Predict a one-to-one phenotype ratio from a heterozygote and recessive homozygote
One T/t contribution combined with t gives equal Tt and tt possibilities.
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
What allele type can the normal tt parent contribute at this gene?
What you will learn
- Construct the Tt crossed with tt square
- Calculate equal genotype and phenotype probabilities
Original teaching model. Pedigrees and records are fictional; allele and probability assumptions are stated.
| Feature | Meaning |
|---|---|
| Parent gametes | T or t; and t |
| Square cells | Tt, Tt, tt, tt |
| Genotype and phenotype | Both predicted ratios are 1:1, with different labels |
| Expected 72 | 36 present and 36 absent; no alternating-order guarantee |
Construct the Tt crossed with tt square
For Tt × tt, the heterozygous parent can contribute T or t at the gene and the recessive homozygous parent contributes t. A two-by-two square can repeat t across its two columns, with T and t beside the rows. Its cells are Tt, Tt, tt and tt. Each offspring receives one allele from each parent; none can be TT because the tt parent has no T contribution.
The genotype grouping gives two Tt positions and two tt positions, simplified to 1:1. Under the pigment expression key, Tt shows pigment and tt does not, so the phenotype ratio is also 1:1. The matching numerical ratios are a feature of this cross and rule; genotype and phenotype are still different concepts and need category labels.
Calculate equal genotype and phenotype probabilities
Each category has probability one half under the standard equal-transmission and independent-combination assumptions. For an expected total of seventy-two offspring, the model gives thirty-six Tt and thirty-six tt, corresponding to thirty-six with pigment and thirty-six without. The total and the denominator are supplied; do not assume the square’s four positions are the actual number of offspring.
A 1:1 prediction does not require alternating phenotypes in birth order or exact equality in every observed sample. Observed records supply actual counts, while the square supplies model probabilities. The same observed half-and-half count can arise in other circumstances, so it does not establish a parental genotype without the model assumptions and other evidence. Supplement lessons will examine the use and limits of a test cross separately.
Pause and explain
Which genotypes are possible in Tt × tt?
Worked example
Predict genotype and phenotype ratios for Tt × tt under the supplied pigment rule.
Show the worked solution
- Combine T/t rows with t/t columns to produce Tt, Tt, tt, tt.
- Count two Tt positions and two tt positions. Tt has pigment and tt lacks pigment, so both labelled ratios are 1:1.
Answer Tt × tt gives two Tt and two tt positions, predicting Tt:tt = 1:1 and pigment present:absent = 1:1.
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.
Fill a square with rows T, t and columns t, t. List the cells and simplify Tt:tt.
Give me a hint
Pair each row allele with each repeated t column.
Compare my reasoning
- Fill the T row with two Tt cells.
- Fill the t row with two tt cells.
- Simplify two:two.
Tt, Tt, tt, tt; genotype ratio Tt:tt = 1:1.
Look for these in your work
- Four correct cells
- Correct labelled 1:1 genotype ratio
For an expected total of 64 offspring under this model, calculate pigment present and absent counts.
Give me a hint
Each phenotype has probability one half.
Compare my reasoning
- Calculate 64 × 1/2 for each category.
- Check the total remains 64.
Expected pigment present 32 and absent 32, a 1:1 phenotype prediction under the supplied rule.
Look for these in your work
- Correct 32 and 32
- Labels counts as expected
Explain why the square cannot give TT and why its 1:1 ratio does not guarantee alternating pigment and absence outcomes.
Give me a hint
Use the gamete types for possibility and probability for order.
Compare my reasoning
- Identify the tt parent’s only t contribution.
- Recognise TT needs two T contributions.
- Distinguish a probability ratio from an event sequence.
TT is impossible in the normal stipulated cross because the tt parent contributes no T. Equal phenotype probabilities do not require an alternating observed sequence.
Look for these in your work
- Explains exclusion of TT
- Rejects alternating-sequence guarantee
Common mistakes
- Predicting TT despite a parent contributing only t
- Calling a 1:1 ratio a guaranteed alternating sequence
- Assuming an observed ratio alone proves the parents
What can you explain now?
Which genotypes are possible in Tt × tt?
Compare with the explanation
Tt and tt
T or t from the first parent combines with t from the second.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Without your notes, construct the tt crossed with tt square and calculate equal genotype and phenotype probabilities. 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.