Heredity Simple Calculation 20, Genetic Simple Calculation

Fig? Anyway, I'll give you the news directly.

First, albinism is an autosomal recessive disorder. If you don't remember, you can also see from the figures 125 and 349 that there is something hidden out of nothing; It is both recessive, and 125 the parents are not sick, but the daughter is sick, so the parents are carriers; If it is assumed to be inherited by companions, it can only be companion X, but if it is companion X, the father must be sick, so the hypothesis is not valid; Conversely, the autosomal hypothesis holds. The first two blanks can be filled, which are implicit aa

Thirdly, according to the fact that 6's parents are carriers and 6 are not diseased, 6 may be carriers or homozygous, but 7 is diseased, so the genotype of the second child may be AA or AA, not necessarily diseased.

Fourth, 10 and 11 are collateral blood relatives within three generations, and if they are married, they are consanguineous marriages.

Fifth, since the paternal genotype of 11 is AA, and 11 itself is disease-free, it means that its genotype is AA; 10 parents are not sick but have brothers who are diseased, indicating that their parents are both aa genotypes, and 10 itself is not diseased, then 10 genotype cannot be AA, the probability of being AA is 1 3, and the probability of being AA is 2 3. Thus, the disease in children of 10 and 11 is only possible if the genotype is AA when 10 is AA, and the probability is 2 3 * 1 4 = 1 6.

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The prevalence of random mating offspring is that the frequency of gene A is the first to be, the frequency of gene A is , the frequency of genotype AA is 2, and the normal frequency of aa in the mating offspring of this population is .

Hemophilia is a recessive genetic disease with X, and the genotype of male patients is XAY, and the genotype of normal men is XAY.

According to the information in the question, the gene frequency of Xa is 1 5000, and the gene frequency of Xa is 4999 5000

Therefore, according to the genetic equilibrium formula, the probability of XAXA in the female population is (1 5000)*(1 5000)=1 250000000

The parent is aaxby aaxbxb, and the two pairs of sexual scum commas can be separated for analysis.

The proportion of offspring producing hairy dominance is 3 4

There are two kinds of xby xby, in which the broad-leaved ones account for 1 2, and the proportion of hairy broad-leaved males in the generation is 3 4*1 2=3 8

Is the answer 1 12?

A normal man and a normal woman can only have a sick child if they are all AA genotypes

aa×aa aa×aa

1 2) aa aa (2 3) aa aa aa (sister).

1 1 2 1 (normal male) aa aa (normal female).

aa aa(1/2) aa

p=1/2×2/3×1/2=1/12

c The probability that the man carries the recessive pathogenic gene is 1 in 2, and the probability of the woman carrying the recessive pathogenic gene is 2/3, and the probability is calculated as 1 in 2 * 2 thirds * 1 in 4 = 1 in 12.

1 The frequency is 198 10000, and the probability of a normal person being a carrier is 198 9999. The probability that this person's brother is a patient and this person is a carrier is 2 3. The probability of disease in offspring is 2 3x198 9999x1 4=1 303