Question

parents are carriers of a recessive disorder, the probability that their child will have the disorder is 25%. explain how this probability is determined using allele combinations.

Explanation:

1. First, define the alleles: let the dominant normal allele = $A$, and the recessive disorder-causing allele = $a$.
2. Carrier parents each have the genotype $Aa$ (they have one normal allele and one disorder allele, but do not show the disorder).
3. Use a Punnett square to map all possible allele combinations a child can inherit:

• Parent 1 can pass on either $A$ or $a$.
• Parent 2 can pass on either $A$ or $a$.
• The four possible offspring genotypes are: $AA$, $Aa$, $Aa$, $aa$.

1. Only the $aa$ genotype (homozygous recessive) results in the disorder, as the trait is recessive.
2. Calculate probability: 1 out of 4 possible genotypes causes the disorder, so $\frac{1}{4} = 25\%$.

Answer:

When both parents are carriers (genotype $Aa$ for a recessive disorder with alleles $A$ (normal) and $a$ (disorder-causing)):

• Each parent can pass either $A$ or $a$ to their child.
• The four possible offspring genotype combinations are $AA$, $Aa$, $Aa$, and $aa$.
• Only the $aa$ genotype (inheriting two recessive alleles) results in the disorder.
• The probability is $\frac{1}{4} = 25\%$, as 1 out of 4 possible allele combinations produces the disorder phenotype.