Populations

In a nutshell

Populations is where genetics meets evolution: instead of tracking alleles through one family, you track them across a whole population and ask whether their frequencies are changing.

The core tool is the Hardy-Weinberg equation, a mathematical model that predicts the frequencies of alleles, genotypes and phenotypes, and lets you detect when a population is evolving.

Almost every mark here is a calculation or a precise definition, so the two things to nail are the exact wording and the difference between an allele frequency and a genotype frequency.

Assumed knowledge: Inheritance and genetic crosses, How genetic diversity arises.

Core content

Species, populations and the gene pool

A species exists as one or more populations.

• Population: a group of organisms of the same species occupying a particular space at a particular time that can potentially interbreed.

That definition has three parts that all carry credit: same species, in the same place at the same time, and able to interbreed. Dropping any one of them is a common way to lose the mark.

Every allele carried by every member of a population, added together, forms the gene pool.

• Gene pool: all the alleles of all the genes present in a population at a given time.
• Allele frequency: the proportion (frequency) of a particular allele of a gene within the population.

Allele frequency is the quantity that evolution acts on. A change in allele frequency in a gene pool from one generation to the next is evolution, so the rest of this subtopic is really about measuring it.

The Hardy-Weinberg principle

The Hardy-Weinberg principle is a mathematical model. It predicts that the frequency of each allele in a population will stay constant from one generation to the next, provided a set of conditions is met.

Learn both halves, because they are credited separately: the principle is about the frequency of alleles, and the claim is that this frequency does not change between generations (no evolution).

If the allele frequencies of a real population do change over time, one or more of the conditions has been broken, which means the population is evolving.

The five conditions

The prediction of constant allele frequencies only holds if all of these are true:

• No mutation, so no new alleles are added to the gene pool.
• No natural selection, so all genotypes are equally fertile and none has a selective advantage.
• The population is large, so allele frequencies are not shifted by chance (genetic drift).
• Random mating, so any individual is equally likely to mate with any other (this is random mating, not random fertilisation).
• No migration, so there is no immigration or emigration, that is no gene flow into or out of the population (it is genetically isolated).

A useful way to remember them: these are exactly the processes that, if they happened, would change allele frequencies. That is why switching them off is what keeps the frequencies the same.

One trap: "no births or deaths" is not a Hardy-Weinberg condition. That belongs to the mark-release-recapture population estimate, and writing it here earns nothing.

The Hardy-Weinberg equations

Two equations work together. Let p be the frequency of one allele (usually the dominant) and q the frequency of the other (usually the recessive) allele of the same gene.

Because the gene has only these two alleles, their frequencies must add up to 1:

$p + q = 1$

The genotype frequencies then follow from combining the alleles at random:

$p^2 + 2pq + q^2 = 1$

Each term is a genotype frequency:

• $p^2$ = frequency of the homozygous dominant genotype (for example AA).
• $2pq$ = frequency of the heterozygous genotype (for example Aa), the carriers.
• $q^2$ = frequency of the homozygous recessive genotype (for example aa).

The single most important fact in this topic: the frequency of the recessive phenotype (the individuals that actually show the recessive characteristic) is $q^2$, not q, because only homozygous recessive individuals show it.

Still don't get it? · why the frequency you are given is q², not q

Imagine each beetle's two body-colour alleles are two coins in its pocket. A black beetle only happens when both coins come up recessive: tails and tails.

Say a single recessive "tails" turns up 2 times in every 10 alleles, a frequency of 0.2. Then getting tails twice over is 0.2 × 0.2 = 0.04, which is 4 in every 100 beetles. So the fraction of beetles that are black is not 0.2, it is 0.2 squared.

Now run it the way an exam asks. You are told 4% of beetles are black. That 4% is the "both coins tails" figure, so it is q², not q. To find the frequency of a single recessive allele, undo the squaring by taking the square root: √0.04 = 0.2. The rule to carry into the exam: the recessive phenotype frequency is q²; square-root it to reach the recessive allele frequency q.

Where the equation comes from

The two equations are not separate facts to memorise, they come straight from combining gametes. Each gamete carries one allele: the frequency of an A gamete is p and of an a gamete is q. Fertilisation pairs them at random:

Adding the boxes gives $p^2 + 2pq + q^2$ (the two heterozygous boxes combine to 2pq), and because every individual is one of the three genotypes, the total is 1.

Still don't get it? · where p² + 2pq + q² = 1 comes from

Think of making a new individual as a lucky dip: you reach into the gene pool and pull out one allele from each parent. The chance of pulling the dominant allele is p, and the recessive allele is q.

A new individual is two draws, one from each parent. There are two ways to build a heterozygote, dominant-then-recessive or recessive-then-dominant, so its chance is pq + pq = 2pq. That is where the 2 comes from, and it is why leaving it out (writing pq) is wrong. Homozygous dominant needs two dominant draws (p × p = p²), and homozygous recessive two recessive draws (q × q = q²).

Every individual has to be one of those three genotypes, so the three chances add to 1: p² + 2pq + q² = 1. In the exam that means the three genotype frequencies always total 1 (or 100%), which is a fast way to check your working.

Data skill: actual versus estimated frequencies

Two different jobs sit behind these questions, and mixing them up is a classic error:

• Actual (observed) frequency: when you can count the alleles directly (the question gives you the numbers), add up all the alleles present and take the proportion that are the allele you want. No Hardy-Weinberg is needed.
• Estimated frequency: when you are only told the frequency of the recessive phenotype, use the Hardy-Weinberg equation to estimate the allele and genotype frequencies.

Only reach for the equation to estimate. Applying it to a small, strongly selected or clearly non-equilibrium population, or when the raw counts are sitting in front of you, is the wrong tool.

Data skill: measuring phenotype frequencies in a population

A common practical task is to estimate allele frequencies from what you can see in a population:

• Sample a large number of individuals from a single population.
• Count how many show each observable phenotype (for example, able or unable to roll the tongue).
• The proportion showing the recessive phenotype estimates $q^2$; from it you get q, then p, then the genotype frequencies.
• A large sample matters, because the model assumes a large population and random mating.

Worked examples

Model calculation 1: from the recessive phenotype to the carrier frequency

In a large, randomly mating population of a beetle, body colour is controlled by one gene, and the recessive allele gives a black body. 4% of the beetles are black. Estimate the percentage that are carriers (heterozygous).

1. Black beetles are homozygous recessive, so their frequency is $q^2$: $q^2 = 0.04$
2. Square-root to get the recessive allele frequency: $q = \sqrt{0.04} = 0.2$
3. Use $p + q = 1$ for the dominant allele frequency: $p = 1 - 0.2 = 0.8$
4. Carriers are heterozygous, frequency $2pq$: $2pq = 2 \times 0.8 \times 0.2 = 0.32$
5. Convert to a percentage: $0.32 \times 100 = 32\%$.

The first mark is won by recognising 4% as $q^2$ rather than q; the final mark needs the answer given as a percentage, not left as 0.32.

Model calculation 2: from an allele frequency to the expected genotypes

In a population meeting the Hardy-Weinberg conditions, the recessive allele has a frequency of 0.3. Estimate the percentage of the population expected to have each genotype.

1. $q = 0.3$, so from $p + q = 1$, $p = 1 - 0.3 = 0.7$.
2. Homozygous dominant: $p^2 = 0.7^2 = 0.49$, which is 49%.
3. Heterozygous: $2pq = 2 \times 0.7 \times 0.3 = 0.42$, which is 42%.
4. Homozygous recessive: $q^2 = 0.3^2 = 0.09$, which is 9%.
5. Check they total 1: $0.49 + 0.42 + 0.09 = 1.00$.

These figures are an estimate: they only hold while the population stays in Hardy-Weinberg equilibrium.

Common exam mistakes

• Treating the recessive phenotype frequency as q instead of q². If 16% show the recessive characteristic, then $q^2 = 0.16$ and $q = 0.4$; using q = 0.16 is the single most common error on this topic.
• Writing pq for the frequency of heterozygotes instead of 2pq. Carriers are $2pq$; pq earns nothing.
• Leaving the answer as a decimal frequency when a percentage was asked for, or the reverse. If a percentage is wanted, multiply by 100.
• Going one step too far: being asked for the recessive allele frequency (q) but carrying on out of habit to work out 2pq, and quoting something like 0.48. Answer the quantity the question actually asks for.
• Using Hardy-Weinberg to find an actual frequency you could simply count. If you are told the numbers of each allele, add them up and take the proportion; the equation is only for estimating.
• Defining a population loosely as "a group of organisms of the same species". The mark also needs the same place and time and that they can interbreed.
• Saying the principle is "about allele frequencies" without adding that the frequencies stay constant from one generation to the next. Both halves are required.
• Listing "no births and deaths" as a condition. That is a mark-release-recapture assumption, not a Hardy-Weinberg one.
• Referring to genes when you mean alleles in the definition of allele frequency, or giving a frequency with no reference to a population.

Key definitions

• Population: a group of organisms of the same species occupying a particular space at a particular time that can potentially interbreed.
• Gene pool: all the alleles of all the genes present in a population at a given time.
• Allele frequency: the proportion (frequency) of a particular allele of a gene within a population.
• Hardy-Weinberg principle: the frequency of an allele in a population stays constant from one generation to the next, provided there is no mutation, no natural selection, a large population, random mating and no migration.
• p + q = 1: the frequencies of the two alleles of a gene add up to 1.
• p² + 2pq + q² = 1: p² is the frequency of the homozygous dominant genotype, 2pq the heterozygous genotype and q² the homozygous recessive genotype.

Specification

• I can state that a species exists as one or more populations, and define a population as a group of organisms of the same species occupying a particular space at a particular time that can potentially interbreed.
• I can explain the concepts of the gene pool and allele frequency.
• I can state that the Hardy-Weinberg principle predicts that allele frequencies do not change from one generation to the next, and give the conditions under which it applies.
• I can use the Hardy-Weinberg equation, $p^2 + 2pq + q^2 = 1$ with $p + q = 1$, to calculate allele, genotype and phenotype frequencies from appropriate data.

Related notes

• Inheritance and genetic crosses
• Evolution may lead to speciation
• How genetic diversity arises