Calculate allele frequencies and test equilibrium claims

Mechanism before a label

For a diploid two-allele locus, allele frequency p equals (2AA+Aa)/(2N), and q=1−p. These counts do not require Hardy–Weinberg equilibrium. The two allele frequencies sum to one because only these two alleles are included.

Random union of gametes gives genotype expectations p², 2pq and q². Maintaining unchanged frequencies across generations additionally requires the idealized absence of selection, mutation and migration and negligible drift. A fit at one locus does not prove every population assumption.

Observed genotype departures can reflect population mixing, nonrandom mating, selection or technical errors. Inferring q by taking the square root of a recessive phenotype frequency requires equilibrium plus a justified phenotype-to-genotype mapping; it is not a universal allele counter.

Predict, compare and calculate

The following record is an original synthetic teaching dataset. Its assigned conditions are not a claim that an actual organism was measured. Write your prediction before revealing the changed condition.

Observed diploid genotypes	Value (individuals)
AA	36
Aa	48
aa	16

N=100, one two-allele locus, complete genotype calls.

Your prediction before comparison

This scratch prediction is not saved automatically. Copy it to the lesson notebook to retain it.

Evidence conditionReference comparisonChanged condition / follow-up

Calculate allele frequencies and test equilibrium claims

Supplied synthetic evidence: N=100, one two-allele locus, complete genotype calls.

Observed diploid genotypes: reference condition

Quantity	Value	Unit
AA	36	individuals
Aa	48	individuals
aa	16	individuals

Task: In a population with AA=18,Aa=24,aa=8, calculate p directly and expected Aa under random union.

An exact synthetic match is not proof of absence of evolutionary forces.

Compare with the conditional interpretation

p=(72+48)/200=0.6; expected counts 36,48,16 match this constructed packet.

Both packets contain120 A copies among200. In the changed packet the observed 20 heterozygotes differ from 2×0.6×0.4×100=48; genotype change and allele-frequency change are not equivalent statements.

Case index for this investigation

0. Connect heritable variation to relative reproductive success
1. Distinguish selection by people from inheritance and trade-offs
2. Separate drift, gene flow and mutation from selection
3. Calculate allele frequencies and test equilibrium claims
4. Combine fossils, homology and molecular evidence
5. Read a longitudinal evolution study without inventing observations
6. Read branching relationships rather than tip order
7. Explain reproductive barriers and divergence
8. Relate variation, changing environments and extinction
9. Separate origin-of-life hypotheses from ancestry evidence

No numerical confidence, laboratory credit or independent mastery is inferred from opening a supplied case. Save your first prediction and comparison in the notebook.

Reset this unsaved comparison

Test your model prediction

Scratch inputs are not saved or graded. Retain your prediction and calculations in the lesson notebook. No simulation output is a measured organism or population.

AA countAa countaa count

N=100; p=(2×36+48)/(2×100)=0.6000; q=0.4000.

Observed and random-union expectations

Genotype	Observed	Expected
AA	36	36.000
Aa	48	48.000
aa	16	16.000

Change to50,20,30. Does p change? Do genotype proportions change? A departure does not uniquely identify selection; a match does not prove all equilibrium assumptions.

Reset practice

Worked reasoning

Both packets contain120 A copies among200. In the changed packet the observed 20 heterozygotes differ from 2×0.6×0.4×100=48; genotype change and allele-frequency change are not equivalent statements.

p=(72+48)/200=0.6; expected counts 36,48,16 match this constructed packet.

Apply to a new case

In a population with AA=18,Aa=24,aa=8, calculate p directly and expected Aa under random union.

Compare your reasoning with the worked response

p=(36+24)/100=0.6. Expected heterozygotes=2×0.6×0.4×50=24. This calculation describes one locus and does not demonstrate that all equilibrium conditions hold.

This public response supplies feedback, not a score for independent reasoning. Record which part of your prediction changed and why.

Evidence and investigation boundary

An exact synthetic match is not proof of absence of evolutionary forces. The deficit does not uniquely identify inbreeding or selection.

For a supervised investigation, predeclare the question, biological unit of replication, treatment/comparison assignment, quantities and units, exclusion rules, graph and uncertainty method. Keep the raw records and the first prediction separate from later revisions. A simulated case does not supply laboratory hours or establish AP authorization. No physical manipulation is required by this lesson.

Source and continuation

Framework topics: 7.5. Current AP Biology framework; supporting scientific reference. Sources checked 30 October 2026; all teaching wording, questions and numbers here are original. Return to AP Biology to compare this mechanism with another scale of organization.