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The pedigree shows the occurrence of a X linked recessive mendelian disease in a human family. Two unaffected family members
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Answer #1

a) In this question, the disease is said to be X-linked recessive, meaning two diseased alleles are needed on the X-chromosome. Now, both the women gentypes for the microsatellite has M1/M1 (homozygous condition) and M1/M2 (heterozygous condition) without the disease. So it is obvious that M1 loci when present in homozygous or heterozygous is not responsible for the disease. The disease microsatellite is M2. The genotype of IV-4 is M2/Y (since he is a man having only one X chromosome)

b) The disease and the microsatellite loci are 10cM apart. So recombination frequency between them in 10.% thus the disease loci and microsatellite loci are linked. IV-4 has unaffected parents but the mother is heterozygous. Let us consider the Punnet square:

M1 M2
M1 M1M1 M1M2
Y M1Y M2Y

So the two possible genotypes is M1Y and M2Y and the possible genotype must be M2Y and the disease loci and microsatellite loci is linked.

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