Natural Selection — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Darwin-Wallace’s theory of evolution by natural selection, biological fitness, selective pressures, modes of natural selection, Hardy-Weinberg modeling of selection, and empirical evidence of selection in wild and experimental populations.

You should already know: Allele and genotype frequencies in diploid populations, Heritable genetic variation from mutation and meiosis, The difference between genotype and phenotype.

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 Natural Selection?

Natural selection is the primary mechanism of adaptive evolution, defined as differential survival and reproduction of individuals within a population due to differences in their heritable phenotypes. Co-developed independently by Charles Darwin and Alfred Russel Wallace, the theory rests on three core observations: 1) populations have variation in heritable traits, 2) more offspring are produced than can survive, leading to competition for limited resources, and 3) traits that improve survival and reproduction will be passed on to more offspring, increasing their frequency in the population over generations. In the AP Biology CED, natural selection is the core topic of Unit 7 (Natural Selection), which contributes 13–20% of the total exam score, with natural selection-specific questions making up 4–8% of the full exam. This topic appears in both multiple-choice (MCQ) questions (testing conceptual understanding of selection modes and fitness) and free-response questions (FRQ, requiring data analysis to support claims about selection). Common synonyms used in exam questions include "differential reproductive success" and "adaptive selection", both referring to the same core process.

2. Biological Fitness and Selective Pressures

Biological fitness (often shortened to fitness) is defined as the relative reproductive success of a genotype compared to other genotypes in the same population. Critically, it is not a measure of physical strength, speed, or longevity alone—what matters for evolution is how many viable, surviving offspring an individual produces. Absolute fitness counts the total number of surviving offspring an individual of a given genotype produces on average. Relative fitness (the value used for all AP Biology calculations) normalizes absolute fitness to the highest value in the population, giving: $$w=\frac{\text{Absolute fitness of genotype}}{\text{Maximum absolute fitness in the population}}$$ Relative fitness ($w$) always ranges from 0 to 1. A selection coefficient ($s$), which measures the strength of selection against a genotype, is calculated as $s=1-w$. A higher $s$ means stronger selection against that genotype. Selective pressures are environmental factors that cause differential reproductive success; they can be abiotic (temperature, drought, toxin exposure) or biotic (predation, competition for mates, parasitism), and they act on existing phenotypic variation, not directly on genotypes.

Worked Example

In a population of wild sunflowers, three genotypes for seed size have the following average number of surviving offspring per individual: AA = 8, Aa = 6, aa = 2. Calculate the relative fitness of each genotype and the selection coefficient against aa.

1. Identify the maximum absolute fitness in the population: 8, from genotype AA, which will have a relative fitness of 1 by definition.
2. Apply the relative fitness formula to each genotype: $w=w_{abs}/w_{max}$.
3. Calculate values: $w_{AA}=8/8=1$, $w_{Aa}=6/8=0.75$, $w_{aa}=2/8=0.25$.
4. Calculate the selection coefficient for aa: $s_{aa}=1-w_{aa}=1-0.25=0.75$.

Exam tip: If you are asked to explain what a fitness of 0 means on the exam, never say "the individual dies young". It means the individual produces zero surviving offspring, so the genotype cannot be passed to the next generation.

3. Modes of Natural Selection

Natural selection is categorized by how it changes the distribution of a continuous (polygenic) phenotype in a population. There are three core modes tested on the AP exam:

1. Directional selection: A single extreme phenotype is favored over all others, shifting the average phenotype of the population over time toward that extreme. For example, selection for smaller body size in marmots in response to climate warming.
2. Stabilizing selection: The intermediate phenotype is favored, and both extreme phenotypes are selected against. This reduces phenotypic variation in the population and keeps the average phenotype near the original optimum. A classic example is human birth weight: intermediate-weight babies have higher survival than very small or very large babies.
3. Disruptive selection: Both extreme phenotypes are favored, and the intermediate phenotype is selected against. This increases phenotypic variation and can lead to sympatric speciation if the two morphs stop interbreeding. For example, beak size in finches where only large and small seeds are available.

Worked Example

A researcher studies body length in a population of garter snakes. After 10 years of a new invasive predator that preys most heavily on medium-sized snakes, the population shows the following changes: original average length = 60 cm, range = 40–80 cm. After selection, 12% of the population is 40–50 cm, 15% is 70–80 cm, and only 3% is 50–70 cm. Identify the mode of selection and predict its long-term effect if the predator remains.

1. Identify which phenotypes have higher frequency after selection: both the short (40–50 cm) and long (70–80 cm) extremes are much more common than the intermediate 50–70 cm.
2. Match to the mode: when both extremes are favored over intermediates, this is disruptive selection.
3. Predict long-term effect: Disruptive selection will maintain or increase phenotypic and genetic variation in the population. If gene flow between the two size morphs is reduced (e.g., large snakes mate with other large snakes and small snakes mate with other small snakes), this can lead to sympatric speciation as the two populations diverge genetically over time.

Exam tip: When identifying a selection mode from a graph, remember the x-axis is phenotype value and the y-axis is frequency. One peak shifted left/right = directional; one narrow peak at the original center = stabilizing; two separate peaks at the extremes = disruptive.

4. Modeling Selection with Hardy-Weinberg Equilibrium

The Hardy-Weinberg principle states that allele and genotype frequencies remain constant from generation to generation only in the absence of evolutionary forces, including natural selection. When selection is acting, we can use the Hardy-Weinberg framework to calculate how allele frequencies change between generations. First, we adjust each initial genotype frequency by its relative fitness, then we normalize the adjusted frequencies by the mean fitness of the population ($\bar{w}$), which is the sum of all adjusted frequencies. The general formula for post-selection genotype frequencies is: $$f(AA{)}_{\text{post-selection}}=\frac{p^2{w}_{AA}}{\bar{w}}$$ where $\bar{w}=p^2{w}_{AA}+2pq{w}_{Aa}+q^2{w}_{aa}$, $p$ is the initial frequency of the A allele, and $q=1-p$ is the initial frequency of the a allele. After calculating post-selection genotype frequencies, the new allele frequency $p^{\prime }$ is $p^{\prime }=f(AA)+\frac{1}{2}f(Aa)$.

Worked Example

In a population, the initial frequency of the recessive a allele is $q=0.5$, and relative fitnesses are $w_{AA}=1$, $w_{Aa}=1$, $w_{aa}=0.25$ (selection against aa). Calculate the new frequency of the a allele ($q^{\prime }$) after one generation of selection.

1. Calculate initial genotype frequencies from Hardy-Weinberg: $p=1-q=0.5$, so $p^2=0.25$, $2pq=0.5$, $q^2=0.25$.
2. Calculate mean fitness $\bar{w}$: $\bar{w}=(0.25)(1)+(0.5)(1)+(0.25)(0.25)=0.8125$.
3. Calculate post-selection genotype frequencies: $f(aa)=\frac{0.25\times 0.25}{0.8125}\approx 0.0769$, $f(Aa)=\frac{0.5\times 1}{0.8125}\approx 0.6154$.
4. Calculate new q: $q^{\prime }=f(aa)+\frac{1}{2}f(Aa)=0.0769+0.5(0.6154)\approx 0.385$.

Exam tip: Always remember to normalize by mean fitness after multiplying by relative fitness. Skipping this step leaves you with total genotype frequencies that do not add up to 1, which will cost points on FRQs.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Defining fitness as "how strong/healthy an individual is" instead of reproductive success. Why: Common language uses "fitness" to mean physical condition, so students default to this incorrect definition. Correct move: Always define fitness as "the relative reproductive success of a genotype, measured by the number of surviving offspring produced" in any FRQ answer.
• Wrong move: Claiming natural selection creates new traits that a population "needs" to survive. Why: Students often anthropomorphize selection, thinking it acts toward a predetermined goal. Correct move: Always state that selection acts on existing heritable variation; new variation comes from random mutation, and selection only increases the frequency of variants that already improve reproductive success.
• Wrong move: Saying selection acts on genotypes instead of phenotypes. Why: Students focus on genetic variation when discussing evolution, so they misstate the level at which selection acts. Correct move: Remember the mantra: "Selection acts on phenotypes, which are coded by genotypes, leading to changes in genotype frequencies over time."
• Wrong move: Confusing disruptive selection with directional selection when a graph shows two peaks. Why: Students remember directional selection favors an extreme, so they mix up the two when both extremes are favored. Correct move: Count the number of peaks after selection: one shifted peak = directional, two peaks at opposite ends = disruptive, one narrow peak at the original center = stabilizing.
• Wrong move: Claiming stabilizing selection changes the average phenotype of a population. Why: Students confuse the effect of stabilizing selection on variation vs. average phenotype. Correct move: Stabilizing selection reduces variation around the existing average, it does not shift the average phenotype of the population.
• Wrong move: Forgetting to divide by mean fitness when calculating post-selection allele frequencies. Why: Students correctly multiply initial frequencies by relative fitness, but stop there, forgetting selection removes genotypes so the total must be renormalized. Correct move: After calculating the sum of (initial frequency × fitness), always divide each adjusted frequency by that sum (mean fitness) to get post-selection frequencies.

6. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

A population of limpets (aquatic snails) has shell color ranging from white to dark brown, with tan being intermediate. Limpets live on intertidal rocks that are either light-colored (covered in white barnacles) or dark-colored (bare rock). Birds (predators) can easily see and eat tan limpets on either background, but rarely see white limpets on barnacles or dark brown limpets on dark rock. Which of the following correctly predicts the mode of natural selection and its effect on shell color variation? A) Directional selection, variation in shell color will decrease B) Stabilizing selection, variation in shell color will increase C) Disruptive selection, variation in shell color will increase D) Disruptive selection, variation in shell color will decrease

Worked Solution: First, identify which phenotypes are favored: the two extreme phenotypes (white and dark brown) are favored, while the intermediate (tan) is selected against. This eliminates A (directional) and B (stabilizing). Disruptive selection removes intermediate phenotypes and preserves the two extreme phenotypes already present in the population, so it increases overall phenotypic variation. This eliminates D, which claims variation decreases. The correct answer is C.

Question 2 (Free Response)

Bacteria develop resistance to antibiotics through natural selection. A researcher studies a population of E. coli growing in medium with a low concentration of the antibiotic rifampicin. She genotypes the rpoB gene (which confers resistance when mutated) and counts the number of viable colonies formed by each genotype after 24 hours. She finds: RR (homozygous resistant) = 120 colonies, Rr (heterozygous) = 90 colonies, rr (homozygous susceptible) = 20 colonies. (a) Calculate the relative fitness of each genotype and the selection coefficient against rr. Show your work. (b) The initial allele frequency of R is p = 0.2. Predict the change in frequency of R after one generation of selection. Justify your prediction. (c) Explain why the frequency of the resistance allele would decrease if the antibiotic is removed from the growth medium. Connect your answer to fitness.

Worked Solution: (a) Absolute fitness is proportional to the number of colonies, so the maximum absolute fitness is 120. Relative fitness $w=w_{abs}/w_{max}$: $w_{RR}=120/120=1$, $w_{Rr}=90/120=0.75$, $w_{rr}=20/120\approx 0.17$. Selection coefficient against rr: $s_{rr}=1-w_{rr}=1-0.17=0.83$. (b) The frequency of R will increase from 0.2 to a higher value. Genotypes carrying the R allele have much higher fitness than rr in the presence of rifampicin, so R alleles are passed on to more offspring. Calculating the new frequency confirms this: $\bar{w}=(0.2^2)(1)+2(0.2)(0.8)(0.75)+(0.8^2)(0.17)=0.3888$, and $p^{\prime }\approx 0.41$, so p increases. (c) Resistance mutations almost always have a fitness cost: the mutated rpoB gene produces a less functional RNA polymerase (the protein it codes for) even in the absence of rifampicin. This means RR and Rr genotypes have lower relative fitness than wild-type rr when there is no antibiotic. Because rr has higher fitness, natural selection will increase the frequency of r and decrease the frequency of R over generations.

Question 3 (Application / Real-World Style)

Climate change is increasing average annual temperatures in mountain habitats. Researchers measured body mass in a population of marmots at 2000m elevation in 1980 and 2020. In 1980, average adult body mass was 3.5 kg, with a range from 2 kg to 5 kg. By 2020, average adult body mass was 3.1 kg, the proportion of the population with mass between 2 kg and 3 kg increased from 25% to 45%, and the proportion between 4 kg and 5 kg decreased from 25% to 8%. Smaller body size increases heat tolerance because it raises the surface area to volume ratio, allowing more efficient heat loss. Identify the mode of selection acting on marmot body mass, calculate the relative fitness of 4–5 kg marmots if 2–3 kg marmots have an absolute fitness of 4.2 surviving offspring per individual and 4–5 kg marmots have an absolute fitness of 1.8, and explain what this fitness value means in context.

Worked Solution: The smaller body mass (one extreme of the original phenotypic range) is favored, while larger body mass (the opposite extreme) is selected against, and the population average has shifted toward smaller size. This is directional selection. The maximum absolute fitness is 4.2 (for 2–3 kg marmots), so relative fitness of 4–5 kg marmots is $w=1.8/4.2\approx 0.43$. A relative fitness of 0.43 means that 4–5 kg marmots produce only 43% as many surviving offspring as the 2–3 kg marmots that are most favored under current warmer temperatures. This matches the observed shift in average body mass over 40 years: directional selection favoring smaller body size in response to the selective pressure of anthropogenic climate change is leading to adaptive change in the marmot population.

7. Quick Reference Cheatsheet

8. What's Next

Natural selection is the core mechanism of adaptive evolution, and it is the foundation for all subsequent topics in Unit 7. Next, you will learn how other evolutionary forces (genetic drift, gene flow, mutation) interact with natural selection to change allele frequencies in populations. Without a solid understanding of fitness, selection modes, and how to model selection, you will not be able to compare the strength of selection to other evolutionary forces or interpret patterns of speciation. Natural selection also connects directly to phylogenetics, where you use shared derived adaptive traits (produced by selection) to build evolutionary trees, and to ecology, where you study how selective pressures shape community interactions.

Other Mechanisms of Evolution Speciation Phylogenetics Hardy-Weinberg Equilibrium