Mendelian genetics: crosses, testcrosses and probability

A familiar ratio is the end of a calculation, not a substitute for checking the cross. Start by defining the alleles and the inheritance model. The examples below assume ordinary Mendelian segregation, equal gamete probabilities and no differences in offspring survival unless stated otherwise.

Genotype and phenotype are different outputs

Let A be completely dominant to a. Each Aa parent produces A and a gametes with probability ½. Combining them gives:

Gamete from one parent	A from the other	a from the other
A	AA	Aa
a	Aa	aa

Each square has probability ¼. AA and Aa share the dominant phenotype in this model, so three squares contribute to that phenotype. With incomplete dominance, the heterozygote has a distinct phenotype and the same genotype distribution instead gives three phenotype classes.

Dominant does not mean common, beneficial or more likely to be transmitted by a heterozygote. It describes the heterozygote’s phenotype relative to the homozygotes. OpenStax distinguishes these inheritance patterns.

What a testcross can reveal

Cross an individual with a dominant phenotype but unknown genotype with aa. If the unknown is AA, all offspring are Aa under this model. If it is Aa, the expected offspring ratio is 1 Aa : 1 aa.

Observing recessive offspring therefore reveals that the unknown parent supplied a. However, a small sample containing only dominant offspring does not prove AA: an Aa parent can produce that sample by chance.

For example, with an Aa × aa cross, the probability that four independent offspring all show the dominant phenotype is (½)⁴ = 1/16. That is a probability of the observation assuming Aa, not the probability that the parent is Aa given the observation. The latter requires additional information.

Two genes: check independence before multiplying

For AaBb × AaBb, independent assortment gives AB, Ab, aB and ab gametes, each with probability ¼. With complete dominance at each locus and no interaction between their effects, the phenotype ratio is 9:3:3:1.

You can calculate a particular outcome without drawing all sixteen squares. For Aabb, the probability is:

P(Aa) × P(bb) = ½ × ¼ = 1/8.

For A–bb, where A– includes either AA or Aa, it is ¾ × ¼ = 3/16. The difference between an exact genotype and a phenotype-compatible genotype changes the answer.

Linked genes may not assort independently; recombination information can then be necessary. See OpenStax’s discussion of segregation and assortment, and connect the gamete step to meiosis.

“At least one” is often easier through its opposite

For an Aa × Aa cross, each offspring has probability ¼ of being aa. Assuming independent offspring, the probability that none of three offspring is aa is (¾)³ = 27/64. Therefore, the probability of at least one aa offspring is 1 − 27/64 = 37/64.

This is an original numerical example of complement reasoning. OpenStax explains the probability rules used in genetic crosses.

The traps to check

• Do not use 3:1 for every one-gene cross: Aa × aa has different parents.
• Do not treat a predicted ratio as a guaranteed count in a small family.
• Do not multiply probabilities across linked loci without justification.
• Do not confuse “at least one” with “exactly one”.

If a result looks familiar, write the assumptions beside it before accepting it. This makes a ratio something you can explain, rather than something you hope matches the question.