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Population Genetics: Selection and mutation as mechanisms of evolution

Population Genetics: Selection and mutation as mechanisms of evolution. Population genetics: study of Mendelian genetics at the level of the whole population. Hardy-Weinberg Equilibrium.

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Population Genetics: Selection and mutation as mechanisms of evolution

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  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 the 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 • The Hardy-Weinberg Equilibrium Principle allows us do this. It enables us to predict allele and genotype frequencies from one generation to the next in the absence of evolution.

  4. Hardy-Weinberg Equilibrium • Assume a gene with two alleles A and a with known frequencies (e.g. A = 0.6, a = 0.4.) • There are only two alleles in the population so their frequencies must add up to 1.

  5. Hardy-Weinberg Equilibrium • Using allele frequencies we can predict the expected frequencies of genotypes in the next generation. • With two alleles only three genotypes are possible: AA, Aa and aa

  6. Hardy-Weinberg Equilibrium • Assume alleles A and a are present in eggs and sperm in proportion to their frequency in population (i.e. 0.6 and 0.4) • Also assume that 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. • Hence the 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 • These frequencies add 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 thatP1 + 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. • Assortment of alleles simply means what occurs during meiosis when only one copy of each pair of alleles enters any given gamete (remember each gamete only contains half the DNA of a body cell).

  15. Conclusions from Hardy-Weinberg Equilibrium • 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.

  16. 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?

  17. Working with the H-W equation • p2 + 2pq + q2 = 1. The terms in the equation represent the frequencies of individual genotypes. [A genotype is possessed by an individual organism so there are two alleles present in each case.] • P and q are allele frequencies. Allelefrequencies are estimates of how common alleles are in the whole population. • It is vital that you understand the difference between allele and genotye frequencies.

  18. Working with the H-W equation • 9 of 100 (frequency = 0.09) of individuals are homozygous for the recessive allele. What term in the H-W equation is that equal to?

  19. 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, which is the frequency of the allele a. • If q=0.3 then p=0.7. Now plug p and q into equation to calculate frequencies of other genotypes.

  20. Working with the H-W equation • p2 = (0.7)(0.7) = 0.49 -- frequency of AA • 2pq = 2 (0.3)(0.7) = 0.42 – frequency of Aa. • To calculate the actual number of heterozygotes simply multiply 0.42 by the population size = (0.42)(100) = 42.

  21. 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?

  22. 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.

  23. 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

  24. 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.

  25. 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.

  26. 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 calculate it by multiplying the number of individuals in the population by two 100*2 =200)

  27. 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

  28. Step 2 Calculate expected number of each genotype • Use the Hardy_Weinberg equation p2 + 2pq + q2 = 1 to calculate what expected genotypes we should have given these observed frequencies of “A” and “a” • 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

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

  30. 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?

  31. 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.

  32. 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.

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

  34. 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 is 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.

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

  36. 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. 5.10

  38. 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

  39. 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.

  40. 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. FIG 5.11

  42. 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. Fig 5.12 Note line colors yellow and red have been flipped in the table.

  44. 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.

  45. 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

  46. 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.

  47. 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 is present. • ADHF enzyme breaks down ethanol twice as fast as ADHS enzyme.

  48. Fig 5.13

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

  50. 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.

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