Explain pure breeding from identical homozygous parents
Pure breeding requires the same homozygous pair in both parents at the tracked gene.
Section 17.4 · Monohybrid inheritance
Core and Extended; Biology 0610, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked
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.
| Feature | Meaning |
|---|---|
| Identical TT parents | T + T gives TT |
| Identical tt parents | t + t gives tt |
| Different homozygous parents | TT × tt gives heterozygous Tt |
| Scope | The 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.
Pause and explain
Which pair meets the identical-homozygous condition?
Worked example
Compare TT × TT, tt × tt and TT × tt under normal one-gene transmission.
Show the worked solution
- For TT × TT each gamete contributes T; for tt × tt each contributes t. The identical homozygotes reproduce their allele pair.
- 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.
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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 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
- Assign T to each parent’s gametes.
- 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
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
- Combine t with t in the first cross.
- Combine T with t in the second.
- 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
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
- List TT and Tt as possible pigment genotypes.
- 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
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.
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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.