NEET Biology Principles of Inheritance and Variation MCQs Set 18

Answer: (a) 35

Explanation:
1. A trihybrid \( F2 \) has \( 2^3 = 8 \) phenotypes and \( 3^3 = 27 \) genotypes.
2. Their total is \( 8 + 27 = 35 \).

Teacher's Note:
1. Remember: \( n \) gene pairs give \( 2^n \) phenotypes and \( 3^n \) genotypes in \( F2 \).

Answer: (c) 1/64

Explanation:
1. The probability of AA from Aa × Aa is 1/4, and the same for BB and CC.
2. Multiplying gives \( 1/4 \times 1/4 \times 1/4 = 1/64 \).

Teacher's Note:
1. Use the product rule for independent genes.

Answer: (a) A, B and D

Explanation:
1. Recessive traits appear only when homozygous.
2. Constricted pod, green seed colour and yellow pod are recessive, whereas axial flower is dominant.

Teacher's Note:
1. Learn the seven contrasting pairs with their dominant member.

Answer: (d) All are correct

Explanation:
1. NCERT lists all four reasons: poor communication, non-acceptance of discrete factors, a new mathematical approach and no physical proof of factors.
2. Hence every statement is correct.

Teacher's Note:
1. Such multi-statement questions are best solved by testing each statement separately.

Answer: (b) 4 : 8 : 16 : 1

Explanation:
1. Their probabilities are \( 1/64, 1/32, 1/16 \) and 1/256 respectively.
2. Multiplying each by 256 gives 4 : 8 : 16 : 1.

Teacher's Note:
1. Multiply the single-gene probabilities (1/4 for homozygous, 1/2 for heterozygous) for each genotype.

Answer: (d) Dominant offspring in a test cross are always homozygous

Explanation:
1. A test cross is a cross of an unknown dominant phenotype with the homozygous recessive parent.
2. Dominant offspring are heterozygous when the tested individual is heterozygous, so they are not always homozygous.

Teacher's Note:
1. A test cross is a special kind of back cross.

Answer: (b) 6.25%

Explanation:
1. P(AA) = 1/4, P(Bb) = 1/2, P(Cc) = 1/2, so the product is 1/16 = 6.25%.
2. Heterozygous genotypes have probability 1/2 and homozygous genotypes 1/4 for each gene in a selfed hybrid.

Teacher's Note:
1. Always convert the fraction to a percentage at the end.

Answer: (b) 8

Explanation:
1. A dihybrid test cross (AaBb × aabb) gives 4 genotypes and 4 phenotypes in the ratio 1 : 1 : 1 : 1.
2. Their total is 8.

Teacher's Note:
1. Note that the 9 genotypes and 4 phenotypes belong to the dihybrid \( F2 \), not to the test cross.

Answer: (c) BbRr, bbRr, bbrr

Explanation:
1. The child bbrr received \( b \) and \( r \) from each parent.
2. So the man must carry \( b \) and \( r \) (BbRr) and the woman must carry \( r \) (bbRr).

Teacher's Note:
1. Always start from the recessive child to find hidden alleles in the parents.

Answer: (a) 4/16

Explanation:
1. RR occurs in 1/4 of the \( F2 \) plants, whatever the other gene is.
2. Out of 16 combinations 4 are RR, i.e.
3. 4/16.

Teacher's Note:
1. Pure round means homozygous for round, regardless of the second character.

Answer: (a) 4

Explanation:
1. Only the heterozygous gene pairs (Cc and Dd) give two kinds of alleles.
2. The number of gamete types is \( 2^2 = 4 \).

Teacher's Note:
1. Homozygous pairs do not add to the number of gamete types.

Answer: (a) 3/8

Explanation:
1. For round seeds, Ww × ww gives 1/2 Ww.
2. For red flowers, Rr × Rr gives \( 3/4 R_ \).

Teacher's Note:
1. Multiplying \( 1/2 \times 3/4 \) gives 3/8.

Answer: (b) 16

Explanation:
1. P(Aa) = 1/2, P(bb) = 1/4 and P(Cc) = 1/2, so P(AabbCc) = 1/16.
2. Of 256 seeds, \( 256/16 = 16 \) plants are expected.

Teacher's Note:
1. Convert the genotype probability into numbers by multiplying with the total.

Answer: (c) 9 : 1

Explanation:
1. The \( F2 \) ratio is 9 : 3 : 3 : 1.
2. Plants showing both dominant characters are 9 parts and those showing both recessive characters are 1 part.

Teacher's Note:
1. Hence the ratio is 9 : 1.

Answer: (a) HbdNMa

Explanation:
1. Pollen grains are haploid and carry one allele of each gene.
2. The plant is nn, so it cannot make a gamete with \( N \).
3. The gamete HbdNMa is therefore impossible.

Teacher's Note:
1. Check each allele of the gamete against the parental genotype.

Answer: (d) 1/8

Explanation:
1. Each allele goes into a gamete with probability 1/2 for independent genes.
2. So P(AbC) = \( 1/2 \times 1/2 \times 1/2 = 1/8 \).

Teacher's Note:
1. Eight equally likely gamete types are formed, so each has probability 1/8.

Answer: (a) 8

Explanation:
1. Three gene pairs are heterozygous (Tt, Rr, Gg), and PP is homozygous.
2. So there are \( 2^3 = 8 \) phenotypes with complete dominance.

Teacher's Note:
1. Homozygous pairs do not increase the number of phenotypes.

Answer: (d) Reciprocal cross

Explanation:
1. In a reciprocal cross the sexes of the parents carrying each trait are interchanged.
2. Different results from the two crosses show sex linkage, while identical results show autosomal inheritance.

Teacher's Note:
1. A test cross only reveals the genotype.

Answer: (b) 1 : 1

Explanation:
1. In \( F2 \) the red plants are 12/16 and the tall plants are 12/16, so the ratio is 1 : 1.
2. Each character separately follows the monohybrid 3 : 1 ratio.

Teacher's Note:
1. Treat each character independently and compare the numbers.

Answer: (a) 0

Explanation:
1. The cross is cc × cc, which can only give cc offspring.
2. An offspring can therefore never be heterozygous for this gene.
3. So the overall probability is zero.

Teacher's Note:
1. One homozygous gene pair is enough to make the product zero.

Answer: (b) DD × dd

Explanation:
1. DD × dd gives only Dd offspring, which are all alike in genotype and phenotype.
2. The other crosses produce more than one genotype.

Teacher's Note:
1. A cross between two different homozygotes gives a uniform \( F1 \).

Answer: (a) 3 : 1

Explanation:
1. Only one gene pair (Bb) is heterozygous, so selfing is a monohybrid cross.
2. The \( F2 \) ratio is 3 : 1.

Teacher's Note:
1. Homozygous gene pairs do not alter the ratio.