Question

a man who is heterozygous for huntingtons has a baby with his wife, who does not have huntingtons. what is the chance their baby will have huntingtons? 100% chance 75% chance 50% chance 25% chance 0% chance

Explanation:

Step1: Define alleles

Let 'H' represent the dominant allele for Huntington's and 'h' represent the recessive allele. The man is heterozygous (Hh) and the wife is homozygous recessive (hh) since she doesn't have the disease.

Step2: Set up Punnett - square

The possible gametes from the man are H and h, and from the wife are all h. The Punnett - square will have two columns (gametes from the man) and two rows (gametes from the wife).

Step3: Fill in Punnett - square

	H	h
h	Hh	hh

Step4: Calculate probability

Out of the four possible genotypes of the offspring in the Punnett - square, 2 are Hh (have Huntington's) and 2 are hh (do not have Huntington's). The probability of having Huntington's (Hh genotype) is $\frac{2}{4}=0.5$ or 50%.

Answer:

50% chance