VS 2017 O Level Pure Biology P2 MS w Feedback

SUGGESTED MARK SCHEME – BIOLOGY 2017 Qn No Answers Mark Feedback 1(a) maintenance of a close to constant internal environment; within narrow and acceptable limits 1 1 1(b) glands which secrete chemical substances called hormones directly into bloodstream , without a duct, in minute amounts; hormone travels to one or more target organs to influence a change in growth, development and activity of the organism, 1 1 1(c)(i) A: glycogen B: glucagon 1 1 1(c)(ii) exercise is an energy-consuming process requires large amount of energy release f or muscle contraction; high respiration rate in muscle cells requires large amounts of glucose since glucose is a respiratory substrate; thus blood glucose concentration decreases 1 1 Most candidates correctly identified the substances A and B ; some did not make the link between respiration and the use of glucose. Incorrect references to ‘producing’ energy were seen in weaker responses. 2(a) contains air for gaseous exchange with the environment for photosynthesis and respiration; OR increase buoyancy of the leaf so that it can float on the surface of water; 1 Many candidates knew that intercellular air spaces have a role in gas exchange. The context was often linked to exchange with the environment. 2(b) exposed to the most amount of light for maximum rate of photosynthesis 1 Good responses referred to maximum light. 2(c)* green pigment found in the chloroplast to help trap light , converting light energy to chemical form 1

splitting water molecules in the light-dependent stage of photosynthesis, releasing oxygen as by-product leading to formation of glucose in later/ light-independent stage 1 1 2(d) 1 1 3(a)* any one from 1. occupies the first trophic level of the food web; 2. autotrophic, converts light energy to chemical energy; 3. source of food for consumers 1

3(b) 74; 10; 2; 1 1 1 A common error was to count the number of species instead of the number of organisms 3(c) size is at least 50% of grid bars proportionate to numbers each trophic level accurately labelled 1 1 1 Good responses included a scale for their drawings and clearly labelled the trophic levels. 4(a) Measure the initial mass of D, E and F before placing the set-ups in its respective environmental conditions. At the end of the experiment, measure the final mass of each set-up. 1

For each set-up, calculate the loss in mass by subtracting final mass from initial mass. 1 4(b) transpiration 1 4(c) Plant D. Plant D has more leaves than F greater surface area from which water can be lost from; Leaves of D were exposed to wind while leaves of E were contained within a humid environment, limiting water loss by transpiration due to decreased water vapour concentration gradient; Thus plant D will lose more water than E and F by transpiration. 1 1 4(d) To ensure that any loss of water from the set-ups can only occur through the leaves by transpiration and not by evaporation from the moist soil. 1 5(a) The process resulting in the production of genetically identical offspring; from one parent, without the fusion of gametes. 1 1 Weaker responses confused asexual reproduction with self-fertilisation.

5(b) 1. Zygote moves down the fallopian tube to uterus (develop) by sweeping action of cilia (describe) of the tube and by peristaltic action (describe); 2. Zygote divides by mitosis to form a ball of cells called embryo; (development) 3. Embryo implants itself in the thickened endometrium+ maintained by progesterone released by corpus luteum; 4. Embryonic villi grow from embryo into endometrium → forming placenta 5. Formation of amniotic sac (survive) and amniotic fluid which ensure survival by protecting the fetus from…… 6. Umbilical cord attaches embryo to placenta, where diffusion of useful substances occurs from maternal blood to foetal blood, and diffusion of waste from foetal to maternal blood occurs (survive). 1 1 1 1 1 Weaker responses did not always correctly differentiate between the terms zygote, embryo and fetus. These weaker responses also showed some misconceptions about the relationship between the placenta, umbilical cord and amniotic sac. 6(a)(i) 1400 1 6(a)(ii) As exercise intensity increased, the volume of each breath increased. 1 6(a)(iii) When exercise intensity = 2 a.u. volume of each breath = 1700 cm 3 Total volume of air breathed in one minute = 1700 x 26 1

= 44200 cm 3 1 6(b) respiration 1 7(a) cardiac muscle tissue contracts to pump blood out of left ventricle relaxes to enable left ventricle to fill with blood 1 1 Most candidates were able to identify the tissue in Fig. 7.1 and describe its function. Weaker candidates identified the tissue as pericardium or left ventricle and did not make it clear that force or pressure was required to move the blood. 7(b)(i) 5.6 1 7(b)(ii) 19 – 7 = 12 minutes 1 7(b)(iii) increase in HR from 2 mins to 5 mins: 168 – 66 = 102 bpm percentage increase: 102/66 x 100% = 155% 3 7(b)(iv)* During exercise, from 2 minutes to 4 minutes, heart rate i ncreased rapidly from 66 to 160 bpm... how many times or by how many units - MANIPULATE THE DATA! from 4 minutes to 6 minutes, the heart rate increases gradually from 160 to the maximum of 170 bpm. From 6 minutes to 7 minutes, the heart remained constant at 170 bpm. 1 1 1 Qn No Answers Mark Feedback

8(a) size is at least 50% of grid provided axes labelled, with units points accurately plotted, allow 1 misplot curve of best fit / point-to-point 1 1 1 1 8(b) when milk fat was digested by lipase, glycerol and fatty acids were produced; fatty acids neutralised the sodium carbonate solution → making it less alkaline, below pH 8; causing phenolphthalein to turn colourless 1 1 1

8(c) at 55˚C, lipase was denatured by the high temperature ; milk fat is not digested and fatty acids are not produced 1 8(d) Optimum temperature of the enzyme is the temperature where enzyme activity is the maximum/highest (at 35˚C → time is shortest at 2.5min) enzyme-substrate complexes formed; Fatty acids produced will be highest at this temperature ; Time taken to cause phenolphthalein to turn colour would be the shortest 1 1 9(a) Total water gain = 1600 + 700 + 200 = 2500 cm 3 per day 1 9(b)* metabolic water is water produced as a result of cellular chemical reactions; such as respiration, condensation reaction in the formation of bonds 1 1 9(c) In a cold environment , less sweat is produced as there is less need to lose latent heat via evaporation of water in sweat; As less water in blood is used to form sweat, water potential of blood increases; Osmoregulation of blood occurs → more water is loss in