Biology · Evolutionary Processes & Adaptation

Alleles, variation and population-level change

A population can contain many alleles even though each diploid individual carries only two copies at one ordinary autosomal locus. To describe population change, first decide what is being counted. A proportion of allele copies answers a different question from a proportion of organisms.

The introductory inheritance chapter distinguishes alleles, genotypes and phenotypes. Meiosis explains segregation and crossing over; the DNA chapter explains sequence changes. Fractions and simple multiplication are enough for the constructed examples here. No example represents a measured natural population unless explicitly identified as published research.

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By the end, you can

Count allele copies separately from individuals and use Hardy-Weinberg proportions as a conditional comparison.

An allele frequency needs a defined population and denominator

At a locus with variants A and a, genotype AA carries two A copies, Aa carries one of each, and aa carries two a copies. In a model of 100 diploid individuals, there are 200 copies of this locus. The number of A copies is twice the number of AA individuals plus the number of Aa individuals.

For 36 AA, 48 Aa and 16 aa individuals, there are 72 + 48 = 120 A copies and 48 + 32 = 80 a copies. The A frequency is 120/200 = 0.60. The a frequency is 80/200 = 0.40. The genotype proportions are instead 0.36, 0.48 and 0.16, with 100 individuals as their denominator.

This counting rule has a stated scope. Sex-linked loci, organelle genomes, different ploidies and copy-number changes may need different denominators. A sample also estimates a population frequency; two small samples can differ even when the underlying population has not changed.

One genotype contributes two locus copies

100 diploid individuals at one autosomal two-allele locus; no copy-number variation. Individual and locus-copy denominators differ.

Read the diagram in words

Representative two-copy genotypes connect to their weighted population counts and then to an allele-copy total.

Genotype	Individuals	Copies of A	Copies of a
AA	36	72	0
Aa	48	48	48
aa	16	0	32

120 A / 200 copies = 0.60; 80 a / 200 copies = 0.40

Genotype	Individuals	Copies of A	Copies of a
AA	50	100	0
Aa	20	20	20
aa	30	0	60

Total locus copies: 200.

Derive expected genotypes from random gamete union

Call the A frequency p and the a frequency q. For this two-allele locus, p + q = 1. Suppose both contributing gamete pools have p = 0.60 and q = 0.40, and gametes unite independently at random with respect to this locus. An A with A pairing has probability p × p = 0.36.

There are two distinct routes to a heterozygote: A from the first pool with a from the second, or a from the first with A from the second. Each has probability 0.24, so their combined probability is 2pq = 0.48. The aa probability is q × q = 0.16. Together p² + 2pq + q² = 1.

These are Hardy-Weinberg genotype expectations for the specified random-union model. In 100 offspring, the expected counts are 36, 48 and 16. Expected does not mean guaranteed: a finite realised group can depart from those counts. The diagram shows population gamete frequencies, not the gametes of one selected pair of parents.

Two gamete routes make a heterozygote

A population random-union expectation, not an Aa × Aa parent cross and not a guaranteed finite offspring count.

Read the diagram in words

A two-by-two area or pairing grid combines equal allele frequencies in two independently sampled gamete pools.

Allele from the first pool	Allele from the second pool	Genotype	Expected probability
A	A	AA	0.36
A	a	Aa	0.24
a	A	Aa	0.24
a	a	aa	0.16

Aa: 0.24 + 0.24 = 0.48

Genotype	Expected count among 100 offspring
AA	36
Aa	48
aa	16

A benchmark is not an inevitable destination

To retain unchanged allele frequencies across generations in the simple Hardy-Weinberg model, assume Mendelian transmission, no selection at the locus, no mutation or gene flow, and a sufficiently large ideal population that drift is neglected. Random mating with the specified gamete frequencies produces the genotype proportions. Real populations may meet some assumptions more closely than others.

The letters p and q are allele-copy frequencies. They are not automatically proportions with a phenotype. Only under an additional complete-dominance assumption would AA and Aa share an A-associated phenotype; in this example their expected combined proportion would be 0.36 + 0.48 = 0.84, not 0.60.

A genotype mismatch with the benchmark prompts investigation, rather than naming a cause. Nonrandom mating can change genotype proportions without changing allele frequencies when the other relevant assumptions hold. Mixing subpopulations, finite sampling and measurement errors also matter. Conversely, agreement at one locus and time does not prove that no evolutionary processes operate.

Worked example

Same allele frequency, different genotype proportions

Now consider a second constructed group of 100 individuals: 50 AA, 20 Aa and 30 aa. Its A count is 100 + 20 = 120; its a count is 20 + 60 = 80. It has the same p = 0.60 and q = 0.40 as the first group.

Its genotype proportions are 0.50, 0.20 and 0.30, which differ from 0.36, 0.48 and 0.16. Counting allele copies does not force them to be randomly paired. These counts alone do not identify inbreeding, selection or another particular explanation; that would require information about mating, population structure, sampling and other conditions.

Check your understanding

In a constructed group of 25 AA, 50 Aa and 25 aa individuals, what are p and q?
Show reasoning
There are 200 locus copies. A contributes 2 × 25 + 50 = 100 and a contributes 50 + 2 × 25 = 100. Thus p = q = 0.50. The observed genotype proportions also equal the random-union expectations 0.25, 0.50 and 0.25.
Why is the expected heterozygote proportion 2pq rather than pq?
Show reasoning
Two ordered gamete combinations give the same Aa genotype: A then a, and a then A. Their mutually exclusive probabilities add: pq + qp = 2pq.
Does complete dominance make an allele spread, or let p alone tell us the phenotype proportion?
Show reasoning
Neither follows. Dominance describes how the heterozygote expresses a character; reproduction and other processes determine allele-frequency change. Under complete A dominance, the A-associated phenotype proportion is AA plus Aa, requiring genotype information or justified genotype expectations.

What to carry forward

Count copies and organisms separately; Hardy-Weinberg expectations are conditional tools for reasoning, not proof of a population’s mechanism or future.

References

Resources supporting this lesson’s explanations and examples.