High school biogenetics probability problem, find specific steps

This is a classic question.

Two pairs of alleles are involved, a and a (on a pair of autosomes), b and b (on sex chromosomes).

When solving this kind of problem, it is necessary to solve it one-to-one, that is, to treat the two pairs as two pairs, and after solving one pair, the other pair will be solved. Finally, the probabilities of two pairs are combined.

Specifically: Study AA first

Known: Female: AA, Male: 2 3AA, 1 3AA

Find: The probability of aa in the offspring.

Analysis: If the male parent is 1 3AA, the offspring will be AA, and AA will not appear

The male parent is 2 3aa, and the offspring will appear aa

Algorithm: aa 2 3aa descendants are 2 3 (1 2aa 1 2aa), so the probability of descendants aa is 2 3 1 2 = 1 3

Re-study the sex chromosomes.

Known: Female 1 2xbxb, 1 2xbxb, male xby

1 2xbxb xby, the offspring is 1 2 (1 2xbxb, 1 2xby), explanation: 1 2 under the premise of 1 2, there are two cases, each accounting for 1 2 in 1 2

1 2xbxb xby, offspring 1 2 (1 4xbxb, 1 4xbxb, 1 4xbxb, 1 4xby).Explanation: Under the premise of 1 2, there are four situations, each accounting for 1 4 of 1 2

The basic calculations are over.

Just now the two pairs are counted separately, and the next step is the last step, the two pairs should be combined!

The genotype of the offspring is aaxbxb: aa is 1 3, xbxb is 1 8, so the probability of the offspring being aaxbxb is 1 3 1 8

The genotype of the offspring is aaxby: aa is 1 3, xby is 1 4 + 1 8, so the probability of the offspring being aaxby is 1 3 1 4 + 1 8).

Keep one and count it yourself.

Understanding must be thorough, not plausible.

Then do three or five questions of the same type, and you will completely open up this joint.

Calculation, with the knowledge of the fourth grade of primary school is enough.

It's just a matter of knowing the method: two pairs, one pair to solve, and then the probability of two pairs is combined together (multiplication).

Solve the problems that high school biology students are prone to.

This is a gene frequency exercise. The general population is split evenly between men and women, so the probability of a male carrier is half that of a carrier in the population. Gene frequencies conform to a formula, if the frequency of gene A is p and the frequency of gene A is q, then their relationship is in the formula where the square of P is the probability of aa, 2pq is the probability of aa, and the square of q is the probability of aa.

Note: I don't know how to mark the square, so I have to write Chinese characters).

Multiply by two because (a+a) = a +a +2aa, which may be either a and a or a combination of a and a. Because you already know that a man is no longer sick, you have to subtract the possibility of his disease, which is 1-1 2500. It's the same as your big string of numbers.

His algorithm adds up the aa and aa groups.

Because there are AAs and AAs among the carriers, they should be counted separately, so they should be multiplied by 2

Because it's a male, the female one should be excluded, but it seems that the carrying rate in the crowd is counted in the question, and if the male one is counted, it will be divided by 2 (half for both men and women). The string that is divided later refers to the probability that there is no albinism in the population.

Formula: p 2 + 2 pq + q 2 = 1

p 2 denotes the frequency of aa.

2pq indicates the frequency of AA.

Q2 indicates the frequency of AA.

From the inscription: p=49 50 q=1 50

So, the probability of aa = 2pq = 2 * 1 50 * 49 50 = 98 2500

Because albinism genes appear in pairs, AA and AA are two different combinations, the probability of calculating AA must be multiplied by 2 to be comprehensive.

You push into the selected fruit flies.

1/3bb2/3bb

The genotype of the black fruit fly is.

bb to get.

The BB must be both parents and both are BB.

Principles of multiplication).

And then bb bb

The probability of getting bb is.

1 4 bar. So 4 9 1 4 = 1 9

OK

There are 4 gamete genotypes, which can be freely combined in pairs.

The probability of suffering from B disease is 1 2 * 1 4 = 1 8The probability of suffering from nail disease is 1 2, and the probability of the occurrence of the two diseases is independent of each other, so the probability of developing one disease is.

Hope!!! Want to explain further?

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

If the question does not mention gender, the gene can be assumed to be autosome.

Easy to get husband: ttaa; Wife: Ttaa

Ideas: If the husband suffers from polydactyly, the genotype is T, the wife is normal, then it is tt, and the child born is normal tt, then the husband must have T in the gene, and the husband's genotype: tt; If the husband and wife are normal (A) and the child has albinism AA, then the husband and wife each have A in their genes, and the husband and wife are both AA).

then polydactyly (t) p: 1 2

Suffering from albinism (AA) q: 1 4

1) Polydactyly: 1 2

2) Suffering from albinism: 1-4

3) Polydactyly: p(1-q)=1 2 3 4=3 8(4)Both diseases: p*q=1 8

Classification and judgment methods of common genetic diseases

Step 1: Determine whether it is a dominant or recessive genetic disorder.

Methods: Look at the total number of patients, if there are many patients in a row and each generation is dominantly inherited. If the number of patients is small, only one generation or alternate generations of patients are recessive. (Out of nothing is implicit, and there is nothing out of the obvious).

Step 2: Determine whether it is an autosomal or X-chromosome inherited disease.

Methods: The number of patients by sex was looked at, and if the number of male and female patients was basically the same, it was an autosomal genetic disease. If there is a significant disparity in the number of men and women, it is an X-chromosome genetic disorder.

In particular, if the number of male patients is much higher than that of female patients, it is considered to be recessive on the X chromosome. Conversely, explicit).

Creating something out of nothing is hidden, and giving birth to a girl with a disease is often hidden; If there is a probability that the middle child is not dominant, and the birth of a girl without disease is often obvious, it depends on what the specific topic is looking for to carry out specific calculations, which can be calculated according to the genotype.

First of all, it is made out of nothing, which is definitely a recessive genetic disease.

If it is with X recessive, then the boy's father must be the patient.

Therefore, this disease is an autosomal recessive disorder.

Let the dominant gene be a and the recessive gene be a

The genotype of the boy's parents is AA

The boy's genotype is AA (1, 3) or.

aa(2/3)

If the boy's wife is normal and the wife's mother is sick, then the boy's wife genotype must be AA If the boy's genotype is AA

AA is combined with AA offspring to have AA

aa) is the probability of carrying the disease-causing gene.

If the boy has a genotype AA

AA is combined with AA offspring to have AA

aaaaaa) son behaves normally and is not aa

then the probability of carrying the disease-causing gene is.

Then the probability of this son carrying the gene of the disease is 1 6 4 9 11 18 Why choose d, I can't say. I can't understand the explanation of the few people downstairs who said that they chose dd!

The disease is autosomal recessive, according to the title, 1It is easy to launch that both the woman and the first husband are heterozygous.

2.It can also be inferred that the gene frequency of a is 1 100, the gene frequency of a is 99 100, and the genotype frequency of a a is 2 1 100 99 100 = 1 50 (approx.) 3The offspring can only be AA or AA if the offspring is a healthy person, and when the genotype of the stephusband is AA (probability 1 50), the offspring can become sick.

So after the couple gets married, aa (woman, probability is 1) aa (husband, probability is 1 50).

The probability of a child's illness (AA) should be 1 4 1 50 = 1 200

dDraw a pedigree chart! It can be judged to be often hidden.

Let the dominant gene be a and the recessive gene be a

The mother is AA and the father is A (AA

2 3aa1 3), the son is identified as a, aa does not exist

The probability of a son aa is 2 3x1 2

1/3x1/2

The probability of aa is 2 3x1 4

1/3x1/2

The probability of carrying the disease-causing gene is 3 5 ......d

Suppose tall A short A red B white B

F2 is AABB, and this kind of problem draws complex free combinations as simple segregation laws, just looking at a single set of genes.

Since the white flower is gone, then BB accounts for 1 3 BB 2 3 offspring safflower ratio is 1- 2 3 * 1 4 (one minus the probability of BB inbred offspring BB) = 5 6

Looking at it separately, the height of the aa:aa:aa=1:2:1 ratio of the dwarf neck of the inbred offspring is 1 4 ( aa) +1 2 * 1 4 (the probability that the aa inbred offspring is aa) = 3 8

Multiply c

Landlord, your workbook is much more difficult than ours, and I've been working on it for a long time (I can't choose C, right?)

I'm too complicated to mislead you, hehe.

This kind of f1, 2, 3 is more troublesome, and it needs to be analyzed from the type and number of genetic factors it produces. Set tall stem d short stem d, red a white a, f2 has the following proportions:

ddaa:(ddaa:ddaa):

ddaa:ddaa):（ddaa:

ddaa:ddaa:ddaa)=1:

Remove the white flower AA and leave it:

ddaa:ddaa):(ddaa:ddaa:ddaa:ddaa)=(2:1):(1:2:2:4)

F3 short stem safflower should be dda, for the convenience of calculation, set each flower can produce all seeds as a unit, select the one that can produce dda, according to the proportion of the corresponding flower and the probability of producing the seeds sought:

1 12 x1 ——— DDAA accounts for 1 12, and the seeds obtained are all sought.

2 12 x3 4 - DDAA accounts for 2 12, and the resulting seed 3 4 is what is sought.

2 12 x1 4 -- and so on.

4/12 x3/16

1/12 x 1+2/12 x 3/4+2/12 x 1/4+4/12 x 3/16=5/16

For the bowl bean problem, you can use the gene frequency to calculate the formula a ten b = 1, (a ten b) square = a square + b square + 2 times 1 3, aa accounts for 2 gene frequency = (1 3 times 2 + 2 3) divided by 2 = 2 3, the frequency of a is 1-2 3 = 1 3. So the offspring aa is 2 3 times 2 3, aa is 1 3 times 1 3, and aa is 2 times 1 3 times 2 by 2 3. Do the same for the next question, but note whether it is natural mating or self-breeding.

Natural mating is calculated by gene frequency, and when separated from self-breeding, it is calculated separately and then added.