Review all six Supplement inheritance requirements with original cross evidence
All six Supplement outcomes connect the genetic model to labelled predictions and bounded evidence.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-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.
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
Which genotype at the supplied ABO locus expresses both A and B contributions?
What you will learn
- Connect test crosses codominance ABO and sex-linkage predictions
- Keep probability assumptions genotype evidence and assessment scope explicit
Original teaching model. Pedigrees and records are fictional; allele and probability assumptions are stated.
| Feature | Meaning |
|---|---|
| Test cross | Recessive partner; informative offspring and finite-sample limits |
| Codominance and ABO | Both contributions; allele-specific genotype and phenotype mapping |
| Sex linkage | Gene location, X/Y contributions and group-specific probabilities |
| Scope | Self-checked practice; teacher, practical and cumulative assessment pending |
Connect test crosses codominance ABO and sex-linkage predictions
The six Supplement outcomes add test crosses, codominance, ABO blood groups, sex-linked characteristics, red-green colour blindness and genetic diagrams for codominant or sex-linked crosses. A test cross uses a known recessive homozygote to investigate a dominant-phenotype unknown. A recessive offspring identifies heterozygosity within the stated normal model, while an all-dominant finite record alone does not prove homozygosity.
Codominance means both alleles contribute to the heterozygous phenotype. In the common ABO model, IA and IB are codominant and each is dominant over Io. The six genotype combinations give A, B, AB or O. IAIo × IBIo gives all four phenotypes in a labelled 1:1:1:1 ratio; IAIB × IAIB gives A:AB:B = 1:2:1. Build the gamete square before grouping phenotypes.
Keep probability assumptions genotype evidence and assessment scope explicit
A sex-linked gene is located on a sex chromosome. In the usual X-linked recessive red-green model, Y has no corresponding allele at the tracked locus. XNXn × XNY gives expressing probability one half among sons but one quarter among all offspring; XNXn × XnY gives one half across all offspring and permits expressing daughters. Carrier genotype status is not automatically trait expression.
The complete Monohybrid inheritance pathway maps all twelve Core and six Supplement requirements to original teaching and self-checked guided, independent and application practice. Expected probabilities require the stated allele, expression and transmission assumptions; finite observations do not guarantee the exact model ratios. Variation has its own teaching pathway. Adaptive features has its own pathway; selection and the wider curriculum need their separate sequences, so finishing this pathway does not establish whole-qualification mastery or a certified examination grade.
Sources for this lesson
Pause and explain
What must accompany a sex-linked percentage?
Worked example
Use genotype-to-phenotype reasoning to compare IAIo × IBIo and XNXn × XNY, then state the limits.
Show the worked solution
- The ABO cross combines IA/Io with IB/Io into four genotypes with A:B:AB:O = 1:1:1:1 in the common model.
- The X-linked cross combines XN/Xn with XN/Y. Only XnY expresses, giving 1/2 among sons and 1/4 among all offspring. Both conclusions require the supplied expression and transmission rules.
Answer Use the chain genotype → gamete → combination → expression key → labelled ratio, then state the evidence and denominator limits.
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.
An unknown pigment parent crossed with tt has a tt offspring. Infer its genotype under the stated rule, and explain why an all-pigment record would give weaker certainty.
Give me a hint
Combine its dominant phenotype with the transmitted t.
Compare my reasoning
- Assign Tt from T expression and t transmission.
- Compare a finite all-pigment sample compatible with TT or Tt.
The unknown parent is Tt within the normal model. A finite all-pigment sample would remain compatible with both TT and Tt, because the heterozygote can transmit T in every recorded event by chance.
Look for these in your work
- Correct informative Tt inference
- Correct all-pigment evidence limit
Give gamete types and the labelled phenotype ratio for IAIo × IBIo, then state why phenotype labels A and B alone do not establish this exact cross.
Give me a hint
The genotype tells you whether each parent carries Io.
Compare my reasoning
- List IA/Io and IB/Io gametes.
- Map the four combinations to A, B, AB and O.
- Keep parental phenotype-genotype alternatives explicit.
Gametes IA or Io, and IB or Io; A:B:AB:O = 1:1:1:1 under the common model. Phenotypes A and B alone also permit homozygous parents, so these heterozygous genotypes must be supplied or supported by evidence.
Look for these in your work
- Correct gametes and labelled ratio
- Retains genotype ambiguity from phenotype alone
For expected 80 from XNXn × XNY, calculate expressing sons, expression among sons and expression among all offspring. State what pathway completion does not establish.
Give me a hint
One expressing position among four, with two son positions overall.
Compare my reasoning
- Calculate twenty expressing sons among forty total sons.
- Calculate 20/40 and 20/80.
- Separate mapped teaching from assessment and wider curriculum completion.
Expected expressing sons 20; expression among sons 50% and among all offspring 25%. Mapped teaching does not establish human teacher approval, supervised practical competence, cumulative assessment, whole-qualification mastery or completion of the wider curriculum.
Look for these in your work
- Correct counts and denominator-specific percentages
- Names assessment and wider-curriculum limits
Common mistakes
- Inferring an unknown genotype with certainty from an all-dominant finite sample
- Using phenotype labels in place of gamete alleles
- Treating this teaching pathway as teacher approval or complete assessment coverage
What can you explain now?
What must accompany a sex-linked percentage?
Compare with the explanation
The parental cross and denominator group
The same expressing category can have different fractions among sons and all offspring.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Without your notes, connect test crosses codominance abo and sex-linkage predictions and keep probability assumptions genotype evidence and assessment scope explicit. 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.