A-LevelBiologyInherited change9700/41 · Oct/Nov 2018 · Q314 Marks

Meiosis is one process that contributes to genetic variation.

(a)(i)State precisely the stage of meiosis where single chromosomes line up on the equator.[1]
(a)(ii)Outline the events taking place during anaphase I of meiosis.[2]
(a)(iii)Describe how crossing over during meiosis leads to genetic variation.[2]
(b)Mutation also causes genetic variation. Some populations of water hemp, Amaranthus tuberculatus, have evolved herbicide resistance as a result of a mutation. This is a problem for farmers as water hemp grows in crop fields, lowering productivity. Two populations of water hemp were tested for resistance to the herbicide mesotrione. One was a population known to be resistant (control) and the other was a test population, whose resistance was unknown. • Leaves were removed and immersed in a radioactively labelled solution of mesotrione. • The leaves absorbed some mesotrione and became radioactive. • Resistant leaves are able to degrade mesotrione by metabolism. • The time for 50% of absorbed mesotrione to degrade was calculated by measuring the radioactivity of the leaves. The results are shown in [Table 3.1]. [Table 3.1: population of water hemp | mean time for 50% of absorbed mesotrione to degrade / hours | standard deviation - test | 27.5 | 4.75; control | 10.1 | 2.34]
(b)(i)Explain how the results in Table 3.1 show that the two populations differ in their resistance to mesotrione.[2]
(b)(ii)Explain why this example of genetic variation is important for natural selection in water hemp populations.[2]
(b)(iii)Farmers can send in a sample of leaves of water hemp from their fields to a laboratory to be tested for resistance to mesotrione or other herbicides. Suggest the benefit of this to a farmer.[1]
(c)The null hypothesis states there is no significant difference between the mean times for 50% of absorbed mesotrione to degrade in the two populations. A t-test can be carried out to compare these two means. The critical value for t at the p = 0.05 significance level is 2.23.
(c)(i)Use the formula in [Fig. 3.1] to calculate the value of t. Show your working. [Fig. 3.1: t = (X̄₁-X̄₂)/√(s₁²/n₁ + s₂²/n₂) where X̄ = mean, s = standard deviation, n₁ = 6 (number of readings for test population), n₂ = 6 (number of readings for control population)] t =[2]
(c)(ii)Use your calculated value of t to explain whether the null hypothesis should be accepted or rejected. accept or reject ... explanation ...[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[1] metaphase, II / 2 ;
[2] any two from 1 spindle (fibres) / microtubules, contract/shorten ; 2 chromosome(s), move / pulled, to pole(s) / centriole(s) ; 3 centromeres, move / lead / pulled, first; 4 homologous chromosomes separate or pair (of chromosomes) / bivalent, pulled apart ;
[2] any two from 1 non-sister chromatids ; 2 swap/exchange, alleles / sections of DNA ; 3 chiasmata; 4 (giving) new / different, allele combinations or recombination ;
[2] any two from 1 test (leaves / population / group) takes, longer / higher time or test group degrades mesotrione more slowly ; 2 so test group, less resistant / fewer have resistance; 3 test standard deviation is, larger / higher, so (population) more variable ;
[2] any two from 1 can, adapt/evolve / survive long term; 2 resistant/mutant, plants / hemp, have selective advantage; 3 in areas where / when, mesotrione is used; 4 resistant/mutant, plants survive and reproduce ;
[1] any one from 1 lets farmer choose, different / correct / effective / best, herbicide ; 2 less money wasted due to, wrong herbicide / reduced crop yield ;
[2] numerator working 17.4; denominator working √3.76 + 0.91 OR √4.67 OR 2.16 ;
[2] any two from 1 reject because, t/ calculated value / it, is, higher / greater / more, than, 2.23/ critical value ; 2 probability of result by chance is less than, 0.05 /5% or p < 0.05 ; 3 there is a significant difference between the, means / populations / test and control;