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Biology · Meiosis & Patterns of Inheritance

Probability, crosses and interpreting pedigrees

Two parents lack a scored trait, but one offspring has it. A family diagram records that pattern. It does not automatically identify a gene or settle the mode of inheritance. We can reason clearly by assigning a simple model, listing what it assumes, and showing what follows within it.

Begin with the cell-division lesson and Heredity, Variation & Evolution: recognise homologous chromosomes, sister chromatids, a locus and an allele. DNA, Replication & Repair explains how copying makes two sister DNA molecules. Gene Expression helps with the later distinction between transmitting a sequence and expressing a trait. Here, fractions and conditional probabilities are explained alongside the cases. All chromosome-count examples use a specified diploid model; the final lessons introduce extensions rather than requiring advanced genetics beforehand.

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By the end, you can

Use product, sum and conditional reasoning in a specified cross, and distinguish a pedigree's records from the inheritance model used to interpret them.

01

Multiply along a path, add alternative paths

For an Aa × Aa cross with equal segregation, one offspring receives a from each parent with probability (1/2)(1/2) = 1/4. Multiplication works here because the parental contributions are assumed independent. If independence is not justified, one needs the probability of the second event given the first, rather than simply multiplying two marginal probabilities.

The genotype Aa can arise through A from the first parent and a from the second, or the opposite path. These paths cannot both describe the same union, so add their probabilities: 1/4 + 1/4 = 1/2. Adding events that overlap requires care; counting the same outcome twice would inflate the result.

Apply the same logic to repeated offspring under a fixed cross. The model assumes each union is independent of previous unions and that neither parental genotype nor the relevant conditions change. A previous aa offspring therefore does not make the next offspring less likely to be aa. Probability is not a quota being used up.

At least one aa includes three distinct paths

At least one aa includes three distinct pathsA two-level tree branches into aa and not aa for each of two independent unions. Edge probabilities and final product probabilities are printed. The three paths with at least one aa are bracketed together; the neither-aa path remains visibly outside.First independent unionaa1/4Not aa3/4Second independent union1/4aa1/163/4Not aa3/161/4aa3/163/4Not aa9/16At least one aa: 7/16Neither aa: 9/16Multiply probabilities along a pathAdd the three disjoint included paths
Fixed Aa × Aa parental genotypes, equal segregation, independent union and no relevant change or loss between outcomes. The two levels show two separate unions, not two loci in one gamete.
Read the diagram in words

The four ordered paths have probabilities 1/16, 3/16, 3/16 and 9/16. The first three include at least one aa, giving 7/16. Equivalently, subtract the neither-aa probability 9/16 from one.

First independent union
OutcomeProbability of this path
aa1/4
Not aa3/4
Multiply probabilities along a path
First independent union → Second independent unionProbability of this pathGrouped event containing this path
aa → aa1/16At least one aa: 7/16
aa → Not aa3/16At least one aa: 7/16
Not aa → aa3/16At least one aa: 7/16
Not aa → Not aa9/16Neither aa: 9/16
  • First independent union
  • Second independent union
  • At least one aa: 7/16
  • Neither aa: 9/16
  • Multiply probabilities along a path
  • Add the three disjoint included paths
02

Separate observed phenotype from inferred genotype

A pedigree represents family relationships and the traits recorded for its members. Label the scoring age, phenotype criteria and missing information. A shaded symbol means the stated phenotype is present, rather than that every possible molecular explanation is known.

Our fictional diagram assigns a fully penetrant autosomal recessive trait to aa. Parents I-1 and I-2 lack it, while offspring II-1 shows it. Exclude new mutation and alternative causes, and assume the recorded parentage. II-1 must then be aa; each parent contributed a. Because each parent lacks the trait, neither is aa, so both are inferred to be Aa.

Offspring II-2 lacks the trait, so II-2 could be AA or Aa. That ambiguity is part of the answer. Without the supplied recessive model, complete penetrance and other assumptions, the small pattern alone would not prove these genotypes or uniquely distinguish an inheritance mechanism.

A recorded phenotype can leave two genotypes unresolved

A recorded phenotype can leave two genotypes unresolvedA two-generation pedigree uses diamonds for individuals whose sex is unspecified, explicit generation/individual IDs, patterned trait-present symbols and unfilled trait-absent symbols. A labelled parental pair joins a sibship line. Below, an Aa×Aa four-cell grid greys out aa after conditioning on trait absence; the three retained equally probable paths include two Aa.I-1: Parent 1Trait absentAaInferred under the modelI-2: Parent 2Trait absentAaInferred under the modelII-1: Offspring 1Trait presentaaAssigned phenotype ruleII-2: Offspring 2Trait absentAA or AaUnresolved genotypeA with AAA1/3A with aAa1/3a with AAa1/3a with aaa0Sex unspecified; irrelevant to this assigned autosomalmodelSpecified trait presentSpecified trait absentAa inferred under the stated modelII-2: AA or Aaaa excluded by the phenotype informationAmong compatible paths: 1/3 AA and 2/3 Aa
Fictional autosomal recessive binary trait, fully penetrant at the scoring age in the stated environment. No new mutation, alternative cause, missing parentage or genotype-dependent loss. Parental genotypes are inferred under this assigned model, not diagnosed from a small pedigree. Relationship lines record both parents of both offspring; the figure does not identify a real family or establish a diagnosis.
Read the diagram in words

I-1 and I-2 lack the fully penetrant recessive trait; their trait-present child II-1 is aa under the assigned rule. The parents are therefore inferred Aa. II-2 lacks the trait and remains AA or Aa, with conditional probabilities 1/3 and 2/3.

A recorded phenotype can leave two genotypes unresolved
Individual IDFamily roleSpecified phenotypeGenotypeGenotype inference status
I-1Parent 1Trait absentAaInferred under the model
I-2Parent 2Trait absentAaInferred under the model
II-1Offspring 1Trait presentaaAssigned phenotype rule
II-2Offspring 2Trait absentAA or AaUnresolved genotype
A recorded phenotype can leave two genotypes unresolved
ParentsOffspring
I-1: Parent 1; I-2: Parent 2; II-1: Offspring 1; II-2: Offspring 2;
Among compatible paths: 1/3 AA and 2/3 Aa
DescriptionGenotypeConditional probability of this pathRetained
A with AAA1/3✓
A with aAa1/3✓
a with AAa1/3✓
a with aaa0×
  • Sex unspecified; irrelevant to this assigned autosomal model
  • Specified trait present
  • Specified trait absent
  • Aa inferred under the stated model
  • II-2: AA or Aa
  • aa excluded by the phenotype information
  • Among compatible paths: 1/3 AA and 2/3 Aa
03

Update the possibilities when information excludes an outcome

Before observing an offspring's phenotype in Aa × Aa, the genotype probabilities are 1/4 AA, 1/2 Aa and 1/4 aa. Learning that the fully penetrant recessive trait is absent excludes aa. The remaining total probability is 3/4.

Among that remaining group, the probability of Aa is (1/2)/(3/4) = 2/3, and of AA is (1/4)/(3/4) = 1/3. This is conditional probability: the denominator describes only the outcomes still compatible with the information. 'One of two genotypes remains' does not make both equally likely.

Real pedigrees can include reduced penetrance, age-dependent expression, shared environment, different molecular causes or uncertain records. These can change the inference. An unknown outsider's genotype should remain unknown unless relevant evidence or an explicit frequency model is supplied; being unrelated does not prove AA.

Worked example

At least one aa among two independent unions

Under the same Aa × Aa conditions, the chance that one offspring is not aa is 3/4. The chance that neither of two independent offspring is aa is (3/4)(3/4) = 9/16. Its complement gives P(at least one aa) = 1 − 9/16 = 7/16.

We can check by adding disjoint paths. Exactly one aa has two orders, each with probability (1/4)(3/4) = 3/16, giving 6/16. Two aa has probability (1/4)(1/4) = 1/16. Thus 6/16 + 1/16 = 7/16. Simply adding 1/4 + 1/4 would count the both-aa outcome twice.

The 2/3 carrier probability for II-2 uses a different question: that offspring is already known to lack the fully penetrant trait. Do not substitute the conditional 2/3 into a calculation about a future union whose phenotype has not been observed.

Check your understanding

  1. The first offspring of the stated Aa × Aa cross is aa. What is the probability that the next offspring is aa?

    Show reasoning

    It remains 1/4 under the fixed independent-union model. The first outcome neither consumes nor compensates for a Mendelian probability.

  2. Why is II-2's probability of being Aa 2/3 rather than 1/2?

    Show reasoning

    The absence of the fully penetrant recessive trait excludes aa. Renormalising gives (1/2)/(3/4) = 2/3. AA retains the other 1/3.

  3. Does the diagram prove that every real family with two trait-absent parents and a trait-present child has a recessive cause?

    Show reasoning

    No. The inference depends on the assigned model and its exclusions. New variants, expression conditions or other causes can produce different explanations; the recorded pattern is evidence to assess.

What to carry forward

Name the event and the information already known before choosing its probability.

References

Resources supporting this lesson’s explanations and examples.

  1. Biology 2e — Inheritance and probability (2018)Mary Ann Clark, Matthew Douglas and Jung Choi · OpenStaxOpen textbook sectionsBiology 2e — Inheritance and probability (2018)Characteristics, traits and dominanceSegregation and independent assortmentopenstax.org
  2. Introduction to Genetics — Chromosomes, crosses and mapping (2023 adaptation of Open Genetics Lectures)Natasha Ramroop Singh · Thompson Rivers UniversityOpen university textbookIntroduction to Genetics — Chromosomes, crosses and mapping (2023 adaptation of Open Genetics Lectures)DNA amount and chromosome numberAssumptions of a dihybrid ratioConditional pedigree probabilitiesAllele phaseLinkage and recombinant fractionsGenetic maps and their limitsTwo-point and three-point crossesTypes of dominanceBiochemical basis of dominanceGene interactions and phenotype groupingPseudoautosomal regionsopengenetics.pressbooks.tru.ca
  3. Help Me Understand Genetics — Inheritance and expression (Live guide)National Library of Medicine · MedlinePlus GeneticsScientific education guideHelp Me Understand Genetics — Inheritance and expression (Live guide)Reduced penetrance and variable expressivityGenomic imprinting and uniparental disomyHeritabilitymedlineplus.gov