# What is the probability that the son will be a carrier of the recessive allele?

## Ectrodactyly (lobster claw) is an inherited physical defect in humans occurring in homozygous recessive individuals. If 2 normal parents had a daughter affected with this condition and a normal son...

Jun 26, 2016

$\frac{2}{3}$ chance or ~67%

#### Explanation:

First we have to know the genotype of the parents. Lets call the dominant allele for ectrodactyly $\textcolor{g r e e n}{\text{E}}$ and the recessive allele $\textcolor{red}{\text{e}}$.

The disease is homozygous recessive, so the daughter must have two recessive alleles, her genotype is $\textcolor{red}{\text{ee}}$. This genotype is only possible if both parents are heterozygous ($\textcolor{g r e e n}{\text{E"color(red)"e}}$).

Knowing this we can make a cross table to show all possible genotypes of the first generation offspring: Since we know that the son is not affected, he is either homozygous dominant ($\textcolor{g r e e n}{\text{EE}}$) or heterozygous ($\textcolor{g r e e n}{\text{E"color(red)"e}}$). The ratio homozygous to heterozygous is 1 : 2.

In this case we are interested in the chance that he is a carrier, which is the case when he is heterozygous. As seen above, this chance is 2 out of 3, which is $\frac{2}{3}$ chance or ~67%.