2024 VJC H2 Bio P2 MS

1 2024 H2 Biology Prelim Paper 2 Answer 1 Amoeba is a single-celled organism of the kingdom Protista. Am oeba gets its nutrients in a heterotrophic manner, taking in food from its surroundings. Fig. 1.1 shows the electron micrograph of an amoeba. Fig. 1.1 (a) With reference to Fig. 1.1 and the information provided, (i) identify structure X and Y and state a structural difference between them. [2] Identify  X: mitochondrion, Y: vesicle / phagocytic vesicle / food vacuole; Structural difference  X is double membrane bound / presence of cristae while Y is si ngle membrane bound / no infolding of membranes; (ii) explain the importance of structure X to the amoeba. [3]  Mitochondrion (X) allows for respiration to take place to produce ATP;  ATP is required for important metabolic processes within amoeba;  ATP is also important for phagocytosis / endocytosis process t o occur for amoeba to take in food from surrounding;

2 (b) Marseillevirus is a virus that infects amoeba. It is taken up by amoeba via phagocytosis. Upon membrane fusion, new progeny viruses are formed through DN A replication and protein synthesis. The progeny viruses are then released as the amoeba is lysed. Fig. 1.2 shows the structure of Marseillevirus. Fig. 1.2 With reference to Fig. 1.2 and the information provided, explai n how Marseillevirus challenges the cell theory. [3]  Identify why Marseillevirus is not considered as a cell making reference to figure 1.2;  Identify how Marseillevirus exhibits living characteristics with reference to the context;  Identify one tenet of the cell theory that it challenges (cell s are the basic unit of life / all living things are made from cells and all cells come from pre existing cells);

3 2 Some structural features of phospholipids are listed in Table 2.1. Table 2.1 features affect role in cell surface membrane synthesised by condensation hydrophobic fatty acid chains √ unsaturated fatty acid chains √ saturated fatty acid chains √ (a) (i) Indicate, with the use of a tick (√), which of the structural features in Table 2.1 affect the role of phospholipids in the cell surface membrane. [1] (ii) Outline how a property of phospholipids allows them to form an effective barrier as the cell surface membrane. [1]  Amphipathic nature of the phospholipid, having both hydrophilic and hydrophobic regions allow the formation of effective lipid bilayer as barrier; (b) Many multicellular organisms produce antimicrobial polypeptide s (APs) that protect them against prokaryotes. Fig. 2.1 shows one type of AP that acts on the cell surface membrane of prokaryotes. Fig. 2.1

4 Fig. 2.2 shows further information about a channel formed in th e cell surface membrane by the APs. Fig. 2.2 Using Fig. 2.2, calculate the cross-sectional area of the chann el through which ions can pass. Use π = 3.14 in your calculation. Give your answer in nm 2 and to 1 decimal place. [2]  Diameter of channel = 4.0-(1.1+1.1) = 1.8nm  Cross sectional area = ¶r2 = 3.14 x (1.8/2)2 = 2.54 = 2.5nm (1 dp) (c) Prokaryotic membranes do not contain cholesterol. APs damage prokaryotic cells but do not damage eukaryotic cells in the organisms that produce them. Explain how that is possible. [2]  Presence of cholesterol in eukaryotic membranes allow them to be more stable and less fluid, hence restricting the ability of the APs to form channels through them;  Presence of higher composition of saturated fatty acids in the phospholipid tails of eukaryotic membranes make them more rigid and hence difficult for APs to form channels;  APs are unable to attach securely to the surface membrane of e ukaryotic cells due to lack of specificity to the glycocalyx layer; (d) To observe the action of APs on prokaryotes, scientists made u se of a microscope that detects fluorescence, as well as monoclonal antibodies. Suggest how the action of APs on prokaryotes can be observed. [2]  Monoclonal antibodies have an epitope / antigen binding region s that are specific to the Aps;  Upon binding, can be detected by the microscope due to the fluorescence they emit;

5 3 Collagen is the most abundant protein found in the human body. Glycine is a common amino acid found in collagen and plays an important role in contributing to the high tensile strength of collagen. (a) Explain how glycine contributes to the high tensile strength o f collagen. [3]  Small size / Small R group / Hydrogen atom as R group of glycine results in the formation of a kinked helix;  This allows the 3 polypeptides to be packed / wound more closely / tightly, (forming tropocollagen);  (The –NH group of) Glycine on one chain forms inter-chain hydrogen bonds with (the –C=O groups of peptide bonds) one of the other two polypeptide chains;  [Idea of increase the tensile strength of collagen] Greater fo rce is required to break the tropocollagen / collagen;  Strand-like / Fibrous structure (of kinked helices) leads to l arge surface area exposed for more hydrogen bond formation; (b) In the human skin, the enzyme collagenase is synthesised and s ecreted by the fibroblast cells to break down collagen in damaged tissues and help in the growth of healthy tissues. Describe one structural difference between collagenase and collagen and exp lain the significance of this difference to the function of the respective proteins. [3] Structural difference:  Collagenase is globular protein whereas collagen is a fibrous protein; OR Collagenase – hydrophobic groups are kept within the globular s tructure and only hydrophilic groups are exposed on the surface of the molecule vs collagen c onsists of many hydrophobic amino acids; Relate significance of difference to function of protein  Thus, collagenase is soluble in water, allows it to carry out its enzymatic function in aqueous medium/be transported in aqueous medium;  vs collagen insoluble in water, and can remain stable to perform its structural function in solution [idea of maintaining function without dissolving];

6 (c) Collagenase has an optimal pH around pH 7. The pH of the extracellular matrix and tissue microenvironment can influence collagenase activity and collagen turnover. For example, inflammation or infection can alter the local pH and impact collagen degradation processes. Explain how alterations in pH can affect the activity of collagenase. [3]  As pH deviates from the optimum (whether higher or lower), the concentration of hydrogen ions would be changed;  This disrupts the ionic bonds and hydrogen bonds between R groups of amino acids in the enzyme collagenase;  denaturing the enzyme molecule, causing it to lose its specific 3D conformation and shape of the active site such that its no longer complementary to the substrate collagen;  The change in concentration of hydrogen ions may also affect the charges of the active site and the substrate such that they may no longer have opposite electrostatic charges;  preventing effective collisions between enzyme and substrate molecules and the enzyme is unable to form enzyme-substrate complex with substrate molecules; (d) Under natural conditions, healthy skin tissues grow under stri ct regulation. Explain why this is significant. [2] Any one  Ensure skin cells only allowed to proceed to the next phase of the cell cycle if it has properly completed the previous phase; OR Halting the cell cycle when the cell has not completed the previous phase properly;  Cells with damaged DNA/ mutated DNA not allowed to divide so that DNA repair can take place or cells undergo apoptosis when the DNA damage is too severe;  ref check for DNA replication, DNA damage and chromosome-to-sp indle attachments; Any one  This helps to prevent the accumulation of mutations in daughter cells;  [Ide