Lesson 3 of 7 / Inheritance and genetic problems
Codominance and multiple alleles in crosses
How do three population alleles fit into a diploid individual?
In this lesson: Solve crosses combining ABO codominance/multiple alleles with another locus.
About 7 min
The key ideaA population can have several alleles, but an individual normally carries two; apply each locus's own expression rules before combining probabilities.
Build from gametes
Combine ABO with an independent locus
IA IO Dd x IB IO dd
8 of 8 equally likely combinations are shown: probability 8/8.
Read every combination as text
- IA D with IB d gives IAIB Dd: AB D_
- IA D with IO d gives IAIO Dd: A D_
- IA d with IB d gives IAIB dd: AB dd
- IA d with IO d gives IAIO dd: A dd
- IO D with IB d gives IBIO Dd: B D_
- IO D with IO d gives IOIO Dd: O D_
- IO d with IB d gives IBIO dd: B dd
- IO d with IO d gives IOIO dd: O dd
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
The simplified ABO model uses IA, IB and IO. IA and IB are codominant: an IA IB individual expresses both A and B antigens. Each is dominant to IO, so IA IO gives group A, IB IO gives group B and IO IO gives group O. Plain-text symbols represent the conventional superscript notation.
Multiple alleles means more than two alternatives exist in the population, not that every person carries all three. Codominance means both specified products or traits are expressed in the heterozygote; it is different from blending into an intermediate product.
For IA IO x IB IO, gametes combine to give IA IB, IA IO, IB IO and IO IO, each with probability 1/4 under the simple model. These are four phenotypes, so a generic 3:1 dominant/recessive ratio is inappropriate.
A second independently assorting locus can be combined by multiplication. If the same parents are Dd and dd at a separate complete-dominance locus, P(D_) = 1/2. Therefore P(group AB and D_) = 1/4 x 1/2 = 1/8. This is a stated genetic model, not a full clinical blood-transfusion compatibility rule.
Step by step
- 1
Use full allele symbols
Phenotype A alone does not specify IA IA or IA IO.
- 2
Solve each locus
Respect codominance at ABO.
- 3
Multiply independent probabilities
State whether the answer is among all offspring or a subset.
Worked example
Work through the evidence
IA IO Dd x IO IO dd: what is the probability of group O with dd?
One way to explain it
P(group O) = 1/2 and P(dd) = 1/2, so the combined probability is 1/4 if the loci assort independently.
Why this answer works
- The O parent always contributes IO.
- The dd parent always contributes d.
Is this true? "Group AB is an intermediate blend in which neither A nor B is expressed."
Both A and B antigen products are expressed in the codominant phenotype.