1、 群體遺傳Population GeneticsBiological EvolutionHuman OriginsmtDNA夏娃(Eve)Y chromosome亞當(dāng)(Adam)Human diversityOverviewsOverviewslMedical population genetics mainly studies the behavior of gene in human population, including the estimation of gene frequencies and heterozygote frequencies(雜合子頻率).l-Broad sen

2、se population(廣義群體): all persons (7 billions) in the world. l-Narrow sense population(狹義群體): Mendel population in some areas.Medical population geneticsMedical 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 th

3、e death of dd affected, d gene will disappear from the population ？TopicslThe Hardy-Weinberg equilibriumlApplication of the Hardy-Weinberg equilibriumlFactors affecting the Hardy-Weinberg equilibriumlInbreeding coefficientThe Hardy-Weinberg EquilibriumThe Hardy-Weinberg EquilibriumlHardy G.H. was an

4、 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.-

5、No mutation, no selection, and no migration.-Random mating at the same time in a population of infinite size.Evidence of the H-W lawEvidence of the H-W lawlAssume one locus with two alleles: A and a, the former being dominant and the latter being recessive. lLet the relative frequencies of the two a

6、lleles 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. Evidence of the H-W lawEvidence of the H-W lawlWith random mating the various gametic combin

7、ations can be represented as: The frequencies of three genotypes among F1 offspring from such matings are p2 (AA), 2pq(Aa), q2(aa).lGene frequencies in F1 offspring: 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 lawEvidence of the H-W

8、lawEvidence of the H-W lawEvidence of the H-W lawlNow 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)AAAAp4AAAa2p3qAAaap2q2Aa(2pq)AaAA2p3qAaAa4p2q2Aaaa2pq3aa(q2)aaAAp2q2aaAa2pq3aaaaq4lGroup together reciprocal mating

9、s, AAAa and AaAA , for example, and sum up their respective frequencies. We thus obtain six mating types and their frequencies.lThe expected proportions of the three genotypes among the progeny of these six matings can be calculated according to a priori segregation ratio.Evidence of the H-W lawEvid

10、ence of the H-W lawMating typeFrequencyAAAaaaAAAAp4p4-AAAa4p3q2p3q2p3q-AaAa4p2q2p2q22p2q2p2q2AAaa2p2q2-2p2q2-Aaaa4pq3-2pq32pq3aaaaq4-q4We 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 =

11、 2p3q+4p2q2 +2pq3 = 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.The Hardy-Weinberg theoremlA population underg

12、oing random mating reaches, in one generation, a distribution of genotype frequencies given by the binomial expansion of p(A)+q(a)2 = p2(AA) + 2pq (Aa) + q2 (aa) (二項(xiàng)式展開(kāi))lOne of the most important feature of the Hardy-Weinberg theorem is that it enables us to express the distribution of genotypes and

13、 phenotypes in a population entirely in terms of the gene frequencies.lFor 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”) =1The Har

14、dy-Weinberg theoremlThe H-W theorem can be extended quite easily to cover multiple alleles.lFor 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+2qrThe Hardy-Weinberg theoremlFor a locus having n alleles (A1, A2,., An) wi

15、th gene frequencies p1, p2,., pn respectively, 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 theoremApplications of the H-W lawApplications of the H-W lawlEstimation of gene frequency and heterozyg

16、ote frequency: lCounting method （基因計(jì)數(shù)法） for estimating frequencies of codominant alleles: such as MN blood group.lPhenotypes: M(+)N(-); M(+)N(+); M(-)N(+)lGenotypes: MM(M); MN(MN); NN(N) p = (2M + 1MN) / 2n = 0.4628 q = 1 p = 0.5372Applications of the H-W lawApplications of the H-W lawlSquare root m

17、ethod (開(kāi)平方根法) for estimating frequency of recessives allele when there is complete dominance:lPopulation incidence of alcaptonuria(尿黑酸尿癥): x = 1/1000000 = q2 so a q = x = 1 / 1000 A p = 1 q = 1 , Aa 2pq = 21 (1/1000) = 1 / 500Applications of the H-W lawApplications of the H-W lawlABO blood group loc

18、us 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 thereforel B + O = q2 + 2qr + r2 = (q + r)2 = (1 - p)2l Then we have p = 1 B + Ol Thus we have q = 1 p r Then, we can estimate IA, IB an

19、d i gene frequency.e.g. q = 0.2115, p = 0.2324, r = 0.5555Applications of the H-W lawApplications of the H-W lawlXg blood group locus is X-linked and has two alleles with Xga being dominant over Xg.lIn male population, the gene frequencies are the same as the phenotype & genotype frequencies.lIn

20、 female population, the gene frequencies can be estimated with square root method.Applications of the H-W lawApplications of the H-W lawNote: the gene frequency from male & female population has not significant difference (P0.05).SexBlood groupGeno-typeNo. of populationFreq. of phenotypeFreq. of

21、 geneMaleXg(a+)Xg(a-)XgaYXgY1881100.6310.369p=0.631q=0.369Female Xg(a+)Xg(a-)Xga XgaXga XgXg Xg260310.8930.107P=0.673q=0.327Some concepts derived from the H-W theoremlFor a rare dominant gene, p2 is negligible, then 2pq / (p2 + 2pq) 1. Nearly all of the affected individuals are heterozygotes.lFor a

22、rare recessive gene, q2 is very much smaller than p which is close to one. 2pq = 2(1-q)q = 2q-2q2 2q. (2pq)/q 2lFor a rare recessive gene, p is close to one. Then, 2pq / q2 = 2p/q = 2 / q Some concepts derived from Some concepts derived from H-W theoremH-W theoremlFor a rare X-linked dominant gene,

23、p is Negligible, so p / (p2 + 2pq) = 1 /(2 p) 1/2lFor a rare X-linked recessive gene, there occur more affected males than affected females, q / q2 = 1/qTest of genetic hypothesislPhenylthiocarbamide (苯硫脲，PTC) tasting .lGiving a genetic hypothesis: alleles T and t, then,lTaster genotype is TT or Tt,

24、 nontaster is ttlPTC tasting with family data:la. Observed family datalb. Expected phenotype distribution among offspring. If it is Taster Taster, the overall proportion of recessives among offspring is p2q2 / p2(1 + q)2 = q / (1 + q)2.Mating typeFreq. of matingFreq. of TasterFreq. of Non-tasterTTTT

25、p4p4-TTTt4p3q4p3q-TtTt4p2q23p2q2p2q2If it is Taster Nontaster, the overall proportion of recessives among offspring is 2pq3 / 2pq2(1+q) = q / (1 + q)Mating typeFreq. of matingFreq. of TasterFreq. of Non-tasterTTtt2p2q22p2q2-Tttt4pq32pq32pq3Test of genetic hypothesisTest of genetic hypothesislEstimat

26、ion of q from population data: q = 0.2981 = 0.5460, p = 0.4540 lChi square test for goodness of fit l- TasterTaster: 2 = 0.0487, df = 1, l P0.05l- TasterNon-taster: 2 = 0.5048, df = 1, l P0.05lThe observed family data supports the hypothesislConsanguineous mating (近親婚配) has a risk higher than that o

27、f random mating of producing a child affected with an AR/XR disorder.lCoefficient of kinship (親緣系數(shù))：The probability that a gene taken at random from an individual A, at a given locus, be identical by descent to a gene taken at random from a second individual B at the same locus.Coefficient of kinshi

28、p & Consanguineous marriageSuppose parent has a gene pair A1A2lK (parent-child or I1-II1) = 1/2lK (sib-sib or II1-II2) =(1/21/2) + (1/21/2) =1/2lK (grandparent-grandchild or I1-III1) =(1/21/2)=1/4Coefficient of kinship = KCoefficient of kinshipInbreeding Coefficient (Inbreeding Coefficient (近婚系數(shù)

29、近婚系數(shù)I or F)I or F)lIC is the probability that an individual receives at a given locus two genes that are identical by descent.lIC for autosomal locus in progeny of first-cousin marriage: I = 1 / 16Evaluating the deleterious effect (Evaluating the deleterious effect (有害效應(yīng)) of consanguineous marriages

30、) of consanguineous marriageslThe 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)q2.lThe probabi

31、lity that a random marriage will produce a homozygous recessive for a gene “a” with frequency q = q2lThe deleterious effect of the first-cousin marriage = (1/16)q + (15/16)q2 / q2lSuggestion from the deleterious effect of consanguineous marriages. lIf a disease is rare, a high incidence of consangui

32、neous 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.Evaluating the deleterious effect Evaluating the deleterious effect of consanguineous marriagesof consanguineous marriagesDeleteriou

33、s effect of Deleterious effect of consanguineous marriagesconsanguineous marriagesGene fre.（q）q2(1/16)pqq2+(pq/ 16)1+15q/ 16q0.200.04000.0100.0501.250.100.01000.00560.0151.560.040.00160.00240.0042.500.020.00040.00120.0024.060.010.00010.00060.00077.190.0010.000000.000060.0000663.5Factor Affecting the

34、 H-W Factor Affecting the H-W lawlawlDeviation from random matinglAssortative mating(選型婚配)(positive or negative): the tendency for human beings to choose partners who share characteristics such as height intelligence and racial. lInbreeding (近親婚配): marriage between closely related individuals.Factor

35、 Affecting the H-W Factor Affecting the H-W lawlawlSelection (s) (選擇) It operates by increasing (position selection) or decreasing (negative selection) the reproductive fitness 生殖適合度(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 雜合子優(yōu)勢(shì))lMutation The