Built to match the topics, formats and difficulty NEET draws from this chapter, in the proportions it draws them. Includes the assertion-reason, statement-based and match-the-column formats that now appear regularly. Answer key with explanations at the end.
1A true-breeding line is one which:
2Which of the following pairs of contrasting traits was not among the seven studied by Mendel in pea?
3In garden pea, the dominant traits for pod colour and seed colour respectively are:
4Mendel's approach was unacceptable to many biologists of his time mainly because he:
5During artificial hybridisation in pea, the removal of anthers from the flower selected as the female parent is termed:
6In a monohybrid cross between TT and tt, the F2 phenotypic and genotypic ratios respectively are:
7Among the tall plants of the F2 generation obtained from Tt × Tt, the fraction expected to be homozygous is:
8A test cross is performed between:
9A violet-flowered pea plant, when crossed with a white-flowered plant, produces offspring of which half are violet and half white. The genotype of the violet parent is:
10The F2 genotypic ratio of a monohybrid cross can be obtained by expanding:
11Consider the following statements regarding the Law of Segregation:
Which of the statements are correct?
12In Antirrhinum, a cross between true-breeding red and true-breeding white plants gives pink F1. The F2 phenotypic and genotypic ratios respectively are:
13Co-dominance is best distinguished from incomplete dominance by the fact that in co-dominance:
14The number of genotypes and phenotypes respectively possible in the human ABO blood group system is:
15The allele i of the ABO system is recessive to both IA and IB because it:
16A woman of blood group AB marries a man of blood group O. The blood groups possible among their children are:
17Assertion (A): Multiple allelism can be demonstrated only through population studies, never within a single individual.
Reason (R): A diploid individual can carry only two alleles of a given gene.
18A diploid organism heterozygous at four loci can produce how many types of gametes, assuming independent assortment?
19How many types of gametes are produced by an individual of genotype AaBBCcDd?
20In the F2 generation of the cross RrYy × RrYy, the number of genotypes and phenotypes respectively is:
21In the F2 of a dihybrid cross, the fraction of individuals expected to be homozygous at both loci is:
22In the cross TtYy × Ttyy, the proportion of offspring expected to be tall with green seeds is:
23Mendel's Law of Independent Assortment was derived from observations on:
24The chromosomal theory of inheritance was put forward by:
25Mendel's results on the inheritance of characters were rediscovered independently in 1900 by:
26Morgan's dihybrid crosses in Drosophila gave F2 ratios deviating significantly from 9:3:3:1 because the two genes concerned were:
27Genes white and yellow showed 1.3 per cent recombination, while white and miniature wing showed 37.2 per cent. It follows that:
28Recombination frequency between gene pairs on the same chromosome was first used as a measure of the distance between genes by:
29When two genes are closely linked, which of Mendel's laws no longer holds?
30Assuming three genes A, B and C control human skin colour additively, which genotype would give a skin colour identical to that of AaBbCc?
31Phenylketonuria is cited in the NCERT text as the example of:
32Assertion (A): Human skin colour occurs as a continuous gradient rather than in a few discrete classes.
Reason (R): Skin colour is governed by three or more genes whose alleles act additively, and it is also influenced by the environment.
33The specific nuclear structure later identified as the X chromosome was traced during spermatogenesis in insects, and named the "X body", by:
34Which statement correctly describes the XO type of sex determination?
35Grasshopper is the classical example of which type of sex determination?
36In many birds, the sex chromosome constitutions of the male and the female respectively are:
37In the honey bee, the drone:
38Which of the following statements about sex determination in the honey bee is incorrect?
39A couple has three daughters. The probability that their fourth child will be a son is:
40Sickle-cell anaemia is the classical example of:
41Insertion or deletion of base pairs within a gene results in:
42Match the items in the two columns:
| (i) Point mutation | (A) UV radiation |
| (ii) Frame-shift mutation | (B) Commonly seen in cancer cells |
| (iii) Mutagen | (C) Sickle-cell anaemia |
| (iv) Chromosomal aberration | (D) Insertion of base pairs |
43In a human pedigree chart, a double horizontal line joining two individuals denotes:
44Myotonic dystrophy is used in the NCERT text as the representative example of:
45Red-green colour blindness occurs in approximately:
46In sickle-cell anaemia, the amino acid substitution responsible for the defect is:
47β-thalassemia is controlled by which gene, on which chromosome?
48Failure of cytokinesis after the telophase stage of cell division results in:
49The karyotype 47, XXY is characteristic of:
50Turner's syndrome represents:
1 Three elements define it: continuous self-pollination, stable inheritance, several generations. Effectively homozygous for the character.
2 Seed weight was not among the seven. The seven characters are stem height, flower colour, flower position, pod shape, pod colour, seed shape and seed colour.
3 The inversion that costs the most marks in this chapter: in pods green is dominant over yellow; in seeds yellow is dominant over green.
4 His use of statistics and mathematical logic in biology was entirely new, and is listed in the text as one of the four reasons his work was ignored until 1900.
5 Emasculation. Bagging is the subsequent step of covering the emasculated flower to prevent stray pollen.
6 Phenotypic 3:1, genotypic 1:2:1. Always read the word immediately before "ratio".
7 Of the 16 F2 proportions, tall plants are 1 TT and 2 Tt out of 3 tall. So 1/3 of the tall plants are homozygous — although they are 1/4 of all F2 plants. The question asks among the tall.
8 Option (a) is a back cross, which is the broader category. A test cross is specifically the cross with the homozygous recessive.
9 A 1:1 outcome in a cross with the recessive parent identifies the unknown as heterozygous, Vv.
10 (½T + ½t)² = ¼ TT + ½ Tt + ¼ tt. The unsquared binomial gives only the gamete frequencies.
11 Statement II is false: a homozygous parent produces only one kind of gamete; it is the heterozygote that produces two kinds in equal proportion.
12 Both ratios are 1:2:1. The genotypic ratio is unchanged from an ordinary monohybrid cross; only the phenotypic ratio changes from 3:1, because the heterozygote now has its own appearance.
13 Incomplete dominance gives an intermediate the parents did not have; co-dominance gives both parental products together, as with A and B sugars in group AB.
14 Six genotypes, four phenotypes. Two genotypes each for groups A and B, one each for AB and O. The reversed order is the standard distractor.
15 It produces no sugar at all, so it has nothing to express when paired with either functional allele. This ties directly to the enzyme explanation of dominance.
16 IAIB × ii gives only IAi (A) and IBi (B). Neither AB nor O is possible — a classic blood-group elimination question.
17 Both true and the reason is the explanation. A gene may have many alleles in a population, but any diploid individual samples only two of them.
18 2n where n is the number of heterozygous loci: 24 = 16.
19 Count only heterozygous loci — Aa, Cc and Dd, so n = 3 and 23 = 8. The BB locus contributes nothing. Answering 16 means counting all four loci.
20 9 genotypes and 4 phenotypes. In general, 3n genotypes and 2n phenotypes. 16 is the number of squares, not of genotypes.
21 The four homozygous genotypes RRYY, RRyy, rrYY and rryy occupy one square each, so 4/16 = 1/4.
22 Solve each gene separately and multiply. Tall = 3/4; green (yy from Yy × yy) = 1/2. So 3/4 × 1/2 = 3/8. This is NCERT exercise 7.
23 The third law came from the dihybrid cross. Dominance and segregation came from the monohybrid cross.
24 Sutton and Boveri noted the parallel; Sutton united it with Mendelian principles and named the theory. Morgan supplied the experimental verification, which is a different contribution.
25 de Vries, Correns and von Tschermak, in 1900, independently.
26 Both genes were on the X chromosome, so they were linked and did not assort independently.
27 Lower recombination means tighter linkage. 1.3 per cent for white and yellow indicates far tighter linkage than 37.2 per cent for white and miniature.
28 Sturtevant, Morgan's student, converted recombination frequency into map distance.
29 Only independent assortment fails. Segregation has no exceptions, which is why it is the answer to "which law is universally applicable".
30 Only the number of dominant alleles matters, not which genes they belong to. AaBbCc has three; AABbcc also has three (A, A, B). AAbbcc has only two.
31 Pleiotropy — one gene, in this case coding for phenyl alanine hydroxylase, producing several phenotypic effects.
32 Both true, and the reason explains the assertion. Additive allele effects plus environmental influence produce a continuous gradient instead of discrete classes.
33 Henking, 1891, in insects, during spermatogenesis. He named it but could not explain its significance.
34 In XO type there is no Y at all, and the male therefore has one chromosome fewer than the female. The "O" denotes absence.
35 Grasshopper — the only example named in the text for XO type.
36 Male ZZ, female ZW. Reversing this is the most common error in the section. Note it is the opposite arrangement to humans, where the female carries the identical pair.
37 The drone is haploid with 16 chromosomes, arises from an unfertilised egg, and produces sperm by mitosis because meiosis cannot halve an already haploid set.
38 A drone cannot have sons: his sperm can only produce fertilised eggs, and every fertilised egg develops into a female. He can, however, have grandsons through his daughters.
39 Each pregnancy is independent, with a 50 per cent probability of either sex. Previous children do not alter it.
40 A single base pair substitution, hence a point mutation. Frame-shift arises from insertions and deletions instead.
41 Frame-shift — the reading frame downstream of the site is thrown out of register.
42 Point mutation is exemplified by sickle-cell anaemia; frame-shift by insertion of base pairs; UV radiation is the mutagen named in the text; chromosomal aberrations are commonly seen in cancer cells.
43 The double line marks a consanguineous mating, which is the pedigree clue pointing towards a recessive disorder.
44 Autosomal dominant. It appears only in the caption of Figure 4.14, which is exactly why it is asked.
45 8 per cent of males against 0.4 per cent of females, because the genes lie on the X chromosome and males have only one X.
46 Glutamic acid is replaced by valine at position 6 of the β-globin chain. The reversed direction and the α-chain are the two standard traps.
47 HBB on chromosome 11. α-thalassemia involves HBA1 and HBA2 on chromosome 16, with four genes in total across both parents.
48 Polyploidy — a whole extra set. Aneuploidy comes from failure of chromatid segregation and involves a single chromosome.
49 Klinefelter's syndrome. Turner's is 45, X0; Down's is 47 with trisomy of chromosome 21.
50 Monosomy of a sex chromosome — one X is absent. Down's syndrome is the trisomy of an autosome.