A-level Biology chi-squared test: calculator and worked questions
Enter the observed and expected numbers. You get χ², the degrees of freedom and the conclusion at the 5% level.
CategoryObserved (O)Expected (E)
χ² = 1.33
Dominant phenotype: (290 − 300)² ÷ 300 = 0.333
Recessive phenotype: (110 − 100)² ÷ 100 = 1
χ² = 0.333 + 1 = 1.33
Degrees of freedom = 2 − 1 = 1; critical value at p = 0.05 is 3.84
1.33 < 3.84, so the difference is not significant: accept the null hypothesis that there is no difference between observed and expected.
Exam-style questions
1 of 4
3 marksCore biology3.7.1
Calculator allowed
AQA 7402
Spec and skills
Paper scope: 7402/2
Paper scope: 7402/3
Spec 3.7.1
A genetic model predicts a 3 : 1 ratio of dominant to recessive phenotypes. A cross gives 132 offspring with the dominant phenotype and 28 with the recessive phenotype.
Use χ2=∑E(O−E)2 to calculate the value of χ2. Give your answer to two decimal places.
[3 marks]
5 marksCore biology3.7.1
Calculator allowed
AQA 7402
Spec and skills
Paper scope: 7402/2
Paper scope: 7402/3
Spec 3.7.1
A chi-squared test compares the observed numbers of two phenotypes from a monohybrid cross with the numbers expected from a 3 : 1 ratio. The calculated value of χ2 is 4.80. The table gives critical values.
Critical values of chi-squared for 1 to 4 degrees of freedom at probabilities of 0.05 and 0.01.
Degrees of freedom
P=0.05
P=0.01
1
3.84
6.63
2
5.99
9.21
3
7.81
11.34
4
9.49
13.28
(a)
Give the number of degrees of freedom and the critical value you would use at P=0.05.
[2 marks]
(b)
What should be concluded about the proposed inheritance model? Explain your answer.
[3 marks]
6 marksUnfamiliar context3.7.1
Calculator allowed
AQA 7402
Spec and skills
Paper scope: 7402/2
Paper scope: 7402/3
Spec 3.7.1
Two plants heterozygous for two genes were crossed. The offspring were: 98 with both dominant features, 24 with only the first dominant feature, 28 with only the second dominant feature and 10 with neither. The student's null hypothesis is that the genes assort independently, giving a 9 : 3 : 3 : 1 ratio. The table gives critical values of χ2.
Critical values of chi-squared for 1 to 4 degrees of freedom at probabilities of 0.05 and 0.01.
Degrees of freedom
P=0.05
P=0.01
1
3.84
6.63
2
5.99
9.21
3
7.81
11.34
4
9.49
13.28
(a)
Calculate χ2 for these results. Give your answer to two decimal places.
[2 marks]
(b)
Use the table to decide whether the null hypothesis should be rejected. Give the degrees of freedom and the critical value you used.
[2 marks]
(c)
A student concludes: "This proves that the two genes are on different chromosomes." Evaluate this conclusion.
[2 marks]
4 marksCore biology3.7.1
Calculator allowed
AQA 7402
Spec and skills
Paper scope: 7402/2
Paper scope: 7402/3
Spec 3.7.1
A test cross between a double heterozygote with the alleles arranged AB/ab and a double recessive, ab/ab, gave 168 AB/ab, 172 ab/ab, 28 Ab/ab and 32 aB/ab offspring. All genotypes survive equally well.
(a)
Calculate the percentage of offspring that came from recombinant gametes of the AB/ab parent.
[2 marks]
(b)
Explain why the recombinant offspring are much less common than the parental types.
χ² = Σ (O − E)² ÷ E: for each category, square the difference between the observed and expected numbers, divide by the expected number, and add them up.
How many degrees of freedom are there?
The number of categories minus one. Two phenotypes give 1 degree of freedom; four phenotypes give 3.
How do I write the conclusion?
Compare χ² with the critical value at p = 0.05. If χ² is smaller, the difference between observed and expected is due to chance and the null hypothesis is accepted. If it is larger, the difference is significant and the null hypothesis is rejected.
