NEET Biology Principles of Inheritance and Variation MCQs Set 20

Answer: (a) 1/16

Explanation:
1. Both parents must be carriers (Aa).
2. Each child has a 1/4 chance of being affected, and fraternal twins are independent events.

Teacher's Note:
1. So the probability for both is \( 1/4 \times 1/4 = 1/16 \).

Answer: (c) Equally, segregation

Explanation:
1. In codominance both alleles are expressed equally in the heterozygote.
2. The alleles still separate in gametes, following the law of segregation.

Teacher's Note:
1. They do not follow dominance because neither masks the other.

Answer: (c) 50% pink and 0% red

Explanation:
1. Pink is the heterozygote (Rr), and the test cross is with the recessive white (rr).
2. The offspring are 1/2 Rr (pink) and 1/2 rr (white), so there are no red plants.

Teacher's Note:
1. Red would need a homozygous RR parent.

Answer: (a) An individual has only two alleles

Explanation:
1. A diploid individual carries only two alleles of a gene.
2. More than two alleles can be seen only by comparing many individuals of a population.

Teacher's Note:
1. Hence multiple alleles are a population-level feature.

Answer: (c) 18 : 6 : 6 : 2 : 9 : 3 : 3 : 1 : 9 : 3 : 3 : 1

Explanation:
1. Multiply (3 : 1) × (3 : 1) × (1 : 2 : 1).
2. This gives 12 phenotype classes: 9, 18, 9, 3, 6, 3, 3, 6, 3, 1, 2, 1 parts.

Teacher's Note:
1. Option \( (c) \) lists exactly these twelve values.

Answer: (c) IAIO and IBIO

Explanation:
1. All four groups appear only when each parent carries the \( O \) allele and one different dominant allele.
2. IAIO × IBIO gives IAIB, IAIO, IBIO and IOIO.

Teacher's Note:
1. No other combination can produce all four groups.

Answer: (d) 2 : 1 : 1

Explanation:
1. Aa × Aa gives 1 AA (disease free), 2 Aa (carriers) and 1 aa (diseased).
2. So the ratio of carrier : free : diseased is 2 : 1 : 1.

Teacher's Note:
1. Read the order asked in the question carefully.

Answer: (b) Incomplete dominance

Explanation:
1. The \( F1 \) shows a phenotype intermediate between the two parents.
2. This is the pattern of incomplete dominance.

Teacher's Note:
1. In complete dominance the \( F1 \) would resemble one parent.

Answer: (b) 5, 6, 16

Explanation:
1. With two gene pairs \( F2 \) shows 5 phenotype classes (4, 3, 2, 1 or 0 contributing alleles) in the ratio 1 : 4 : 6 : 4 : 1 out of 16.
2. The mulatto classes with 2 dominant alleles number 6.

Teacher's Note:
1. Total \( F2 \) combinations are 16.

Answer: (b) 4 : 2 : 2 : 1

Explanation:
1. The AA and BB zygotes die.
2. The survivors are 2 Aa : 1 aa for the first gene and 2 Bb : 1 bb for the second.

Teacher's Note:
1. Multiplying gives 4 : 2 : 2 : 1.

Answer: (c) 9

Explanation:
1. Pleiotropy does not change the genetic ratio.
2. Two heterozygous gene pairs give \( 3 \times 3 = 9 \) genotypes in \( F2 \).

Teacher's Note:
1. Pleiotropy affects phenotypes, not genotype counts.

Answer: (c) 14

Explanation:
1. Two polygenes give 5 phenotype classes and \( 3 \times 3 = 9 \) genotypes in \( F2 \).
2. Their total is \( 5 + 9 = 14 \).

Teacher's Note:
1. Compare with the \( 2^n \) and \( 3^n \) rules of ordinary dihybrid crosses.

Answer: (d) 0% with small starch grains

Explanation:
1. All \( F1 \) plants are heterozygous (Bb) and have intermediate-sized starch grains.
2. So none have small grains.

Teacher's Note:
1. The cross gives a uniform \( F1 \), as in the law of dominance.

Answer: (d) 3 coloured and 5 white

Explanation:
1. Coloured flowers need \( C_ \) and \( P_ \).
2. In CcPp × Ccpp, \( C_ \) has probability 3/4 and Pp × pp gives \( P_ \) with probability 1/2, so coloured = 3/8.

Teacher's Note:
1. Out of 8, 3 are coloured and 5 are white.

Answer: (d) 7.5 cm

Explanation:
1. The difference of 30 cm is due to four capital alleles.
2. Each allele therefore adds \( 30 / 4 = 7.5 \) cm.

Teacher's Note:
1. This is additive gene action.

Answer: (b) 48.5 cm

Explanation:
1. The plant has three capital alleles, each adding 7.5 cm to the base height of 26 cm.
2. Height = \( 26 + 3 \times 7.5 = 48.5 \) cm.

Teacher's Note:
1. Count contributing alleles and multiply.

Answer: (b) The female parent was heterozygous

Explanation:
1. XWXw × XwY gives red and white eyes in equal numbers.
2. A pure red female (XWXW) would give all red progeny.

Teacher's Note:
1. So the female must be heterozygous.

Answer: (c) 10% or 6%

Explanation:
1. The order \( b-c-a \) is fixed by the first three distances.
2. Locus \( d \) is 2 units from a, so it can be between \( c \) and a (cd = \( 8 - 2 = 6 \)) or beyond a (cd = \( 8 + 2 = 10 \)).

Teacher's Note:
1. Both are possible with the data given.

Answer: (b) 2X/3A - female

Explanation:
1. A ratio of \( 2X/3A \) is 0.67, which gives an intersex, not a normal female.
2. The other matches are correct: 1.5 is metafemale, 0.5 is male and 0.33 is supermale.

Teacher's Note:
1. Normal females have a ratio of 1.0.

Answer: (a) Males and females respectively

Explanation:
1. This is XO type sex determination.
2. Males are XO (17) and females are XX (18).

Teacher's Note:
1. The male is therefore heterogametic, producing gametes with and without the \( X \).

Answer: (a) 0%

Explanation:
1. A son receives the \( Y \) chromosome from his father and the \( X \) from his mother.
2. An X-linked gene of the father therefore cannot pass to a son.

Teacher's Note:
1. The father passes it to all his daughters.

Answer: (c) 50%

Explanation:
1. The lady is XCXc and the man is XhY(hypertrichosis).
2. Sons XcY(hyper) carry colourblindness plus hypertrichosis (25% of all offspring), and daughters XcXh carry two X-linked alleles (25%).

Teacher's Note:
1. Total = 50%.