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...

1 Answer
Jun 26, 2016

Answer:

#2/3# chance or ~#67%#

Explanation:

First we have to know the genotype of the parents. Lets call the dominant allele for ectrodactyly #color(green)"E"# and the recessive allele #color(red)"e"#.

The disease is homozygous recessive, so the daughter must have two recessive alleles, her genotype is #color(red)"ee"#. This genotype is only possible if both parents are heterozygous (#color(green)"E"color(red)"e"#).

Knowing this we can make a cross table to show all possible genotypes of the first generation offspring:

enter image source here

Since we know that the son is not affected, he is either homozygous dominant (#color(green)"EE"#) or heterozygous (#color(green)"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 #2/3# chance or ~#67%#.