Given genotype frequencies calculate allele frequencies in a

Given genotype frequencies, calculate allele frequencies in a gene pool ! Alleles = A, a Genotypes = AA, Aa, aa Frequency of allele A: f (A) = f (AA) + 1/2 f (Aa) Frequency of allele a: f (a) = f (aa) + 1/2 f (Aa) f(A) + f(a) = 1. 0 p + q = 1. 0 (allele frequencies) p = 1 - q or, q = 1 - p p 2 + 2 pq + q 2 = 1 f [AA] + 2 [f(Aa)] + f [aa] = 1 A a A AA Aa aa 1 26 a

Hardy-Weinberg Equilibrium Parental generation: 2 alleles, r and R f (R) + f (r) = 1. 0 p + q = 1. 0 p = 0. 1, q = 0. 9 In the next generation (F 1): p 2 + 2 pq + q 2 = 1 predicts allele freqs. F 1 genotype Genotype Allele freq. p 2 (. 01) RR p = 0. 01+. 18/2=. 1 q 2 (. 81) rr q = 0. 81+. 18/2 =. 9 2 pq (. 18) Rr 2 27 a-1

Hardy-Weinberg Equilibrium Parental allele frequencies p and q predict F 1 generation genotype frequencies, by the formula p 2 + 2 pq + q 2 = 1 Note: parental generation genotype frequencies do NOT predict F 1 generation genotype frequencies!! 3 27 a-2

Hardy-Weinberg Equilibrium Conclusions: 1)Allele frequencies are conserved (i. e. , the same) from one generation to the next. 2) genotype frequency reaches Hardy. Weinberg equilibrium in one generation 4 27 a-2

Hardy-Weinberg’s caveats: But Hardy-Weinberg Law: allele frequencies in a population remain constant from generation to generation …. . Only • IF random mating • IF all genotypes are equally viable • IF not disturbed by mutation, selection or whatever Bottom line: Only in an IDEAL population is genetic diversity conserved forever. 5

Sex = Random sampling of a gene pool Population of 10 individuals (N = 10) Phenotypes Red White Genotypes RR, Rr, rr Allele frequencies R = 0. 6 r = 0. 4 Parental gene pool R r r R RR R r r Rr 10 genotypes r R R R r R R 20 alleles Parental gametes Probability of F 1 r =. 4; R =. 6 6 25 A -1

F 1 genotypes and phenotypes Genotype Frequency Rr or r. R. 48 rr. 16 RR. 36 ------- Phenotype Frequency R_. 48+. 36=. 84 rr. 16 ------- =1 7 25 A -2

If allele frequencies are P and Q in the parental generation, how do we calculate what they will be in the F 1 generation? 8

If allele frequencies are P and Q in the parental generation, how do we calculate what they will be in the F 1 generation? Genotype frequencies in F 1 are calculated by: p 2 + 2 pq + q 2 = 1 From which we can calculate p and q for F 1 9

If the fraction of the population with allele “A” at a given locus is. 7, and the fraction of the population with “a” at the locus is. 3, what will be the expected genotype frequencies in F 1, , the next generation? 10

If the fraction of the population with allele “A” at a given locus is. 7, and the fraction of the population with “a” at the locus is. 3, what will be the expected genotype frequencies in the F 1? Allele frequencies in P (parental generation): A =. 7 = p a =. 3 = q Expected genotypes and their frequencies in F 1: AA = p 2 =. 49 aa = q 2 =. 09 Aa = 2 pq =. 42 11

What will be the expected phenotypes and their ratios in this example? 12

What will be the expected phenotypes and their ratios in this example? Allele frequencies in P (parental generation): A =. 7 = p a =. 3 = q Expected genotypes in F 1: AA = p 2 =. 49 aa = q 2 =. 09 Aa = 2 pq =. 42 Expected phenotypes in F 1: A‑ =. 49 +. 42 =. 91 aa =. 09 13

Heterozygosity defined H = % heterozygous genotypes for a particular locus = % heterozygous individuals for a particular locus = probability that a given individual randomly selected from the population will be heterozygous at a given locus 14 29 f

H =? Aa aa aa AA aa a. A aa AA a. A AA aa aa AA aa a. A H = 4/15 15

Heterozygosity defined H (“H-bar”) = average heterozygosity for all loci in a population. H estimated = % heterozygous loci those examined H = 2 pq 16 29 f

Calculating H (assuming simple dominance and Hardy-Weinberg Eq. ) calculate H if q 2 = 0. 09 f (a) = 0. 3 = q f (A) = 0. 7 = p q 2 = 0. 09 p 2 = 0. 49 2 pq = 0. 42 H = 2 pq = 0. 42 (for only 2 alleles) 17 29 a - 1

Calculating H (but if……) Codominance Genotype AA Aa aa Phenotype N Red 50 Pink 22 White 10 total = 82 H = 22/82 (don’t need Hardy - Weinberg) 18 29 a - 1

Calculating H for 3 alleles: p, q, r p = 0. 5 q = 0. 4 r = 0. 1 H = 2 pq + 2 qr + 2 pr =. 40 +. 08 +. 10 =. 58 p q p pp pq pr q pq qq qr r pr qr rr r 19 29 a

1% of golden lion tamarins have diaphragmatic hernias, a condition expressed only in the homozygous recessive genotype. Calculate the number of heterozygous individuals in the wild population (N = 508). Assume Hardy-Weinberg equilibrium and simple dominance. Genotypes: AA F 1 generation p 2 = ? 2 pq = ? q 2 =. 01 Aa aa = f (AA) = f (Aa) = H = f (aa) 20 29 ez - 1

q 2 =. 01 q= = 0. 1 p + q = 1 so p = 1 - q p = 1 -. 1 =. 9 H = 2 pq = 2 x. 9 x. 1 = 0. 18 Nheterozygous =. 18 x 508 = 91 21 29 ez - 2

H and P H = heterozygosity = the percent of heterozygous genotypes in the population for that locus H = 2 pq (for 2 allele case) H = 2 pq + 2 qr + 2 pr (for 3 allele case) P = allelic diversity = percent polymorphism = percent of loci for which alternative alleles exist in the population 22

Gene pools Population 1 N = 8 (7 homozygous) 2 N = 16 alleles f (blue) = 1/16 f (red) = 15/16 Population 2 N=8 2 N = 16 alleles f (blue) = 16/16 typical population P = approximately 0. 25 individual H = approximately 0. 07 Polymorphic locus Monomorphic locus 23

The relationship of P to H Possible alleles = a, b, c aa bb ab ab cc ac ac ac H = 1/4 N alleles = 3 H = 4/4 N alleles = 3 Conclusion: H is not sensitive to the number of different alleles for the locus. 24 30 f

The relationship of H to P Population A, locus X alleles a b frequency. 5. 5 H = 2 pq =. 50 P is low Population B, locus X alleles frequency a. 7 b. 05 c. 05 d. 05 e. 05 f. 05 g. 05 H =. 495, P is high 25 30 A

Uses of molecular genetics* in conservation 1) Parentage and kinship 2) Within-population genetic variability 3) Population structure and intraspecific phylogeny 4) Species boundaries, hybridization phenomena, and forensics 5) Species’ phylogenies and macroevolution *e. g. , electrophoresis protein sequencing DNA fingerprinting immunological techniques 26 15 -2

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Western Pyrénées National Park, France 28

Effective population size N = population size = total number of individuals Ne = effective population size = ideal population size that would have a rate of decrease in H equal to that of the actual population (N) number of individuals contributing gametes to the next generation 29

Effective population size Predictable loss of heterozygosity (H) in each generation for non-ideal populations GLT Ne =. 32 N; N = 100, Ne = 32 Loss of H(N) = loss of H (Ne) If Ne/N 1, then rate of loss of H is minimum. The larger the Ne, the lower the rate of loss of H. 1 Rate of loss of H defined: 2 Ne per generation 30 32 b

Examples of effective population size Taxon Drosophila Humans a snail species plants golden lion tamarins Ne. 48 to. 71 N. 69 to. 95 N. 75 N lower. 32 N (94 of 290) 31 32 A-2

Assumptions of an ideal population • • • Infinitely large population random mating no mutation no selection no migration 32 31 a -1

5 causes of microevolution 1) genetic drift - stochastic variation in inheritance Expected F 2: 9 - 3 - 1 Random deviation Observed F 2: 9 - 3 - 2. 8 - 1. 2 2) Assortative (nonrandom) mating 3) Mutation 4) Natural selection 5) Migration (gene flow) 33 31 a-2

Sampling Error F 1 allele frequencies = Parental allele frequencies Caused by, for example: Behavioral traits producing assortative mating Genetic stochasticity Results in Genetic Drift = random deviation from expected allele frequencies 34 34 A-2

Fixation of alleles Fn Parental generation for many populations A = 1. 0 a=0 time Genetic drift fragmentation A =. 5 a =. 5 p = q =. 5 fixed lost p = 1. 0 A=0 a = 1. 0 lost fixed q = 1. 0 35 34 A-1

What is a formula for calculating the effect of unequal numbers of males and females (non-random breeding)on Ne? Ne = 4 MF M+F M = # of breeding males F = # of breeding females Population A M = 50 F = 50 Population B M = 10 F = 90 N = 100 Ne = 4 x 50 50 + 50 = 100 = 4 x 10 x 90 10 + 90 = 36 36 10 f

The effect of non-random mating on H Given 2 cases, with N = 150 and * Ht=1 = 1 - 1 2 Ne Ne = 100 (population A) Ne = 36 (population B) Ht = the proportion of heterozygosity remaining in the next (t=1) generation Population A: Ht = 1 - 1 2 x 100 Population B: Ht = 1 - 1 2 x 36 = 1 -. 005 =. 995 = 1 -. 014 =. 986 %H remaining after t=1 generations 37 36 A-1

Generalized equation: * Ht = H 0 1 - 1 2 Ne t t = # of generations later H 0 = original heterozygosity What is H after 5 more generations? Population A: H 5 = H 0 (. 995) 5 =. 995 (. 995)5 =. 970 Population B: H 5 = H 0 (. 986) 5 =. 995 (. 986)5 =. 919 38 36 A-2

Formulae for calculating H and Ne 1 = proportion of H 0 lost at each generation 2 Ne 1 - 1 = proportion of H 0 remaining after the first generation 2 Ne Ht = H 0 1 - 1 2 Ne Ne = 4 MF M+F t = the absolute amount of H 0 remaining after t generations 1) unequal sex ratios or 2) nonrandom breeding decrease Ne 39 37 A

Mutation Nondisjunctive point mutations over short term: not important in changing allele frequencies f (A 1) = 0. 5 mutation rate A 1 --> A 2 = 1 105 over 2000 generations, f (A 1) = 0. 49 If f (A 2) increases rapidly, selection must be involved Long-term, over evolutionary time mutation is critical - providing raw material for natural selection Mutation rate is independent of H, P, Ne but mutation can increase H and increase P 40 36 A 1