CBSE previous Year Solved Papers Class 12 Biology Outside Delhi 2013
Time allowed : 3 hours Maximum Marks: 70
General Instructions :
- There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
- Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
- Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
- Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
- Section D contains question number 23, Value Based Question of four marks.
- Section E contains question number 24 to 26, Long Answer type questions of five marks each.
- There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.
Question.1. Name an organism where cell division in itself is mode of reproduction.
Answer : The division of cell in itself is a mode of reproduction found in amoeba and paramoecium.
Question.2. When does a human body elicit a anamnestic response ?
Answer : When our body attacked by pathogens for the second time the memory cells which were formed during the first attack produces a highly intensified secondary or anamnestic response.
Question.3. Name any two disease the ‘Himgiri’ variety of wheat is resistant to.
Answer : The ‘Himgiri’ variety of wheat is resistant to leaf and stripe rust and hill bunt disease.
Question.4. State the role oi transposons in silencing of mRNA in eukaryotic cells.
Answer : Role of transposons : Silencing of a gene is done in order to prevent translation of mRNA, where transposons act as a complementary RNA that is used to stop translation.
Question.5. Why are green algae not likely to be found in the deepest strata of the ocean ?
Answer : Green algae not likely to be found in the deepest strata of the ocean because deep inside the sea presence of sufficient light for photosynthesis and brackish water are not available so green algae are not present at this level, instead algae inhabits littoral zone of water.
Question.6. State what does‘standing crop’of a trophic level represent.
Answer : ‘Standing crop’ of a trophic level represent certain j mass of a living material at a particular time.
Question.7. Why is the use of unleaded petrol recommended for motor vehicles equipped with catalytic converters ?
Answer : The use of unleaded petrol- recommended for 1 motor vehicles equipped with catalytic converters because lead in petrol inactivates the catalysts which convert harmful pollutants (CO, unburnt hydrocarbons, nitric oxide) to lesser ,harmful pollutants (C02, HzO, N2).
Question.8. Name the type of biodiversity represented by the following :
(i) 1000 varieties of mangoes in India
(ii) Variations in terms of potency and concentration of reserpine in Rauwolfia vomitoria growing in different
regions of Himalayas.
Answer : (i) Genetic Biodiversity : (ii) Genetic Biodiversity
SECTION – B
Question.9. In angiosperm, zygote is diploid while primary endosperm cell is triploid. Explain.
Answer : (i) Fertilisation of haploid egg cell by one haploid male gamete to form diploid zygote also called syngamy.
(iii) Fertilisation of two (diploid) polar nuclei by the other haploid male gamete to form triploid primary endosperm nucleus also called triple fusion.
Question.10. A cross between red flower bearing plant and a white flower bearing plant of Antirrhinum produced all plants having pink flowers. Work out a cross to explain how this is possible.
Phenotypic ratio: 1:2:1
Genotypic ratio: 1:2:1
R(Red) factor is not completely dominant over r(white) factor is incomplete dominance.
Question.11. List the two main propositions of Oparin and Haldane.
Answer : The two main propositions of Oparin and Haldane are : (i) The first form of life could have come from preexisting non-living organic molecules (e.g., RNA, protein etc.) i.e., first abiogenesis and biogenesis later.
(ii) The first form of life was preceded by chemical evolution i. e., formation of organic molecules from inorganic molecules like CH4, NH3 etc.
Question.12. Write the events that take place when a vaccine for any disease is introduced into the human body.
Answer : Vaccines are a non-virulent form of pathogens, when administered into the body, the body start making antibodies against the antigens present in the vaccine. , The vaccine also generate memory – B and T – cells, that recognize the pathogen quickly on. subsequent second exposure and wipe out the invaders with a massive production of antibodies.
Why is a person with cuts and bruises following an accident administered tetanus antitoxin ? Give reasons. Answer : Tetanus antitoxins neutralize and provide passive immunity to the bacterial toxin. The antitoxin contain antibody against pathogen it attach and inactivate pathogen.
Question.13. Name the bacterium responsible for the large holes seen in ‘Swiss Cheese’. What are these holes due to ?
Answer : Propionibacterium sharmanii is responsible for the large holes seen in Swiss Cheese. The large holes are due to production of a large amount of C02 by a bacterium.
Question.14. Name the source of the DNA polymerase used in PCR technique. Mention why it is used.
Answer : Thermus aquaticus because it is heat stable DNA polymerase. Polymerase chain reaction (PCR) is a method in which the desired gene is synthesised in vitro in following steps:
(a) Denaturation : The double-stranded DNA is denatured ’ by applying high temperature of 95°C for 15 seconds. Each separated single stranded now acts as template for DNA synthesis.
(b) Annealing : Two sets of primers are added which anneal to the 3′ end of each separated strand. Primers act as initiators of replication.
(c) Extension : DNA polymerase extends the primers by adding nucleotides complementary to the template provided. in the reaction. A thermostable DNA polymerase (Taq polymerase) is used in the reaction which can tolerate the high temperature of the reaction. All these steps are repeated many times to obtain several copies of desired DNA.
Question.15. Write any four ways used to introduce a desired DNA segment into bacterial cell in recombinant technology experiment.
Answer : The four ways used to introduce a desired DNA segment into bacterial cell in recombinant. technology experiment as following:
(i) Chemical Method : Poration by divalent cation such as calcium.
(ii) Micro injection : Recombinant DNA is directly injected into the nucleus of an animal cell.
(iii) Biolistic or gene gun: Plant cells are bombarded with hi gh velocity micro-particles of gold or tungsten coated with DNA.
(iv) Disarmed pathogen vectors : when allowed to infect the cell, transfer the recombinant DNA into the host.
Question.16. Why is proinsulin so called ? How is insulin different from it?
Answer : Proinsulin is a protein molecule and like a pro¬enzyme. It contain an extra strech of C peptide so it need to be processed to become fully mature and functional hormone like insulin, it is a mature hormone and is produced by the beta cells. Proinsulin is different from insulin because it serves as a precursor hormone to insulin.
Question.17. Where would you expect more species biodiversity in tropics or in polar region ? Give reasons in support of your answer.
Answer : High species or biodiversity lies in tropical areas because tropics are :
(i) Undistributed habitats since millions of year in comparison to temperate and polar region which face frequent glaciation. It favours speciation, as speciation is product of time.
(ii) Less seasonal variation than polar areas.
(iii) High availability of solar radiatipns than polar area, which harbours more plant species.
Question.18. “It is possible that a species may occupy more than one trophic level in the same ecosystem at the same time.” Explain with the help of one example.
Answer : Yes, as the trophic level of a species represents the functional role of organism in energy flow which is determined by the food intake. The availability of food is depends on what the organism want to eat, so have more than one trophic level at a time.
Ex. Sparrow is a Primary consumer, when eating seeds where as Secondary consumer, when eating insects.
Question.19. Explain the steps in the formation of an ovum from an oogonium in humans.
Answer : The steps in the formation of an ovum from an oogonium in humans involves in oogenesis process. It can be divided into three stages :
(a) Multiplication phase (b) Growth phase
(c) Maturation phase
(a) Multiplication phase : In this stage primordial germ cells or ovum mother cells are repeatedly divided by mitosis to form large number of diploid oogonia. This process completes in embryo stage of female in most higher animals.
(b) Growth phase : In this process oogonia grow in size and form primary oocytes. The growth phase is the longest phase oogenesis (except humans). During growth phase size of egg increases many times.
(c) Maturation phase : Oogenesis takes place in the ovaries. In contrast to males the initial steps in egg production occur prior to birth. By the time the foetus is 25 weeks old, all the oogonia that she will ever produce, are already formed by mitosis. Hundreds of these diploid cells develop into primary oocytes, begin the first steps of the first meiotic division, proceed up to diakinesis, and then stop any further development. The oocytes grows much larger and complete the meiosis I, forming a large secondary oocyte and a small polar body that receives very little amount of cytoplasm but one full set of chromosomes.
Suggest and explain any three Assisted Reproductive Technologies (ART) to an infertile couple.
Answer : Three assisted reproductive technologies (ART) to an infertile couple are :
(i) In vitro fertilisation (IVF-fertilisation outside the body in almost similar conditions as that in the body) followed by embryo transfer (ET) is one of such methods. In this method, popularly known as,, test tube baby programme, ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under simulated conditions-in the laboratory. The zygote or early embryos (with upto 8. blastomeres) could then be transferred into the fallopian tube (ZIFT- zygote intra fallopian transfer) and embryos with more than 8 blastomeres, into the uterus (IUT —intra uterine transfer), to complete its further development.
(ii) Embryos formed by in-vivo fertilization (fusion of gametes within the female) also could be used for such transfer to assist those females who cannot conceive. Transfer of an ovum collected from a donor into the fallopian tube (GIFT – gamete infra fallopian transfer) of another female who cannot produce one, but can provide suitable environment for fertilisation and further development is another method attempted.
(iii) Intra cytoplasmic sperm injection (ICSI) is another ‘ specialised procedure to form an embryo in the laboratory in which a sperm is directly injected into the ovum.
Infertility cases either due to inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates, could be corrected by Artificial Insemination (AI) technique. In this technique, the semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus (IUI-Intrauterine Insemination) of the female.
Question.20. Why a human females are rarely haemophilic ? Explain. How do haemophilic patients suffer ?
Answer : Haemophilia is sex linked recessive disease; it is transmitted from unaffected female carrier to a male child with haemophilia. Y has no allele for this. If male inherits Xh from the mother, he will be haemophilic (with the genotype XhY). If female inherits XhXh, one from the carrier tnother and one from her haemophilic father, then she can be haemophilic. Simple cut will results to increased bleeding time in haemophilie patients.
Question.21. In a maternity clinic, for some reasons the authorities are not able to hand over the two new-borns to their respective real parents. Name and describe the technique that you would suggest to sort out the matter.
Answer: DNA Fingerprinting or DNA test is the technique that suggested to describe the parental identification of these two new born babies in a maternity clinic.
The procedure of finger printing is as follows :
(i) Isolation of DNA,
(ii) Digestion of DNA by restriction endonucleases,
(iii) Separation of DNA fragments by electrophoresis,
(iv) Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,
(v) Hybridisation using labelled VNTR probe, and
(vi) Detection of hybridised DNA fragments by autoradio-graphy. Half of the band of child will resemble to father and half to mother.
Question.22. Explain the increase in the number of melanic (dark winged) moths in the urban areas of post – industrialisation period in England.
Answer: In England, before industrial revolution the environ-ment was unpolluted. The white-winged moths were more and lichens on the barks of trees were pale. The white-winged moths could easily camouflage, while the dark winged were spotted out by the birds for food. Hence, they could not * survive. After industrial revolution the lichens became dark (due to soot-deposit). This favoured the dark-winged moths while the white-winged were picked by birds. The population of the former which was naturally selected increased.
Question.23. Describe how biogas is generated from activated sludge. List the components of biogas.
Answer : Biogas .can be produced by anaerobic digestion or fermentation of biodegradable materials. Bio wastes are collected and slurry of dung is mixed, a floating cover is placed over the slurry. The slurry having gas outlet is placed which keeps on rising as the gas is produced in the tank due to microbial activity of methanogens like Methanobacterium. Anaerobic fermentation of waste biomass can be visualized in three stages:
(i) The facultative anaerobic microbes degrade the complex polymers to simple monomers by enzymatic action. The Polymer like cellulose, hemicellulose, proteins and lipids get degraded into monomers but lignins and inorganic salts are left as residue because they do not degrade.
(ii) In second stage, monomers are converted in to organic acids by microbial action under partially aerobic conditions which are finally converted to acetic acid.
(iii) In third stage acetic acid is oxidized in to methane by the activity of anaerobic methanogenic bacteria. These bacteria are commonly found in the anaerobic sludge during sewage treatment. In this whole process digestion of cellulose takes place at very slow rate so that it is the “rate limiting factor in biogas production.”
Question.24. Name the pest that destroys the cotton balls. Explain the role of Bacillus thuringiensis in protecting the cotton crop against the pest to increase the yield.
Answer : Cotton ballworm is die pest that destroy the cotton balls. Bt toxin protein is produced by a soil bacterium called Bacillus thuringiensis in inactive prototoxin and crystalline form. The prototoxin form does not kill the bacteria. It becomes active and toxic when it is consumed by insects such as lepidopterans (armyworm), coleopterans (beetles) and dipterans (flies/ mosquitoes) due to presence of alkaline pH in the gut. The activated toxin (delta endotoxins) binds to the epithelial cells in the midgut of an insect and creates pores that cause lyses and swelling, eventually killing the insect.
Question.25. (a) Write the importance of measuring the size of a population in a habitat or an ecosystem.
(b) Explain with the help of an example how the percentage cover is a more meaningful measure of population size than more numbers.
Answer : (a) The importance of measuring the size of a population cover is more meaningful measure of population size than numbers. The size of the population tell us alot about its status in the habitat.
(b) Percentage cover is more meaningful measure of popula-tion size than more numbers because the’ relative abundance of a species is not only determined by number of individual w but by both i. e., the relative abundance in number and relative abundance in biomass.
Ex. In unit area the number of a grass species individuals or relative abundance in number is high but not in relative abundance of biomass. If the same area has one or two ficus benghalensis (Bargad) tree as it is very low in relative abundance in number while high in relative abundance of biomass.
Question.26. Differentiate between two different types of pyramids of biomass with the help of example of each.
Answer : Difference between two types of-pyramids of biomass:
Pyramid of Biomass is graphic representation of amount of biomass per unit area in the trophic levels with producers at the base & top carnivore at the apex. Biomass is maximum in producers. Only 10% of biomass is passed to next level. This is in accordance to the 10% law by Lindeman, (1942). Thus,.the biomass at higher trophic levels become smaller and smaller.
The pyramid of biomass is upright in grassland ecosystem.
For an aquatic system, the pyramid of biomass may be inverted or spindle-shaped. This is because the diatoms & other phytoplankton have a small standing crop, but they have a high annual productivity & high turn over rate. The reason or small standing crop is their short life span.
Question.27. (a) Describe the endosperm development in coconut.
(b) Why is tender coconut considered a healthy source of nutrition ?
(c) How are pea seeds different from castor seeds with respect to endosperm ?
Answer : Endosperm is ,a nutritive triploid tissue formed from mitotic divisions in primary endosperm nucleus (PEN). The cells of this tissue are filled with reserve food material & are used for the nutrition of the developing embryo.
In cococut, the type of endosperm formed is cellular endosperm, For this, the PEN divides many times and each division is followed by wall formation.
SECTION – D
Question.28 (a) Draw a L.S. of a pistil showing pollen tube entering the embryo-sac in an angiosperm and level any six parts other than stigma, style and ovary.
(b) Write the changes a fertilized ovule undergoes within the ovary in an angiosperm plant.
(a) Draw a diagrammatic sectional view of a human seminiferous tubules and label sertoli cells, primary spermatocyte, spermatogonium and spermatozoa in it.
(b) Fertilised ovule forms seed which contains embryonic plant, reserve food and protective coat. A seed contains/ consists of two parts—seed contains/consists
(i) Seed Coat: It is the outer covering of the seed, functioning as a protective coat. It is formed from the integuments of the ovule, which hardens after fertilisation. The outer seed coat is called Testa & is formed from outer integument. The inner seed coat is called Tegumen is formed from inner integument. The micropyle remains as a small pore in the seed coat. It helps in the entry of oxygen & H20 into the seed during germination. Above the micropyle, hilum is visible as a depression, being remanant of the attachment point.
(ii) Embryo : The embryo is made up of embryonal axis Cotyledon/s and endosperm. Embryonal axis is the central axis, also called ‘Tigellum. One end of Tigellum bears Radicle (future root) & the other end bears plumule (future shoot). Cotyledon is present at the node of embryo axis. It is a fleshy structure, used for storage of food. Endosperm is the nutritive layer. Present inside the seed coat providing nutrition to growing embryo.
(a) Spermatogenesis is under the control of endocrine hormones.
(i) Hypothalamus produces Gonadotropin Releasing Hormone (GnRH).
(ii) GnRH acts on anterior pituitary to produce gonadotropins,
ICSH (Interstitial Cell stimulating Hormone) & FSH. * ICSH acts on interstitial or leydig cells which produce testosterone.
(iii) FSH stimulates sertoli cells to develop ABP (Androgen Binding Protein) which help in concentrating testosteron in seminiferous tubules.
(iv) Excess testosterone inhibits LH/ICSH production by anterior pituitary & subsequendy GriRH production by Hypothalamus.
(v) Sertoli cells also produce a glycoprotein called Inhibition which suppresses FSH synthesis by anterior pituitary and GnRH by Hypothalamus.
This is called Negative Feedback control for release of testosterone.
Question.29. (a) Write the conclusion drawn by Griffith at the end of his experiment with streptococcus pneumonia.
(b) How did O. Avery, C. MacLeod and M. McCarty prove the DNA was the genetic material ?
Answer : (a) Griffith transformation experiment : Griffith performed his experiment in 1928 on Streptococcus pneumoneae bacteria which cause pneumonia in mice.
He used two strains of bacteria.
(i) Rough strain : Non capsulate non-virulent rough colonies on culture media.
(ii) Smooth strain : Capsulated virulent form smooth colonies on media.
Experiment: (i) Mice + Smooth strain bacteria —> Dead mice.
(ii) Mice + Rough strain bacteria —> Living mice.
(iii) Mice + Heat killed bacteria + Rough bacteria —> Dead Bacteria.
On the basis of third experiment he proposed that rough bacteria absorb some heat stable material from dead smooth bacteria and transformed into smooth bacteria which killed the mice.
(b) O Avery, C. Macleod and M. McCarty prove the DNA was the genetic material by purifying biochemicals (proteins, DNA, RNA etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed. They also discovered that protein – digesting enzymes (proteases) and RNA – digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material.
(a) Explain the mechanism of sex-determination in humans.
(b) Differentiate between male heterogamety and female heterogamety with the help of an example of each.
Answer : (a) Sex determination in humans :
(i) The males have 22 pairs autosomes and A pair of XY-
(ii) The females have 22 pairs autosomes and a pair of XX- chrornosomes
(iii) In male, 50% of sperms carry X- chromosome and other 50% carry Y – chromosomes.
(iv) In females, all ova contains X-chromosomes.
(v) The sex of an individual is determined by the type of sperm fertilizing the ovum.
(vi) If the ovum is fertilized by Y-chromosome, the zygote (XY) develops info a male and if the ovum in fertilized by X-chromosome, Zygote (XX) develops into a female.
(b) There are two types of sex determining mechanisms, i.e., XO type and XY type. But in both cases males produce two different types of gametes,
(i) Either with or without X-chromosome or,
(ii) Some gametes with X-chromosome and some with Y-chromosome.
Such types of sex determination mechanism is designated to be an example of male heterogamety. In some other organisms, e.g., birds a different mechanism of sex determination is observed. In this case the total number of chromosome is same in both males and females. But two . different types of gametes in terms of the sex chromosomes, are produced by females, i.e., female heterogamety. The two different sex chromosomes of a female bird has been designated to be the Z and W chromosomes. In these organisms the females have one Z and one W chromosome, ‘ whereas males have a pair of Z-chromosomes besides the autosomes.
Question.30. A person in your colony has recently been diagnosed with AIDS. People/Residents in the colony want him to leave the colony for the fear spread of AIDS.
(a) Write your views on the situation, living reasons.
(b) List the possible preventive measures that you would suggest to the residents of your locality in a meeting organized by you so that they understand the situation.
(c) Write the symptoms and the causative agent of AIDS.
Answer: (a) AIDS is infectious but not contagious it does not spread by shaking hand and use of common utensils so there is no need of fear to live with AIDS patient.
(b) Making blood (from blood banks) safe from HIV, ensuring the use of only disposable needles and syringes in public and private hospitals and clinics, free distribution of condoms, controlling drug abuse, advocating safe sex and promoting regular check-ups for HIV in susceptible populations, are some such steps taken up.
(c) Symptoms and the causative agent of AIDS :
AIDS is caused by the Human Immuno Deficiency Virus (HIV), a member of a group of viruses called retrovirus, which have an envelope enclosing the RNA genome. These target the T lymphocytes, due to which the person starts suffering from infections that could have been otherwise overcome such as those due to bacteria especially Mycobacterium, viruses, fungi and even parasites like Toxoplasma. The patient becomes so immuno-deficient that he/she is unable to protect himself/herself against these infections.
Question.3. Write the basis on which an organism occupies a space in its community/natural surroundings.
Answer: It is based on the feeding relationships of that organism with other organisms and source of their nutrition or food.
Question.8. Name an algae that reproduces asexually through zoospores. Why are these reproductive units so called.
Answer : Algae that reproduces asexually .through zoospores is “Chlamydomonas”. They are microscopic and motile, due to flagella so known as zoospores.
Question.12. “Stability of a community depends on its species richness”. Write how did David Tilman show this experimentally.
Answer : (i) David Tilman s long term ecosystem experiment – shows that plots with more species show less year-to-year variation in total biomass.
(ii) He also showed in his experiment that, increased diversity contributed to higher productivity.
Question.14. Name the haploid cells present in an unfertilized mature embryosac of a flowering plant. Vfite the total number of cells in it.
Answer : In an unfertilized mature embryo sac of a flowering
k plant 6 haploid cells are present, three antipodal cells, two synergids and one egg cell.
Question.15. In a typical monohybrid cross the F2-population ratio is written as 3 : 1 for phenotype but expressed as 1 : 2 : 1 for genotype. Explain with the help of an example.
Genotype 1 : 2 : 1- One is homozygous dominant and 2 are heterozygous but dominant and one is homozygous recessive .
Question.16.Mention the contribution of S. L. Miller’s experiment on Origin of Life.
Answer : Miller carried out the experiment as a proof of oparin-Haldane hypothesis. The experiment was as follows :
(i) Miller sealed a minture of water vapour (H20), NH3, CH4 8H2 in a spark chamber, which was provided with electrodes to provide electrical discharges.
(ii) The electrical sparks were of 75,000 volts and the ratio of CH4 : NH3 : H2 as 2 : 1 : 2 and water vapour at 800°C.
(iii) The spark chamber was connected to another falsk with arrangement for boiling H20 for evaporation.
(iv) The other end of spark chamber was connected to a condenser for condensation and collection of equeous solution.
(v) A trap was connected with flask for boiling HzO.
(vi) The apparatus was a controlled one, without any energy source.
(vii) After 18 days, he analysed the products after cooling (hem and proposed that HGN was formed from methane and NH3 and reacted with other compounds of gas to born amino acids like Alanine, Glycine and As partic Acid.
He gare the conclusion that similar synthesis could have occured in the prinitire atmsopheric condition.
Question.20. (a) Explain “birth rate” in a population by taking a suitable example.
(b) Write the other two chlracteristics which only a population shows but an individual cannot.
Answer : (a) It is the average number of new individuals added per unit population per year due to births, hatchings and germinations. If in a pool there were 20 lotus plants last year and plants reproduced and gave rise to 8 new plants, the total population is now 28. Birth rate will be equal to 8/20=0.4 offspring per lotus per year (b) (i) Death rate : per capita death
(ii) Sex Ratio : An individual is either a male or female, but a population has a sex ratio, e.g. 60% of the population of females and 40% of males.
Question.25. A burglar in a huff forgot to wipe off his blood – stains from the place of crime where he was involved in a theft. Name the technique which can help in identifying the burglar from the blood stains. Describe the technique.
Answer: The technique which help in identifying the burglar from the blood stain is DNA fingerprinting. The procedure of DNA fingerprinting is as follows :
(i) Isolation of DNA : Isolation of DNA from the blood stain .
(ii) Cutting, sizing, and sorting : Special enzymes called restriction enzymes are used to cut the DNA at specific places. The DNA pieces are sorted according to size by a sieving technique called electrophoresis. This technique is the biotechnology equivalent of screening sand through progressively finer mesh screens to determine particle sizes. The DNA fragments contain VNTRs (Variable Nun her of Tandem Repeats)
(iii) Transfer of DNA to nylon : The distribution of DNA pieces is transferred to a nylon sheet by placing the sheet on the gel and soaking them overnight.
(iv) Probing : Adding radioactive or coloured probes to the nylon sheet produces a pattern called the DNA fingerprint. Each probe typically sticks in only one or two specific places on the nylon sheet.
(v) Radiograph : The -final DNA fingerprint is built by using several probes (5-10 or more) simultaneously. These places are marked as dark bands when X-ray film is developed. This process is called auto radiography.
Question.28. (a) Write the specific features of the genetic code AUG.
(b) Genetic codes can be universal and degenerative. Write about them, giving one example of each.
(c) Explain aminoacylation of tRNA.
Answer: (a) Features of the genetic code AUG are as follows:
(i) AUG has dual function.
(ii) AUG codes for Methionine and acts a initiator codon.
(b) (i) Universal: From bacteria to human UUU would code for Phenylalanine. Some exceptions are there in mitochondrial codons and in some protozoans.
(ii) Degenerate : Some amino acids are coded by more than one codon. E.g. Phenylalanine is coded by UUU and UUC.
(c) Aminoacylation of tRNA : (i) In the presence of enzyme Amino-acyl-tRNA synthetase from DHU loop of tRNA, specific amino acid binds with ATP.
(ii) The AA-AMP-E complex formed in first step reacts with specific tRNA. Thus, amino acid is transferred to tRNA. The enzyme & AMP are then released.
(a) Differentiate between dominance and co-dominance.
(b) Explain co-dominance taking an example of human blood groups in population.
(b) In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three alleles IA, IBand i Hence, referred to as multiple allelism. A person possesses any two of the three alleles. IA and IB dominate over i. But with each other, IAand IB are co-dominant.
Question.1. What is detritus food chain made up of ? How do they meet their energy and nutritional requirements ?
Answer : Detritus food chain made up of decomposers which are heterotrophic organisms, mainly fungi and bacteria. They get energy and nutrients by decomposing dead organic matter or detritus.
Question.3. Name the phenomenon and one bird where the female gamete directly develops into a new organism.
Answer : Phenomenon : Parthenogenesis Bird : Turkey
Question.10. What is meant by “alien species” invasion ? Name one plant one animal alien species that are a threat to our Indian species.
Answer : An alien species whose introduction does or is likely to pose threat to the survival of many native species and cause their extinction.
Plants ; Parthenium, Lantana, Eicchomia Animal: African catfish, Clarias gariepinus
Question.14. Work out a cross to find the genotype of a tall pea plant. Name the type of cross.
Answer : (i) Test cross is used to find out the genotype of any trait. In this cross F hybrid of pure tall plant and a pure dwarf plant is crossed with a dwarf plant, e.g., Tt x tt Plants of F, Tt (Tall) x tt (Recessive)
The progeny consists of tall and dwarf plants in the ratio of 1:1
(ii) If the dominant plant are homozygous i.e. TT, then the progeny will have all tall plants :
Question.15. Write the Oparin and Haldanes hypothesis about the origin of life on earth. How does meteorite analysis favour this hypothesis.
Answer : Oparin-Haldane Hypothesis : Alexander I Oparin (1894-1980) a Russian biochemist &J.B.S. Haldane (1892-,1964), a British Scientist, put forward the concept that the first living organism evolved from non-living material. According to Oparin & Haldane (1929). Spontaneous generation of early molecules might have taken place through a series of chemical reactions from the earth’s primordial soup in a reducing atmosphere. The compounds of soup could be expected to react with one another producing a variety of chemical substances like amino acids, sugars, N2 bases. These precursor molecules then combined resulting in the appearance of proteins, polysacharides & nucleic acids. Energy required for these reactions was provided by UV radiations, cosmic rays, electric discharges etc.
Question.19 (i) Explain DNA polymorphism as the basis of genetic mapping of human genome.
(ii) State the role of VNTR in DNA fingerprinting.
Answer : (i) Polymorphism (variation at genetic level) are tlie result of mutation. Variation in allele sequence is DNA
polymorphism i.e., if more than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01. In other term we can say that if an inheritable mutation is observed in a population at high frequency, it is referred as DNA polymorphism. There is a variety of different types of polymorphism ranging from single nucleotide change to very large scale changes. For evolution and specification, such polymorphism are important. DNA polymorphism is also used in genetic and physical maps on the human genome. In genetic mapping, information on polymorphism of restriction endonuclease recognition sites are used.
(ii) Role of VNTR in DNA Finger printing : Radiolabelled VNTR (Variable Number of Tandem Repeat) is used as a probe in DNA finger printing. A tandem repeat is a short sequence of DNA that is repeated in a head-to-tail fashion at a specific chromosomal locus. Tandem repeats are interspersed throughout the human genome. Some sequences are found at only one site – a single locus – in the human genome. For many tandem repeats, the number of repeated units vary between individuals. Such loci are termed VNTRs.
Question.30. How does the process of natural selection affect Hardy- Weinberg equilibrium ? Explain. List other four factors that disturb the equilibrium.
Answer : Population or Mendelian population is a group of individuals present in ageographical area which share acommon pool. Gene frequency is the percentage of an allele in relation to the total alleles of a gene in an interbreeding population. All the genes & their alleles together constitute gene pool. Normally the alleles tend to maintain an equilibrium with reference to one another over the generations. It is referred as genetic equilibrium. Such a population is referred as non¬evolving populaiton. G.H. Harady & withelm weinberg proposed a principle independently in 1908, about the genetic structure of a non-evolving population. It is known as Hardy-weinberg equilibrium. The allelic frequencies in non-evolving population are stable and remain constant from generation to generation.
The Hardy-Weinberg Equilibrium can change by process of natural selection. Natural or survival of the fittest is a major factor that adds variation in the population, change the gene frequies of the gene pool leading to evolution & formation of a new distinctive gene pool. It operates through differential or non-random reproduction. If same type of selection continues for a few generation; the gene popl will undergo change the alleles having the advanage of surviving. Thus it leads to change of gene pool.
(a) Explain Mendel’s Law of independent assortment by taking a suitable example.
(b) How did Morgan show the deviation in inheritance pattern in Drosophila with respect to this law.
Answer : (a) Dihybrid cross is based on Law of Independent Assortment. This law states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.
The image shows a true-breeding plant with the dominant traits of green pod colour (GG) and yellow seed colour (YY) being cross-pollinated with a true-breeding plant with yellow pod colour (gg) and green seeds (yy). The resulting offspring are all heterozygous for green pod colour and yellow seeds (GgYy). If the offspring are allowed to self pollinate, a 9 : 3 : 3 : 1 ratio will be seen in the next generation. About 9 plants will have green pods and yellow seeds, 3 will have green pods and green seeds, 3 will have yellow pods and yellow seeds and 1 will have a yellow pod and green seeds.
(b) Morgan and his group observed that when the two genes in a dihybrid cross were located on the same chromosome, the proportion of parental gene combinations in the progeny were much higher than the non-parental or new combinations -(recombinants) of genes.