Lesson 4 of 8 / Inheritance and phenotype
Two genes with an intermediate phenotype
How does one changed dominance rule alter the cross?
In this lesson: Solve a dihybrid cross involving incomplete dominance or codominance.
About 6 min
The key ideaSolve the segregation at each locus, then apply its own phenotype rule. An intermediate or codominant locus produces three distinguishable genotype classes.
Explore the idea
Keep pink as a separate phenotype
RWTt x RWTt
RR red, RW pink, WW white. T is dominant for tall growth; the two loci assort independently. Row and column labels are gametes, not parent genotypes.
| Gamete | RT | Rt | WT | Wt |
|---|---|---|---|---|
| RT | RRTT | RRTt | RWTT | RWTt |
| Rt | RRTt | RRtt | RWTt | RWtt |
| WT | RWTT | RWTt | WWTT | WWTt |
| Wt | RWTt | RWtt | WWTt | WWtt |
Explanation
In an original flower model, RR is red, RW is pink and WW is white, so colour shows incomplete dominance. At a separate independently assorting locus, T is dominant for tall growth and t gives short growth. Consider RWTt x RWTt.
Each parent makes RT, Rt, WT and Wt gametes with equal expected probability. Colour gives 1/4 RR, 1/2 RW and 1/4 WW. Height gives 3/4 T_ and 1/4 tt. These probability distributions differ, so replacing everything with 9:3:3:1 would lose the pink category.
The six joint phenotype probabilities are red tall 3/16, pink tall 6/16, white tall 3/16, red short 1/16, pink short 2/16 and white short 1/16. Their sum is 16/16, a useful check that all mutually exclusive outcomes have been counted.
If RW instead showed both red and white patches, the same allele combinations would produce a codominant phenotype. The genetic segregation can be unchanged while the phenotype labels change. Always read the problem's description of the heterozygote.
Step by step
- 1
Keep the two phenotype maps separate
RR/RW/WW give three colours, while TT/Tt share the tall phenotype.
- 2
List gametes
Each gamete carries a colour allele and a height allele.
- 3
Multiply and check
Compute joint probabilities under independence and check they sum to one.
Worked example
Find pink and short
What fraction of RWTt x RWTt offspring are expected to be pink and short?
One way to explain it
P(RW) x P(tt) = 1/2 x 1/4 = 1/8, or 2/16. RW determines pink, while tt is required for short growth.
Why this answer works
- An intermediate colour requires the heterozygote.
- Short height requires the recessive homozygote.
- Multiply the two independent probabilities.
Is this true? "Incomplete dominance prevents allele segregation."
Alleles still segregate. The change is how genotypes appear as phenotypes, not whether gametes carry separate alleles.