Human Origins mtdna 夏娃 (Eve) Y chromosome 亚当 (Adam) Biological Evolution

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1 Population Genetics 群体遗传

2 Human Origins mtdna 夏娃 (Eve) Y chromosome 亚当 (Adam) Biological Evolution

3 Human diversity

4 Overviews Medical population genetics mainly studies the behavior of gene in human population, including the estimation of gene frequencies and heterozygote frequencies( 杂合子频率 ). --Broad sense population( 广义群体 ): all persons (6 billions) in the world. --Narrow sense population( 狭义群体 ): Mendel population in some areas.

5 Medical population genetics F1:DD x dd ~Dd (Dominant) F2:Dd x Dd 3/4 will be DD/Dd (Dominant) The frequency of D gene in population?while d gene?with the death of dd affected, d gene will disappear from the population?

6 Topics The Hardy-Weinberg equilibrium Application of the Hardy-Weinberg equilibrium Factors affecting the Hardy- Weinberg equilibrium Inbreeding coefficient

7 The Hardy-Weinberg Equilibrium Hardy G.H. was an English mathematician and Weinberg W. was a German physician. In 1908, they developed independently a mathematical model. They made the following assumptions: ---The genotypes could be distinguished unequivocally, meaning that the frequencies of the phenotypes were the same as those of the genotypes. ---No mutation, no selection, and no migration. ---Random mating at the same time in a population of infinite size.

They made the following assumptions: ---The genotypes could be distinguished unequivocally, meaning that the

8 Evidence of the H-W law Assume one locus with two alleles: A and a, the former being dominant and the latter being recessive. Let the relative frequencies of the two alleles be p and q respectively. As there are only two alleles, p + q = 1. Then the sperms produced by the male population and eggs produced by the female population will contain them in the same proportions.

Let the relative frequencies of the two alleles be p and q respectively.

9 Evidence of the H-W law With random mating the various gametic combinations can be represented as: The frequencies of three genotypes among F1 offspring from such matings are p 2 (AA), 2pq(Aa), q 2 (aa).

10 Evidence of the H-W law Gene frequencies in F1 offspring: AA: p 2 ; Aa: pq+qp=2pq; aa: q 2 A = p 2 +1/2(2pq)=p 2 +pq=p(p+q)=p a = q 2 +1/2(2pq)=q 2 +pq=q(p+q)=q

11 Evidence of the H-W law Now consider that F1 offspring mate with each other. The frequencies of various matings can be represented as: AA(p 2 ) Aa(2pq) aa(q 2 ) AA(p 2 ) AA AA AA Aa AA aa p 4 2p 3 q p 2 q 2 Aa(2pq) Aa AA Aa Aa Aa aa 2p 3 q 4p 2 q 2 2pq 3 aa(q 2 ) aa AA aa Aa aa aa p 2 q 2 2pq 3 q 4

12 Group together reciprocal matings, AA Aa and Aa AA, for example, and sum up their respective frequencies. We thus obtain six mating types and their frequencies. The expected proportions of the three genotypes among the progeny of these six matings can be calculated according to a priori segregation ratio. Mating type Frequency AA Aa aa AA AA p 4 p AA Aa 4p 3 q 2p 3 q 2p 3 q - Aa Aa 4p 2 q 2 p 2 q 2 2p 2 q 2 p 2 q 2 AA aa 2p 2 q 2-2p 2 q 2 - Aa aa 4pq 3-2pq 3 2pq 3 aa aa q q 4

The expected proportions of the three genotypes among the progeny of these six matings can be calculated according to a

13 Evidence of the H-W law We have the predicted frequency distribution of genotype among the F2 progeny of all matings. Frequency of AA = p 4 +2p 3 q+p 2 q 2 =p 2 (p 2 + 2pq+ q 2 ) = p 2 ; Frequency of Aa = 2p 3 q+4p 2 q 2 +2pq 3 = 2pq Frequency of aa = p 2 q 2 +2pq 3 +q 4 =q 2 The relative frequencies of three genotypes among F2 progeny are the same as those among F1 generation: p 2 (AA), 2pq(Aa) and q 2 (aa). So are gene frequencies. Frequency of A = p; Frequency of a = q.

Frequency of aa = p 2 q 2 +2pq 3 +q 4 =q 2 The relative frequencies of three genotypes among F2 progeny are the same as

14 The Hardy-Weinberg theorem A population undergoing random mating reaches, in one generation, a distribution of genotype frequencies given by the binomial expansion of [p(a)+q(a)] 2 =p 2 (AA) + 2pq (Aa) + q 2 (aa) One of the most important feature of the Hardy-Weinberg theorem is that it enables us to express the distribution of genotypes and phenotypes in a population entirely in terms of the gene frequencies.

(Aa) + q 2 (aa) One of the most important feature of the Hardy-Weinberg theorem is that it enables us

15 For a locus of two alleles, A and a, with gene frequencies p and q respectively: Relative genotype frequencies: p 2 +2pq+q 2 =1 Relative phenotype frequencies, assuming complete dominance of A over a: (p 2 +2pq)( A )+q 2 ( a ) =1 The H-W theorem can be extended quite easily to cover multiple alleles. For three alleles of the ABO locus, I A,I B and i, with gene frequencies p, q and r respectively: then (p + q + r) 2 =p 2 +q 2 +r 2 +2pq+2pr+2qr

a ) =1 The H-W theorem can be extended quite easily to cover multiple alleles.

16 The Hardy-Weinberg theorem For a locus having n alleles (A 1, A 2,..., A n ) with gene frequencies p 1, p 2,..., p n respectively, the genetic equilibrium of this locus under random mating can be expressed as: [p 1 (A 1 )+p 2 (A 2 )+..., +p n (A n )] 2 = pi 2 (AiAi) + 2pipj(AiAj)

.., p n respectively, the genetic equilibrium of this locus under

17 Applications of the H-W law Estimation of gene frequency and heterozygote frequency: Counting method for estimating frequencies of codominant alleles: such as MN blood group. Phenotypes: M(+)N(-); M(+)N(+); M(-)N(+) Genotypes: MM(M); MN(MN); NN(N) p = (2M + 1MN) / 2n = q = 1 p =

18 Applications of the H-W law Square root method for estimating frequency of recessives allele when there is complete dominance: Population incidence of alcaptonuria( 尿黑酸尿症 ): x = 1/ = q 2 so [a] q = x = 1 / 1000 [A] p = 1 q = 1, [Aa] 2pq = 2 1 (1/1000) = 1 / 500

Population incidence of alcaptonuria( 尿黑酸尿症 ): x = 1/1000000 =

19 Applications of the H-W law ABO blood group locus have three allele: according to H-W law, we assume I A (p), I B (q), i (r), then (p + q + r) 2 = p 2 + 2pq + q 2 + 2qr + r 2 + 2pr we have r = O, B = q 2 + 2qr, O = r 2 therefore B + O = q 2 + 2qr + r 2 = (q + r) 2 = (1 - p) 2 Then we have p = 1 B + O Thus we have q = 1 p r Then, we can estimate I A, I B and i gene frequency. e.g. q = , p = , r =

2qr, O = r 2 therefore B + O = q 2 + 2qr + r 2 = (q + r) 2 = (1 - p) 2 Then we have p = 1 B + O Thus we

20 Applications of the H-W law Xg blood group locus is X-linked and has two alleles with Xg a being dominant over Xg. In male population, the gene frequencies are the same as the phenotype & genotype frequencies. In female population, the gene frequencies can be estimated with square root method. Note: the gene frequency from male & female population has not significant difference (P>0.05).

In male population, the gene frequencies are the same as the phenotype & genotype frequencies.

21 Applications of the H-W law Sex Blood group Genotype No. Of popula -tion Freq.of gene Male Xg (a+) Xg a Y Xg (a-) XgY Freq. of phenotype Female Xg (a+) Xg a Xg a Xg a Xg Xg (a-) XgXg

22 Some concepts derived from the H-W theorem For a rare dominant gene, p 2 is negligible, then 2pq / p 2 + 2pq 1. Nearly all of the affected individuals are heterozygotes. For a rare recessive gene, q 2 is very much smaller than p which is close to one. 2pq = 2(1-q)q = 2q-2q 2 2q. q (2pq)/2 =2 or (2pq)/q 2 For a rare recessive gene, p is close to one. Then, 2pq / q 2 = 2p/q = 2 / q

23 Some concepts derived from H-W theorem For a rare X-linked dominant gene, p is Negligible, so p / (p 2 + 2pq) = 1 /(2 p) 1/2 For a rare X-linked recessive gene, there occur more affected males than affected females, q / q 2 = 1/q

24 Test of genetic hypothesis Phenylthiocarbamide (PTC) tasting. Giving a genetic hypothesis: alleles T and t, then, Taster genotype is TT or Tt, nontaster is tt PTC tasting with family data: a. Observed family data b. Expected phenotype distribution among offspring.

25 If it is Taster Taster, the overall proportion of recessives among offspring is p 2 q 2 / p 2 (1 + q) 2 = (q / 1 + q) 2. Mating type Freq.of mating Freq. of Taster TT TT p 4 p 4 - TT Tt 4p 3 q 4p 3 q - Freq. of Nontaster Tt Tt 4p 2 q 2 3p 2 q 2 p 2 q 2

26 If it is Taster Nontaster, the overall proportion of recessives among offspring is 2pq 3 / 2pq 2 (1+q) = q / 1 + q Mating type Freq. of mating Freq. of Taster TT tt 2p 2 q 2 4p 2 q 2 - Freq. of Nontaster Tt tt 4pq 3 2pq 3 2pq 3

27 Test of genetic hypothesis Estimation of q from population data: q = = , p = Chi square test for goodness of fit --- Taster Taster: χ 2 = O-E 2 / E = , P> Taster non-taster: χ 2 = , df = 1, P>0.05 The observed family data supports the hypothesis

28 Inbreeding Coefficient (I or F) IC is the probability that an individual receives at a given locus two genes that are identical by descent. IC for autosomal locus in progeny of first-cousin marriage: I = 1 / 16

29 Evaluating the deleterious effect of consanguineous marriages The total probability that a first-cousin marriage will produce a gene a with frequency q = The probability that the genes come together from their common ancestor + the probability that the genes come together from some other source = (1/16)q + (15/16)q 2. The probability that a random marriage will produce a homozygous recessive for a gene a with frequency q = q 2

30 Evaluating the deleterious effect of consanguineous marriages The deleterious effect of the first-cousin marriage = [(1/16)q + (15/16)q 2 ] / q 2 Suggestion from the deleterious effect of consanguineous marriages. If a disease is rare, a high incidence of consanguineous marriages among the parents can be considered quite conclusive evidence. Namely: i. That it is inherited. ii. That the responsible gene is an autosomal recessive.

31 Deleterious effect of consanguineous marriages Gene fre. (q) q2 (1/16)pq q2+(pq/ 16) 1+15q/ 16q

32 Factor Affecting the H-W law Deviation from random mating Assortative mating (positive or negative): the tendency for human beings to choose partners who share characteristics such as height intelligence and racial. Inbreeding: marriage between closely related individuals. Mutation The rate of mutation for any particular gene is probably fairly constant.

33 Factor Affecting the H-W law Selection (s) It operates by increasing (position selection) or decreasing (negative selection) the reproductive fitness (f). s + f = 1. Relatively high frequencies of some mutant genes are believed to be maintained by a selective advantage in heterozygotes (hybrid vigour, heterosis 杂合子优势 ) Migration: It is referred to as gene flow. Random genetic drift: It occurs small & isolated population. A gene may became extinguished or fixed.

34 Well bear and well rear in China is not Eugenics 隐性基因频率为 q 去除 aa 后的隐性基因频率 q = pq/p(1+q) = q/(1+q) n 代后 :q n =q 0 /(1+nq 0 ) n = 1/q n 1/q >200 0 代两种相反的因素可稳定平衡

35 Conclusion H-W law is probably only an approximation, because it is certainly hard to believe that all assumptions, on the basis of which the principle is derived, are realized in any real population. However, the effects of mutation, selection, migration, and random genetic drift are very rarely large enough to be detectable among genotype distributions. The sample must be very large, or conditions very unusual, for a deviation from equilibrium to be detectable.

36 思考题等显性遗传时采用什么方法估计基因频率? 一对基因有显隐性时如何估计基因频率和杂合子频率? 请用群体遗传学知识, 解释为什么在 XD 遗传时女性患者多于男性患者? 为什么 XR 遗传时男性患者多于女性患者?

37 在某个群体中测得 PTC 非尝味者频率为 16%, 试问 : 1)PTC 尝味者与非尝味者的基因频率分别是多少? 2) 在尝味者与尝味者婚配类型的合并资料中, 你期望子代中尝味者与非尝味者各占多少? 3) 尝味者与非尝味者的合并资料中, 子代中尝味者与非尝味者各占多少?

38 试计算下列家系中 A 和 B 的亲缘系数 (K) 以及 C 的近婚系数 (F)

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Human Origins mtdna 夏 娃 (Eve) Y chromosome 亚 当 (Adam) Biological Evolution

Human Origins mtdna 夏娃 (Eve) Y chromosome 亚当 (Adam) Biological Evolution