Question

The genotype of F₁ individuals in tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F₂ offspring will have the following genotypes?

(a) aabbccdd

(b) AaBbCcDd

(c) AABBCCDD

(d) AaBBccDd

(e) AaBBCCdd

Short answer

(a) The probability of F2 offspring with the genotype aabbccdd is 1/256.

(b) The probability of F2 offspring with the genotype AaBbCcDd is 1/16.

(c) The probability of F2 offspring with the genotype AABBCCDD is 1/256.

(d) The probability of F2 offspring with the genotype AaBBccDd is 1/64.

(e) The probability of F2 offspring with the genotype AaBBCCdd is 1/128.

Step-by-step solution

Pattern of inheritance and probability of inheritance of traits

Inheritance is the transfer of genes from the parents to the offspring. The inheritance occurs via sexual reproduction, and the offspring generated from the parents are known as F1 offspring.

F2 offspring are obtained by crossing the F1 hybrid.

In general, the probability of occurrence of a homozygous trait “XX or xx” is ¼, and the probability of occurrence of a heterozygous trait “Xx” is ½.

Explanation of part “a”

The given offspring is aabbccdd. All the traits are homozygous for the four recessive traits. Thus, the probability of the F2 offspring is as follows:

$\text{Probability}=\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}=\frac{1}{256}$

Explanation of part “b”

The given F2 offspring is AaBbCcDd. All the given traits are heterozygous for each trait. Thus, the probability of the F2 offspring is as follows:

$\text{Probability}=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{16}$

Explanation of part “c”

F2 offspring given in the question is AABBCCDD. The genotypic combination provided is homozygous for four dominant traits. Thus, the probability of the F2 offspring is as follows:

$\text{Probability}=\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}=\frac{1}{256}$

Explanation of part “d”

F2 offspring given in the question is AaBBccDd. Aa and Dd are heterozygous for the traits, so they have a probability of ½ each. BB and cc are homozygous for dominant and recessive traits, respectively, so they have a probability of ¼ each.

Thus, the overall probability of the F2 offspring is as follows:

$\text{Probability}=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}=\frac{1}{64}$

Explanation of part “e”

F₂given in the question isAaBBCCdd. In this genotype, Aa is heterozygous for a trait with a probability is 1/2. BB, CC, and dd are homozygous for the traits, each with the probability of ¼.

Thus, the overall probability of the F₂ offspring is as follows:

$\text{Probability}=\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}=\frac{1}{128}$