NEET Biology Reproduction Genetics and Biotechnology MCQs Set 01

Answer: (d) Natural selection is known to affect Hardy-Weinberg equilibrium

Explanation:
1. The first cellular life is thought to have originated about 2000 million years ago, not 200 million.
2. Geological and biological history of the earth are correlated (fossil record). Melanic moths were more common in industrialised areas where dark tree trunks hid them from predators.
3. Natural selection is one of the factors that disturbs Hardy-Weinberg equilibrium (others: gene flow, genetic drift, mutation, recombination).

Teacher's Note:
1. Five factors disturb Hardy-Weinberg equilibrium: gene migration, genetic drift, mutation, genetic recombination and natural selection.
2. Industrial melanism in the peppered moth is a classic example of natural selection.

Answer: (c) ampulla

Explanation:
1. The ovum released at ovulation is picked up by the infundibulum and moves into the oviduct.
2. Sperms swim up the female tract and meet the ovum in the ampullary region of the fallopian tube (near the ampullary-isthmic junction). Fertilisation takes place here, and the zygote then moves towards the uterus.

Teacher's Note:
1. The oviduct has three parts: infundibulum, ampulla and isthmus.
2. Implantation occurs in the uterus, not the oviduct.

Answer: (b) male sterilisation

Explanation:
1. Vasectomy is a surgical method in which a small portion of the vas deferens is removed or tied.
2. It stops sperm from reaching the urethra, so it is a permanent method of male sterilisation.

Teacher's Note:
1. Tubectomy is the female counterpart (cutting the fallopian tubes).
2. Sterilisation is a terminal contraceptive method.

Answer: (c) Eight

Explanation:
1. A heterozygous gene pair gives two kinds of gametes. For three independently assorting heterozygous gene pairs the number of gamete types is \( 2^n = 2^3 = 8 \).
2. The gametes are ABC, ABc, AbC, Abc, aBC, aBc, abC, abc.

Teacher's Note:
1. Number of gamete types is \( 2^n \), where n is the number of heterozygous gene pairs.
2. Homozygous pairs do not add to the variety.

Answer: (a) Four

Explanation:
1. Genotypes \( I^AI^A \) and \( I^Ai \) give blood group A; \( I^BI^B \) and \( I^Bi \) give B; \( I^AI^B \) gives AB (co-dominance); ii gives O.
2. Hence four phenotypes: A, B, AB and O.

Teacher's Note:
1. \( I^A \) and \( I^B \) are co-dominant, and both are dominant over i.
2. Six genotypes but only four phenotypes.

Answer: (d) 10 cM (trans)

Explanation:
1. Parental types are the most numerous classes: \( +b \) and \( a+ \) (45 each). So the F\(_1\) parent had \( +b \) on one chromosome and \( a+ \) on the other: the trans (repulsion) phase.
2. Recombinants are \( ++ \) and \( ab \): \( 5 + 5 = 10 \) of 100 offspring.
3. Recombination frequency \( = 10\% \), so the distance is 10 cM (map units).

Teacher's Note:
1. 1% recombination equals 1 map unit (centimorgan).
2. The most frequent classes are parental; the least frequent are recombinant.

Answer: (a) 23S rRNA

Explanation:
1. In bacteria the 23S rRNA of the large subunit acts as a ribozyme: it catalyses peptide bond formation (peptidyl transferase) while also serving as a structural component of the ribosome.
2. 5.8S and 18S rRNAs occur in eukaryotes.

Teacher's Note:
1. Ribozyme means catalytic RNA.
2. In eukaryotes the corresponding catalytic RNA is 28S rRNA.

Answer: (b) gene transfer without a vector

Explanation:
1. PEG makes the plasma membrane of protoplasts permeable (and induces fusion), so that naked DNA can enter the cell.
2. It is therefore a direct gene transfer method that needs no vector.

Teacher's Note:
1. Other direct gene transfer methods: electroporation, microinjection and the gene gun.
2. Vectors such as Ti plasmid belong to indirect methods.

Answer: (d) EcoRV

Explanation:
1. EcoRI, HindIII and BanI cut the two strands at different positions and leave single-stranded (sticky) overhangs.
2. EcoRV cuts in the middle of its recognition sequence GAT|ATC and cuts both strands at the same point, giving blunt ends.

Teacher's Note:
1. Sticky ends join easily through base pairing; blunt ends need more ligase.
2. EcoRV is a standard blunt cutter.

Answer: (a) mutation breeding

Explanation:
1. Mutations were induced in mung bean by gamma rays and the mutants were screened.
2. Plants resistant to yellow mosaic virus and powdery mildew were selected and multiplied. This is mutation breeding.

Teacher's Note:
1. Mutation breeding creates new alleles using radiation or chemicals.
2. Resistant varieties of mung bean and Abelmoschus (bhindi) were produced this way.

Answer: (d) Clostridium butylicum - lipase - removal of oil stains

Explanation:
1. (a), (b) and (c) are correct matches.
2. Clostridium butylicum produces butyric acid, not lipase. The detergent enzyme lipase for removing oil stains is obtained from Candida lipolytica (and similar microbes).

Teacher's Note:
1. Streptokinase is called a clot buster.
2. Statins are produced by Monascus purpureus (yeast).

Answer: (a) Statements (A) and (B)

Explanation:
1. (A) is wrong: Rosie was the first transgenic cow, not buffalo.
2. (B) is wrong: DNA is separated from other macromolecules with enzymes such as lysozyme, cellulase, chitinase (to break the cell), ribonuclease and protease; restriction enzymes are used to cut DNA at specific sites.
3. (C) and (D) are correct, so the statements with mistakes are A and B.

Teacher's Note:
1. Rosie (1997) produced human alpha-lactalbumin-enriched milk.
2. Downstream processing means separation and purification of the product.

Answer: (c) cross-breeding

Explanation:
1. Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri ewes with Merino rams.
2. Mating of superior males of one breed with superior females of another breed is cross-breeding.

Teacher's Note:
1. Cross-breeding combines desirable qualities of two breeds.
2. Out-crossing is within the same breed but with no common ancestor for 4-6 generations.

Answer: (a) 8, 8, 16, 24

Explanation:
1. Radicle is sporophytic (2n = 16), so n = 8.
2. Tube nucleus and antipodal cells are haploid: 8 each.
3. Definitive nucleus (secondary nucleus) is formed by the fusion of two polar nuclei: \( 2n = 16 \). Endosperm (formed by triple fusion) is \( 3n = 24 \).

Teacher's Note:
1. Embryo-sac cells are all haploid; the central cell has two polar nuclei that fuse.
2. Endosperm is triploid in angiosperms.

Answer: (c) dehiscence

Explanation:
1. The anther wall has four layers: epidermis, endothecium, middle layers and tapetum.
2. The endothecium is a layer of cells with hygroscopic thickenings; its shrinking and curving ruptures the anther at the stomium, helping dehiscence and release of pollen.
3. Nourishment of the developing pollen is the function of the tapetum.

Teacher's Note:
1. Tapetum nourishes pollen; endothecium helps in anther dehiscence.
2. Epidermis protects the anther.

Answer: (c) karyotype

Explanation:
1. The karyotype is the characteristic chromosome complement of a species: the number, size and shape of its chromosomes.
2. Polyploidy is more than two sets of chromosomes; polyteny is the presence of many chromatids side by side in giant chromosomes.

Teacher's Note:
1. Karyotype is the chromosome set of a species; an idiogram is only its diagram.
2. Karyotyping is used to detect chromosomal disorders.

Answer: (c) Klinefelter's syndrome

Explanation:
1. Klinefelter's syndrome has the karyotype 47, XXY: a male with an extra X chromosome.
2. Characteristic features: overall masculine development but development of breasts (gynaecomastia) and sterility.

Teacher's Note:
1. Turner's syndrome is 45, X0 (female); Jacob's syndrome is 47, XYY.
2. Extra X in a male causes feminine traits.