①a gene frequency = total number of A genes/(total number of A genes+total number of A genes) =(2n 1+n2)/2N or n 1/N+n2/2N.
②AA genotype frequency = number of AA genotype individuals/total diploid population = n1/n.
⑵ The calculation relationship between gene frequency and genotype frequency is deduced from the above ① and ②: A gene frequency = n 1/n+ 1/2 N2/n = AA genotype frequency+1/2 AA genotype.
The types of gene frequency calculation and its formula deduce that the ideal population is a population in genetic balance and follows Hardy-Weinberg equilibrium law. Genetic balance means that the gene frequency and genotype frequency of a population remain stable and balanced between generations, without mutation and natural selection and migration.
For the large population of Aa genotype (zero generation or a generation in genetic balance), the frequency of gene A is P(A)=p, the frequency of gene A is P(a)=q, the sum of gene frequency of dominant gene A and gene frequency of recessive gene A is p+q= 1, and the frequency of male and female individuals transmitting gene A gametes to their offspring is p, corresponding to the frequency of transmitting recessive gene A gametes.
Table 1 male gamete
Female gamete A (P) A (Q) A (P) Aa (P 2) Aa (PQ) A (Q) Aa (Q 2) As can be seen from the above table, there are three genotypes AA, AA and AA in the offspring of this population, and the frequency of occurrence of the three genotypes is P (AA) = P× P = respectively. p(Aa)= 2×p×q = 2pq = H； P(aa)= q×q = q^2=R。 And the sum of their frequencies is p 2+2pq+q 2 = (p+q) 2 = 1. Its gene frequency is the frequency of A gene P (a) = D+1/2h = P2+PQ = P (P+Q) = P; The frequency of a gene p (a) = r+1/2h = Q2+pq = q (p+q) = q shows that the frequency of the offspring gene is the same as that of the parents. Therefore, in all offspring, if there are no factors such as mutation, migration and selection, the genetic components of this population will always be in a balanced state of P 2+2Pq+Q 2. The calculation of genetic balance of sex-linked genes and multialleles still follows the above rules. Using this rule, gene frequency can be obtained from the known genotype frequency; On the contrary, genotype frequencies can be obtained from known gene frequencies. According to the investigation, the incidence of this disease is about110000. What is the gene frequency of recessive pathogenic gene (A) and the genotype frequency of carriers (Aa) carrying this recessive gene in the population?
A. 1% and 0.99% B. 1% and 1.98% C. 1%+0% and 3.96% D. 1% and 0.1.
Analysis: phenylketonuria is an autosomal recessive genetic disease. Because the genotype of the disease is aa, namely Aa = 0.000 1, A = 0.0 1, A =1-A =1-0.0/= 0.99, the frequency of the carrier genotype is AA = 2×
Answer: b
Variant 1. On an island, 500 men out of every 10,000 people suffer from red-green color blindness. What is the number of female carriers per 10,000 people on the island? (Suppose the ratio of male to female is 1: 1) (b)
1000
Variation 2: human ABO blood type is determined by three alleles: ia, IB and I. Through sampling survey, it is found that the blood type frequency (genotype frequency): iaia (iai) = 0.45; Type b (IBIB, ibi) = 0.13; Type AB (IAIB)= 0.06；; Type o (ii)=0.36. Try to calculate the frequencies of IA, IB and I alleles.
A: The IA frequency is 0.3, the IB frequency is 0. 1, and the I frequency is 0.6. For the population living in nature, ideal conditions can't exist at the same time, and the gene frequency of the population can't be balanced, but it is constantly changing and developing. This unbalanced population often uses the data obtained from sampling survey to calculate its gene frequency, which can be divided into two types according to the location of genes.
2. Calculation of1autosomal gene frequency
According to the definition, the frequency of a gene = the number of a gene/the total number of alleles of this gene × 100%. If a diploid organism has a pair of alleles A and A at a certain locus on its autosome, their gene frequencies are P and Q, respectively, and three genotypes of aa, aa and Aa can be formed. The genotype frequencies are D, H and R, respectively. The total number of individuals is n, the number of Aa individuals is n 1, the number of AA individuals is n2, and the number of AA individuals is n3 and N 1+. So:
Genotype frequency AA =D=n 1/N, n1= nd;
Genotype frequency Aa =H=n2/N, N2 = NH；;
Genotype frequency aa =R=n3/N, N3 = NR；;
The frequency of gene a is p (a) = (2n1+N2)/2n = (2nd+NH)/2n = d+1/2h = p.
The frequency of gene a is p (a) = (2n3+N2)/2n = (2nr+NH)/2n = r+1/2h = q.
Because p+q= 1, d+1/2h+r+1/2h = d+r+h =1.
As can be seen from the above inference,
① Basic calculation formula of autosomal gene frequency:
The frequency of gene A =(2× the number of homozygotes of the gene+1× the number of heterozygotes) /2× the total number of individuals in the population survey.
② Derivation and calculation formula of autosomal gene frequency:
Frequency of gene = homozygous frequency of gene+1/2 heterozygous frequency.
Example: 100 individuals were randomly selected from the population, and 30, 60 and 10 AA genotype individuals were detected. Find the gene frequency of this pair of alleles.
Solution 1:
First, find out the total number of alleles and the number of A or A in this population. 100 individuals * * * has 200 genes; There are 2×30+60= 120 A gene and 2× 10+60=80 A gene. Then it is calculated from the basic formula of autosomal gene frequency:
The frequency of gene A is 120÷200=60%.
The frequency of gene A is 80÷200=40%.
Solution 2:
According to the meaning of the question, the frequencies of aa, Aa and AA genotypes are 30%, 60% and 10%, respectively. Through the deduction and calculation of autosomal gene frequencies, it is concluded that:
The frequency of gene A is 30%+ 1/2×60%=60%.
The frequency of gene A is 10%+ 1/2×60%=40%.
Variant 1: It is known that brown (a) is dominant to blue (a) in human eyes, which belongs to the inheritance controlled by genes on autosomes. Among the population of 30,000, there are 3,600 people with blue eyes and 26,400 people with brown eyes, of which12,000 people are homozygous. Then, the gene frequencies of A and A genes in this population are (E) respectively.
A.64% and 36% B.36% and 64% C.50% and 50% D.82% and 18% E.58% and 42%.
Variant 2: A certain number of individuals are randomly selected from a population, of which 40% are bb genotype individuals, 50% are Bb genotype individuals, and 10% are BB genotype individuals, so the frequencies of genes B and B are (b) respectively.
A.90%， 10% B. 65%，35% C. 50%，50% D. 35%，65%
2.2 or the calculation of Y chromosome genetic frequency
For sex-linked inheritance, the gene frequency calculation of genes located in X and Y homologous fragments is the same as that of autosomes. However, the gene located in the non-homologous segments of X and Y is inherited with the X chromosome, and there is no such gene and its allele on the Y chromosome. Similarly, Y chromosome inheritance, there is no equivalent gene on X chromosome. Therefore, when calculating the total number of genes, we only need to consider the total number of genes on the X chromosome (or Y chromosome). If a diploid organism has a pair of alleles B and B at a certain locus on the X chromosome, and their gene frequencies are P and Q respectively, five genotypes XBxbxbXBXB, XBXBxb, XBXBXB, XbY and XBY can be formed, and the genotype frequencies are E, F, G, H and I respectively, and the total number of individuals is n, the number of individuals in XBXBXB is n 1, and the number of individuals in XBXBXB is xb.
E=n 1 /N、F=n2 /N、G=n3 /N、H=n4 /N、I = n5/N；
p(B)=(2n 1+N2+n4)/[2(n 1+N2+n3)+(n4+n5)]=(2n 1+N2+n4)/ 1.5N = 2/3(2E+F+H)
p(b)=(2n 3+N2+n5)/[2(n 1+N2+n3)+(n4+n5)]=(2n 3+N2+n5)/ 1.5N = 2/3(2G+F+I)
As can be seen from the above inference,
① Basic formula for calculating gene frequency of X chromosome:
Gene frequency =(2× number of female homozygotes+number of female heterozygotes+number of men containing the gene) /(2× number of female individuals+number of male individuals)
② Deduction and calculation formula of X chromosome gene frequency:
Gene frequency of a gene =2/3(2× female homozygote frequency+female heterozygote frequency+male genotype frequency of a gene) (when the number of male and female individuals is equal)
Example: 100 individuals were randomly selected from a population, and the individuals with xbxbxbxbxb, xbxbxb, XbY and XBY were 44, 5, 1 and 43, 7 respectively. Find the gene frequency of Xb and XB.
Solution 1:
As far as this pair of alleles is concerned, each female contains 2 genes and each male contains 1 gene (there is no allele on the Y chromosome). Then, 100 individuals * * * have 150 genes, of which females have 2×(44+5+ 1)= 100 genes and males have 43+7=50 genes. There are 44×2+5+43= 136 in Xb gene and 5+ 1×2+7= 14 in XB gene. Then according to the basic formula of X chromosome gene frequency, the calculation is as follows:
The gene frequency of XB is 136 ÷ 150 ≈ 90.7%.
The gene frequency of Xb is 14 ÷ 150 ≈ 9.3%.
Solution 2:
According to the meaning of the question, the genotype frequencies of xbxbxbxb, xbxbxb, XbY, XBY and xby, xby are 44%, 5%, 1% and 43%, 7%, respectively. Because the genotype frequencies of female and male individuals account for 50% respectively, it is calculated from the deduction formula of X chromosome gene frequency:
The gene frequency of XB gene is 2/3× (2× 44%+5%+43%) ≈ 90.7%.
The gene frequency of Xb gene is 2/3× (2×1%+5%+7%) ≈ 9.3%.
Variant 1: A factory has 200 male and 200 female employees. According to the survey, there are 5 female color-blind gene carriers 15, and 5 male patients 1 1 person. Then the frequency of color blindness gene in this population is (b)
A.B. 6% C. 9% D. 7.8%
Solution: The number of color-blind genes (A recessive) is 5 * 2+15+1,and the sum of non-color-blind genes (A dominant) and color-blind genes is 200 * 2+200, so the frequency of color-blind genes is 36/600=0.06.
Variant 2: A genetic survey of students in a European school found that hemophilia patients accounted for 0.7% (male: female = 2:1); Hemophilia carriers account for 5%, so the frequency of X in this population is (C)
A.2.97% B.0.7% C.3.96% D.3.2%
Analysis:
Method 1: First, make sure that 2: 1 is the male-female ratio of patients, and 1: 1 is human. Suppose the total number of people is 3000. Male patients were 3000× 0.7 %× 2/3 = 14, and female patients were 3000× 0.7 %× 1/3 = 7. The carrier is 3000×5%= 150. Then the frequency of X = (14+7× 2+150)/(1500× 2+1500) = 3.96%.
Method 2: The ratio of male to female in the population is 1: 1, which is calculated according to the deduction formula of X chromosome gene frequency:
The frequency of x = 2/3 (0.7 %×1/3× 2+0.7 %× 2/3+5%) = 3.96%.
Answer: choose C.
In a word, although the calculation types of gene frequency are complex and varied, and the thinking methods vary widely, as long as we master the conditions and methods of gene frequency calculation, find out the reasons and use them flexibly, we can accurately calculate the correct answer.
Main references
1. Li Nan. The process of evolution Beijing: Higher Education Press, 1990.9: 244-276.
2. Zhu Zhengwei, Zhao Zhanliang. Biology compulsory course 2: heredity and evolution. Beijing: People's Education Press, 2007: 1 15.
This law is also called "the law of gene balance". 1908, the British mathematician Godfrey Harold Hardy first discovered and proved this law. 1909, the German doctor Wilhelm Weinberg also independently proved this law, hence the name Hardy-Weinberg Law.
It is mainly used to describe the relationship between allele frequency and genotype frequency in a population. The content is:
① Infinite populations randomly mate under ideal conditions, and after many generations, the gene frequency and genotype frequency can still maintain a stable balance.
② In the case of a pair of alleles, the gene frequency relationship between gene P (dominant) and gene Q (recessive) is as follows:
(p+q)^2= 1
Binomial expansion: p 2+2pq+q 2 = 1.
It can be seen that in the formula, p 2 is the proportion of dominant homozygote, 2 Q is the proportion of heterozygote and q 2 is the proportion of recessive homozygote.
Hardy-Weinberg law can also be applied to more complex situations, such as polyploidy.
[Example 1] In a population, aa individuals account for 30%, Aa individuals account for 60%, and AA individuals account for 10%. Calculate the frequency of a and a genes.
[Resolution] The frequency of a gene is 30%+ 1/2×60%=60%.
The frequency of gene A is 10%+ 1/2×60%=40%.
[Answer] 60% 40%
Conclusion: The sum of allele frequencies in the population is equal to 1, and the sum of genotype frequencies in the population is also equal to 1. The change of gene frequency leads to the change of population gene pool, so biological evolution is essentially the process of population gene frequency change.