High School Biology Genetics Math Frequency

There are two solutions to this problem: First, push directly. According to the topic, the second generation of children has bb, bb, bb, where gray is bb, bb, the ratio is 1 3 and 2 3, and the free mating of the second generation is:

1 3bb 1 3bb ,1 3bb 2 3bb ,1 3bb 2 3bb , 2 3bb 2 3bb , so the frequencies of bb, bb and bb in the three generations are 1 3*1 3+1 3*2 3*1 2+1 3*2 3*1 2+2 3*2 3*2 3*1 4=4 9,1 3*2 3*1 2+1 3*2 3*1 2+2 3*2 3*1 2=4 9,2 3*2 3*1 4=1 9.

Second, using the theory of genetic equilibrium, that is, in an environment where there is no obvious natural selection, individuals freely mate and reproduce for one generation, and the gene frequency does not change, i.e., (p+q) 2=p 2+2pq+q 2. Among them, p and q represent gene frequencies, p 2, 2pq, and q 2 represent genotype frequencies respectively, and the frequency of b gene in the gray of the second generation of children is 1 3+2 3 * 1 2=2 3, and the frequency of b gene is 2 3*1 2=1 3, so the frequency of bb in the third generation of children is 2 3*2 3=4 9, bb is 2*2 3*1 3=4 9, and bb is 1 3*1 3=1 9.

Heredity is not probability, but integral.

I'm right, here's how.

In the second generation of children, bb:bb=1:2 so (1 3bb+2 3ab)*(1 3bb+2 3bb).

The above formula removes the brackets and then reorganizes and merges.

Solution: bb1 9 bb4 9 bb4 9

I wonder if it's wrong here, it should be the BB BB multiplied twice, because the BB BB and the BB BB are not the same.

That's it, it can be bb as the paternal parent and bb as the female parent or vice versa as the female parent and bb as the male parent.

So here you have to multiply by 2

The second generation of free mating is not only bb*bb, bb*bb, but also bb*bb, so it is very troublesome to multiply one-to-one like your method, here I say a conventional method I use to do the question, all the relationship between a population and a certain generation of free mating for the next generation, are to be calculated by calculating the gene frequency. For example, in this question, if the black body is removed from the second generation, then bb:bb=1:

2 then the gene frequency of b in this generation is 2 3, the gene frequency of b is 1 3, and the proportion of the next generation of bb is the square of the b gene frequency, that is, 2 3 * 2 3 = 4 9, and the proportion of bb is 1 3 * 1 3 = 1 9, and bb needs to use the gene frequency of 2 times b multiplied by the gene frequency of b, that is, 2 * 2 3*1 3 = 4 9 This is simple and not easy to make mistakes.

It's your wrong way of thinking, the various proportions are right after removing the black in the second generation, but that's free mating, which can't be counted like you, it's become a matter of gene frequency. Think of it this way: one-third of the homozygous gray (bb), two-thirds of the heterozygous gray (bb), two-thirds of the dominant gene (b) and one-third of the recessive (b), according to the Hardy-Weinberg theorem

Pure gray (BB) is 4/9, miscellaneous ash (BB) is 4/9, and pure black (BB) is 1/9.

The previous one is right, starting from the second generation bb (1 3) . bb(2 3) can be calculated as b (2 3). b(1/3)。

So in the three generations, bb is 4, 4 is 4, and 1 9. You can see it, right? In fact, from this point of view, it is more simple and straightforward.

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How genotype frequencies are calculated.

Gamete ratio:

a:30%*1+60%*1/2+10%*0=60%;

a:1-60%=40%

Frequency of offspring genotype:

aa:60%*60%=36%;

aa:60%*40%+40%*60%=48%;

aa:40%*40%=16%

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 calculated that it was also 1 9Why do you have 1 6 algorithms below aa and aa, aa and aa together? Isn't AA and AA okay? I firmly believe that it is 1

If you don't understand upstairs, don't talk nonsense.

Because rust resistance is controlled by dominant genes.

P stands for AA and AA

After inbreeding, the F1 generation is AAAAAAA

AA excludes self-inbred after AA.

Assuming that each plant produces m offspring, then the aa inbred offspring is m rust-resistant plants.

AA inbred offspring have 2m offspring.

Of these, AA accounts for a quarter.

That's m 2.

And because the F2 generation has a total of 3m individuals.

So divide m2 by 3m

It's m 6, so choose B

Hope it helps.

If you still don't understand, you can send me a message.

Hehe. The gene frequencies you count are for the F1 generation.

So the result is not correct.

What kind of rotten analysis is this, the author is brain-dead, right? Or is the Chinese ...... taught by the physical education teacherYou're only asking about the calculation of gene frequencies, right?

Very simple. If aa accounts for 1 3, aa accounts for 2 3.

You can think of it as a population of 3 individuals, 1 AA and 2 AA.

There are 2 A's in AA, 1 A and 1 A in AA.

So, there are a total of 4 A's and 2 A's for these 3 individuals.

The gene frequency of a = the number of a (the number of a + the number of a) = 4 6 = 2 3, the gene frequency of a = the number of a (the number of a + the number of a) = 2 6 = 1 3.

If the words "free mating", "random mating", etc., are emphasized, they can be calculated using gene frequency.

First, the frequency of gene A = frequency of aa + frequency of aa 2, frequency of gene a = frequency of aa + frequency of aa 2

Then, according to the "Hardy-Weinberg" law, it can be seen:

aa=a·a；aa=2·a·a；aa=a·a。

I can only snipe sons, e.g. aa2 3 aa, 1 3 then the gene frequency of a a is, 1 3 2 3 * 1 2 = 2 3 a is 2 3 * 1 2 = 1 3

First of all, according to the knowledge in the textbook, you can know that the inheritance of red and white eye color in fruit flies is accompanied by the X chromosome, according to what you said below, one out of 5,000 fruit flies is a white-eyed female fly, and one is a white-eyed male fly, so out of 5,500 fruit flies, 2,500 female flies, that is, there are 5,500 genes, and there are 2,500 male flies, which means that there are 2,500 genes, so there are 7,500 genes, and there is a white-eyed female fly among the 5,000 fruit flies, and there is a white-eyed female fly, and there are two white-eyed male flies, which means that there is a white-eyed gene So for a long time, there are three white eye genes, and the total number of white eye genes is three.

Then the quest is 3 7500, which is 1 2500

But if you want to ask this question, there is still a vague point, that is, there should be heterozygous in the remaining red-eyed fruit flies, because the red face appears to be dominant, and the white eye is recessive, so there is a possibility that there are heterozygous in the red eye female fly, so there are more than three genes for the white eye, so the answer is uncertain.

1 50 because the probability of gene frequency b and b combining is 1 50 * 1 50 2500

Upstairs there will also be alleles on the XY chromosome, and the genes that control the eye color of the flies are indeed on the sex chromosomes.

I think it's important to make it clear whether the fly is a male or a female, and if it's a female, then the probability of getting the white eye gene b is 1 50, and if it's a male, it's 1 2500