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IGCSE · Cambridge (CIE) · 0973

Construct and interpret a Punnett square for a one-gene cross

A square combines gamete contributions before counting genotypes and phenotypes.

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

What belongs on a row or column label here?

What you will learn

  • Combine one row allele and one column allele in each cell
  • Read the cells as equally likely possibilities under the model

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

Construct and interpret a Punnett square for a one-gene crossA square combines gamete contributions before counting genotypes and phenotypes.Tt × TtColumns and rows show gamete allelesTtTtTTpresentTtpresentTtpresentttabsentPossible positions, not four guaranteed offspringTT:Tt:tt = 1:2:1Pigment present:absent = 3:1T is dominant for pigment; tt givesabsence
A square combines gamete contributions before counting genotypes and phenotypes.
Allele, phenotype and evidence key
FeatureMeaning
Column gametesT and t from one parent
Row gametesT and t from the other
Internal cellsTT, Tt, Tt, tt
Model interpretationFour equal positions; three distinct genotype names

Combine one row allele and one column allele in each cell

A Punnett square organises possible combinations of parental gamete alleles. For Tt × Tt, put T and t from one parent above the columns, and T and t from the other beside the rows. Each internal cell contains one row contribution and one column contribution. The four combinations are TT, Tt, Tt and tt, using the standard order for the same allele pair.

The labels outside the square describe gametes at the tracked gene; the internal pairs describe possible offspring genotypes. Do not put the full parental Tt record on each row or column. Keep an explicit allele key beside the model: T is the pigment-producing dominant allele and t is the recessive absence allele in these original examples.

Read the cells as equally likely possibilities under the model

Under equal gamete probabilities and independent combination, each of the four positions has probability one quarter. The repeated Tt genotype occupies two positions, so its probability is two quarters rather than one quarter. Distinct genotype names and equally likely positions are different counts: three genotype names occur among four positions.

Convert genotype to phenotype only after filling the square. TT and both Tt positions show pigment; tt does not, giving three pigment positions to one absence position. The square describes possible events and model probabilities, not four actual offspring that must occur together. If the parent genotypes or expression rule change, rebuild the combinations rather than copying the previous ratio.

Sources for this lesson

Pause and explain

Why does Tt have probability two quarters in Tt × Tt?

Put the idea to work

Worked example

Construct the four equal-probability positions for Tt × Tt.

Show the worked solution
  1. Place T and t above the columns and T and t beside the rows. Each entry represents one gamete contribution.
  2. Combine each row with each column: TT, Tt, Tt, tt. Count both Tt positions when computing probabilities.

Answer Rows T/t and columns T/t give TT, Tt, Tt, tt. Count Tt twice in the probability because it occupies two equal-probability positions.

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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

Fill the square with rows T, t and columns T, t, recording the cells row by row.

Give me a hint

Join the row allele with the column allele.

Compare my reasoning
  1. Fill the T row as TT and Tt.
  2. Fill the t row as tT and tt.
  3. Write tT in standard order Tt.

The cells are TT, Tt, Tt and tt. They represent four possible equal-probability positions under the standard model.

Look for these in your work

  • Four correct combinations
  • tT and Tt recognised as the same genotype
2 · Independent

Count the distinct genotype names and the positions belonging to each in that square. Give genotype probabilities.

Give me a hint

Do not confuse distinct names with repeated positions.

Compare my reasoning
  1. Count one TT, two Tt and one tt.
  2. Divide each count by four positions.

Three distinct genotypes: TT has 1/4, Tt has 2/4 or 1/2, and tt has 1/4.

Look for these in your work

  • Correct 1:2:1 genotype position counts
  • Correct probabilities with denominator four
3 · Transfer

Using T as dominant pigment, convert each square position to phenotype and correct the claim that four actual offspring must match the four cells exactly.

Give me a hint

Apply the expression key and distinguish possibilities from observations.

Compare my reasoning
  1. Assign pigment to TT and both Tt positions.
  2. Assign absence to tt.
  3. State the probabilistic interpretation.

Three positions show pigment and one shows absence, predicting a 3:1 phenotype ratio. Four actual offspring are not guaranteed to match the square exactly; its cells represent possible combinations and model probabilities.

Look for these in your work

  • Correct 3:1 phenotype prediction
  • Rejects guaranteed four-offspring matching

Common mistakes

  • Putting two-allele parent genotypes in gamete labels
  • Counting three distinct genotype names as three equally probable outcomes
  • Reading four square cells as a guaranteed set of four offspring
Recall without your notes

What can you explain now?

Why does Tt have probability two quarters in Tt × Tt?

Compare with the explanation

It occurs in two equally likely square positions

T+t and t+T give the same genotype through two positions.

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

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Make it stick

Without your notes, combine one row allele and one column allele in each cell and read the cells as equally likely possibilities under the 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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