Albinism in humans is known to be controlled by the autosomal recessive gene A

Because the causative gene for albinism is controlled by the autosomal recessive gene A, the female patient's genotype is AA and her gametes are only A. The child's genes come from both parents, so one of the female's gametes A must be passed on to the offspring, so whether it is a daughter or a son, her genotype must include A.

Whether the offspring is sick or not depends on what the father's genotype is, because the father is not sick, so the father's genotype is AA or AA. If the father is AA, the offspring genotype may be AA (not diseased) or AA (diseased); If the father is AA, the offspring genotype is only AA.

To sum up, the genotypes of both of them can only be AA and AA.

Males are normal, so the genotype may be AA or AA, while females are diseased and AA, and the offspring born are not diseased, according to the law of gene segregation and free combination, only AA can be produced, not AA

It can only be AA and AA, because males (normal) are AAs, and females (diseased) are AAs, that is, women do not have dominant A-genes, so AAs do not appear, and the result is AA and AA. Hope it helps.

Because the female patient has the genotype AA, if the offspring are normal, they must be heterozygous, because the female patient must pass on her A gene to her children.

Because their mother, as a patient, must be AA, and the genes she provides must be A, so the child must have A.

Since it is controlled by the recessive gene A, and the female is the diseaser, the female gene must be AA

If the wife's younger brother is albino, the younger brother is AA and both parents are AA, and the wife's genotype may be 1 3 AA or 2 3 AA. The husband's mother is AA, and if the husband is normal, he must be AA.

Only when the wife is 2 3aa, can give birth to a sick child, aa and aa have a sick child probability of 1 4, then the couple has a sick child: 1 4 2 3 = 1 6, then the probability of the child is not sick 1-1 6 = 5 6.

I hope it helps you, if you don't understand, you can ask me again.

There are a lot of situations in this question, so you might as well think about it backwards, calculate the probability of being born with a sick person, and then subtract it by 1.

If both parents are normal and sick, both parents must be heterozygous AA.

The husband's mother is sick AA, so the husband must be AA;

The wife's younger brother is sick AA, and the wife is not sick, so the wife's parents must be biheterous AA, and the wife may be AA (2 3) or AA (1 3);

When a wife and husband are a hybrid AA, the probability of having a sick child is 1 4.

Then the probability of a sick child is 1 (the husband is AA) times 2 3 (the wife is AA) multiplied by 1 4 (the sick child is double heterozygous) = 1 6

Then the probability of the child's normal = 1-1 6 = 5 6

Choose d 5 6

The husband's mother is albinism, it can be determined that the husband's gene is AA, and the mother's younger brother is albinism, indicating that the mother's mother and father are AA and AA, so the probability that the mother is a carrier is 2 3, if you want the child to have a normal complexion, it is 1-abnormal probability, then if it is not normal, it is the probability of aa, the husband is aa, the mother is 2 3aa, then if you want AA to be born, the probability is 1 2 times 2 3 times 1 2 is 1 6

1-1 6 is the normal probability of 5 6 . o Pick d I've been out of high school for 3 years and I can still remember playing like this haha.

The husband is AA. Two-thirds of wives are AA and one-third are AA. The child's normal 1 minus (2/3 times 1/4) equals 5 6

Can you read it?

According to the information in the stem, the genotic of the deduced father is AADD, and the genotype of the mother is AADD. Depending on the genotype of the parent, it's easy to do the following four questions!

1) Normal may be 1 8

2) The possibility of suffering from only one disease 1 2

3) The possibility of suffering from two diseases at the same time 3 8

4) Likelihood of illness 7 8

Answer D analysis: The traits of organisms are controlled by genes, and genes are divided into dominant and recessive; When there is a pair of genes that control a certain trait in the cell, one is dominant and the other is recessive, only the trait controlled by the dominant gene will manifest itself; Recessive traits are exhibited when the genes that control a trait are recessive genes

Answer: Albinism is a genetic disease controlled by recessive genes on autosomes, so the genetic composition of skin color control of patients with albinism must be recessive genes, and people with normal skin color may be carriers of albinism genes

a. If both parents have albinism, the genes that control skin disorder and withering of their children are also a pair of recessive genes, that is, they are also albinos and cannot be normal people; Wu is not in line with the topic

b. If both parents are carriers of the albinism gene, they behave as normal people, but their children have a 25% chance of developing albinism; Therefore, it does not fit the topic

c. If one of the parents is a normal person, and the pair of genes that control skin color are dominant genes, and the other parent is an albinism patient, the children born in this case are normal, but they are all carriers of albinism genes; Therefore, it does not fit the topic

d. If both parents are carriers of albinism genes and appear to be normal people, but the children born to them are 1 4 likely to have albinism; In this case, it may be that "the child is sick, and the parents are normal people", so it is in line with the topic

Comment: This question examines the dominant recessiveness of genes and their relationship with trait performance; Traits controlled by recessive genes do not manifest in the presence of dominant genes

If polydactyly is controlled by the B gene and albinism is controlled by the A gene, then the normal genotypes of the father syndactyly and the mother are A b and A BB respectively, and they know that they have a child with albinism but normal fingers (AABB), so the couple's genotypes are: AAB and ABB, so the probability of having another child without disease = 34 1

2 38, so the probability of having another sick child is 58

Therefore, d

(3) Probability of suffering from only one disease = 1 - (probability of suffering from both diseases + probability of not suffering from both diseases) Because the father has syndactyly and the mother is normal, they have a child with albinism (BB) but normal fingers (aa), so the genotypes of the parents can be deduced are ABB and AABB respectively, so the probability of both diseases is 1 2 times 1 4 = 1 8

The probability of normal fingers is = 1 2, and the probability of non-albino is 3 4Therefore, the probability of not suffering from both diseases is 1 2 times 3 4 =

So the probability of suffering from only one disease is 1-1 8-3 8=1 2 The prevalence of the last child = 1-normal rate = 1-3 8 = 5 8 I hope it can help you, satisfied.

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The probability of having a child born to the couple with normal skin color is 5 to 6.

Analysis: If the husband's mother is an albinist, then the husband's genotype is AA, the wife's younger brother is an albinism, the probability that the wife is AA is 2 3, and the probability that the next generation is albinism is 1*2 3*1 4=1 6, then the probability of the next generation is normal is 5 6.

Because the wife's younger brother is an albinist, the wife's younger brother's parents, who are also grandparents, must be AA to have children with albinism.

Analysis: 1. The husband's mother and the wife's younger brother are albinism, and it is inferred that the husband is a carrier of the albinism gene, and his genotype is AA.

The wife's younger brother is an albinist with both the mother and the mother of the albinism gene, i.e. AA and AA. The wife herself has a normal phenotype, so the wife's genotype is: 1 3aa or 2 3aa.

Since the husband's genes are AA, the wife's genotype is: 1 3AA or 2 3AA. They give birth to patients with albinism only AA and 2 3 AA.

The odds of a person with albinism are 2 3 1 4 1 6. The probability of a child's normal is 1 1 6 5 6.

2. Because the wife's younger brother is an albinist patient, it is inferred that the wife's father and mother are carriers of the albinism gene, that is, AA and AA. That is, his grandparents' genotypes are AA and AA, and his grandfather cannot be AA.

The father is definitely AA; The mother is AA probability 2 3, AA probability 1 3 mother is AA child can be albinism, and the probability of albinism is 2 3 * 1 4 = 1 6, that is, the probability of normal skin color is 5 6

Also, if the grandparents are AA or AA, then the wife's younger brother will not have AA and will not be able to become an albinist.

Over, I hope you understand.

Because my grandparents gave birth to a son with albinism, i.e. AA, according to Mendel's law of separation, you will get two genotypes, AA and AA, and the phenotypes of both genotypes are normal, and you will not be able to give birth to a son with albinism.

If the wife's skin color is normal, then her genotype is A?His younger brother is an albinist and has the genotype AA

If the phenotype of the parents is normal, then the genotype of both parents is A? a?

Such a map:

a? x a?

a?AA is complete, and the genotype of my grandparents is AA AA

In this case, the wife's brother cannot be an albinist.

If the genotype of my grandfather and grandmother is AA AA, then how can my wife's younger brother be an albinist?

Answer] B Answer Analysis] Test Question Analysis: Albinism is a familial hereditary disease, which is autosomal recessive inheritance, and often occurs in people who are close relatives who marry.

b. If the husband and wife are carriers of the recessive pathogenic gene of albinism, although the husband and wife are normal, their children can be sick, so it is in line with the topic;

d. If one of the spouses is sick and the other is not a carrier of the albinism gene, the children are normal and therefore do not meet the topic;

Therefore, choose B. Test Center: This question examines the transmission of genes between parents and children.