1. Fraction
What percentage is 1/4?
Help beside the question
Read or point first. Multiply by 100.
Answer guidance
25%.
Inheritance probability: predict and evaluate
Learning objective
Convert genetic probabilities between fractions, ratios and percentages, and explain why observations vary.
40 minutes · one route · start from what you show today. Point, speak, sign, draw, write or direct an adult. An adult may read or scribe your exact words. Choose one route; do not complete all three.
LAUNCH · GCSE BIOLOGY FOUNDATION · W13 · LESSON 3 · 4 MINUTES
Four clues to start
Supported · 4 questions
2. Genotypes
What is the genotype ratio for Aa × Aa?
Help beside the question
Read or point first. Count AA, Aa and aa.
Answer guidance
1 AA : 2 Aa : 1 aa.
3. Phenotypes
With complete dominance, what is the phenotype ratio?
Help beside the question
Read or point first. AA and Aa show the dominant phenotype.
Answer guidance
3 dominant : 1 recessive.
4. Independence
Does an aa offspring force the next offspring to be AA?
Help beside the question
Read or point first. Each fertilisation is a new event.
Answer guidance
No.
Standard · 4 questions
1. Fraction
What percentage is 1/4?
Help beside the question
Multiply by 100.
Answer guidance
25%.
2. Genotypes
What is the genotype ratio for Aa × Aa?
Help beside the question
Count AA, Aa and aa.
Answer guidance
1 AA : 2 Aa : 1 aa.
3. Phenotypes
With complete dominance, what is the phenotype ratio?
Help beside the question
AA and Aa show the dominant phenotype.
Answer guidance
3 dominant : 1 recessive.
4. Independence
Does an aa offspring force the next offspring to be AA?
Help beside the question
Each fertilisation is a new event.
Answer guidance
No.
Stretch · 4 questions
1. Fraction
What percentage is 1/4?
Help beside the question
Multiply by 100.
Answer guidance
25%.
2. Genotypes
What is the genotype ratio for Aa × Aa?
Help beside the question
Count AA, Aa and aa.
Answer guidance
1 AA : 2 Aa : 1 aa.
3. Phenotypes
With complete dominance, what is the phenotype ratio?
Help beside the question
AA and Aa show the dominant phenotype.
Answer guidance
3 dominant : 1 recessive.
4. Independence
Does an aa offspring force the next offspring to be AA? Give a reason.
Help beside the question
Each fertilisation is a new event.
Answer guidance
No. Accept a scientifically correct reason that uses the relevant model or evidence.
LAUNCH · GCSE BIOLOGY FOUNDATION · W13 · LESSON 3 · 4 MINUTES
Choose a first idea
Aa × Aa predicts a 25% chance of aa. Which statement follows?
Think or discuss first. No answer chosen yet.
LAUNCH · GCSE BIOLOGY FOUNDATION · W13 · LESSON 3 · 5 MINUTES
Watch the model
For this one-gene, complete-dominance model, both parents are heterozygous: Aa × Aa.
Worked explanation · reveal after thinking
This is a simple monohybrid complete-dominance model. Dominant does not mean common, stronger, healthier or better. Treat each fertilisation as an independent event. Avoid deterministic single-gene examples such as eye colour or tongue rolling.
32-second silent captioned explainer · optional
No sound or autoplay. Use instead of part of the modelling time, not as an extra activity.
Full transcript
Describe the parents. For this one-gene, complete-dominance model, both parents are heterozygous: Aa × Aa.
Separate into gametes. Each parent can produce gametes carrying A or a. Put one allele in each margin box.
Combine one from each. Each offspring box combines one allele from the top with one from the side: AA, Aa, Aa and aa.
Read the probability. The chance of aa is one out of four, or 25%, for each fertilisation. Four offspring do not have to follow this pattern.
LAUNCH · GCSE BIOLOGY FOUNDATION · W13 · LESSON 3 · 5 MINUTES
Sort using the evidence
Choose a card, then choose a group. Explain one placement using its clue.
Drag, or use Tab and Enter to select a card and its group. Pointing or adult-directed placement works too.
Model expectation
Observed simulated result
Unjustified certainty
Choose a card first.
Read the card clues
Expected aa in 20 is 5: 20 × 1/4 = 5.
Expected dominant phenotype is 75%: Three of four model boxes show it.
A simulated sample contains 7 aa: This is a sample outcome, not the theoretical probability.
Another sample contains 3 aa: Samples can differ by chance.
Every four offspring must include one aa: The model does not guarantee a sequence.
The next offspring is certain to balance the counts: Independent events do not remember earlier outcomes.
Would a small change help? · optional
Ask the lesson teacher for clearer labels, fewer cards, reading help, an adult-directed model or more thinking time. Your teacher will explain what can change, try it and check its effect. Ask privately or pass; no extra form is needed.
Group explanations · reveal after sorting
Expected aa in 20 is 5 → Model expectation. 20 × 1/4 = 5.
Expected dominant phenotype is 75% → Model expectation. Three of four model boxes show it.
A simulated sample contains 7 aa → Observed simulated result. This is a sample outcome, not the theoretical probability.
Another sample contains 3 aa → Observed simulated result. Samples can differ by chance.
Every four offspring must include one aa → Unjustified certainty. The model does not guarantee a sequence.
The next offspring is certain to balance the counts → Unjustified certainty. Independent events do not remember earlier outcomes.
LAUNCH · GCSE BIOLOGY FOUNDATION · W13 · LESSON 3 · 4 MINUTES
Work through one example
Worked example
A model gives a 50% chance of an outcome. Find the expected count in 30 trials.
Step 1
Convert 50% to 0.5.
Step 2
Expected count = 0.5 × 30 = 15.
Step 3
This is a model expectation, not a required pattern.
Step 4
Independent trials can give an observed count above or below 15.
LAUNCH · GCSE BIOLOGY FOUNDATION · W13 · LESSON 3 · 4 MINUTES
Expected versus observed
All samples model Aa × Aa. Calculate the expected aa count, then compare it with the simulated result. Do not change the theoretical probability because of one sample.
| Sample | Total | Expected aa | Observed aa |
|---|---|---|---|
| A | 20 | ___ | 7 |
| B | 20 | ___ | 3 |
| C | 80 | ___ | 18 |
Worked explanation · reveal after thinking
Expected aa counts are 5, 5 and 20. Observed aa percentages are 35%, 15% and 22.5%. Variation is compatible with random sampling; these small datasets alone do not establish a different inheritance model.
LAUNCH · GCSE BIOLOGY FOUNDATION · W13 · LESSON 3 · 10 MINUTES
Show your understanding
Supported · choose one route
1. Calculate
Complete the expected aa counts for A, B and C.
For each sample, write total ___. Find one quarter of that total for the expected aa count.
2. Convert
Convert observed 7/20 and 18/80 to percentages.
Write two calculations: 7 ÷ 20 × 100 and 18 ÷ 80 × 100. Keep the correct total each time.
3. Explain
Explain why 7/20 does not automatically disprove the one-in-four model.
Compare predicted probability with this small sample. Use: expected ___, observed ___; chance can produce ___.
Standard · choose one route
1. Calculate
Complete the expected aa counts for A, B and C.
Total × 0.25.
2. Convert
Convert observed 7/20 and 18/80 to percentages.
Count ÷ total × 100.
3. Explain
Explain why 7/20 does not automatically disprove the one-in-four model.
Expected proportions need not match small samples exactly.
Stretch · choose one route
1. Calculate
Complete the expected aa counts for A, B and C.
Total × 0.25.
2. Convert
Convert observed 7/20 and 18/80 to percentages.
Count ÷ total × 100.
3. Keep events independent
Three offspring are not aa. What is the predicted chance that the next is aa in this model? Explain.
Do not subtract previous outcomes from the four-box diagram.
LAUNCH · GCSE BIOLOGY FOUNDATION · W13 · LESSON 3 · 4 MINUTES
Two small checks
Supported · 2 questions
1. Calculate
For Aa × Aa, how many aa offspring are expected in a model sample of 80?
Help beside the question
Expected number = probability × total. Write: working ___; answer ___; unit or percent ___.
Answer guidance
1/4 × 80 = 20; this is an expectation, not a guarantee.
2. Interpret
Why might a sample of 20 contain 7 aa rather than exactly 5?
Help beside the question
Use: my answer is ___ because ___. Point to a model or card first. Small samples vary by chance.
Answer guidance
Random variation means observed counts need not equal the expected proportion exactly.
Standard · 2 questions
1. Calculate
For Aa × Aa, how many aa offspring are expected in a model sample of 80?
Help beside the question
Expected number = probability × total.
Answer guidance
1/4 × 80 = 20; this is an expectation, not a guarantee.
2. Interpret
Why might a sample of 20 contain 7 aa rather than exactly 5?
Help beside the question
Small samples vary by chance.
Answer guidance
Random variation means observed counts need not equal the expected proportion exactly.
Stretch · 2 questions
1. Calculate
For Aa × Aa, how many aa offspring are expected in a model sample of 80?
Help beside the question
Expected number = probability × total.
Answer guidance
1/4 × 80 = 20; this is an expectation, not a guarantee.
2. Keep events independent
Three offspring are not aa. What is the predicted chance that the next is aa in this model? Explain.
Help beside the question
Do not subtract previous outcomes from the four-box diagram.
Answer guidance
25%. Each fertilisation is independent, so earlier outcomes do not use up or force a later genotype.
Keep one useful science explanation. No additional participation record is required.