Question

Assuming that everyone is married into the pedigree is either heterozygous or homozygous, what is the probability that the first child of the couple A and B will have the disease?

Lo

Answer 1

Lets cosider the parentage of A and B seperately.

The first cross in the parentage of A produces 3 progenies (2 male and a female) in which a male isdiseased.Here the parents are not diseased and a male offspring shows disase and one male does'nt have disease.From this we can assess that it is an X-linked recessive trait (also no disease in female parent). The female progeny in first cross may be a carrier or a normal. In the second cross, that female parent may be carrier or a normal.But in the case of third cross in A section, a disease free male parent is crossed with female which may be a carrier or normal.so the progenies of the third coss in left are females may be carries or normal. So A may be a carrier or normal.

In considering the parentage of B,In the first cross the male is normal and the female may be carrier or normal.The secnd cross is with a normal male with n female which may be normal or carrier.The mother in secoond case is a carrier because there is a diseased female in the progenies.In the third cross the male is normall (Because X- linked recessive) and female may be normal or carrier. And B is normal.

When cross between A and B is considered,there is two possibilities that A may be normal or carrier.

If A is a carrier there is a possibility of 4 progenies(2 males and 2 females) and one male progency should be diseased .

If A is normal there is a chance of two progenies which are normal.

So out of this 6 chances, only one chance for disease.

Therefore probablity for diseased is 1/6.
hx hx hx xx/xx XX

ХХК XX

Ах в ТВ - ху A xdx А х В ху xdx ХХ ху diseased 4 A X X А х В total xx хуу So , one dreased XX XY No dixeosed. -