For problems 10-12, assume the genetic disonder indicated in the pedigree \" ?0·

Question

For problems 10-12, assume the genetic disonder indicated in the pedigree " ?0· , 55b- (right) is very rare and fully-penetrant. 10) what probable mode of interi ance is indicated by his pedigree, 5 111 11) III-3 and III-4 are about to have a sixth child (who will become individual "IV-6" in the pedigree). What is the probability that this child will be affected by the disorder? IV 12) Consider individual "II-I1" in the pedigree (ie, the oldest sibling o the father indicated in generation III) What is the probability that II-1 is a heterozygous carrier of the mutation? Problems 13 and 14 refer to the second pedigree diagram (right), which shows the inheritance of a rare, fully penetrant, X-linked recessive disorder called favism. 13) How many of the females indicated in the pedigree are "known" carriers (ie., virtually certain to be heterozygous for the favism gene/mutation)? 14) What is the probability that a child (indicated by the ") borm to the indicated couple in the 3" generation will have favism?

Explanation / Answer

10. The disorder is Autosomal recessive disorder( As both males and females are infected and the transfer is not from father to daughter or mother to son so it is not X-linked and recessive because the infected person have parents who are not infected which means they are heterogenous and the gene is recessive so not showing phenotype)

11. Now for the child to be infected, both the parents needs to be heterozygous ( lets say genotype Aa)

Table 1

Probability that the child will be infected that is have a genotype aa is 1/4= 0.25

12. The mother of III-1 being heterozygous has a probability of 0.5 as both her parents ( generation I) was heterozygous ( see table 1 Aa= 2/4=0.5)

Now III1's father also can be either homozygous(AA) or heterozygous (Aa).

Now two combinations are possible as one of the parent has to be heterozygous

Aa X AA or AA X Aa

In each cases,

the probability of III1 being heterogenous is 4/2= 0.5.

So total probability of the III1 individual being heterogenous is 0.5 X 0.5= 0.25

13. Only 1 female in the 2nd generation whose father was infected is certain to be a carrier. Other females have different probalities of being carrier depending upon the genotype of their parents.

A a A AA Aa a Aa aa

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