1. 11/24/08
Lecture 33 – Evolution 2
− Population Genetics: A field of genetics that studies genetic variation and genetic evolution of populations.
− Results usually in proportion giving in decimals.
− Population: Localized group of individuals of the same species that can successfully interbreed and produce
fertile offspring.
− Gene pool: the total of all the genes in a population.
− Hardy-Weinberg Theorem
− Based the mathematics expression of population genetics.
− A binomial expansion.
− Hardy-Weinberg Equilibrium
− A population is in the Hardy-Weinberg equilibrium if the proportions of genotypes remains unchanged
generation after generation.
− 5 conditions must be met in the population for the Hardy-Weinberg equilibrium to be established:
− Population must be large.
− Mating must be random.
− Natural selection cannot take place.
− No gene flow (no exchange of genes between population.
− The Hardy-Weinberg equilibrium is totally hypothetical.
− For example, Gene A (Possible genotype within the population: AA, Aa, aa)
− Given 0.25AA, 0.50Aa, 0.25aa (this population is in the Hardy-Weinberg equilibrium)
− p = let p equal the proportion of the dominant allele in the population.
− q = let q equal the proportion of the recessive allele in the population.
− There are two ways to find the proportions of the offspring.
− Long Method
− Let the population be 100 individuals (total alleles = 100)
− p = (50 + 50) / 200 = 100/200 = 0.50
− q = (50 + 50) / 200 = 100/200 = 0.50
− p + q = 1.0
− Short Method
− p = proportion of dominant individuals + 0.5 props of heterozygous
− 0.25 + 0.25 = 0.50
q = 1 – p = 0.50
− What will be the offspring?

2. − p
2
= proportion of homozygous dominant individuals in the population
− 2pq = proportion of heterozygous in the population.
− p
2
= proportion of homozygous recessive in the population.
− Natural selection (if applied)
− Contain genotypes are more likely to leave offspring than others.
− Extreme example (complete selection against the homozygous recessive allele).
− Generation 0
− (0.25AA 0.50Aa) 0.25aa (no genes passed on)
− 1 AA : 2Aa
− p = 4/6 = 0.66...
− q = 2/6 = 0.33...
− Generation 1
− (0.66)
2
+ 2(0.66)(0.33) + (0.33)
2
− (0.444) + (0.444) + (0.111)
− Increase, Decrease, None