Biological genetic disease probability calculation, biological genetic probability calculation metho

Are you saying that the grandparents are both AA, or the father's siblings have a patient, and the father is normal, then there is a 2 3 probability that it is AA and a 1 3 probability is AA, so when it is AA, there is a 1 4 * 1 2 probability of having a sick boy, and when it is AA, it is not possible.

There is 2 3 * 1 4 * 1 2 = 1 12

The father is 1/2 likely to be AA or AA, and when it is AA, 1 in 4 is likely to have a sick child, which is 1 in 8 boys who are sick, and when it is a father AA, it is impossible to have a sick child. So the answer is 1 in 16.

If the father is AA, the child must not have the disease, because it is recessive.

If it is AA, the probability of having a sick child is 1 4, or a boy, the probability should be 1 8

The probability that the father is AA is 1 2

The probability of AA (father) and AA (mother) giving birth to AA (sick boy) is 1 4

So the probability that they will have a sick boy is 1 2 * 1 4 = 1 8.

Father: aa aa- a a a a

Female: aa - a a

aa - 3/8

aa - 1/2

aa - 1/8

Boys have a chance of 1 2

So probability = 1 2 1 8 1 = 1 16

If the father is AA to be sick, then the probability of AA is 1 2

The probability of AA being combined with AA to get AA is 1 4, so the probability of having a child and getting sick is 1 2 1 4 = 1 8

And the probability of having a boy is 1 2, and the final score is 1 16

Previous generation genes.

Copy library if a is the percentage.

If the ratio is x, then the percentage of relative a is (1-x), then aa is the square of x, aa is 2x(1-x), and aa is (1-x) squared. In the same way, if B accounts for 100%, then B is 0, so BB accounts for 0*0=0, BB accounts for 0*100%=0, BB accounts for 100%*100%=100%.

In the following question, first of all, there is only the B gene, so the offspring must only have BB, so you don't have to think about it. And know aa:aa=1:

1, then in the gametes, a accounts for 3 4, and a accounts for 1 4, so according to the algorithm I mentioned earlier, the offspring aa accounts for 3 4 * 3 4 = 9 16, aa accounts for 2 * 3 4 * 1 4 = 6 16, and aa accounts for 1 4 * 1 4 = 1 16. Among them, if AA and AA only mate with the same type, they will only give birth to the same type of offspring AA and AA, which belongs to stable inheritance, and AA mating will cause trait separation and cannot be stably inherited (I don't know if stable inheritance is understood in this way). Therefore, the individuals who can be stably inherited account for 9 16 + 1 16 = 10 16 = 5 8, choose b.

In the case of independent inheritance, multiple pairs of traits are decomposed into a single relative trait, and then analyzed separately according to the law of gene segregation, and finally the analysis results of each pair of relative traits are multiplied, which is based on the multiplication theorem in probability theory.

The multiplication theorem is that if the occurrence of one event does not affect the occurrence of another, the probability that the two events will occur at the same time is equal to the product of the probability of their separate occurrence.

The law of free combination of genes involves the independent inheritance of many pairs of genes, therefore, according to the multiplication theorem in probability theory, the performance of multiple pairs of genes is the product of the performance of each of the equivalent genes when they are inherited separately.

Six ways to calculate the probability of biological inheritance.

Here's how the genetic probability of a creature is calculated:

Gene frequency = number of genes Total number of all alleles * 100%.

Take a, a as an example, the gene frequency is the frequency of a or the coarse closure frequency of a, and the sum of the two is 1. The frequency of the genotype is AA or AA or AA, and the sum of the three is 1In addition, if the frequency of AA is 16%, then the frequency of A is 40%, because the frequency of AA genotype rock crack is the approximate auction rate of 40%*40%=16% of the occurrence of A.

Suppose there are two alleles A and A on a locus in a diploid organism, and assuming that there are n individuals in the population, and the number of individuals in the three genotypes of AA, AA, and AA are N1, N2, and N3 respectively, then the frequencies of gene A and the frequencies of AA genotypes in the population are respectively

Frequency of genes a= total number of genes a(total number of genes a+total number of genes a) = (2n1+n2) 2n or n1 n+n2 2n

Frequency of aa genotype = number of individuals of aa genotype Total number of this diploid population = n1 n.

The relationship between gene frequency and genotype frequency is deduced from the above: the frequency of gene a = n1 n + 1 2·n2 n = frequency of aa genotype + frequency of 1 2·aa genotype.

Gene frequency uses.

The sum of the frequencies of the various genes at a certain locus and the frequencies of the various genotypes in a population is equal to 1. For a population, ideally, the gene frequencies of the population remain stable across generations.

But under natural conditions, it is affected by genetic mutations, genetic recombination, natural selection, migration, and genetic drift. The gene frequencies of populations are constantly changing, which allows organisms to constantly develop and evolve. Therefore, calculating the gene frequencies of a population can help to understand the evolution of that population.

First empty: 15 types.

Solution: The combination of the three genotypes of autosomes AA, AA, and AA and the five genotypes of sex chromosomes, XBXB, XBXB, XBXB, XBY, and XBY, is 15.

Second empty: 2 types.

Solution: Females are aaxbxb, aaxbxb, males can only be aaxby, because only in this way can white individuals be produced, so there are two combinations of parents.

Third empty: 9 16.

Solution: The genotypes of the parents are aaxbxb and aaxby.

In the offspring, A- accounts for 3 4, and the dominant traits XBXB, XBXB, and XBY on the sex chromosomes account for 3 4, 3 4*3 4=9 16.

Since we are in a hurry today, please explain if you answer incorrectly.