Population Genetics — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Allele and genotype frequency calculations, the Hardy-Weinberg Equilibrium (HWE) principle and its assumptions, factors that alter allele frequencies, genetic drift, and selection modeling for evolving populations.

You should already know: Basic Mendelian inheritance for diploid organisms, definitions of alleles and genotypes, what a biological population is.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Biology style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Population Genetics?

Population genetics is the quantitative study of the distribution and change in allele and genotype frequencies within biological populations, and it forms the core mathematical foundation for the study of evolution in AP Biology. Per the AP Biology Course and Exam Description (CED), this topic accounts for ~15% of the total exam weight for Unit 7 (Natural Selection), and questions on population genetics appear regularly in both the multiple-choice (MCQ) and free-response (FRQ) sections of the exam. Standard notation used consistently across AP Biology problems is: $p$ = frequency of the dominant allele in the population, $q$ = frequency of the recessive allele, and uppercase terms $p^2$, $2pq$, $q^2$ represent the frequencies of the homozygous dominant, heterozygous, and homozygous recessive genotypes, respectively. Population genetics frames evolution as a measurable change in allele frequency over generations, rather than just a qualitative change in trait distribution, allowing researchers to test hypotheses about what evolutionary forces are acting on a population.

2. Hardy-Weinberg Equilibrium (HWE)

The Hardy-Weinberg Equilibrium is a null model that describes a population that is not evolving, meaning allele and genotype frequencies remain constant from generation to generation. For a biallelic (two-allele) autosomal locus, the model is derived from the rules of probability: if alleles are sampled independently from the gene pool, the probability of inheriting two copies of the dominant allele is $p\times p=p^2$, two copies of the recessive allele is $q\times q=q^2$, and one of each allele is $(p\times q)+(q\times p)=2pq$. This gives the core HWE equations: $$p+q=1$$ $$p^2+2pq+q^2=1$$ For HWE to hold, five assumptions must be met: (1) no new mutations, (2) random mating, (3) no natural selection, (4) very large population size (no genetic drift), and (5) no gene flow (no immigration or emigration). If any assumption is violated, frequencies will change, indicating evolution is occurring.

Worked Example

A population of 500 wildflowers has 120 red-flowered (RR), 280 pink-flowered (Rr), and 100 white-flowered (rr) individuals. Calculate allele frequencies for R and r, then determine if the population is in HWE for this locus.

1. Count total alleles for this diploid population: $500\text{ individuals}\times 2=1000$ total alleles.
2. Count R alleles: $(120\times 2)+280=520$, so $p=520/1000=0.52$. Count r alleles: $(100\times 2)+280=480$, so $q=480/1000=0.48$. Confirm $p+q=0.52+0.48=1$, which checks out.
3. Calculate expected HWE genotype frequencies: $p^2=(0.52{)}^2=0.2704$, $2pq=2(0.52)(0.48)=0.4992$, $q^2=(0.48{)}^2=0.2304$. Sum to 1, as expected.
4. Compare to observed genotype frequencies: Observed $RR=120/500=0.24$, $Rr=280/500=0.56$, $rr=100/500=0.20$. Observed and expected frequencies differ significantly, so the population is not in HWE.

Exam tip: Always count alleles correctly for diploid organisms: multiply the number of homozygotes by 2 before adding heterozygotes to get the total allele count. Never use individual counts directly for $p$ and $q$.

3. Mechanisms of Evolution (HWE Violations)

HWE is a null model: when we observe deviations from HWE predictions, we can infer that one or more evolutionary mechanisms are acting on the population. Each deviation corresponds to a violation of one of the five HWE assumptions, and each mechanism alters allele and genotype frequencies in predictable ways:

• Mutation: Creates new alleles, but changes frequency very slowly over time; it is the ultimate source of genetic variation.
• Non-random mating (e.g., inbreeding): Increases homozygosity by genotype, but does not change overall allele frequency on its own.
• Natural selection: Differential survival and reproduction of individuals based on phenotype causes consistent, adaptive changes in allele frequency. It is modeled using relative fitness, which is the survival rate of a genotype divided by the maximum survival rate in the population.
• Genetic drift: Random change in allele frequency due to sampling error in small populations; changes are non-adaptive.
• Gene flow: Movement of alleles between populations reduces genetic differentiation between groups.

Worked Example

A population of lizards has three genotypes for body size: $BB$ (large), $Bb$ (medium), $bb$ (small). Survival rates for each genotype are: $BB$: 60%, $Bb$: 80%, $bb$: 40%. Current allele frequencies are $p=0.5$, $q=0.5$. Calculate relative fitness for each genotype, and find the new frequency of $b$ after selection.

1. Relative fitness ($w$) is calculated by dividing each genotype's survival rate by the highest survival rate (80% for $Bb$): $w_{BB}=0.6/0.8=0.75$, $w_{Bb}=0.8/0.8=1.0$, $w_{bb}=0.4/0.8=0.5$.
2. Pre-selection genotype frequencies are: $p^2=0.25$ ($BB$), $2pq=0.5$ ($Bb$), $q^2=0.25$ ($bb$).
3. Calculate unnormalized post-selection frequencies by multiplying pre-selection frequency by relative fitness: $BB=0.25\times 0.75=0.1875$, $Bb=0.5\times 1.0=0.5$, $bb=0.25\times 0.5=0.125$. The sum of these values is the mean fitness $\bar{w}=0.8125$.
4. Normalize by mean fitness to get actual post-selection frequencies: $BB\approx 0.231$, $Bb\approx 0.615$, $bb\approx 0.154$. New $q=f(bb)+0.5f(Bb)=0.154+(0.5\times 0.615)\approx 0.46$.

Exam tip: When calculating relative fitness, always divide by the highest survival rate, not the sum of survival rates. This scales all fitness values between 0 and 1 for easy comparison.

4. Genetic Drift: Founder and Bottleneck Effects

Genetic drift is random fluctuation in allele frequency caused by sampling error in small populations, and it is a frequently tested HWE violation on the AP exam. Unlike natural selection, genetic drift causes non-adaptive changes: allele frequency shifts are random and unrelated to the fitness of the allele. Two specific, commonly tested scenarios of genetic drift are:

1. Founder effect: Occurs when a small subset of a larger population founds a new, isolated population. The founder group is not a genetically representative sample of the original population, so allele frequencies change randomly.
2. Bottleneck effect: Occurs when a large population is drastically reduced in size by a random event (e.g., natural disaster, habitat destruction, overhunting). The surviving population has reduced genetic diversity and altered allele frequencies due to random sampling.

Worked Example

A main population of bats has an allele frequency of $p=0.8$ for the dominant $A$ allele (fog echo-location ability) and $q=0.2$ for the recessive $a$ allele. A small group of 10 bats is blown off course to a new isolated island, and by chance 1 bat is $aa$, 8 are $AA$, and 1 is $Aa$. Is this founder effect or bottleneck effect? What is the new frequency of $a$?

1. Classify the event: This is founder effect, not bottleneck. A new isolated population is founded by a small subset of the original population, which matches the definition of founder effect. Bottleneck effect refers to reduction of an existing population, not founding of a new one.
2. Calculate total alleles: 10 diploid bats = $10\times 2=20$ total alleles.
3. Count $a$ alleles: 1 $aa$ contributes 2 $a$ alleles, 1 $Aa$ contributes 1 $a$ allele, so total $a=2+1=3$.
4. New allele frequency: $q=3/20=0.15$, compared to the original $q=0.2$, showing random allele frequency change from genetic drift.

Exam tip: On AP exam questions, the most common distinction tested is founder effect vs bottleneck effect: remember: founder effect = new population, bottleneck effect = existing population reduced. Both are types of genetic drift.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Counting individuals instead of alleles when calculating $p$ and $q$. For example, using 100 recessive individuals out of 500 total as $q=0.2$, instead of calculating the actual allele count. Why: Students confuse genotype frequency (count of individuals) with allele frequency (count of alleles) for diploid organisms. Correct move: Always multiply the number of homozygous individuals by 2, add the number of heterozygous individuals, then divide by twice the total number of individuals to get allele frequency.
• Wrong move: Assuming $q^2$ equals the frequency of the recessive allele. Why: Students mix up notation for allele frequency ($q$) and homozygous recessive genotype frequency ($q^2$), leading to algebra errors in HWE problems. Correct move: Always label variables first; $q=\sqrt{q^2}$ when the only observed data is the number of recessive phenotypes in a HWE population.
• Wrong move: Claiming non-random mating (like inbreeding) changes allele frequencies. Why: Students confuse genotype frequency change with allele frequency change. Correct move: Non-random mating changes how alleles are arranged into genotypes, not total allele counts. Exclude it from lists of mechanisms that change allele frequency unless selection is also acting.
• Wrong move: Calling genetic drift "natural selection that happens in small populations". Why: Students mix up adaptive (selection) and non-adaptive (drift) evolutionary change. Correct move: Always note that genetic drift causes random changes in allele frequency that are not tied to allele fitness, unlike natural selection.
• Wrong move: Claiming a population is in HWE because most assumptions are met. Why: Students memorize the assumptions but forget all must hold for no evolution. Correct move: If any one HWE assumption is violated, the population is evolving, so allele frequencies will change between generations.

6. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

In a population of humans, the frequency of the recessive autosomal disorder cystic fibrosis is 1 in 2500 births. Assuming the population is in Hardy-Weinberg Equilibrium, what is the frequency of carriers (heterozygotes) in this population? A) ~0.02 B) ~0.04 C) ~0.96 D) ~0.98

Worked Solution: Cystic fibrosis is recessive, so only homozygous recessive individuals have the disorder, meaning the disorder frequency equals $q^2$. We calculate $q^2=1/2500=0.0004$, so $q=\sqrt{0.0004}=0.02$. The dominant allele frequency $p=1-q=0.98$. Carriers are heterozygotes, with frequency $2pq=2(0.98)(0.02)=0.0392\approx 0.04$. The correct answer is B.

Question 2 (Free Response)

A population of snapdragons has flower color controlled by one locus with two alleles: $C^R$ (red) and $C^W$ (white). Heterozygotes $C^R{C}^W$ have pink flowers. You survey a population of 200 snapdragons and count 50 red-flowered, 80 pink-flowered, 70 white-flowered plants. (a) Calculate the current allele frequencies of $C^R$ and $C^W$ in this population. Show your work. (b) Calculate the expected genotype frequencies if this population is in Hardy-Weinberg Equilibrium. (c) A researcher observes that pollinators prefer pink flowers over red or white, leading to higher reproductive success for pink-flowered plants. Predict how this will affect genotype frequencies over time, and explain why this violates an assumption of HWE.

Worked Solution: (a) Total alleles = $200\times 2=400$. Number of $C^R$ alleles = $(50\times 2)+80=180$, so $p=f(C^R)=180/400=0.45$. Number of $C^W$ alleles = $(70\times 2)+80=220$, so $q=f(C^W)=220/400=0.55$. Confirm $0.45+0.55=1$, which is correct. (b) Expected HWE frequencies: $p^2=(0.45{)}^2=0.2025$ ($C^R{C}^R$ red), $2pq=2(0.45)(0.55)=0.495$ ($C^R{C}^W$ pink), $q^2=(0.55{)}^2=0.3025$ ($C^W{C}^W$ white). Sum of frequencies equals 1, as expected. (c) Over time, the frequency of heterozygous pink-flowered plants will increase, while the frequency of homozygous red and white plants will decrease. This violates the HWE assumption of no natural selection: differential reproductive success based on flower color means the $C^R$ and $C^W$ alleles have lower fitness when homozygous than when heterozygous, so genotype frequencies will deviate from HWE expectations and evolution will occur.

Question 3 (Application / Real-World Style)

Researchers studying endangered greater prairie chickens in Illinois found that after habitat fragmentation reduced the population from 100,000 birds to just 50 birds in the 1990s, the percentage of eggs that hatched dropped from 80% to 30%. Genetic analysis found that many harmful recessive alleles that were rare in the original population became common in the small 1990s population. (a) Name the evolutionary mechanism that caused this change, and explain how it occurred. (b) If the original population had 12 distinct alleles at a locus involved in immune function, and the surviving 50 birds only have 3 alleles remaining, what percentage of alleles were lost due to this event?

Worked Solution: (a) This is the bottleneck effect, a type of genetic drift. Habitat fragmentation caused a drastic, random reduction in population size, leading to random loss of genetic variation and an increase in the frequency of rare harmful recessive alleles that were previously present at very low frequency in the large original population. Increased homozygosity for harmful recessive alleles reduces survival and reproductive success. (b) Original alleles = 12, remaining alleles = 3. Alleles lost = $12-3=9$. Percentage lost = $(9/12)\times 100=75\%$. In context, this extreme loss of genetic diversity from genetic drift in the bottlenecked population increases extinction risk by reducing the population's ability to adapt to new environmental stressors like disease or climate change.

7. Quick Reference Cheatsheet

8. What's Next

Population genetics provides the quantitative framework for all subsequent study of evolution in AP Biology Unit 7. Next, you will apply the core concepts of allele frequency change and the null model of HWE to analyze patterns of natural selection (directional, stabilizing, disruptive) and the process of speciation. Without mastering the material in this chapter, you will not be able to correctly interpret experimental data on evolutionary change, which is a regularly tested skill in both MCQ and FRQ sections of the AP exam. This topic also connects back to Mendelian genetics and molecular genetics, and feeds into the big idea that evolution is a measurable change in genetic variation over time. The following topics build directly on this chapter:

Patterns of Natural Selection Speciation Macroevolution Mendelian Inheritance