Question

question 2 (1 point)
tay - sachs is a recessive genetic disease in humans. if one parent is homozygous
normal and the other is a carrier (heterozygous), what are the chances that their child
will have this disease?
a) 0%
b) 25%
c) 50%
d) 75%
e) 100%

Explanation:

Step1: Determine the genotypes

Let the normal allele be \(A\) and the disease - causing allele be \(a\). One parent is homozygous normal (\(AA\)), and the other is a carrier (\(Aa\)).

Step2: Set up the Punnett square

The possible gametes from the homozygous normal parent (\(AA\)) are all \(A\). The possible gametes from the carrier parent (\(Aa\)) are \(A\) and \(a\). The Punnett square is:

	\(A\)	\(a\)

Step3: Analyze the genotypes of offspring

The genotypes of the offspring are \(AA\) (homozygous normal) and \(Aa\) (heterozygous carrier). There is no \(aa\) (affected by the recessive disease) genotype.

Answer:

A. 0%