Population Genetics

Population Genetics Unit 4 AP Biology

Population Genetics • Study of the frequency of particular alleles and genotypes in a population • Ex. Suppose you want to determine the % of alleles and genotypes for Alu in the PV92 region.

Sample Data- Alu in the PV92 region • 100 AP Biology students total 45 students are +/+ 20 students are +/- 35 students are -/-

Allelic Frequencies & Genotype Frequencies • Genotype Frequency • % of a particular genotype that is present in a population • Allelic Frequency • % of particular allele in a population

Sample Problem • 100 AP Biology students total • 45 students are +/+ • 20 students are +/- • 35 students are -/- • Calculate the genotype frequencies • Calculate the allelic frequencies

Solution- Genotype Frequencies • Genotype frequency of +/+ 45 (+/+) / 100 students total = 0.45 • Genotype Frequency of +/- 20 (+/-) / 100 students total = 0.20 • Genotype Frequency of -/- • 35 (-/-) / 100 students total = 0.35

Solution- Allelic Frequencies • Step 1: Calculate the total # of alleles • 100 students x 2 alleles each = 200 total alleles

Solution- Allelic Frequencies • Allelic Frequency of “+” allele • Each student with +/+ contributes 2 “+” alleles each  2 x 45 = 90 “+” alleles • Each student with +/- contributes 1 “+” allele each  1 x 20 = 20 “+” alleles • 90 + 20 = 110 “+” alleles • Allelic Frequency = 110 “+” alleles /200 total = 0.55

Solution- Allelic Frequencies • Allelic Frequency of “-” allele • Each student with -/- contributes 2 “-” alleles each  2 x 35 = 70 “-” alleles • Each student with +/- contributes 1 “-” allele each  1 x 20 = 20 “-” alleles • 70 + 20 = 90 “-” alleles • Allelic Frequency = 90 “-” alleles /200 total = 0.45

Practice Problem #1 • 325 ants total: • 140 ants have the genotype GG • 75 ants have the genotype Gg • 110 ants have the genotype gg • Calculate the genotype and allelic frequencies.

Solution #1 • Genotype Frequencies: • GG = 140 / 325 = 0.43 • Gg = 75 / 325 = 0.23 • gg = 110 / 325 = 0.34

Solution #1 • How many alleles total? 325 X 2 = 650 alleles total • Allelic Frequencies: • G : ( (140 x 2) + 75) / 650 = 0.55 • g : ( (110 x 2) + 75 ) / 650 = 0.45

Question… • What should all of the allelic frequencies add up to? • They should add up to 1 • What should all of the genotype frequencies add up to? • They should add up to 1

Gene Pool • All the alleles in a population • Alleles in the gene pool can change as a result of several different factors (mutations, immigration, natural selection)

Evolution • If the genotype and allelic frequencies are NOT changing from generation to generation then the population is NOT evolving.

Hardy Weinberg Equilibrium • Mathematical model to describe a nonevolving population • In real life, populations are almost never in Hardy Weinberg Equilibrium • 2 variables: p = allelic frequency of one allele q = allelic frequency of other allele

Conditions of Hardy Weinberg • Very large population size • No Gene Flow (no moving in or out) • No Mutations (no new alleles introduced) • No Sexual Selection • No Natural Selection (no allele causes the individual to survive better or worse) • Not meeting these conditions would cause the allelic and genotype frequencies to change

Hardy Weinberg Equations • p and q are frequencies of alleles in a population • p + q = 1 (the 2 alleles make up 100% of the alleles) • From this, there are 3 genotypes possible: • Homozygous dominant (Ex. HH) • Homozygous recessive (Ex. hh) • Heterozygous (Ex. Hh)

Hardy Weinberg Equations • Using the allelic frequencies (p and q), genotype frequencies can be calculated for each genotype • p2 = genotype frequency for homozygous dominant (HH) • 2pq = genotype frequency for heterozygous genotype (Hh) • q2 = genotype frequency for homozygous recessive (hh)

Hardy Weinberg Equations • Equation: p2 + 2pq + q2 = 1 • Why is it 2pq? • Because there are two heterozygous combinations possible (Hh and hH)

Hardy Weinberg Equations • Based on probability • p = 0.8 (allelic frequency for CR) • q = 0.2 (allelic frequency for CW) • Chance of CRCR = 0.8 x 0.8 = 0.64

Population in HW Equilibrium? • You can use calculations to determine if a population is in Hardy Weinberg Equilibrium • Let’s take practice problem #1: • Actual allelic frequencies are: • p = 0.55 q = 0.45 • Actual genotype frequencies are: • 0.43 (GG), 0.23 (Gg), 0.34 (gg)

If the population is in HW Equilibrium… • If the population is in Hardy Weinberg equilibrium, then the calculated expected genotype frequencies should match the actual genotype frequencies. • Using the Hardy Weinberg Equation: • Expected GG Genotype frequency = p2 • Expected Gg Genotype frequency = 2pq • Expected gg genotype frequency = q2

Expected Genotype Frequencies • p = 0.55, q = 0.45 • Expected Genotype frequencies: • GG = p2 = (0.55)2 = 0.30 • Gg = 2pq = 2 (0.55)(0.45) = 0.50 • gg = q2 = (0.45)2 = 0.20 • Actual genotype frequencies are: • 0.43 (GG), 0.23 (Gg), 0.34 (gg) • The actual genotype and expected genotype frequencies do not match  the population is NOT in HW Equilibrium.

One more Practice Problem • You are studying a ferret population for their nose sizes. B is the allele for a long nose, b is the allele for a short nose. • 78 ferrets are BB • 65 ferrets are Bb • 21 ferrets are bb • What are the actual allelic and genotype frequencies? Is the population in HW Equilibrium?

Solution • Actual genotype frequencies: • BB = 78 / 164 = 0.47 • Bb = 65 / 164 = 0.40 • bb = 21 / 164 = 0.13

Solution • Actual Allelic frequencies: • Total alleles = 2 x 164 = 328 • Allelic frequency of B = p = ( (2 x 78) + 65) /328 = 0.67 • Allelic frequency of b = q = ( (2 x 21) + 65) / 328 = 0.33

Solution • Expected genotype frequencies: • BB = p2 = (0.67)2 = 0.45 • Bb = 2pq = 2(0.67)(0.33) = 0.44 • bb = q2 = (0.33)2 = 0.11 • These do not match the actual frequencies, so the population does not appear to be in HW Equilibrium.

Population Genetics