Predict the ABO phenotypic ratio from two AB model parents
Codominance makes AB a distinct category, so label the one-to-two-to-one ratio carefully.
Section B16.3 · Monohybrid inheritance
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Recall chromosomes, genes and alleles, meiosis, haploid gametes and fertilisation restoring two chromosome sets.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Are IAIB and IBIA different unordered genotype combinations?
What you will learn
- Construct the IAIB crossed with IAIB square
- Explain a one-to-two-to-one ratio without an O outcome
Original teaching model. Pedigrees and records are fictional; allele and probability assumptions are stated.
| Feature | Meaning |
|---|---|
| Parents | IAIB × IAIB |
| Genotypes | IAIA, IAIB, IAIB, IBIB |
| Phenotypes | A:AB:B = 1:2:1 |
| O outcome | Not predicted because neither parent supplies Io |
Construct the IAIB crossed with IAIB square
Each AB parent has genotype IAIB in the common model and can contribute IA or IB at the gene. The square gives IAIA, IAIB, IAIB and IBIB, normalising the order of the equivalent IBIA record to IAIB. A repeated mixed genotype occupies two equal-probability positions; it is one genotype category with total probability one half.
The phenotype key gives A from IAIA, AB from the two IAIB positions, and B from IBIB. The labelled phenotype ratio A:AB:B is 1:2:1. Do not report a 3:1 complete-dominance grouping or combine AB with A: both allele contributions form the distinct AB phenotype in this rule.
Explain a one-to-two-to-one ratio without an O outcome
O is not predicted because neither parent has Io to contribute. For forty expected offspring under the equal-transmission model, the expected phenotype counts are ten A, twenty AB and ten B. These counts sum to forty and reflect the one-quarter, one-half and one-quarter probabilities. The model does not require those exact counts in every finite observed sample.
A ratio’s order matters. A:AB:B = 1:2:1 is different in appearance from A:B:AB = 1:1:2, but both describe the same probabilities if the labels are kept. Follow the question’s requested category order and verify the genotype-to-phenotype mapping before simplifying. The supplied model is an inheritance exercise, not a clinical interpretation of a person’s blood group.
Sources for this lesson
Pause and explain
Which ratio follows IAIB × IAIB in A:AB:B order?
Worked example
Predict the common-model phenotypes for IAIB × IAIB and explain whether O is possible.
Show the worked solution
- Combine IA/IB with IA/IB to obtain IAIA, IAIB, IAIB and IBIB.
- Assign A, AB, AB and B, giving A:AB:B = 1:2:1. Neither parent supplies Io, so the IoIo genotype required for O cannot be formed.
Answer IAIB × IAIB gives A:AB:B = 1:2:1; no Io is available, so O is not predicted in the common normal model.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
List the four square positions from IA/IB rows and IA/IB columns, then assign phenotypes.
Give me a hint
Combine one allele from each contribution and normalise order.
Compare my reasoning
- List IAIA, IAIB, IAIB and IBIB.
- Assign A, AB, AB and B.
IAIA → A; IAIB → AB; IAIB → AB; IBIB → B.
Look for these in your work
- Correct four positions
- Correct phenotype mapping
For 80 expected offspring from two AB model parents, calculate A, AB and B counts in that order.
Give me a hint
Use one quarter, one half and one quarter.
Compare my reasoning
- Calculate eighty times each probability.
- Check the expected counts sum to eighty.
Expected A 20, AB 40 and B 20, with A:AB:B = 1:2:1.
Look for these in your work
- Correct twenty forty twenty
- Expected wording and labelled ratio
Explain why O is excluded and express the same prediction in A:B:AB order.
Give me a hint
Look for Io, then reorder the existing categories rather than change probabilities.
Compare my reasoning
- Identify that neither parent contributes Io.
- Reorder one-quarter, one-quarter and one-half categories.
O is not predicted because neither IAIB parent has Io. In A:B:AB order the phenotype ratio is 1:1:2, the same probabilities with different labels.
Look for these in your work
- Correct absence of Io explanation
- Correct reordered labelled ratio
Common mistakes
- Treating IBIA as a fourth different genotype
- Grouping AB into a dominant A category
- Reporting numbers without the phenotype order
What can you explain now?
Which ratio follows IAIB × IAIB in A:AB:B order?
Compare with the explanation
1:2:1
One A position, two AB positions and one B position occur.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Without your notes, construct the iaib crossed with iaib square and explain a one-to-two-to-one ratio without an o outcome. Use the supplied allele key and state what the evidence can establish.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.