High School Biological Probability Calculation Questions, How to Calculate Biological Probability Ca

As shown in the figure, the female parents should be two heterozygous, so the ratio of AA to AA is 1:2, the male is AA, the probability of the child of these two people suffering from the disease controlled by A is 1 6, and H is inherited with sex, and the probability of disease is 1 2, so the probability of not suffering from both diseases is the product of the probability of not suffering from both diseases.

If I'm not mistaken, the questions should be incomplete.

This question should have a diagram.

It can be seen that the female has no symptoms of concomitant genetic diseases, but it does not mean that the probability of the two genotypes aaxhxh and aaxhxh is 1:1, but should be 1:2

It is calculated in two cases.

aaxhy × aaxhxh

The normal probability is (1 3) * 1 * (1 2) = 1 6aaxhy aaxhxh

The probability is (2 3) * (3 4) * (1 2) = 1 4 plus 5 12

We assume that AA is nail disease; XH is disease B.

Male aaxhy, female aaxhxh or aaxhxh, the probability of their offspring suffering from nail disease is 1 6 (here the probability of aa is calculated as 2 3), the probability of suffering from disease B is 1 2, the probability of offspring suffering from disease is: 1 6 + 1 2-1 6 * 1 2 = 7 12, then the probability of the offspring not suffering from disease is: 5 12.

1 (hypothetical gray body dominant gene A black body recessive gene A).

The gray body taken from F2 has (AA, AA, AA), then the frequency of the A gene is 2 3, and the frequency of the A gene is 1 3

In F3, AA type accounts for 2 3 2 3 = 4 9, AA type accounts for 1 3 1 3 = 1 9, and AA type accounts for 4 9

The gray body (aa, aa) accounts for a total of 8 9, and the black body (aa) accounts for 1 9, and the ratio is 8:1

2. The child has AA and AA

Then the parental genotype must be AA

The chance of giving birth to a genotype AA is 1 4, and the chance of AA is also 1 4 If it is required to have two such children in succession, it should be 1 4 1 4 = 1 16

I think the title is that it doesn't have to be in order, it's 1 8

Let's choose C for 3. In the case of inbreeding, without genetic mutations, it stands to reason that gene frequencies do not change

An autosomal recessive disorder occurs in 1% of the population", assuming that the patient with the disease is AA, the frequency of the AA genotype is 1, that is, Q2

1%, then, a gene frequency q = 1 10. "There is a well-behaved couple whose wife is a carrier of the autosomal gene and the color blindness gene that causes the disease", then the wife's genotype can be determined to be AAXBXB. Then the husband's genotype can only be either aaxby or aaxby.

p+q)2=p2+2pq+q2=1, it can be deduced that the gene frequency of a is q=1 10, then the gene frequency of a is p=9 10, the frequency of aa genotype is p2=81, and the frequency of aa genotype is 2pq

18%。Now the husband's performance is normal, so the possibility that his genotype is AA is ruled out. Thus, the probability that the husband's genotype is aaxby 2pq (p2+2pq) 2 11 leads to the result:

The probability that the wife's genotype is aaxbxb and the husband's genotype is aaxby 2 11 The probability that the offspring of a couple with aaxby genotype and aaxbxb will have both genetic diseases at the same time 1 4x1 4=1 16 and then add the probability of the husband's genotype aaxby 2 11, i.e., 1 16x2 11=1 88

I don't know if you can understand it, if you don't understand, you can ask me, it should be right.

This diagram means that there is no recessive (ee) at all, right?

1.It can be seen from the figure that B and B genes are located on autosomes, D and D genes are located on Z sex chromosomes, and two pairs of genes are located on non-homologous chromosomes, and their inheritance follows the law of free combination of genes.

Hybridization with bbzdzd can be used by gene segregation, bb bb, offspring b bb 3 4 1 4, zdw zdzd, offspring zdz zdw zdw 2 4 1 4 1 1 4, according to the suggestion that as long as there is a b gene, the trait is a white cocoon, and bb zdw is also a white cocoon, so the probability of a white cocoon in offspring is 3 4 1 1 1 4 1 4 13 16.

3.According to the title, the genotypes of the white cocoon are b zdz, b zdw, b zdzd, b zdw, bbzdzd, bbzdw, and the genotypes of the yellow cocoon are bbzdz and bbzdw, if you want to make the offspring have a white cocoon and a yellow cocoon ratio of 3 1, in addition to bbzdzd bbzdw, it can also be bbzdzd bbzdw

4.；If d is a recessive homozygous lethal gene, then the genotypes of the yellow cocoon moth have bbzdzd, bbzdzd, bbzdw, if a certain group of this moth starts with a male-to-female ratio of 1 1, b and b frequency ratio of 1 1, d and d frequency ratio 3 1, the gene frequency will not change after free mating of the population, then the genotype frequency is bb 1 4, bb 2 4, bb 1 4, that is, b bb 3 4 1 4, in female individuals zdw 3 4, zdw 1 4, in male individuals zdzd 9 16, zdzd 6 16, zdzd 1 16, that is, zdz zdzd 15 16 1 16, all offspring (including the deceased individual) have yellow cocoons: bbzdz (1 4) (15 16) 1 2 15 128, bbzdw (1 4) (3 4) 1 2 3 32, together is 27 128, there is no death in the moth that forms yellow cocoons, and the dead individuals are:

b zdzd (3 4) (1 16) 1 2 3 128, b zdw (3 4) (1 4) 1 2 3 32, bbzdzd (1 4) (1 16) 1 2 1 128, bbzdw (1 4) (1 4) 1 2 1 32, together is 20 128, and all surviving offspring accounted for (15 128) (1 20 128) 1 4.

Albinism is often hidden, and color blindness is x-hidden.

The woman's parents are normal, but she gave birth to a child with albinism, indicating that she is AA, AA is normal but has given birth to a son with color blindness, indicating that it is X(B)X(B), X(B)YIn summary, the mother is AAX(B)X(B), and the father is AAX(B)Y, and she herself is not sick, indicating that she has A and X(B).

The aa probability is 1 3 and the aa probability is 2 3

When she is AA, the child must not have albinism.

When she is AA, because her husband is AA, there is a probability of 1 2 left with a child who does not have albinism (AA).

That is, the probability that the child will not have albinism is 1 3 plus (2 3) (1 2) to get 2 3

x(b)x(b) probability is 1 2, x(b)x(b) probability is 1 2x(b)x(b) when the child must not be sick, and the probability of having a son is 1 2x(b)x(b) when the son is not sick if x(b)y, the probability is 1 4, so the total probability is the sum of (1 2) (1 2) and (1 2) (1 4), giving 3 8

So the probability is (2 3) (3 8) = 1 4

Suppose the causative gene for albinism is A, and color blindness is B.

The woman's younger brother is AAXBY, so she is a XBX. It can be inferred that the genotype of the father is AAXBY and that of the mother is AAXB. Then, as can be seen from the question, her husband's genotype is AAXBY.

Here, we should analyze the alleles one-to-one, which is a trick.

Albinism genes are analyzed first.

The possible genotype of the woman is 1 3aa or 2 3aa. The probability that his son will not have albinism is 1 3 2 3 1 2 = 2 3.

Then analyze the causative genes of color blindness. then her genotype probability is 1 2xbxb or 1 2xbxb.

The probability that his son will not suffer from the disease is 1 2

So the normal probability of its child is.

Genotype algorithm: albinism gene A color blindness XB; Both parents have normal sons with albinism and color blindness, so the genotypes of both parents: mother aaxbxb, father aaxby; The woman is normal then her genotype:

1/6aaxbxb ,1/6aaxbxb,1/3aaxbxb,1/3aaxbxb；Albinism color vision normal male genotype: aaxby, after marriage:

1 6aaxbxb is definitely normal and because it is a son, it is 1 12, 1 6aaxbxb in albinism is completely normal, sex chromosomes are normal 3 4, son 1 3 so normal son 1 24, 1 3aaxbxb in color vision is completely normal, albinism 1 2 son 1 2, so normal son 1 12, 1 3aaxbxb albinism 1 2,, color vision 3 4 son 1 3, so normal son 1 24;

General: Normal sons are: 1 4.

Gene probability algorithm: the female genotype is obtained from the above: 1 6aaxbxb, 1 6aaxbxb, 1 3aaxbxb, 1 3aaxbxb;

Because the male albinism gene: aa can be disregarded, a probability 2 3, a probability 1 3, xb probability 3 4: y probability 1 2 albinism normal 2 3, normal color vision 3 4, son 1 2, so normal son:

The idea is as follows, if you have any questions, please point it out:

Albinism is autosomal recessive, and if present, the genotype should be: AA; Color blindness is recessive on the X chromosome, so it can be concluded that the genotype of the girl's younger brother should be: AAXBY.

Therefore the father of the girl should be: aaxby; The mother is AAXBXB.

The paternal heredity should be: axb, ay, axb, ay; Mother's: axb, axb, axb, axb;

Because the girl is normal: her genes should be: aaxbxb (1 6), aaxbxb (1 6), aaxbxb (2 6), aaxbxb (2 6).

The genotype of the woman's husband in the question is certain, which is: aaxby, and the genetic children are: axb(1 2), ay(1 2).

Therefore: 1. When the woman is: aaxbxb, the genetic child is only: axb, and the probability of giving birth to a normal son at this time: 1x1 2=1 2;

2. When the woman is: aaxbxb, the genetic child is: axb(1 2), axb(1 2), and the probability of giving birth to a normal son: 1 2x1 2=1 4;

3. When the woman is: aaxbxb, the genetic child is axb(1 2), axb(1 2), and the probability of giving birth to a normal son: 1 2x1 2=1 4;

4. When the woman is: aaxbxb, the genetic child is axb (1 4), axb (1 4), axb (1 4), axb (1 4) The probability of giving birth to a normal son at this time: 1 4x1 2=1 8;

To sum up: if a woman marries a man with albinism and has a normal monochrome, the probability of having a normal son is: 1 6x1 2+1 6x1 4+2 6x1 4+1 8x2 6=1 4.

It seems to be too detailed. ）

Her parents are normal but have a brother with albinism, so her parents are AA (A is normal A albino), she herself is normal but not necessarily AA, it may also be AA, where the probability of AA is two-thirds, the probability of AA is one-third, her husband is albino, and the probability of giving birth to an albino child is two-thirds (when the female is AA) In addition, the girl's brother is colorblind XBY, indicating that the female may be one-half xbxb and one-half xbxb, and if you want not to be colorblind, you need to provide XB, so the probability is three-quarters, that is to say, the probability of not being colorblind is three-quarters, the probability of non-albino is two-thirds, and the probability of a boy is one-in-two, so the probability of having a normal son is the product of these three numbers, and the answer is one-quarter.