Question

a. Determine the most probable mode of inheritance: autosomal or sex-linked, dominant or recessive, of the disorders in the t
b. If individual IV-1 from the first pedigree marries an affected woman, what is the probability that their first child is af
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Answer #1

Qn b.

Mode of inheritance in First pedigree is Autosomal recessive .

Because all the affected individuals have phenotypically normal parents. It means both of them are carrier of that trait . So it is recessive trait. Also there is no gender biasness . It is autosomal trait.

Genotype of affected individual = aa

Genotype of phenotypically normal individual =Aa/AA

In generation IV , 2 individuals IV2 and IV4 are affected. So their genotype is 'aa' . Both of their parents are phenotypically normal, they must be a carrier. So genotype of their parents 'Aa' and 'Aa'.

Individual IV1 marries an affected woman(aa). To produce an affected child, the individual IV1 have to be a Carrier (Aa). If his genotype is AA then all children will be phenotypically normal.

So the chance of IV1 to be a carrier is 1/2

Both parents are recessive, Aa x Aa A AA alAa) The chance of being a carries (Aa) = 2 2 = 1

The Child of IV1 and an affected woman would be disordered, if IV1 transmit his recessive allele to his offspring .

Genotype of IV1 is Aa. He can produce 2 type of gametes either 'A' or 'a'. So the chance of transmitting allele 'a' is 1/2.

So the Probability of the child is affected by the disorder = 1/2 × 1/2 = 1/4 or 25%.

Qn c.

The second pedigree is a Sex-linked recessive trait.

Affected offsprings have phenotypically normal parents and affected parents have phenotypically normal offsprings. Affected father(II5) has no affected offsprings (Gen III) while affected mother (III11) has all male affected offsprings . Gender biasness is also present.

Individual IV3 is phenotypically normal. Her father is normal (XY).Her mother(XXc) and all brothers (XcY) are affected.

She have to be a carrier to produce affected offspring. So the Probability of individual IV3 will be a carrier female is 1/2.

xx xx -xx x. Ах

If she marries a normal man (XY) then the chance of their first child to be affected son is 1/2 × 1/2 =1/4

(Affected) XX X X X

If they have a second son , then the chance that he is affected is 1/2 .

Because the first affected Male child confirms that the mother is a carrier. So the Probability of the mother (IV3) for being carrier will not considered here.

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