The genotype frequencies of AA Aa and aa in a certain biological population are 0 3, 0 4 and 0 3, re

1）；No.

Explanation: The conditions given in the question are in line with Hardweinberg's equilibrium theory, so that ideally, the frequencies of each gene and genotype of the offspring of the population are the same as those of the parents, which is the content of the modern biological evolution theory in Chapter 7 of Compulsory 2.

2) Whether the conditions given in this question are incomplete, it should be stated that the ratio of males to females in the two genotypes is 1:1, if this is the case, then the genotype that ABB can produce is AB or AB male and female gametes, and ABB individuals can produce male and female gametes of ABB, since the ratio of ABB to ABB is 1:1 given in the question, according to which the genotype of gametes and their proportions are:

ab=1/2x1/2+1/2=3/4;ab=1 2x1 2=1 4, then the individual genotype that can be stably inherited in the offspring should be aabb=3 4x3 4=9 16; aabb=1 4x1 4=1 16, so that the proportion of offspring that can be stably inherited is 9 16 + 1 16 = 5 8

Explanation: Because peas are mostly closed pollination under natural conditions, that is, self-pollination, they are all self-inbred combinations, AA's own offspring are all AA, 1 4 of AA's inbred offspring are AA, and AA's own offspring do not have AA, so the probability of AA in the offspring should be.

I personally understand and hope it can help you.

The aa genotype frequency is aa genotype frequency + 1 2aa genotype frequency = if there is no natural selection, and the gene mutation is not considered, the gene frequency of the population does not change, and the genotype frequency is calculated as the aa genotype frequency = the square of the a gene frequency, that is, 25%.

Free mating, where genes A and genes B are calculated and multiplied separately, i.e., aa:aa=1:1, bb:

BB = 1:1, because B gene is homozygous and always stable inheritance, as long as the stable genetic homozygosity after mating of gene A can be calculated, A gene frequency = 3 4, A gene frequency = 1 4, after free mating AA genotype frequency + AA genotype frequency = square of A gene frequency + square of A gene frequency = 5 8

If the population is pea, that is, it can only be self-bred, then the proportion of AA in the offspring = 1 2AA in the inbred offspring + 1 2 The proportion of AA in the inbred offspring = 1 2*1 4 + 1 2 = 5 8

Summary. Because he gave you the frequency of his parents, he had to calculate it according to what he gave, and then he later said that the offspring would mate randomly, that is, the formula of gene frequency was used to bring in the calculation.

Good. I don't understand these options.

Teacher, does the ontology conform to the law of genetic equilibrium?

I choose C for this topic

How do you find the gene frequencies of safflower, white flower, and a and a in random mating offspring?

Pay attention to the random mating here, random mating should be based on the law of distribution before the turn of the gene, you can't simply use self-breeding to calculate, don't reply first, I'll write you Hui Zheng's options.

Teacher, is the gene frequency of parent a?

That's right, it's the same as what you wrote above.

Therefore, d is not the same as the parent.

You must remember the above question for this question.

Because he gave you the frequency of his parents, he had to calculate according to what he gave, and then he later said that the descendants of the faction after the fight were randomly mated, that is, the formula of gene frequency was used to bring in the calculation.

Free mating, in accordance with the Hardy-Weinberg law.

a=p,a=q,then a+a=p+q=1,aa+aa+aa=p2+2pq+q2=1).

The gene frequency of a is the frequency of the aa genotype + 1 2 the frequency of the aa genotype = 5 the frequency of the aa genotype is the square of the a gene =

In the parent, the gene frequency of a = aa genotype frequency + (1 2) * aa genotype frequency =

Progeny aa genotype frequency = (gene frequency of a) 2=

Calculate probabilities mathematically:

Count with biological methods:

There should be a formula on this question material, I forgot, you look at the information, there are several situations, what is this called? Forgot,

(1) From the inscription, it can be seen that aa=, aa=, aa=, the frequency of a gene in the population is; a=

2=(2) If the population is large enough, does not mutate, does not select, does not move in or out, and there is random mating between individuals in the population, the heredity of the population conforms to the Hardy-Weinberg equilibrium formula, and the genotype frequency of AA in the offspring of the population is AA=, and if random mating continues, the genotype frequency of AA in its offspring does not change

3) Assuming that the animal population meets the above four basic conditions, but does not mate randomly, and only mates between the same genotypes, that is, the mating mode is AA and AA, AA and AA, and AA and AA for mating, theoretically, the genotype frequencies of AA, AA, and AA in the offspring of the population are: AA=

4=，aa=

2=，aa=

4=；If the offspring also only mate between the same genotype, the genotype frequencies of aa, aa, and aa in their offspring are calculated as described above, aa=

4=0，45，aa=

2=，aa=0，4+

4=, so if the offspring also only mate with the same genotype, the genotype frequencies of aa, aa, and aa in their offspring will change

So the answer should be:

1) and (2) Large enough genetic mutation (or mutation) selection will not.

3) Yes.

Because aa is, the frequency of the a gene in aa is half =

Whereas, the genotype frequency of AA is:

So the frequency of the a gene =

The A in AA counts twice.

AA counts once the last (

(1) In a known plant population, 30% of individuals with AA genotype and 20% of individuals with AA genotype account for 50% of the population The gene frequency of the population is as follows: A=30%+1

2×50%=55%，a=20%+1

2) The individual self-inbred offspring of this population, the individual self-inbred offspring of aa and aa are all homozygous, and the genotype of the individual inbred offspring of aa is aa:aa:aa=1:

2:1, so if the population is self-bred, the genotype frequency of AA in the offspring is 30% + 50% 1

4=, the genotype frequency of AA is 20%+50% 1

4=, since the frequencies of the inbred genes of a and a do not change Therefore, the plants of this population are inbred, and the frequencies of individuals of the genotype of aa and aa and the frequencies of the genes of a and a are respectively.

3) The essence of biological evolution is the change of gene frequency of the population, and from the above calculations, it can be seen that the gene frequency of the plant has not changed in two years, so the organism has not evolved

4) The modern theory of biological evolution believes that the basic unit of biological evolution is the population, the raw materials of evolution are provided by mutation and genetic recombination, the direction of biological evolution is determined by natural selection, and the essence of evolution is the directional change of gene frequency, so the answer is:

3) No evolution has occurred and there has been no change in gene frequency.

4) Population: Directed changes in gene frequencies in populations.

aa=20 aa=50 aa=30% a=(20x2+30) 200=35% a=65%.

There was no change in gene frequency after inbreeding a=35% a=65%, but there was a change in genotype frequency.

aa=20%+30%x1/4= aa=50%+30%x1/4=

aGene frequency.

aGene frequency.

AA genotype frequency.

AA genotype frequency.

AA genotype frequency.

aa & aa aa & aa aa & aa

aa aa aa aa

The probability of producing aa is.

The probability of generating AA is understood.

The gene frequencies of a and a are root number and root number, respectively. Let the gene frequency of a be p and the gene frequency of a be q, then:

p（aa）= p×p= p^2=d；p（aa）=2×p×q=2pq=h； p（aa）= q×q = q2=r。So the gene frequencies of a and a are root number and root number respectively.