SCIENCE · INVESTIGATE AND EXPLAIN

Build an inheritance prediction from gametes

Combine parental allele probabilities before applying a dominance assumption.

Learning outcome: Calculate genotype probabilities and distinguish genotype, phenotype and allele frequency.

Starting recap

AA and aa are homozygous; Aa is heterozygous. In this one-locus model, an Aa parent contributes A or a with equal probability.

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Try

Aa × Aa offers four equally weighted gamete-pair slots: AA, Aa, aA and aa. The middle two have the same genotype label.

Open the experiment

Predict

Predict genotype and phenotype probabilities for Aa × aa under complete dominance of A.

Your prediction and reason

Experiment

Compare Aa × Aa, Aa × aa and AA × aa. Trace every cell in the probability grid and capture two crosses that have different genotype distributions.

Combine parental gamete probabilities

Worked teaching model · synthetic values · One autosomal locus has alleles A and a. Each parent contributes one allele with equal segregation, fertilization is random, and offspring outcomes are independent with no selection or mutation. Complete dominance is assumed only for the displayed phenotype prediction.

Parent 1 genotype: Aa Parent 2 genotype: Aa

Reset keeps your prediction, captures and practice history.

Four equally weighted gamete-pair slots

Each square has probability 1/4; repeated labels are combined
Parent 1 ↓ / Parent 2 →	A (slot 1)	a (slot 2)
A (slot 1)	AA	Aa
a (slot 2)	Aa	aa

A homozygous parent repeats its allele in both slots. Two heterozygote routes, A/a and a/A, both count as Aa. This grid predicts probabilities rather than ordering actual offspring.

Calculated values on the same fixed scale
Row	Quantity	Value (%)
1	AA offspring probability	25
2	Aa offspring probability	50
3	aa offspring probability	25
4	A-dominant phenotype probability	75

Aa × Aa: AA 25%, Aa 50%, aa 25%.

Multiply the independent parental gamete probabilities, then add the two routes to Aa. These are probabilities per offspring, not a promise that every group of four has the displayed ratio. Dominant does not mean more frequent, stronger or more beneficial.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Start with parental contributions

For Aa × Aa, each parent supplies A with probability 1/2 and a with probability 1/2. Independence gives probability 1/4 for each paired contribution: A/A, A/a, a/A and a/a. The two mixed routes both yield Aa, so genotype probabilities are 1/4 AA, 1/2 Aa and 1/4 aa.

Under the separate assumption of complete dominance, AA and Aa share the A-dominant phenotype, giving phenotype probabilities 3/4 and 1/4. Genotype describes allele composition; phenotype describes an observed characteristic. Dominance does not mean an allele is common or advantageous. Background: OpenStax, inheritance and traits.

Change a parent and reconstruct the grid

For Aa × aa, the second parent always supplies a. The first supplies A or a equally, yielding 1/2 Aa and 1/2 aa with no AA. For AA × aa, every paired contribution is A/a, so all modeled offspring are Aa.

The grid uses two equally weighted slots for each parent. A homozygous parent has the same allele in both slots; duplicated labels do not create a second allele type. Every square has probability 1/4, and matching genotype outcomes are added.

Use phenotype evidence carefully

An individual with the A-dominant phenotype could be AA or Aa. Crossing it with aa distinguishes their probability predictions: AA predicts all Aa; Aa predicts half Aa and half aa. A small sample containing only dominant-phenotype offspring does not prove the unknown parent is AA, because chance can produce that outcome under Aa too.

The model covers one autosomal locus, equal segregation and independent outcomes. It omits linkage, incomplete dominance, multiple genes, selection and environmental influences. Use it to derive a conditional prediction, then state which assumptions require evidence in a real inheritance study.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain why an A-dominant phenotype does not distinguish AA from Aa. Propose a cross that could help distinguish them and identify the sampling limitation.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Count chromosome sets separately from DNA copies Next: Compare offspring counts with an inheritance model

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