Lesson 5 of 8 / Inheritance and phenotype
More alleles in the population, two in a person
How do multiple alleles fit a two-gene diagram?
In this lesson: Solve a dihybrid problem involving multiple alleles and codominance.
About 6 min
The key ideaA population can have more than two alleles at a locus, but an ordinary diploid individual has two. Solve that locus before combining it with a second independent locus.
Explore the idea
Four ABO groups, two marker classes
IAiMm x IBimm
IA and IB are codominant, both dominant to i. M/m is an invented independent marker, not a clinical trait. Row and column labels are gametes, not parent genotypes.
| Gamete | IBm | im |
|---|---|---|
| IAM | IAIBMm | IAiMm |
| IAm | IAIBmm | IAimm |
| iM | IBiMm | iiMm |
| im | IBimm | iimm |
Explanation
The ABO blood-group model has three alleles: IA, IB and i. IA and IB are codominant with each other, and each is dominant to i. IAIA or IAi gives group A; IBIB or IBi gives B; IAIB gives AB; ii gives O.
Multiple alleles describes the range in the population, not three copies in one ordinary diploid individual. For IAi x IBi, the ABO offspring genotypes are IAIB, IAi, IBi and ii, each expected at 1/4.
Now add an invented, independently assorting marker locus: M is dominant to m. In the exercise IAiMm x IBimm, the first parent makes IAM, IAm, iM and im gametes; the second makes IBm and im. A four-by-two genetic diagram has eight equally likely combinations.
The marker cross Mm x mm gives 1/2 M_ and 1/2 mm. Thus AB with mm has probability 1/4 x 1/2 = 1/8, as does each specified ABO group with one marker class. The marker is a teaching model, not a real disease or clinical interpretation of blood group.
Step by step
- 1
Define all alleles
State codominance and recessiveness at the multiple-allele locus.
- 2
List complete gametes
Include one ABO allele and one marker allele in every gamete.
- 3
Combine the loci
Use independence to calculate the required joint probability.
Worked example
Blood group O with a dominant marker
For IAiMm x IBimm under the stated model, calculate P(group O and M_).
One way to explain it
Group O requires ii, with probability 1/4. M_ occurs with probability 1/2 from Mm x mm. The joint probability is 1/8.
Why this answer works
- Use the genotype required for O.
- Apply the second-locus cross, not an assumed 3:1 ratio.
- Multiply under the stated independence.
Is this true? "An AB person carries IA, IB and i because ABO has three alleles."
An AB individual carries IA and IB. The third allele exists in the population but need not be present in that person.