BIOL 2970 — l19 20 hwe mutation migration

Hardy–Weinberg equilibrium: what it is

A model population with NO evolutionary forces. Allele frequencies stay at p and q, and genotype frequencies go to p², 2pq, q² and stay there.

Hardy and Weinberg

G.H. Hardy (mathematician) and W. Weinberg (physician).

Why use a model that's never exactly true?

It's the baseline: a deviation tells you some force is acting, and every later model (mutation, migration, drift, inbreeding) is HWE plus one force.

The five forces HWE assumes are absent

Finite (small) population size, non-random mating, mutation, migration, selection.

Population vs. gene pool

Population: all individuals, described by GENOTYPE frequencies. Gene pool: all alleles in the breeding members, described by ALLELE frequencies.

Allele frequency from genotype COUNTS

p = (2·N_AA + N_Aa) / 2N and q = (2·N_aa + N_Aa) / 2N. Divide by 2N because each diploid carries two copies.

Allele frequency from genotype FREQUENCIES; this always works, even if a population is not in Hardy-Weinberg Equilibrium

p = f(AA) + ½·f(Aa); q = f(aa) + ½·f(Aa).

Can you always get allele frequencies from genotype frequencies?

Yes, always, whether or not the population is at HWE.

Can you always get genotype frequencies from allele frequencies?

No. Only if you assume random mating (HWE); then they're p², 2pq, q².

Why is the heterozygote frequency 2pq, not pq?

There are two ways to make one: A egg + a sperm (pq), or a egg + A sperm (qp).

How many generations of random mating to reach HW genotype frequencies?

One, from any starting genotype frequencies (if the assumptions hold).

Practice: AA 36, Aa 48, aa 16 (N = 100). p and q?

p = (72 + 48)/200 = 0.6; q = 0.4.

Practice: genotype frequencies AA 0.50, Aa 0.20, aa 0.30. p? At HWE?

p = 0.50 + 0.10 = 0.6. HWE predicts 0.36 / 0.48 / 0.16, so no: there's a big heterozygote deficit.

Trait with complete dominance, population at HWE. Where do you start?

The recessive phenotype is only aa, so its frequency = q². Take √ to get q, then p = 1 − q.

Why can't you just count alleles when there's complete dominance?

AA and Aa look the same, so you can't count heterozygotes. You have to assume HWE and use q².

Practice: 1 in 2,500 newborns has a recessive disease. q? Carrier frequency? Assume Hardy-Weinberg Equilibrium.

q = √(1/2500) = 0.02; p = 0.98; carriers = 2pq ≈ 0.039 (about 1 in 26).

Practice: 91% show the dominant phenotype (HWE). AA? Aa?

aa = 0.09, so q = 0.3 and p = 0.7. AA = 0.49; Aa = 0.42.

Trap: q vs. carrier frequency

q is the allele frequency. Carriers are heterozygotes: 2pq (about 2q when q is small). Different numbers.

Multiple alleles at HWE

Each homozygote = (its frequency)². Each heterozygote = 2 × (product of its two allele frequencies). Everything sums to 1.

Practice: three alleles at 0.6, 0.3, 0.1. Total heterozygote frequency?

1 − (0.36 + 0.09 + 0.01) = 0.54.

Chi-square test of HWE: the steps

1) p and q from the observed counts. 2) Expected COUNTS = p²N, 2pqN, q²N. 3) χ² = Σ(O − E)²/E. 4) Compare to the critical value for your df.

Degrees of freedom for an HWE χ² (course formula)

df = (number of genotype classes) − (number of alleles). 2 alleles: 3 − 2 = 1. 3 alleles: 6 − 3 = 3. 4 alleles: 10 − 4 = 6.

Why is df = 1, not 2, with two alleles?

You estimated p from the same data (q follows), which costs one more degree of freedom: 3 categories − 1 − 1.

Critical χ² value for df = 1 at p = 0.05

3.84. Above 3.84 → reject HWE. Below → fail to reject.

Null hypothesis for an HWE χ² test

The observed genotype numbers DO fit Hardy–Weinberg expectations.

What does rejecting HWE tell you?

Some evolutionary force (drift, migration, selection, non-random mating…) may be acting at this locus. It doesn't tell you which one.

Trap: what goes into (O − E)²/E?

Counts, never frequencies. Multiply expected frequencies by N first.

Course example: Pueblo MN blood groups

83 MM, 46 MN, 11 NN (N = 140) → p ≈ 0.76, q ≈ 0.24. χ² is small (≈1.2–1.6 depending on rounding) with df = 1, so it fits HWE.

Mixing two genetically different populations (before they mate)

The combined group is NOT at HWE. It has too many homozygotes. One generation of random mating restores HW genotype frequencies.

Germline vs. somatic mutations

Germline (in cells that make eggs/sperm): heritable, matter for evolution. Somatic: not passed on, but can cause cancer.

In what sense are mutations 'random'?

Random with respect to what the organism NEEDS: they don't arise because they'd help. Not random in rate, in effect (most are harmful or neutral), or in cause (UV, chemicals).

Luria & Delbrück (1943)

Counts of phage-resistant bacteria per culture were clumped (some huge 'jackpots'), so resistance mutations arose BEFORE phage exposure, not in response to it.

Lederbergs' replica plating (1952)

Colonies never exposed to penicillin were stamped onto penicillin plates. The resistant ones traced back to master-plate colonies that were already resistant: mutation came before exposure.

Typical mutation rates (course numbers)

Per cell per locus: ~10⁻⁴ (microsatellites) down to ~10⁻⁹ (a single SNP site). Per gene, phenotypic: ~10⁻⁵ per generation.

Why is the reverse mutation rate (ν) usually much lower, around 10⁻⁸?

There are many ways to break a gene and very few ways to fix one.

μ in the one-way mutation model

Probability that a B allele mutates to b in a generation, which is also the fraction of B alleles that mutate. The fraction that don't is 1 − μ.

One-way mutation equation

p_n = p0 · (1 − μ)^n, where p is the allele being mutated AWAY from (it shrinks each generation).

Solve the one-way mutation equation for n

n = ln(p_n / p0) / ln(1 − μ). Any log base works if you use it on both.

Trap: which allele goes in the one-way mutation equation?

The one that DECLINES. If the new allele must reach 0.1, set p_n = 0.9 for the old allele.

How weak a force is mutation?

At μ = 10⁻⁵, going from p = 1 to 0.99 takes about 1,000 generations, and it slows down as fewer source alleles remain.

Practice: p0 = 0.8, μ = 10⁻⁴, n = 1,000. p_n?

0.8 × (0.9999)^1000 ≈ 0.724.

Two-way mutation equilibrium

At equilibrium p·μ = q·ν, so q̂ = μ / (μ + ν), where μ is the rate TOWARD the q allele.

Sanity check for q̂ = μ / (μ + ν)

If the rate toward q is the bigger one, q̂ must be above 0.5. A small q̂ means you swapped the rates.

Practice: B→b at 3 × 10⁻⁶, b→B at 1 × 10⁻⁶. Equilibrium frequency of b?

3 / (3 + 1) = 0.75.

If mutation is such a weak force, why does it matter?

It supplies the raw material: every new variant starts as a mutation. No mutation, no evolution.

Could recurrent mutation alone explain the fast rise of lactase persistence?

No. Even at a high 10⁻⁵, reaching 0.1 takes about 10,500 generations, far too slow.

Migration (gene flow) in population genetics

Movement of GENES into another population where they reproduce: individuals, seeds, or pollen. Seasonal back-and-forth travel doesn't count.

When does migration change allele frequencies?

Only when the source and recipient populations differ genetically.

The four migration models

Mainland–island (one-way), island (every population feeds one shared migrant pool), stepping stone (only to neighbors), isolation by distance (continuous population; dispersal drops with distance).

Which migration model is one-way? Which has no separate subpopulations?

One-way: mainland–island. No discrete subpopulations: isolation by distance.

Island model: what is the migrant pool's frequency?

The average of all the populations (the focal one included), since every population contributes to it.

m (migration rate)

Fraction of the population each generation that are immigrants. 1 − m are natives.

One generation of migration

q1 = (1 − m)·q0 + m·qm: a weighted average of natives and migrants.

t generations of migration

qt = (q0 − qm)(1 − m)^t + qm, or equivalently (qt − qm) = (q0 − qm)(1 − m)^t.

Plain-English meaning of the migration equation

The difference between the population and the migrant pool shrinks by a factor of (1 − m) every generation.

Solve the migration equation for t

t = ln[(qt − qm)/(q0 − qm)] / ln(1 − m).

Solve the migration equation for m

m = 1 − [(qt − qm)/(q0 − qm)]^(1/t).

Solve the migration equation for qm

qm = [qt − q0(1 − m)^t] / [1 − (1 − m)^t].

Sanity check for any migration answer

qt must land between q0 and qm, moving toward qm.

Overall effect of gene flow

Homogenizing: it makes populations more alike. It works against drift and divergent selection.

Mainland–island problem: what is qm?

The mainland's frequency, which stays fixed because the small island barely affects it.

Practice: q0 = 0.2, qm = 0.7, m = 0.1. q after 1 generation?

0.9(0.2) + 0.1(0.7) = 0.25.

Practice: q0 = 0.2, qm = 0.7, m = 0.1. q after 5 generations?

0.7 + (0.2 − 0.7)(0.9)^5 ≈ 0.405.

Practice: m = 0.1. Generations to halve the gap between a population and the migrant pool?

(0.9)^t = 0.5 → t = ln 0.5 / ln 0.9 ≈ 6.6 generations.

Two-way (island) vs. one-way migration at the same m

Two-way moves the recipient more slowly (the pool is now a mix), and both populations change, converging on an intermediate value.

Isolation by distance

Genetic difference grows with geographic distance because individuals don't move far, even with no barriers. The most common pattern in humans.

European SNP PCA (Activity 21): what did it show?

The genetic plot of individuals looked like a map of Europe. Barriers (Alps, Pyrenees, Adriatic) showed as gaps; elsewhere, isolation by distance.

Lactase persistence: did the −13910T allele reach Europe from the Middle East?

Probably not. That allele isn't found in the Middle East, where a different mutation causes lactase persistence.

You see: genotype counts. Asked for allele frequencies.

p = (2·N_AA + N_Aa)/2N.

You see: allele frequencies + 'random mating.' Asked for genotypes.

p², 2pq, q² (times N for numbers).

You see: 'Is this population in HWE?' with counts.

χ² test, df = genotypes − alleles (= 1 for two alleles), critical value 3.84.

You see: one mutation rate and a number of generations.

p_n = p0(1 − μ)^n, using the declining allele.

You see: forward AND backward mutation rates; asked the long-run frequency.

q̂ = μ/(μ + ν), with μ the rate toward q.

You see: a migration rate m plus a mainland/migrant-pool frequency.

qt = (q0 − qm)(1 − m)^t + qm (rearrange for t, m, or qm).