医学遗传与胚胎发育 Chapter 11 群体遗传

1、 群体遗传Population Genetics1Biological EvolutionHuman OriginsmtDNA夏娃(Eve)Y chromosome亚当(Adam)2Human diversity3OverviewsMedical population genetics mainly studies the behavior of gene in human population, including the estimation of gene frequencies and heterozygote frequencies.-Broad sense population(广

2、义群体): all persons (7 billions) in the world. -Narrow sense population(狭义群体): Mendel population in some areas.4Medical population geneticsF1：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 disapp

3、ear from the population ？5The Hardy-Weinberg EquilibriumHardy 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, meaning that the frequencies o

4、f 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.6TopicsThe Hardy-Weinberg equilibriumApplication of the Hardy-Weinberg equilibriumFactors affecting the Hardy-Weinberg equilibrium7Evid

5、ence of the H-W lawAssume 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 produce

6、d by the female population will contain them in the same proportions. 8Evidence of the H-W lawWith random mating the various gametic combinations can be represented as:The frequencies of three genotypes among F1 offspring from such matings are p2(AA), 2pq(Aa), q2(aa).9Gene frequencies in F1 offsprin

7、g: AA: p2; Aa: pq+qp=2pq; aa: q2 A = p2+1/2(2pq)=p2+pq=p(p+q)=p a = q2+1/2(2pq)=q2+pq=q(p+q)=qEvidence of the H-W law10Evidence of the H-W lawNow consider that F1 offspring mate with each other. The frequencies of various matings can be represented as:AA (p2)Aa (2pq)aa (q2)AA (p2)AAAAp4AAAa2p3qAAaap

8、2q2Aa (2pq)AaAA2p3qAaAa4p2q2Aaaa2pq3aa (q2)aaAAp2q2aaAa2pq3aaaaq411Group together reciprocal mating, 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.

9、Mating typeFrequencyAAAaaaAAAAp4p4-AAAa4p3q2p3q2p3q-AaAa4p2q2p2q22p2q2p2q2AAaa2p2q2-2p2q2-Aaaa4pq3-2pq32pq3aaaaq4-q412We have the predicted frequency distribution of genotype among the F2 progeny of all matings.Frequency of AA = p4 +2p3q+p2q2 =p2(p2+ 2pq+ q2 ) = p2;Frequency of Aa = 2p3q+4p2q2 +2pq3

10、 = 2pqFrequency of aa = p2q2 +2pq3+q4 = q2The relative frequencies of three genotypes among F2 progeny are the same as those among F1 generation: p2(AA), 2pq(Aa) and q2 (aa). So are gene frequencies. Frequency of A = p; Frequency of a = q.Evidence of the H-W law13The Hardy-Weinberg theoremA populati

11、on undergoing random mating reaches, in one generation, a distribution of genotype frequencies given by the binomial expansion (二项式展开) of p(A)+q(a)2 = p2(AA) + 2pq (Aa) + q2 (aa)One of the most important feature of the Hardy-Weinberg theorem is that it enables us to express the distribution of genot

12、ypes and phenotypes in a population entirely in terms of the gene frequencies.14For a locus of two alleles, A and a, with gene frequencies p and q respectively: Relative genotype frequencies: p2+2pq+q2 = 1 Relative phenotype frequencies, assuming complete dominance of A over a: (p2+2pq)(”A”)+q2 (”a”

13、) = 1The H-W theorem can be extended quite easily to cover multiple alleles. For three alleles of the ABO locus, IA, IB and i, with gene frequencies p, q and r respectively: then (p + q + r)2 = p2+q2+r2+2pq+2pr+2qr15For a locus having n alleles (A1, A2,., An) with gene frequencies p1, p2,., pn respe

14、ctively, the genetic equilibrium of this locus under random mating can be expressed as: p1(A1)+p2(A2)+., +pn(An)2 =pi 2(AiAi) + 2pipj(AiAj)The Hardy-Weinberg theorem16Applications of the H-W lawEstimation of gene frequency and heterozygote frequencyTest of genetic hypothesis.Evaluating the deleterio

15、us effect (有害效应) of consanguineous marriages17Estimation of gene frequency and heterozygote frequencyCounting method （计数法） 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 = 0.4628 q =

16、 1 p = 0.537218Square root method (开平方根法) for estimating frequency of recessives allele when there is complete dominance:Population incidence of alcaptonuria(尿黑酸尿症): x = 1/1000000 = q2 so a q = x = 1 / 1000 A p = 1 q 1 , Aa 2pq = 21 (1/1000) 1 / 500Estimation of gene frequency and heterozygote frequ

17、ency19ABO blood group locus have three alleles: according to H-W law, we assume IA (p), IB (q) , i (r), then (p + q + r)2= p2 + 2pq + q2+ 2qr + r2 + 2pr we have r = O, B = q2 + 2qr, O = r2 therefore B + O = q2 + 2qr + r2 = (q + r)2 = (1 - p)2 Then we have p = 1 B + O Thus we have q = 1 p r Then, we

18、can estimate IA, IB and i gene frequency.e.g. q = 0.129, p = 0.281, r = 0.590 (page 120)Estimation of gene frequency and heterozygote frequency20Xg blood group locus is X-linked and has two alleles with Xga being dominant over Xg.In male population, the gene frequencies are the same as the phenotype

19、 & 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 (P0.05). Estimation of gene frequency and heterozygote frequency21SexBlood groupGenotypeNo. of populationF

20、req. of phenotypeFreq. of geneMaleXg(a+)XgaY1880.6310.631Xg(a-)XgY1100.3690.369FemaleXg(a+)XgaXga2600.8930.673XgaXgXg(a-)XgXg310.1070.327Xg blood group locus is X-linked and has two alleles with Xga being dominant over Xg22Some concepts derived from the H-W theoremFor a rare dominant gene, p2 is neg

21、ligible, then 2pq / (p2 + 2pq) 1. Nearly all of the affected individuals are heterozygotes.For a rare recessive gene, q2 is very much smaller than p which is close to one. 2pq 2q; then, (2pq)/q 2For a rare recessive gene, p is close to one. Then, 2pq / q2 = 2p/q = 2 / q 23Some concepts derived from

22、H-W theoremFor a rare X-linked dominant gene, p is negligible, so p / (p2 + 2pq) = 1 /(2 p) 1/2For a rare X-linked recessive gene, there occur more affected males than affected females, q / q2 = 1/q24Test of genetic hypothesisPhenylthiocarbamide (苯硫脲, PTC) tasting.Giving a genetic hypothesis: allele

23、s T and t, then,Taster genotype is TT or Tt, nontaster is ttPTC tasting with family data:a.Observed family datab.Expected phenotype distribution among offspring. 25If it is Taster Taster, then Mating typeFreq. of matingFreq. of TasterFreq. of Non-tasterTTTTp4p4-TTTt4p3q4p3q-TtTt4p2q23p2q2p2q2The ove

24、rall proportion of recessives among offspring is p2q2 / p2(1 + q)2 = q / (1 + q)2.26If it is Taster Nontaster, thenMating typeFreq. of matingFreq. of TasterFreq. of Non-tasterTTtt2p2q22p2q2-Tttt4pq32pq32pq3The overall proportion of recessives among offspring is 2pq3 / 2pq2(1+q) = q / (1 + q)27Test o

25、f genetic hypothesisEstimation of q from population data: q = 0.2981 = 0.5460, p = 0.4540Chi square test for goodness of fit TasterTaster: 2 = |O-E|2 / E = 0.0487, df = 1, p 0.05 Tasternon-taster: 2 = 0.5048, df = 1, p 0.05The observed family data supports the hypothesis28Inbreeding Coefficient (近婚系

26、数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 or F = 1 / 1629Evaluating the deleterious effect (有害效应) of consanguineous marriagesThe total probability that a first-cousi

27、n 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)q2.The probability that a random marriage will produce a homozygous recessive for a g

28、ene “a” with frequency q = q230So, the deleterious effect of the first-cousin marriage = (1/16)q + (15/16)q2 / q2Suggestion 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

29、 evidence. Namely: i. That it is inherited. ii. That the responsible gene is an autosomal recessive.Evaluating the deleterious effect of consanguineous marriages31Deleterious effect of consanguineous marriagesGene fre.（q）q2(1/16)pqq2+(pq/ 16)1+15q/ 16q0.200.04000.0100.0501.250.100.01000.00560.0151.5

30、60.040.00160.00240.0042.500.020.00040.00120.0024.060.010.00010.00060.00077.190.0010.000000.000060.0000663.532Factor Affecting the H-W lawDeviation from random matingAssortative mating(选型婚配)(positive or negative): the tendency for human beings to choose partners who share characteristics such as height intelligence and racial. Inbreeding (近亲婚配): marriage be