By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Complete Guide For GCSE/A-Level Biology (AQA, Edexcel, OCR, WJEC)
"Mastering population growth models can earn you 10–15% of your Biology exam marks—and help you predict real-world crises like pandemics, endangered species recovery, or even zombie outbreaks. Today, you’ll learn the exact steps to solve any question on r/K-selection, logistic growth, or predator-prey cycles."
Formula: [ \frac{dN}{dt} = rN \left( \frac{K - N}{K} \right) ]
Variables: - ( \frac{dN}{dt} ) = population growth rate (change in population over time) - ( r ) = intrinsic growth rate (max growth rate without limits) - ( N ) = current population size - ( K ) = carrying capacity (max sustainable population)
MEMORISE THIS – You’ll need to interpret graphs and calculate growth rates.
MEMORISE THIS TABLE – Examiners love comparing traits.
MEMORISE THIS PATTERN – You’ll be asked to sketch or interpret these graphs.
Question: A population of rabbits has: - ( r = 0.2 ) per year - ( K = 500 ) rabbits - Current population ( N = 100 )
Calculate the growth rate ( \frac{dN}{dt} ).
Solution:1. Identify model: Logistic growth (given ( K )).2. Extract values: ( r = 0.2 ), ( N = 100 ), ( K = 500 ).3. Apply formula: [ \frac{dN}{dt} = 0.2 \times 100 \times \left( \frac{500 - 100}{500} \right) ] [ = 20 \times \left( \frac{400}{500} \right) ] [ = 20 \times 0.8 = 16 ]4. Answer: The population grows by 16 rabbits per year.
What we did and why: - Used the logistic growth formula because the question gave a carrying capacity (( K )). - Multiplied ( r ), ( N ), and the fraction ( \frac{K - N}{K} ) to find the growth rate.
Question: A species of fish produces 1,000 eggs per year but only 5 survive to adulthood. Is this an r-selected or K-selected species? Explain.
Solution:1. Identify traits: - Many offspring (1,000 eggs) → r-selected trait. - Low survival (only 5 adults) → r-selected trait.2. Compare to table: - r-selected species have many offspring, low survival. - K-selected species have few offspring, high survival.3. Answer: This is an r-selected species because it produces many offspring with low survival rates.
What we did and why: - Matched the given data to the r/K-selection table. - Justified the answer using specific traits (offspring number and survival).
Question: The graph below shows lynx and hare populations over 20 years. a) Which line represents the predator? Explain. b) Why does the predator population peak after the prey?
Solution:1. Identify predator/prey: - The higher peak is the prey (hares). - The lower peak is the predator (lynx). - Answer (a): The red line (lower peak) is the predator because predators are less abundant than prey.2. Explain lag time: - Predators reproduce more slowly than prey. - When prey increase, predators take time to find food, reproduce, and increase. - Answer (b): The predator population peaks after the prey because it takes time for predators to reproduce and respond to increased food availability.
What we did and why: - Used graph trends (peaks and timing) to identify predator/prey. - Explained the biological reason (reproduction delay) for the lag time.
"Here’s what you need to remember for your exam:1. Logistic growth = S-shaped curve, carrying capacity (( K )). Use the formula ( \frac{dN}{dt} = rN \left( \frac{K - N}{K} \right) ).2. r-selected species = many babies, low survival (e.g., insects). K-selected = few babies, high survival (e.g., elephants).3. Predator-prey graphs = prey peaks first, predators follow. Always mention lag time.4. Common mistakes: Mixing up ( r ) and ( K ), forgetting units, and ignoring lag time.5. Exam traps: Examiners love hiding ( K ) in graphs or asking for trends without labels.
Now go practice—you’ve got this!"
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