Genetics Quiz: Hardy-Weinberg Principle

Definitions

• Allele: Variations of genes that determine specific traits.
• Locus: The specific location of a gene on a chromosome.

Hardy-Weinberg Principle

• Five Assumptions:
  • Random mating occurs within the population.
  • There are no mutations affecting alleles.
  • No migration into or out of the population changes allele frequencies.
  • Natural selection does not favor any particular genotype.
  • The population is sufficiently large to avoid genetic drift.

Hardy-Weinberg Equilibrium

• Equations:

  • Allele frequency: ( p + q = 1 )
    • ( p ): Frequency of the dominant allele
    • ( q ): Frequency of the recessive allele
  • Genotype frequency: ( p^2 + 2pq + q^2 = 1 )
    • ( p^2 ): Frequency of homozygous dominant genotype
    • ( 2pq ): Frequency of heterozygous genotype
    • ( q^2 ): Frequency of homozygous recessive genotype

• Allele Frequency Calculation:

  • Total number of alleles = Total number of chromosomes x 2
  • Allelic frequencies calculated by dividing the number of dominant or recessive alleles by the total number of alleles.

Genotype Frequency Calculation

• Observed Genotypes:

  • HH: 40
  • Hh: 45
  • hh: 50

• Total Genotypes: 135

• Genotype Frequencies:

  • HH: ( 40/135 = 0.30 )
  • Hh: ( 45/135 = 0.33 )
  • hh: ( 50/135 = 0.37 )

• Allele Frequencies:

  • Total alleles = ( 135 \times 2 = 270 )
  • ( f(H) = (2 \times 40 + 45)/270 = 0.46 )
  • ( f(h) = 0.54 ) (derived from ( p + q = 1 ))

Expected Genotype Frequencies under H-W Equilibrium

• Calculated from allele frequencies:
  • HH: ( 0.2116 \times 135 = 28.57 )
  • Hh: ( 0.4968 \times 135 = 67.07 )
  • hh: ( 0.2916 \times 135 = 39.37 )

Chi-Square Test for H-W Equilibrium

• Formula:
  ( \chi^2 = \frac{(O - E)^2}{E} )
  • O: Observed number of genotypes
  • E: Expected number of genotypes
• Calculations:
  • For HH: ( \frac{(40 - 28.57)^2}{28.57} = 4.5 )
  • For Hh: ( \frac{(45 - 67.07)^2}{67.07} = 7.2 )
  • For hh: ( \frac{(50 - 39.37)^2}{39.37} = 2.8 )
• ( \chi^2 = 4.5 + 7.2 + 2.8 = 14.57 )

Sickle-Cell Anemia and Malaria Resistance

• Scenario: 9% (0.09) of a population has severe sickle-cell anemia (genotype ss).

• Calculating q:

  • ( q^2 = 0.09 \rightarrow q = 0.3 )

• Calculating p:

  • ( p + q = 1 \rightarrow p = 0.7 )

• Heterozygous Frequency:

  • ( 2pq = 2 \times 0.7 \times 0.3 = 0.42 \rightarrow 42% )

• For a 16% (0.16) Population with ss:

  • ( q^2 = 0.16 \rightarrow q = 0.4 )
  • ( p + q = 1 \rightarrow p = 0.6 )

Test your understanding of the Hardy-Weinberg Principle and its assumptions. This quiz will cover key concepts like allele frequencies and genotype calculations. Perfect for students studying genetics or related fields.