There were 20,000 fruit flies on a small island, of which 15 were 55 and 30 were of genotype VV vv a

Let's simplify the numbers, assuming that there were 20 fruit flies on the island, and 2 from the outer islands. Then in the original fruit flies on the island, the gene frequency of v is , and the gene frequency of v is .

Coupled with the invasion outside the island, we need to know that no matter what kind of mating method, the total gene frequency is unchanged, so just calculate the gene frequency of the original V outside the island and the total on the island.

Then it is (, which is about 48%.

Or maybe so, this method is what I used to use before, just for convenience.

There are 20 of them on the island, so there are 40 genes, right?

Then there are 3 vvs, so there are 6 Vs, and vvs have 11, then there are 11 Vs, then there are 17 Vs in total, and 4 outside the island, so there are 21 Vs in total, and the genes are 44 in total, so the frequency of V is 21 44=

I don't know if you can understand me writing like this.

WW has 20000*, WW has 20000*, WW has 20000*, 2000 WW comes in, there are 22000 lizards in total, of which WW has 3000 + 2000 = 5000, WW is still 11000, and the gene frequency of W is 5000 22000+ is approximately equal.

There are 20,000* in VV, 20,000* in VV, 20,000* in VV, 2,000 VVs coming in, and a total of 22,000 lizards, of which VV has 3,000 + 2,000 = 5,000, VV is still 11,000, and the gene frequency of V is 5,000 22,000 + which is about equal to choose B

This requires a re-priming of the gene frequency.

Originally, there were 20,000 fruit flies, and the individuals with AA basal type were 20,000*15%=3,000.

then a gene. The frequency is: (3000*2+20000*55%) (22000*2)=

The gene frequencies of a are:

Since there is no genetic mutation in F1, it is a random mating, which is in line with Hardy. Weinberg's Law. There is no change in gene frequency. i.e. the gene frequency of a is equal to that of the parent, yes.

Choice B analysis] Before entering the rapid vertical invasion of birds, there were 3000 VV individuals, 11000 VV individuals, 6000 VV individuals, after the invasion, VV became 3000 + 2000 = 5000, other types of individuals remained unchanged, but the total number became 22000, then the V gene frequency was (5000 2 + 11000) (22000 2) mu age = 48%.

The answer is: b.

Since it is free mating, the gene frequency should be constant, and it should be the same before mating as it is after mating.

Answer: The gene frequency of V in F1 is approximately.

Solution: The original vv = 20000 * 15% = 3000, pure and simple.

vv=20000*55%=11000;

vv=20000*30%=6000pcs.

When 2000 Drosophila with the genotype vv of the state trouser stool are introduced, the number of flies becomes 22000, then v=(5000*2+11000) 22000*2=.

Since all Drosophila are mated randomly, the gene frequency of V in F1 should be the same as that of the parental traveling generation, too.

1. There were 20,000 original fruit flies, of which the genotype VV, VV and VV fruit flies accounted for % and 30% respectively, then VV=20 000 15%=3000, VV=20000 55%=11000 Due to the invasion of 2000 fruit flies with genotype VV from outside the island, now VV=3000+2000=5000

2. According to the formula for calculating gene frequency: the frequency of a gene in the population = the total number of genes in the population and the total number of alleles in the population 100%, calculate the gene frequency of v = the total number of v genes v and the total number of v genes 100% = (5 000 2 + 11 000) (20 000 2 + 2 000 2) 48%.

3. According to the random mating of individuals in the population, the gene frequency of V in the F1 generation is the same as that of its parents, which is also about 48%.

Untie; After the migration of Drosophila, the number of individuals of different genotypes of Drosophila was as follows: mm=20000 15%+2000=5000, mm=20000 55%=11000, mm=6000, and the total number of the species was 20000+2000=22000. The gene frequency of m in this population is: m=m (m+m) 100%=(11000+2 6000) (22000 2) 100% 52%, according to the title, all fruit flies are randomly mated, the natural environment is not selected, if the sudden change is not considered, the hail plexus gene frequency is f1

The generation is unchanged at 52%.

Therefore, c

After migration, random mating, the gene frequency of the population remains unchanged, so the frequency of A is (160*2+320+0) 800*2=40%, the gene frequency of A is 1-40%=60%, and the proportion of Aa is 2*40%*60%=48% according to the theory of gene balance after random mating. Pick B

The proportion of migrated genes a: (160+320 2) 800=40% a: (320+320 2) 800=60%, so the proportion of aa is 2

Gene frequency of a = number of genes of a Total number of genes = (80 2 + 150 1) (250 2) = 62%.

Gene frequency of a = number of genes of a Total number = (20 2 + 150 1) (250 2) = 38% (or 1-62% = 38%)

a=32%+60%/2=62%

a=8%+60%/2=38%

The calculation process is like this, whether it is necessary to explain the rules for doing such questions.

Example 1: 100 individuals are randomly selected in a population, and 10 individuals with genotype AA, AA, and AA are detected, respectively. What is the frequency of genes a and a? Analysis: In the case of the alleles A and A.

It is assumed that the V gene frequency in the population is x

Then the frequency of v when producing gametes is x, the frequency of v is 1-x, the genotype of the offspring stump wing, the number of stub wing flies is equal to x*x = 4%, and the solution is x=

Therefore, the frequency of v in the original population is v is.

Number of remnant winged flies: 4% Homozygous number of long-winged flies.

Heterozygous long-winged Drosophila population 1-4%-64%=32%.

After the addition of homozygous Vv flies, the number of genes for V did not change, but the population size doubled at a frequency of 20%*20000 40000=10%.

The number of v genes increases, and the simplified calculation can directly use 1-10%=90% as its gene frequency, so this question b is incorrect.

The original group is newly joined.

v vv vv 800 vv 3200

v vv 3200 vv 12800 + 20000vv

Uh, it's a little hard, no.