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2. 2 Hardy-Weinberg Equilibrium Lecture 5

3. 3 The Hardy-Weinberg Equilibrium

4. 4 Hardy-Weinberg Theorem Hardy-Weinberg - original proportions of genotypes in a population will remain constant from generation to generation - Sexual reproduction (meiosis and fertilization) alone will not change allelic (genotypic) proportions.

5. 5 Assumptions of the H-W Theorem 1.Large population size -small populations can have chance fluctuations in allele frequencies (e.g., fire, storm, floods).

6. 6 Assumptions of the H-W Theorem 2.No migration - immigrants can change the frequency of an allele by bringing in new alleles to a population. 3.No net mutations -if alleles change from one to another, this will change the frequency of those alleles

7. 7 Assumptions of the H-W Theorem 4.Random mating - if certain traits are more desirable, then individuals with those traits will be selected and this will not allow for random mixing of alleles. 5.No natural selection - if some individuals survive and reproduce at a higher rate than others, then their offspring will carry those genes and the frequency will change for the next generation.

8. 8 Allele freq. f(A) = p f(a) = q p + q = 1 Sum of all alleles = 100% Genotypic freq. f(AA) = p 2 Dominant homozygous f(Aa) = 2 pq Heterozgous f(aa) = q 2 Recessive homozygous p 2 + 2 pq + q 2 = (p+q) 2 = 1 Sum of all genotypes = 100% GametesA (p)a(q) A(p)AA (pp) Aa (pq) a(q)aA (qp) aa (qq) AA = p*p = p 2 Aa = pq + qp = 2pq aa = q*q = q 2 Hardy-Weinberg Theorem (2 alleles at 1 locus)

9. 9 The gene pool of a non-evolving population remains constant over multiple generations; i.e., the allele frequency does not change over generations of time. The Hardy-Weinberg Equation: binomial expansion (p+q) 2 = p 2 + 2pq + q 2 = 1.0 p 2 = frequency of AA homozygous genotype; 2pq = frequency of Aa plus aA heterozygous genotypes; q 2 = frequency of aa homozygous genotype Hardy-Weinberg Principle

10. 10 Assumptions of the H-W Equilibrium

11. 11 Assumptions of the H-W Equilibrium

12. 12 for a population with genotypes: 100 GG 160 Gg 140 gg Genotype frequencies Phenotype frequencies Allele frequencies calculate: Assumptions of the H-W Equilibrium

13. 13 for a population with genotypes: 100 GG 160 Gg 140 gg Genotype frequencies Phenotype frequencies Allele frequencies 100/400 = 0.25 GG 160/400 = 0.40 Gg 140/400 = 0.35 gg 260/400 = 0.65 green 140/400 = 0.35 brown 2*100 + 160/800 = 0.45 G 2*140 + 160/800 = 0.55 g 0.65260 calculate: Assumptions of the H-W Theorem

14. 14 another way to calculate allele frequencies: 100 GG 160 Gg 140 gg Genotype frequencies Allele frequencies 0.25 GG 0.40 Gg 0.35 gg 360/800 = 0.45 G 440/800 = 0.55 g OR [0.25 + (0.40)/2] = 0.45 [0.35 + (0.40)/2] = 0.65 G g G g 0.25 0.40/2 = 0.20 0.35 p (G) = D + H/2 q (g) = R + H/2 Assumptions of the H-W Theorem

15. 15 Hardy-Weinberg Equilibrium Population of cats n=100 16 white and 84 black aa = white A_ = black Can we figure out the allelic frequencies of individuals AA and Aa?

16. 16 p 2 + 2pq + q 2 and p+q = 1 (always two alleles) 16 cats white = 16 (aa) then (q 2 = 16/100 = 0.16) This we know we can see and count!!!!! If p + q = 1 then we can calculate p from q 2 q = square root of q 2 = q √0.16 q = 0.4 p + q = 1 then p = 0.6 (0.6 +0.4 = 1) P 2 = 0.36 All we need now are those that are heterozygous (2pq) (2 x 0.6 x 0.4)=0.48 0.36 + 0.48 + 0.16 = 1 Hardy-Weinberg Equilibrium

17. 17 Hardy-Weinberg Equilibrium

18. 18 Example use of H-W theorem 1000-head sheep flock. No selection for color. Closed to outside breeding. 910 white (B_); 90 black (bb) Start with known: f(black) = f(bb) = 0.09 =q 2 Then, p = 1 – q = 0.7 = f(B) f(BB) = D = p 2 = (0.7) 2 = 0.49 f(Bb) = H = 2pq = 2 * 0.7 * 0.3 = 0.42 f(bb) = R= q 2 = (0.3) 2 = 0.09

19. 19 In summary: Allele freq. f(B) = p = 0.7 (est.) f(b) = q = 0.3 (est.) Phenotypic freq. f(white) = 0.91 (actual) f(black) = 0.09 (actual) Genotypic freq. f(BB) = p 2 = 0.49 (est.) f(Bb) = 2pq = 0.42 (est.) f(bb) = q 2 = 0.09 (actual)