Population Genetics: Selection and mutation as mechanisms of evolution

1. Population Genetics: Selection and mutation as mechanisms of evolution Population genetics: study of Mendelian genetics at the level of the whole population.

2. Hardy-Weinberg Equilibrium To understand conditions under which evolution can occur, it is necessary to understand the population genetic conditions under which it will not occur.

3. Hardy-Weinberg Equilibrium Hardy-Weinberg Equilibrium Principle allows us to predict allele and genotype frequencies from one generation to the next in the absence of evolution.

4. Hardy-Weinberg Equilibrium Assume two alleles A and a with known frequencies (e.g. A = 0.6, a = 0.4.) Only two genes in population so their allele frequencies add up to 1.

5. Hardy-Weinberg Equilibrium Can predict frequencies of genotypes in next generation using allele frequencies. Possible genotypes: AA, Aa and aa

6. Hardy-Weinberg Equilibrium Assume alleles A and a enter eggs and sperm in proportion to their frequency in population (i.e. 0.6 and 0.4) Assume sperm and eggs meet at random (one big gene pool).

7. Hardy-Weinberg Equilibrium Then we can calculate expected genotype frequencies. AA: To produce an AA individual, egg and sperm must each contain an A allele. This probability is 0.6 x 0.6 = 0.36 (probability sperm contains A times probability egg contains A).

8. Hardy-Weinberg Equilibrium Similarly, we can calculate frequency of aa. 0.4 x 04 = 0.16.

9. Hardy-Weinberg Equilibrium Probability of Aa is given by probability sperm contains A (0.6) times probability egg contains a (0.4). 0.6 x 04 = 0.24.

10. Hardy-Weinberg Equilibrium But, there’s a second way to produce an Aa individual (egg contains A and sperm contains a). Same probability as before: 0.6 x 0.4= 0.24. Overall probability of Aa = 0.24 + 0.24 = 0.48.

11. Hardy-Weinberg Equilibrium Genotypes in next generation: AA = 0.36 Aa = 0.48 Aa= 0.16 Adds up to one.

12. Hardy-Weinberg Equilibrium General formula for Hardy-Weinberg. Let p= frequency of allele A and q = frequency of allele a. p2 + 2pq + q2 = 1.

13. Hardy Weinberg Equilibrium with more than 2 alleles If three alleles with frequencies P1, P2 and P3 such that P1 + P2 + P3 = 1 Then genotype frequencies given by: P12 + P22 + P32 + 2P1P2 + 2P1 P3 + 2P2P3

14. Conclusions from Hardy-Weinberg Equilibrium Allele frequencies in a population will not change from one generation to the next just as a result of assortment of alleles and zygote formation. If the allele frequencies in a gene pool with two alleles are given by p and q, the genotype frequencies will be given by p2, 2pq, and q2.

15. Working with the H-W equation You need to be able to work with the Hardy-Weinberg equation. For example, if 9 of 100 individuals in a population suffer from a homozygous recessive disorder can you calculate the frequency of the disease causing allele? Can you calculate how many heterozygotes are in the population?

16. Working with the H-W equation p2 + 2pq + q2 = 1. The terms in the equation represent the frequencies of individual genotypes. P and q are allele frequencies. It is vital that you understand this difference.

17. Working with the H-W equation 9 of 100 (frequency = 0.09) of individuals are homozygotes. What term in the H-W equation is that equal to?

18. Working with the H-W equation It’s q2. If q2 = 0.09, what’s q? Get square root of q2, which is 0.3. If q=0.3 then p=0.7. Now plug p and q into equation to calculate frequencies of other genotypes.

19. Working with the H-W equation p2 = (0.7)(0.7) = 0.49 2pq = 2 (0.3)(0.7) = 0.42 Number of heterozygotes = 0.42 times population size = (0.42)(100) = 42.

20. Working with the H-W equation: 3 alleles There are three alleles in a population A1, A2 and A3 whose frequencies respectively are 0.2, 0.2 and 0.6 and there are 100 individuals in the population. How many A1A2 heterozygotes will there be in the population?

21. Working with the H-W equation: 3 alleles Just use the formulae P1 + P2 + P3 = 1 and P12 + P22 + P32 + 2P1P2 + 2P1 P3 + 2P2P3 = 1 Then substitute in the appropriate values for the appropriate term 2P1P2 = 2(0.2)(0.2) = 0.08 or 8 people out of 100.

22. Other examples of working with HW equilibrium: is a population in HW equilibrium? In a population there are 100 birds with the following genotypes: 44 AA 32 Aa 24 aa How would you demonstrate that this population is not in Hardy Weinberg equilibrium

23. Three steps Step 1: calculate the allele frequencies. Step 2: Calculate expected numbers of each geneotype (i.e. figure out how many homozygotes and heterozygotes you would expect.) Step 3: compare your expected and observed data.

24. Step 1 allele frequencies Step 1. How many A alleles are there in total? 44 AA individuals = 88 A alleles (because each individual has two copies of the A allele) 32 Aa alleles = 32 A alleles Total A alleles is 88+32 =120.

25. Step 1 allele frequencies Total number of a alleles is similarly calculated as 2*24 + 32 = 80 What are allele frequencies? Total number of alleles in population is 120 + 80 = 200 (or you could caculate it by multiplying the number of individuals in the population by two 100*2 =200)

26. Step 1 allele frequencies Allele frequencies are: A = 120/200= 0.6. Let p = 0.6 a = 80/200 = 0.4. Let q = 0.4

27. Step 2 Calculate expected number of each genotype Use the Hardy_Weinberg equation p2 + 2pq + q2 = 1 Expected frequency of AA = p2 = 0.6 * 0.6 = 0.36 Expected frequency of aa = q2 = 0.4*0 .4 =0.16 Expected frequency of Aa = 2pq = 2*.6*.4 = 0.48

28. Step 2 Calculate expected number of each genotype Convert genotype frequencies to numbers by multiplying by population size of 100 AA = 0.36*100 = 36 aa = 0.16*100 = 16 Aa = 0.48*100 = 48

29. Step 3 Compare Observed and Expected values Observed population is: 44 AA 32 Aa 24 aa Expected population is: 36AA 48Aa 16aa These numbers are not the same so the population is not in Hardy-Weinberg equilibrium. An assumption of the Hardy Weinberg equilibrium is being violated. What are those assumptions?

30. Assumptions of Hardy-Weinberg 1. No selection. When individuals with certain genotypes survive better than others, allele frequencies may change from one generation to the next.

31. Assumptions of Hardy-Weinberg 2. No mutation If new alleles are produced by mutation or alleles mutate at different rates, allele frequencies may change from one generation to the next.

32. Assumptions of Hardy-Weinberg 3. No migration Movement of individuals in or out of a population will alter allele and genotype frequencies.

33. Assumptions of Hardy-Weinberg 4. No chance events. Luck plays no role. Eggs and sperm collide at same frequencies as the actual frequencies of p and q. When assumption violated and by chance some individuals contribute more alleles than others to next generation allele frequencies may change. This mechanism of allele frequency change is called Genetic Drift.

34. Assumptions of Hardy-Weinberg 5. Individuals select mates at random. If this assumption is violated allele frequencies will not change, but genotype frequencies may.

35. Hardy-Weinberg Equilibrium Hardy Weinberg equilibrium principle identifies the forces that can cause evolution. If a population is not in H-W equilibrium then one or more of the five assumptions is being violated.

37. Can selection change allele frequencies? Two alleles B1 and B2 Frequency of B1= 0.6 and frequency of B2 = 0.4. Random mating gives genotype frequencies 0.36 B1B1 0.48B1B2 0.16B2B2

38. Can selection change allele frequencies? Assume 100 individuals 36 B1B1 48 B1B2 16 B2B2 Incorporate selection. Assume all B1B1 survive, 75% of B1B2 survive and 50% of B2B2 survive.

39. Can selection change allele frequencies? Population now 80 individuals: 36 B1B1 36 B1B2 8 B2B2 Allele frequencies now: B1 = 72 + 36/160 = 0.675 B2 = 36+16/160 = 0.325 Selection resulted in allele frequency change.

41. Can selection change allele frequencies? Selection in previous example very strong. What patterns expected with weaker selection. Initial frequencies B1 = 0.01, B2 = 0.99.

43. Can selection change allele frequencies? Rate of change of B1is rapid when selection pressure is strong, but much slower, although still steady, under weak selection.

44. Empirical examples of allele frequency change under selection Clavener and Clegg’s work on Drosophila. Two alleles for ADH (alcohol dehydrogenase breaks down ethanol) ADHF and ADHS

45. Empirical examples of allele frequency change under selection Two Drosophila populations maintained: one fed food spiked with ethanol, control fed unspiked food. Populations maintained for multiple generations.

46. Empirical examples of allele frequency change under selection Experimental population showed consistent long-term increase in frequency of ADHF Flies with ADHF allele have higher fitness when ethanol present. ADHF enzyme breaks down ethanol twice as fast as ADHS enzyme.

48. Empirical examples of allele frequency change under selection: Jaeken syndrome Jaeken syndrome: patients severely disabled with skeletal deformities and inadequate liver function.

49. Jaeken syndrome Autosomal recessive condition caused by loss-of-function mutation of gene PMM2 codes for enzyme phosphomannomutase. Patients unable to join carbohydrates and proteins to make glycoproteins at a high enough rate. Glycoproteins involved in movement of substances across cell membranes.

50. Jaeken syndrome Many different loss-of-function mutations can cause Jaeken Syndrome. Team of researchers led by Jaak Jaeken investigated whether different mutations differed in their severity. Used Hardy-Weinberg equilibrium to do so.

51. Jaeken syndrome People with Jaeken syndrome are homozygous for the disease, but may be either homozygous or heterozygous for a given disease allele. Different disease alleles should be in Hardy-Weinberg equilibrium.

52. Jaeken syndrome Researchers studied 54 patients and identified most common mutation as R141H. Dividing population into R141H and other alleles. Allele frequencies are: Other: 0.6 and R141H: 0.4.

53. Jaeken syndrome If disease alleles in H-W equilibrium then predict genotype frequencies of Other/other: 0.36 Other/R141H: 0.48 R141H/R141H: 0.16

54. Jaeken syndrome Observed frequencies are: Other/Other: 0.2 Other/R141H: 0.8 R141H/R141H : 0 Clearly population not in H-W equilibrium.

55. Jaeken syndrome Researchers concluded that R141H is an especially severe mutation and homozygotes die before or just after birth. Thus, there is selection so H-W assumption is violated.

56. Using H-W to predict potential spread of CCR5-?32 allele Will AIDS epidemic cause CCR5-?32 (delta32) allele to spread? Offers protection against HIV infection. In principle it could, but models suggest it probably will not in any real population.

57. Spread of CCR5-?32 allele Scenario 1. Initial allele frequency 20%. 25% of heterozygotes and those homozygous for normal allele die of AIDS. Homozygous ?32 individuals do not die of AIDS. Over 40 generations allele increases to almost 100%.

59. Spread of CCR5-?32 allele In human population with high HIV infection rate and high frequencies of ?32 allele, ?32 could spread rapidly.

60. Spread of CCR5-?32 allele Scenario 2. In areas with low HIV infection rates (under 1%), but high levels of ?32 (as found in Europe), selection too weak to raise delta 32 frequency much.

62. Spread of CCR5-?32 allele Scenario 3. In sub-Saharan Africa HIV infection rate about 25%. However, ?32 allele almost absent Under these conditions ?32 frequency will hardly change because most copies of allele are in heterozygotes, which are not protected from HIV.

64. Testing predictions of population genetics theory Theory predicts that if an average individual carrying an allele has higher than average fitness that the frequency of that allele will increase from one generation to the next. Obviously, the converse should be true and a deleterious allele should decrease in frequency if its bearers have lower fitness.

65. Testing predictions of population genetics theory Mathematical treatment of effect of selection on gene frequencies is given in Box 6.3 (page 186) of your text. Main point is that if the average fitness of an allele A when paired at random with other alleles in the population is higher than the average fitness of the population, then it will increase in frequency.

66. Tests of theory Dawson (1970). Flour beetles. Two alleles at locus: + and l. +/+ and +/l phenotypically normal. l/l lethal.

67. Dawson’s flour beetles Dawson founded two populations with heterozygotes (frequency of + and l alleles thus 0.5). Expected + allele to increase in frequency and l allele to decline over time.

69. Dawson’s flour beetles Predicted and observed allele frequencies matched very closely. l allele declined rapidly at first, but rate of decline slowed.

70. Dawson’s flour beetles Dawson’s results show that when recessive allele is common, evolution by natural selection is rapid, but slows as recessive allele becomes rarer. Hardy-Weinberg explains why.

71. Dawson’s flour beetles When recessive allele (a) common e.g. 0.95 genotype frequencies are: AA (0.05)2 Aa (2 (0.05)(0.95) aa (0.95)2 0.0025AA 0.095Aa 0.9025aa With more than 90% of phenotypes being recessive, if aa is selected against expect rapid population change.

72. Dawson’s flour beetles When recessive allele (a) rare [e.g. 0.05] genotype frequencies are: AA (0.95)2 Aa 2(0.95)(0.05) aa (0.05)2 0.9025AA 0.095Aa 0.0025aa Fewer than 0.25% of phenotypes are aa recessive. Most a alleles are hidden from selection as heterozygotes. Expect only slow change in frequency of a.

73. Maintaining multiple alleles in gene pool Dawson’s beetle work shows that deleterious rare alleles may be very hard to eliminate from a gene pool because they remain hidden from selection as heterozygotes. (this only applies if the allele is not dominant.)

74. Maintaining multiple alleles in gene pool One way in which multiple alleles may be maintained in a population is through heterozygote advantage. Classic example is sickle cell allele.

75. Sickle Cell Anemia Sickle cell anemia is a condition common among West Africans (and African Americans of West African ancestry). In sickle cell anemia red blood cells are sickle shaped. Usually fatal by about age 10.

84. Maintaining multiple alleles in gene pool Another way in which multiple alleles are maintained is frequency-dependent selection. Frequency-dependent selection occurs when rare alleles have a selective advantage.

85. Frequency-dependent selection Color polymorphism in Elderflower Orchid Two flower colors: yellow and purple. Offer no food reward to bees. Bees alternate visits to colors. How are two colors maintained in the population?

86. Gigord et al. hypothesis: Bees tend to visit equal numbers of each flower color so rarer color will have advantage (will get more visits from pollinators). Frequency-dependent selection

87. Experiment: provided five arrays of potted orchids with different frequencies of yellow orchids in each. Monitored orchids for fruit set and removal of pollinaria (pollen bearing structures) Frequency-dependent selection

88. As predicted, reproductive success of yellow varied with frequency. Frequency-dependent selection

90. It is obvious that selection is a very powerful evolutionary force but how strong is mutation alone as an evolutionary force? To check: Two alleles A and a. Frequency of A = 0.9, a = 0.1. Mutation as an evolutionary force

91. Assume A mutates to a at rate of 1 copy per 10,000 per generation (high rate, but within observed range) and all mutations occur in gametes. How much does this change gene pool in next generation? Mutation as an evolutionary force

92. Hardy Weinberg genotypes in current generation: 0.81 AA, 0.18 Aa, 0.01 aa With no mutation allele frequency in gene pool 0.9 A, 0.1 a Mutation as an evolutionary force

93. But mutation reduces frequency of A and increases frequency of a A a 0.9 - (0.0001)(0.9) 0.1 + (0.0001)(0.9) 0.89991A 0.10009a Mutation as an evolutionary force

95. Not a big change. After 1000 generations frequency of A = 0.81. Mutation as an evolutionary force

97. Mutation alone clearly not a powerful evolutionary force. But mutation AND selection make a very powerful evolutionary force. Mutation as an evolutionary force

98. Lenski’s E. coli work Lenski et al. studied mutation and selection together in an E. coli strain that did not exchange DNA (hence mutation only source of new variation). Bacteria grown in challenging environment (low salts and low glucose medium) so selection would be strong.

99. Lenski’s E. coli work 12 replicate populations tracked over about 10,000 generations. Fitness and cell size of populations increased over time. Pattern of change interesting: steplike. Why is it steplike?

101. Lenski’s E. coli work Step-like pattern results when a new mutation occurs and sweeps through the population as mutant bacteria out-reproduce competitors. Remember, without mutation evolution would eventually cease. Mutation is ultimate source of genetic variation.

102. Mutation-selection balance Most mutations are deleterious and natural selection acts to remove them from population. Deleterious alleles persist, however, because mutation continually produces them.

103. When rate at which deleterious alleles are being eliminated is equal to their rate of production by mutation we have mutation-selection balance. Mutation-selection balance

104. Equilibrium frequency of deleterious allele q = square root of µ/s where µ is mutation rate and s is the selection coefficient (measure of strength of selection against allele; ranges from 0 to 1). See Box 6.10 page 215 for derivation of equation. Mutation-selection balance

105. Equation makes intuitive sense. If s is small (mutation only mildly deleterious) and µ (mutation rate) is high than q (allele frequency) will also be relatively high. If s is large and µ is low, than q will be low too. Mutation-selection balance

106. Spinal muscular atrophy is a generally lethal condition caused by a mutation on chromosome 5. Selection coefficient estimated at 0.9. Deleterious allele frequency about 0.01 in Caucasians. Inserting above numbers into equation and solving for µ get estimated mutation rate of 0.9 X 10-4 Mutation-selection balance

107. Observed mutation rate is about 1.1 X10-4, very close agreement in estimates. High frequency of allele is accounted for by observed mutation rate. Mutation-selection balance

108. Is frequency of Cystic fibrosis maintained by mutation selection balance? Cystic fibrosis is caused by a loss of function mutation at locus on chromosome 7 that codes for CFTR protein (cell surface protein in lungs and intestines). Major function of protein is to destroy Pseudomonas aeruginosa bacteria. Bacterium causes severe lung infections in CF patients.

109. Cystic fibrosis Very strong selection against CF alleles, but CF frequency about 0.02 in Europeans. Can mutation rate account for high frequency?

110. Assume selection coefficient (s) of 1 and q = 0.02. Estimate mutation rate µ is 4.0 X 10-4 But actual mutation rate is only 6.7 X 10-7 Is there an alternative explanation? Cystic fibrosis

111. Cystic fibrosis May be heterozygote advantage. Pier et al. (1998) hypothesized CF heterozygotes may be resistant to typhoid fever. Typhoid fever caused by Salmonella typhi bacteria. Bacteria infiltrate gut by crossing epithelial cells.

112. Cystic fibrosis Hypothesized that S. typhi bacteria may use CFTR protein to enter cells. If so, CF-heterozygotes should be less vulnerable to S. typhi because their gut epithilial cells have fewer CFTR proteins on cell surface.

113. Cystic fibrosis Experimental test. Produced mouse cells with three different CFTR genotypes CFTR homozygote (wild type) CFTR/?F508 heterozygote (?F508 most common CF mutant allele) ?F508/?F508 homozygote

114. Cystic fibrosis Exposed cells to S. typhi bacteria. Measured number of bacteria that entered cells. Clear results

116. Cystic fibrosis ?F508/?F508 homozygote almost totally resistant to S. typhi. Wild type homozygote highly vulnerable Heterozygote contained 86% fewer bacteria than wild type.

117. Cystic fibrosis Further support for idea ?F508 provides resistance to typhoid provided by positive relationship between ?F508 allele frequency in generation after typhoid outbreak and severity of the outbreak.