Evolution and Genetics: Hardy-Weinberg Equilibrium

Announcements

  • Lecture Assignment 2 is due on Friday.
  • The class on Friday requires a laptop, iPad/tablet, or internet-connecting device (not a phone).
  • Laptops can be checked out from the library: https://libraries.psu.edu/about/libraries/robert-e-eiche-library-altoona-campus/study-robert-e-eiche-library-altoona-campus
  • A session for students interested in health professions (Allopathic Medicine (MD), Osteopathic Medicine (DO), Physician Assistant, Pharmacy, Dentistry, Optometry, Physical Therapy, Veterinary Science, Occupational Therapy, Genetic Counseling, Podiatric Medicine, Public Health, Chiropractic Medicine, Medical Technology, Audiology, or other health-related fields) will be held on February 4th at 12:10-1:00 PM in 150 Hawthorn.

Hardy-Weinberg Principle

  • Evolution: Change in allele frequency over time within a population.
  • Hardy-Weinberg (H-W) Principle: In the absence of evolutionary agents, allele and genotype frequencies remain constant over time in a population.
  • The H-W principle serves as a null hypothesis for testing evolutionary change on a particular trait.

Assumptions of the H-W Equilibrium Principle

  • The H-W Equilibrium Principle assumes no evolutionary agents are acting on the population.
    • No selection: Everyone survives and reproduces equally.
    • No mutation: No new alleles are introduced, and existing alleles do not change.
    • No migration (gene flow): There is no movement of individuals into or out of the population.
    • No chance events (genetic drift): This assumes a very large population size to avoid random fluctuations in allele frequencies.
    • No mate preference (random mating): Individuals mate randomly with one another, without any preference for certain genotypes.

H-W Formulas

  • The Hardy-Weinberg formula uses allele frequencies to predict genotype (and phenotype) ratios in a population that is in H-W equilibrium.
  • Genotype Frequency: The sum of genotype frequencies for all possible genotypes at a locus equals 1.0.
    • Genotype\ frequency\ GG + genotype\ frequency\ Gg + genotype\ frequency\ gg = 1.0
    • p^2 + 2pq + q^2 = 1.0
  • Allele Frequency: The sum of allele frequencies for all possible alleles at a locus equals 1.0.
    • Allele\ frequency\ G + allele\ frequency\ g = 1.0
    • p + q = 1.0

Example Problem: Floppy Ear Gene in Rabbits

  • Scenario: Consider a rabbit population with floppy ears (recessive phenotype) and not floppy ears (dominant phenotype).
    • 100 rabbits have floppy ears (recessive phenotype).
    • 300 rabbits do not have floppy ears (dominant phenotype).
  • Goal: Calculate the genotype and allele frequencies in the population, assuming it is in Hardy-Weinberg equilibrium.

H-W Problem: Steps

  1. Write out the H-W equations
    • p^2 + 2pq + q^2 = 1.0
    • p + q = 1.0
  2. Evaluate which values you have and which are unknown
    • q (frequency of the recessive allele)
    • p (frequency of the dominant allele)
    • q^2 (frequency of the homozygous recessive genotype)
    • p^2 (frequency of the homozygous dominant genotype)
    • Creating a table can help with organization.
  3. Identify the EASIEST of the unknown variables to find
    • Often, this involves squaring a variable or finding the square root of a variable.
  4. Plug values into the H-W equations
  5. Repeat steps 2-3 until you have all allele and genotype frequencies.
  6. Check your work by plugging all values into the H-W equations to ensure the math works out.

H-W Problem Table Example

  • Initial Table Setup:
    • Allele Frequencies:
      • (F) p = ?
      • (f) q = ?
    • Genotype Frequencies:
      • (FF) p^2 = ?
      • (Ff) 2pq = ?
      • (ff) q^2 = 0.25 (given that 100 out of 400 rabbits have floppy ears, so q^2 = 100/400 = 0.25)
    • Equations:
      • p^2 + 2pq + q^2 = 1.0
      • p + q = 1.0
  • Filled Table:
    • Allele Frequencies:
      • (F) p = 1 - q = 0.5
      • (f) q = \sqrt{0.25} = 0.5
    • Genotype Frequencies:
      • (FF) p^2 = (0.5)^2 = 0.25
      • (Ff) 2pq = 2(0.5)(0.5) = 0.5
      • (ff) q^2 = 0.25
    • Equations:
      • p^2 + 2pq + q^2 = 1.0
      • p + q = 1.0

H-W Problem: Common Mistakes

  • Calculation errors
    • Use your calculator and double or triple check your work.
  • Forgetting what the variables stand for
    • Memorize these! Use flashcards, mnemonic devices, songs, etc.
  • Forgetting the H-W equations
    • See above; memorize them.
  • Trouble identifying which value is given in the problem
    • Stay calm! Focus on the numbers given and how they fit together.

Practice Problem 1

  • Scenario: Assuming a population is in Hardy-Weinberg equilibrium, calculate genotype and allele frequencies for flower color, where B = blue flower color and b = red flower color, N = 1000.
    • There are 40 individuals in the population with red flowers.
  • Solution:
    • Allele Frequencies:
      • (B) p = 0.80
      • (b) q = 0.20
    • Genotype Frequencies:
      • (BB) p^2 = 0.64
      • (Bb) 2pq = 0.32
      • (bb) q^2 = 40/1000 = 0.04
    • Equations:
      • p^2 + 2pq + q^2 = 1.0
      • p + q = 1.0

Review Resources