Question

if two parent that were heterozygous for a recessive trait (carriers) had four children, what is the chances of them having a child that has the recessive disorder? (remember you have to have two recessive alleles to get the disorder)
a. 0%
b. 10%
c. 25%
d. 50%
e. 75%
f. 100%

Explanation:

Step1: Define alleles

Let the dominant allele be \(A\) and the recessive allele be \(a\). The parents are heterozygous, so their genotypes are \(Aa\).

Step2: Create Punnett - square

The possible gametes from each parent are \(A\) and \(a\). The Punnett - square has 4 possible combinations: \(AA\), \(Aa\), \(aA\), \(aa\).

Step3: Calculate probability

Out of the 4 possible genotypes (\(AA:Aa:aA:aa = 1:2:1\)), the genotype \(aa\) (which results in the recessive disorder) is 1 out of 4. The probability is \(\frac{1}{4}=0.25\) or 25%.

Answer:

c. 25%