Calculating and analysing outcomesSpec 3.16
In short
The outcomes of a monohybrid cross can be given as a ratio, a probability or a percentage. From a Punnett square, count the boxes with that outcome out of four. For two carrier parents (Ff × Ff), the probability of an affected child is 1/4, 0.25 or 25%. Each child is an independent event, so the probability stays the same.
You need to calculate and analyse the outcomes of monohybrid crosses and pedigrees for dominant and recessive traits. Outcomes can be given as a ratio, a probability or a percentage.
| Outcome | Boxes out of 4 | Probability | Percentage |
|---|---|---|---|
| Short plant (tt) | 1 | 1/4 or 0.25 | 25% |
| Tall plant (TT or Tt) | 3 | 3/4 or 0.75 | 75% |
As a ratio, this is 3 tall : 1 short. A 3 : 1 ratio means 1 in 4 offspring are short, not 1 in 3.
Probability for a recessive disorder
A disorder is caused by a recessive allele (f). Two parents do not have the disorder but both are carriers (Ff). What is the probability that their child has the disorder?
- Gametes from each parent: F or f.
- Possible offspring: FF, Ff, Ff and ff.
- Only ff has the disorder. That is 1 out of 4 equally likely outcomes.
Answer: Probability = 1/4 = 0.25 = 25%.
Each child is an independent event. If the first child has the disorder, the probability for the next child is still 1 in 4. Probabilities describe chance. They do not guarantee the actual numbers in a family.
Show the Punnett square, count the favourable boxes out of the total, then convert to a ratio, fraction or percentage as the question asks.
Written and checked against the Edexcel GCSE Biology (1BI0) specification · Updated October 2026