PART – A
I. Select the correct alternative from the choices given below: 1 x 15 = 15
1. Haploid condition is not observed in which of the following cells.
a) Synergids and Egg
b) Zygote and PEN
c) Antipodal and Egg
d) Antipodal and Synergids
Ans: b) Zygote and PEN
2. Statement I: Formation of fruit without fertilization is called apomixis
Statement II: In some species of Asteraceae and grasses seeds are formed without fertilization
a) Both Statement I and Statement II are correct
b) Both Statement I and Statement II are incorrect
c) Statement I is correct and Statement II is incorrect
d) Statement I is incorrect and Statement II is correct
Ans: d) Statement I is incorrect and Statement II is correct
3. During gestation, the foetus develops limbs and digits by the end
a) First month
b) Second month
c) Third month
d) Fifth month
Ans: b) Second month
4. The secondary oocyte after ovulation is covered by a non-cellular layer of
a) Cumulus oophorus
b) Corona radiata
c) Zona pellucida
d) Cortical layer
Ans: c) Zona pellucida
5. An example of hormone-releasing IUD among the following
a) Cu - 7
b) Lippes loop
c) LNG - 20
d) Multiload 375
Ans: c) LNG - 20
6. Which of the following is a foetal sex determination test?
a) ZIFT
b) GIFT
c) MTP
d) Amniocentesis
Ans: d) Amniocentesis
7. Which of the following Mendelian gene disorder is the representation of autosomal recessive trait?
a) Hemophilia
b) Thalassemia
c) Sickle cell anemia
d) Myotonic dystrophy
Ans: c) Sickle cell anemia
8. The process of removal of introns and joining of exons in a defined order in a primary transcripts occurs in
a) Prokaryotes
b) Eukaryotes
c) Prokaryotes and Eukaryotes
d) Prokaryotes and Protista
Ans: b) Eukaryotes
9. A type of Natural selection in which more individuals acquire mean character value is called
a) Stabilizing selection
b) Disruptive selection
c) Directional selection
d) Dominant selection
Ans: a) Stabilizing selection
10. Drug called "Heroin" is synthesized by
a) Methylation of Morphine
b) demethylation of Morphine
c) Acetylation of Morphine
d) deacytalation of Morphine
Ans: Acetylation of Morphine
11. The fungus not used in the production of any Industrial product is
a) Penicillium
b) Aspergillus
c) Trichoderma polysporum
d) Glomus
Ans: d) Glomus
12. Significance of Insertional inactivation method in Recombinant DNA technology is to
a) Introduce the recombinants
b) Isolate gene of Interest
c) Select the recombinants
d) Select the gene of interest
Ans: a) Introduce the recombinants
13. Which of the following organisms are studied by Cornell’s in his elegant field experiments to study competition
a) Warbler species
b) Chathamalus and Balanus
c) Cucko and Crow
d) Cattle egret and grazing cattle
Ans: b) Chathamalus and Balanus
14. The correct sequence in the process of decomposition is
a) Humification----Leaching----Catabolism---- Mineralisation ----Fragmentation
b) Catabolism----Leaching----Fragmentation----Humification---- Mineralisation
c) Leaching----Fragmentation ----Catabolism----Humification---- Mineralisation
d) Fragmentation ----Leaching----Catabolism-----Humification----Mineralisation
Ans: d) Fragmentation ----Leaching----Catabolism-----Humification----Mineralisation
15. Western Ghats have a greater diversity of
a) Amphibians
b) Reptiles
c) Aves
d) Mammals
Ans: a) Amphibians
II. Fill in the blanks by choosing the appropriate word/Words from those given below: 1 x 5 = 5
(Commensalism, Alveoli, Ammensalism, Panspermia, Codominance, Perisperm)
16. The residual, persistent nucellus is called Perisperm
17. The cells of Alveoli secrete milk in the mammary glands.
18. AB blood group inheritance is an example for Codominance
19. Panspermia is the theory that proposes that units of life called spores were transferred to different planets including earth
20. A population interaction in which one species is harmed and the other species is unaffected is Ammensalism
PART - B
III. Answer any FIVE of the following questions in 3 – 5 sentences wherever applicable: 2 x 5 = 10
21. List any four complications a person suffers from untreated sexually transmitted infections?
Ans: Complications include pelvic inflammatory diseases (PID), abortions, still births, ectopic pregnancies, infertility or even cancer of the reproductive tract.
22. State the two medical grounds on which a pregnancy can be terminated according to the amended Medical termination of pregnancy act 2017.
Ans: MTP is done in the following situations:
Failure of contraceptive used during coitus or rapes.
When continued pregnancy is harmful or fatal to either mother or foetus or both.
In the case of foetal abnormalities.
23. Give the phenotypes of the parental Drosophila that has produced 1.3% and 37.2% recombinants respectively in T. H. Morgan Dihybrid cross experiment.
Ans: It was Morgan (1910) who clearly proved and defined linkage on the basis of his breeding experiments in fruitfly Drosophila melanogaster.
In his experiments involving dihybrid cross to study the pattern of inheritance of body color and eye color in Drosophila, he crossed yellow bodied, white eyed female flies with wild type male flies. The percentage of offsprings with recombinant phenotypes for the characters in the F2 generation was found to be 1.3%. This included the offspring flies having brown body, white eyes or yellow body, red eyes.
The remaining 98.7% of the F2 offsprings had parental phenotypes, that is, yellow body, white eyes or brown body, red eyes.
24. Differentiate divergent evolution from convergent evolution.
Ans:
25. List any two differences between active and passive immunity.
Ans:
26. What are primary lymphoid organs? Give two examples
Ans: Lymphoid organs are the organs which provide micro-environments for the development
and maturation of T-lymphocytes. And sites where immature lymphocytes will differentiate into antigen specific lymphocytes.
Exampls: Bone marrow and Thymus.
27. Baculoviruses are excellent biocontrol agents in Integrated Pest Management. Comment.
Ans: The majority of baculoviruses used as biological control agents are in the genus Nucleopolyhedrovirus.
These viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications.
They have been shown to have no negative impacts on plants, mammals, birds, fish or even on non-target insects.
This is especially desirable when beneficial insects are being conserved to aid in an overall integrated pest management (IPM) programme, or when an ecologically sensitive area is being treated.
28. Ecological pyramids have limitations. Justify the statement with two reasons.
Ans:
It never takes into account the same species belonging to two or more trophic levels.
It assumes a simple food chain, which never exists in nature.
In spite of the vital role played by saprophytes/decomposers, they are not given any position in ecological pyramids.
PART - C
IV. Answer any FIVE of the following questions in 40 – 80 words each wherever applicable: 3 x 5 = 15
29.
a) Why is bagging of emasculated flowers essential during hybridization experiment?
b) Mention the cells of the mature pollen grain.
c) Give the scientific name of the plant that has the viability record of 10,000 years.
Ans:
a) The emasculated flower is covered with a paper bag to prevent contamination from unwanted pollens.
b) Larger vegetative cell and the smaller gererative cell.
c) Lupinus arcticus excavated from Arctic Tundra.
30. Explain the changes that occur in ovary and uterus during luteal phase of menstrual cycle.
Ans: During Luteal phase, the remaining parts of the Graafian follicle transform as the corpus luteum. The corpus luteum secretes large amounts of progesterone which is essential for maintenance of the endometrium. Such an endometrium is necessary for implantation of the fertilised ovum and other events of pregnancy.
31. Draw a diagrammatic sketch of the Lac operon when lactose is present in the medium.
Ans:
32. With respect to the evolution of man, name a, b, c, d, e, and f
Ans:
a - Ethiopia
b - 1.5 million years ago
c - Homo erectus
d - East and central Asia
e - Neanderthalensis
f - 75,000-10,000 years ago
33. Mention the three critical areas of biotechnology.
Ans:
Providing best catalyst as improved organism, usually a microbe or pure enzyme.
Creating optimal conditions by engineering for a catalyst to act, and
Downstream processing technologies to purify the protein/organic compound.
34. What is gene therapy? Explain the steps involved in curing ADA deficiency by gene therapy.
Ans: Gene therapy is a collection of methods that allows correction of gene defects, diagnosed in a child or embryo.
ADA is caused due to deletion of gene for adenosine deaminase.
In some cases, it can be cured by bone marrow transplantation and enzyme replacement therapy, but it is not fully curative.
Lymphocytes from patient's blood were grown in a culture and functional ADA, cDNA was introduced in these lymphocytes using a retroviral vector.
The lymphocytes were transferred into the patient's body. Periodic infusion of such genetically engineered lymphocytes is done because these cells are mortal.
For permanent cure, gene isolated from the bone marrow cells producing ADA, at early embryonic stage can be a possible cure.
Other diseases like cystic fibrosis, haemophilia, cancer, Parkinson's, etc., are also treated by gene therapy.
35.
a) Co-extinctions lead to loss of biodiversity. Justify the statement with two examples.
b) What are hot spots of biodiversity?
Ans:
a) When a species becomes extinct, the plant and animal species associated with it in an obligatory manner, also become extinct.
For example, if the host fish species becomes extinct, all those parasites exclusively dependent on it, will also become extinct; in plant—pollinator mutualism also, extinction of one result in the extinction of the other.
b) Biodiversirt hotspots are regions of high levels of species richness and high degree of endemism.
There are 34 hot spots in the world. In India, the three hot spots are Western Ghats and Sri Lanka, Indo-Burma and Himalaya.
36. Describe the components of an aquatic ecosystem taking pond as an example.
Ans: A pond is a shallow water body in which all the four basic components of an ecosystem are well exhibited.
The abiotic component is the water with all the dissolved inorganic andorganic substances and the rich soil deposit at the bottom of the pond.
The solar input, the cycle of temperature, day-length and other climatic conditions regulate the rate of function of the entire pond.
The autotrophic components include the phytoplankton, some algae and the floating, submerged and marginal plants found at the edges.
The consumers are represented by the zooplankton, the free swimming and bottom dwelling forms.
The decomposers are the fungi, bacteria and flagellates especially abundant in the bottom of the pond.
PART – D
Section - I
V. Answer any FOUR of the following questions in about 200 – 250 words each wherever applicable:
5 x 4 = 20
37. Draw a neat labeled diagram of human male reproductive system.
Ans:
38. Mention the chromosomal disorders that are due to trisomy, represent their karyotype and two symptoms each.
Ans: Down’s syndrome is caused by an extra copy of chromosome number 21 (trisomy of 21).
This disorder was first discovered by Langdon Down (1866).
Karyotype: 47,XX,+21 or 47,XY,+21
Symptoms:
Short statured with small round head.
Partially open mouth with protruding furrowed tongue.
Palm is broad with characteristic palm crease.
Physical, psychomotor and mental development is retarded.
39. With the help of schematic representation illustrate how an infected animal cell can survive while viruses are being replicated and released.
Ans:
40. With reference to DNA finger printing define the following terms:
a) Repetitive DNA
b) Satellite DNA
c) DNA polymorphism
d) VNTR
e) Probe
Ans:
a) Repetitive DNA: Repetitive DNA are DNA sequences that contain small segments, which are repeated many times.
b) Satellite DNA: Satellite DNA are DNA sequences that contain highly repetitive DNA.
c) DNA polymorphism: More than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01 is called DNA polymorphism.
d) VNTR: or the Variable Number of Tandem Repeats are the repeated DNA sequences at a defined locus.
e) Probe: A probe is a single-stranded sequence of DNA or RNA used to search for its complementary sequence in a sample genome.
41. What is genetic code? Explain any four salient features of genetic code.
Ans: The relationship between the sequence of nucleotides on mRNA and sequence of amino acids in the polypeptide is called genetic code.
The codons are triplet. Out of 64 codons, 61 code for 20 amino acids and 3 codons (UAA,
One codon codes for only one particular amino acid, hence the code is unambiguous and specific.
Some amino acids are code by more than one codon, hence the code is degenerate.
The codon is read on mRNA in a continuous fashion, i.e., without punctuations and thus the code is comma less.
Genetic code is nearly universal,i.e., a particular codon codes for the same amino acid in all organisms except in mitochondria and few protozoa.
AUG is a dual function codon, it codes for methionine (met) and it is also acts as initiator codon.
42. Describe the biological treatment of primary effluent.
Ans: After primary treatment, primary effluent is passed into large aeration tanks with constant mechanical agitation and air supply.
Useful aerobic microbes grow rapidly and form flocs.
Flocs are masses of bacteria associated with fungal filaments to form mesh-like structures.
The growing microbes consume organic matter and thus reduce the biochemical oxygen demand (BOD).
When BOD of sewage has reduced, the effluent is passed into settling tank.
Here, the bacterial flocs settle, and the sediment is called activated sludge.
A small part of the sludge is used as an inoculum in the aeration tank and the remaining part is passed into large tanks called anaerobic sludge digesters.
In the digesters, other kinds of bacteria, which grows anaerobically, digest bacteria and fungi in sludge.
During this digestion, bacteria produce mixture of gases such as methane, hydrogen
sulphide and C02 which form the biogas.
43.
a) Explain the process of Polymerase chain reaction in amplification of desired DNA.
b) Draw a labeled diagram of pBR322 vector.
Ans:
a) The Polymerase Chain Reaction (PCR) is a reaction in which amplification of specific DNA sequences is carried out in vitro.
This technique is used in labs to make billions of copies of the desired gene for research, diagnostic and therapeutic purposes.
b)
44.
a) Study the population growth curve given below and answer the questions that follows;
i) Identify the growth curves ‘a’ and ‘b’
ii) Mention the conditions responsible for the curves ‘a’ and ‘b’ respectively.
b) Explain the mechanism of sexual deceit in relation to mutualism.
Ans:
a)
i) a is Exponential growth and the b is Logistic growth.
ii) The Exponential growth (a) occurs when responses are not limiting the growth. The Logistic growth (b) occurs when responses are limiting the growth.
b) Sexual deceit is the process in which petal of orchid flower bears an uncanny resemblance to the female of the bee in size, colour and markings to attract the male bee for pollination.
The male bee pseudocopulates with it and during this process of pseudocopulation, the pollen grains are dusted on the body of male bees. With such pollen dusts, male bee pseudocopulates to another flower of the same species and pollination takes place.
Section – II
VI. Answer any ONE of the following questions in about 200 – 250 words each wherever applicable:
5x 1= 5
45. Double fertilization is the unique feature of angiosperms and the products of this double fertilization is zygote and PEN. In context of this when a hexaploid plant is pollinated by a tetraploid plant find out the ploidy of zygote and PEN through a schematic illustration.
Ans:
Pollen grains are taken from a tetraploid (4n) plant. So, the male gamete will be 2n.
The polar nuclei belong to a hexaploid (6n) plant. So, they will be 3n.
The PEN (primary endosperm nucleus) is formed by the fusion of a male gamete with two polar nuclei inside the ovule.
So, the ploidy of the PEN will be 2n + 3n + 3n = 8n.
The Ploidy of the zygote will be 2n + 3n = 5n.
46. ABO blood grouping provides a good example of multiple alleles and are controlled by the gene ‘I’. This gene product is responsible for the production of a sugar polymer that protrudes from its surface. The ‘I’ gene has three alleles they all follow a specific pattern of in,
a) What are the probable number of phenotypes and genotypes for ABO blood group in human population?
Ans:
Phenotype: There are four main blood group phenotypes based on the presence or absence of the A and B antigens:
Type A, Type B, Type AB, Type O.
There are four main phenotypes (A, B, AB, and O) and six possible genotypes that can result in these phenotypes.
b) Mention the genotypes of all the blood group phenotypes.
Ans:
Genotype: The possible genotypes and their corresponding phenotypes are as follows:
Type A - IA IA and IA i
Type B - IB IB and IA i
Type AB - IA IB
Type O - ii
c) Name the type of blood groups of the parental combination in which both their blood group is not inherited to their children.
Ans: In the case of a Type AB parent and a Type O parent, their children would all have Type A or Type B blood, but not Type AB or Type O. So, the parental combination is when one parent has Type AB blood and the other has Type O blood.
47. Five patients suffering from certain diseases visit a local primary health centre. The Doctor does a thorough check and prepares the report of the five patients and is indicated in the below given table. Analyse the table and diagnose the disease they are suffering from and causative agent of the diseases.
Patient 1 - High fever, constipation, stomach ache, loss of appetite, headache
Patient 2 - Chills and high fever recurring every 3 – 4 days
Patient 3 - Constipation, mucous and blood clots in stool, abdominal pain and cramps
Patient 4 - Internal bleeding, blockage in the internal passage, muscular pain, fever
Patient 5 - Dry, scaly lesions on skin, nails and scalp
Ans:
Patient 1 - Typhoid - Salmonella typhi
Patient 2 - Malaria - Plasmodium vivax, Plasmodium falciparum
Patient 3 - Amoebiasis - Entamoeba histolytica
Patient 4 - Ascariasis - Acaris lumbricoides
patient 5 - Ringworm - Microsporum, Trychophyton, Epidermophyton