# Natural Selection

> AP Biology · Unit 7: Natural Selection
> Source: https://www.owlsprep.com/study/ap-biology-u7-natural-selection/

This study guide covers core principles of evolution by natural selection for AP Biology, including fitness, selective pressures, selection modes, Hardy-Weinberg selection modeling, and exam strategies for MCQ and FRQ questions.

**Prerequisites:** Allele and genotype frequencies in diploid populations; Heritable genetic variation from mutation and meiosis; The difference between genotype and phenotype

## Learning objectives

- Explain the core principles of evolution by natural selection
- Calculate relative fitness and selection coefficients for genotypes
- Identify the three modes of natural selection from empirical data
- Model changes in allele frequency due to selection using Hardy-Weinberg framework
- Avoid common exam mistakes related to natural selection concepts

## Core Principles of 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
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 contributes 13–20% of the total Unit 7 exam score, with natural selection-specific questions making up 4–8% of the full exam. It appears in both multiple-choice (MCQ) and free-response (FRQ) questions.

**Exam command terms**

Common exam synonyms for natural selection have specific meanings:

- **Differential reproductive success** — Synonym for natural selection, refers to differing survival/reproduction rates of different phenotypes

- **Adaptive selection** — Refers to selection that increases the frequency of traits that improve fitness

## Biological Fitness and Selective Pressures

**Biological Fitness** — The relative reproductive success of a genotype compared to other genotypes in the same population. Fitness is measured by the number of viable surviving offspring produced, not physical strength or longevity.

*Notation:* $w$ (relative fitness)

*Example:* A genotype that produces 8 surviving offspring has higher fitness than one that produces 2.

**Selection Coefficient** — Measures the strength of selection against a genotype, calculated as $s = 1 - w$. Higher values indicate stronger selection against the genotype.

*Notation:* $s$

Selective pressures are environmental factors that cause differential reproductive success. They can be abiotic (temperature, drought, toxin exposure) or biotic (predation, competition, parasitism), and 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 has a relative fitness of 1 by definition.
2. Apply the relative fitness formula:
3. $$w = \frac{\text{Absolute fitness of genotype}}{\text{Maximum absolute fitness in the population}}$$
4. Calculate relative fitness for all genotypes:
5. $$w_{AA} = 8/8 = 1, \quad w_{Aa} = 6/8 = 0.75, \quad w_{aa} = 2/8 = 0.25$$
6. Calculate the selection coefficient for aa:
7. $$s_{aa} = 1 - w_{aa} = 1 - 0.25 = 0.75$$

> **Exam tip:** If asked to explain what a fitness of 0 means, never say 'the individual dies young' — it means the individual produces zero surviving offspring, so the genotype cannot be passed to the next generation.

## 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:

- **Directional selection**: A single extreme phenotype is favored over all others, shifting the average phenotype of the population over time toward that extreme. Example: selection for smaller body size in marmots responding to climate warming.
- **Stabilizing selection**: The intermediate phenotype is favored, and both extreme phenotypes are selected against. This reduces phenotypic variation and keeps the average phenotype near the original optimum. Example: human birth weight, where intermediate-weight babies have higher survival.
- **Disruptive selection**: Both extreme phenotypes are favored, and the intermediate phenotype is selected against. This increases phenotypic variation and can lead to sympatric speciation. Example: beak size in finches when 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 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 range.
2. Match to the correct 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, 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: x-axis = phenotype value, y-axis = frequency. One shifted peak = directional; one narrow peak at the original center = stabilizing; two separate peaks at extremes = disruptive.

## 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 use the Hardy-Weinberg framework to calculate how allele frequencies change between generations.

First, adjust each initial genotype frequency by its relative fitness, then 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 mean fitness is calculated as:

$$\bar{w} = p^2 w_{AA} + 2pq w_{Aa} + q^2 w_{aa}$$

$p$ = initial frequency of the A allele, $q = 1-p$ = initial frequency of the a allele. After calculating post-selection genotype frequencies, the new allele frequency is:

$$p' = f(AA) + \frac{1}{2}f(Aa)$$

**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 ($q'$) after one generation of selection.

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

> **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.

## AP-Style Practice Questions

**Worked example:** 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?<br>A) Directional selection, variation in shell color will decrease<br>B) Stabilizing selection, variation in shell color will increase<br>C) Disruptive selection, variation in shell color will increase<br>D) Disruptive selection, variation in shell color will decrease

1. 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 options A (directional) and B (stabilizing).
2. Disruptive selection removes intermediate phenotypes and preserves the two extreme phenotypes already present in the population, so it increases overall phenotypic variation. This eliminates option D, which claims variation decreases. The correct answer is C.

**Worked example:** 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.<br>(a) Calculate the relative fitness of each genotype and the selection coefficient against rr. Show your work.<br>(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.<br>(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.

1. (a) Absolute fitness is proportional to the number of colonies, so the maximum absolute fitness is 120. Using $w = w_{abs}/w_{max}$:
2. $$w_{RR} = 120/120 = 1, \quad w_{Rr} = 90/120 = 0.75, \quad w_{rr} = 20/120 ≈ 0.17$$
3. Selection coefficient against rr: $s_{rr} = 1 - w_{rr} = 1 - 0.17 = 0.83$.
4. (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. Calculations confirm this: $\bar{w} = (0.2^2)(1) + 2(0.2)(0.8)(0.75) + (0.8^2)(0.17) = 0.3888$, and $p' ≈ 0.41$, so p increases.
5. (c) Resistance mutations almost always have a fitness cost: the mutated *rpoB* gene produces a less functional RNA polymerase 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.

**Worked example:** 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.

1. 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.
2. The maximum absolute fitness is 4.2 (for 2–3 kg marmots), so relative fitness of 4–5 kg marmots is:
3. $$w = 1.8 / 4.2 ≈ 0.43$$
4. 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.

## Common pitfalls

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

## Cheatsheet

| Category | Formula/Definition | Notes |
| --- | --- | --- |
| Relative Fitness | $w = \frac{\text{Absolute fitness of genotype}}{\text{Maximum absolute fitness in population}}$ | Ranges 0–1; 1 = highest fitness, 0 = no surviving offspring |
| Selection Coefficient | $s = 1 - w$ | Higher $s$ = stronger selection against a genotype |
| Mean Population Fitness | $\bar{w} = p^2 w_{AA} + 2pq w_{Aa} + q^2 w_{aa}$ | Used to normalize post-selection genotype frequencies |
| Post-selection AA frequency | $f(AA) = \frac{p^2 w_{AA}}{\bar{w}}$ | All post-selection frequencies sum to 1 after normalization |
| Post-selection A allele frequency | $p' = f(AA) + \frac{1}{2}f(Aa)$ | $q' = 1 - p'$ always holds |
| Directional Selection | One extreme phenotype favored | Shifts average phenotype in the population |
| Stabilizing Selection | Intermediate phenotype favored | Reduces variation, does not shift average |
| Disruptive Selection | Both extreme phenotypes favored | Increases variation, can lead to speciation |

## What's next

Natural selection is the foundation of all evolutionary biology, and understanding it prepares you to explore more advanced topics in AP Biology Unit 7, including speciation, evidence for evolution, and population genetics. Beyond the exam, natural selection explains real-world phenomena like the evolution of antibiotic resistance in bacteria, pesticide resistance in agricultural pests, and adaptive responses to climate change in wild populations. Solid understanding of fitness and selection modes is also required for interpreting phylogenetic trees and understanding how macroevolutionary patterns emerge from microevolutionary processes. Use the links below to continue your study of Unit 7 topics.

- [Evidence for Evolution](https://www.owlsprep.com/study/ap-biology-u7-evidence-of-evolution/)
- [Hardy-Weinberg Equilibrium Basics](https://www.owlsprep.com/study/ap-biology-u7-hardy-weinberg-equilibrium/)
- [Artificial Selection](https://www.owlsprep.com/study/ap-biology-u7-artificial-selection/)

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