Biology Genetics - Inheritance Question

Question

assume that ischemic heart disease is conveyed by homozygous recessive alleles, hh, at a single locus. Individuals with at least one copy of the dominant allele H have normal blood vessels. Dwayne and Lekesia who plan to start a family: Dwayne was married once before, and he and his first wife had a child who has ischemic heart disease. Lekesia’s brother died from complications associated with ischemic heart disease however Dwayne and Lakesia were found to be disease free. 1. What is the probability that Dwayne and Lakesia will have a child with severe hypercholesterolemia? 2. Dwayne and Lakesia plan to have 3 children. What is the likelihood that they would have 2 normal girls and a boy with ischemic heart disease? Answer should be shown as a percent.

Answer 1

First, we determine the possible genotypes of Dwayne and Lakesia.

Dwayne: Dwayne does not have ischemic heart disease. However, he had a child which did. This means the genotype of the child is hh. This means that Dwayne contributed one h allele to his child. However, since he is healthy, we conclude that he has one dominant H allele and his genotype is Hh.

Lakesia: Lakesia does not have ischemic heart disease. However, her brother did, which means his genotype was hh. This is only possible if he was the product of an HhxHh cross. This means that both of his parents (and Lakesia's parents) are Hh. There are three possible genotypes in a monohybrid cross (HhxHh). These genotypes are HH:Hh:hh and they exist in a 1:2:1 ratio. Since Lakesia is healthy, we conclude her genotype is not hh. There is a 1/3 probability that she is HH and a 2/3 probability that she is Hh.

Next we determine the probability that they produce a child with ischemic heart disease. This means that we need to determine the probability that the child is hh. In order for this to happen, both of his parents must be Hh. In a monohybrid cross, the probability of being homozygous recessive (hh) is 1/4. However, this probability is only possible if we KNOW that both parents are Hh. It is possible that Lakesia is not Hh. So the probability of the child being hh is (1/4)*(2/3)=2/12=1/6. We use 1/4 because this is the probability that both parents contribute an h allele. We use 2/3 because this is the probability that Lakesia has an h allele to give.

Lastly, we attack question 2 which asks us to determine the probability that Dwayne and Lakesia have 2 normal daughters and one affected son. We will look at the probability of each event individually.

Event 1: 1 Normal Daughter: (1/2)*(5/6)= 5/12. (1/2)= probability of having a girl and (5/6)= probability of having an unaffected child. This value is derived from the previous question. Probabilities sum to 1, and if the probability of having an affected child is 1/6, having an unaffected child is 5/6.

Event 2: Same as Event 1

Event 3: Affected son: (1/2)*(1/6)=1/12

Multiply the probability of all three events and convert to percentage.