Question

1. What is the risk of Fabry disease in Mary's unborn child (IV.2 in the pedigree), provided that her nephew (IV.1) does have Fabry disease, and how is it calculated? 2 II III IV 1 A. The risk is 1/8: 7 that III.7 is a heterozygous carrier, reduced by 50% to the next generation, and with a 50% chance of a boy = /*/*/ -1/8 B The risk is 1/4: 7 that III.7 is a heterozygous carrier, reduced by 50% to the next generation = ²/- % C The risk is 1/6: 2/3 that III.7 is a heterozygous carrier, reduced by 50% to the next generation, and with a 50% chance of a boy - 2/3*\*% - 1/6 D. The risk is 1/16: 7 that II.4 is a heterozygous carrier, reduced by 50% for each of the two generations, and with a 50% chance of a boy = y*%**%-1/16

Answer 1

Answers

A

Explanation

Given pedigree is X-linked recessive inheritance.

Now, we need to assign the genotypes to the pedigree. It is the X linked recessive inheritance. Therefore, to get 3rd males of II generation. The parent's genotype would be XcX (carrier female) and XY (normal male). Male 1st of IV generation is affected therefore, 4th female of II generation must be carrier (XcX). Accordingly, the genotypes of the pedigree can be assigned.

0 2) (TV 【X ] GO 2 国人又 2xx Xcx Q Q 口区 G口以作 XX (S g XY T - 人X 又又 心 X口 2 I 人X 0 XX

Now let us concentrate genotype of female 7th of III generation. This female can be either XX or XcX. The disease will be transmitted only through XcX genotype of the female 7th of III generation. Therefore, genotype of this female will be XcX. Now in the mating of 7th female of III generation and 8th male of III generation, only males can be affected. Sex of the 2th of IV generation is not declared. Therefore there are 1/2 change to be male. Now calculate the frequency of the male affected in the cross between XcX (7th of III generation) and XY (8th of III generation).

Xclx x xx/xx rperlyy. XcX » No risk (Nama) XX - Normal XcY=> infected (1/4) xy =) Nürnal

Only 1/4 probability for the infection to male. Therefore final calculation is 1/4* (probability of male(1/2)). Hence the answer is 1/8.

Explanation for the wrong answer

B. As sex is unspecified in the 2th of IV generation. Therefore, the probability of having male (infected sex) must be considered.

C. III 7th is not infected therefore we can not consider its probability in the calculation.

D. II 4th is not infected therefore we can not consider its risk in the calculations.