2010 Biology Compulsory 2 Review Syllabus

1.Start by determining the type of genetic disorder:

From II, there is no mesogenesis, and the nail disease is dominant, and the disease is inconsistent from II7 and III11, and the nail disease is not accompanied by X dominant, but autosomal dominant;

From II, out of nothing, it is known that B disease is recessive, and it is known that II3 is not a carrier of B disease genes, and B disease is known to be recessive with X.

2.Next, the III9 and III12 genotypes are determined

Since iii knows that ii genotypes are aaxby and aaxbxb, so iii9 genotype is 1 2aaxbxb or 1 2aaxbxb

From III11 to know that II genotypes are AAXBY and AAXB respectively, so III12 genotype is 1 3AAXBY or 2 3AAXBY.

3.Finally, calculate the probability:

The probability of disease A is 1 3 + 2 3 * 1 2 = 2 3, the probability of disease B is 1 8, and the probability of both diseases is 2 3 * 1 8 = 1 12, so the probability of disease is 2 3 + 1 8-1 12 = 17 24

Because black hair is dominant to brown hair, brown hair female B genotype is DD.

Because the black yak has 3 4 is heterozygous and 1 4 is homozygous. Therefore, the parent genotype of the black bull A has the following possible combinations:

dd + dd: The probability is 1 4 x 3 4 + 3 4 x 1 4 = 6 16;

dd + dd：1/4 x 1/4 =1/16；

dd + dd：3/4 x 3/4 = 9/16。

The probabilities of the genotype of the offspring of the various parent combinations are as follows:

dd + dd ——1/2 dd + 1/2 dd

dd + dd—— 1 dd

dd + dd ——1/3 dd + 2/3 dd

So the probability that the black haired male bull A genotype is DD is: 6 16 x 1 2 + 1 16 x 1 + 9 16 x 1 3 = 7 16;The probability of genotype DD is: 6 16 x 1 2 + 1 16 x 0 + 9 16 x 2 3 = 9 16.

If the black male A is DD, the probability of him giving birth to a brown cow is 0 with the brown female B.

If the black-haired male A is DD, the probability of giving birth to a brown-haired calf with the brown-haired female B is 1 2, and the probability of giving birth to a brown-haired cow is 1 4.

Therefore, the total probability of a black male A and a brown female B giving birth to a brown cow is 7 16 x 0 + 9 16 x 1 4 = 9 64.

I think you can take a look at this.

It was very detailed.