A-LevelBiologyInherited change9700/43 · May/June 2018 · Q313 Marks

Meiosis is described as reduction division. Explain why meiosis is necessary in the life cycle of a sexually reproducing organism.

(a)Meiosis is described as reduction division. Explain why meiosis is necessary in the life cycle of a sexually reproducing organism.[3]
(b)(i)Name two examples of environmental conditions that affect plant phenotype, other than mineral deficiency. For each example, describe how it affects the phenotype.[2]
(b)(ii)Explain why mutations are important in selection.[2]
(c)A study compared root growth of thale cress, Arabidopsis thaliana, in two different soil types: • full nutrient • low phosphate. Two different populations of thale cress were used: • thale cress with a functional enzyme X • thale cress, with a non-functional enzyme X. 30 seedlings from each population were placed in each type of soil and left to grow for seven days. At the start all seedlings had a root of the same length. After seven days, the length of this root was measured again for each seedling. The mean final root length and standard deviation was calculated for each population of thale cress. The mean final root lengths are shown in [Fig. 3.1]. With reference to [Fig. 3.1], describe the effect of the low phosphate soil type compared to the full nutrient soil type on root growth, for both populations of thale cress.[2]
(d)(i)The null hypothesis states there is no significant difference between the mean final root lengths of the two populations of thale cress grown in low phosphate soil type. 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 2.00. | population grown in low phosphate soil type | mean final root length / mm | standard deviation | |:--------------------------------------------|:----------------------------|:-------------------| | functional enzyme X | 21 | 0.5 | | non-functional enzyme X | 33 | 0.8 | [Fig. 3.2] shows the formula for calculating the value of t. ``` |X₁ - X₂| t = ————————— √(s₁²/n₁) + (s₂²/n₂) ``` x = mean s = standard deviation n = sample size (number of measurements) Use the formula in [Fig. 3.2] to calculate the value of t. Show your working. t = ..............................................................[2]
(d)(ii)Use your calculated value of t to explain whether the null hypothesis should be accepted or rejected. accept or reject ........................................................ explanation ......................................................................................................................................................................................................................................................................................................................................................[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 (May/June 2018). The mark or mark type is shown in brackets.

3(a) three from:
1 idea of halving the number of chromosomes ;
2 (to make) haploid / n, gametes ;
3 (so) on fertilisation, (restores) diploid number/2n;
4 maintains constant chromosome number / prevent chromosome number doubling ;
3(b)(i) two from:
1 ref. to light intensity and change in size of, plant/ part of plant;
2 wind speed and size of plant;
3 waterlogged soil and leaves / flowers, above water;
4 pH and flower colour ;
5 AVP; environmental condition and effect
6 AVP; environmental condition and effect
3(b)(ii) two from:
1 genetic variation / new alleles ;
2 (therefore) new phenotypes;
3 (new phenotype) has a selective advantage / is selected for ;
4 more likely to survive (selection pressure);
3(c) two from:
1 in full nutrient there is little difference (between two populations);
2 in low phosphate non-functional X root growth is, longer / ora ;
3 in low phosphate both populations grow less well than in full nutrient / ora ;
4 comparative data quote to support mp1/mp2/mp3; A processed data
3(d)(i) 69.67 to 70.6 ;; max 2 d.p
allow 1 working mark for
21-33 = (-)12 (for calculating difference between means)
√0.52 / 30 + 0.82/30 (for calculating the standard error in the difference)
3(d)(ii) reject null hypothesis
t/69.67, is higher than, 2.00 / the critical value (at p = 0.05);
there is a significant difference between, the means of the functional X and non-functional X populations;
allow ecf from calculated value of t