Presentation on theme: "Basic Principles of Population Genetics Lecture 4"— Presentation transcript:
1Basic Principles of Population Genetics Lecture 4 Background Readings: Chapter 1, Mathematical and statistical Methods for Genetic Analysis, 1997, Kenneth Lang.This slide show follows closely Chapter 1 of Lang’s book. Prepared by Dan Geiger..
2Founders’ allele frequency 231A1/A2B1/B2A’1/A’2B’1/B’2A”1/A”2B”1/B”2In order to write down the likelihood function of a data given a pedigree structure and a recombination value , one need to specify the probability of the possible genotypes of each founder. Assuming random mating we have,Pr(G1,G2)=Pr(A1/A2, B1/B2) Pr(A’1/A’2, B’1/B’2)The likelihood function also consists of transmission matrices that depend on and penetrances matrices to be discussed later.
3Hardy-Weinberg and Linkage Equilibriums The task at hand is to establish a theoretical basis for specifying the probability Pr(A1/A2, B1/B2) of a multilocus, from allele frequencies. We will derive under various assumptions the following two rules which are widely used in genetic analysis (Linkage & Association) and which ease computations a great deal. Of course, the assumptions are not satisfied for all genetic analyses.Hardy-Weinberg (HW) Equilibrium: Pr(A1/A2) = PA1· PA2, namely, the probability of an ordered genotype A1/A2 is the product of the frequencies of the alleles constituting that genotype.A1A2B1B2Linkage Equilibrium: Pr(A1B1) = PA1· PB1, namely, the probability of a haplotype A1,B1 is the product of the frequencies of the alleles constituting that haplotype.These rules imply: Pr(A1/A2, B1/B2)=PA1· PA2 · PB1 · PB2
4A simple setup to study HW equilibrium Consider a bi-allelic locus A with alleles A1, A2 .Let u,v, and w be the frequencies of unordered genotypes A1/A1, A1/A2, A2/A2. Clearly, u+v+w=1.How are these frequencies related to allele frequencies p1 and p2of A1 and A2 ,respectively ?Answer: p1 = u + ½v and p2 = ½v + wBut, the Hardy-Weinberg equilibrium states that alsou = p12v = 2 p1 p (The factor 2 because A1/A2 genotypes are not ordered.)w = p22(p1+p2)2=1Clearly these relations do not hold for arbitrary frequencies u,v,w ; only for those values in the image of this polynomial mapping.
5Assumptions made to Justify HW Infinite population sizeDiscrete generationsRandom matingNo selectionNo migrationNo mutationEqual initial genotype frequencies in the two sexesHW equilibrium can be shown to hold under more relaxed sets of assumptions as well. These assumption are clearly not universal.
6What happens after one generation ? Mating Type-Unordered genotypeNature of Offspringand segregation ratiosFrequency of matesA1/A1 x A1/A1A1/A1u2A1/A1 x A1/A2½ A1/A1 + ½ A1/A22uvA1/A1 x A2/A2A1/A22uwA1/A2 x A1/A2¼ A1/A1 + ½ A1/A2 + ¼ A2/A2v2A1/A2 x A2/A2½ A1/A2 + ½ A2/A22vwA2/A2 x A2/A2A2/A2w2(u+v+w)2=1Frequency of A1/A1 after one generation:u’=u2+ ½(2uv)+ ¼v2= (u+ ½v)2 = p12
7After one generation …So, after one generation the genotype frequencies u,v,w change to u’,v’,w’ as follows (using the previous table):Frequency of A1/A1: u’=u2+uv+ ¼v2= (u+ ½v)2 = p12Frequency of A1/A2:v’=uv+2uw + ½v2 + vw = 2(u+½v)(½v+w) = 2p1p2Frequency of A2/A2: w’=¼v2 + vw + w2 = (½v+w)2 = p22Hardy-Weinberg seems to be established after one generation, butu’,v’,w’ are frequencies for the second generation while p1 and p2 are defined as the allele frequencies of the first generation. Are these also the allele frequencies of the second generation ?Yes ! Because p’1= u’+ ½v’ = p12+p1p2=p1 and similarly p’2= p2.
8After yet another generation … Have we reached equilibrium ? Let’s look at one more generation and see that genotype frequencies are now fixed.Frequency of A1/A1: u”=(u’+ ½v’)2 = (p12+p1p2)2 = p12Frequency of A1/A2: v”= 2(u’+ ½v’)(½v’+w’)= 2(p12+p1p2 )(p22+p1p2 )= 2p1p2Hardy-Weinberg is indeed established after one generation; allele and genotype frequencies do not change under the assumptions we have made. Can you trace where each assumption is used ?Frequency of A2/A2: w”=(½v’+w’)2 = (p22 + p1p2)2 = p22
9Use of Assumptions in the derivation Infinite population sizeDiscrete generations (mating amongst ith generation members only)Random matingNo selectionNo migrationNo mutationEqual initial genotype frequencies in the two sexesSegregation ratios below assume 1,2,3,7Mating Type-Unordered genotypeNature of Offspringand segregation ratiosFrequency of matesA1/A1 x A1/A2½ A1/A1 + ½ A1/A22uvFrequency formula of A1/A1 after one generation: u2+ ½(2uv)+ ¼v2 assume 4,5,6.
10An alternative justification Previously, we started with arbitrary genotype frequencies u,v,w and showed that they are modified after one generation to satisfy HW equilibrium.Now we start with arbitrary allele frequencies p1 and p2.Random mating is equivalent to random pairing of alleles; each person contributes one allele with the prescribed frequencies.So the frequency of A1/A1 in the new generation is p12 , that of A1/A2 is 2p1p2 , and that of A2/A2 is p Argument completed ?The frequency p’1 of A1 in this new generation is p12+ ½(2p1p2 )= p’1 and the frequency of A2 in this new generation is p22+ ½(2p1p2 )=p’2. So after one generation allele frequency is fixed and satisfies the HW equilibrium .Exercise: Generalize the argument to k-allelic loci.
11HW equilibrium at X-linked loci Consider an allele at an X-linked locus. At generation n, let qn denote that allele’s frequency in females and rn denote that allele’s frequency in males.More explicitly,Questions:What is the frequency pn of the allele in the population ?Does pn converge and to which value p ?Does qn and rn converge to the same value ?
12Argument Outline Assuming equal number of males and females, we have pn = 2/3 qn + 1/3 rn for every n.Let p = p0 = 2/3 q0 + 1/3 r0. We will now show that both qn and rn converge quickly to p (but not in one generation as before).Having shown this claim, the female genotype frequency of A1/A1 must be p2 , that of A1/A2 is 2p(1-p) , and that of A2/A2 is (1-p)2, satisfying HW equilibrium.For male, genotypes A1 and A2 have frequencies p and 1-p.
13The recursion equations Because a male always gets his X chromosome from his mother, and his mother precedes him by one generation,rn = qn (Eq. 1.1)Similarly, females get half their X-chromosomes from females and half from males,qn = ½ qn-1+ ½ rn (Eq. 1.2)Eqs 1.1 and 1.2 imply:2/3 qn+1/3 rn =2/3(½ qn-1+ ½ rn-1 ) + 1/3 qn-1=2/3 qn-1 + 1/3 rn-1It follows that the allele frequency pn= 2/3 qn + 1/3 rn never changes and remains equal to p0= p. To see that qn converges to p, we need to relate the difference qn-p with the difference qn-1-p.
14The fixed point solution qn-p = qn- 3/2 p + ½ p= ½ qn-1+ ½ rn-1 -3/2 (2/3 qn-1 + 1/3 rn-1) + ½ p= - ½ qn-1+ ½ p (just cancel terms)= - ½ (qn-1- p)Continuing in this manner,qn-p= - ½ (qn-1- p) = (- ½)2 (qn-2- p) = …= (- ½)n (q0- p) 0So in each step the difference diminishes by half and qn approaches p in a zigzag manner. Hence, rn = qn-1 also converges to p. What does this mean ?Having shown this claim, the female genotype frequency of A1/A1 must be p2 , that of A1/A2 is 2p(1-p) , and that of A2/A2 is (1-p)2, satisfying HW equilibrium. For male, genotypes A1 and A2 have frequencies p and 1-p. HW equilibrium is not reached in one generation but gets there fast (quite there in 5 generations).
15Linkage equilibrium Let Ai be allele at locus A with frequency pi Let Bj be allele at locus B with frequency qjDenote the recombination between these loci by f and m for females and males, respectively.Let = (f + m )/2.Linkage equilibrium means that Pr(Ai Bj) = piqjAiA’iBjB’jWe use the same assumptions employed earlier to demonstrate linkage equilibrium, namely, to show that Pn(Ai Bj) converges to piqj at a rate that is fastest when the recombination is the largest.
16Convergence Proof= ½ [ (1-f )Pn-1(Ai Bj) + f piqj ] +No recombinationrecombinationPn(Ai Bj) = ½ [gamete from female] + ½ [gamete from male]½ [gamete from male]= ½ [ (1-f )Pn-1(Ai Bj) + f piqj ] + ½ [ (1-m )Pn-1(Ai Bj) + m piqj ]= (1- )Pn-1(Ai Bj) + piqjSo, Pn(Ai Bj) - piqj = (1- ) [Pn-1(Ai Bj) – piqj]=…= (1- )n[P0(Ai Bj) – piqj]In short, we have established, For loci on different chromosomes, the deviation from linkage is halved each generation. For close loci with small , convergence is slow.Exercise: Repeat this analysis for three loci (Problem 7, with guidance, in Kenneth Lang’s book).
17Ramifications for Association studies Many diseases are thought to been caused by a single random mutation that survived and propagated to offspring, generation after generation.MarkerMutated locusSuppose there is a close marker:Would we see association at random population samples?If the mutation happened many generations ago, no trace will be significant. Allele frequency will reach linkage equilibrium ! We need a combination of close markers and recentallele age of the disease. Association studies like that are also called linkage disequilibrium mapping or LD mapping in short.
18Selection and FitnessFitness of a genotype is the expected genetic contribution of that genotype to the next generation, or to how many offspring it contributes an allele. Let the fitness of the three genotypes of an autosomal bi-allelic locus be denoted by wA/A, wA/a and wa/a .If pn and qn are the allele frequencies of A and a, then the average fitness under HW equilibrium, is wA/Apn2 + wA/a 2pnqn + wa/a qn2.Conventions: Since only the ratios of fitness of various genotypes matter, namely, wA/A /wA/a and wa/a /wA/a, we arbitrarily set wA/a =1 and define wA/A = 1-r, wa/a = 1-s, where r 1 and s 1.Interpretation: When s=r=0, there is no selection.When r is negative A/A has advantage over A/a. Similarly with negative s. When r is positive (must be fraction), A/A has a disadvantage over A/a. When both s and r are positive, there is a heterozygous advantage.
19Assuming selection exists … Our goal is to study the equilibrium of allele frequencies under various selection possibilities (namely, different values for r and s).In our new notations the average fitness wn at generation n is given by wn (1-r)pn2 + 2pnqn + (1-s)qn2 = 1-rpn2 -sqn2A/AA/aa/aFirst, note that pn+1 = [(1-r)pn2 + pnqn] / wnTo find equilibrium we study the difference pn pn+1 - pnpn pn+1 - pn = [(1-r)pn2 + pnqn] / wn - pn= [(1-r)pn2 + pnqn- (1-rpn2 -sqn2)pn] / wn= [pnqn (s- (r+s) pn)] / wn
20Interpretation when r>0 and s0 We just derived pn = [pnqn (s- (r+s) pn)] / wnConvergence occurs when pn=0, namely, when pn=0, pn=1 (i.e., qn=0) or pn=s/(r+s). Where should it converge to ?Claim: When (r>0 and s 0), pn 0, i.e., allele A disappears. In the opposite case (r0 and s>0), allele a should be driven to extinction.(Why is this extinction process sometimes halted in real life ? )Proof: When (r>0 and s 0), the linear function g(p)=s-(r+s) psatisfies g(0) 0 and g(1) < 0, hence it is negative at (0,1).Thus, pn monotonically decreases at each step. So pn must approach 0 at equilibrium. Similarly, with the other case.
21when r and s have the same sign Conclusion I (for negative sign): If r and s are negative, (pn ) > 1, so pn 0 for p0 above s/(r+s), and pn 1 for p0 below s/(r+s). In other words, s/(r+s) is an unstable equilibrium.
22when r and s are both positive If both r and s are positive (Heterozygous advantage), thenHencehas a constant sign and declines in magnitude.Conclusion II: If both r and s are positive, pn s/(r+s) and this point is a stable equilibrium.Conclusion III (rate of convergence): If p0 s/(r+s), namely the starting point is near equilibrium, then,and we get (locally) a geometric convergence
23Heterozygous advantage If we observe a recessive disease that is maintained in high frequency, how can we explain it ? Intuition says that it should disappear.However, if the A/a genotype has an advantage over other genotypes, then the defective allele would be kept around.Technically, if both r and s are positive, then the A/a genotype has the best fit.The best evidence for such phenomena is the sickle cell anemia.In some part of Africa, this anemia, despite being a recessive disease, is kept in high frequency. It turns out that the A/a genotype appears to provide protection against malaria ! (so it has high fit in swamp-like areas).
24Sickle cell anemiaאנמיה חרמשית - Medical EncyclopediaRed blood cells, sickle cellSickle cell anemia is an inherited autosomal recessive blood disease in which the red blood cells produce abnormal pigment (hemoglobin).The abnormal hemoglobin causes deformity of the red blood cells into crescent or sickle-shapes, as seen in this photomicrograph.The sickle cell mutation is a single nucleotide substitution (A T) at codon 6 in the beta-hemoglobin gene, resulting in the following substitution of amino acids: GAG (Glu) GTG (Val).Source (Edited):
25Facts about Sickle cell Disease Sickle Cell Disease is much more common in certain ethnic groups affecting approximately one out of every 500 African Americans.Because people with sickle trait were more likely to survive malaria outbreaks in Africa than those with normal hemoglobin, it is believed that this genetically aberrant hemoglobin evolved as a protection against malaria.Although sickle cell disease is inherited and present at birth, symptoms usually don't occur until after 4 months of age.Sickle cell anemia may become life-threatening when damaged red blood cells break down (and other circumstances). Repeated crises can cause damage to the kidneys, lungs, bones, eyes, and central nervous system.Blocked blood vessels and damaged organs can cause acute painful episodes. These painful crises, which occur in almost all patients at some point in their lives. Some patients have one episode every few years, while others have many episodes per year. The crises can be severe enough to require admission to the hospital for pain control.
26Balance of Mutation and Selection Most mutations are neutral or deleterious. We discuss balance between deleterious mutations and selection. Let denote the mutation rate from a to A. Suppose the equilibrium frequency of allele A is p and of a is q=1-p.When is a balance achieved between selection (say, preferring allele a ) and mutation that changes allele a back to allele A ?The frequencies p and q must satisfy the equilibrium condition:(r>0, s=0)(recessive disease)This yields 1- rp2 = 1- and thus p2 = /r and a balance is achieved that retains both alleles.
27Finite Population Genetic Drift Alelle 10Alelle 5Source: Gideon GreenspanAfter 800 generations, by simulation, from the ten alleles only two remain: numbered 5 and number 7.