Fitness — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Darwinian fitness in the context of cellular energetics, absolute vs relative fitness definitions, selection coefficients, calculation of fitness from survival and reproductive data, and links between metabolic efficiency and fitness.

You should already know: Natural selection and heritable trait variation; Enzyme kinetics and metabolic efficiency; Hardy-Weinberg equilibrium principles.

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 Fitness?

In AP Biology Unit 3: Cellular Energetics, fitness (often called Darwinian fitness) describes an organism’s ability to survive to reproductive age, find a mate, and produce viable offspring, directly tied to how efficiently its cellular processes generate energy for growth and reproduction. Unlike the common usage of "fitness" as physical strength, biological fitness is a population-level relative measure of reproductive success. This topic is allocated ~2-4% of the total exam weight per the Course and Exam Description (CED), and questions can appear on both multiple-choice (MCQ) and free-response (FRQ) sections. It is most often integrated with concepts of metabolic adaptation, enzyme function, or natural selection. Two standard notations are used: $w$ for relative fitness, and $W$ for absolute fitness. Synonyms you may see on the exam include Darwinian fitness and selective fitness.

2. Absolute vs Relative Fitness

Absolute fitness ($W$) is the total number of viable offspring produced by an individual of a given genotype over its lifetime. For a whole genotype, it is calculated as the product of the survival rate (proportion of individuals of that genotype that survive to reproduction) multiplied by the average number of offspring per surviving individual of that genotype: $$W=(\text{survival rate})\times (\text{average offspring per survivor})$$ Relative fitness ($w$), by contrast, normalizes absolute fitness to the maximum absolute fitness observed in the population, so all values of $w$ fall between 0 and 1. This is the more commonly used measure on the AP exam, because it directly describes how fit a genotype is relative to the most successful genotype in the population, which lets us predict the rate of evolutionary change. In the context of cellular energetics, differences in relative fitness almost always stem from differences in metabolic efficiency: a genotype that produces a more efficient enzyme for ATP production will have more energy available for reproduction, leading to higher fitness.

Worked Example

In a population of desert wildflowers, two genotypes for the RuBisCO enzyme are observed. Genotype AA has a 70% survival rate to reproduction, and surviving plants produce an average of 8 viable seeds each. Genotype Aa has an 80% survival rate and produces an average of 9 viable seeds. Genotype aa has a 90% survival rate and produces an average of 10 viable seeds. Calculate the absolute fitness of each genotype, then the relative fitness of each.

1. Recall the formula for absolute fitness: $W=s\times \bar{n}$, where $s$ = survival rate and $\bar{n}$ = average offspring per survivor.
2. Calculate absolute fitness for each genotype: $W_{AA}=0.70\times 8=5.6$, $W_{Aa}=0.80\times 9=7.2$, $W_{aa}=0.90\times 10=9.0$.
3. Identify the maximum absolute fitness in the population: $W_{max}=9.0$ for genotype aa.
4. Calculate relative fitness as $w_i=W_i/W_{max}$: $w_{AA}=5.6/9.0\approx 0.62$, $w_{Aa}=7.2/9.0=0.80$, $w_{aa}=9.0/9.0=1.0$.

Exam tip: If the question gives you total viable offspring per genotype (not separate survival and average offspring counts), use the total number of offspring directly as absolute fitness, no extra calculation needed.

3. Selection Coefficients and Fitness

A selection coefficient ($s$) measures the strength of natural selection against a particular genotype. It is directly calculated from relative fitness, because it quantifies how much less fit a genotype is compared to the most fit genotype in the population. The formula for the selection coefficient against genotype $i$ is: $$s_i=1-w_i$$ If $s_i=0$, there is no selection against the genotype: it has equal fitness to the most fit genotype. If $s_i=1$, selection against the genotype is complete: all individuals of that genotype either die before reproducing or produce no viable offspring, so their fitness is 0. In cellular energetics, selection coefficients tell us how strongly negative or positive selection acts on a given metabolic trait. For example, a high selection coefficient against a genotype with a defective ATP synthase enzyme confirms that inefficient energy production has strong negative effects on survival and reproduction.

Worked Example

Using the relative fitness values calculated in the previous example ($w_{AA}=0.62$, $w_{Aa}=0.80$, $w_{aa}=1.0$), calculate the selection coefficient against each genotype, and interpret what the selection coefficient against AA means in terms of RuBisCO efficiency.

1. Recall that the selection coefficient for any genotype is defined as $s_i=1-w_i$.
2. Calculate each coefficient: $s_{AA}=1-0.62=0.38$, $s_{Aa}=1-0.80=0.20$, $s_{aa}=1-1.0=0$.
3. Interpret the result: A selection coefficient of 0.38 against AA means AA individuals have a 38% reduction in relative fitness compared to the most fit aa genotype.
4. Contextualize for RuBisCO function: This confirms that the RuBisCO variant produced by the AA genotype is less efficient at carbon fixation than the variant from aa, leading to lower glucose production, less energy for growth and reproduction, and lower fitness.

Exam tip: Always remember that the selection coefficient measures selection against a genotype, not for it. If the question asks for the strength of selection for a genotype, it will equal $w_i$, not $s_i$.

4. Metabolic Efficiency and Fitness

This is the core connection between fitness and Unit 3 Cellular Energetics: differences in cellular metabolic processes directly translate to differences in fitness because all cellular work (growth, reproduction, damage repair, homeostasis) requires ATP. For example, in endotherms living in cold environments, mutations that increase the efficiency of cellular respiration produce more ATP per glucose molecule, leaving more energy available for thermoregulation and offspring production, which increases relative fitness. Similarly, in plants, mutations that reduce photorespiration by RuBisCO increase net glucose production for growth and seed development, leading to higher fitness in high-light, high-temperature environments. On the AP exam, you will often be given data on metabolic rate or enzyme efficiency and asked to predict or calculate relative fitness from that data.

Worked Example

Two genotypes of rainbow trout (an aquatic ectotherm) are compared for efficiency of cellular respiration at warm water temperatures, which are becoming more common due to climate change. Genotype X produces a hexokinase variant that yields 32 ATP per glucose molecule. Genotype Y produces a variant that yields 27 ATP per glucose molecule. Both genotypes consume the same mass of glucose per day, and all individuals of both genotypes survive to reproduction. Average number of offspring is directly proportional to total ATP available. (a) Predict which genotype has higher relative fitness. (b) Calculate the relative fitness of each genotype.

1. Because all individuals survive and offspring number is proportional to ATP output per glucose, we can use ATP production per glucose as a proxy for absolute fitness.
2. Absolute fitness values are $W_X=32$ and $W_Y=27$. The maximum absolute fitness is 32 for Genotype X, so X is predicted to have higher fitness.
3. Calculate relative fitness by dividing each absolute fitness by the maximum absolute fitness: $w_X=32/32=1.0$, $w_Y=27/32\approx 0.84$.
4. Final interpretation: Genotype X, with the more efficient hexokinase, has a 16% higher relative fitness than Genotype Y at warm temperatures.

Exam tip: On FRQs that ask you to connect metabolic efficiency to fitness, always explicitly state that lower efficiency reduces energy available for reproduction, leading to lower fitness. This connection is almost always required for full points.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the highest frequency genotype as the denominator for relative fitness calculations, instead of the highest absolute fitness genotype. Why: Students confuse allele frequency with fitness, assuming common genotypes are always the most fit. Correct move: Always first identify the maximum value of absolute fitness in the dataset, then use that value as the denominator regardless of genotype frequency.
• Wrong move: Interpreting a fitness of 0 to mean all individuals of that genotype die immediately after birth. Why: Students forget fitness counts viable offspring, not just survival. Correct move: Remember fitness of 0 can also occur if individuals survive to adulthood but produce no viable offspring, not just if they die before reproduction.
• Wrong move: Calculating selection coefficient as $s_i=w_i-1$ instead of $1-w_i$, leading to negative selection coefficients. Why: Students mix up the order of subtraction when recalling the formula from memory. Correct move: Remember selection coefficient measures reduced fitness, so it ranges from 0 (no reduction) to 1 (complete reduction), so always subtract relative fitness from 1.
• Wrong move: Assuming that higher metabolic rate always means higher fitness. Why: Students associate more energy with higher fitness, but ignore that higher metabolic rate means more glucose consumed to produce the same ATP. Correct move: Fitness depends on efficiency (ATP produced per unit resource), not just total metabolic rate. If two organisms use the same amount of resources, the one with higher ATP output has higher fitness.
• Wrong move: Confusing biological fitness with physical strength or individual health. Why: The common English usage of "fitness" interferes with the precise biological definition. Correct move: Always remember biological fitness is measured exclusively by the number of viable offspring produced, not how strong or healthy an individual organism is.

6. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

Cyanobacteria living in the open ocean experience variable levels of light. A researcher measures three genotypes of the enzyme RuBisCO in a population, with the following data:

Worked Solution: First, calculate absolute fitness for each genotype by multiplying survival rate by average offspring per survivor. This gives $W_{G1G1}=0.6\times 12=7.2$, $W_{G1G2}=0.8\times 12=9.6$, $W_{G2G2}=0.9\times 11=9.9$. Next, find the maximum absolute fitness in the population, which is 9.9 for G2G2. Relative fitness for G1G2 is calculated as $W_{G1G2}/W_{max}=9.6/9.9\approx 0.96$. The correct answer is B.

Question 2 (Free Response)

In a population of wild mice, two alleles for the ATP synthase gene exist: $A$ and $a$. Researchers collect the following data for each genotype living in a cold mountain environment:

Worked Solution: (a) Absolute fitness $W=\text{survival}\times \text{average offspring}$: $W_{AA}=0.75\times 6=4.5$, $W_{Aa}=0.85\times 6=5.1$, $W_{aa}=0.95\times 5=4.75$. Maximum absolute fitness $W_{max}=5.1$ (Aa). Relative fitness $w_i=W_i/W_{max}$: $w_{AA}\approx 0.88$, $w_{Aa}=1.0$, $w_{aa}\approx 0.93$.

(b) Selection coefficient $s_i=1-w_i$: $s_{AA}=0.12$, $s_{Aa}=0$, $s_{aa}=0.07$. A selection coefficient of 0.07 means there is weak negative selection against the aa genotype: aa individuals have a 7% reduction in relative fitness compared to the most fit Aa genotype.

(c) Leaky proton flow in aa ATP synthase reduces the efficiency of oxidative phosphorylation, so less ATP is produced per glucose molecule. This reduced energy production leads to fewer offspring per surviving aa individual, lowering aa fitness. Because the $A$ allele is associated with higher fitness across all genotypes, the frequency of the $a$ allele will decrease over time due to natural selection.

Question 3 (Application / Real-World Style)

Climate change is increasing average ocean temperatures, which alters metabolic demand in reef-building corals. Researchers measured fitness for two coral symbiont (Symbiodinium) genotypes that differ in the efficiency of their photosynthetic pathway at elevated temperatures. Genotype 1 has an 82% survival rate at 2°C above ambient, and surviving colonies produce an average of 3.1 new viable colonies per year. Genotype 2 has a 65% survival rate at 2°C above ambient, and surviving colonies produce an average of 4.2 new viable colonies per year. Calculate the relative fitness of each genotype, predict which genotype will become more common in warming oceans, and explain what this means for the long-term survival of coral reefs.

Worked Solution: First calculate absolute fitness for each genotype: $W_1=0.82\times 3.1\approx 2.54$, $W_2=0.65\times 4.2=2.73$. The maximum absolute fitness is 2.73 for Genotype 2. Relative fitness values are $w_1=2.54/2.73\approx 0.93$, $w_2=2.73/2.73=1.0$. Genotype 2 has higher relative fitness at elevated temperatures, so it will become more common over time as oceans warm. This means coral reefs are more likely to persist in warming oceans if they host the more heat-tolerant Genotype 2 symbiont, since the higher fitness of this symbiont allows coral colonies to continue surviving and reproducing at higher temperatures.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of fitness in the context of cellular energetics is a critical prerequisite for understanding metabolic adaptation, which you will explore next in Unit 3, and later for population genetics in Unit 7 (Natural Selection). Without understanding how differences in cellular energy production translate to differences in fitness, you cannot predict how natural selection will act on metabolic traits or how populations will adapt to changing environmental conditions like climate change. This topic also connects to the bigger core concept of how genotype determines phenotype, and how phenotype ultimately determines evolutionary success across all biological systems. Next topics you will move to after this: Metabolic Adaptation, Enzyme Regulation, Natural Selection, Population Genetics