Population Genetics

Population Genetics — Deep Dive

Part 1 of 4 — Allele Frequencies & Hardy-Weinberg Equilibrium

From Individuals to Populations

Population genetics tracks allele frequencies in a gene pool rather than the genotypes of single crosses. For a two-allele gene:

$p = \text{frequency of allele } A, \qquad q = \text{frequency of allele } a, \qquad p + q = 1$

Counting alleles directly (when you know genotype counts): each individual carries 2 alleles, so in a population of $N$ individuals with $n_{AA}$, $n_{Aa}$, $n_{aa}$:

$p = \frac{2n_{AA} + n_{Aa}}{2N}$

Homozygotes contribute two copies, heterozygotes one — a step MCAT questions love to test.

The Hardy-Weinberg Principle

If a population meets five conditions, allele frequencies do not change across generations, and genotype frequencies are locked to allele frequencies:

$p^2 + 2pq + q^2 = 1$

The Five Conditions (know them cold)

1. No mutation
2. No natural selection (equal survival and reproduction across genotypes)
3. Random mating (no assortative mating, no inbreeding)
4. No gene flow (no migration in or out)
5. Very large population (no genetic drift)

Violating any one of these is what we MEAN by evolution at the population level: a change in allele frequencies over generations.

The Standard Solving Recipe

Almost every H-W problem is solved in the same order:

1. Start from the ONE quantity you can trust: the frequency of the recessive phenotype equals $q^2$ (recessive phenotype = one genotype). You cannot start from the dominant phenotype — it lumps $p^2$ and $2pq$ together.
2. $q = \sqrt{q^2}$
3. $p = 1 - q$
4. Compute whatever is asked: carriers $= 2pq$, homozygous dominant $= p^2$, etc.

Sanity Checks

• When $q$ is small, carriers ($2pq \approx 2q$) vastly outnumber affected individuals ($q^2$). Example: $q = 0.01$ → affected $= 10^{-4}$, carriers $\approx 0.02$ — a 200-fold difference.
• Heterozygosity $2pq$ is maximized at $p = q = 0.5$, where $2pq = 0.5$.
• One generation of random mating is enough to establish H-W genotype proportions at an autosomal locus, given the allele frequencies.

Worked Example — From Phenotype Counts to Carrier Frequency

Problem. In a population of 10,000 people in Hardy-Weinberg equilibrium, 36 have an autosomal recessive condition. How many people are expected to be carriers?

Step 1 — Recessive phenotype gives $q^2$. $q^2 = \frac{36}{10000} = 0.0036$

Step 2 — Take the square root. $q = \sqrt{0.0036} = 0.06$

Step 3 — Find $p$. $p = 1 - 0.06 = 0.94$

Step 4 — Carrier frequency and count. $2pq = 2(0.94)(0.06) = 0.1128$

Carriers $= 0.1128 \times 10000 = 1128$ people.

Fast MCAT approximation. When $q$ is small, $p \approx 1$, so $2pq \approx 2q = 0.12$ → about 1200 carriers. Notice the exact answer (1128) is close; on a timed exam, $2q$ gets you to the right answer choice.

Reverse-direction check. If a question instead gives you "84% of the population shows the dominant phenotype," then the recessive phenotype is $16\%$: $q^2 = 0.16$, $q = 0.4$, $p = 0.6$ — never take $\sqrt{0.84}$.

Hardy-Weinberg Basics 🎯

Key Takeaways — Part 1

• $p + q = 1$; genotypes at equilibrium: $p^2 + 2pq + q^2 = 1$
• Allele counting: $p = (2n_{AA} + n_{Aa})/2N$ — homozygotes count twice
• Five H-W conditions: no mutation, no selection, random mating, no migration, infinite (very large) population
• Solving recipe: recessive phenotype $= q^2$ → $q$ → $p$ → whatever is asked; NEVER square-root the dominant phenotype frequency
• Rare alleles hide in heterozygotes: carriers $\approx 2q$ dwarf the $q^2$ affected class; $2pq$ peaks at 0.5 when $p = q = 0.5$

Population Genetics — Deep Dive

Part 2 of 4 — Breaking Equilibrium: Drift, Founder Effects & Gene Flow

Genetic Drift: Chance, Not Fitness

Genetic drift is random fluctuation in allele frequencies from generation to generation due to sampling error in which gametes happen to form the next generation. Key properties:

• Strength scales inversely with population size: dramatic in small populations, negligible in huge ones
• Direction is random — drift is not adaptive and can fix harmful alleles or eliminate beneficial ones
• Given enough time, drift drives every allele to fixation (frequency 1) or loss (frequency 0); a neutral allele's probability of eventual fixation equals its current frequency
• Drift always reduces genetic variation (heterozygosity) within a population, while making isolated populations diverge from each other

Two Named Drift Scenarios

Bottleneck effect: a catastrophe (disease, hunting, natural disaster) randomly kills most of the population. The survivors' allele frequencies — determined by luck, not fitness — become the new gene pool. Example: northern elephant seals, hunted to about 20 individuals, retain very low genetic diversity even after rebounding to over 100,000.

Founder effect: a small group leaves to establish a new population. The founders carry only a subset of the source population's variation, and any allele over-represented among them (by chance) becomes common in the descendants. Examples: Ellis-van Creveld syndrome in the Old Order Amish; Tay-Sachs and other alleles in historically small, endogamous populations.

Both are drift; the difference is HOW the small sample arises (survivors vs. emigrants).

Gene Flow (Migration)

Movement of alleles between populations via migrating, breeding individuals:

• Homogenizes populations — makes allele frequencies more similar, opposing divergence by drift or local selection
• Can introduce new alleles (acts a bit like mutation from the receiving population's perspective)
• One generation of migration at rate $m$ shifts the recipient's allele frequency toward the donor's: $p' = (1 - m)\,p_{\text{resident}} + m\,p_{\text{migrant}}$

Mutation

The ultimate SOURCE of all new alleles, but per-locus mutation rates (roughly $10^{-5}$ to $10^{-6}$ per gamete per generation) are too small to change frequencies much on their own. Mutation supplies variation; selection and drift determine its fate.

Effective Population Size (concept level)

The effective population size $N_e$ is the size of an idealized population that would drift as much as the real one. Unequal sex ratios, variance in offspring number, and past bottlenecks all make $N_e$ smaller than the census count — often much smaller. Conservation passages use $N_e$ to explain why a species with thousands of individuals can still be drifting rapidly.

Worked Example — Reading a Drift Experiment

Passage-style problem. A researcher establishes 100 replicate fly populations, each with exactly 10 males and 10 females, all starting at frequency 0.5 for a neutral eye-color allele $bw^{75}$. A parallel set of 100 populations is founded with 500 flies each. After 19 generations:

• Small populations: many are FIXED for $bw^{75}$ (frequency 1) or have LOST it (frequency 0); few remain near 0.5. The distribution of frequencies is U-shaped.
• Large populations: nearly all remain close to 0.5.

Question 1 — Why the U-shape only in small populations? With 20 breeding flies, each generation is a small random sample of gametes, so frequencies take large random steps — this is drift. Once a population hits 0 or 1, it stays (no mutation in the experiment): fixation and loss are absorbing states. Large populations take tiny steps and stay near the start.

Question 2 — What fraction of small populations should ultimately fix $bw^{75}$? For a neutral allele, the probability of eventual fixation equals its starting frequency: $0.5$. So about half the small populations fix it and half lose it — matching the roughly symmetric U.

Question 3 — What happens to variation? WITHIN each small population, heterozygosity collapses toward zero (drift removes variation). ACROSS populations, variance in allele frequency grows (populations diverge). This within/between contrast is the fingerprint that distinguishes drift from selection in MCAT data: selection would push replicate populations in the SAME direction; drift scatters them in both.

Drift, Founder & Bottleneck 🎯

Key Takeaways — Part 2

• Drift = random sampling error in allele transmission; strong in small populations, directionless, non-adaptive
• Neutral allele's fixation probability = its current frequency; drift ends in fixation or loss and erodes heterozygosity within populations while scattering frequencies between them
• Bottleneck = random survivors after a crash; founder effect = small emigrant sample — both are drift, and they explain lost diversity (elephant seals) and disease-allele hotspots (Amish)
• Gene flow homogenizes populations: $p' = (1-m)p_{\text{resident}} + m\,p_{\text{migrant}}$
• Mutation creates variation but is too slow to shift frequencies alone
• Drift vs. selection in data: scattered, size-dependent changes = drift; consistent directional, size-independent changes = selection

• Low-yield extras: effective population size $N_e$ is the size of an idealized population that drifts as much as the real one; unequal sex ratios, offspring-number variance, and past bottlenecks push it far below the census count

Population Genetics — Deep Dive

Part 3 of 4 — Selection in Populations: Fitness, Selection Modes & Balancing Selection

Fitness Made Quantitative

Relative fitness $w$ scores each genotype's reproductive output against the best genotype (which gets $w = 1$). The selection coefficient $s = 1 - w$ measures the reproductive penalty. If aa individuals leave 70 offspring for every 100 left by AA, then $w_{aa} = 0.7$ and $s = 0.3$. A lethal or sterile genotype has $w = 0$, $s = 1$.

Selection changes allele frequencies by weighting each genotype's contribution to the next gene pool by its fitness — it is Hardy-Weinberg with the "no selection" condition deliberately broken.

Three Modes of Selection on a Quantitative Trait

The population-genetic signature to memorize: directional moves the mean; stabilizing and disruptive move the variance (down and up, respectively).

Balancing Selection: Keeping Both Alleles

Heterozygote advantage (overdominance) maintains a stable polymorphism because the fittest genotype, Aa, cannot breed true — every generation of $Aa \times Aa$ matings regenerates both homozygotes. Classic case: sickle cell. Where malaria is endemic, HbA/HbS heterozygotes resist severe malaria, outreproducing both HbA/HbA (malaria-susceptible) and HbS/HbS (sickle cell disease). Selection therefore holds HbS at high frequency in malaria regions and lets it decay where malaria is absent — allele frequency maps mirror malaria maps.

Heterozygote disadvantage (underdominance) is unstable: whichever allele is rarer spends more of its time in low-fitness heterozygotes, so it is driven out. The population tips toward fixing whichever allele starts on the favored side of an unstable equilibrium (the more common allele when the two homozygotes are equally fit).

Negative frequency-dependent selection gives a genotype higher fitness when it is RARE (predators overlook uncommon prey morphs; rare self-incompatibility alleles in plants find more compatible mates). It actively protects polymorphism, pushing frequencies back toward an interior equilibrium.

Inbreeding: Mating, Not Selection

Inbreeding (mating between relatives) redistributes genotypes — heterozygotes become rarer and both homozygote classes become more common — but by itself it does not change allele frequencies. It is not evolution at the allele level; it changes WHO carries the alleles. Inbreeding depression follows because rare deleterious recessives are suddenly exposed as homozygotes.

The inbreeding coefficient $F$ is the probability that an individual's two alleles are identical by descent. Heterozygosity falls from $2pq$ to $2pq(1 - F)$, with the deficit split equally between the two homozygote classes.

Why Rare Recessives Resist Selection

Selection can only "see" a recessive allele in aa homozygotes. When $q$ is small, only $q^2$ of the population is exposed while the fraction $2pq \approx 2q$ hides in unaffected carriers. At $q = 0.01$, carriers outnumber affected individuals by about 200 to 1 — so even a lethal recessive declines very slowly, and eugenic "removal of the affected" is mathematically futile. Selection against a dominant deleterious allele, by contrast, is fast: every copy is exposed.

Worked Example — One Generation of Selection Against aa

Problem. A large, randomly mating insect population has $p = 0.6$, $q = 0.4$. A new pesticide kills half of all aa individuals before reproduction ($w_{AA} = w_{Aa} = 1$, $w_{aa} = 0.5$, so $s = 0.5$). What is $q$ in the next generation?

Step 1 — Genotypes at birth (H-W). $p^2 = 0.36, \qquad 2pq = 0.48, \qquad q^2 = 0.16$

Step 2 — Weight each genotype by fitness and find mean fitness. $\bar{w} = (0.36)(1) + (0.48)(1) + (0.16)(0.5) = 0.36 + 0.48 + 0.08 = 0.92$

Step 3 — New frequency of a. Each surviving Aa contributes one a allele of its two; each surviving aa contributes two of two: $q' = \frac{\tfrac{1}{2}(0.48)(1) + (0.16)(0.5)}{\bar{w}} = \frac{0.24 + 0.08}{0.92} = \frac{0.32}{0.92} \approx 0.348$

So one generation of strong selection moves $q$ from 0.40 only to about 0.35.

Step 4 — Contrast with a RARE recessive. Now suppose aa is lethal ($s = 1$) and $q = 0.01$. Selection can remove only the aa class, $q^2 = 0.0001$ of the population, while about $2q = 0.02$ of the population carries the allele unseen. The share of all a copies sitting in aa individuals is $2q^2 / 2q = q$ — just 1% — so even a lethal removes only about 1% of the allele's copies per generation: $q$ falls from 0.0100 to about 0.0099. Rare recessives are nearly immortal in the gene pool.

The general recursion for a recessive lethal is $q' = q/(1 + q)$, which gives $q_t = q_0/(1 + t\,q_0)$ after $t$ generations: halving a rare lethal's frequency takes about $1/q_0$ generations (about 100 generations at $q_0 = 0.01$).

Balanced-polymorphism check. In a malaria zone, suppose $w_{AS} = 1$, $w_{AA} = 0.85$ (malaria deaths), $w_{SS} = 0.2$ (sickle cell disease). Both homozygotes lose to the heterozygote, so neither allele can be eliminated: at equilibrium the S allele settles where the two homozygote penalties balance — high enough that S can reach frequencies near 0.1 to 0.2 despite being lethal when homozygous.

Selection & Balancing Polymorphism 🎯

Key Takeaways — Part 3

• Relative fitness $w$ is scaled to the best genotype; selection coefficient $s = 1 - w$
• Directional selection shifts the mean; stabilizing shrinks variance; disruptive inflates variance (bimodality)
• Heterozygote advantage (sickle cell in malaria zones) is a stable balanced polymorphism; heterozygote disadvantage is unstable and fixes whichever allele starts on the favored side (the more common one when homozygotes are equally fit)
• Negative frequency-dependent selection favors rare types and preserves polymorphism
• Inbreeding raises homozygosity and lowers heterozygosity without changing allele frequencies; inbreeding depression comes from exposed deleterious recessives
• Selection against a rare recessive is inefficient because the allele hides in heterozygotes ($2pq$ dwarfs $q^2$); a lethal at frequency $q$ removes only about the fraction $q$ of its copies per generation

• Low-yield extras: the inbreeding coefficient $F$ (probability two alleles are identical by descent) lowers heterozygosity to $2pq(1-F)$; a recessive lethal follows $q' = q/(1+q)$, so halving a rare lethal takes about $1/q_0$ generations

Population Genetics — Deep Dive

Part 4 of 4 — MCAT Integration: X-Linked, Multi-Allele & Passage-Style H-W

X-Linked Hardy-Weinberg

Males are hemizygous: one X means one allele, so a male's phenotype directly reports his genotype.

Two exam-ready consequences:

• The frequency of affected MALES equals the allele frequency itself — the fastest way to read $q$ from a passage.
• The male-to-female ratio of an X-linked recessive condition is $q / q^2 = 1/q$, so the rarer the allele, the more male-biased the condition. Color blindness at $q = 0.08$ gives a ratio of $1/0.08 = 12.5$.

Multiple Alleles: ABO Blood Groups

With three alleles ($I^A$ at frequency $p$, $I^B$ at $q$, $i$ at $r$): $p + q + r = 1$, and genotypes expand as

$(p + q + r)^2 = p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1$

Solving order: type O is the only pure class, so start with $r = \sqrt{r^2}$, then use type A (or B) plus $r$ to extract $p$ (or $q$).

Testing H-W Fit: Chi-Square Logic

1. From OBSERVED genotype counts, compute allele frequencies by allele counting.
2. Use those frequencies to compute EXPECTED H-W counts ($p^2 N$, $2pqN$, $q^2 N$).
3. $\chi^2 = \sum (O - E)^2 / E$ over the three genotype classes.
4. Degrees of freedom = (number of classes) − (number of independent allele frequencies estimated) − 1 = $3 - 1 - 1 = 1$ for a biallelic locus — NOT 2, because $p$ was estimated from the same data.
5. Compare to the critical value (3.84 at df = 1, 5% level): larger means the population deviates from H-W.

Diagnosing WHICH Assumption Broke

The chi-square only says "not H-W" — the pattern of deviation names the culprit:

• Heterozygote deficit (too few Aa): inbreeding or positive assortative mating; OR unrecognized population substructure — pooling two subpopulations with different allele frequencies produces a heterozygote shortfall even if each subpopulation is itself in H-W.

• This pooling artifact is named the Wahlund effect.

• Heterozygote excess: overdominant selection acting before sampling, or negative assortative mating.
• Allele frequencies shifting across generations (genotypes still fitting H-W within each generation): selection, drift, or migration — use population size and directionality across replicates to separate them.

Combining H-W with Pedigrees

Passages love hybrid problems: get the CARRIER probability of an outside partner from population data ($2pq$, or the shortcut $2q$ for rare alleles), get the family member's carrier probability from Mendelian logic (e.g., an unaffected sibling of an affected child is a carrier with probability $2/3$), then multiply: risk of an affected child $= P(\text{both carriers}) \times \tfrac{1}{4}$.

Worked Example — X-Linked Color Blindness, Both Sexes

Problem. In a large population, 8% of males are red-green color blind (X-linked recessive). Predict (a) the frequency of color-blind females, (b) the frequency of carrier females, (c) the male-to-female ratio of the condition, and (d) the probability that a random woman's first son is color blind.

(a) Males read out the allele frequency directly. $q = 0.08, \qquad p = 0.92$

Color-blind females need two copies: $q^2 = (0.08)^2 = 0.0064 = 0.64\%$

(b) Carrier females. $2pq = 2(0.92)(0.08) = 0.1472 \approx 14.7\%$

Note the asymmetry: nearly 15% of women carry the allele but only 0.64% express it.

(c) Sex ratio of the condition. $\frac{0.08}{0.0064} = \frac{1}{q} = 12.5$

Twelve to thirteen affected males for every affected female — the hallmark of X-linked recessive inheritance at the population level.

(d) A random woman's son. A son inherits one of his mother's two X chromosomes at random, so the chance his X carries the allele equals the allele frequency among female X chromosomes, which is $q = 0.08$. (Longer route that must agree: mothers are $q^2 = 0.0064$ affected, passing the allele with probability 1, plus $2pq = 0.1472$ carriers passing it with probability $\tfrac{1}{2}$: $0.0064 + 0.0736 = 0.08$.) ✓

Assumption check. This calculation quietly assumed allele frequencies are equal in the two sexes and that mating is random with respect to color vision — exactly the kind of assumption a passage will violate on purpose, e.g., by describing a founder population or assortative mating, and then asking why the observed female frequency exceeds $q^2$.

Passage-Style Hardy-Weinberg 🎯

Key Takeaways — Part 4

• X-linked recessive: affected males $= q$ (hemizygosity reads out the allele frequency), affected females $= q^2$, carriers $= 2pq$; male:female ratio $= 1/q$
• A random woman's son is affected with probability $q$ — the two-path calculation (affected mothers + half of carriers' sons) must collapse to $q$
• ABO: $p + q + r = 1$; start from type O ($r^2$), the only single-genotype class; AB $= 2pq$
• H-W fit test: expected counts from allele-counted frequencies; $\chi^2 = \sum (O-E)^2/E$; df $= 3 - 1 - 1 = 1$ for two alleles
• Heterozygote deficit → inbreeding, assortative mating, or pooled subpopulations (hidden substructure); heterozygote excess → overdominance or negative assortative mating; shifting allele frequencies → selection, drift, or migration
• Hybrid pedigree problems: population gives the outsider's carrier risk ($\approx 2q$), Mendel gives the relative's (unaffected sib of affected $= 2/3$), multiply both by $\tfrac{1}{4}$

• Low-yield extras: the heterozygote deficit from pooling subpopulations is named the Wahlund effect