Class 12

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 11 Biotechnology: Principles and Processes Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

PSEB 12th Class Biology Guide Biotechnology: Principles and Processes Textbook Questions and Answers
Question 1.
Can you list 10 recombinant proteins which are used in medical ; practice? Find out where they are used as therapeutics (use the internet).
Answer:

Recombinant proteins	Therapeutic uses
(a) Insulin	Used for diabetes mellitus
(b) OKT-3	Therapeutic antibody, used for reversal
(c) DNase	Treatment of cystic fibrosis
(d) Reo Pro	Prevention of blood clots
(e) Blood clotting factor VIII	Treatment of haemophilia A
(f) Blood clotting factor IX	Treatment of haemophilia B
(g) Tissue plasminogen activator	For acute myocardial infarction
(h) Interferon alpha (INF alpha)	Used for hepatitis C
(i) Interferon beta (INF beta)	Used for multiple sclerosis
(j) Interferon gamma (INF gamma)	Used for granulomatous disease

Question 2.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.
Answer:

Question 3.
From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
The molecular size of DNA molecules is more than that of enzymes. It is because an enzyme (protein) is synthesised from a segment of DNA called the gene.

Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
The molar concentration of human DNA in a human diploid cell is as follows:
Total number of chromosomes × 6.023 × 10²³
= 46 × 6.023 × 10²³
= 2.77 × 1018moles

Question 5.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
No. Eukaryotic cells do not have restriction endonucleases. All the restriction endonucleases have been isolated from the various strains of bacteria and they are also named according to the genus and species of prokaryotes. The first letter of the enzyme comes from the genus and the second two letters come from the species of the prokaryotic cell from which they have been isolated. For example, EcoRI comes from Escherichia coliRY 13. In EcoRI, the letter ‘R’ is derived from the name of the strain. Roman numbers following the names indicate the order in which the enzymes were isolated from that strain of bacteria.

Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:
Shake flasks are used for growing microbes and mixing the desired materials on a small scale in the laboratory. However, the large-scale production of a desired biotechnological product requires large stirred tank bioreactors.

Besides better aeration and mixing properties, bioreactors have the following advantages:

• It has an oxygen delivery system.
• It has a foam control, temperature and pH control system.
• Small volumes of culture can be withdrawn periodically.

Question 7.
Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Answer:
Some palindromic DNA sequences and the restriction enzymes which act on them are as follows :

• 5′-AGCT-3′ Alul (Arthrobacter luteus)
  3′-TCGA-5′
• 5′-GAATTC-3′ EcoRI (Escherichia coli)
  3′-CTTAAG-5′
• 5′-AAGCTT-3′ Hindlil (Haemophilus influenzae)
  3′-TTCGAA-5′
• 5′-GTCGAC-3′ Sail (Streptomyces albus)
  3′-CAGCTG-5′
• 5′-CTGCAG-3′ PstI (Providencia stuartii)
  3′-GACGTC-5′

Question 8.
Can you recall meiosis and indicate at what stage a recombinant DNA is made?
Answer:
A recombinant DNA is made in the pachytene stage of prophase I by f crossing over.

Question 9.
Can you think and answer how a reporter enzyme can be used to
monitor transformation of host cells by foreign DNA in addition to a selectable marker?
Answer:
Reporter enzyme can differentiate recombinants from non-recombinants on the basis of their ability to produce a specific colour in the presence of a chromogenic substrate. DNA is inserted within the coding sequence of the enzyme p-galactosidase. This results into inactivation of the enzyme which is referred to as insertional inactivation.

The presence of a chromogenic substrate gives blue-coloured colonies if the plasmid in the bacteria does not have an insert. The presence of the insert results in insertional inactivation of α-galactosidase and the colonies do not produce any colour. These are identified as recombinant colonies.

Question 10.
Describe briefly the following:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing
Answer:
(a) Origin of Replication: This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells. This sequence is also responsible for controlling the copy number of the linked DNA. So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin supports a high copy number.

(b) Bioreactors: Bioreactors are vessels in which raw materials are biologically converted into specific products, individual enzymes etc. using microbial plant, animal or human cells. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen). The most commonly used bioreactors are of stirring type. A biogas plant is a good example of a bioreactor.

(c) Downstream Processing: After completion of the biosynthetic stage, the product is subjected through a series of processes before it is ready for marketing as a finished product. The processes include separation and purification, which are collectively referred to as downstream processing. The product has to be formulated with suitable preservatives. Such formulation has to undergo thorough clinical trials as in the case of drugs. Strict quality control testing for each product is also required. Downstream processing and quality control testing vary from product to product.

Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase
Answer:
(a) PCR: PCR stands for Polymerase Chain Reaction. In this reaction multiple copies of the gene (or DNA) of interest is synthesised in vitro using two sets of primers and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as template. The segment of DNA can be amplified to approximately billion times if the process of replication of DNA is repeated many times.

(b) Restriction Enzymes and DNA: Restriction enzymes are synthesised by microbes as a defence mechanism and are specifically endonucleases which cleave the double-stranded DNA with the desired genes. This activity occurs at a limited number of sites depending on the number of recognition sequences in DNA. Lysing enzymes, synthesising enzymes (DNA polymerase and reverse transcriptase) and ligases are also tools of genetic engineering.

(c) Chitinase: During the isolation of DNA in the processes of recombinant DNA technology, the fungal cell is heated with an enzyme called chitinase. The chitinase enzyme dissolves the chitin membrane to open the cell for release of DNA along with other macromolecules such as RNA, proteins, polysaccharides and lipids.

Question 12.
Discuss with your teacher and find out how to distinguish between
(a) Plasmid DNA and Chromosomal DNA
(b) RNA and DNA
(c) Exonuclease and Endonuclease
Answer:
(a)

Plasmid DNA	Chromosomal DNA
Plasmid DNA is the naked double-stranded DNA which forms a circle with no free ends. It is associated with few proteins but contains RNA polymerase enzyme. They are smaller than the host chromosomes and can be easily separated.	Chromosomal DNA is a double-stranded linear DNA molecule associated with large proteins. This DNA exists in relaxed and supercoiled forms and provides a template for replication and transcription. It has free ends represented as 3′-5′.

(b)

DNA	RNA
(i) It is mainly confined to the nucleus. A small quantity occurs in mitochondria and chloroplasts.	It mainly occurs in the cytoplasm. A small quantity is found in the nucleus.
(ii) Its quantity is constant in each cell of a species.	Its quantity varies in different cells.
(iii) It contains deoxyribose sugar.	It contains ribose sugar.
(iv) Its pyrimidines are adenine and thymine.	Its pyrimidines are adenine and uracil.
(v) The amount of adenine is equal to the amount of thymine. Also, the amount of cytosine is equal to the amount of guanine.	Adenine and uracil are not necessarily in equal amounts, nor are cytosine and guanine necessarily in equal amounts.
(vi) It can replicate itself.	It cannot replicate itself. It is formed by DNA. Some RNA viruses (paramyxo virus) can produce RNA from an RNA template.

(c) Exonucleases are nucleases which cut off the nucleotides from the 5′ or 3′ ends of a DNA molecule, whereas endonucleases are nucleases which cleave the DNA duplex at any point except at the ends.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 10 Microbes in Human Welfare Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare

PSEB 12th Class Biology Guide Microbes in Human Welfare Textbook Questions and Answers

Question 1.
Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?
Answer:
Curd can be used as a sample for the study of microbes. Curd contains numerous lactic acid bacteria (LAB) or Lactobacillus. These bacteria produce acids that coagulate and digest milk proteins. A small drop of curd contains millions of bacteria, which can be easily observed under a microscope.

Question 2.
Give examples to prove that microbes release gases during metabolism.
Answer:
The examples of microbes that release gases during metabolism are as follows :
(a) Bacteria and fungi carry out the process of fermentation and during this process, they release carbon dioxide. Fermentation is the process of converting a complex organic substance into a simpler substance with the action of bacteria or yeast. Fermentation of sugar produces alcohol with the release of carbon dioxide and very little energy.

(b) The dough used for making idli and dosa gives a puffed appearance. This is because of the action of bacteria which releases carbon dioxide. This CO₂ released from the dough gets trapped in the dough, thereby giving it a puffed appearance.

Question 3.
In which food would you find lactic acid bacteria? Mention some of their useful applications.
Answer:
Lactic acid bacteria can be found in curd. It is this bacterium that promotes the formation of milk into curd. The bacterium multiplies and increases its number, which converts the milk into curd. They also increase the content of vitamin B₁₂ in curd.
Lactic acid bacteria are also found in our stomach where it keeps a check on the disease-causing micro-organisms.

Question 4.
Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve use of microbes.
Answer:

1. Wheat Product: Bread, cake etc.
2. Rice Product: Idli, dosa etc.
3. Bengal Gram Product: Dhokla, khandvi etc.

Question 5.
In which way have microbes played a m^jor role in controlling diseases caused by harmful bacteria?
Answer:
Several micro-organisms are used for preparing medicines. Antibiotics are medicines produced by certain micro-organisms to kill other disease-causing micro-organisms. These medicines are commonly obtained from bacteria and fungi. They either kill or stop the growth of disease-causing micro-organisms. Streptomycin, tetracycline, and penicillin are common antibiotics. Penicillium notatum produces chemical penicillin, which checks the growth of Staphylococci bacteria in the body. Antibiotics are designed to destroy bacteria by weakening their cell walls. As a result of this weakening, certain immune cells such as the white blood cells enter the bacterial cell and cause cell lysis. Cell lysis is the process of destroying cells such as blood cells and bacteria.

Question 6.
Name any two species of fungus, which are used in the production of the antibiotics.
Answer:
Antibiotics are medicines that are produced by certain micro-organisms to kill other disease-causing micro-organisms. These medicines are commonly obtained from bacteria and fungi.
The species of fungus used in the production of antibiotics are as follows:

Antibiotic	Fungus source
1. Penicillin	Penicillium notatum
2. Cephalosporin	Cephalosporium acremonium

Question 7.
What is sewage? In which way can sewage be harmful to us?
Answer:
Sewage is the municipal waste matter that is carried away in sewers and drains. It includes both liquid and solid wastes, rich in organic matter and microbes. Many of these microbes are pathogenic and can cause several water borne diseases. Sewage water is a major cause of polluting drinking water. Hence, it is essential that sewage water is properly collected, treated, and disposed. If untreated sewage is disposed into rivers and streams, it will pollute the water bodies.

Question 8.
What is the key difference between primary and secondary sewage treatment?
Answer:

Primary sewage treatment	Secondary sewage treatment
1. It is a mechanical process involving the removal of coarse solid materials.	It is a biological process involving the action of microbes.
2. It is inexpensive and relatively less complicated.	It is a very expensive and complicated process.

Question 9.
Do you think microbes can also be used as source of energy? If yes, how?
Answer:
Yes, microbes can be used as a source of energy. Bacteria such as Methane bacterium is used for the generation of gobar gas or biogas.

The generation of biogas is an anaerobic process in a biogas plant, which consists of a concrete tank (10-15 feet deep) with sufficient outlets and inlets. The dung is mixed with water to form the slurry and thrown into the tank. The digester of the tank is filled with numerous anaerobic methane-producing bacteria, which produce biogas from the slurry. Biogas can be removed through the pipe which is then used as a source of energy, while the spent slurry is removed from the outlet and is used as a fertiliser.

Question 10.
Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished.
Answer:
Microbes play an important role in organic farming, which is done without the use of chemical fertilisers and pesticides. Bio-fertilisers are living organisms which help increase the fertility of soil. It involves the ’ selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. Bio-fertilisers are introduced in seeds, roots, or soil to mobilise the availability of nutrients. Thus, they are extremely beneficial in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azpirillum and Azotobacter are free living nitrogen-fixing bacteria whereas Anabaena, Nostoc and Oscillatoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilisers are cost effective and eco-friendly.

Microbes can also act as bio-pesticides to control insect pests in plants.
An example of bio-pesticides is Bacillus thuringiensis, which produces a toxin that kills the insect pests. Dried bacterial spores are mixed in water and sprayed in agricultural fields. When larvae of insects feed on crops, these bacterial spores enter the gut of the larvae and release r toxins, thereby it. Similarly, Trichoderma are free living fungi. They live in the roots of higher plants and protect them from various pathogens.

Baculoviruses is another bio-pesticide that is used as a biological control I agent against insects and other arthropods.

Question 11.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A, B and C were recorded as 20mg/L, 8mg/L and 400mg/L, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?
Answer:
Biological oxygen demand (BOD) is the method of determining the amount of oxygen required by micro-organisms to decompose the waste present in the water supply. If the quantity of organic wastes in the water supply is high, then the number of decomposing bacteria present in the water will also be high. As a result, the BOD value will increase. Therefore, it can be concluded that if the water supply is more polluted, then it will have a higher BOD value. Out of the above three samples, sample C is the most polluted since it has the maximum BOD value of 400 mg/L.

After untreated sewage water, secondary effluent discharge from a sewage treatment plant is most polluted. Thus, sample A is secondary effluent discharge from a sewage treatment plant and has the BOD value of 20 mg/L, while sample B is river water and has the BOD value of 8 mg/L.
Hence, the correct label for each sample is:

Label	BOD value	Sample
A.	20 mg/L	Secondary effluent discharge from a sewage treatment plant
B.	8 mg/L	River water
C.	400 mg/L	Untreated sewage water

Question 12.
Find out the name of the microbes from which Cyclosporin A (an immunosuppressive drug) and Statins (blood cholesterol lowering agents) are obtained.
Answer:

Drug	Function	Microbe
1. Cyclosporin-A	Immuno suppressive drug	Trichoderma polysporum
2. Statin	Blood cholesterol lowering agent	Monascus purpureus

Question 13.
Find out the role of microbes in the following and discuss it with your teacher.
(a) Single cell protein (SCP)
(b) Soil
Answer:
(a) Single Cell Protein (SCP): A single cell protein is a protein obtained from certain microbes, which forms an alternate source of proteins in animal feeds. The microbes involved in the preparation of single cell proteins are algae, yeast, or bacteria. These microbes are grown on an industrial scale to obtain the desired protein. For example, Spirulina can be grown on waste materials obtained from molasses, sewage, and animal manures. It serves as a rich supplement of dietary nutrients such as proteins, carbohydrate, fats, minerals, and vitamins. Similarly, micro-organisms such as Methylophilus and methylotrophus have a large rate of biomass production. Their growth can produce a large amount of proteins.

(b) Soil: Microbes play an important role in maintaining soil fertility. They help in the formation of nutrient-rich humus by the process of decomposition. Many species of bacteria and cyanobacteria have the ability to fix atmospheric nitrogen into usable form. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillum and Azotobacter are free living nitrogen-fixing bacteria, whereas Anabaena, Nostoc, and Oscillatoria are examples of nitrogen-fixing cyanobacteria.

Question 14.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer.
Biogas, Citric acid, Penicillin and Curd
Answer:
The order of arrangement of products according to their decreasing importance is :
Penicillin > Biogas > Citric acid > Curd
Penicillin is the most important product for the welfare of human society. It is an antibiotic, which is used for controlling various bacterial diseases. The second most important product is biogas. It is aneco-friendly source of energy. The next important product is citric acid, which is used as a food preservative. The least important product is curd, a food item obtained by the action of Lactobacillus bacteria on milk.

Question 15.
How do biofertilisers enrich the fertility of the soil?
Answer:
Bio-fertilisers are living organisms which help in increasing the fertility of soil. It involves the selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. These are introduced to seeds, roots, or soil to mobilise the availability of nutrients by their biological activity. Thus, they are extremely beneficial jf in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillum and Azotobacter are free living nitrogen-fixing bacteria, whereas Anabaena, Nostoc, and Oscillatoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilisers are cost effective and eco-friendly.

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:

Crop plant	Hybrid variety
Wheat	Sonalika and Kalyan Sona
Rice	Jaya and Ratna
Cauliflower	Pusa Shubhra and Pusa Snowball K-1
Cowpea	Pusa Komal
Mustard	Pusa Swarnim

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:

Disease	Causative organism	Mode of transmission
a. Amoebiasis	Entamoeba histolytica	It is a vector-borne disease that spreads by the means of contaminated food and water. The vector involved in the transmission of this disease is the housefly.
b. Malaria	Plasmodium sp.	It is a vector-borne disease that spreads by the biting of the female Anopheles mosquito.
c. Ascariasis	Ascaris lumbricoides	It spreads via contaminated food and water.
d. Pneumonia	Streptococcus pneumoniae	It spreads by the sputum of an infected person.

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

Innate immunity	Acquired immunity
1. It is a non-pathogen specific type of defense mechanism.	It is a pathogen specific type of defense mechanism.
2. It is inherited from parents and protects the individual since birth.	It is acquired after the birth of an individual.
3. It operates by providing barriers against the entry of foreign infectious agents.	It operates by producing primary and secondary responses, which are mediated by B-lymphocytes and T-lymphocytes.
4. It does not have a specific memory.	It is characterised by an immunological memory.

(b) Active and passive immunity

Active immunity	Passive immunity
1. It is a type of acquired immunity in which the body produces its own antibodies against disease-causing antigens.	It is a type of acquired immunity in which readymade antibodies are transferred from one individual to another.
2. It has a long lasting effect.	It does not have long lasting effect.
3. It is slow. It takes time in producing antibodies and giving responses.	It is fast. It provides immediate relief.
4. Injecting microbes through vaccination inside the body is an example of active immunity.	Transfer of antibodies present in the mother’s milk to the infant is an example of 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:

Normal cell	Cancerous cell
1. Normal cells show the property of contact inhibition. Therefore, when these cells come into contact with other cells, they stop dividing.	Cancerous cells lack the property of contact inhibition. Therefore, they continue to divide, thereby forming a mass of cells or tumor.
2. They undergo differentiation after attaining a specific growth.	They do not undergo differentiation.
3. These cells remain confined at a particular location.	These cells do not remain confined at a particular location. They move into neighbouring tissues and disturb its function.

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:

Animal	Fossil
1. Man	Ramapithecus
2. Horse	Eohippus
3. Dog	Leptocyon
4. amel	Protylopus
5. Elephant	Moerithers
6. Whale	Protocetus
7. Fish	Arandaspis
8. Tetrapods	Icthyostega
9. Bat	Archaeonycteris
10. Giraffe	Palaeotragus

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

Repetitive DNA	Satellite DNA
Repetitive DNA are DNA sequences that contain small segments, which are repeated many times.	Satellite DNA are DNA sequences that contain highly repetitive DNA.

(b) mRNA and tRNA

mRNA	tRNA
1. mRNA or messenger RNA acts as a template for the process of transcription.	tRNA or transfer RNA acts as an adaptor molecule that carries a specific amino acid to mRNA for the synthesis of polypeptide.
2. It is a linear molecule.	It has clover leaf shape.

(c) Template strand and Coding strand

Template strand	Coding strand
1. Template strand of DNA acts as a template for the synthesis of mRNA during transcription.	Coding strand is a sequence of DNA that has the same base sequence as that of mRNA (except thymine that is replaced by uracil in DNA).
2. It runs from 3′ to 5′.	It runs from 5′ to 3′.

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

Dominance	Recessive
1. A dominant trait expresses itself in the presence or absence of a recessive trait.	A recessive trait is able to express itself only in the absence of a dominant trait.
2. For example, tall plant, round seed, violet flower, etc. are dominant traits in a pea plant.	For example, dwarf plant, wrinkled seed, white flower, etc. are recessive traits in a pea plant.

(b) Homozygous and Heterozygous

Homozygous	Heterozygous
1. It contains two similar alleles for a particular trait.	It contains two different alleles for a particular trait.
2. Genotype for homozygous possess either dominant or recessive, but never both the alleles. For example, RR or rr	Genotype for heterozygous possess both dominant and recessive alleles. For example, Rr
3. It produces only one type of gamete.	It produces two different types of gametes.

(c) Monohybrid and Dihybrid

Monohybrid	Dihybrid
1. Monohybrid involves cross between parents’, which differs in only one pair of contrasting characters.	Dihybrid involves cross between parents, which differs in two pairs of contrasting characters.
2. For example, the cross between tall and dwarf pea plant is a monohybrid cross.	For example, the cross between pea plants having yellow wrinkled seed with those having green round seeds is a dihybrid cross.

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.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 4 Reproductive Health Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 4 Reproductive Healths

PSEB 12th Class Biology Guide Reproductive Health Textbook Questions and Answers

Question 1.
What do you think is the significance of reproductive health in a society?
Answer:
Reproductive health is the total well being in all aspects of reproduction. It includes physical, emotional, behavioural, and social well being. Sexually transmitted diseases such as AIDS, gonorrhoea, etc. are transferred from one individual to another through sexual contact. It can also lead to unwanted pregnancies.

Hence, it is necessary to create awareness among people, especially the youth, regarding various reproduction related aspects as the young individuals are the future of the country and they are most susceptible of acquiring sexually transmitted diseases. Creating awareness about the available birth control methods, sexually transmitted diseases and their preventive measures, and gender equality will help in bringing up a socially conscious healthy family. Spreading. awareness regarding uncontrolled population growth and social evils among young individuals will help in building up a reproductively healthy society.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Answer:
Reproductive health is the total well being in all aspects of reproduction. The aspects which have to be given special attention in the present scenario are as follows:
1. Counselling and creating awareness among people, especially the youth, about various aspects of reproductive health, such as sexually transmitted diseases, available contraceptive methods, case of pregnant mothers, adolescence, etc.

2. Providing support and facilities such as medical assistance to people during pregnancy, STDs, abortions, contraceptives, infertility, etc. for building a reproductively healthy society.

Question 3.
Is sex education necessary in schools? Why?
Answer:
Yes, introduction of sex education in schools is necessary. It would provide right information to young individuals at the right time about various aspects of reproductive health such as reproductive organs, puberty, and adolescence related changes, safe sexual practices, sexually transmitted diseases, etc.

The young individual or adolescents are more susceptible in acquiring various sexually transmitted diseases. Hence, providing information to them at the right time would help them to lead a reproductively healthy life and also protect them from the myths and misconceptions about various sex related issues.

Question 4.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.
Answer:
Yes, the reproductive health has tremendously improved in India in the last 50 years. The areas of improvement are as follows:

1. Massive child immunisation programme, which has lead to a decrease in the infant mortality rate.
2. Maternal and infant mortality rate, which has been decreased drastically due to better post natal care.
3. Family planning, which has motivated people to have smaller families.
4. Use of contraceptive, which has resulted in a decrease in the rate of sexually transmitted diseases and unwanted pregnancies.

Question 5.
What are the suggested reasons for population explosion?
Answer:
The human population is increasing day by day, leading to population explosion. It is because of the following two major reasons:
(a) Decreased death rate
(b) Increased birth rate and longevity
The death rate has decreased in the past 50 years. The factor leading to decreased death rate and increased birth rate are control of diseases, awareness and spread of education, improvement in medical facilities, ensured food supply in emergency situation, etc. All this has resulted in an increase in the longevity of an individual.

Question 6.
Is the use of contraceptives justified? Give reasons.
Answer:
Yes, the use of contraceptives is absolutely justified. The human population is increasing tremendously. Therefore, to regulate the population growth by regulating reproduction has become a necessary demand in the present times. Various contraceptive devices have been devised to reduce unwanted pregnancies, which help in bringing down the increased birth rate and hence, in checking population explosion.

Question 7.
Removal of gonads cannot be considered as a contraceptive option. Why?
Answer:
Contraceptive devices are used to prevent unwanted pregnancy and to prevent the spreading of STDs. There are many methods, such as natural, barrier, oral, and surgical methods, that prevent unwanted pregnancy. However, the complete removal of gonads cannot be a contraceptive option because it will lead to infertility and unavailability of certain hormones that are required for normal functioning of accessory reproductive parts. Therefore, only those contraceptive methods can be used that prevent the chances of fertilisation rather than making the person infertile forever.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Answer:
Amniocentesis is a pre-natal diagnostic technique that is used to determine the sex and metabolic disorders of the developing foetus in the mother’s uterus through the observation of the chromosomal patterns. This method was developed so as to determine any kind of genetic disorder present in the foetus. However, unfortunately, this technique is being misused to detect the sex of the child before birth and the female foetus is then aborted. Thus, to prevent the increasing female foeticides, it is necessary to ban the usage of amniocentesis technique for determining the sex of a child.

Question 9.
Suggest some methods to assist infertile couples to have children.
Answer:
Infertility is the inability of a couple to produce a baby even after unprotected intercourse. It might be due to abnormalities present in either male or female, or might be even both the partners. The techniques used to assist infertile couples to have children are as follows:
(a) Test Tube Babies: This involves in-vitro fertilisation where the sperms meet the egg outside the body of a female. The zygote, hence produced, is then transferred in the uterus or fallopian tube of a normal female. The babies produced from this method are known as test tube babies.

(b) Gamete Intra Fallopian Transfer (GIFT): It is a technique that involves the transfer of gamete (ovum) from a donor into the fallopian tube of the recipient female who is unable to produce eggs, but has the ability to conceive and can provide right conditions for the development of an embryo.

(c) Intra Cytoplasmic Sperm Injection (ICSI): It is a method of injecting sperm directly into the ovum to form an embryo in laboratory.

(d) Artificial Insemination: Artificial insemination is a method of transferring semen (sperm) from a healthy male donor into the vagina or uterus of the recipient female. It is employed when the male partner is not able to inseminate the female or has low sperm counts.

Question 10.
What are the measures one has to take to prevent from contracting STDs?
Answer:
Sexually transmitted diseases (STDs) get transferred from one individual to the other through sexual contact. Adolescents and young adults are at the greatest risk of acquiring these sexually transmitted diseases. Hence, creating awareness among the adolescents regarding its after-effects can prevent them from contracting STDs. The use of contraceptives, such as condoms, etc. while intercourse, can prevent the transfer of these diseases. Also, sex with unknown partners or multiple partners should be avoided as they may have such diseases. Specialists should be consulted immediately in case of doubt so as to assure early detection and cure of the disease.

Question 11.
State True/False with explanation.
(a) Abortions could happen spontaneously too. (True/False)
(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
(c) Complete lactation could help as a natural method of contraception. (True/False)
(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people. (True/False)
Answer:
(a) False
Abortion is term given for medical termination of pregnancy.

(b) False
Infertility is defined as the inability of the couple to produce baby even after unprotected coitus. It might occur due to abnormalities/defects in either male or female or both.

(c) False
Complete lactation or lactational amenorrhea is a natural method of contraception. Flowever, it is limited till lactation period, which continues till six months after parturition.

(d) True.

Question 12.
Correct the following statements:
(a) Surgical methods of contraception prevent gamete formation.
(b) All sexually transmitted diseases are completely curable.
(c) Oral pills are very popular contraceptives among the rural women.
(d) In E. T. techniques, embryos are always transferred into the uterus.
Answer:
(a) Surgical methods of contraception prevent the flow of gamete during intercourse.
(b) Some of the sexually transmitted diseases are curable if they are detected early and treated properly. AIDS is still an incurable disease.
(c) Oral pills are very popular contraceptives among urban women.
(d) In embryo transfer technique, 8 celled embryos are transferred into the fallopian tube while more than 8 celled embryos are transferred into the uterus.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 3 Human Reproduction Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 3 Human Reproduction

PSEB 12th Class Biology Guide Human Reproduction Textbook Questions and Answers

Question 1.
Fill in the blanks:
(a) Humans reproduce …………………. .(asexually/sexually)
Answer:
sexually.

(b) Humans are …………………… .(oviparous/viviparous/ovoviviparous)
Answer:
viviparous.

(c) Fertilisation is ………………………. in humans, (external/internal)
Answer:
internal

(d) Male and female gametes are …………………. .(diploid/haploid)
Answer:
haploid.

(e) Zygote is ………………….. .(diploid/haploid)
Answer:
diploid.

(f) The process of release of ovum from a mature follicle is called ……………………. .
Answer:
ovulation.

(g) Ovulation is induced by a hormone called …………………… .
Answer:
luteinising hormone.

(h) The fusion of male and female gametes is called ……………………… .
Answer:
fertilisation.

(i) Fertilisation takes place in ………………… .
Answer:
fallopian tube (ampullary-isthmic junction).

(j) Zygote divides to form ……………….., which is implanted in uterus.
Answer:
blastocyst

(k) The structure which provides vascular connection between foetus and uterus is called …………………… .
Answer:
placenta.

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

Question 3.
Draw a labelled diagram of female reproductive system.
Answer:

Question 4.
Write two major functions each of testis and ovary.
Answer:
Functions of the testis
(a) They produce male gametes called spermatozoa by the process of spermatogenesis.
(b) The leydig cells of the seminiferous tubules secrete the male sex hormone called testosterone. Testosterone aids the development of secondary sex characteristics in males.

Functions of the ovary
(a) They produce female gametes called ova by the process of oogenesis.
(b) The growing Graafian follicles secrete the female sex hormone called estrogen. Estrogen aids the development of secondary sex characteristics in females.

Question 5.
Describe the structure of a seminiferous tubule.
Answer:
The production of sperms in the testes takes place in a highly coiled structure called the seminiferous tubules. These tubules are located in the testicular lobules. Each seminiferous tubule is lined by germinal epithelium. It is lined on its inner side by two types of cells namely spermatogonia aid sertoli cells respectively. Spermatogonia are male germ cells which produce primary spermatocytes by meiotic divisions. Primary spermatocytes undergo further meiotic division to form secondary spermatocytes and finally, spermatids. Spermatids later metamorphoses into male gametes called spermatozoa. Sertoli cells are known as nurse cells of the testes as they provide nourishment to the germ cells. There are large polygonal cells known as interstitial cells or leydig cells just adjacent to seminiferous tubules. These cells secrete the male hormone called testosterone.

Question 6.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Answer:
Spermatogenesis is the process of the production of sperms from the immature germ cells in males. It takes place in seminiferous tubules present inside the testes. During spermatogenesis, a diploid spermatogonium (male germ cell) increases its size to form a diploid primary spermatocyte. This diploid primary spermatocyte undergoes first meiotic division (meiosis I), which is a reductional division to form two equal haploid secondary spermatocytes. Each secondary spermatocyte then undergoes second meiotic division (meiosis II) to form two equal haploid spermatids. Hence, a diploid spermatogonium produces four haploid spermatids. These spermatids are transformed into spermatozoa (sperm) by the process called spermiogenesis.

Question 7.
Name the hormones involved in regulation of spermatogenesis.
Answer:
Follicle-stimulating hormones (FSH) and luteinising hormones (LH) are secreted by gonadotropin releasing hormones from the hypothalamus. These hormones are involved in the regulation of the process of spermatogenesis. FSH acts on sertoli cells, whereas LH acts on leydig cells of the testis and stimulates the process of spermatogenesis.

Question 8.
Define spermiogenesis and spermiation.
Answer:
Spermiogenesis : It is the process of transforming spermatids into matured spermatozoa or sperms.
Spermiation : It is ‘the process when mature spermatozoa are released from the sertoli cells into the lumen of seminiferous tubules.

Question 9.
Draw a labelled diagram of sperm.
Answer:

Question 10.
What are the major components of seminal plasma?
Answer:
Semen (produced in males) is composed of sperms and seminal plasma. The major components of the seminal plasma in the male reproductive system are mucus, spermatozoa, and various secretions of accessory glands. The seminal plasma is rich in fructose, calcium, ascorbic acid, and certain enzymes. It provides nourishment and protection to sperms.

Question 11.
What are the major functions of male accessory ducts and glands?
Answer:
The male accessory ducts are vasa efferentia, epididymis, vas deferens, and rete testis. They play an important role in the transport and temporary storage of sperms. On the contrary, male accessory glands are seminal vesicles, prostate glands, and bulbourethral glands. These glands secrete fluids that lubricate the reproductive system and sperms. The sperms get dispersed in the fluid which makes their transportation into the female body easier. The fluid is rich in fructose, ascorbic acid, and certain enzymes. They also provide nutrients and activate the sperm.

Question 12.
What is oogenesis? Give a brief account of oogenesis.
Answer:
Oogenesis is the process of the formation of a mature ovum from the oogonia in females. It takes place in the ovaries. During oogenesis, a diploid oogonium or egg mother cell increases in size and gets transformed into a diploid primary oocyte. This diploid primary oocyte undergoes first meiotic division i.e., meiosis I or reductional division to form two unequal haploid cells. The smaller cell is known as the first polar body, while the larger cell is known as the secondary oocyte. This secondary oocyte undergoes second meiotic division i.e., meiosis II or equational division and gives rise to a second polar body and an ovum. Hence, in the process of oogenesis, a diploid oogonium produces a single haploid ovum while two or three polar bodies are produced.

Question 13.
Draw a labelled diagram of a section through ovary.
Answer:

Question 14.
Draw a labelled diagram of a Graafian follicle.
Answer:

Question 15.
Name the functions of the following:
(a) Corpus luteum
(b) Endometrium
(c) Acrosome
(d) Sperm tail
(e) Fimbriae
Answer:
(a) Corpus Luteum: It is formed from the ruptured Graafian follicle. It secretes progesterone hormone during the luteal phase of the menstrual cycle. A high level of progesterone inhibits the secretions of FSH and LH, thereby preventing ovulation. It also allows the endometrium of the uterus to proliferate and to prepare itself for implantation.

(b) Endometrium: It is the innermost lining of the uterus. It is rich in glands and undergoes cyclic changes during various phases of the menstrual cycle to prepare itself for the implantation of the embryo.

(c) Acrosome: It is a cap-like structure present in the anterior part of the head of the sperm. It contains hyaluronidase enzyme, which hydrolyses the outer membrane of the egg, thereby helping the sperm to penetrate the egg at the time of fertilisation.

(d) Sperm Tail: It is the longest region of the sperm that facilitates the movement of the sperm inside the female reproductive tract.

(e) Fimbriae: They are finger-like projections at the ovarian end of the fallopian tube. They help in the collection of the ovum (after ovulation), which is facilitated by the beating of the cilia.

Question 16.
Identify True/False statements. Correct each false statement to make it true.
(a) Androgens are produced by Sertoli cells. (True/False)
(b) Spermatozoa get nutrition from Sertoli cells. (True/False)
(c) Leydig cells are found in ovary. (True/False)
(d) Leydig cells synthesise androgens. (True/False)
(e) Oogenesis takes place in corpus luteum. (True/False)
(f) Menstrual cycle ceases during pregnancy. (True/False)
(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience. (True/False)
Answer:
(a) Androgens are produced by Sertoli cells.
False Correct : Leydig cells.

(b) Spermatozoa get nutrition from Sertoli cells.
True

(c) Leydig cells are found in ovary.
False Correct : spermatogonia.

(d) Leydig cells synthesise androgens.
True

(e) Oogenesis takes place in corpus luteum.
False Correct : ovaries

(f) Menstrual cycle ceases during pregnancy.
True

(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience.
True

Question 17.
What is menstrual cycle? Which hormones regulate menstrual cycle? ,
Answer:
The menstrual cycle is a series of cyclic physiologic changes that take place inside the female reproductive tract in primates. The whole cycle takes around 28 days to complete. The end of the cycle is accompanied by the breakdown of uterine endothelium, which gets released in the form of blood and mucus through the vagina. This is known as menses.

The follicle stimulating hormone (FSH), luteinising hormone (LH), L estrogen, and progesterone are the various hormones that regulate the menstrual cycle. The level of FSH and LH secreted from the anterior pituitary gland increases during the follicular phase. FSH secreted under the influence of RH (releasing hormone) from the hypothalamus , stimulates the conversion of a primary follicle into a graafian follicle.

The level of LH increases gradually leading to the growth of follicle and f secretion of estrogen. Estrogen inhibits the secretion of FSH and stimulates the secretion of luteinising hormone. It also causes the thickening of the uterine endometrium. The increased level of LH causes the rupturing of the graafian follicle and release the ovum into the fallopian tube. The ruptured graafian follicle changes to corpus luteum and starts secreting progesterone hormone during the luteal phase.

Progesterone hormone helps in the maintenance and preparation of endometrium for the implantation of the embryo. High levels of progesterone hormone in the blood decrease the secretion of LH and FSH, therefore inhibiting further ovulation.

Question 18.
What is parturition? Which hormones are involved in induction of parturition?
Answer:
Parturition is the process of giving birth to a baby as the development of the foetus gets completed in the mother’s womb. The hormones involved in this process are oxytocin and relaxin. Oxytocin leads to the contraction of smooth muscles of myometrium of the uterus, which directs the full term foetus towards the birth canal. On the other hand, relaxin hormone causes relaxation of the pelvic ligaments and prepares the uterus for child birth.

Question 19.
In our society the women are often blamed for giving birth to [ daughters. Can you explain why this is not correct?
Answer:
All human beings have 23 pairs of chromosomes. Human males have 22 pairs of autosomes and contain one or two types of sex chromosome. They are either X or Y. On the contrary, human females have 22 pairs of autosomes and contain only the X sex chromosome. The sex of an individual is determined by the type of the male gamete (X or Y), which fuses with the X chromosome of the female. If the fertilising sperm is X, then the baby will be a girl and if it is Y, then the baby will be a boy.
Hence, it is incorrect to blame a woman for the gender of the child.

Question 20.
How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins bom were fraternal?
Answer:
An ovary releases an egg every month. When two babies are produced in succession, they are called twins. Generally, twins are produced from a single egg by the separation of early blastomeres resulting from the first zygotic cleavage. As a result, the young ones formed will have the same genetic make-up and are thus, called identical twins.

If the twins born are fraternal, then they would have developed from two separate eggs. This happens when two eggs’ (one from each ovary) are released at the same time and get fertilised by two separate sperms. Hence, the young ones developed will have separate genes and are therefore, called non-identical or fraternal twins.

Question 21.
How many eggs do you think were released by the ovary of a female dog which gave birth to 6 puppies?
Answer:
Dogs and rodents are polyovulatory species. In these species, more than one ovum is released from the ovary at the time of ovulation. Hence, six eggs were released by the ovary of a female dog to produce six puppies.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 15 Biodiversity and Conservation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation

PSEB 12th Class Biology Guide Biodiversity and Conservation Textbook Questions and Answers

Question 1.
Name the three important components of biodiversity.
Answer:
Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Three important components of biodiversity are as follows :

• Genetic diversity
• Species diversity
• Ecological diversity

Question 2.
How do ecologists estimate the total number of species present in the world?
Answer:
The diversity of living organisms present on the Earth is very vast. According to an estimate by researchers, it is about seven million. the total number of species present in the world is calculated by statistical comparison between species richness of a well-studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the Earth.

Question 3.
Give three hypotheses for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypotheses proposed by scientists for explaining species richness in the tropics.

1. Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.
2. Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.
3. Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Answer:
The slope of regression (z) has a great significance in order to find a species-area relationship. It has been found that in smaller areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic- group or the region. However, when a similar analysis is done in larger areas, then the slope of regression is much steeper.

Question 5.
What are the major causes of species losses in a geographical region?
Answer:
Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Biodiversity around the world is declining at a very fast pace. The following are the major causes for the loss of biodiversity around the world :

(i) Habitat Loss and Fragmentation: Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanisation. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

(ii) Over-exploitation: Due to over-hunting and over-exploitation of various plants and animals by humans, many species have become endangered or extinct (such as the tiger and the passenger pigeon).

(iii) Alien Species Invasions: Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

(iv) Co-extinction: In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 6.
How is biodiversity important for ecosystem functioning?
Answer:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. As we all know, various trophic levels are connected through food chains. If anyone organism or all organisms of any one trophic level is killed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food.

If all deer’s are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 7.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are tracts of forest which are regenerated around places of worship. Sacred groves are found in Rajasthan, Western Ghats of Karnataka, and Maharashtra, Meghalaya, and Madhya Pradesh. Sacred grows help in the protection of many rare, threatened, and endemic species of plants and animals found in an area. The process of deforestation is strictly prohibited in this region by tribals. Hence, the sacred grove biodiversity is a rich area.

Question 8.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Answer:
The biotic components of an ecosystem include the living organisms such as plants and animals. Plants play a very important role in controlling floods and soil erosion. The roots of plants hold the soil particles together, thereby preventing the top layer of the soil to get eroded by wind or running water. The roots also make the soil porous, thereby allowing groundwater infiltration and preventing floods. Hence, plants are able to prevent soil erosion and natural calamities Fucii as floods and droughts. They also increase the fertility of soil and biodiversity.

Question 9.
The species diversity of plants (22 percent) is much less than that of animals (72 percent). What could be the explanations to how animals achieved greater diversification?
Answer:
72 percent of species recorded on the Earth are animals and only 22 percent species are plants. There is quite a large difference in their percentage This is because animals have adapted themselves to ensure their survival in changing environments in comparison to plants. For example, insects and other animals have developed a complex nervous system to control and coordinate their body structure. Also, repeated body/ segments with paired appendages and external cuticles have made insects versatile and have given them the ability to survive in vain JUS habitats as compared to other life forms.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the Earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them.

Scientists have been able to eliminate smallpox virus from the world through the use of vaccinations. This shows than humans deliberately want to make these species extinct. Several other eradication programs such as polio and Hepatitis B vaccinations are aimed to eliminate these disease-causing microbes.