IB Biology HLNatural SelectionPaper 1 & 2~13 min read
Hardy-Weinberg Principle
This is the only calculation in the topic, and it always starts in the same place: the recessive phenotype. Find that one number and everything else falls out of two short equations.
📚 What you need to know
The Hardy-Weinberg principle states that if certain conditions are met, allele frequencies will not change from one generation to the next.
There are seven conditions that must be met for the principle to hold.
p + q = 1, where p is the frequency of the dominant allele and q the frequency of the recessive allele.
p2 + 2pq + q2 = 1, giving the frequencies of the three genotypes.
p2 is homozygous dominant, 2pq is heterozygous, q2 is homozygous recessive.
Always start from the recessive phenotype, because it is the only one whose genotype you know for certain.
If real genotype frequencies do not fit the equation, one or more conditions is not being met.
The principle is the idea of equilibrium; the equations are the tool for estimating frequencies.
What the principle actually says
The Hardy-Weinberg principle says that if certain conditions are met, the allele frequencies of a gene within a population will not change from one generation to the next. In other words, the population is in equilibrium and is not evolving.
That may sound useless — why model a population that is not changing? Because it gives you a baseline. Once you can calculate what a non-evolving population should look like, any difference between that prediction and the real data tells you something is going on.
The equation that comes with the principle lets you calculate allele and genotype frequencies within a population, and predict how those frequencies will change in future generations.
The two equations
Assume a trait is controlled by a single gene with only two alleles, B and b. Every individual is then one of three genotypes: homozygous dominant BB, heterozygous Bb, or homozygous recessive bb.
All frequencies here are proportions of 1, not percentages. If every individual were BB, its frequency would be 1; if half were, it would be 0.5.
Allele frequencies
p + q = 1
Since there are only two alleles at the locus, their frequencies must total 1. In a population of 100 individuals there are 200 alleles. If 120 of them are the dominant allele, then p = 120 ÷ 200 = 0.6, and q = 1 − 0.6 = 0.4.
Genotype frequencies
p2 + 2pq + q2 = 1
Where does the second equation come from? Think about how a zygote is made.
To be homozygous dominant, the offspring must inherit a dominant allele from both parents: p × p = p2.
To be homozygous recessive, it must inherit a recessive allele from both parents: q × q = q2.
To be heterozygous, it can inherit the dominant allele from the father and the recessive from the mother, or the other way round. Two routes, so 2pq.
Those are the only three possibilities, so together they must add up to 1.
If you ever forget the equation, draw this grid. It rebuilds itself in about fifteen seconds.
Working through the calculation
Almost every Hardy-Weinberg question gives you the recessive phenotype, because that is the only phenotype whose genotype you can be certain about. Anything showing the dominant phenotype could be BB or Bb, so it tells you nothing on its own.
Questions can hand you any box in this chain. Identify which one you have been given before you write anything down.
🧩 The method
Find the proportion of individuals showing the recessive phenotype. That is q2.
Take the square root to get q.
Use p = 1 − q to get p.
Square p for the homozygous dominant frequency.
Work out 2pq for the heterozygous frequency.
Check: p2 + 2pq + q2 should come to 1. If it does not, go back.
If the question wants numbers of individuals rather than frequencies, multiply each frequency by the population size at the very end.
Worked examples
WE 1
Find all three genotype frequencies
In a population of beetles, 9% of individuals show the recessive phenotype of a pale shell. Calculate the frequencies of all three genotypes. (4 marks)
Step 1: identify what you have
Pale shell is the recessive phenotype, so those individuals must be homozygous recessive. q² = 0.09Step 2: find q
q = √0.09 = 0.30Step 3: find p
p = 1 − 0.30 = 0.70Step 4: find the genotype frequencies
p² = 0.70 × 0.70 = 0.49 and 2pq = 2 × 0.70 × 0.30 = 0.42
Step 5: check
0.49 + 0.42 + 0.09 = 1.00
Homozygous dominant 0.49, heterozygous 0.42, homozygous recessive 0.09convert the percentage to a proportion before doing anything else — 9% becomes 0.09
WE 2
Turn frequencies into numbers of individuals
A population of 400 plants contains 64 with white flowers, the recessive phenotype. Calculate how many plants are expected to be heterozygous. (4 marks)
Step 1: find q²
q² = 64 ÷ 400 = 0.16Step 2: find q and p
q = √0.16 = 0.40, so p = 1 − 0.40 = 0.60
Step 3: find 2pq
2pq = 2 × 0.60 × 0.40 = 0.48Step 4: convert to individuals
0.48 × 400 = 192 plants
Step 5: check
p² = 0.36, so 144 + 192 + 64 = 400 ✓
192 heterozygous plantsthe marks are for the working, so write every line even if your calculator does it in one go
WE 3
When the data does not fit
A population of 1000 animals contains 560 AA, 280 Aa and 160 aa. The allele frequencies are p = 0.70 and q = 0.30. Compare the observed numbers with those expected under Hardy-Weinberg and suggest an explanation. (4 marks)
Step 1: expected frequencies
p² = 0.49, 2pq = 0.42, q² = 0.09
Step 2: expected numbers
490 AA, 420 Aa and 90 aa in a population of 1000.
Step 3: compare
There are far fewer heterozygotes than expected (280 against 420) and more of both homozygotes.
Step 4: suggest why
One or more Hardy-Weinberg conditions is not being met — most likely mating is not random, for example if individuals of similar phenotype breed together, or selection is acting against heterozygotes.
A poor fit is evidence that something is acting on the populationthis is what the principle is really for: the mismatch is the interesting result, not the calculation
The seven conditions
For the principle to apply correctly to a population, a series of conditions or assumptions has to be met.
Condition
What it means
Organisms are diploid
Each individual carries two alleles at the locus
Reproduction is sexual only
No asexual reproduction contributing to the next generation
No overlap between generations
Parents do not mate with their own offspring
Mating is random
No mate choice based on phenotype, so no sexual selection
The population is large
So that genetic drift has a negligible effect
No migration, mutation or selection
No individuals entering or leaving, no new alleles made, and no natural or artificial selection
Allele frequencies are equal in both sexes
The same p and q apply to males and females
If the genotype frequencies in a real population do not fit the equation, that indicates one or more of these conditions is not being met. The principle is useful for building models and making predictions, but the assumptions listed are very rarely, if ever, all present in nature.
Do not mix these two up. The equations are used to estimate allele and genotype frequencies in a population. The principle suggests there is an equilibrium between allele frequencies, with no change between generations. A question asking for the principle wants the idea, not the algebra.
Begin every question by asking two things on paper: what have I been given, and what am I being asked for? Do I know q2, or do I know p? Am I calculating 2pq, or a number of individuals? Once both ends of the journey are written down, the steps in between are obvious.
💡 Exam tips
Always start from the recessive phenotype. It is the only genotype you can read straight off a phenotype.
Write q2 = … as your first line. It anchors the whole answer.
Convert percentages to proportions before calculating, and convert back only at the end if asked.
Do the final check that the three genotype frequencies add to 1.
Learn all seven conditions — a question is as likely to ask for them as for a calculation.
If asked why real populations do not fit, name a specific condition being broken rather than saying “the conditions were not met”.
⚠ Common mistakes
Treating the dominant phenotype as p2. The dominant phenotype is p2 + 2pq, because it includes heterozygotes.
Forgetting to square root. q2 is a genotype frequency; q is an allele frequency.
Forgetting the 2 in 2pq. There are two ways to make a heterozygote.
Leaving the answer as a frequency when the question asked for a number of individuals.
Confusing the principle with the equations. One is an idea about equilibrium, the other is a calculation.
Assuming a real population is in Hardy-Weinberg equilibrium. Almost none are, which is exactly why the model is useful.
Up next: Artificial Selection. Everything so far has been the environment doing the choosing. On the last page of this topic, humans take over the job — and the results are faster, stranger, and not always kind to the organism.
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