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Hardy -- Weinberg When will a population evolve? The Hardy-Weinberg principle states that the frequencies of alleles and genotypes in a population’s gene pool remain constant over the generations unless acted upon by agents other than recombination from sexual reproduction; populations do NOT evolve, unless they are NOT in H-W equilibrium H-W equilibrium conditions and mechanisms of change The Mechanisms of Change are the agents of evolutionary change Gene Flow Mutation Non-random mating Selection Genetic Drift Populations & gene pools Concepts a population is a localized group of interbreeding individuals The gene pool is collection of alleles in the population remember difference between alleles & genes! allele frequency is how common an allele is in the population Ex: 70% of individuals carry at least one dominant allele Evolution of populations Evolution = change in allele frequencies in a population hypothetical: what conditions would cause allele frequencies to not change? Let’s imagine a non-evolving population To stop evolution, REMOVE all agents of change very large population size (no genetic drift) 2. no migration (no gene flow in or out) 3. no mutation (no genetic change) 4. random mating (no sexual selection) 5. no natural selection (everyone is equally fit) *This is Hardy-Weinberg equilibrium* 1. Hardy-Weinberg theorem Counting total Alleles (regardless of genotype) assume 2 alleles = B, b frequency of dominant allele = p frequency of recessive allele = q frequencies must add to 1 (100%), since all individuals have one or both so: p+q=1 BB Bb bb Hardy-Weinberg theorem Counting Individual genotypes frequency of homozygous dominant genotype: p x p = p2 frequency of homozygous recessive genotype: q x q = q2 frequency of heterozygotes: (p x q) + (q x p) = 2pq Again, frequencies of all individuals must add to 1 (100%), so: p2 + 2pq + q2 = 1 BB Bb bb NOTICE THE DIFFERENCE!!! Alleles: p+q=1 B Genotypes: p2 + 2pq + q2 = 1 BB BB b Bb Bb bb bb Using Hardy-Weinberg equation Population of 100 cats. Black is dominant to q2 (bb): 16/100 = white in fur color. 84 are .16 black and 16 are white. q (b): √.16 = 0.4 Can I determine the p (B): 1 - 0.4 = 0.6 number of each genotype? p2=.36 BB 2pq=.48 Bb q2=.16 bb Must assume population is in H-W What are the genotype frequencies? equilibrium! Online Practice http://bio1151.nicerweb.com/med/QUIZ/hw_q.html http://nhscience.lonestar.edu/biol/hwe.html http://www.hinsdale86.org/staff/kgabric/HWanswe rs.htm http://natureboy.uta.edu/hwprobs.htm ff f F FF Ff