2 Population GeneticsStudy of the frequency of particular alleles and genotypes in a populationEx. Suppose you want to determine the % of alleles and genotypes for Alu in the PV92 region.
3 Sample Data- Alu in the PV92 region 100 AP Biology students total45 students are +/+20 students are +/-35 students are -/-
4 Allelic Frequencies & Genotype Frequencies Genotype Frequency% of a particular genotype that is present in a populationAllelic Frequency% of particular allele in a population
5 Sample Problem 100 AP Biology students total 45 students are +/+20 students are +/-35 students are -/-Calculate the genotype frequenciesCalculate the allelic frequencies
6 Solution- Genotype Frequencies Genotype frequency of +/+45 (+/+) / 100 students total =Genotype Frequency of +/-20 (+/-) / 100 students total =Genotype Frequency of -/-35 (-/-) / 100 students total =
7 Solution- Allelic Frequencies Step 1: Calculate the total # of alleles100 students x 2 alleles each= 200 total alleles
8 Solution- Allelic Frequencies Allelic Frequency of “+” alleleEach student with +/+ contributes 2 “+” alleles each 2 x 45 = 90 “+” allelesEach student with +/- contributes 1 “+” allele each 1 x = 20 “+” alleles= 110 “+” allelesAllelic Frequency = 110 “+” alleles /200 total=
9 Solution- Allelic Frequencies Allelic Frequency of “-” alleleEach student with -/- contributes 2 “-” alleles each 2 x 35 = 70 “-” allelesEach student with +/- contributes 1 “-” allele each 1 x = 20 “-” alleles= 90 “-” allelesAllelic Frequency = 90 “-” alleles /200 total=
10 Practice Problem #1 325 ants total: 140 ants have the genotype GG75 ants have the genotype Gg110 ants have the genotype ggCalculate the genotype and allelic frequencies.
12 Solution #1 How many alleles total? Allelic Frequencies: 325 X = alleles totalAllelic Frequencies:G : ( (140 x 2) + 75) / =g : ( (110 x 2) ) / =
13 Question… What should all of the allelic frequencies add up to? They should add up to 1What should all of the genotype frequencies add up to?
14 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)
15 EvolutionIf the genotype and allelic frequencies are NOT changing from generation to generation then the population is NOT evolving.
16 Hardy Weinberg Equilibrium Mathematical model to describe a nonevolving populationIn real life, populations are almost never in Hardy Weinberg Equilibrium2 variables:p = allelic frequency of one alleleq = allelic frequency of other allele
17 Conditions of Hardy Weinberg Very large population sizeNo Gene Flow (no moving in or out)No Mutations (no new alleles introduced)No Sexual SelectionNo 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
18 Hardy Weinberg Equations p and q are frequencies of alleles in a populationp + 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)
19 Hardy Weinberg Equations Using the allelic frequencies (p and q), genotype frequencies can be calculated for each genotypep2 = genotype frequency for homozygous dominant (HH)2pq = genotype frequency for heterozygous genotype (Hh)q2 = genotype frequency for homozygous recessive (hh)
20 Hardy Weinberg Equations p pq q2 = 1Why is it 2pq?Because there are two heterozygous combinations possible (Hh and hH)
21 Hardy Weinberg Equations Based on probabilityp = 0.8 (allelic frequency for CR)q = 0.2 (allelic frequency for CW)Chance of CRCR = 0.8 x 0.8 = 0.64
22 Population in HW Equilibrium? You can use calculations to determine if a population is in Hardy Weinberg EquilibriumLet’s take practice problem #1:Actual allelic frequencies are:p = q =Actual genotype frequencies are:0.43 (GG), (Gg), (gg)
23 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 = p2Expected Gg Genotype frequency = 2pqExpected gg genotype frequency = q2
24 Expected Genotype Frequencies p = , q =Expected Genotype frequencies:GG = p2 = (0.55)2 =Gg = 2pq = 2 (0.55)(0.45) = 0.50gg = q2 = (0.45)2 =Actual genotype frequencies are:0.43 (GG), (Gg), (gg)The actual genotype and expected genotype frequencies do not match the population is NOT in HW Equilibrium.
25 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 BB65 ferrets are Bb21 ferrets are bbWhat are the actual allelic and genotype frequencies? Is the population in HW Equilibrium?
27 Solution Actual Allelic frequencies: Total alleles = 2 x 164 = 328 Allelic frequency of B = p= ( (2 x 78) + 65) / =Allelic frequency of b = q= ( (2 x 21) + 65) / =
28 Solution Expected genotype frequencies: BB = p2 = (0.67)2 = 0.45Bb = 2pq = 2(0.67)(0.33) = 0.44bb = q2 = (0.33)2 = 0.11These do not match the actual frequencies, so the population does not appear to be in HW Equilibrium.