NEET Biology Principles of Inheritance and Variation MCQs Set 19

Answer: (b) AaBB × aaBB

Explanation:
1. AaBB × aaBB gives AaBB and aaBB in equal numbers, so the phenotypes are 1 : 1.
2. AaBb × aabb gives 1 : 1 : 1 : 1.

Teacher's Note:
1. The other two crosses give a single phenotype.

Answer: (b) TtCc × ttCc

Explanation:
1. A tall probability of 1/2 needs Tt × tt.
2. A coloured probability of 3/4 needs Cc × Cc.
3. Combining them gives TtCc × ttCc.

Teacher's Note:
1. Break a two-gene problem into two single-gene problems.

Answer: (b) 16

Explanation:
1. \( 81 = 3^4 \) genotypes means four heterozygous gene pairs.
2. The number of phenotypes is \( 2^4 = 16 \).

Teacher's Note:
1. Genotypes follow \( 3^n \) and phenotypes follow \( 2^n \).

Answer: (b) 1/16

Explanation:
1. Each gene gives a homozygous offspring with probability 1/2 (Aa × aa, Bb × BB, Cc × CC, Dd × dd).
2. So the probability is \( (1/2)^4 = 1/16 \).

Teacher's Note:
1. Multiply the probabilities for the four genes.

Answer: (c) 0

Explanation:
1. The male is CC, so he cannot form a sperm with allele \( c \).
2. The abcS combination is therefore impossible, and the probability is zero.

Teacher's Note:
1. Check for homozygous genes first, since they can make a combination impossible.

Answer: (b) Factors

Explanation:
1. Mendel called the units of inheritance factors.
2. The word gene was introduced later by Johannsen.

Teacher's Note:
1. Factors are what we now call genes.

Answer: (b) The father is homozygous for blood group A

Explanation:
1. The \( O \) child is ii, so each parent must carry \( i \).
2. The father is therefore IAi (heterozygous), and the mother is IBi.

Teacher's Note:
1. Hence statement \( (b) \) is incorrect, while \( (a) \) and \( (c) \) are correct.

Answer: (c) 25%

Explanation:
1. Green is dominant, so \( F1 \) is all heterozygous green.
2. In \( F2 \) the ratio is 3 green : 1 yellow, so yellow plants are 25%.

Teacher's Note:
1. The recessive trait reappears in one quarter of the \( F2 \).

Answer: (d) Both (a) and (b)

Explanation:
1. A true-breeding line shows stable inheritance after continuous selfing.
2. Mendel used 14 true-breeding varieties arranged as 7 pairs of contrasting traits.

Teacher's Note:
1. So statements \( (a) \) and \( (b) \) are correct and \( (c) \) is wrong.

Answer: (a) Law of dominance

Explanation:
1. In the monohybrid \( F2 \) the dominant and recessive traits appear in a 3 : 1 ratio because one allele dominates the other.
2. The law of segregation explains why the recessive trait reappears.

Teacher's Note:
1. Independent assortment applies to two or more genes.

Answer: (d) More than one phenotype influenced by the same gene

Explanation:
1. When one gene influences more than one phenotype (pleiotropy), dominance depends on the phenotype under consideration.
2. The same allele can be dominant for one effect and recessive for another.

Teacher's Note:
1. Dominance is thus not an independent property of the gene.

Answer: (c) Sutton and Boveri

Explanation:
1. Sutton and Boveri proposed the chromosomal theory of inheritance.
2. They noted that chromosome behaviour parallels the behaviour of Mendel's factors.

Teacher's Note:
1. Morgan later gave experimental proof using Drosophila.

Answer: (a) ABO blood group

Explanation:
1. The ABO blood group is controlled by three alleles (IA, IB and \( i \)) of one gene.
2. Sickle-cell anaemia has two alleles.

Teacher's Note:
1. Remember that an individual carries only two alleles, so multiple alleles are seen only in a population.

Answer: (d) 4/16

Explanation:
1. Wrinkled seeds (rr) form 1/4 of the \( F2 \), which is 4/16.
2. The wrinkled plants include yellow wrinkled (3/16) and green wrinkled (1/16).

Teacher's Note:
1. Add them: \( 3/16 + 1/16 = 4/16 \).

Answer: (d) All of the above

Explanation:
1. The law of dominance has three parts: characters are controlled by discrete factors, factors occur in pairs, and in a dissimilar pair one factor dominates.
2. All three statements belong to it.

Teacher's Note:
1. So the answer is all of the above.

Answer: (c) Down's syndrome

Explanation:
1. Down's syndrome is trisomy of chromosome 21, an autosome.
2. Klinefelter's (XXY) and Turner's (XO) involve sex chromosomes, and phenylketonuria is a gene (enzyme) defect.

Teacher's Note:
1. Aneuploidy means a gain or loss of single chromosomes.

Answer: (b) Codominant genes

Explanation:
1. In codominance both alleles of a pair express themselves in the heterozygote, for example IA and IB.
2. They still segregate during gamete formation.

Teacher's Note:
1. Codominance is therefore consistent with the law of segregation.

Answer: (d) 19.9 ft

Explanation:
1. There are 6 alleles shared over a 20 ft range, so each contributing allele adds about 3.3 ft.
2. AaBbCc has three contributing alleles, giving \( 10 + 3 \times 3.3 = 19.9 \) ft.

Teacher's Note:
1. It is the mid-value, as expected for a heterozygote.

Answer: (a) 37.5%

Explanation:
1. Broad leaves (dominant) occur in 3/4 of the \( F2 \).
2. Pink flowers (heterozygotes) occur in 1/2 of the \( F2 \).

Teacher's Note:
1. The combined probability is \( 3/4 \times 1/2 = 3/8 \) = 37.5%.

Answer: (a) 25% of the offspring will have yellow fruits

Explanation:
1. The offspring are WwYy and Wwyy (white, 50%), wwYy (yellow, 25%) and wwyy (green, 25%).
2. So 25% have yellow fruits.

Teacher's Note:
1. This is dominant epistasis, where \( W \) masks the effect of \( Y \).

Answer: (c) 50%

Explanation:
1. The \( F2 \) genotypic ratio is 1 BB : 2 Bb : 1 bb.
2. Heterozygotes (Bb) have intermediate starch grains, so they are 50%.

Teacher's Note:
1. Here the heterozygote shows an intermediate phenotype.

Answer: (c) The number of phenotypes in multiple alleles is calculated by the formula 2n + 1

Explanation:
1. Multiple alleles are alternative forms of one gene at the same locus, seen only in a population.
2. The number of phenotypes does not follow a fixed formula like \( 2n + 1 \), because it depends on dominance relations.

Teacher's Note:
1. Hence statement \( (c) \) is wrong.