Population GeneticsPopulation Genetics

Unit 4AP 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.