1. Population Genetics
Unit 4
AP Biology

2. 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.

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

4. Allelic Frequencies & Genotype Frequencies
Genotype Frequency
% of a particular genotype that is present in a population
Allelic 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 frequencies
Calculate 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 alleles
100 students x 2 alleles each
= 200 total alleles

8. 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 “+” alleles
= 110 “+” alleles
Allelic Frequency = 110 “+” alleles /200 total =

9. 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 “-” alleles
= 90 “-” alleles
Allelic Frequency = 90 “-” alleles /200 total =

10. 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.

11. Solution #1 Genotype Frequencies: GG = 140 / 325 = 0.43

12. Solution #1 How many alleles total? Allelic Frequencies:
325 X = alleles total
Allelic 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 1
What 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. Evolution
If 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 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

17. 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

18. 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)

19. 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)

20. Hardy Weinberg Equations
p pq q2 = 1
Why is it 2pq?
Because there are two heterozygous combinations possible (Hh and hH)

21. 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

22. 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 = 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 = p2
Expected Gg Genotype frequency = 2pq
Expected 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) = 0.50
gg = 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 BB
65 ferrets are Bb
21 ferrets are bb
What are the actual allelic and genotype frequencies? Is the population in HW Equilibrium?

26. Solution Actual genotype frequencies: BB = 78 / 164 = 0.47

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 = 0.45
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.