Interpret finite test-cross evidence without confusing likelihood and genotype certainty
An unlikely finite record under one genotype is not automatically a genotype probability.
Extended only (Supplement); 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 probability does the Tt × tt model give each pigment outcome?
What you will learn
- Calculate a finite all-dominant record probability under a stipulated heterozygote
- Separate the probability of a record from the probability of an unknown genotype
Original teaching model. Pedigrees and records are fictional; allele and probability assumptions are stated.
| Feature | Meaning |
|---|---|
| Four events given Tt | (1/2)^4 = 1/16 = 6.25% all pigment |
| Six events given Tt | (1/2)^6 = 1/64 = 1.5625% all pigment |
| Eight events given Tt | (1/2)^8 = 1/256 ≈ 0.39% all pigment |
| Interpretation | Record likelihood under Tt, not an automatic posterior genotype probability |
Calculate a finite all-dominant record probability under a stipulated heterozygote
In the stipulated Tt × tt model, each offspring has pigment probability one half and absence probability one half. If events are independent, the probability that all four recorded offspring show pigment is one half multiplied by itself four times: one sixteenth. This is six point two five per cent. A possible all-pigment record does not require the unknown parent to be TT.
For eight independent events under the same heterozygote model, all pigment has probability one divided by two to the eighth power, one two-hundred-and-fifty-sixth, approximately zero point three nine per cent. The larger sample makes this particular record less likely under Tt, but not impossible. Exact probabilities depend on the stipulated normal model, equal allele transmission and independent events.
Separate the probability of a record from the probability of an unknown genotype
This calculation is the probability of an all-pigment record assuming the parent is Tt. It is not automatically the probability that the unknown parent is Tt after observing the record. To calculate that different question, additional information about the competing hypotheses and their prior probabilities would be needed. Do not relabel one sixteenth as certainty that the genotype is TT or as a posterior probability for Tt.
A recessive tt offspring remains informative within the normal model because it establishes a transmitted t from the unknown dominant-phenotype parent. A record with no recessive offspring may support the TT hypothesis, especially as the sample grows, but conclusions must reflect the evidence supplied. These calculations are original teaching-model examples, not instructions for a clinical genetic test or an assessment of a real family.
Pause and explain
What does the calculated all-pigment probability assume?
Worked example
If the unknown parent is Tt in a Tt × tt cross, find the chance of all-pigment records of four and eight independent offspring.
Show the worked solution
- Each offspring has pigment probability 1/2 under the supplied model. Multiply independent probabilities: (1/2)^4 = 1/16 and (1/2)^8 = 1/256.
- Convert to percentages: 6.25% and 0.390625% (about 0.39%). These probabilities describe the record assuming Tt; they are not automatically the probability of Tt given that record.
Answer Under Tt × tt with independent equal-probability events: all pigment among four has probability 1/16 = 6.25%; among eight, 1/256 ≈ 0.39%. These are record probabilities conditional on Tt.
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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.
Assuming Tt × tt and independent events, calculate the probability that all four offspring show pigment and convert it to a percentage.
Give me a hint
Multiply one half four times.
Compare my reasoning
- Calculate (1/2)^4.
- Multiply the resulting fraction by one hundred.
1/16, or 6.25%, for an all-pigment four-offspring record under the stipulated Tt model.
Look for these in your work
- Correct one-sixteenth
- Correct 6.25% with conditional wording
Under the same model, calculate the probability of all pigment among six offspring. State whether it is zero.
Give me a hint
Use six independent half-probability events.
Compare my reasoning
- Calculate (1/2)^6.
- Convert one sixty-fourth to a percentage.
- Distinguish a small probability from impossibility.
1/64 = 1.5625%; it is small but not zero under the stipulated heterozygote model.
Look for these in your work
- Correct one-sixty-fourth and percentage
- Recognises nonzero probability
A student says an all-pigment four-offspring record means the parent has probability exactly 1/16 of being Tt. Explain the distinction.
Give me a hint
Ask whether the calculation conditions on genotype or on the observed record.
Compare my reasoning
- Identify 1/16 as the record probability assuming Tt.
- Recognise the different question about unknown genotype after data.
- Name the need for competing hypotheses and prior information.
1/16 is the probability of that record assuming Tt, not automatically the probability of Tt given the record. The latter would need additional information about the competing genotype hypotheses and their prior probabilities.
Look for these in your work
- Correct conditional-probability distinction
- Names missing hypothesis or prior information
Common mistakes
- Calling a record probability the probability of the parent genotype
- Treating unlikely as impossible
- Ignoring the independence and equal-transmission assumptions
What can you explain now?
What does the calculated all-pigment probability assume?
Compare with the explanation
That the parent is Tt and events follow the stated independent equal-probability model
It is conditional on the model rather than a genotype conclusion from data.
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Make it stick
Without your notes, calculate a finite all-dominant record probability under a stipulated heterozygote and separate the probability of a record from the probability of an unknown genotype. 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.