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IGCSE · Cambridge (CIE) · 0973

Explain pure breeding from identical homozygous parents

Pure breeding requires the same homozygous pair in both parents at the tracked gene.

Monohybrid inheritance · Cambridge IGCSE 0973 · B16.3
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

Which genotype is homozygous?

What you will learn

  • Predict the one-gene outcome of identical homozygous parents
  • Limit pure breeding to the stated characteristic and model

Original teaching model. Pedigrees and records are fictional; allele and probability assumptions are stated.

Explain pure breeding from identical homozygous parentsPure breeding requires the same homozygous pair in both parents at the tracked gene.Identical TT parentsT + T gives TTIdentical tt parentst + t gives ttDifferent homozygous parentsTT × tt gives heterozygous TtScopeThe tracked gene andcharacteristic, not the entiregenome
Pure breeding requires the same homozygous pair in both parents at the tracked gene.
Allele, phenotype and evidence key
FeatureMeaning
Identical TT parentsT + T gives TT
Identical tt parentst + t gives tt
Different homozygous parentsTT × tt gives heterozygous Tt
ScopeThe tracked gene and characteristic, not the entire genome

Predict the one-gene outcome of identical homozygous parents

Two identical homozygous individuals that breed together are pure-breeding for the stated characteristic in the model. At the gene followed here, TT crossed with TT gives offspring with TT, while tt crossed with tt gives offspring with tt. Each parent’s normal gametes carry only the allele form in its matching pair, so combining the contributions reproduces that allele pair.

Identical refers to the relevant homozygous genotype, not necessarily every feature or every gene in the parents. Two TT plants can differ in height because of another gene or growing conditions. The predicted offspring allele pair at the T gene remains TT under the supplied normal transmission rule. Keep the pure-breeding conclusion tied to the specific gene and characteristic.

Limit pure breeding to the stated characteristic and model

Do not substitute two homozygous parents of different genotypes. TT crossed with tt gives Tt rather than reproducing either parental genotype. Each parent is homozygous, but the parents are not identical at this gene; their offspring are heterozygous. Thus having two homozygous parents is not enough without the same allele pair in both.

Pure breeding describes the model’s inherited trait pattern, not a guarantee of identical whole organisms or fixed appearance under every environment. The model assumes normal transmission, the stated expression rule and no change to the tracked alleles. A shared dominant phenotype does not prove the parents are TT: the phenotype could also belong to Tt. Establish the genotype condition before applying the rule.

Sources for this lesson

Pause and explain

Which pair meets the identical-homozygous condition?

Put the idea to work

Worked example

Compare TT × TT, tt × tt and TT × tt under normal one-gene transmission.

Show the worked solution
  1. For TT × TT each gamete contributes T; for tt × tt each contributes t. The identical homozygotes reproduce their allele pair.
  2. For unlike homozygotes TT × tt, one parent contributes T and the other t. The offspring pair is Tt, which is different from either parental pair.

Answer TT × TT gives TT; tt × tt gives tt. TT × tt gives Tt and does not meet the identical-homozygote condition.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

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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.

1 · Guided

List the gamete allele and offspring genotype for TT crossed with TT.

Give me a hint

Each parent has only T to contribute at this gene.

Compare my reasoning
  1. Assign T to each parent’s gametes.
  2. Combine T with T.

Each parent contributes T at this gene, giving TT offspring under the normal model.

Look for these in your work

  • Correct gamete T from each parent
  • Correct offspring TT
2 · Independent

Compare tt crossed with tt and TT crossed with tt. Which reproduces the identical parental genotype at this gene?

Give me a hint

Check both parents before combining contributions.

Compare my reasoning
  1. Combine t with t in the first cross.
  2. Combine T with t in the second.
  3. Apply the identical-parent condition.

tt × tt reproduces tt. TT × tt gives Tt and does not reproduce either homozygous parent genotype; its parents are not identical at this gene.

Look for these in your work

  • Correct tt and Tt outcomes
  • Distinguishes identical from different homozygous parents
3 · Transfer

Two pigment-producing plants are proposed as pure-breeding parents. Under the rule that TT and Tt both show pigment, explain what genotype condition must first be established.

Give me a hint

The same phenotype need not identify the same allele pair.

Compare my reasoning
  1. List TT and Tt as possible pigment genotypes.
  2. Require the same homozygous genotype in both parents.

Both must be established as TT at this gene for the stated identical-homozygote pure-breeding rule. Pigment alone leaves TT or Tt possible and does not establish that condition.

Look for these in your work

  • Names TT condition in both parents
  • Explains phenotype ambiguity

Common mistakes

  • Calling any two homozygous parents identical
  • Inferring TT from dominant appearance alone
  • Extending pure breeding at one gene to identical whole organisms
Recall without your notes

What can you explain now?

Which pair meets the identical-homozygous condition?

Compare with the explanation

tt crossed with tt

Both parents have the same homozygous genotype.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Without your notes, predict the one-gene outcome of identical homozygous parents and limit pure breeding to the stated characteristic and model. 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.

Keep learning Sciences — Co-ordinated (9-1)

Course scope and assessment

This 9–1 pathway maps all 152 detailed subject-content sections in the published 0973 syllabus for exams in 2028 and 2029. The course map follows the same 2028–2029 section codes as Cambridge's A*–G 0654 syllabus, while keeping 0973's grading and entry rules separate. It adapts lessons from the related single-science courses and excludes clear single-science-only topics. A 0973-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The separate 2025–2027 syllabus has not been reconciled here. Cambridge lists 0973 only in a limited number of administrative zones: June in zone 3 and centres in Oman and the UAE; November in zone 3 only. Confirm entry with your centre.

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