Basic Principles of Population Genetics Lecture 2

1. Basic Principles of Population GeneticsLecture 2 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. .

2. A1/A2 B1/B2 A’1/A’2 B’1/B’2 1 2 A”1/A”2 B”1/B”2 3 Founders’ allele frequency In 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.

3. A1 A2 B1 B2 Hardy-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 genotypeA1/A2 is the product of the frequencies of the alleles constituting that genotype. Linkage 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

4. Answer: p1 = u + ½v and p2 = ½v + w But, the Hardy-Weinberg equilibrium states that also u = p12 v = 2 p1 p2 (The factor 2 because A1/A2 genotypes are not ordered.) w = p22 ------------- (p1+p2)2=1 Clearly these relations do not hold for arbitrary frequencies u,v,w ; only for those values in the image of this polynomial mapping. A 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 unorderedgenotypesA1/A1, A1/A2, A2/A2. Clearly, u+v+w=1. How are these frequencies related to allele frequencies p1 and p2 of A1 and A2 ,respectively ?

5. Assumptions made to Justify HW • Infinite population size • Discrete generations • Random mating • No selection • No migration • No mutation • Equal initial genotype frequencies in the two sexes HW equilibrium can be shown to hold under more relaxed sets of assumptions as well. These assumption are clearly not universal.

6. What happens after one generation ? (u+v+w)2=1 Frequency of A1/A1after one generation: u’=u2+ ½(2uv)+ ¼v2 = (u+ ½v)2 = p12

7. After 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 = p12 Frequency of A1/A2: v’= uv+2uw + ½v2 + vw = 2(u+½v)(½v+w) = 2p1p2 Frequency of A2/A2: w’=¼v2 + vw + w2 = (½v+w)2 = p22 Hardy-Weinberg seems to be established after one generation, but u’,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.

8. Frequency of A1/A2: v”= 2(u’+ ½v’)(½v’+w’) = 2(p12+p1p2 )(p22+p1p2 )= 2p1p2 Hardy-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 ? After 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= p12 Frequency of A2/A2: w”=(½v’+w’)2 = (p22 + p1p2)2= p22

9. Use of Assumptions in the derivation • Infinite population size • Discrete generations (mating amongst ith generation members only) • Random mating • No selection • No migration • No mutation • Equal initial genotype frequencies in the two sexes Segregation ratios below assume 1,2,3,6,7 Frequency formula of A1/A1after one generation: u2+ ½(2uv)+ ¼v2 assume 4,5.

10. 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. An alternative justification Previously, we started with arbitrary genotype frequenciesu,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 p22. Argument completed ?

11. HW 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 ?

12. Argument 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.

13. Eqs 1.1 and 1.2 imply: 2/3 qn+1/3 rn = The recursion equations Because a male always gets his X chromosome from his mother, and his mother precedes him by one generation, rn = qn-1(Eq. 1.1) Similarly, females get half their X-chromosomes from females and half from males, qn = ½ qn-1+ ½ rn-1(Eq. 1.2) 2/3(½ qn-1+ ½ rn-1 ) + 1/3 qn-1= 2/3 qn-1 + 1/3 rn-1 It 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.

14. The 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)  0 So in each step the difference diminishes by half and qn approaches p in a zigzag manner. Hence, rn = qn-1also 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).

15. Ai A’i Bj B’j Linkage equilibrium Let Ai be allele at locus A with frequency pi Let Bj be allele at locus B with frequency qj Denote the recombination between these loci by fand mfor females and males, respectively. Let = (f + m )/2. Linkage equilibrium means that Pr(Ai Bj) = piqj We 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.

16. No recombination recombination = ½ [ (1-f )Pn-1(Ai Bj) + f piqj] + In short, we have established, . For loci far on a chromosome, the deviation from linkage is halved each generation. For close loci with small , convergence is slow. Convergence Proof Pn(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) + piqj So, Pn(Ai Bj) - piqj = (1-) [Pn-1(Ai Bj) – piqj]= …= (1-)n[P0(Ai Bj) – piqj] Exercise: Repeat this analysis for three loci (Problem 7, with guidance, in Kenneth Lang’s book).

17. Suppose there is a close marker: Marker Mutated locus Ramifications for Association studies Many diseases are thought to been caused by a single random mutation that survived and propagated to offspring, generation after generation. 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 recent allele age of the disease. Association studies like that are also called linkage disequilibrium mapping or LD mapping in short.

18. 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. Selection and Fitness Fitness 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. 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.

19. First, note that pn+1 = [(1-r)pn2 + pnqn] / wn (multiply by 2 to see why) Assuming 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 -sqn2 a/a A/a A/A To find equilibrium we study the difference pn  pn+1 - pn pn  pn+1 - pn = [(1-r)pn2 + pnqn] / wn - pn = [(1-r)pn2 + pnqn- (1-rpn2 -sqn2)pn] / wn = [pnqn (s- (r+s) pn)] / wn

20. Interpretation when r>0 and s0 We just derived pn = [pnqn (s- (r+s) pn)] / wn Convergence 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 (r0 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) p satisfies g(0)  0 and g(1) < 0, hence it is negative at (0,1). Thus, pn monotonicallydecreases at each step. So pn must approach 0 at equilibrium. Similarly, with the other case.

21. when r and s have the same sign Conclusion I (for negative sign): If r and s are negative, (pn ) > 1, so pn 1 for p0 above s/(r+s), and pn 0 for p0below s/(r+s). In other words, s/(r+s) is an unstable equilibrium.

22. Hence has a constant sign and declines in magnitude. 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 when r and s are both positive If both r and s are positive (Heterozygous advantage), then Conclusion II: If both r and s are positive, pn s/(r+s) and this point is a stable equilibrium.

23. Heterozygous 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).

24. Sickle cell anemiaאנמיה חרמשית - Medical Encyclopedia Red blood cells, sickle cell Sickle 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): http://www.nlm.nih.gov/medlineplus/ency/imagepages/1212.htm

25. Facts about Sickle cell Disease • Sickle Cell Disease is much more common in certain ethnic groups affecting approximately one out of every 500 African Americans. • Although sickle cell disease is inherited and present at birth, symptoms usually don't occur until after 4 months of age. • 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. • 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.

26. (r>0, s=0) (recessive disease) Balance 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: This yields 1- rp2 = 1- and thus p2 = /r and a balance is achieved that retains both alleles.

27. Finite Population Genetic Drift Alelle 10 Alelle 5 Source: Gideon Greenspan After 800 generations, by simulation, from the ten alleles only two remain: numbered 5 and number 7.