Question

both parents of a newborn are heterozygous for an autosomal dominant genetic disorder. what are the odds that the newborn would also display the disorder? 0/4 1/4 1/2 3/4

Explanation:

Step1: Determine the genotypes of parents

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

Step2: Create a Punnett square

The possible gametes from each parent are \(A\) and \(a\). The Punnett square is:

	\(A\)	\(a\)
\(a\)	\(Aa\)	\(aa\)

Step3: Analyze the genotypes for the disorder

For an autosomal - dominant disorder, individuals with \(AA\) or \(Aa\) will display the disorder.
The genotypes from the Punnett square are \(AA:Aa:aa = 1:2:1\).
The proportion of individuals with the disorder (\(AA + Aa\)) is \(\frac{1 + 2}{4}=\frac{3}{4}\)

Answer:

\(\frac{3}{4}\)