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IGCSE · Cambridge (CIE) · 0654

Use a test cross to investigate an unknown dominant-phenotype genotype

The recessive partner’s fixed contribution makes the unknown allele contribution informative.

Section B16.3 · Monohybrid inheritance

Monohybrid inheritance · Cambridge IGCSE 0654 · B16.3

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 genotypes can fit the unknown pigment-producing parent under this key?

What you will learn

  • Compare an unknown dominant-phenotype parent with a recessive homozygote
  • Use a recessive offspring to identify a heterozygote in the stated model

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

Use a test cross to investigate an unknown dominant-phenotype genotypeThe recessive partner’s fixed contribution makes the unknown allele contribution informative.Two possible unknown-parent genotypesKnown recessive partner: ttTT × ttColumns and rows show gamete allelesttTTTtpresentTtpresentTtpresentTtpresentFour possible equal-probability positionsTt × ttColumns and rows show gamete allelesttTtTtpresentTtpresentttabsentttabsentFour possible equal-probability positionsA recessive offspring identifies Ttin this normal model
The recessive partner’s fixed contribution makes the unknown allele contribution informative.
Allele, phenotype and evidence key
FeatureMeaning
Unknown pigment parentTT or Tt under the supplied rule
Known partnertt; supplies t only
Predicted TT caseOnly Tt offspring; all possible outcomes show pigment
Predicted Tt caseTt and tt equally; a tt record identifies Tt in the normal model

Compare an unknown dominant-phenotype parent with a recessive homozygote

A test cross investigates the unknown genotype of an individual showing the dominant phenotype. In the original pigment model, an unknown pigment-producing plant can be TT or Tt. Cross it with a known recessive homozygous tt plant. That partner contributes only t at the gene, making any T or t supplied by the unknown parent informative about its allele pair.

If the unknown parent is TT, its normal gametes carry T and every possible offspring genotype is Tt, showing pigment. If it is Tt, its normal gametes can carry T or t; the square gives equal Tt and tt positions. Under the standard assumptions, the heterozygote case predicts pigment present to absent as 1:1. Keep predicted possible outcomes separate from a particular observed sample.

Use a recessive offspring to identify a heterozygote in the stated model

An observed pigment-absent tt offspring receives t from the known tt partner and t from the unknown parent. The unknown parent also has T because it shows pigment under the supplied complete-dominance rule. These two facts identify it as Tt in the normal one-gene model. The inference depends on the stated trait rule, parentage, accurate records and normal allele transmission.

An observed finite sample showing only pigment does not logically prove TT: a Tt parent can by chance supply T to all the recorded offspring. A larger suitable sample with no recessive outcome can support the TT hypothesis, but uncertainty remains unless the task supplies stronger evidence. The test-cross method predicts contrasting distributions and uses informative offspring; it does not turn a small all-dominant record into a guarantee.

Sources for this lesson

Pause and explain

Which partner makes the normal test cross informative here?

Put the idea to work

Worked example

How can an unknown pigment-producing T_ parent be investigated using a known tt partner?

Show the worked solution
  1. Consider both possible pigment-parent genotypes. TT supplies only T and gives Tt; Tt supplies T or t and gives Tt or tt.
  2. If a normal-model tt offspring is observed, the unknown parent contributed t while also having T for pigment. This identifies Tt under the stated assumptions.

Answer Unknown pigment parent × known tt: TT predicts only Tt; Tt predicts Tt and tt equally. A tt offspring identifies the unknown parent as Tt within the stated normal model.

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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

Compare the possible offspring from TT × tt and Tt × tt. Give genotypes and the pigment phenotype predictions.

Give me a hint

Use the known partner’s t contribution in both cases.

Compare my reasoning
  1. Combine T from TT with t to give Tt only.
  2. Combine T or t from Tt with t.
  3. Apply the pigment expression rule.

TT × tt predicts only Tt, all possible offspring showing pigment. Tt × tt predicts Tt and tt in a 1:1 ratio, with pigment present:absent = 1:1.

Look for these in your work

  • Correct two genotype patterns
  • Correct contrasting phenotype predictions
2 · Independent

An unknown pigment-producing plant crossed with known tt gives at least one pigment-absent offspring. Infer the unknown genotype under the stated normal model.

Give me a hint

The absent offspring is tt and needs a t contribution from each parent.

Compare my reasoning
  1. Identify tt for the absent offspring.
  2. Trace t to the unknown parent.
  3. Combine that with T required for its own pigment phenotype.

The unknown parent is Tt: it shows pigment through T and supplied t to the tt offspring.

Look for these in your work

  • Correct Tt inference
  • Explains phenotype and offspring contribution
3 · Transfer

Eight recorded offspring all show pigment after an unknown pigment parent is crossed with tt. Explain why this observation alone does not prove the unknown parent is TT.

Give me a hint

A heterozygote can transmit T in each recorded event by chance.

Compare my reasoning
  1. Compare the prediction from TT with the possible finite record from Tt.
  2. Separate an all-pigment sample from all possible outcomes.

Both TT and Tt are compatible with this finite all-pigment record. TT predicts only pigment outcomes, but a Tt parent can by chance supply T in all eight events; absence of a recorded recessive outcome alone is not proof of TT.

Look for these in your work

  • Keeps both possibilities compatible
  • Distinguishes finite record from predicted possibilities

Common mistakes

  • Crossing the unknown parent with another unknown dominant phenotype
  • Confusing predicted all-possible outcomes with a finite observed record
  • Calling a dominant-phenotype unknown parent tt
Recall without your notes

What can you explain now?

Which partner makes the normal test cross informative here?

Compare with the explanation

A known tt recessive homozygote

It contributes t only, exposing the unknown parent’s contribution.

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

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Make it stick

Without your notes, compare an unknown dominant-phenotype parent with a recessive homozygote and use a recessive offspring to identify a heterozygote in the stated 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 (Double)

Course scope and assessment

This 2028–2029 pathway maps all 152 detailed subject-content sections in the published 0654 syllabus to original teaching and staged practice. It adapts teaching from the related single-science pathways and excludes clear 0610-only topics such as classification systems, human excretion, tropic responses, food supply, population dynamics, detailed pollution and biotechnology. Some explanations may include wider single-science context. A 0654-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 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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