A high school genetic probability problem

(1) Out of thin air, girls' diseases are often hidden. It can be seen from this formula that galactosemia is an autosomal recessive genetic disease.

Color blindness is a single gene X recessive genetic disease (this must be remembered). Suppose that galactosemia alleles are A and A, and color blindness alleles are B and B.

Then the genotype of 12 is A-XBY, that is, it is only necessary to infer whether the galactosemia genotype of 12 is heterozygous or homozygous.

3(Aa) 4(Aa)

↘↙

8 (2/3Aa, 1/3AA) Because 8 is not sick, the genotype of 8 must be A-, and the genotype of 1 1 is also (2/3Aa, 1/3AA).

1(AA) 2(Aa or AA)

↘↙

7(Aa)

So all genotypes of 12 are 2/3Aa×Aa×? = 1/6AA，2/3Aa×Aa×？ = 1/6aa，2/3Aa×Aa×？ =2/6Aa，

1/3AA×Aa×？ = 1/6AA， 1/3AA×Aa×？ = 1/6Aa. Because 12 is not sick, we should remove all of 1/6aa, that is, 1/5aa, 1/5aa.

1 1, 12 The probability of getting sick after giving birth is ﹙ 2/5aa ﹢ 1/5aa ﹚× 2/3aa×? = 1/ 10

(2) The genotype of 5 is aaxby, and the genotype of 6 is AaXBX-. It can be seen that the genotypes of 1 1 are 2/3aaaxbxbb and1/3aaaxbb respectively. According to (1), all genotypes of 12 are 2/5aaaxby and 3/5aaaxby.

1 1, 12 The probability of giving birth to a child without galactosemia is:1-10 = 9/10.

Born in11(xbxbxbb) ×12 (xby)

↘↙

﹙? XBXb，？ XBXB，？ XBY，？ XbY﹚

It is known that the probability that 1 1 and 12 have children without color blindness is: 1-? =?

To sum up, the probability that 1 1 and 12 have children without getting sick is 9/ 10×? =27/40

It took me a day to come up with it, and I'd love to ~ ~ ~