in the treatment of many disorders such as cold, flu, and 1 dysentery. Other commercial products obtained from honey bees include bee’s wax and bee pollen. Bee’s wax is used for making cosmetics, polishes, and is even used in several medicinal preparations. Therefore, to meet the increasing demand of honey, people have started practicing bee-keeping on a large scale. It has become an f income generating activity for farmers since it requires a low investment and is labour intensive.

Question 6.
Discuss the role of fishery in enhancement of food production.
Answer:
Fishery is an industry which deals with catching, processing, and marketing of fishes and other aquatic animals that have a high economic value. Some commercially important aquatic animals are prawns, crabs, oysters, lobsters, and octopus. Fisheries play an important role in the Indian economy. This is because a large part of the Indian population is dependent on fishes as a source of food, which is both cheap and high in animal protein. Fishery is an employment generating industry especially for people staying in the coastal areas. Both fresh water fishes (such as Catla, Rohu, etc.) and marine fishes (such as tuna, mackerel, pomfret, etc.) are of high economic value.

Question 7.
Briefly describe various steps involved in plant breeding.
Answer:
Plant breeding is the process in which two genetically dissimilar varieties are purposely crossed to produce a new hybrid variety. As a result, characteristics from both parents can be obtained in the hybrid plant variety. Thus, it involves the production of a new variety with the desired characteristics such as resistance to diseases, climatic adaptability, and better productivity. The various steps involved in plant breeding are as follows:

(a) Collection of Genetic Variability: Genetic variability from various wild relatives of the cultivated species are collected to maintain the genetic diversity of a species. The entire collection of the diverse alleles of a gene in a crop is called the germplasm collection.

(b) Evaluation of Germplasm and Selection of Parents: The germplasm collected is then evaluated for the desirable genes. The selected plants with the desired genes are then used as parents in plant breeding experiments and are multiplied by the process of hybridisation.

(c) Cross-hybridisation between Selected Parents: The next step in plant breeding is to combine the desirable characters present in two different parents to produce hybrids. It is a tedious job as one has to ensure that the pollen grains collected from the male parent reach the stigma of the female parent.

(d) Selection of Superior Hybrids: The progenies of the hybrids having the desired characteristics are selected through scientific evaluation. The selected progenies are then self-pollinated for several generations to ensure homozygosity.

(e) Testing, Release, and Commercialisation of New Cultivars: The selected progenies are evaluated for characters such as yield, resistance to diseases, performance, etc. by growing them in research fields for at least three growing seasons in different parts of the country. After thorough testing and evaluation, the selected varieties are given to the farmers for growing in fields for a large-scale production.

Question 8.
Explain what is meant by biofortification.
Answer:
Biofortification is a process of breeding crops with higher levels of vitamins, minerals, proteins, and fat content. This method is employed to improve public health. Breeding of crops with improved nutritional quality is undertaken to improve the content of proteins, oil, vitamins, minerals, and micro-nutrients in crops. It is also undertaken to upgrade the quality of oil and proteins. An example of this is a wheat variety known as Atlas 66, which has high protein content in comparison to the existing wheat. In addition, there are several other improved varieties of crop plants such as rice, carrots, spinach etc. which have more nutritious value and more nutrients than the existing varieties.

Question 9.
Which part of the plant is best suited for making virus-free plants and why?
Answer:
Apical and axillary meristems of plants is used for making virus-free plants. In a diseased plant, only this region is not infected by the virus as compared to the rest of the plant region. Hence, the scientists remove axillary and apical meristems of the diseased plant and grow it in vitro to obtain a disease-free and healthy plant.
Virus-free plants of banana, sugarcane, and potato have been obtained using this method by scientists.

Question 10.
What is the major advantage of producing plants by micropropagation?
Answer:
Micropropagation is a method of producing new plants in a short duration using plant tissue culture.
Some major advantages of micropropagation are as follows:

• Micropropagation helps in the propagation of a large number of plants in a short span of time.
• The plants produced are identical to the mother plant.
• It leads to the production of healthier plantlets, which exhibit better disease-resisting powers.

Question 11.
Find out what the various components of the medium used for propagation of an explant in vitro are?
Answer:
The major components of medium used for propagation of explants in vitro are carbon sources such as sucrose, inorganic salts, vitamins, amino acids, water, agar-agar, and certain growth hormones such as auxins, gibberellins and cytokinins etc.

Question 12.
Name any five hybrid varieties of crop plants which have been developed in India.
Answer:
The five hybrid varieties of crop plants which have been developed in India are as follow:

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 8 Human Health and Disease Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 8 Human Health and Disease

PSEB 12th Class Biology Guide Human Health and Disease Textbook Questions and Answers

Question 1.
What are the various public health measures, which you would suggest as safeguard against infectious diseases?
Answer:
Public health measures are preventive measures which are taken to check the spread of various infectious diseases. These measures should be taken to reduce the contact with infectious agents.
Some of these methods are as follows:
1. Maintenance of Personal and Public Hygiene: It is one of the most important methods of preventing infectious diseases. This measure includes maintaining a clean body, consumption of healthy and nutritious food, drinking clean water, etc. Public hygiene includes proper disposal of waste material, excreta, periodic cleaning, and disinfection of water reservoirs.

2. Isolation: To prevent the spread of air-borne diseases such as pneumonia, chicken pox, tuberculosis, etc., it is essential to keep the infected person in isolation to reduce the chances of spreading these diseases.

3. Vaccination: Vaccination is the protection of the body from communicable diseases by administering some agent that mimics the microbe inside the body. It helps in providing passive immunisation to the body. Several vaccines are available against many diseases such as tetanus, polio, measles, mumps, etc.

4. Vector Eradication: Various diseases such as malaria, filariasis, dengue, and chikungunya spread through vectors. Thus, these diseases can be prevented by providing a clean environment and by preventing the breeding of mosquitoes. This can be achieved by not allowing water to stagnate around residential areas. Also, measures like regular cleaning of coolers, use of mosquito nets and insecticides such as malathion in drains, ponds, etc. can be undertaken to ensure a healthy environment. Introducing fish such as Gambusia in ponds also controls the breeding of mosquito larvae in stagnant water.

Question 2.
In which way has the study of biology helped us to control infectious diseases?
Answer:
Various advancements that have occurred in the field of biology have helped us gain a better understanding to fight against various infectious diseases. Biology has helped us study the life cycle of various parasites, pathogens, and vectors along with the modes of transmission of various diseases and the measures for controlling them. Vaccination programmes against several infectious diseases such as small pox, chicken pox, tuberculosis, etc. have helped to eradicate these diseases. Biotechnology has helped in the preparation of newer and safer drugs and vaccines. Antibiotics have also played an important role in treating infectious diseases.

Question 3.
How does the transmission of each of the following diseases take * place?
(a) Amoebiasis
(b) Malaria
(c) Ascariasis
(d) Pneumonia
Answer:

Question 4.
What measure would you take to prevent water-borne diseases?
Answer:
Water-borne diseases such as cholera, typhoid, hepatitis B, etc. spread by drinking contaminated water. These water-borne diseases can be prevented by ensuring proper disposal of sewage, excreta, periodic cleaning. Also, measures such as disinfecting community water reservoirs, boiling drinking water, etc. should be observed.

Question 5.
Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines.
Answer:
‘A suitable gene’ refers to a specific DNA segment which can be injected into the cells of the host body to produce specific proteins. This protein kills the specific disease-causing organism in the host body and provides immunity.

Question 6.
Name the primary and secondary lymphoid organs.
Answer:

• Primary lymphoid organs include the bone marrow and the thymus.
• Secondary lymphoid organs include the spleen, lymph nodes, tonsils, Peyer’s patches of small intestine, and appendix.

Question 7.
The following are some well-known abbreviations, which have been used in this chapter. Expand each one to its full form:
(a) MALT
(b) CMI
(c) AIDS
(d) NACO
(e) HIV
Answer:
(a) MALT: Mucosa-Associated Lymphoid Tissue
(b) CMI: Cell-Mediated Immunity
(c) AIDS: Acquired Immuno Deficiency Syndrome
(d) NACO: National AIDS Control Organisation
(e) HIV: Human Immuno Deficiency Virus

Question 8.
Differentiate the following and give examples of each:
(a) Innate and acquired immunity
(b) Active and passive immunity
Answer:
(a) Innate and acquired immunity

(b) Active and passive immunity

Question 9.
Draw a well-labelled diagram of an antibody molecule.
Answer:

Question 10.
What are the various routes by which transmission of human i immuno-deficiency virus takes place?
Answer:
AIDS (Acquired Immuno-Deficiency Syndrome) is caused by the Human I Immuno-deficiency Virus (HIV).
It has the following modes of transmission:

• Unprotected sexual contact with an infected person.
• Transfusion of blood from an infected person to a healthy person.
• Sharing infected needles and syringes.
• From an infected mother to a child through the placenta.

Question 11.
What is the mechanism by which the AIDS virus causes . deficiency of immune system of the infected person?
Answer:
AIDS (Acquired Immuno-Deficiency Syndrome) is caused by the Human immuno-deficiency virus (HIV) via sexual or blood-blood contact. After entering the human body, the HIV virus attacks and enters the macrophages. Inside the macrophages, the RNA of the virus replicates ’ with the help of enzyme reverse transcriptase and gives rise to viral
DNA. Then, this viral DNA incorporates into the host DNA and directs the synthesis of virus particles. At the same time, HIV enters helper T-lymphocytes. It replicates and produces viral progeny there. These newly formed progeny viruses get released into the blood, attacking p other healthy helper T-lymphocytes in the body. As a result, the number of T-lymphocytes in the body of an infected person decreases progressively, thereby decreasing the immunity of a person.

Question 12.
How is a cancerous cell different from a normal cell?
Answer:

Question 13.
Explain what is meant by metastasis.
Answer:
The property of metastasis is exhibited by malignant tumors. It is the pathological process of spreading cancerous cells to the different parts of the body. These cells divide uncontrollably, forming a mass of cells called tumor. From the tumor, some cells get sloughed off and enter into the blood stream. From the blood stream, these cells reach distant parts of the body and therefore, initiate the formation of new tumors by dividing actively.

Question 14.
List the harmful effects caused by alcohol/drug abuse.
Answer:
Alcohol and drugs have several adverse effects on the individual, his family, and the society.
A. Effects of Alcohol
Effects on the Individual: Alcohol has an adverse effect on the body of an individual. When an individual consumes excess alcohol, it causes damage to the liver and the nervous system. As a result, other symptoms such as depression, fatigue, aggression, loss of weight and appetite may also be observed in the individual. Sometimes, extreme levels of alcohol consumption may also lead to heart failure, resulting coma and death. Also, it is advisable for pregnant women to avoid alcohol as it may inhibit normal growth of the baby.

Effects on the Family: Consumption of excess alcohol by any family member can have devastating effects on the family. It leads to several domestic problems such as quarrels, frustrations, insecurity, etc. Effects on the Society:

• Rash behaviour
• Malicious mischief and violence
• Deteriorating social network
• Loss of interest in social and other activities

B. Effects of Drugs
Effects on the Individual: Drugs have an adverse effect on the central nervous system of an individual. This leads to the malfunctioning of several other organs of the body such as the kidney, liver, etc. The spread of HIV is most common in these individuals as they share common needles while injecting drugs in their body. Drugs have long-term side effects on both males and females. These side effects include increased aggressiveness, mood swings, and depression.

Effects on the Family and Society: A person addicted to drugs creates problems for his family and society. A person dependant on drugs becomes frustrated, irritated, and anti-social.

Question 15.
Do you think that friends can influence one to take alcohol/ drugs? If yes, how may one protect himself/herself from such an influence?
Answer:
Yes, friends can influence one to take drugs and alcohol. A person cart take the following steps for protecting himself/herself against alcohol/ drug abuse:
(a) Increase your willpower to stay away from alcohol and drugs. One should not experiment with alcohol for curiosity and fun.

• Avoid the company of friends who take drugs.
• Seek help from parents and peers. ‘
• Take proper knowledge and counselling about drug abuse. Devote your energy in other extra-curricular activities.
• Seek immediate professional and medical help from psychologists and psychiatrists if symptoms of depression and frustration become apparent.

Question 16.
Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher. Ans. Drug and alcohol consumption has an inherent addictive nature associated with euphoria and a temporary feeling of well-being. Repeated intake of drugs increases the tolerance level of the body’s receptors, leading to more consumption of drugs.

A. Effects of Alcohol
Effects on the Individual: Alcohol has an adverse effect on the body of an individual. When an individual consumes excess alcohol, it causes damage to the liver and the nervous system. As a result, other symptoms such as depression, fatigue, aggression, loss of weight and appetite may al“so be observed in the individual. Sometimes, extreme levels of alcohol consumption may also lead to heart failure, resulting coma and death. Also, it is advisable for pregnant women to avoid alcohol as it may inhibit normal growth of the baby.

Effects on the Family: Consumption of excess alcohol by any family member can have devastating effects on the family. It leads to several domestic problems such as quarrels, frustrations, insecurity, etc. Effects on the Society:

• Rash behaviour
• Malicious mischief and violence
• Deteriorating social network
• Loss of interest in social and other activities

B. Effects of Drugs
Effects on the Individual: Drugs have an adverse effect on the central nervous system of an individual. This leads to the malfunctioning of several other organs of the body such as the kidney, liver, etc. The spread of HIV is most common in these individuals as they share common needles while injecting drugs in their body. Drugs have long-term side effects on both males and females. These side effects include increased aggressiveness, mood swings, and depression.

Effects on the Family and Society: A person addicted to drugs creates problems for his family and society. A person dependant on drugs becomes frustrated, irritated, and anti-social.

Question 15.
Do you think that friends can influence one to take alcohol/ drugs? If yes, how may one protect himself/herself from such an influence?
Answer:
Yes, friends can influence one to take drugs and alcohol. A person cart take the following steps for protecting himself/herself against alcohol/ drug abuse:

• Increase your willpower to stay away from alcohol and drugs. One should not experiment with alcohol for curiosity and fun.
• Avoid the company of friends who take drugs.
• Seek help from parents and peers.
• Take proper knowledge and counselling about drug abuse. Devote your energy in other extra-curricular activities.
• Seek immediate professional and medical help from psychologists and psychiatrists if symptoms of depression and frustration become apparent.

Question 16.
Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher.
Answer:
Drug and alcohol consumption has an inherent addictive nature associated with euphoria and a temporary feeling of well-being. Repeated intake of drugs increases the tolerance’ level of the body’s receptors, leading to more consumption of drugs.

Question 17.
In your view what motivates youngsters to take to alcohol or drugs and how can this be avoided?
Answer:
Many factors are responsible for motivating youngsters towards alcohol or drugs. Curiosity, need for adventure and excitement, experimentation are the initial causes of motivation. Some youngsters start consuming drugs and alcohol in order to overcome negative emotions (such as stress, pressure, depression, frustration) and to excel in various fields. Several mediums like television, internet, newspaper, movies etc. are also responsible for promoting the idea of alcohol to the younger generation. Amongst these factors, reasons such as unstable and unsupportive family structures and peer pressure can also lead an individual to be dependant on drugs and alcohol.

Preventive measures against addiction of alcohol and drugs are as follows:

• Parents should motivate and try to increase the willpower of their child.
• Parents should educate their children about the ill-effects of alcohol. They should provide them with proper knowledge and counselling regarding the consequences of addiction to alcohol.
• It is the responsibility of the parent to discourage a child from experimenting with alcohol. Youngsters should be kept away from the company of friends who consume drugs.
• Children should be encouraged to devote their energy in other extra¬curricular and recreational activities.
• Proper professional and medical help should be provided to a child if sudden symptoms of depression and frustration are observed.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 7 Evolution Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 7 Evolution

PSEB 12th Class Biology Guide Evolution Textbook Questions and Answers

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory.
Answer:
Darwinian selection theory states that individuals with favourable variations are better adapted than individuals with less favourable variation. It means that nature selects the individuals with useful variation as these individuals are better evolved to survive in the existing environment. An example of such selection is antibiotic resistance in bacteria. When bacterial population was grown on an agar plate containing antibiotic penicillin, the colonies that were sensitive to penicillin died, whereas one or few bacterial colonies that were resistant to penicillin survived.

This is because these bacteria had undergone chance mutation, which resulted in the evolution of a gene that made them resistant to penicillin drug. Hence, the resistant bacteria multiplied quickly as compared to non-resistant (sensitive) bacteria, thereby increasing their number. Hence, the advantage of an individual over other helps in the struggle for existence.

Question 2.
Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.
Answer:
Fossils of dinosaurs have revealed the evolution of reptiles in Jurassic period. As a result of this, evolution of other animals such as birds and mammals has also been discovered. However, two unusual fossils recently unearthed in China have ignited a controversy over the evolution of birds. Confuciusomis is one such genus of primitive birds that were crow sized and lived during the Creataceous period in China.

Question 3.
Attempt giving a clear definition of the term species.
Answer:
Species can be defined as a group of organisms, which have the capability to interbreed in order to produce fertile offspring.

Question 4.
Try to trace the various components of human evolution (hint: brain size and function, skeletal structure, dietary preference, etc.)
Answer:
The various components of human evolution are as follows:
(i) Brain capacity
(ii) Posture
(iii) Food/dietary preference and other important features

Question 5.
Find out through internet and popular science articles whether animals other than man has self-consciousness.
Answer:
There are many animals other than humans, which have self-consciousness. An example of an animal being self-conscious is dolphins. They are highly intelligent. They have a sense of self and they also recognise others among themselves and others. They communicate with each other by whistles, tail-slapping, and other body movements.

Not only dolphins, there are certain other animals such as crow, parrot, chimpanzee, gorilla, orangutan, etc., which exhibit self-consciousness.

Question 6.
List 10 modern-day animals and using the internet resources link it to a corresponding ancient fossil. Name both.
Answer:
The modern-day animals and their ancient fossils are listed in the following table:

Question 7.
Practise drawing various animals and plants.
Answer:
Ask your teachers and parents to suggest the names of plants and animals and practise drawing them. You can also take help from your book to find the names of plants and animals.

Question 8.
Describe one example of adaptive radiation.
Answer:
Adaptive radiation is an evolutionary process that produces new species from a single, rapidly diversifying lineage. This process occurs due to natural selection. An example of adaptive radiation is Darwin finches, found in Galapagos Island. A large variety of finches is present in Galapagos Island that arose from a single species, which reached this land accidentally. As a result, many new species have evolved, diverged, and adapted to occupy new habitats. These finches have developed different eating habits and different types of beaks to suit their feeding habits. The insectivorous, blood sucking, and other species of finches with varied dietary habits have evolved from a single seed eating finch ancestor.

Question 9.
Can we call human evolution as adaptive radiation?
Answer:
No, human evolution cannot be called adaptive radiation. This is because adaptive radiation is an evolutionary process that produces new species from a single, rapidly diversifying lineage, which is not the case with human evolution. Human evolution is a gradual process that took place slowly in time. It represents an example of anagenesis.

Question 10.
Using various resources such as your school Library or the internet and discussions with your teacher, trace the evolutionary stages of any one animal, say horse.
Answer:
The evolution of horse started with Eohippus during Eocene period. It involved the following evolutionary stages:

1. Gradual increase in body size
2. Elongation of head and neck region
3. Increase in the length of limbs and feet
4. Gradual reduction of lateral digits
5. Enlargement of third functional toe
6. Strengthening of the back
7. Development of brain and sensory organs
8. Increase in the complexity of teeth for feeding on grass

The evolution of horse is represented as follows:

Eohippus: It had a short head and neck. It had four functional toes and a splint of 1 and 5 on each hind limb and a splint of 1 and 3 in each forelimb. The molars were short crowned that were adapted for grinding the plant diet.

Mesohippus: It was slightly taller than Eohippus. It had three toes in each foot.

Merychippus: It had the size of approximately 100 cm. Although it still had three toes in each foot, but it could run on one toe. The side toe did not touch the ground. The molars were adapted for chewing the grass.

Pliohippus: It resembled the modern horse and was around 108 cm tall. It had a single functional toe with splint of 2nd and 4th in each limb.

Equus: Pliohippus gave rise to Equus or the modem horse with one toe in each foot. They have incisors for cutting grass and molars for grinding food.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 6 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

PSEB 12th Class Biology Guide Molecular Basis of Inheritance Textbook Questions and Answers

Question 1.
Group the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous Bases: Adenine, thymine, uracil, and cytosine.
Nucleosides: Cytidine and guanosine.

Question 2.
If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer:
According to Chargaffs rule, the DNA molecule should have an equal ratio of pyrimidine (cytosine and thymine) and purine (adenine and guanine). It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules.
% A = % T and % G = % C
If double stranded DNA has 20% of cytosine, then according to the law, it would have 20% of guanine.
Thus, percentage of G + C content = 40%
The remaining 60% represents both A + T molecule. Since adenine and guanine are always present in equal numbers, the percentage of adenine molecule is 30%.

Question 3.
If the sequence of one strand of DNA is written as follows: 5-ATGCATGCATGCATGCATGCATGCATGC-3′
Write down the sequence of complementary strand in 5′ → 3′ direction.
Answer:
The DNA strands are complementary to each other with respect to base sequence. Hence, if the sequence of one strand of DNA is
5′- ATGCATGCATGCATGCATGCATGCATGC – 3′
Then, the sequence of complementary strand in 5′-3′ direction will be
3′- TACGTACGTACGTACGTACGTACGTACG – 5′
Therefore, the sequence of nucleotides on DNA polypeptide in 5′-3′ direction is
5′- GCATGCATGCATGCATGCATGCATGCAT – 3′

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows:
5′ – ATGCATGCATGCATGCATGCATGCATGC-3 Write down the sequence of mRNA.
Answer:
If the coding strand in a transcription unit is
5′ – ATGCATGCATGCATGCATGCATGCATGC-3′
Then, the template strand in 3′ to 5′ direction would be
3′ – TACGTACGTACGTACGTACGTACGTACG-5′
It is known that the sequence of mRNA is same as the coding strand of DNA.
However, in RNA, thymine is replaced by uracil.
Hence, the sequence of mRNA will be
5′ – AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer:
Watson and Crick observed that the two strands of DNA are f anti-parallel and complementary to each other with respect to their base sequences. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. It means that the double stranded DNA molecule separates and then, each of the separated strand acts as a template for the synthesis of a new complementary strand. As a result, each DNA molecule would have one parental strand and a newly synthesised daughter strand.

Since only one parental strand is conserved in each daughter molecule, it is known as semi-conservative mode of replication.

Quetion 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:
There are two different types of nucleic acid polymerases.

1. DNA-dependent DNA polymerases
2. DNA-dependent RNA polymerases

The DNA-dependent DNA polymerases use a DNA template strand for synthesising a new strand of DNA, whereas DNA-dependent RNA polymerases use a DNA template strand for synthesising a new strand of RNA.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic material. They used different radioactive isotopes to label DNA and protein coat of the bacteriophage.

They grew some bacteriophages on a medium containing radioactive phosphorus (³²) to identify DNA and some on a medium containing radioactive sulphur (³⁵S) to identify protein. Then, these radioactive labelled phages were allowed to infect E.coli bacteria. After infecting, the protein coat of the bacteriophage was separated’from the bacterial cell by blending and then subjected to the process of centrifugation. Since the protein coat was lighter, it was found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge tube. Hence, it was proved that DNA is the genetic material as it was transferred from virus to bacteria.

Question 8.
Differentiate between the following :
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand Arts,
Answer:
(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

(c) Template strand and Coding strand

Question 9.
List two essential roles of ribosome during translation.
Answer:
The important functions of ribosome during translation are as follows :
(a) Ribosome acts as the site where protein synthesis takes place from individual amino .acids. It is made up of two subunits.
The smaller subunit comes in contact with mRNA and forms a protein synthesising complex whereas the larger subunit acts as an amino acid binding site.

(b) Ribosome acts as a catalyst for forming peptide bond. For example, 23s r-RNA in bacteria acts as a ribozyme.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer:
Lac operon is a segment of DNA that is made up of three adjacent structural genes, namely, an operator gene, a promoter gene, and a regulator gene. It works in a coordinated manner to metabolise lactose into glucose and galactose.

In lac operon, lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promoter region. Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose so that lactose is metabolised into glucose and galactose. After sometime, when the level of inducer decreases as it is completely metabolised by enzymes, it causes synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence, the transcription is stopped. This type of regulation is known as negative regulation.

Question 11.
Explain (in one or two lines) the function of the following:
(a) Promoter
(b) tRNA
(c) Exons
Answer:
(a) Promoter: Promoter is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.

(b) tRNA: tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.

(c) Exons: Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.

Question 12.
Why is the Human Genome project called a mega project?
Answer:
Human genome project was considered to be a mega project because it had a specific goal to sequence every base pair present in the human genome. It took around 13 years for its completion and got accomplished in year 2006. It was a large scale project, which aimed at developing new technology and generating new information in the field of genomic studies. As a result of it, several new areas and avenues have opened up in the field of genetics, biotechnology, and medical sciences. It provided clues regarding the understanding of human biology.

Question 13.
What is DNA fingerprinting? Mention its application.
Answer:
DNA fingerprinting is a technique used to identify and analyse the variations in various individuals at the level of DNA. It is based on variability and polymorphism in DNA sequences.
Applications

1. It is used in forensic science to identify potential crime suspects.
2. It is used to establish paternity and family relationships.
3. It is used to identify and protect the commercial varieties of crops and livestock.
4. It is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.

Question 14.
Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics
Answer:
(a) Transcription: It is the process of synthesis of RNA from DNA template. A segment of DNA gets copied into mRNA during the process. The process of transcription starts at the promoter region of the template DNA and terminates at the terminator region. The segment of DNA between these two regions is known as transcription unit. The transcription requires RNA polymerase enzyme, a DNA template, four types of ribonucleotides, and certain cofactors such as Mg²⁺.
The three important events that occur during the process of transcription are as follows:

1. Initiation
2. Elongation
3. Termination

The DNA-dependent RNA polymerase and certain initiation factors bind at the double stranded DNA at the promoter region of the template strand and initiate the process of transcription. RNA polymerase moves along the DNA and leads to the unwinding of DNA duplex into two separate strands. Then, one of the strands, called sense strand, acts as template for mRNA synthesis. The enzyme, RNA polymerase, utilises nucleoside triphosphates (dNTPs) as raw material and polymerises them to form mRNA according to the complementary bases present on the template DNA«. This process of opening of helix-and elongation of polynucleotide chain continues until the enzyme reaches the terminator region. As RNA polymerase reaches the terminator region, the newly synthesised mRNA transcripted along with enzyme is released. Another factor called terminator factor is required for the termination of the transcription.

(b) Polymorphism: It is a form of genetic variation in which distinct nucleotide sequence can exist at a particular site in a DNA molecule. This heritable mutation is observed at a high frequency in a population. It arises due to mutation either in somatic cell or in the germ cells. The germ cell mutation can be transmitted from parents to their offsprings. This results in accumulation of various mutations in a population, leading to variation and polymorphism in the population. This plays a very important role in the process of evolution and tracing human history.

(c) Translation: It is the process of polymerising amino acid to form a polypeptide chain. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain.
The process of translation involves following three steps:

1. Initiation
2. Elongation
3. Termination

During the initiation of the translation, tRNA gets charged when the amino acid binds to it using ATP. The start (initiation) codon (AUG) present on mRNA is recognised only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in a large subunit for the attachment of subsequent amino acids. The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the process of translation. During the elongation process, the ribosome moves one codon downstream along with mRNA so as to leave the space for binding of another charged tRNA. The amino acid brought by tRNA gets linked with the previous amino acid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codon (VAA, UAG, and UGA), the process of translation gets terminated. The polypeptide chain is released and the ribosomes get detached from mRNA.

(d) Bioinformatics: It is the application of computational and statistical techniques to the field of molecular biology. It solves the practical problems arising from the management and analysis of biological data. The field of bioinformatics developed after the completion of human genome project (HGP). This is because enormous amount of data has been generated during the process of HGP that has to be managed and stored for easy access and interpretation for future use by various scientists. Hence, bioinformatics involves the creation of biological databases that store the vast information of biology.

It develops certain tools for easy and efficient access to the information and its utilisation. Bioinformatics has developed new algorithms and statistical methods to find out the relationship between the data, to predict protein structure and their functions, and to cluster the protein sequences into their related families.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 5 Principles of Inheritance and Variation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation

PSEB 12th Class Biology Guide Principles of Inheritance and Variation Textbook Questions and Answers

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:
Mendel selected pea plants to carry out his study on the inheritance of characters from parents to offspring. He selected a pea plant because of the following features:

• Peas have many visible contrasting characters such as tall/dwarf plants, round/wrinkled seeds, green/yellow pod, purple/white flowers, etc.
• Peas have bisexual flowers and therefore undergo self pollination easily. Thus, pea plants produce offsprings with same traits generation after generation.
• In pea plants, cross pollination can be easily achieved by emasculation in which the stamen of the flower is removed without affecting the pistil.
• Pea plants have a short life span and produce many seeds in one generation.

Question 2.
Differentiate between the following:
Dominance and Recessive Homozygous and Heterozygous
(a) Dominance and Recessive
(b) Homozygous and Heterozygous
(c) Monohybrid and Dihybrid.
Answer:
(a) Dominance and Recessive

(b) Homozygous and Heterozygous

(c) Monohybrid and Dihybrid

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer:
Locus is a fixed position on a chromosome, which is occupied by a single or more genes. Heterozygous organisms contain different alleles for an allelic pair. Hence, a diploid organism, which is heterozygous at four loci, will have four different contrasting characters at four different loci.

For example, if an organism is heterozygous at four loci with four characters, say Aa, Bb, Cc, Dd, then during meiosis, it will segregate to form 8 separate gametes.

If the genes are not linked, then the diploid organism will produce 16 different gametes. However, if the genes are linked, the gametes will reduce their number as the genes might be linked and the linked genes will be inherited together during the process of meiosis.

Question 4.
Explain the Law of Dominance using a monohybrid cross.
Answer:
Mendel’s law of dominance states that a dominant allele expresses itself in a monohybrid cross and suppresses the expression of recessive allele. However, this recessive allele for a character is not lost and remains hidden or masked in the progenies of F₁ generation and reappears in the next generation.

For example, when pea plants with round seeds (RR) are crossed with plants with wrinkled seeds (rr), all seeds in F₁ generation were found to be round (Rr). When these round seeds were self fertilised, both the round and wrinkled seeds appeared in F₂ generation in 3 : 1 ratio. Hence, in F₂ generation, the dominant character (round seeds) appeared and the recessive character (wrinkled seeds) got suppressed, which reappeared in F₂ generation.

Question 5.
Define and design a test-cross.
Answer:
Test cross is a cross between an organism with unknown genotype and a recessive parent. It is used to determine whether the individual is homozygous or heterozygous for a trait.
If the progenies produced by a test cross show 50% dominant trait and 50% recessive trait, then the unknown individual is heterozygous for a trait. On the other hand, if the progeny produced shows dominant trait, then the unknown individual is homozygous for a trait.

Question 6.
Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Answer:
In guinea pigs, heterozygous male with black coat colour (Bb) is crossed with the homozygous female having white coat colour (bb). The male will produce two types of gametes, B and b, while the female will produce only one kind of gamete, b. The genotypic and phenotypic ratio in the progenies of Fx generation will be same i.e., 1:1.

Question 7.
When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to he
(a) tall and green.
(b) dwarf and green.
Answer:
A cross between tall plant with yellow seeds and tall plant with green seeds will produce
(a) three tall and green plants
(b) one dwarf and green plant

Question 8.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dihybrid cross?
Answer:
When two individual heterozygous for two loci (Yy Rr) are crossed and the two loci are linked, the distribution of the phenotypic feature of F₁ generation will be in the ratio of 3:1 \(\frac{3}{4}\) of the individuals will show
both the dominant traits and \(\frac{1}{4}\) of the individuals will show both the
recessive traits. It is because the genes for both the traits are present on the same chro

Question 9.
Briefly mention the contribution of T.H. Morgan in genetics.
Answer:
Morgan’s work is based on fruit flies (Drosophila melanogaster). He formulated the chromosomal theory of linkage. He defined linkage as the co-existence of two or more genes in the same chromosome and performed dihybrid crosses in Drosophila to show that linked genes are inherited together and are located on X-chromosome. His experiments have also proved that tightly linked genes show very low recombination while loosely linked genes show higher recombination.

Question 10.
What is pedigree analysis? Suggest how such an analysis, can be useful.
Answer:
Pedigree analysis is a record of occurrence of a trait in several generations of a family. It is based on the fact that certain characteristic features are heritable in a family, for example, eye colour, skin colour, hair form and colour, and other facial characteristics. Along with these features, there are other genetic disorders such as Mendelian disorders that are inherited in a family, generation after generation. Hence, by using pedigree analysis for the study of specific traits or disorders, generation after generation, it is possible to trace the pattern of inheritance. In this analysis, the inheritance of a trait is represented as a tree, called family tree. Genetic counselors use pedigree chart for analysis of various traits and diseases in a family and predict their inheritance patterns. It is useful in preventing haemophilia, sickle cell anaemia, and other genetic disorders in the future generations.

Question 11.
How is sex determined in human beings?
Answer:
Human beings exhibit male heterogamy. In humans, males (XY) produce two different types of gametes, X and Y. The human female (XX) produces only one type of gametes containing X chromosomes. The sex of the baby is determined by the type of male gamete that fuses with the female gamete. If the fertilising sperm contains X chromosome, then the baby produced will be a girl and if the fertilising sperm contains Y chromosome, then the baby produced will be a boy. Hence, it is a matter of chance that determines the sex of a baby. There is an equal probability of the fertilising sperm being an X or Y chromosome. Thus, it is the genetic make up of the sperm that determines the sex of the baby.

Question 12.
A child has hlood group O. If the father has hlood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Answer:
The blood group characteristic in humans is controlled by three set of alleles, namely, I^A,I^B and i. The alleles, I^A and I^B, are equally dominant whereas allele, i, is recessive to the other alleles. The individuals with genotype, I^A I^A and I^A i, have blood group A whereas the individuals with genotype, I^B I^B and I^B i, have blood group B. The persons with genotype I^A I^B have blood group AB while those with blood group O have genotype ii.
Hence, if the father has blood group A and mother has blood group B, then the possible genotype of the parents will be
Father
I^AI ^{or A}I^Ai

Mother
I^BI^B or I^Bi
A cross between homozygous parents will produce progeny with AB blood group.

A cross between heterozygous parents will produce progenies with AB blood group (I^A I^B) and O blood group (ii).

Question 13.
Explain the following terms with example
(a) Co-dominance
(b) Incomplete dominance
Answer:
(a) Co-dominance: Co-dominance is the phenomenon in which both the alleles of a contrasting character are expressed in heterozygous condition. Both the alleles of a gene are equally dominant. ABO blood group in human beings is an example of co-dominance. The blood group character is controlled by three sets of alleles, namely, I^A, I^B, and i. The alleles, I^A and I^B, are equally dominant and are said to be co-dominant as they are expressed in AB blood group. Both these alleles do not interfere with the expression of each other and produce their respective antigens. Hence, AB blood group is an example of co-dominance.

(b) Incomplete Dominance: Incomplete dominance is a phenomenon in which one allele shows incomplete dominance over the other member of the allelic pair for a character. For example, a monohybrid cross between the plants having red flowers and white flowers in Antirrhinum species will result in all pink flower plants in Fj generation. The progeny obtained in Fx generation does not resemble either of the parents and exhibits intermediate characteristics. This is because the dominant allele, R, is partially dominant over the other allele, r. Therefore, the recessive allele, r, also gets expressed in the generation resulting in the production of intermediate pink flowering progenies with Rr genotype.

What is point mutation? Give one example.
Answer:
Point mutation is a change in a single base pair of DNA by substitution, deletion, or insertion of a single nitrogenous base. An example of point mutation is sickle cell anaemia. It involves mutation in a single base pair in the beta-globin chain of haemoglobin pigment of the blood. Glutamic acid in short arm of chromosome II gets replaced with valine at the sixth position.

Question 15.
Who had proposed the chromosomal theory of the inheritance?
Answer:
Sutton and Boveri proposed the chromosomal theory of inheritance in 1902. They linked die inheritance of traits to the chromosomes.

Question 16.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
Two autosomal genetic disorders are as follows:
1. Sickle Cell Anaemia: It is an autosomal linked recessive disorder, which is caused by point mutation in the beta-globin chain of haemoglobin pigment of the blood. The disease is characterised by sickle shaped red blood cells, which are formed due to the mutant haemoglobin molecule. The disease is controlled by Hb^A and Hb^S allele. The homozygous individuals with genotype, Hb^SHb^S, show the symptoms of this disease while the heterozygous individuals with genotype, Hb^A Hb^S, are not affected. However, they act as carriers of the disease.

Symptoms: Rapid heart rate, breathlessness, delayed growth and puberty, jaundice, weakness, fever, excessive thirst, chest pain, and decreased fertility are the major symptoms of sickle cell anaemia disease.

(b) Down’s Syndrome: It is an autosomal disorder that is caused by the trisomy of chromosome 21.
Symptoms : The individual is short statured with round head, open mouth, protruding tongue, short neck, slanting eyes, and broad short hands. The individual also shows retarded mental and physical growth.