Are there any weird genetic cases or questions for the test ?

Abstract Non-hereditary variation: Variation caused by the influence of the external environment, the genetic material has not changed, and cannot be passed on to offspring. For example, cats and dogs have acquired tail docking, and their offspring are just as healthy and have tails.

Non-hereditary variation: Variation caused by the influence of the external environment, the genetic material has not changed, and cannot be passed on to future generations. For example, cats and dogs have acquired tail docking, and their offspring are just as healthy and have tails.

Hehe, I'm a little proud, I've graduated from high school, and I'm selling well, and it's very interesting to give you a question.

And the problem here is, you have to figure out the concept of heterozygotes, and heterozygotes mean that they are heterozygous, that is, whether it's sexually dyed, chromosomes, or autosomes, if one is heterozygous, then it's heterozygous, and you've calculated the probability that both are heterozygous, so you get the idea? Hehe.

In a couple with normal skin color, the husband's mother and the wife's brother are albinos, and the rest of the family members are normal, the probability of having a child with normal skin color is: 5 6

Correct Answer Analysis:

Albinism is inherited recessively on the autosomal basis. According to the title, it can be seen that the husband's mother's genotype is AA, and the husband's genotype is AA; The genotype of the wife's younger brother is AA, and it can be seen that the genotype of the wife's parents is AA, and the wife's phenotype is normal, so there are two possibilities for her genotype, which are 1 3AA and 2 3AA.

Based on the above analysis, it can be seen that the probability of the child born to the couple having a normal skin color is 1 1 3 + 1 2 3 3 4 5 6

Albinism is often hidden, the husband's mother is AA, the husband is AA, and the wife's younger brother is AA It can be seen that the wife's parents are AA, and the wife is normal, so 1 3AA, 2 3AA, the probability that the child born to the couple has a normal skin color is 1 2 3 1 2 1 2 5 6

Well! The husband's mother is albinist, AA. Then this husband must be AA.

The wife's younger brother is an albinist, indicating that both her parents are AA. Then she's 1 2 . (She can't be AA) Then the probability that their son has a normal skin color is 1 3+1 2 2 3=2 3 That's it!!

Albino is often hidden, paternal gene behavior AA, maternal genotype is 1 3 AA, 2 3 AA, so is 5 6

The husband's mother is albino, AA. Then this husband must be AA. The wife's younger brother is an albinist, which means that her parents are AA (not mentioning that her parents have albinism means that there is no albinism by default).

Then she is 1 2, she can't be AA, she doesn't have albinism) The probability that their son has a normal skin color is 1 3 + 1 2 2 3 = 2 3

The conditions are not very specific, and here he does not specifically mention that the phenotype is normal.

If a male patient with an autosomal recessive disease is AA, his parents should be AA and AA, and if his sister is normal, there is a 1 3 probability that it is AA and a 2 3 probability that it is AA. The probability of his brother-in-law is unknown, but generally in this kind of topic, those who are not from the family of this genetic disease should be counted as AA, and the same is true for his wife.

Therefore, his son is AA (AA*AA), and the probability of the son's aunt and cousin AA is 2 3*1 2=1 3 (only half of the male patient's sister is AA when he has AA). The incidence probability of the child after the son and the aunt and cousin get married is 1 3 * 1 4 = 1 12 (when the aunt and cousin are AA, there is a 1 4 probability of giving birth to diseased offspring).

Premise: Individuals who are not specified are of a normal phenotype.

Therefore, both parents in the male patient (AA) have the genotype AA

The male patient's sister is 2 3 and his daughter may be 2 3 1 2 = 1 3

The male patient's son was 100% genotype

Therefore, the incidence probability of their children after marriage is 1 4 1 3 = 1 12

Choose method b, b accounts for b.

Then bb is 2*