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 • 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.

3. 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.

4. 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

5. 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).

6. 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. • ______________________________________________________________________________

7. Hardy-Weinberg Equilibrium • Similarly, we can calculate frequency of aa. • ___________________

8. Hardy-Weinberg Equilibrium • Probability of Aa is given by probability sperm contains A (0.6) times probability egg contains a ___________________________________________________

9. 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: ________________ • Hence the overall probability of Aa = ____________________________

10. Hardy-Weinberg Equilibrium • Genotypes in next generation: • AA = _________ • Aa = __________ • aa= ____________ • These frequencies add up to one.

11. Hardy-Weinberg Equilibrium • General formula for Hardy-Weinberg. • Let p= frequency of allele A and q = frequency of allele a. • _________________.

12. 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.

13. 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?

14. 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.

15. 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?

16. Working with the H-W equation • It’s q2. • ____________________________________________________________________________________ • If _________________. Now plug p and q into equation to calculate frequencies of other genotypes.

17. Working with the H-W equation • _______________________________ • _____________________________________________________________ • To calculate the actual number of heterozygotes simply multiply 0.42 by the population size = __________________.

18. Other examples of working with HW equilibrium: is a population in HW equilibrium? • In a population there are 100 birds with the following genotypes: • 49 AA • 32 Aa • 24 aa • How would you demonstrate that this population is not in Hardy Weinberg equilibrium

19. 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.

20. Step 1 allele frequencies • Step 1. How many “A” alleles are there in total? • 49 AA individuals = 98 “A” alleles (because each individual has two copies of the “A” allele) • 32 Aa alleles = 32 “A” alleles • Total “A” alleles is 98+32 =130.

21. 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)

22. 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

23. 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

24. 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