Question

Two parents learn they are each carriers for Cystic Fibrosis, each having the genotype (Aa).

Using a punnett square, perform a genetic cross to determine the probability that their children will inherit two recessive alleles and have Cystic Fibrosis (aa).

In your answer, explain the probability using either percentages (%) or odds out of 4 (1/4, 3/4, etc).

Answer 1

When two carriers of Cystic Fibrosis (Aa) are crossed, the probability that their children will inherit two recessive alleles and have Cystic Fibrosis (aa) is 25%. This can be determined by performing a Punnett square, which shows all the possible gamete combinations and the resulting offspring genotypes. The Punnett Square would look like this:

A | a

—————

A | AA | Aa

a | Aa | aa

As you can see, out of the four boxes, one of them contains two recessive alleles, so the probability of their children having Cystic Fibrosis is one out of four, or 25%.