Chapter 26, Problem 1CQ

What is the probability of each of the following sets of parents producing the given genotypes in their offspring?

To determine

The probability of parents having genotype AA and Aa will produce 3 offsprings with genotype aa.

Answer to Problem 1CQ

A cross between parents having genotype AA and Aa will have zero probability of producing three offsprings with genotype ’aa’.

Explanation of Solution

Punnett square is used in genetics to determine the probability of a progeny’s genotype and phenotype. It is a box figure on which the alleles of each parent are listed. A cross between the parents gives the genotype or phenotype of each progeny.

In this case,

The genotype of one parent is AA and the genotype of other parent is given as Aa. A cross between these parents can be depicted through Punnett square:

	A	A
A	AA	AA
a	Aa	Aa

The probability that the cross will produce three progeny with genotype aa is zero.

To determine

The probability of parents having genotype Aa and Aa will produce 3 offsprings with genotype Aa.

Answer to Problem 1CQ

A cross between parents having genotype Aa and Aa has probability of $\frac{1}{8}$ for producing three offsprings with genotype Aa.

Explanation of Solution

The genotype of one parent is Aa and the genotype of other parent is given as Aa  A cross between these two parents can be depicted through Punnett square:

	A	a
A	AA	Aa
a	Aa	aa

The probability of offsprings having genotype Aa is $\frac{2}{4}$. Therefore, the probability that three offsprings will have genotype Aa is:

$\begin{array}{c}\frac{2}{4}\times \frac{2}{4}\times \frac{2}{4}=\frac{8}{64}\\ =\frac{1}{8}\end{array}$

To determine

The probability of parents having genotype Aa and aa will produce 3 offsprings with genotype Aa.

Answer to Problem 1CQ

A cross between parents having genotype Aa and aa has $\frac{1}{8}$ probability of producing three offsprings with genotype Aa.

Explanation of Solution

The genotype of one parent is Aa and the genotype of other parent is given as aa A cross between these two parents can be depicted through Punnett square:

	A	a
a	Aa	aa
a	Aa	aa

The probability of offsprings having genotype Aa is:

$\begin{array}{c}\frac{2}{4}\times \frac{2}{4}\times \frac{2}{4}=\frac{8}{64}\\ =\frac{1}{8}\end{array}$