Question

This group of questions refers to the probabilities below. Assume that the alleles referred to all assort independently. (A) 0 (B) 1/16 (C) 1/4 (D) 1/2 (E) 3/4

Answer 1

Answer:

Option D

Explanation:

The complete question is

Assume that the alleles referred to all assort independently.

(A) 0

(B) 1/16

(C) 1/4

(D) 1/2

(E) 3/4

Probability that the genotype Aa will be produced by the parents Aa x Aa

Solution -

Given -

Genotype of both the parents is heterozygous i.e Aa

Also, it is given that the allele assort independently

Now a cross between Aa and Aa will produce following offspring -

Aa * Aa

AA, Aa, Aa and aa

Out of four offspring, two offspring have Aa genotype.

Thus, probability is equal to total number of offspring with genotype Aa with the total number of offspring.

$\frac{2}{4} \\= \frac{1}{2}$

nearly 50% of the offspring have Aa genotype

Hence, option D is correct