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1 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 AAAa aa A a Aa p 2 + 2pq + q 2 = 1 f [AA] + 2 [f(Aa)] + f [aa] = 1 26a

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2 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 + 2pq + 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 2pq (.18) Rr 27a-1

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3 Hardy-Weinberg Equilibrium Parental allele frequencies p and q predict F1 generation genotype frequencies, by the formula p 2 + 2pq + q 2 = 1 27a-2 Note: parental generation genotype frequencies do NOT predict F1 generation genotype frequencies!!

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4 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 27a-2

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

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6 Sex = Random sampling of a gene pool Population of 10 individuals (N = 10) White PhenotypesRed White GenotypesRR,Rr,rr Allele frequenciesR = 0.6r = 0.4 Parental gene pool 10 genotypes 20 alleles Parental gametes Probability of F 1 r =.4; R =.6 R r r R R R R r r R r r R R R r R r R R 25A -1

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

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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?

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9 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 F1 are calculated by: p 2 + 2pq + q 2 = 1 From which we can calculate p and q for F1

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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 F 1,, the next generation?

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11 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 F1: AA = p 2 =.49 aa = q 2 =.09 Aa = 2pq =.42

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

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13 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 F1: AA = p 2 =.49 aa = q 2 =.09 Aa = 2pq =.42 Expected phenotypes in F1: A ‑ =.49 +.42 =.91 aa =.09

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14 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 29f

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15 Aa aa aa AA aa aA aa AA aA AA aa aa AA aa aA H = 4/15 H = ?

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

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17 Calculating H (assuming simple dominance and Hardy-Weinberg Eq.) calculate H if q 2 = 0.09 f (a) = 0.3 = qq 2 = 0.09 f (A) = 0.7 = pp 2 = 0.49 2 pq = 0.42 H = 2pq = 0.42 (for only 2 alleles) 29a - 1

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

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19 Calculating H for 3 alleles: p, q, r p = 0.5 q = 0.4 r = 0.1 H = 2pq + 2qr + 2pr =.40 +.08 +.10 =.58 pppq pr r r q p pq pq qq qr prqrrr 29a

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20 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 Aa aa F1 generation p 2 = ? = f (AA) 2pq = ? = f (Aa)= H q 2 =.01 = f (aa) 29ez - 1

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21 q == 0.1 p + q = 1 so p = 1 - q p = 1 -.1 =.9 H = 2pq = 2 x.9 x.1 = 0.18 N heterozygous =.18 x 508 = 91 29ez - 2 q 2 =.01

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22 H and P H = heterozygosity = the percent of heterozygous genotypes in the population for that locus H = 2pq (for 2 allele case) H = 2pq + 2qr + 2 pr (for 3 allele case) P = allelic diversity = percent polymorphism = percent of loci for which alternative alleles exist in the population 29A

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23 Gene pools Population 1 N = 8 (7 homozygous) 2N = 16 alleles f (blue) = 1/16 f (red) = 15/16 Population 2 N = 8 2N = 16 alleles f (blue) = 16/16 Polymorphic locus Monomorphic locus 5f population P = approximately 0.25 individual H = approximately 0.07 typical

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

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

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26 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 15 -2 *e.g., electrophoresis protein sequencing DNA fingerprinting immunological techniques

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27 From Encarta

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28 Western Pyrénées National Park, France From Encarta

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29 Effective population size N = population size = total number of individuals N e = 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 32A-1

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

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

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

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33 5 causes of microevolution 1) genetic drift - stochastic variation in inheritance Expected F 2 : 9 - 3 - 3 - 1 Observed F 2 : 9 - 3 - 2.8 - 1.2 2) Assortative (nonrandom) mating 3) Mutation 4) Natural selection 5) Migration (gene flow) 31a-2 Random deviation

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34 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 34A-2

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35 Fixation of alleles Parental generation for many populations A =.5 a =.5 p = q =.5 fragmentation FnFn A = 1.0 a = 0 fixed lost fixed A = 0 a = 1.0 lost q = 1.0 p = 1.0 Genetic drift 34A-1 time

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

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

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38 36A-2 Generalized equation: H t = H 0 1 - 1 t t = # of generations later 2N e 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 *

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39 Formulae for calculating H and N e 37A 1 = proportion of H 0 lost at each generation 2N e 1 - 1 = proportion of H 0 remaining after the first generation 2 N e H t = H 0 1 - 1 t = the absolute amount of H 0 remaining after 2N e t generations N e = 4 MF 1) unequal sex ratios or M + F2) nonrandom breeding decrease N e

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40 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 10 5 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, N e but mutation can increase H and increase P 36A1

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