Evolution: Hardy-Weinberg Equilibrium Biology 11H Evolution: Hardy-Weinberg Equilibrium

Background Genetics Refresher Remember you have 2 copies of each gene (1 from each parent) Genes can have different versions EX. gene for hair colour comes in many versions: brown, blonde, red, black, grey The versions are called ALLELES and you can get a different one from each your parents EX. mom gave you brown hair (B) and dad gave you blonde (b) The genes you actually possess are called your GENOTYPE (Bb) but which do you express?? Usually there is a dominant (B) and a recessive version (b) of the gene The physical appearance of those genes is called PHENOTYPE EX. brown hair even though you are carrying a blonde gene you’re not expressing it Genotypes: BB, Bb, bb Phenotypes: Brown (BB and Bb), blonde (bb)

Genotype details… BB: homozygous dominant Bb: heterozygous (AKA carrier) Bb: homozygous recessive

Hardy - Weinberg Equilibrium Let the frequency for normal body, the dominant allele, be designated “p” and the recessive yellow allele as “q” Then p + q = 1 When an offspring is conceived, its genotype is determined by (p+q) x (p+q) =1 Simplified, this is p2 + 2pq + q2 = 1

Hardy-Weinberg Equilibrium p2 + 2pq + q2 = 1 This is the Hardy-Weinberg equilibrium which predicts a situation in a population which is NOT evolving.

Conditions for HW Equilibrium HW Conditions: Population must be large enough to make chance alone unable to change its gene frequencies The population must be completely isolated from all other populations - no immigration or emigration Mutations must not occur in the population Mating must be totally random, i.e. no mating preferences are shown by anyone All members of the population have an equal chance of survival

Conditions for HW Equilibrium Since none of these conditions ever occur in any natural population, the HW equilibrium provides us with a tool to measure the rate of evolution in any population with observable traits.

Solving a HW problem Albinism is a rare genetically inherited trait that is only expressed in the phenotype of homozygous recessive individuals (aa).  The most characteristic symptom is a marked deficiency in the skin and hair pigment melanin.  This condition can occur among any human group as well as among other animal species.  The average human frequency of albinism in North America is only about 1 in 20,000. 1. Start with the frequency of the recessive: q2 = 1/20,000 q2 = 0.00005 2. Calculate the q value: √q2 = √ 0.00005 q = 0.007 3. Use the second equation: p + q = 1 to solve for p p + 0.007 = 1 p = 0.993 4. Use p value to solve p2 and 2pq p2 = (0.993) 2 2pq = 2(0.993)(.007) p2 = 0.986 2pq = 0.014 5. Double check full equation = 1 p2 + 2pq = q2 = 1 0.986 + 0.014 + 0.00005 = 1 6. Answer any questions!!