4-Population Genetics Notes

Population genetics: focuses on genetic changes within an interbreeding population. Gene pool = _______________________________ Allele Frequency = all the genes in a population Shows frequency of each allele in the gene pool (0.625 B, 0.375 b)

Changing the Gene Pool Microevolution: a change in a population’s gene pool over a succession of generations (a brief period of geologic time). Microevolution may be due to natural selection or other factors such as genetic drift. Genetic Drift: random change of allele frequencies in a small population due to chance , not natural selection.

Founder Effect:______________________ ___________________________ Example: ___________________________ migration of a small group to a new area can lead to genetic drift. Amish in America (extra digits)

Bottleneck Effect: ________ ___________ Example: _________________ an event that drastically lowers the population numbers endangered species

Genetic Equilibrium: Allele frequencies are not changing = ________________ no evolution

The Hardy Weinberg principle describes a hypothetical situation in which there is no change in the gene pool (frequencies of alleles), hence no evolution. The frequencies of alleles will remain unchanged generation after generation (= NO EVOLUTION) if the following conditions are met: Large population Random mating No mutation No migration No natural selection

The Hardy-Weinberg Equation: To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation. p2 + 2pq + q2 = 1 and p + q = 1 Using the equation: p = the dominant allele frequency (represented here by A). q = the recessive allele frequency (represented here by a) For a population in genetic equilibrium: p + q = 1.0 (The sum of the frequencies of both alleles is 1.)

The Hardy-Weinberg Equation: (p + q)2 = 1 so p2 + 2pq + q2 = 1 The three terms of this binomial expansion indicate the frequencies of the three genotypes: p2 = genotype frequency of AA (homozygous dominant) 2pq = genotype frequency of Aa(heterozygous) q2 = genotype frequency of aa (homozygous recessive)

SAMPLE PROBLEM #1: In pigs, black coat is recessive to white. What is the percentage of heterozygotes in this population? Calculate q2: Count the individuals that are homozygous recessive in the illustration to the left. Calculate the frequency of the total population they represent. This is q2. q2= 4/16 = 0.25

SAMPLE PROBLEM #1: In pigs, black coat is recessive to white. What is the percentage of heterozygotes in this population? 2. Find q: Take the square root of q2 to obtain q, the frequency of the recessive allele. q = √0.25 = 0.5

SAMPLE PROBLEM #1: In pigs, black coat is recessive to white. What is the percentage of heterozygotes in this population? 3. Find p: The sum of the frequencies of both alleles = 100%, p + q = 1. You know q, so what is p, the frequency of the dominant allele? p = 1 – q p = 1 - 0.5 = 0.5

SAMPLE PROBLEM #1: In pigs, black coat is recessive to white. What is the percentage of heterozygotes in this population? 4. Find 2pq: The frequency of the heterozygotes is represented by 2pq. Multiply by 100 to calculate the percent of the population that is heterozygous for white coat: 2pq = 2(0.5)(0.5) 2pq = 0.5 0.5 * 100= 50% Heterozygous