Hardy–Weinberg & Population Genetics – Comprehensive Study Notes

Population Dynamics & Gene Pools

  • Goal of this section:
    • Quantitatively track how genetic makeup (dominant/recessive alleles, homozygous/heterozygous states) changes within populations over time.
    • Use calculated values to infer why a recessive or dominant allele may be trending upward or downward (environmental pressure, mating success, etc.).
  • Gene pool concept:
    • Imagine collecting every allele carried by individuals in a defined population (e.g., the classroom).
    • Greater allelic variation → higher probability that at least some individuals possess genotypes conferring survival & reproductive success under environmental change (biodiversity advantage).
    • Natural sources of variation reviewed in earlier units: meiosis (independent assortment, crossing-over) & random fertilization.

Three Frequencies We Can Measure

  • Genotype frequency – proportion of individuals with each genotype (e.g., BB, Bb, bb).
  • Phenotype frequency – proportion showing each physical trait (e.g., black vs white coat).
  • Allele frequency – proportion of each allele in the gene pool (e.g., B vs b).
  • Snapshot nature: Each set of frequencies represents a single moment; comparing successive snapshots lets us infer evolutionary trends.

Hardy–Weinberg Principle (H–W)

  • Co-formulated (1908) by Godfrey Hardy (mathematician) & Wilhelm Weinberg (physician).
  • Provides two algebraic relationships that connect allele & genotype frequencies if five idealized conditions are met for the instant of sampling.
  • 5 equilibrium conditions (rarely all true in nature, but assumed for calculations):
    1. Large population size – random events don’t swamp the whole population.
    2. Random mating – no sexual selection or mating preferences.
    3. No mutation – alleles themselves are stable in that moment.
    4. No migration – closed population (no gene flow in/out).
    5. No natural selection – every phenotype equal for survival & reproduction.
  • Even though these assumptions are unrealistic in the long term, H–W still works as a null model. Deviations between successive snapshots highlight evolutionary forces.

Symbol & Formula Summary

  • Symbols (must be memorised):
    • p = frequency of the dominant allele.
    • q = frequency of the recessive allele.
    • By definition p + q = 1(100 %).
  • Genotype-frequency expansion (squaring the binomial): (p + q)^2 = p^2 + 2pq + q^2 = 1 where
    • p^2 → homozygous dominant frequency.
    • 2pq → heterozygous frequency.
    • q^2 → homozygous recessive frequency.
  • Complete dominance & one trait at a time only – the scheme fails for incomplete dominance, codominance, multiple alleles, or linked loci.

Step-by-Step Calculation Workflow (Recommended)

  1. Identify data type in the prompt: Are given numbers genotype counts, phenotype counts, or direct allele frequencies?
  2. Start with the recessive phenotype whenever only phenotype data are supplied, because only one genotype ( q^2 ) produces it.
  3. Convert counts → frequencies (divide by total population size; express as decimals).
  4. Solve sequentially:
    • If recessive phenotype available → q^2. Take square root to get q.
    • Compute p = 1 – q.
    • Plug p & q into p^2 or 2pq as needed.
  5. (Optional) Convert frequency to expected number of individuals: multiply by population size.
  6. Keep answers to 3 decimal places (course convention).

Worked Example 1 – Genotype vs Phenotype (Mice)

  • Data: 200 mice; 72 BB (black), 96 Bb (black), 32 bb (white).
  • Genotype frequencies (divide by 200):
    • BB = 72/200 = 0.360
    • Bb = 96/200 = 0.480
    • bb = 32/200 = 0.160
  • Phenotype frequencies:
    • Black = 72 + 96 = 168/200 = 0.840
    • White = 32/200 = 0.160
  • Allele frequencies:
    • p = (2×72 + 96) / (2×200) = 0.600
    • q = 1 – 0.600 = 0.400
      (confirms bb frequency q^2 = 0.16).

Worked Example 2 – Direct Allele Frequency Provided

  • Prompt: Dominant allele R has frequency 20 \%.
  • Convert: p = 0.20 ⇒ q = 1 – 0.20 = 0.80.
  • Genotype frequencies:
    • p^2 = (0.20)^2 = 0.040 (4 % homozygous dominant)
    • 2pq = 2(0.20)(0.80) = 0.320 (32 % heterozygous)
    • q^2 = (0.80)^2 = 0.640 (64 % homozygous recessive)
  • If population size = 1{,}000: expect 640 homozygous recessive, 320 heterozygous, 40 homozygous dominant.

Worked Example 3 – Birds with Red vs Blue Tails

  • 16 red-tailed (recessive) & 34 blue-tailed (dominant phenotype) in a flock of 50.
  • Step 1: Use recessive phenotype.
    • q^2 = 16/50 = 0.320 ⇒ q = \sqrt{0.320} = 0.566 (3 dp).
  • Step 2: p = 1 – 0.566 = 0.434.
  • Step 3: Requested values:
    • Frequency red allele (already q = 0.566).
    • Frequency blue allele p = 0.434.
    • Heterozygotes: 2pq = 2(0.434)(0.566) = 0.491.
    • Homozygous blue: p^2 = (0.434)^2 = 0.188.
  • Common mistake flagged by instructor: Starting with 34/50 would incorrectly assume all blue are one genotype; always start from recessive phenotype.

Worked Example 4 – Cystic Fibrosis among Caucasian Newborns

  • Incidence given: 1 in 1 700 newborns affected (recessive disease).
  • q^2 = 1 / 1700 = 0.000588.
  • q = \sqrt{0.000588} = 0.0243.
  • p = 1 – 0.0243 = 0.9757.
  • Carrier (heterozygote) frequency: 2pq = 2(0.9757)(0.0243) = 0.0474 ≈ 4.7 % of Caucasian newborns are carriers.

Genetic Equilibrium vs Micro-Evolution

  • If successive snapshots show no significant change in p or q → population is in genetic equilibrium (no evolution detectable during period).
  • Detectable change (even small) over time → micro-evolution is occurring.
  • Researchers plot allele-frequency trajectories + environmental data (climate, predators, human activity) to hypothesize selective forces.

Real-World Illustrations Mentioned

  • Alberta has Canada’s highest proportion of blood-type O donors – hypothesis: migration & subsequent reproductive patterns created “stocking of the pond”.
  • Shifts between dominance of deer antler size, peacock trains, etc. result from changing sexual-selection pressures.

Practical & Exam Tips

  • A calculator is essential; provided formula sheet lists only:
    p + q = 1 and (p + q)^2 = p^2 + 2pq + q^2 = 1 plus amino-acid table for later genetics.
  • Most exam items ask for carrier (heterozygous) frequency because it requires full use of both equations.
  • Always label what each calculated number represents ( p,q,p^2,2pq,q^2 ) to avoid mixing them up.
  • Maintain three decimal places unless question specifies otherwise.
  • Check that computed p^2 + 2pq + q^2 ≈ 1.000 (rounding validation).

Common Pitfalls Highlighted by Instructor

  • Using dominant phenotype count as though it equals one genotype (forgetting two genotype possibilities).
  • Forgetting to convert counts to decimal frequencies before manipulation.
  • Trying to apply H–W to traits with incomplete dominance, codominance, or >2 alleles.
  • Ignoring population-size requirement; tiny samples produce misleading allele estimates.

Connections to Earlier Units & Future Topics

  • Relies on prior understanding of meiosis, segregation, crossing-over (source of variation).
  • Bridges to upcoming Molecular Genetics unit where mutation mechanisms & DNA-level changes will be quantified.
  • Chemical-equilibrium analogy (Le Châtelier) used to conceptually frame “genetic equilibrium”.

Ethical / Philosophical / Practical Angles

  • Blood donation: Knowing regional allele distributions (e.g., type O) helps public-health planning.
  • Carrier screening (e.g., cystic fibrosis) – population-frequency data underpins genetic-counselling programs.
  • Conservation biology: Detecting low heterozygosity alerts managers to inbreeding & extinction risk.
  • Intellectual caution: H–W is a null expectation; real populations almost always break ≥1 assumption, so significant deviations must be interpreted, not ignored.