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Unformatted text preview: modifications of the frequency of certain genes within a population changes over time only through populations can you actually see patterns of genetic change Hardy-Weinberg: mathematical equation describing how equilibrium occurs when everything is random—in this way determines whether evolution occurring. they measure observed genotype frequencies within a species, then compare them against predicted genotype frequencies to see how alleles combine. Two alleles: dominant A, recessive a with Hardy-Weinberg, A becomes p and a becomes q p and q are the FREQUENCY of A and a when half of the genes in gene pool are A, p = 0.5 the chances of creating genotype AA are p^2 Aa = p * q and aA = p * q so heterozygous = 2pq aa = q^2 HARDY-WEINBERG= p^2 + 2pq + q^2 heads= p, tails=q HH, p^2 = 0.5*0.5=0.25 HT, p*q= 0.5*0.5=0.25 TH, p*q= 0.5*0.5=0.25 0.50 total TT, q^2= 0.5*0.5=0.25 Ex. have 100 cats, the trait is hair color. have black, orange, and calico as alleles. the genotypes are BB, OO, BO (i.e. calico is black and orange together, codominant) 25 black cats, BB, 25% of pop 40 calico cats, BO, 40% of pop 35 orange cats, OO, 35% of pop 100 cats=200 alleles BB cats have two of B so 50 B alleles total, and 0 orange alleles BO cats have 40 B and 40 O alleles OO cats have 70 O and no B B=90 total alleles, O=110 total alleles 90/200=.45 (frequency of B allele in gene pool) 110/200=.55 (frequency of O) when there is no evolution: BB = .45*.45= .2025 BO = .45*.55= .4950 (the frequency of black, orange and calico cats in pop.) OO = .55*.55= .3025 these must add to 1. back to data, we counted that 25 black cats, 40 calico and 35 orange. so this means evolution definitely happening. looks like have more black and less orange…so more BB alleles? ...
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This note was uploaded on 03/02/2011 for the course ANTH 001 taught by Professor Mcfarlin during the Spring '11 term at GWU.
- Spring '11
- Biological Anthropology