Lesson 2 of 7 / Inheritance and genetic problems
Dihybrid crosses and test crosses
When is a 9:3:3:1 ratio justified?
In this lesson: Solve independent dihybrid and test-cross problems with stated assumptions.
About 8 min
The key ideaGenerate gametes from the model first; independent assortment and complete dominance are assumptions behind familiar ratios.
Build from gametes
Which gamete combinations produce this class?
AaBb x AaBb
16 of 16 equally likely combinations are shown: probability 16/16.
Read every combination as text
- A B with A B gives AA BB: A_B_
- A B with A b gives AA Bb: A_B_
- A B with a B gives Aa BB: A_B_
- A B with a b gives Aa Bb: A_B_
- A b with A B gives AA Bb: A_B_
- A b with A b gives AA bb: A_bb
- A b with a B gives Aa Bb: A_B_
- A b with a b gives Aa bb: A_bb
- a B with A B gives Aa BB: A_B_
- a B with A b gives Aa Bb: A_B_
- a B with a B gives aa BB: aaB_
- a B with a b gives aa Bb: aaB_
- a b with A B gives Aa Bb: A_B_
- a b with A b gives Aa bb: A_bb
- a b with a B gives aa Bb: aaB_
- a b with a b gives aa bb: aabb
Green = selected class. Each locus segregates normally; the two loci assort independently and offspring classes have equal viability in this model. An underscore means either allele can fill that position. IA, IB and IO name ABO alleles. XR and Xr mean an X chromosome bearing R or r; Y is a chromosome, not a recessive allele.
Explanation
For AaBb with independently assorting loci, gametes AB, Ab, aB and ab each have probability 1/4. In AaBb x AaBb, combining the gametes gives 16 equally likely cells in a Punnett square. Group genotypes into phenotypes only after applying the stated dominance relationships.
With complete dominance at both loci and no gene interaction affecting the classes, A_B_, A_bb, aaB_ and aabb occur in expected proportions 9/16, 3/16, 3/16 and 1/16. The underscore means either allele can occupy the second position, not an unknown extra gene.
A test cross pairs an individual showing dominant phenotypes with a homozygous recessive tester. For AaBb x aabb under independent assortment, the offspring directly reveal the four gamete types from the heterozygote and are expected in a 1:1:1:1 ratio. An all-dominant small sample does not prove homozygosity with certainty.
Use multiplication for independent events: P(aa and bb) = 1/4 x 1/4 = 1/16 in the heterozygote cross. Linkage or selection among offspring can invalidate simple independence assumptions. Expected ratios describe probabilities; observed small samples need not match exactly.
Step by step
- 1
List gametes with probabilities
Do not jump directly to a memorised ratio.
- 2
Combine parental contributions
Use a square or independent probabilities.
- 3
Group the requested phenotypes
Check whether dominance or interaction changes the classes.
Worked example
Work through the evidence
In AaBb x aabb with independent loci, what fraction is aaB_?
One way to explain it
1/4: the heterozygote contributes aB with probability 1/4, while the tester always contributes ab.
Why this answer works
- The tester contributes no uncertainty at either locus.
- aaB_ is aaBb in this particular cross.
Is this true? "Every cross involving two genes gives 9:3:3:1."
That ratio needs a specific parental cross and genetic assumptions. Test crosses and linked or interacting loci differ.