Derive cross probabilities and distinguish expectation from outcome

Mechanism before a label

An allele is one version of a locus. In an ideal heterozygote Aa, segregation gives each gamete an A or a with probability one half. Combining independent gametes from Aa × Aa yields AA, Aa and aa probabilities one quarter, one half and one quarter.

If A is completely dominant for the scored phenotype under the specified environment, AA and Aa share the dominant category. The resulting 3:1 phenotype ratio is conditional on complete dominance, equal survival and unbiased detection; genotype probabilities do not depend on the visual label chosen.

Use the multiplication rule for independent events and the addition rule for mutually exclusive routes to an outcome. The two routes A from parent1/a from parent2 and a from parent1/A from parent2 both contribute to Aa. Small samples can depart from expected ratios without contradicting segregation.

Read a pedigree as a set of constrained family records

A pedigree represents relationships and scored phenotypes across generations. Conventional symbols distinguish sex categories and shading marks the specified phenotype; shading does not reveal every genotype. State the inheritance model, penetrance, new-mutation assumptions and family relationships before assigning alleles.

Original fictional pedigree in accessible text: Two unshaded parents I-1 and I-2 have three children: unshaded daughter II-1, shaded son II-2 and unshaded son II-3. Under an autosomal, fully penetrant recessive aa-only model with no new mutation, assign the parents and shaded child’s genotypes. Can II-1 be assigned one genotype? Then compare a fully penetrant X-linked recessive model, with the father unaffected and no new mutation. This is not a real family or personal-risk task. OpenStax inheritance rules, checked 31 October 2026.

Check the family-model inference

Autosomal model: both parents must be Aa, shaded II-2 is aa, and unshaded II-1 can be AA or Aa. Conditional on being unshaded, her carrier probability is (1/2)/(3/4)=2/3. X-linked model: the mother must carry the recessive allele and the father supplies a normal X to daughters; an unshaded daughter can still be a carrier. The same small pedigree fits both models. An affected daughter with an unaffected father would contradict the stipulated simple X-linked recessive model, but would not contradict an autosomal Aa×Aa model. Model assumptions, not shading alone, support the distinction.

Construct: draw the pedigree from the text, then draw a 2×2 Aa×Aa Punnett square with one gamete from each parent per axis. For independent AaBb×AaBb loci, multiply the two 3/4 dominant-phenotype probabilities to obtain 9/16 for AB; the other categories are 3/16, 3/16 and 1/16. Linkage, survival differences or a different phenotype rule can invalidate that phenotype ratio without invalidating arithmetic.

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.

Expected genotype	Value (offspring among 160)
AA	40
Aa	80
aa	40

Expected values from Aa × Aa, not simulated random draws.

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

Derive cross probabilities and distinguish expectation from outcome

Supplied synthetic evidence: Expected values from Aa × Aa, not simulated random draws.

Expected genotype: reference condition

Quantity	Value	Unit
AA	40	offspring among 160
Aa	80	offspring among 160
aa	40	offspring among 160

Task: For two independent offspring of Aa × Aa, calculate the probability that exactly one is aa.

Expectations are not a guarantee for a finite observed cohort.

Compare with the conditional interpretation

The dominant phenotype expectation is 120 when dominance is complete.

P(Aa)=P(A then a)+P(a then A)=1/4+1/4=1/2. For two independent Aa × Aa offspring, P(both aa)=1/4 ×1/4=1/16; this is not the probability that one of two is aa.

Case index for this investigation

0. Track homologs and chromatids through meiosis
1. Explain crossing over and assortment without mixing mechanisms
2. Derive cross probabilities and distinguish expectation from outcome
3. Use a testcross to estimate recombination and retain limits
4. Test an inheritance null with chi-square reasoning
5. Distinguish dominance, codominance and multiple alleles
6. Interpret chromosome-linked and cytoplasmic inheritance
7. Separate genotype, environment and their interaction

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

Worked reasoning

P(Aa)=P(A then a)+P(a then A)=1/4+1/4=1/2. For two independent Aa × Aa offspring, P(both aa)=1/4 ×1/4=1/16; this is not the probability that one of two is aa.

The dominant phenotype expectation is 120 when dominance is complete.

Apply to a new case

For two independent offspring of Aa × Aa, calculate the probability that exactly one is aa.

Compare your reasoning with the worked response

There are two mutually exclusive orders: aa then not-aa, or not-aa then aa. Their total is 2×(1/4)×(3/4)=3/8. State the independence assumption.

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

Evidence and investigation boundary

Expectations are not a guarantee for a finite observed cohort. Differential viability would change observed offspring proportions.

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: 5.3. 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.