CBSE Class 12 Biology Question Paper (Outside Delhi 2019) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Outside Delhi 2019) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Outside Delhi 2019) with Solutions

Time Allowed: 3 Hours
Maximum Marks : 70

General Instructions:

1. There are total 27 questions and four sections in the question paper. All questions are compulsory.
2. Section A contains questions number 1 to 5, very short answer type questions of one mark each.
3. Section B contains questions number 6 to 12, short answer type-I questions of two marks each.
4. Section C contains questions number 13 to 24, short answer type-II questions of three marks each.
5. Section D contains question number 25 to 27, long answer type questions of five marks each.
6. There is no overall choice in the question paper, however, an internal choice is provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks. In these questions, an examinee is to attempt any one of the two given alternatives.
7. Wherever necessary, the diagram drawn should be neat and properly labelled.

Section – A

Question 1.
British geneticist R.C. Punnett developed a graphical representation of a genetic cross called “Punnett Square”. Mention the possible result this representation predicts of the genetic cross carried.
Answer:
The Punnett square helps in understanding the production of gametes by the parents, and formation of the zygotes in F ₁ and F ₂ generation. The Punnett square is a graphical representation for the calculation of probability of all the possible genotypes of offsprings in a genetic cross.

Note
The Punnett square was developed by a British geneticist, Reginald C. Punnett. Phenotypes refer to the observable characteristics of organisms that involve appearance, development and behaviour of organisms. Genotypes refer to the genetic constitution of organisms.

Question 2.
State the two principal outcomes of the experiments conducted by Louis Pasteur on origin of life.
Answer:
The two principle outcomes of the experiments conducted by Louis Pasteur on origin of life are as follows:

• He showed that in pre-sterilised flasks, life did not come from killed yeast.
• Whereas in another flask open to air, new living organisms arose from ‘killed yeast’.

Question 3.
Name the layer of the atmosphere that is associated with ‘good ozone’.
OR
Mention the term used to describe a population interaction between an orchid growing on a forest tree.
Answer:
Stratosphere is also called as ‘good’ ozone and is found in the upper part of the atmosphere.

Note
The stratosphere acts as a shield for absorbing ultraviolet radiation from the sun.

OR

The term used to describe a population interaction between an orchids growing on a forest tree is called commensalism. Commensalism is a biological interaction between two organisms in which one organism gets benefits from others while the other organism is neither benefited nor harmed.

An orchid growing on the branch of a forest tree is an epiphyte (plants that are growing on other plants). In this biological interaction, the orchid gets support, more sunlight for photosynthesis and nutrients from the mango tree. Whereas mango tree remains unaffected.

Question 4.
What are ‘floes’, formed during secondary treatment of sewage ?
OR
Write any two places where methanogens can be found.
Answer:
The secondary treatment plant allows the vigorous growth of useful microbes into floes. Floes are the masses of bacteria associated with fungal filaments to form mesh like structures. They are formed when the primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and then air is pumped into it.

OR

Methanogens are commonly found in the guts of animals, hydrothermal vents, wetlands and deep layers of marine sediments.

Question 5.
At what stage does the meiosis occur in an organism exhibiting haploidic life cycle and mention the fate of the products thus produced.
Answer:
The haploidic stage is a multi-cellular and a diploid stage. It is a single celled stage in which meiosis take place during zygote formation at the time of fertilization.

Section – B

Question 6.
You are conducting artificial hybridization on papaya and potato. Which one of them would require the step of emasculation and why ? However for both you will use the process of bagging. Justify giving one reason.
Answer:
Potato require emasculation because it has bisexual flower whereas papaya would require only bagging for artificial hybridisation as papaya has unisexual flowers. After pollinating with the desired pollen grain it is required to bag the plant in order to prevent the plant from pollination by undesirable pollen grains.

Note
The process of removal of anthers from the flower bud before the dehiscence of anther by a pair of forceps is called Emasculation. The emasculated flower is covered with a bag of suitable size which is made up of butter paper in order to prevent from contamination of stigma with unwanted pollen and this process is called bagging.

Question 7.
How would the gene flow or genetic drift affect the population in which either of them happen to take place ?
Answer:
If gene migration occurs multiple times then it is known as gene flow. The Hardy-Weinberg law states that the gene pool remains constant. Gene flow is the transfer of genetic information from one population to another. If the same change takes place by chance then it is called genetic drift. When migration of a section of population to another place and population occurs, gene frequencies change in the original as well as in the new population.

New genes or alleles are added to the new population and are lost from old population. There would be a gene flow if this gene migration happens multiple times. If the same change occurs by chance, then it is called genetic drift.

Note
Gene pool refers to the total number of genes and their alleles in a population.

Question 8.
Differentiate between the roles of B-lymphocytes and T-lymphocytes in generating immune responses.
OR
Principle of vaccination is based on the property of “memory” of the immune system.
Taking one suitable example, justify the statement.
Answer:

B-lymphocytes	T-lymphocytes
(i) The B-lymphocytes produces an army of proteins in response to pathogen into the blood and these proteins are called antibodies.	(i) The T-lymphocytes do not secrete any antibodies but help B-cells to produce antibodies.
(ii) The response produced by B-lymphocytes is also called as humoral immune response.	(ii) The response produced by T-lymphocytes is also called as Cell-mediated immune response.
(iii) B-cells mature in the bone marrow	(iii) T-cells mature in the Thymus.

Note
Both B-lymphocytes and T-lymphocytes are formed in the bone marrow.

OR

In case of polio vaccination, an antigenic protein preparation of pathogen is introduced into the body which is inactivated or weakened form. The antibodies produced in the body against polio antigens would neutralise the pathogenic agents which responsible for causing polio. The vaccines also generate memory B-and T-cells which recognises the pathogen on subsequent exposure and encounter the pathogen by the production of antibodies.

Question 9.
Explain the relevance of “Totipotency” and “Somaclones” in raising healthy banana plants from virus infected banana plants.
Answer:
Totipotency is the capacity to generate a whole plant from any cell or explants whereas somaclones are formed by micropropagation and the plants produced are genetically identical to their original plant from which they were grown. In the virus infected banana plant, the meristem (apical or axillary) is free from virus and one can remove the meristem and grow it In-vitro to obtain virus free banana plant.

Question 10.
How is a continuous culture system maintained in bioreactors and why ?
Answer:
A continuous culture system is maintained in a bioreactor by continuously and regularly feeding with culture medium steadily and by providing optimum growth conditions such as pH, temperature, vitamins, salts, oxygen. The used medium is drained out from one side of the bioreactor and the fresh medium is added from the other side. Continuous culture system is maintained in bioreactors in order to maintain the cells in their physiologically most active log/ exponentially phase. This type of culturing method produces a large biomass results in higher yield of the desired protein.

Question 11.
List any four ways by which GMO’s have been useful for enhanced crop output.
Answer:
The four ways by which GMO’s have been useful for enhanced crop output are as follows:

• GMOs made crop more tolerant to abiotic stress such as cold, drought, salt and heat.
• It reduces dependency on chemical pesticides such as pest-resistant crops.
• It helped in the reduction post harvest losses.
• GMOs increases the efficiency of mineral usage by plants and this helps in the prevention of early exhaustion of fertility of soil.
• GMOs also enhance nutritional value of food such as vitamin ‘A’ enriched rice.

Question 12.
Mention four significant services that a healthy forest ecosystem provide.
OR
Substantiate with the help of one example that in an ecosystem mutualists
(i) tend to co-evolve and
(ii) are also one of the major causes of biodiversity loss.
Answer:
Ecosystems are communities formed by the interaction between living (plants, animals, microbes) and non-living components (air, water, mineral soil) of the environment. The benefits are known as ecosystem services. A healthy forest ecosystem contains all the biotic and abiotic components of the environment.

The four significant services provided by healthy forest ecosystems are:

• Ecological functions such as carbon storage, nutrient cycling, water and air purification, as well as the maintenance of wildlife habitat.
• Maintenance of biodiversity.
• Provides goods such as timber, food, fuel and bioproducts.
• Social and cultural benefits such as recreation, traditional resource uses and spirituality. (2 Marks)

OR

(i) In nature, mutualists often co-evolve such as in Mediterranean orchid Ophrys. Ophrys employs sexual deceit to get pollinated by a species of bee. One petal of flower resemble to female bee. If female bee changes its colour pattern ever slightly the success of pollination will be reduced unless orchid flower co evolves to maintain resemblance with female bee.

(ii) Co-extinction is one of the ‘Evil Quartet’ in which organisms with obligatory relationship like plant pollinator mutualism will result in extinction of one partner if other is eliminated in nature.

Section – C

Question 13.
Pollen banks are playing a very important role in promo! in r plant breeding programme the world over. How are pollens preserved in the pollen banks ? Explain. How are such banks benefiting our farmer ? Write any two ways.
Answer:
Pollen grains are preserved in a pollen bank in liquid nitrogen at temperature -196 C and this process are called cryopreservation. Pollen grains in plants are used for transferring haploid male genetic material from the anther to the stigma of another flower in case of cross-pollination. The pollen grains are preserved to introduce the desired characters into hybrid varieties of plant.

Pollen banks are used to store pollen grains for a short as well as a very long period of time in viable conditions. These pollen grains can be used in various crop breeding programs, biochemical and physicochemical studies. The very important application of pollen bank is to preserve the biodiversity in the form of preservation of the value of genetic resources.

The pollen grains can be stored by two methods such as:
The pollen grains can be stored by two methods:

(i) Short term storage of pollen grain: It involves the storage of pollen grains under low temperature and low humidity, and storage in an organic solvent (the simple method by which pollen grains are dried over silica and stored in organic solvents and maintained in the refrigerator or deep freezer)

(ii) Long term storage of pollen: It involves the storage of pollen grains in freeze or vacuum-dried conditions [in which pollen grains are stored at subzero temperature (-60 degrees C to -80 degrees C)] and cryopreservation (pollen grains are dried by using a ‘Pollen Drier’ containing air of 20 degrees C and 20—40 % humidity to bring their water content below a threshold level and stored in liquid nitrogen (-196 degrees C)). This method is very effective for storage of pollen grains of a number of species, including cereals, even for over 10 years).

The pollens preserved in pollen banks are benefitting farmers in the following two ways:

• By using these pollens the plants that are facing extinction can be reproduced.
• Pollen grains can be later used in plant breeding programs.
• By conserving agricultural biodiversity.

Question 14.
Draw a labelled diagram to show interrelationship of four accessory ducts in a human male reproductive system.
OR
Draw a sectional view of the human ovary showing the different stages of developing follicles, corpus luteum and ovulation.
Answer:
The internal organs of the male reproductive system are called accessory organs. They include the vas deferens, seminal, vesicles, prostate gland, and bulbourethral glands.
Diagrammatic representation of interrelationship of four accessory ducts in a human male reproductive system:

OR

Diagrammatic representation of human ovary showing the different stages of developing follicles, corpus luteum and ovulation:

Question 15.
Compare in any three ways the chromosomal theory of inheritance as proposed by Sutton and Bovery with that of experimental results on pea plant presented by Mendel.
OR
(a) Explain linkage and recombination as put forth by T.H. Morgan based on his observations with Drosophila melanogaster crossing experiment.
(b) Write the basis on which Alfred Sturtevant explained gene mapping.
Answer:
The chromosomal theory of inheritance given by Sutton and Boveri and experimental results presented by Mendel can be compared in the following ways:

1. In a diploid organism, the factors (genes) and chromosomes occur in pairs.
2. Both chromosomes as well as genes segregate at the time of gamete formation such that only one of each pair is transmitted to a gamete. So, a gamete contains only one chromosome of a type and only one of the two alleles of a trait.
3. Each pair of chromosome and gene segregates independent of another pair.
4. The paired condition of both chromosomes as well as Mendelian factors is restored during fertilisation.

Note
The chromosomal theory of inheritance was proposed by Sutton and Boveri states that the genes are located at specific loci on the chromosome that segregate and assort independently during the process of meiosis and then recombine at the time of fertilization in the zygote.

OR

(a) T.H. Morgan studied X-linked genes in Drosophila and observed that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations is much higher than the non-parental type. He attributed this due to the physical association or linkage of the two genes on a chromosome and coined the term linkage and the term recombination describes the generation of non-parental gene combinations.

(b) Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of distance between them and mapped their position on the chromosome.

Note
Genes present on the same chromosome are separated during meiosis and the new combination of the gene could be formed and this phenomenon is called recombination of the gene. They are attached to one another like beads on a string in a linear organisation. The distance between the linked genes indicates the strength of linkage.

Question 16.
Explain the mechanism of DNA replication with the help of a replication fork. What role does the enzyme DNA-ligase play in a DNA replication fork ?
OR
Construct and label a transcription unit from which the RNA segment given below has been transcribed. Write the complete name of the enzyme that transcribed this RNA.

Answer:
DNA replication is semiconservative. The replication occurs at origin of replication within a small opening of DNA helix called as ‘replication fork’. It requires helicase, topoisomerase, single strand binding proteins, DNA-dependent DNA polymerase for polymerisation of deoxyribonucleotides, deoxyribonucleotides as substrate and source of energy for polymerisation reaction, primase for the synthesis of RNA primer, and DNA ligase to join DNA fragments.

The various steps of DNA replication are as follows:

(i) Helicase unwinds the double helix by breaking the hydrogen bonds between complementary base pairs, while single strand binding proteins helps to stabilize the single strands prevent them from rejoining. The enzyme topoisomerase enzyme releases tension generated due to unwinding and super coiling at the end of DNA opposite to the replication fork.

(ii) The DNA-dependent DNA polymerases cannot initiate the process of replication on their own. So, primase enzyme forms short sequences of RN A called primers that provide a starting point for elongation.

(iii) DNA polymerase catalyses polymerisation only in one direction i.e. 5′ → 3′ so on one strand (the template with polarity 3′ 5′) the replication is continuous while on other (the template with polarity 5′ 3′), it is discontinuous. These discontinuously synthesised fragments are called ‘Okazaki fragments’. Okazaki fragments are later joined by enzyme DNA ligase.

Diagrammatic representation of replication fork:

Note
DNA (Deoxyribonucleic acid) is a biological process the formation of two identical copies of DNA from parent DNA.

OR

The process of copying genetic information from one strand of DNA into RNA is called transcription. The enzyme catalyse the polymensation only in one direction that is 5’→3’, the strand that has polarity 3’→5’ acts as a template and is called template strand. Whereas the other strand is called coding strand and has polarity 3’→5.

The promoter is located towards 5’-end (upstream) of the structural gene. Promoter provides binding site for RNA polymerase. Terminator is located towards 3 ‘-end (downstream) of the coding strand and terminates the process of transcription.

For the given RNA, the transcription unit will be:

Diagrammatic representation of transcription unit:

DNA dependent RNA polymerase enzyme is required for the process of transcription. (3 Marks)

Note
The mRNA molecule have uracil (U) in place of thymine (T).

Question 17.
(a) Write two differences between Homo erectus and Homo habilis.
(b) Rearrange the following from early to late geologic periods:
Carboniferous, Silurian, Jurassic.
Answer:
Difference between Homo erectus and Homo habilis.

S. No.	Character	Homo erectus	Homo habilis
(i)	Brain capacity	900 cc	650-800 cc
(ii)	Eating habit	They probably ate meat	They probably did not eat meat

(b) The correct sequence from early to late geological period is: Silurian period Carboniferous period and Jurassic period.

Question 18.
Name the group of bacteria involved in setting milk into curd. Explain the process they carry in doing so. Write another beneficial role of such bacteria.
Answer:
Lactic acid bacteria (LAB) are involved in setting milk into the curd.
Process of curd formation from milk: Milk is converted into curd by the process of fermentation. Milk consists of globular proteins called casein. The curd forms because of the chemical reaction between the lactic acid bacteria and casein. During fermentation, the bacteria use enzymes to produce energy (ATP) from lactose.

Under suitable condition, these bacteria multiply and produce acids which coagulate and partially digest the milk proteins and change the milk into curd. Lactobacilli bacteria present in curd multiply in milk and convert the lactose sugar into lactic acid.

Beneficial role played by lactic acid bacteria are:

1. they are used in food fermentation.
2. also found in the human stomach where these bacteria prevent the growth of certain disease-causing microbes.
3. improve lactose digestion and increases the content of vitamin B]7 in curd.
4. play a role in preventing and treating diarrhoea and act on the immune system, helping the body to resist and fight infection.

Question 19.
Bee keeping practice is a good income generating industry. Write the different points to be kept in mind for successful bee keeping. Write the scientific name of the most common Indian species used for the purpose.
Answer:
Apiculture (also called bee keeping) is a process of rearing and management of honey bees for commercial production of bee wax, honey and royal jelly. The rearing of bees can be done in the area with plantation such as orchards, farms, etc., where sufficient pollen and nectar are available. Such places are called apiary sites.

The following points should be kept in mind for beekeeping:

(i) Since pollen and nectar are available only in the flowering season so seasonal management is very important. This season is known as honey flow season. In this season the pollination efficiency as well as honey production, both get increased. The nectar collected by the bees is passed through the pre-oral cavity and tongue to ripen it. Honey is then deposited in the cells. Hence, more space for storage of honey should be provided during this season.

(ii) During the summer season, adequate shade and water should be provided by artificial constructions or by keeping the beehives under the trees. Proper ventilation, sugar syrup and pollen supplements should also be provided as a food source.

(iii) During winter season, disease-free colonies should be maintained. The hives should be provided with new queens.

(iv) During the rainy season, dampness should be avoided and since bees cannot go out to collect pollen and nectar, sugar syrup should be provided as food.

(v) One should wear protective clothing and gloves while handling the honey bees and collecting honey. Smokers are also used by the bee-keepers to keep them calm while handling them.

Apis irtdica (Indian bee) is the most common species of bee reared in India. The other important species that can also be reared are Apis mellifera (Italian bee), Apis dorsata (rock bee) and Apis jiorea (little bee).

Question 20.
(a) Match the microbes listed under Column-A with the products mentioned under Column-B.

(b) Why does ‘Swiss Cheese’ develop large holes ?
Answer:
(i)

Column A	Column B	Correct Options
(H) Penicillium notatum	(i) Statin	(H) – (iii)
(I) Trichoderma polysporum	(ii) ethanol	(I) – (>v)
(J) Monascus purpurae	(iii) antibiotic	(J)-(0
(K) Saccharomyces cerevisae	(iv) cyclosporin-A	(K) – (ii)

(ii) The development of large holes in ‘Swiss cheese’ is because of the production of large amount of carbon dioxide by the bacterium Propionibacterium-sharmanii.

Question 21.
Describe the formation of recombinant DNA by the action of EcoRI.
OR
Describe the process of amplification of “gene of interest” using PCR technique.
Answer:
The various steps involved in the formation of recombinant DNA by the action of EcoRI are as follows:

• Both vector DNA and foreign DNA is cut with EcoRI.
• Both DNAs will possess smaller fragments with overhanging stretches called sticky ends on each strand.
• After digestion, both vector DNA and foreign DNA are mixed and allowed to join together with the help of DNA ligase enzyme. This results in the formation of recombinant DNA molecules.

Diagrammatic representation of steps involved in the formation of recombinant DNA by action of restriction endonuclease enzyme-EcoRI:

OR

With the help of recombinant DNA technology called Polymerase Chain Reaction technique (PCR) multiple copies of the ‘gene of interest’ are obtained in Vitro. A single PCR amplification cycle involves three steps which are as follows:
(a) Denaturation: This is the first step of PCR, in which the target DNA is heated at high temperature such as 94-96°C. It facilitates the separation of two strands of DNA. Each separated strand of DNA acts as a template for synthesis of DNA.

(b) Annealing: This is the second step of PCR, in which two oligonucleotide primers are used to hybridize each single-stranded template DNA. The sequence of primers is complementary to 3′ end of the template DNA strand.
This step of PCR occurs at low temperature 4O-60°C than denaturation. The annealing temperature depends upon the length and sequence of the primers.

(c) Extension: This is the third and the last step of PCR, in which enzyme TaqDNA polymerase synthesizes the DNA between the primers. This step also requires dNTPs and Mg2++. The optimum temperature for an extension is 72°C.
Diagrammatic representation of steps involved in PCR:

Note
Enzyme used in PCR is a DNA polymerase such as Taq polymerase. This enzyme is stable at high temperature as it is isolated from thermostable bacteria Thermus aquaticus.

Question 22.
Two children, A and B aged 4 and 5 years respectively visited a hospital with a similar genetic disorder. The girl A was provided enzyme-replacement therapy and was advised to revisit periodically for further treatment. The girl, B was, however, given a therapy that did not require revisit for further treatment.
(a) Name the ailments the two girls were suffering from ?
(b) Why did the treatment provided to girl A required repeated visits ?
(c) How was the girl B cured permanently ?
Answer:
(a) The two girls are suffering from a genetic disorder resulting in adenosine deaminase (ADA) deficiency due to deletion of its gene that codes for adenosine deaminase enzyme

(b) Girl A was treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. This technique is not completely curative as it requires repeated infusion.

(c) Girl B was treated using gene therapy where the gene isolate from marrow cells producing ADA was introduced into cells at an early embryonic stage in order to provide permanent cure.

Note
ADA deficiency is caused because of the deletion of the gene that codes for adenosine deaminase enzyme. Severe combined Immunodeficiency disorder (SCID) is caused because of the defect in gene which codes for adenosine deaminase enzyme.

Question 23.
List six advantages of “ex-situ” approach to conservation of biodiversity.
Answer:
Ex-situ (‘off site’) conservation is a set of”Conservation techniques -involving the transfer of a target species away from its native habitat to a place of safety, such as a zoological garden, botanical garden or seed bank. Ex-situ techniques include: seed storage, captive breeding, slow- growth storage, DNA storage.

Question 24.
While on a visit to a pond in the city-neighbourhood, the visitors were delighted to find large expanse of water covered with colourful algal mass.
(a) As a student of biology, do you agree with their delight ? Give reasons in support of your answer.
(b) Explain the cause of such algal growth.
Answer:
(a) As a student of biology, I disagree to the delight of visitors to find large expense of water covered with colorful algal mass because algal mass is the cause of deterioration of water quality (decline in dissolved oxygen) leading to death of fish and other aquatic organisms. Some bloom-forming algae are extremely also toxic to human beings and animals.

(b) The cause of algal growth is because of excessive growth of planktonic (free-floating) algae, called an algal bloom due to the addition of large amounts of nutrients (nitrogen and phosphorous) in water bodies through human activities such as agricultural runoff, excessive use of fertilisers, industrial effluents, sewage treatment and soil erosion.

Section – D

Question 25.
(a) Explain one application of each one of the following :
(A) Amniocentesis
(B) Lactational amenorrhea
(C) ZIFT
(b) Prepare a poster for the school programme depicting the objectives of : “Reproductive and Child Health Care Programme”.

OR

(a) Explain any two ways by which apomictic seed can develop.
(b) List one advantage and one disadvantage of a apomictic crop.
(c) Why do farmers find production of hybrid seeds costly ?
Answer:
(a)
(A) Amniocentesis: This is also called amniotic fluid test as it is used for the detection of chromosomal abnormalities in the developing foetus. This test is done after the 16 weeks of pregnancy. In this, a small quantity of sample is taken from the amniotic sac and DNA is examined for genetic abnormalities.

(B) Lactational amenorrhea: It is a type of natural contraceptive method is characterized by absence of menstruation. In this method, the ovulation and menstrual cycle do not occur during the period of intense lactation followed by parturition.

(C) ZIFT (Zygote Intra-fallopian Transfer): It is an assisted reproductive technology. In this method, ova is collected from the female (wife/ donor) and sperms from the male (husband/ donor). Both the ova and sperms are induced to form zygote in the laboratory. The zygote is allowed to form 8 blastomeres stage and is transferred into fallopian tube of female.

(b) Poster presentation on “Reproductive and Child Health Care Programme”

Note
The goal of “Reproductive and child health Care programme ‘ is to provide facilities and support for building up a reproductively healthy society

OR

(a) The different ways by which apomictic seeds are developed involved:

1. Formation of diploid eg aš embryo without undergoing reduction division or fertilization fòr example: Asteraceae.
2. In several citrus fruits and mango varieties, nuclear cells surrounding the embryo sac start dividing and project into the embryo sac. It gets further developed into the embryos. (2 Marks)

(b) Advantages of apornictic crop: Apornixis reduces the cost of the hybrid production so that new varieties of seeds are produced more quickly and at a cheaper rate.

Disadvantages of apomictic crop:
Apomictic seeds reduce the genetic diversity because of a lack of variations.

(c) Hybrid seeds are produced by cross-pollination of plants and in order to produce desirable hybrid character, plant breeders and scientists are trying thousands of combination to produce such hybrids. So hybrid seeds are produced every years and it requires a lots of scientific research for the production of such hybrid seeds. It is expensive and hence the cost of hybrid becomes too expensive for the farmers.

Question 26.
Differentiate between incomplete dominance and co-dominance. Substantiate your answer with one example of each.
OR
(a) Write the contributions of the following scientists in deciphering the genetic code. Georce Gamow ; Hargobind Khorana ; Marshall Nirenberg ; Severo Ochoa
(b) State the importance of a Genetic code in protein biosynthesis.
Answer:
The difference between incomplete and co-dominance are as follows:

Incomplete dominance	Codominance
(i) It is a phenomena in which none of the alleles of a gene is dominant over each other and a new phenotype is formed. The new phenotype is intermediate between the independent expression of two alleles.	(i) Codominance is a phenomena in which both the alleles of a gene expresses themselves independently in a heterozygote.
(ii) New phenotypes are always formed as a result of incomplete dominance.	(ii) In case of codominance, no new phenotype is formed.
(iii) Examples are snapdragon and mirabilis jalapa	(iii) Examples are Roan character in cattles and blood grouping in humans.

OR

(a) George gamow: He worked on radioactive decay that affects the nucleus of atomon a stellar nucleosynthesis and star formation.

Hargobind Khorana: He synthesized copolymers of nucleotides such as UGUGUGUG. They observed that they have stimulated tite formation of polypeptides having alternatively similar amino acid sequence such as cysteinc-valine-eysteine.

Marshall Nirenberg: He discovered the first “triplet”-a sequence of three bases of DNA that codes for one of the twenty amino acids which serves as the building blocks of the proteins.

Severo Ochoa: He investigated how DNA and RNA are formed as well as which enzymes control this process. He discovered polynucleotide phosphorylase enzyme which helps in polymerization of ribonucleotides in template independent manner.

(b) Genetic code is the biochemical basis of heredity consisting of codons in DNA and RNA that determine the specific amino acid sequence in proteins and appear to be uniform for nearly all known forms of life.

Importance of genetic code in protein biosynthesis:

• Genetic code is important because it provides information encoded in genetic material (DNA or RNA sequences) to be translated into proteins (amino acid sequences) by living cells.
• All organisms on Earth utilise proteins in chemical reactions to facilitate physiological functions.
• Differences in genetic code result in coding of different proteins, and it leads to cause genetic variations in organisms.

Question 27.
(a) What is “population” according to you as a biology student ?
(b) “The size of a population for any species is not a static parameter.” Justify the statement with specific reference to fluctuations in the population density of a region in a given period of time.
OR
(a) What is hydrarch succession ?
(b) Compare the pioneer species and climax communities of hydrarch and xerarch succession respectively.
(c) List the factors upon which the type of invading pioneer species depend in secondary hydrarch succession. Why is the rate of this succession faster than that of primary succession ?
Answer:
(a) Population is the number of people or animals of the same group or species, living in a particular geographical area, and have the capability of interbreeding.

(b) The size of a population for any species is not a static parameter because it keeps changing with time, depending on various factors including food availability, predation, pressure and adverse weather. The density of a population in a given area during a period fluctuates due to changes in four basic processes two of which are natality and immigrationcontribute to an increase in population density and two are mortality and emigration to a decrease.

1. Natality: Number of births during a given period in the population that are added to the initial density.
2. Mortality: Number of deaths in the population during a given period.
3. Immigration: Number of individuals of the same species that have come into the habitat from elsewhere during given time period.
4. Emigration: Number of individuals of the population who had left the habitat and gone elsewhere during given time period.

OR

(a) Hydrarch succession occurs in the wetter areas and the successional series progress from hydric to the mesic conditions. (1 Mark)
(b) Comparison of pioneer species and climax communities of hydrarch and xerarch succession:

Characteristics	Hydrarch succession	Xerarch succession
Pioneer species	Pioneer communities are small phytoplanktons for example: Diatoms.	Pioneer communities are usually lichens. For example: Lichens secretes acids to dissolve rocks and also helps in weathering as well as in soil formation.
Climax communities	Climax community would be a forest, the water body is converted into land with time. It also results in mesic conditions.	The climax community would be a forest. Xerarch succession results in mesic conditions.

Note
The species that invade a hare rocks are called pioneer species and Climax community involves the changes that lead finally to a community which is in near equilibrium with the environment.

(c) Factors on which invading pioneer species depend in a secondary hydrarch succession are as follows:
Condition of soil, presence of seeds or propagules in the environment, and availability of water

Secondary succession is the ecological succession that occurs after the initial succession has been disrupted and some plants and animals still exist. Rate of secondary hydrarch succession is faster than that of primary succession because of the following reasons:

1. Some soil and nutrients were already present due to which climax reaches more quickly. So there is no need for pioneer species to colonize the land nor does decomposition need to occur to create a layer of topsoil.
2. Seeds, roots and underground vegetative organs of plants may still survive in the soil.