100 points, biology question an autosomal recessive gene b controlled hereditary disease,

If two normal parents have a daughter and a normal son, what is the probability that the son will carry the recessive gene?

Answer] 2 3

Analysis] The parents are normal, and a daughter is sick, then the disease is a recessive genetic disease, both parents are carriers AA AA--- the genotype of a normal son is: AA and AA two possibilities: AA is 1 3, AA is 2 3 (at this time, there is no need to consider AA, because the son has been told to be normal).

If the normal son is married to a normal woman, and the first child they have has the disease, what is the probability that the second child will also have the disease?

Answer] 1 4

Analysis] Since the first child born to a normal son and a normal woman is sick, it can be deduced that the son's genotype is AA (unrelated to the inference in the previous question), and the genotype of the normal woman is also AA, then.

The probability of AA AA --- sick child (AA) is 1 4

Note: Whether the first child is sick or not does not affect the second child).

If the son is married to a normal woman and the woman's brother has the disease, what is the probability that they will have one child with the disease?

Answer] 1 9

Analysis] The probability that the son is a carrier (aa) is 2 3, and since his wife's parents are carriers (which can be inferred from his brother-in-law's illness), the probability that his wife is a carrier (aa) is also 2 3.

aa aa - probability of sick child (aa) = 2 3 2 3 1 4 = 1 9

If (3) after marriage, the first two children are sick, then what is the probability that the third child is normal?

Answer] 3 4

Analysis] At this time, it can be inferred that both the son and the woman must be AA according to the meaning of the title.

aa aa - the probability of a sick child (aa) = 1 4, then the probability of having a normal child is: 1 - 1 4 = 3 4

Note: The illness of the first two children does not affect the third child).

By the mantra: "Something out of nothing is hidden, and giving birth to a girl is often hidden." It is known to be an autosomal recessive disorder. The genotypes of both parents are AA and AA.

They gave birth to a normal son, and his genotype is possible in two ways, as well as the proportions 1 3aa, 2 3aa [note: since it has been stated that the performance is normal, the proportion can only be found in the normal performance]. Therefore, the probability that this son carries this recessive gene is 2 3.

If the normal son is married to a normal woman, and the first child they have has the disease, it means that both husband and wife are carriers, i.e. heterozygous. The probability that the second child will also have this disease is 1 4.

Note: This question is a different assumption from the second question and cannot be linked] If the son is married to a normal woman, and the woman's brother has this disease, it means that both parents of the woman are carriers, i.e., heterozygous, and in the same way, [explained by the first question] the probability that the woman is a carrier is 2 3aa, and the probability that this son is a carrier is also 2 3aa. Then, the probability of one of their children suffering from this disease = 2 3 * 1 2 * 2 3 * 1 2 = 1 9

If, (3) after marriage, the first two children are sick, indicating that both husband and wife are carriers, i.e., heterozygous, then the probability that the third child is normal is 3 4

First of all, the parents' genes are judged to be bb, and the daughter's genes are bb

1. The probability of the gene has nothing to do with whether the person has been born with the disease, so the son gene is bb or bb bb (the latter two are carriers) The probability of carrying 2 3

2. There is a child who is sick, so the son and the woman have the genes of bb bb and the second child has the genes of bb bb bb bb

Probability of illness 1 4

3. The woman's brother is sick, so the woman's parents have a gene of bb bb, and the probability of the woman carrying it is also 2 3 as 1 says

When both are carried, there is a 1 4 probability that the child will be sick, so 2 3 * 2 3 * 1 4 1 9

4. As mentioned earlier, it has nothing to do with whether or not a sick person has been born, but a sick person has been born, so it is determined that two people are carriers, so if 2, the probability of being sick is 1 4, and the normal probability is 3 4

pbbxbb

f1bb(1/4)

bb(1/2)

bb(1/4)

1.The probability that a son carries a recessive gene is (1 2) (1 4 + 1 2) = 2 32"Making something out of nothing" can introduce the genotype of a normal son and a normal woman as bbxbb, then the probability of having a second child with this disease is 1 4.

3.The probability that the normal woman is bb is 2 3, because the probability that the son's genotype is bb is 2 3

It can be concluded that the probability of having a child with the disease is: (2 3) (2 3) (1 4) 1 9

4 From the illness of the two children, it can be seen that this son and this normal woman are both recessive gene carriers, that is, they are both BB, and the normal probability of the third child is 3 4

Two normal parents have a daughter and a normal son, then the daughter bb boy bb or bb parents have at least one bb then bb 1 3 bb 1 2 so be it too few points The idea is written in thousands of words It is very troublesome to discuss the situation separately.

Settle -- I just found the idea to solve the problem.

And I'm only going to do the first two questions, because I've memorized it, and I'm crazy.

I haven't studied biology in years.

I just came over and took a look.

In fact. = =.I'm just coming to take a look. You're a fast learner.

Wow, that's a complicated question.

B is correct, the mother has no dominant gene, the son can only be X(D)ya, it is not certain whether the mother is diseased, as long as the mother has the dominant gene, it is not necessarily diseased C, and the maternal grandfather's recessive gene may not necessarily be passed from the mother to the son (same as a) D, (same as a).

b The mother is sick and the son must be sick Characteristics The woman must suffer from the man, and the father and son of the female patient must be sick Because if a girl is a patient, her genotype must be XAXA, and one of her two sex chromosomes must be from her father, so her father's gene must be XAY, and it must be a patient. The son of this woman must have received an X chromosome from her, (the father of the son can only give the son Y), so the son must be sick.

I choose BThe mother is diseased, the genotype is XBXB, and only one type of egg cell is produced, which is XB

The son's X chromosome is from the mother, the Y chromosome is from the father, and the genotype is XBY, so he will be sick.

Assuming that the disease is inherited dominantly with X, the disease on sex chromosome X of the second generation of No. 4 must have been passed on to his daughter, and his daughter in the figure does not have this disease, so B is not correct.

If it is with X dominant, Instance 3 in 2 will not be affected.

The X chromosome is dominant, which means that the probability of having a chromosome directly determines the chance of disease, and the female has 2 X chromosomes and a male one, and the female has 2 times the chance of being male.

Suppose the genetic disease is controlled by a pair of alleles a and a, according to the title, aa = 181, so a = 1

9, a=89 so aa:aa=89 8

9 = 4:1, so the probability of heterozygous AA in normal couples is 15 and the probability of disease in children is 15 1

Therefore, b

No. 9 is diseased, indicating that the genotype of the parents is BB and the genotype of the 8 is not sick, then the genotype may be BB, BB, according to the fact that the parents are not diseased, the probability of their genotype is 2 3, when they marry a sick (BB) woman (2 3BBXBB), the probability of the child born is 2 3x1 2=1 3 regardless of male and female

You may struggle with the question of a boy being born, because it is already clear that a boy is born, so you don't have to think about multiplying the sex by 1 2, and if you change the question to "What is the probability that the child they will have will be a sick boy?" the answer is 1 in 6.

If you have any questions, you can continue to ask.

The following figure is the genetic pedigree of a certain family of human genetic diseases A and B genetic diseases (let the A genetic disease and a and a and a so the nail disease have a middle birth, and the mother is diseased, and the son is not diseased, it is an autosomal dominant genetic disease.

The genetic disease is obviously controlled by recessive genes, the 9th disease indicates that the genes of 3 and 4 are obviously bb, and the gene types of children 3 and 4 are bb, bb, bb, bb four kinds, the 8th is not diseased, and the bb is excluded, and the 8th is because the probability of bb is 1 3, the probability of being bb is 2 3, and the probability of marrying the sick woman on the 8th is 2 3 times 1 2 (the probability of giving birth to a boy) is equal to 1 3

6- Patient aa probability 2 3 aa probability 1 3, should be said to be a boy in the title, so 2 3 times 1 2 equals 1 3