In the family tree shown below, both P and Q are carriers of a recessive allele which causes a condition. Only individuals R and X have the condition. What is the percentage likelihood of S, T and U each being a carrier?

Answer:

17) E

Let the recessive allele be ‘t’

We know P and Q have the recessive allele. Since the condition is caused by recessive allele, the person must have be homozygous recessive. Therefore, the person with the condition must have the genotype ‘tt’

However, we are told that only persons R and X have the condition, meaning that P and Q must not be homozygous recessive. Hence, P and Q must be heterozygous. Their genotype must be ‘Tt’

          T                                t

T TT(25%) Tt (50%)

t Tt (50%) tt (25%)

We are asked to find out the percentage probability of S, T and U being carriers of the recessive allele. They cannot have the genotype ‘tt’ as we know that they do not have the condition themselves.

Ideally, we would think that all three of them have 50% probability of being a carrier.

However, probability of U being a carrier has to be 100%. Since X has the condition, U must definitely have the recessive allele.

Probability of S and T carrying the recessive allele is 50%.

P(S) = 50%
P(T) = 50%
P(U) = 100%

Hence, the answer is E.