A-LevelBiologyThe mitotic cell cycle9700/42 · May/June 2018 · Q313 Marks

Within a population, the variation for one characteristic is usually the result of genetic and environmental causes.

(a)(i)Meiosis is one source of genetic variation. (i) Describe the events that take place during prophase I of meiosis in an animal cell.[4]
(a)(ii)Explain how independent assortment of homologous chromosomes leads to genetic variation during meiosis I.[2]
(b)(i)Both genetic and environmental factors contribute to variation in body mass in a bird population. One cause of variation in body mass in chickens is infection with the gut protoctist Eimeria. When fed with the same diet, infected chickens have a lower gain in body mass than healthy chickens in the same population. (i) Suggest reasons why infected chickens have a lower gain in body mass than healthy chickens.[2]
(b)(ii)Suggest one environmental factor that affects the growth and reproduction of Eimeria.[1]
(c)(i)A study was carried out to investigate the effect of treating infected chickens with extracts from a plant, Bidens pilosa. B. pilosa is used in traditional medicine for the treatment of some infectious diseases. • Three populations, each containing 25 chickens infected with Eimeria were observed. • The body mass of each chicken was measured at the start of the study. • Each population was given a different treatment for 56 days: Ο standard diet with no B. pilosa extract Ο standard diet with low dose of B. pilosa extract Ο standard diet with high dose of B. pilosa extract. • The body mass of each chicken was measured at the end of the study and the gain in body mass over the 56 days was calculated. • The mean gain in body mass was calculated for each population. The results are shown in Table 3.1. Table 3.1 treatment mean gain in body mass / g no B. pilosa 1773.0±23.9 low dose of B. pilosa 2093.1 ±34.2 high dose of B. pilosa 2033.3 ±29.9 A statistical analysis of the results of the study of these three populations confirmed that there was a significant difference in mean gain in body mass between low dose B. pilosa and high dose B. Pilosa extract treatments. (i) Name a statistical test that could be used to analyse the results of this study.[1]
(c)(ii)Sketch on Fig. 3.1 curves to show the pattern of variation for gain in body mass in chickens treated with low dose B. pilosa extract and in chickens treated with high dose B. pilosa extract. Label one curve 'low dose' and the other curve 'high dose'. [Fig. 3.1][3]

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 (May/June 2018). The mark or mark type is shown in brackets.

[4] any 4 of: 1 chromosomes become visible; 2 condensation / coiling / supercoiling; A description 3 nuclear envelope or nuclear membrane, disintegrates / disappears; 4 nucleolus, disintegrates / disappears ; 5 centrioles migrate to (opposite) poles ; I centrosomes 6 spindle forms / microtubules assemble ; 7 synapsis / bivalents form / homologous chromosomes pair up; 8 chiasmata formation / crossing over;
[2] any 2 of: 1 (homologous chromosomes / bivalents) line up randomly on the, equator / metaphase plate; 2 produces (many) different chromosome combinations ; 3 number of possible combinations is 2ⁿ; 4 new combinations of maternal and paternal alleles;
[2] any 2 of: 1 protoctist uses up nutrients ; 2 loss of appetite / inability to feed ; 3 idea of impaired absorption / diarrhoea ; 4 (so) less nutrients available for, growth / gain of body mass / increase in biomass ; 5 AVP; e.g. idea of energy used to fight infection / ref. to immune response
[1] (chicken) body temperature / competition with microbes / presence of drugs / presence of plant extract;
[1] t-test ;
[3] any 3 of: 1 two labelled bell-shaped curves ; 2 high dose curve shifted to left of low dose curve ; 3 high dose curve peak to left of 2063 ; 4 high dose curve, narrower / peak higher ;