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Three systems of vocabulary Alleles and Genotypes in Populations that Mate at Random Alan R. Rogers September 12, 2014 Illustration of classical usage Those organisms (homozygotes) which received like genes, in any pair of corresponding loci, from their two parents, would necessarily hand on genes of this kind to all of their offspring alike; whereas those (heterozygotes) which received from their two parents genes of different kinds. . . (Fisher, 1930, p. 8) Position on chromosome Protein-coding locus Physical copy of DNA at locus One of several variants at a locus 1 locus gene gene allele 2 locus gene allele allele 3 locus gene gene copy allele 1 is classical usage, 2 is Gillespie’s, and we try to keep to 3. Transferrin genotype frequencies in a baboon troop G’type CC CD DD Total Number of baboons C D 80 160 0 15 15 15 5 0 10 100 175 25 Relative frequency x̂CC = 80/100 = 0.80 x̂CD = 15/100 = 0.15 x̂DD = 5/100 = 0.05 p̂ = 175/200 = 0.875 Note: “hat” indicates values describing sample rather than population. I’ll often ignore this distinction. Alternative calculation of p p̂ = x̂CC + x̂CD /2 = 0.80 + 0.15/2 = 0.875 The sample allele frequency p̂ is an estimate of the population allele frequency p. The population allele frequency is also the probability that a gene drawn at random from the population is a copy of allele C . Expected genotype frequencies What is the probability that a random baboon will have genotype CD? Depends on characteristics of population. To describe these effects, we need a model. Model: random mating, no selection Event CC Event CD can be decomposed as follows: Gene copy from Mom Dad C D D C Sum: Probability p × (1 − p) (1 − p) × p 2p(1 − p) Why multiply? Why multiply? Why add? Hardy-Weinberg result Genotype CC CD DD Gene copy from Mom Dad C C Sum: Probability p×p p2 Why multiply? Observed versus expected g’type freqs Relative frequency xCC = p 2 xCD = 2pq xDD = q 2 I Random mating does not change p. I Given allele frequency, we know genotype frequencies. Heterozygosity on human chromosome 1 Genotype CC CD DD Relative frequency Observed Expected xCC = 0.80 p 2 = 0.77 xCD = 0.15 2pq = 0.22 xDD = 0.05 q 2 = 0.02 Observed: relatative frequency of genotype in data Expected: Hardy-Weinberg formula What if males and females have different allele frequencies? Sex ♂ ♀ Sex ♂ ♀ Genotype frequencies A1 A1 A1 A0 A0 A0 x11 x10 x00 y11 y10 y00 Allele frequency pm = x11 + x10 /2 pf = y11 + y10 /2 An autosomal locus in a nuclear family Summary Probabilities that gametes carry A1 Dad Mom • • • • ........................ ........................ ... ... ..... ..... .. .. .. ... ... ... .. . .... . .. . .. . .. . . ... ... . .. . . . . ...... . . . ........ ....... ........................ . . . . . . . . . . . . . ... ... ... ... ... .. ... .. . . . ....... .............. ................................ ..... ..... ... .... .. .. ... .. .. ... .. . . ...... . ..................... ? ? • • Child ♂ ♀ x11 + x10 /2 = pm y11 + y10 /2 = pf Child genotype probabilities I At equilibrium under random mating, allele frequencies determine genotype frequencies. I Hermaphrodites reaches equilibrium in 1 generation. I Autosomal loci in sexual populations reach equilibrium in 2 generations. I X-linked loci in reach equilibrium only gradually. 0 x11 = pm pf 0 x10 = pm (1 − pf ) + pf (1 − pm ) 0 x00 = (1 − pm )(1 − pf ) The sexes now have equal allele frequencies. 0 0 p 0 = x11 + x10 /2 = (pm + pf )/2