2 Population Genetics:The study of evolution from a genetic point of view (remember evolution is a gradual change in the genetic material of a population). Within a population, individualsmay vary in observable traits.Some individuals show extremetraits, while most show averagetraits. This is shown in abell curve.
3 Allele Frequencies & the Gene Pool Gene pool: the total genetic information available in a population.Allele frequency: the % of an allele in a gene pool.For example, suppose there are 10 tall (5 TT and 5 Tt) and 10 short pea plant (tt).Find the allele frequency for T: __________________Find the allele frequency for t: __________________***NOTE : the sum of the frequencies of T and t should add to 1.0 (or 100%)***10T (in 5TT) and 5T (in 5Tt)= 15T/40alleles = 0.3755t (in 5Tt) and 20t (in 10tt)= 25t/40alleles= 0.625
4 Phenotype frequency:The number of individuals with a particular phenotype divided by the total # of individuals in the population. (To find the phenotype frequency, divide the # of individuals with one phenotype by the total of individuals present.)For example, suppose there are 10 tall and 10 short pea plants.Find the phenotype frequency for tall: ______________Find the phenotype frequency for short: ____________***NOTE : the sum of the frequencies of tall and short should add to 1.0 (or 100%)***10tall/20 plants= 0.510short/20 plants= 0.5
5 Genotype frequencyNumber of individuals with a particular genotype divided by the total # of individuals in the population. (ex- RR, Rr, rr) (To find the genotype frequency, divide the # of individuals with one genotype by the total of individuals present.)For example, suppose there are 6 tall (TT), 4 tall (Tt) and 10 short (tt) pea plants.Find the genotype frequency for TT: _________________Find the genotype frequency for Tt: _________________Find the genotype frequency for tt: _________________***NOTE : the sum of the frequencies of TT, Tt, and tt should add to 1.0 (or 100%)***6TT/20plants=0.34Tt/20 plants= 0.210tt/20 plants=0.5
6 Hardy-Weinberg Equilibrium says that allele frequencies in a population tend to remain the same from generation to generation unless acted upon by outside influences (see assumptions on next slide).The Hardy-Weinberg equilibrium is based on a set of assumptions about an ideal hypothetical population that is NOT evolving.The H-W law says that if no evolution is occurring, then the allele frequencies will remain the same in each succeeding generation of sexually reproducing individuals.
7 In order for equilibrium to remain in effect (i. e In order for equilibrium to remain in effect (i.e. that no evolution is occurring) then the following five conditions must be met:No mutations, so no change in allele frequency.No immigration or emigration within the population.Population size is large.Random mating occurs (no sexual selection).Natural selection does not occur (all organisms are equally fit for the environment).
8 Remember, this is a theoretical state for a hypothetical population! Real populations do not follow all of these necessary conditions.Some or all of these types of forces all act on living populations at various times and evolution at some level occurs in all living organisms.However, by showing how genetic equilibrium is maintained, Hardy-Weinberg genetic equilibrium can allow us to consider what forces disrupt equilibrium.
9 Respond to the following in terms of how they would cause disruption in a population’s equilibrium: Mutation:Migration:Genetic Drift:Nonrandom Mating:Natural Selection:
10 Hardy-Weinberg Formulas: They allow scientists to determine whether evolution has occurred.Any changes from generation to generation in the gene/allele frequencies in the population over time can be detected, thus allowing a simplified method of determining whether evolution is occurring.
11 Hardy-Weinberg Formulas: There are two formulas that must be known:p2 + 2pq + q2 = 1 and p + q = 1p = frequency of the dominant allele in the populationq = frequency of the recessive allele in the populationp2 = percentage of homozygous dominant individualsq2 = percentage of homozygous recessive individuals2pq = percentage of heterozygous individuals
12 Practice Problem #1: p2 + 2pq + q2 = 1 and p + q = 1 If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes (2pq)?q2= homozygous recessive individuals =98/200= 0.49q= √q2 = √0.49 = 0.7p= 1-q = = 0.32pq= heterozygotes= 2(0.7)(0.3)= 0.42Answer: 42% of the population are heterozygous for this trait!
13 Practice Problem #2: p2 + 2pq + q2 = 1 and p + q = 1 Lets say that brown fur coloring is dominant to gray fur coloring in mice. If you have 168 brown mice in a population of 200 mice…What is the predicted frequency of homozygous recessive (q2)?q2= homozygous recessive individuals =32/200= 0.16What is the predicted frequency of homozygous dominant (p2)?First we have to find q and then p…q= √q2 = √0.16 = 0.4p= 1-q= 1-0.4= 0.6p2= homozygous dominant individuals= 0.62= 0.36What is the predicted frequency of heterozygotes (2pq)?2pq= heterozygotes= 2(0.6)(0.4)= 0.48