The shape of the lesson
By the end of the lesson, learners can
- Observed against expected: Press Grow 160 offspring several times.
- Degrees of freedom and the critical value: The simulator uses 3 degrees of freedom and a critical value of 7.82.
- State the null hypothesis: no significant difference.
- Expected numbers = total × fraction for each class.
- Degrees of freedom = number of classes − 1.
- Compare with the critical value at p = 0.05, then conclude.
One picture every idea returns to
The chi-squared (χ2) test compares observed (O) numbers in each category with the expected (E) numbers predicted by a genetic ratio. Start with a null hypothesis: there is no significant difference between observed and expected results, and any difference is due to chance. Calculate E from the total and the ratio, then χ2 = Σ (O − E)2 / E (the formula is provided). Find the degrees of freedom = number of categories − 1, and compare χ2 with the critical value at p = 0.05 from the table. If χ2 is less than the critical value, the difference is not significant and the null hypothesis is accepted; if it is greater, the difference is significant (probability less than 0.05 of arising by chance), so the null hypothesis is rejected and something else, such as linkage or epistasis, may be operating.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Chance, or a real difference?. Real crosses never give perfect ratios. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Observed against expected | Press Grow 160 offspring several times. Project tab 4 of the lab (Inheritance lab, the chi-squared test) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Degrees of freedom and the critical value | The simulator uses 3 degrees of freedom and a critical value of 7.82. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Degrees of freedom and the critical value, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Always use the actual numbers of offspring for O and E, never percentages or ratios. | And E does not have to be a whole number: 100 × 3/4 = 75, but 50 × 3/4 = 37.5 is fine. |
| A non-significant result does not prove the hypothesis is true; it means the data are consistent with it. | Say "the difference is not significant" rather than "there is no difference". |
The chi-squared test: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
State the null hypothesis used in a chi-squared test on the results of a genetic cross, and explain what is meant by a significant difference.
A cross is expected to give a 9 : 3 : 3 : 1 ratio. The observed numbers are 315, 108, 101 and 32 (total 556). Calculate χ2 (give the expected values to one decimal place).
In a monohybrid cross 400 offspring are produced: 280 dominant, 120 recessive. Test the fit to a 3 : 1 ratio. The critical value for 1 degree of freedom at p = 0.05 is 3.84.
A test cross of a double heterozygote gives χ2 = 31.5 against an expected 1 : 1 : 1 : 1 ratio, with two classes much larger than the other two. Explain these results.
Explain why the chi-squared test should be carried out on the numbers of offspring and not on percentages, and why a large sample is preferred.
A student calculates χ2 = 4.62 for a cross with three phenotype classes. Critical values at p = 0.05: 1 df 3.84; 2 df 5.99; 3 df 7.82. State the degrees of freedom and the conclusion, and explain one mistake the student could make with the table.
Answer key and marking notes
- Null hypothesis: there is no significant difference between the observed and expected results; any difference is due to chance.
- A significant difference is one with a probability of less than 0.05 (5%) of arising by chance, so it is unlikely to be due to chance alone.
- E = 556 × 9/16 = 312.75 (312.8), 556 × 3/16 = 104.25 (104.3) twice, 556 × 1/16 = 34.75 (34.8).
- (O − E)2/E: 0.016, 0.135, 0.101, 0.218.
- χ2 = 0.47 (to 2 significant figures).
- Degrees of freedom = 3; 0.47 is less than 7.82, so the difference is not significant.
- E = 300 dominant and 100 recessive.
- χ2 = (280 − 300)2/300 + (120 − 100)2/100 = 1.33 + 4.00 = 5.33.
- 5.33 > 3.84, so the difference is significant: reject the null hypothesis; the results do not fit 3 : 1.
- 31.5 is far above the critical value (7.82 for 3 degrees of freedom), so the difference is significant / the null hypothesis is rejected.
- The genes are linked (on the same chromosome), so the parental allele combinations are inherited together, giving the two large classes.
- The two small classes are recombinants produced by crossing over in prophase I.
- The size of χ2 depends on the actual numbers; percentages would give a value unrelated to sample size and the wrong result.
- Chance deviations have a smaller effect in large samples, so the test is more reliable (small expected numbers, below about 5, make it unreliable).
- Degrees of freedom = 3 − 1 = 2; critical value 5.99.
- 4.62 < 5.99, so the difference is not significant; accept the null hypothesis: the results fit the expected ratio.
- Using 3 degrees of freedom (the number of classes) instead of 2, or using the 1 df value, which would give the wrong conclusion (4.62 > 3.84).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What does the chi-squared test assess in a genetics experiment?
- Awhether the difference between observed and expected results is significant or due to chance
- Bwhether two sets of observed results are correlated with each other
- Cthe mean and standard deviation of the observed numbers
- Dwhich allele is dominant in the cross
A cross is expected to give a 1 : 2 : 1 ratio. 240 offspring are produced. What are the expected numbers?
- A60, 120, 60
- B80, 80, 80
- C60, 180, 60
- D120, 240, 120
Observed 70 and 30 against expected 75 and 25. What is the value of χ2?
- A1.33
- B0.33
- C1.00
- D50.0
How many degrees of freedom are used for a chi-squared test on a cross with five phenotype classes?
- A4
- B5
- C6
- D1
A dihybrid cross gives χ2 = 9.1. The critical value at p = 0.05 for 3 degrees of freedom is 7.82. What should be concluded?
- Athe difference is significant, so the null hypothesis is rejected
- Bthe difference is not significant, so the null hypothesis is accepted
- Cthe result proves that the genes are linked
- Dthe result is due to chance, with a probability above 0.05
In a test cross, 400 offspring are expected in a 1 : 1 : 1 : 1 ratio. The observed numbers are 112, 88, 95 and 105. What is χ2 and the conclusion? (critical value for 3 df at p = 0.05 is 7.82)
- A3.38, not significant: the data fit 1 : 1 : 1 : 1
- B3.38, significant: the genes are linked
- C0.85, not significant: the data fit 1 : 1 : 1 : 1
- D13.5, significant: the genes are linked
A student tests 9 : 3 : 3 : 1 data and finds χ2 = 6.4. They use 4 degrees of freedom (critical value 9.49) by mistake. Does correcting the error change the conclusion?
- Ano: with 3 df the critical value is 7.82, and 6.4 is still not significant
- Byes: with 3 df the difference becomes significant
- Cyes: with 3 df the null hypothesis must be rejected
- Dno: degrees of freedom do not affect the critical value
A dihybrid F2 of 320 plants gives 236 : 6 : 4 : 74 against an expected 9 : 3 : 3 : 1. What is the most likely explanation?
- Athe genes are linked, so recombinant phenotypes are rare
- Bthe genes are on different chromosomes and assort independently
- Cone gene is epistatic to the other, giving 9 : 3 : 4
- Dthe sample is too small for the chi-squared test to work
Quiz key
The chi-squared test compares observed and expected numbers to judge whether any difference is significant. Correlation needs a different test. Mean and standard deviation describe spread, not fit to a ratio. Dominance comes from the cross itself, not from the test.
240 ÷ 4 = 60 per part, so E = 60, 120 and 60. 80 each ignores the ratio. 60, 180, 60 does not add up to 240. 120, 240, 120 multiplies the ratio by 120 and doubles the total.
(70 − 75)2/75 = 0.33 and (30 − 25)2/25 = 1.00, so χ2 = 1.33. 0.33 and 1.00 are single classes only. 50.0 adds the squared differences (25 + 25) without dividing each by E.
Degrees of freedom = classes − 1 = 4. 5 is the number of classes. 6 adds one instead of subtracting. 1 is the value for a two-class cross.
9.1 is greater than 7.82, so p < 0.05: the difference is significant and the null hypothesis is rejected. Accepting it would need a value below 7.82. A statistical test suggests an explanation such as linkage; it does not prove it. A probability above 0.05 would belong to a value below the critical value.
(144 + 144 + 25 + 25) / 100 = 3.38, below 7.82, so not significant. Calling 3.38 significant reads the comparison the wrong way round. 0.85 divides by 400 instead of 100. 13.5 divides by 25 instead of 100.
Four classes give 3 df, critical value 7.82. 6.4 is below 7.82, so the conclusion (not significant) is unchanged. The difference would become significant only if 6.4 exceeded 7.82, which it does not, and so the null hypothesis is still accepted. Degrees of freedom do change the critical value, which is why the error matters in general.
Expected 180 : 60 : 60 : 20. The two parental classes are far above expectation and the two mixed classes far below, the pattern of linkage with few recombinants. Independent assortment would give close to 9 : 3 : 3 : 1. Epistasis would merge classes, not shrink the two middle ones to almost zero. 320 is a large sample.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.