Population Genetics and Hardy-Weinberg Equilibrium

Population Genetics

Focus on organisms with diploid genetics.
Considers one gene with two alleles.
Asks whether allelic and genotypic frequencies in a group of individuals indicate random mating.
Compares observed genotypic frequencies to expected frequencies using the chi-square test.

Describes expected genotypic frequencies when gametes form zygotes in all possible combinations.
p = dominant allele frequency.
q = recessive allele frequency.
Genotype frequencies:
- AA = p^2
- Aa = 2pq
- aa = q^2

Example: M/N blood types among 600 individuals.
- M (L^ML^M) = 200
- MN (L^ML^N) = 300
- N (L^NL^N) = 100
Calculation of allele frequencies:
- Number of L^M alleles = (200 \, \text{x} \, 2) + 300 = 700
- Number of L^N alleles = (100 \, \text{x} \, 2) + 300 = 500
- Total number of alleles = 1200
- Frequency of L^M \,(p) = \frac{700}{1200} = 0.583
- Frequency of L^N \,(q) = \frac{500}{1200} = 0.417
- p + q = 1.000

Using the sample of 600 individuals to test if the blood group gene is in HWE.
Observed genotype frequencies:
- L^ML^M = 200
- L^ML^N = 300
- L^NL^N = 100
Expected genotype numbers based on calculated allele frequencies:
- Frequency of L^M = p = 0.583
- Frequency of L^N = q = 0.417
- Expected L^ML^M = p^2 \, \text{x} \, 600 = 0.340 \, \text{x} \, 600 = 204
- Expected L^ML^N = 2pq \, \text{x} \, 600 = 0.486 \, \text{x} \, 600 = 292
- Expected L^NL^N = q^2 \, \text{x} \, 600 = 0.174 \, \text{x} \, 600 = 104
Chi-square calculation:
- \chi^2 = \sum \frac{(O-E)^2}{E} (where O is Observed and E is Expected)
- \chi^2 = \frac{(200-204)^2}{204} + \frac{(300-292)^2}{292} + \frac{(100-104)^2}{104} = 0.075 + 0.219 + 0.154 = 0.451
Degrees of freedom (DF) = # genotypes - # alleles = 3 - 2 = 1
Critical value for \chi^2 with 1 DF at 0.05 significance level = 3.84
Since \chi^2 = 0.451 < 3.84, we accept the null hypothesis.
The blood group gene is likely in Hardy-Weinberg Equilibrium.

Allele and genotype frequencies do not change from generation to generation if:
- Random mating in a large population (panmictic population).
- No new mutations.
- No migration (closed population).
- No selection (equal chance of survival for all genotypes).
If any assumption is not met, the population may not be in HWE.

Predict allele frequencies from genotype frequencies.
Calculate carrier frequency of recessive disease-causing alleles (e.g., cystic fibrosis).
Example: Frequency of disease = 0.09% of population (homozygous recessive).
- q^2 = 0.0009
- q = \sqrt{0.0009} = 0.03
- p = 0.97 (frequency of normal allele)
- Carrier frequency = 2pq = 2(0.03)(0.97) = 0.058 or 5.8%
If you are a carrier, the chance of having a child with the disease is 5.8% (chance that mate is a carrier) \
Probability calculation: 0.058 (your chance of being a carrier ) x 0.25 (chance of passing on affected allele) = 0.0145 or 1.45% chance

Frequency of X-linked genotype corresponds with:
- Homogametic sex (XX): p^2 + 2pq + q^2
- Heterogametic sex (XY): p + q
Example: If 10% of males have hemophilia allele (recessive):
- X^hY (males with) = 0.10 = q
- X^+Y (males without) = 0.9 = p
- X^hX^h (females with) = q^2 = 0.01
- X^hX^+ (female carriers) = 2pq = 0.18
- X^+X^+ (female homozygous normal) = p^2 = 0.81

A population is composed of 50 Aa individuals and 50 aa individuals. Are these genotypes in HWE?
Allele frequencies:
- a frequency = p = 50/200 = 0.25
- A frequency = q = 150/200 = 0.75
Expected genotype numbers (total individuals = 100):
- AA = p^2 \, \text{x} \, 100 = 0.0625 \, \text{x} \, 100 = 6.25
- Aa = 2pq \, \text{x} \, 100 = 0.375 \, \text{x} \, 100 = 37.5
- aa = q^2 \, \text{x} \, 100 = 0.5625 \, \text{x} \, 100 = 56.25
Chi-square test:
- \chi^2 = \frac{(50-6.25)^2}{6.25} + \frac{(50-37.5)^2}{37.5} + \frac{(0-56.25)^2}{56.25} = 277.76
Since \chi^2 \,\, \text{is much greater than}\, 3.84, we reject the hypothesis that the genotypes are in Hardy-Weinberg Equilibrium (HWE).