Population Genetics Unit 4 AP Biology Population Genetics Study of the frequency of particular...
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Transcript of Population Genetics Unit 4 AP Biology Population Genetics Study of the frequency of particular...
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