Lesson 3 of 8 / Inheritance and phenotype
Build a two-gene cross from gametes
When is 9:3:3:1 justified?
In this lesson: Solve dihybrid crosses with explicit independent-assortment and dominance assumptions.
About 6 min
The key ideaList complete gametes before combining them. AaBb can produce AB, Ab, aB and ab equally when the loci assort independently.
Explore the idea
Build and classify the 16 combinations
AaBb x AaBb
Assumes independent assortment, complete dominance at both loci, and equal gamete/offspring viability. Row and column labels are gametes, not parent genotypes.
| Gamete | AB | Ab | aB | ab |
|---|---|---|---|---|
| AB | AABB | AABb | AaBB | AaBb |
| Ab | AABb | AAbb | AaBb | Aabb |
| aB | AaBB | AaBb | aaBB | aaBb |
| ab | AaBb | Aabb | aaBb | aabb |
Explanation
A dihybrid problem follows two loci. For AaBb, each gamete contains one allele from the A locus and one from the B locus. With independent assortment, the four types AB, Ab, aB and ab each have expected probability 1/4. AA and Bb are not complete gametes for this two-locus problem.
Crossing AaBb with AaBb gives 16 equally weighted gamete combinations. At each locus, complete dominance gives a 3/4 dominant-phenotype and 1/4 recessive-phenotype probability. Independence allows the two locus probabilities to be multiplied.
The joint phenotypes are A_B_ = 9/16, A_bb = 3/16, aaB_ = 3/16 and aabb = 1/16. The underscore means either allele can occupy that second position while the dominant allele is present. This gives the familiar 9:3:3:1 expectation only under the stated model.
Small observed samples need not match the exact ratio. Different parental genotypes, dominance patterns, linkage or viability effects can also change expectations. H1 requires solving the stated crosses, not assuming every two-gene question has the same answer.
Step by step
- 1
Define each locus
State which phenotype is dominant and whether independent assortment is assumed.
- 2
List complete gametes
For a dihybrid heterozygote, combine one allele from each locus.
- 3
Combine and classify
Use a genetic diagram and then translate genotypes into the requested joint phenotype.
Worked example
One dominant, one recessive
For AaBb x AaBb with independent assortment and complete dominance, what is the chance of A_bb?
One way to explain it
(3/4 for A_) x (1/4 for bb) = 3/16. The A locus may be AA or Aa, while the B locus must be bb.
Why this answer works
- Specify the acceptable genotypes at each locus.
- Multiply only because the model gives independence.
Is this true? "A dihybrid gamete can be Aa or Bb."
A gamete must have one allele from each locus, such as AB or ab, not both alleles from just one locus.