A topic related to finding gene frequencies in living organisms

A total of 900 people, 1800 genes, 23 people with AB (IAIB) blood type, 23 IA, 23 IB; There are 441 people with O(II) blood group and 882 I genes; There are 371 people with blood group B (IBI), 371 IB genes and 371 I genes; There are 65 people with blood type A (IAI), 65 more from IA, and 65 from I.

So, there are 88 IAs in total. There are 394 IB and 1318 i, so the IA frequency is 88 1800=, the IB frequency is 394 1800=, and the I frequency is.

b×b＋2bb＋bb≠1 ?(Explain that this population is not a population of the law of genetic equilibrium, and the population of the so-called law of genetic equilibrium must meet the following conditions: :(1) in a large population; (2) random rather than selective marriage; (3) there was no natural selection; (4) no mutations occurred; (5) There is no large-scale migration.

The gene frequencies and genotypic frequencies of the population remain constant throughout the reproduction passage from generation to generation.

In this way, b b 2bb bb is equal to 1, the genotype frequency of the dominant homozygous is p squared, and the genotype of the recessive individual is q squared, and the gene frequency can be found by the method of genotype frequency square.

You can look at this question this way.

Assuming that there are 100 individuals, then according to the question conditions, there are a total of 78 + 18 * 2 = 114 B alleles and 86 B alleles.

The frequencies of b and b are 114 200 and 86 200 respectively, and you can think of a prescription that indicates that you have a certain basis of probability, but you do not have a thorough understanding of probability.

Because 200 males only have 200 color vision genes, the total number of color vision genes in these 400 people is (200*2+200).

Because the color blindness gene is inherited with X recessive, the male Y chromosome does not have this gene and cannot be counted, so 200 males are only counted as 200, and females need to be added 400.

One color blind is with X recessive genetic disease, note that the male is XY

Mathematically calculate the ratio by multiplying the gametes of genes.

Let me give you an example.

In guinea pigs, black cats are dominant to white hair, and if 90% of the gene pool is dominant B and 10% are recessive B, the frequencies of genotypes bb, bb, and bb in the population are respectively.

If p is used to denote the frequency of a gene, then there is: p(b)=90%, p(b)=10%, so p(bb)=p(b)*p(b)=81%;

p(bb)=p(b)*p(b)+p(b)*p(b)=2*10%*90%=18%;

p(bb)=p(b)*p(b)=1%

Solution 1:

First, find the gene frequencies of 20,000+2,000 fruit flies.

vv1 ：（20000*15%+2000*100%）/(20000+2000) = a

vv1 ：20000*55%/22000 = b

vv1 ：20000*30%/22000 = c

Then mating freely;

vv2 ：a*a+1/2*a*b+1/4*b*b

vv2 ：a*c+1/2*a*b+1/2*b*b+1/2*b*c

vv2 ：1/4*b*b+c*c

Explanation: Generate vv2, which can be vv*vv, which is 100%, or vv*vv, which is only 50%, or vv*vv, which is only 25%; VV and VV are calculated in the same way.

Solution 2: In the parent, the frequency of v is (2000*100%*2+20000*15%*2+20000*55%*1) (2200*2)= x

The frequency of v is (20000*55%*1+20000*30%*2) (2200*2)=y

vv2=x*x

vv2=x*y

vv2=y*y

Explanation: vv is v*v.

First, find the frequencies of each gene in each population.

Same approach for group 2. Gene frequency of a = a= result aa = aa = aa = aa = 2*

From 30% of individuals with genotype AA and 10% of individuals with genotype AA, it can be concluded that AA accounts for 1-30%-10% = 60%.

Through the ratio of the three, the gene frequency of A is 60%, and the gene frequency of A is 40%.

a：(30%*2+60%)/2=60%

a：(10%*2+60%)/2=40%

Genotype probability of offspring:

aa：60%*60%=36%

aa：40%*40%=16%

aa：60%*40%*2=48%

It is derived from a+a=1, (a+a) 2=aa+2aa+aa=1.

This question should take into account the situation and number of each genotype and should be calculated using the overall genotype. Individuals with genotype AA accounted for 30% and individuals with genotype AA accounted for 10%. So 60% of individuals with genotype AA are aa.

then a is 30%+60% 2=60%, and a is 10%+60% 2=40%. So aa=60% 60%=36%, aa=40% 40%=16%.

Homozygous for a gene can lead to the death of an embryo of a certain organism, and it is known that the frequency of gene dust in a natural population of the organism is close to 50%, and the heterozygote in the population accounts for about 99%.

Because there are basically three kinds of gametes in the question, but because aa is lethal, so there are only two genes, we assume that in all cases the proportion of aa is x, so aa accounts for 1-x, so we can get the gene frequency is x+(1-x)2=50%=a's frequency =(1-x) 2 The above equation can only be true when x is 0, and because the frequency of aa is close to 50%, the proportion of aa is 99%.