2 Context Evolution appears to be the typical situation Sometimes evolution is not occurringHardy-Weinberg EquilibriumConditions that specify if evolution is NOT happeningEquilibrium = no evolutionMore of a theoretical case than a real testBut affords comparison to evolution
3 DefinitionsPopulation Genetics – study of how populations change genetically over timePopulation – Group of same species individuals that can and do interbreed successfullyGene pool – all of the alleles at all loci in all the members of a populationFixed – only one allele exists for a locus in the populationMore fixed alleles = lower species diversity
4 Hardy-Weinberg If gene pools are NOT Evolving, can use Hardy- Weinberg Describes a population that is NOT evolvingAllelic frequencies will remain constantGene frequencies remain constant throughout the generations
5 Conditions for Hardy-Weinberg 1. No mutations2. Random mating3. No natural selections4. Large population size5. No gene flowImmigration or emigrationGenes coming into or leaving the population
6 Math in Biology?p = Frequency of the dominant allele = f(A) q = Frequency of the recessive allele = f(a) There are only 2 alleles (dominant + recessive), so p + q = 1 square both sides: (p + q)2 = 12 expand the binomial: p2 + 2pq + q2 = 1
7 Hardy-Weinberg Math (Page 2) p2 = pp = f(aa) = Frequency of homozygous dominantpq = f(Aa) = frequency of heterozygoteHowever, remember from Punnett Square, there were 2 heterozygotes for the F1 generation; So2pq = f(Aa)q2 = qq = f(aa) = frequency of homozygous recessive
8 In Summary Summary: p2 + 2pq + q2 = 1 f(AA) + f(Aa) + f(aa) = 1 f(Homo. Dom.) + f(Hetero.) + f(Homo. Recess.) = 1Are there any other genotypes?Makes sense?
9 Problem (Part 1)A trait has two characters, dominant (A) and recessive (a). In a population of 500 individuals in Hardy- Weinberg equilibrium, 25% display the recessive phenotype (aa).a) What is the frequency of the dominant allele in the population?b) What is the frequency of the recessive allele in this population?
10 Answer f(aa) = 0.25 OR q2 = 0.25 q = √(0.25) = 0.5 OR f(a) = 0.5 So the frequency of the recessive allele = 0.5From before: p + q = 1 AND q = 0.5, sop = 0.5 = f(A) So the frequency of the dominant allele = 0.5
11 Part 2 of Problem What are the frequencies of a) homozygous dominant genotype?b) Heterozygous genotype?c) homozygous recessive genotype?
12 Part 2 of Problem (Answer) What are the frequencies ofa) homozygous dominant genotype?Since p = 0.5, p2 = (0.5)2 = 0.25b) Heterozygous genotype?Since p = 0.5 & q = 0.5, 2pq = 2(0.5)(0.5) = 0.5c) homozygous recessive genotype?Yep, it was given, but if we must: q = 0.5, q2 = (0.5)2 = 0.25
13 Part 3 of Problem How many individuals have the a) homozygous dominant genotype?b) Heterozygous genotype?c) homozygous recessive genotype?
14 Part 3 of Problem (Answer) How many individuals have thea) homozygous dominant genotype?500 members & p2 = 0.25, (0.25)(500) = 125b) Heterozygous genotype?500 members & p = 0.5, q = 0.5 2pq = 0.5So 500 (0.5) = 250 heterozygotesc) homozygous recessive genotype?500 members & q2 = 0.25, (0.25)(500) = 125
15 Part 4 of Problem How many individuals have the a) Dominant phenotype? b) Recessive phenotype?
16 Part 4 of Problem (Answer) How many individuals have thea) Dominant phenotype?= (Homo. Dom.) + (Hetero.) = = 375b) Recessive phenotype?= Homo. Recess. = 125