2022 JC2 H1 BT P1 annotated answers

Civics Group Index Number Name (use BLOCK LETTERS) ST. ANDREW’S JUNIOR COLLEGE 2022 JC2 BLOCK TEST H1 BIOLOGY 8876/1 Paper 1: Multiple Choice Thursday 7th July 2022 1 hour Additional Materials: Multiple Choice Answer Sheet Soft clean eraser (not supplied) Soft pencil (type B or HB is recommended) READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, civics group and index number on the multiple choice answer sheet in the spaces provided. There are 30 questions in this paper. Answer all questions. For each question, there are four possible answers, A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate multiple choice answer sheet. INFORMATION TO CANDIDATES Each correct answer will score one mark. A mark will not be deducted for wrong answer. Any rough working should be done in this booklet. At the end of the examination, submit both question paper and multiple choice answer sheet. This document consists of 19 printed pages. [Turn over H1

2 SAJC / H2 Biology 9744/1 JC2 BT 2022 1 A scientist carried out an experiment to separate cell structures in animal cells. The cells were lysed to release the cell structures. This extract was transferred into a centrifuge tube and then spun in a centrifuge. The heaviest cell structure sank to the bottom, forming pellet 1, as shown in the diagram. The liquid above pellet 1 was poured into a clean centrifuge tube and spun in the centrifuge at a higher speed to separate the next heaviest cell structure. This cell structure sank to the bottom, forming pellet 2. This procedure was repeated twice more to obtain pellet 3 and pellet 4, each containing a single type of cell structure. Which row shows the order in which the cell structures were collected? Pellet 1 Pellet 2 Pellet 3 Pellet 4 A Nucleus Lysosomes Mitochondria Ribosomes B Nucleus Mitochondria Lysosomes Ribosomes C Ribosomes Lysosomes Mitochondria Nucleus D Ribosomes Mitochondria Lysosomes Nucleus Explanation: Pellet 1 should contain the largest cell organelle (nucleus) and cell debris Students should work from pellet 4 backwards to figure out the cell structures in pellet 3 and 2 respectively Pellet 4 should contain ribosomes (smallest organelle) Hence Pellet 2 contains mitochondria (ave size 0.5–3m) and Pellet 3 contains lysosomes (ave size 0.1–1.2 m)

3 SAJC / H2 Biology 9744/1 JC2 BT 2022 2 The figure below shows an electron micrograph with two plant cells. Which of the following statements correctly describe the labelled structures? 1. R contains circular DNA and is found in both prokaryotic and eukaryotic cells. 2. P has a fluid mosaic structure and regulates the movement of substances between the two plant cells. 3. S acts as a selective permeable barrier. 4. Q contains enzymes which play an important role in cell division. A 1 and 2 only B 2 and 3 only C 3 and 4 only D 1, 2 and 3 only Explanation: R is a mitochondrion → does have circular DNA, but only found in eukaryotic cells. P is the cellulose cell wall, not the phospholipid bilayer. S is the plasma membrane, hence is selectively permeable. Q is the nucleus, it contains enzymes like DNA polymerase, helicase, ligase which are involved in DNA replication, which has to occur prior to cell division. It also contains RNA polymerase which transcribes genes whose products trigger the start of cell division.

4 SAJC / H2 Biology 9744/1 JC2 BT 2022 3 The graph shows how the rate of entry of glucose into a cell changes as the concentration of glucose outside the cell changes. What is the cause of the plateau at X? A The cell has used up its supply of ATP. B All the carrier proteins are saturated with glucose. C The carrier proteins are denatured and no longer able to function. D The concentrations of glucose inside and outside the cell are equal. Explanation: Graph shows that as concentration of glucose outside the cell increases, the rate of entry ot glucose into the cell increases until a plateu. This implies that the movement of glucose is a passive process (i.e. does not require ATP), hence option A is wrong. Glucose is a polar molecule, and require a hydrophilic channel to pass through the hydrophobic core of the membrane. Hence, the rate of entry will reach a maximum when the carrier/channel protein becomes saturated with glucose when glucose concentration outside the cell becomes sufficien tly high. Option C is wrong because if carrier proteins are denatured, the rate of glucose entry into the cell will decrease drastically instead of being at maximum. Option D is wrong because the rate of glucose entry into the cell will become zero when th e concentration of glucose inside and outside the cell are equal. 4 The diagram shows the structure of the cell membrane with molecules labelled 1 to 6. Which row correctly identifies function of two of the numbered molecules?

5 SAJC / H2 Biology 9744/1 JC2 BT 2022 molecule function molecule function A 1 Cell-to-cell adhesion 4 stabilizes the membrane B 2 acts as a receptor Glycolipid functions as a receptor for cell-cell communications 5 active transport no hydrophilic channel present for transport C 3 facilitated diffusion phospholipid bilayer acts as a barrier to prevent diffusion of charged and polar molecules 4 regulates the fluidity of the membrane D 6 active transport 5 acts as a receptor receptors have to face the exterior surface of the plasma membrane ANS: A 5 The graphs show the effects of temperature and pH on enzyme activity. Which of the following is true? A At P, hydrophobic interactions are formed between the enzyme and substrate. B At Q, the kinetic energy of substrate and enzyme is the highest. C At R, hydrogen bonds in the enzyme break. D At S, peptide bonds in the enzyme break. Explanation: Option B is wrong because kinetic energy of substrate and enzyme continues to increase beyond Q as temperature increases. Option C is wrong because R is the optimum pH of the enzyme, where all the bonds between R groups are intact. Option D is wrong because denaturation of proteins (whether it is due to high temperature, or huge deviation from optimum pH) only breaks the non -covalent bonds between R groups (e.g. ionic bonds, hydrophobic interactions and hydrogen bonds). Peptide bonds are not broken during denaturation.

6 SAJC / H2 Biology 9744/1 JC2 BT 2022 6 Graph Y shows the progression of an enzyme -catalysed reaction carried out at 20°C, with 1.0 moldm-3 of substrate, 0.2 moldm-3 of enzyme, and at its optimum pH. Graphs X and Z show the progression of the same enzyme-catalysed reaction, but with one or more changes in the experimental condition. Which of the following correctly identifies the change (s) in experimental condition resulting in Graphs X and Y? Graph X Graph Z A 0.4 moldm-3 of enzyme is used Temperature is decreased to 10°C B 2.0 moldm-3 of substrate is used Temperature is decreased to 10°C C Temperature is increased to 30°C 0.5 moldm-3 of substrate is used D 0.1 moldm-3 of enzyme is used pH is increased by 2. Explanation: Graph X has a higher rate of enzyme activity, but produced the same concentration of product. Hence, it can be due to more enzyme being added, or increased temperature, but not increase in substrate concentration. Graph Z has a rate of enzyme reaction, or produced a lower concentration of products. Hence, it can be due to complete denaturation of enzyme due to increased pH before all the substrate has been converted into product, or due to decreased concentration of substrate. A decrease in tem perature alone will only slow down the rate of reaction, but the graph should sti