Lesson 7 of 8 / Inheritance and phenotype
Use a recessive tester to reveal gametes
How can offspring reveal an unknown genotype?
In this lesson: Solve single- and two-locus test crosses and interpret finite offspring samples.
About 5 min
The key ideaA test cross uses a homozygous recessive tester so offspring phenotypes reveal the alleles contributed by the unknown parent under the stated dominance model.
Explore the idea
A recessive tester reveals each contribution
AaBb x aabb
The tester contributes only ab. The four expected classes are equal under independent assortment and complete dominance. Row and column labels are gametes, not parent genotypes.
| Gamete | ab |
|---|---|
| AB | AaBb |
| Ab | Aabb |
| aB | aaBb |
| ab | aabb |
Explanation
An individual with a dominant phenotype may be homozygous or heterozygous. Crossing it with a homozygous recessive tester distinguishes the possibilities: AA x aa gives all Aa, while Aa x aa gives an expected 1 Aa : 1 aa.
For two independently assorting loci, AaBb x aabb gives AB, Ab, aB and ab gametes from the unknown parent and only ab from the tester. Offspring are AaBb, Aabb, aaBb and aabb in an expected 1:1:1:1 ratio. Each phenotype reflects one of the unknown parent's gametes.
An observed recessive offspring demonstrates that the unknown parent supplied the recessive allele in the simple model. However, an all-dominant small sample does not prove homozygosity: a heterozygote can produce several dominant offspring by chance.
Use the exact cross rather than memorising a fixed ratio. A parent heterozygous at only one of the two loci gives a different distribution. State independent assortment and complete dominance when these are needed for phenotype-based inference.
Step by step
- 1
Choose the tester
Use a homozygous recessive individual at the relevant loci.
- 2
List the possible unknown gametes
Each proposed genotype makes a particular gamete distribution.
- 3
Match offspring and qualify
A finite sample supports an inference but may not prove a missing class has zero probability.
Worked example
Five dominant offspring
An unknown A_ parent crossed with aa produces five dominant offspring. Does this prove AA?
One way to explain it
No. If the unknown parent is Aa, each offspring has probability 1/2 of the dominant phenotype; five in a row can occur with probability (1/2)^5 = 1/32. More data can strengthen an inference, but the small sample is not proof.
Why this answer works
- Keep a possible heterozygous explanation.
- Quantify the chance under that explanation.
Is this true? "The tester should be homozygous dominant to expose recessive alleles."
A dominant tester can mask the allele contributed by the unknown parent. A homozygous recessive tester makes that contribution visible under the model.