Chapter B, Problem 1P

Summary Introduction

To analyze:

Describe following on the basis of an autosomal recessive condition:

The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele

Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.

If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.

Introduction:

In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.

Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Explanation of Solution

Pedigree chart of cross between both heterozygous parents $\left(\text{Aa × Aa}\right)$ is as follows:

♂ ♀	A	a
A	AA	Aa
a	Aa	aa

The phenotypic ratio is $\text{3:1}$(3 normal and 1 diseased), and genotypic ratio is $\text{1:2:1}$.

Homozygous dominant (AA) is $\frac{\text{1}}{\text{4}}$,

Heterozygous carrier (Aa) is $\frac{\text{2}}{\text{4}}$,

Homozygous recessive (aa) is $\frac{\text{1}}{\text{4}}$.

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is $\frac{\text{1}}{\text{4}}$.

In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.

The chance of having normal (AA / Aa) child is $\frac{\text{3}}{\text{4}}$ and out of this, the chance of the child being a heterozygous carrier (Aa) for the condition is $\frac{\text{2}}{\text{3}}$. So, the chance of a normal child being a carrier for the condition is $\frac{\text{2}}{\text{3}}$.

Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is $\frac{\text{1}}{\text{4}}$.

The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is $\frac{\text{2}}{\text{3}}$.

Conclusion

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is $\frac{\text{1}}{\text{4}}$.

The chance of the normal child being a carrier for the condition is $\frac{\text{2}}{\text{3}}$.

The chance that the second child of the couple will be homozygous recessive (aa) for the mutation is $\frac{\text{1}}{\text{4}}$.

The chance that the second child of the couple will be a heterozygous carrier (Aa) for the mutation is $\frac{\text{2}}{\text{3}}$.