A-LevelBiologyInherited change9700/42 · Feb/Mar 2022 · Q415 Marks

(a) Meiosis is an important process that contributes to genetic variation in a population. Fig. 4.1 shows a pair of homologous chromosomes during prophase I of meiosis. [Figure 4.1]

(a)(i)On Fig. 4.1, use the letter C with a label line to identify where crossing over occurs.[1]
(a)(ii)Describe how crossing over produces genetic variation in a population.[4]
(b)(i)Wing pattern in the butterfly species Heliconius melpomene is controlled by genes on autosomal chromosomes. The gene for banding pattern on the upper wing has two alleles: • a dominant allele coding for a full band • a recessive allele coding for a broken band. The gene for ray pattern on the lower wing has two alleles: • a dominant allele coding for rays • a recessive allele coding for no rays. Scientists crossed a butterfly that was homozygous dominant for both genes with a butterfly that was homozygous recessive for both genes. The scientists wanted to check whether the phenotypic ratio for offspring in the F2 generation agreed with the expected phenotypic ratio of 9 : 3 : 3 : 1. The results of this genetic cross are shown in Fig. 4.2. [Figure 4.2] Explain the term F1 generation.[2]
(b)(ii)The scientists used the chi-squared (x²) test to compare their data to the expected phenotypic ratio of 9 : 3 : 3 : 1. The formula for the chi-squared test is shown in Fig. 4.3. [Figure 4.3] Complete Table 4.1 and use the chi-squared formula in Fig. 4.3 to calculate the x² value for these data.[3]
(b)(iii)The critical value at the 0.05 probability level and three degrees of freedom is 7.81. Using the result of your calculation in (b)(ii), explain whether the results of the study agree with the expected ratio of phenotypes for the F2 generation.[3]
(b)(iv)Suggest two reasons why phenotypic ratios in the F2 generation do not always match the expected ratios.[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.

[1] label line to where white chromatid crosses over grey chromatid ;
[4] any four from: 1 (crossing over is) between non-sister chromatids ; 2 (non-sister chromatids have) different combinations of alleles ; 3 exchange alleles; 4 (chromatids have) new combinations of alleles ; 5 linkage groups broken ; 6 idea that gametes have unique combination of alleles ; 7 random fusion of gametes (during fertilisation) ;
[2] any two from: 1 (first) generation of offspring (from parental cross) ; 2 (parents are) homozygous / true bred / pure bred / pure breeding; 3 F1 generation will be heterozygous ;
[3] phenotype | expected ratio | O | E | O-E | (O – E)² | (O – E)² / E full band and rays | 9 | 99 | 76.5 | 22.5 | 506.25 | 6.62 full band and no rays | 3 | 21 | 25.5 | -4.5 | 20.25 | 0.79 broken band and rays | 3 | 8 | 25.5 | -17.5 | 306.25 | 12.01 broken band and no rays | 1 | 8 | 8.5 | -0.5 | 0.25 | 0.03 1 mark for E and O – E correct; 1 mark for (O – E)² and (O – E)² / E correct; x² = 19.45; ecf for 1 mistake in one box
[3] any three from: 1 19.45/chi-squared value, is greater than, 7.81 / critical value (at p = 0.05); 2 results are / observed ratio is, significantly different to expected ratio; 3 so differences are not due to chance; 4 due to, increase in full band phenotypes / decrease in broken band phenotypes;
[2] any two from: 1 epistasis; 2 linkage; 3 lethal allele; 4 chance deviations; 5 environmental effects; 6 mutations ; 7 small sample size ;