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A2 · Practice questions · The chi-squared test

O, E, and the critical value.

Six original Cambridge-style questions on testing genetic data with chi-squared, from expected numbers to the final conclusion.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
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Five steps, every time.

01
Recall
[2 marks]

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.

  • 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. ✓
02
Calculation
[4 marks]

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).

  • 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. ✓
03
Calculation
[3 marks]

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.

  • 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. ✓
04
Explain
[3 marks]

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.

  • 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. ✓
05
Analysis
[2 marks]

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.

  • 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). ✓
06
Data
[3 marks]

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.

  • 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). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.