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A-LevelBiologyInherited change9700/42 · Oct/Nov 2023 · Q59 Marks

The fruit fly, Drosophila melanogaster, has autosomal genes for body colour and wing shape. Gene B/b is involved in the production of body colour: • B = dominant allele for brown body colour • b = recessive allele for black body colour. Gene D/d is involved in wing shape: • D = dominant allele for straight wing • d = recessive allele for curved wing. A dihybrid test cross was carried out between flies heterozygous for body colour and for wing shape and flies homozygous recessive for body colour and for wing shape. (a) Table 5.1 shows the number of offspring of each phenotype obtained in the test cross. [Table 5.1]

  1. (a)Use Table 5.1 to calculate the expected number of each phenotype if the two genes are on different autosomes. Write your answers in the table.[1]
  2. (b)A chi-squared (x2) test was carried out to compare the observed results with the results that would be expected from a dihybrid cross involving genes on different autosomes. The value of x2 = 2097.836. Table 5.2 shows the critical values for the x² distribution. [Table 5.2] Explain how the value of x² and Table 5.2 can be used to assess the significance of the difference between the observed results and the expected numbers in Table 5.1.[4]
  3. (c)Suggest explanations for the observed results in Table 5.1.[4]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

  1. [1] (a) All expected numbers (brown body colour, straight wings: 1827; brown body colour, curved wings: 1827; black body colour, straight wings: 1827; black body colour, curved wins: 1827) are correct.
  2. [1] (b) ref. to 3 degrees of freedom ;
  3. [1] (b) the, x² value / 2098, is greater than 7.815 (at p=0.05);
  4. [1] (b) there is a significant difference ;
  5. [1] (b) reject the null hypothesis ;
  6. [1] (b) any difference is not due to chance / less than 5% (probability) that difference is due to chance;
  7. [1] (b) the, x² value / 2098, is greater than, 11.345 at p=0.01 / 16.266 at p=0.001 or less than, 1% / 0.1% (probability), that difference is due to chance ;
  8. [1] (c) greater numbers of parental phenotypes / fewer numbers of recombinants;
  9. [1] (c) genes are linked / autosomal linkage / genes are on same chromosome ;
  10. [1] (c) alleles inherited together ;
  11. [1] (c) B inherited with D and b inherited with d or B and D in the same gamete and b and d in the same gamete;
  12. [1] (c) no independent assortment;
  13. [1] (c) crossing over (produces recombinants);
  14. [1] (c) during, meiosis / prophase I/gamete formation;
  15. [1] (c) AVP; e.g. idea that genes are far enough apart on chromosome for recombination to occur

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About this question

Topic
Inherited change, A-Level Biology (syllabus 9700). Structured question, 9 marks.
Source
A-Level Biology, Oct/Nov 2023 Paper 4 Variant 2, question 5.

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