Population Genetics

Introduction to Population Genetics

• Definition: Study of how allele and genotype frequencies change over generations.

• Fundamental question: If we know the population composition in one generation, what will it be in the next generation?

• Key factors influencing population composition:

  • Mating patterns

  • Offspring survival

  • Generation dynamics

Terminology in Population Genetics

• Alleles: Different forms of the same gene (e.g., A and a).

• Gene Frequency: Proportion of each gene/allele in the population.

• Genotype: Genetic combination of alleles at a given locus. Types of genotypes:

  • Homozygous Dominant: AA

  • Heterozygous: Aa

  • Homozygous Recessive: aa

• Genotype Frequency: Proportion of individuals with each genotype in the population.

• Phenotype: Observable physical traits that result from genotype.

Genes and Genotypes Examples

The Hardy-Weinberg Principle

• Definition: Allele and genotype frequencies in a population will not change from generation to generation when no evolutionary forces are acting upon it.

• Founders: G. H. Hardy & W. Weinberg.

Hardy-Weinberg Equilibrium

• Formula: $p + q = 1$ where:

  • $p$ is the frequency of allele A

  • $q$ is the frequency of allele a

• Genotype frequencies: $p^2 + 2pq + q^2 = 1$ where:

  • $p^2$ = Frequency of homozygous dominant (AA)

  • $2pq$ = Frequency of heterozygous (Aa)

  • $q^2$ = Frequency of homozygous recessive (aa)

• Occurs in the absence of evolution.

Hardy-Weinberg Equations Overview

• Genotype Frequency Calculation:

  • Genotypes are AA, Aa, aa and their respective frequencies are P, Q, R.

  • Hardy-Weinberg gives the relationships:

  • $p^2 = P$

  • $2pq = Q$

  • $q^2 = R$

• Requirement: The sum of frequencies should equal 1: $p^2 + 2pq + q^2 = 1$ and $p + q = 1$.

Assumptions for Hardy-Weinberg Equilibrium:

1. No mutation

2. Random mating

3. No gene flow (in or out)

4. Very large population size

5. No natural selection

• If any assumption is violated, equilibrium is disrupted.

Example: Calculating Observed Genotype Frequencies

• In a population of 8 individuals:

  • 3 are AA, 3 are Aa, 2 are aa.

• Calculations:

  • Frequency of AA (P) = $3/8 = 0.375$

  • Frequency of Aa (Q) = $3/8 = 0.375$

  • Frequency of aa (R) = $2/8 = 0.25$

• Check Sum:

  • $P + Q + R = 0.375 + 0.375 + 0.25 = 1$

Calculating Observed Allele Frequencies

• Total number of genes in a population defines its gene pool.

• In the same population of 8 individuals:

  • Consider alleles:

  • AA, AA, AA, Aa, Aa, Aa, aa, aa

• Calculate:

  • Frequency of A (p) = $9/16 = 0.5625$

  • Frequency of a (q) = $7/16 = 0.4375$

• Check Sum:

  • $p + q = 1$ = $0.5625 + 0.4375 = 1$.

Allele Frequencies from Genotype Frequencies

• Formula:

  • $p = P + rac{1}{2}Q$

  • $q = R + rac{1}{2}Q$

• Example:

  • If a population consists of 40% AA, 30% Aa, and 30% aa:

  • Calculations:

  • $p = 0.4 + rac{1}{2}(0.3) = 0.4 + 0.15 = 0.55$

  • $q = 0.3 + rac{1}{2}(0.3) = 0.3 + 0.15 = 0.45$

• Check Sum:

  • $p + q = 0.55 + 0.45 = 1$

Calculating Expected Allele Frequencies

• To calculate $q$ from the homozygous recessive genotype frequency $q^2$:

  • Example: If 7% of individuals display the recessive phenotype:

  • Calculation of $q$:

  • $ext{Recessive allele frequency} = q = ext{sqrt}(0.07) = 0.265$

• Note: Calculations should always start with the recessive genotype/allele.

Practical Example

• Given:

  • $p = 0.7$ (frequency of dominant allele for dark fur)

  • $q = 0.3$ (frequency of recessive allele for tan fur)

• Algebraically:

  • $p + q = 1$

  • Total genotype frequencies:

  • $p^2 + 2pq + q^2 = 1$

• Breakdown:

  • Genotype Frequencies:

  • DD: 0.49

  • Dd: 0.42

  • dd: 0.09

• Gamete Frequencies:

  • Offspring gametes generated at random from parental genotypes:

  • $p^2 = (0.7)^2 = 0.49$

  • $pq = (0.7)(0.3) = 0.21$

  • $q^2 = (0.3)^2 = 0.09$

Application of Hardy-Weinberg Principle

• Example: 3 in every 1234 individuals have a recessive disease:

  • Afflicted individuals have homozygous recessive genotype:

  • Calculate allele and gene frequency:

  • Frequency of afflicted, $q^2 = 3/1234 = 0.002431$

  • $q$ = Sqrt$0.002431 = 0.049306$

  • $p = 1 - q = 1 - 0.049306 = 0.950694$

  • Frequency of carriers (heterozygous): $2pq = 2(0.049306)(0.950694) = 0.093750$

  • Number of carriers = $0.09375 imes 1234 = 115.68$

Population Genetic Studies Examples

Blood Groups Analysis

• Data Collection: For different populations:

  • Observed frequencies and expected proportions among groups:

  • African American:

    • Observed: 79 LmLm, 138 LmLn, 61 LnLn; Total = 278

    • Expected proportions: 0.284 for LmLm, 0.499 for LmLn, 0.219 for LnLn.

  • European American:

  • Observed: 1787 LmLm, 3039 LmLn, 1303 LnLn; Total = 6129

  • Native American:

  • Observed: 123 LmLm, 72 LmLn, 10 LnLn; Total = 205

• Expected frequencies were determined using Hardy-Weinberg equations for respective allele frequencies.

Jaeken Syndrome

• Definition: Carbohydrate-Deficient Glycoprotein Syndrome, inherited recessive disease.

• Symptoms include:

  • Skeletal deformities

  • Inadequate liver function

  • Abnormal fat distribution

• High mortality rate among patients under 5 years of age.

• Mutation identified in PMM2 gene which is significant for glycoprotein synthesis.

• Patients are homozygous, harboring two mutant alleles.

• Mutations Identified: List of numerous point mutations and their corresponding amino acid changes.

Frequency Analysis in Jaeken Syndrome Patients

• Among 54 patients, most frequent mutation was R141H. Observed genotypes:

  • 0/54 R141H/R141H = R

  • 11/54 other/other = P

  • Resulting frequencies:

  • other/other: 0.2037

  • R141H/other: 0.7963

  • R141H/R141H: 0

• Calculation of expected genotype frequencies based on Hardy-Weinberg Equilibrium derived from observed frequencies.

Statistical Analysis: Chi-Square Test

• Used to determine if significant differences exist between observed and expected genotype frequencies.

• Formula to calculate chi-square value:

  • $ext{c}^2 = rac{( ext{observed} - ext{expected})^2}{ ext{expected}}$

• Application of the chi-square test to Jaeken Syndrome patient genotype frequencies:

  • Observed and expected frequencies calculation and comparison:

  • Genotypes (R141H/R141H, R141H/other, other/other)

  • Resulting chi-square values were computed to assess significance.

Conclusion and Implications

• The analysis concluded that the investigated alleles within the PMM2 locus are not in Hardy-Weinberg equilibrium in the Jaeken Syndrome population, indicating potential evolutionary influences.