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Question 28

In Drosophila fruit flies, white eye color is a sex-linked recessive trait. The allele for white eyes is carried on the X chromosome.

A female carrier of the white-eye allele and a male with wild-type red eyes were crossed and produced 16 offspring. The expected ratio of phenotypes in the offspring of this cross is 2 red-eyed females : 1 red-eyed male : 1 white-eyed male.

a.

A student performed a chi-squared (χ2\chi^2χ2) test on the phenotypes of the actual 16 offspring. Part of the calculation is shown in the table below.

PhenotypesObserved number (OOO)Expected number (EEE)O−EO-EO−E(O−E)2(O-E)^2(O−E)2(O−E)2E\frac{(O-E)^2}{E}E(O−E)2​
Red-eyed female2.04.00.50
Red-eyed male-1.01.00.25
White-eyed male-1.01.00.25
χ2=\chi^2 =χ2=1.00

Complete the table by finding the value of the observed (OOO) and expected (EEE) numbers for each phenotype.

[2]
b.

Using the portion of the χ2\chi^2χ2 critical values table below, identify the critical value for this test at the 5%5\%5% significance level (p=0.05p = 0.05p=0.05).

Degrees of freedom (ν\nuν)10%10\%10% level (p=0.10p=0.10p=0.10)5%5\%5% level (p=0.05p=0.05p=0.05)1%1\%1% level (p=0.01p=0.01p=0.01)
12.7063.8416.635
24.6055.9919.210
36.2517.81511.340
[2]
c.

State whether the student should accept or reject the null hypothesis (that there is no significant difference between the observed and expected ratios), and justify your answer using the calculated χ2\chi^2χ2 value and the critical value.

[2]
Markscheme

Analysis Questions

  1. A Level
  2. /Biology
  3. /Analysis

45 exam-style questions on OCR A A Level Biology Analysis. Each one has a worked solution and a mark scheme showing where the marks go.

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