2020 EJC Promo JC1 P2 Bio (A)

©EJC 2020 9744/02/J1H2PROMO/2020 [Turn over EUNOIA JUNIOR COLLEGE JC1 Promotional Examinations 2020 General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME ANSWER KEY CIVICS GROUP 2 0 - REGISTRATION NUMBER H2 Biology Paper 2 Structured Questions & Free Response Questions 9744/02 02 October 2020 2 hours Candidates are to answer questions in Section A in this question booklet. Candidates are to answer questions in Section B in the answer booklet provided. Additional Materials: 8-page Answer Booklet READ THESE INSTRUCTIONS FIRST Write your name, civics group and registration number on all the work you hand in. There are two sections in this paper, Section A and Section B. You are advised to plan your time appropriately to complete both Sections. Answer all questions. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue, or correction tape/fluid. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 16 printed pages and 1 blank page. For Examiner’s Use 1 2 3 4 5 6 7 8 Total 80

2 ©EJC 2020 9744/02/J1H2PROMO/2020 For Examiner’s Use Section A Answer all the questions in this section. 1 Emperor penguins are the only species of penguin that breed during the Antarctic winter and they may stand still for multiple days to incubate their eggs. In order t o survive in freezing temperatures, the feet of emperor penguins contain a high percentage of unsaturated fats. (a) Explain the significance of unsaturated fatty acids (fatty acid tails) in cell membranes. [2] 1. Presence of C=C bonds causes kinks in unsaturated fatty acid tails [R: unsaturated fats] of phospholipids [R: kinks in membrane]; 2. This prevents close packing of phospholipids at low temperatures , reducing hydrophobic interactions between fatty acid tails, thus maintaining membrane fluidity [R: increasing fluidity] OR preventing cell membranes [R: unsaturated fatty acid tails] from freezing/solidifying; Teacher’s comments:  Please take note of the edit in the question. Due to the errata, marks were still awarded if students wrote fatty acids instead of fatty acid tails. However, note that fatty acids are different from fa tty acid tails. Fatty acids are long -chained carboxylic acids, consisting of a long hydrocarbon chain and a carboxyl group. Fatty acid tails are fatty acids that have formed ester bonds with glycerol in phospholipids (or triglycerides).  Fats were rejected as fats are esters of fatty acids but the specific type of lipids in cell membranes is phospholipids.  Quite a large number of students mentioned the presence of kinks but did not state that they are caused by C=C bonds, which are only p resent in unsaturated fatty acid tails, but absent in saturated fatty acid tails.  Unsaturated fatty acid tails only confer higher membrane fluidity in comparison to saturated fatty acid tails at the same temperature. At lower temperatures, the kinetic energy of phospholipids decreases and phospholipids are more closely packed together due to stronger hydrophobic interactions. Presence of unsaturated fatty acid tails in phospholipids helps to MAINTAIN membrane fluidity by reducing the extent at which the membrane loses its fluidity, but not increasing the fluidity. (b) Fig. 1.1 shows the structure of a lipoprotein in the blood of emperor penguin s. Lipoproteins transport fats from the liver to other tissues via the bloodstream. The membrane proteins of lipoproteins allow for fats to be deposited to the target tissue. Fig. 1.1

3 ©EJC 2020 9744/02/J1H2PROMO/2020 [Turn over] For Examiner’s Use With reference to Fig 1.1, describe how the structure of lipoproteins allow s for fats to be transported from the liver to a specific tissue via the bloodstream. [3] 1. Phospholipids form a single layer / monolayer [R: bilayer] to serve as boundary between blood and (hydrophobic) fats; 2. (Non-polar) hydrophobic hydrocarbon / fatty acid tails of phospholipids face inwards and interact with hydrophobic fats in the lipoprotein core OR exclude water; 3. (Charged) hydrophilic phosphate heads of phospholipids face outwards and interact with (polar) water molecules in blood / aqueous environment in blood, allowing the lipoprotein to be soluble in blood; 4. Membrane protein recognises and binds to receptors of target cells [R: target tissue] via complementary conformation and charge; Points 1-3 (max 2) Teacher’s comments:  This application question is generally not so well done. Question requires student to make the conceptual link between structure and function of diffe rent components of lipoprotein (phospholipids and membrane protein).  Some students wrongly identified the phospholipid monolayer as bilayer, which suggests lack of / incorrect interpretation of Fig 1.1.  To obtain points 2 and/or 3 , students must address the orientation of the hydrophilic phosphate heads and/or hydrophobic fatty acid tails, AND suggest what the y interact with respectively, based on the information given in the question.  Lipoprotein is a macromolecular complex of lipids and proteins. It is NOT a cell, so it is wrong to refer the phospholipid monolayer as cell membrane, or suggest that membrane protein is involved in cell-cell recognition.  Membrane protein of lipoprotein, also known as apolipoprotein (FYI), is NOT transport protein (channel protein / carrier protein) / enzyme / receptor protein. It serves as a ligand to bind to lipoprotein receptors on cell surface membrane.  With reference to Fig 1.1, fats inside the lipoprotein have 3 tails, which differ from the surrounding phospholipids with 2 tails. The fats are triglycerides (glycerol attached to 3 fatty acid chains).  Fats (triglycerides) are non-polar and hydrophobic, NOT polar and hydrophilic. (c) Explain how the structure of haemoglobin allows it to transport oxygen efficiently in red blood cells. [3] 1. [Binding of 4 oxygen molecules] Haemoglobin has 4 subunits (2 -globin subunits and 2 -globin subunits), each containing a polypeptide chain (globin) and a haem group [A: ref. to Fe2+ ion in haem group], which allows one haemoglobin to carry up to 4 oxygen molecules at the same time; 2. [Reversible binding of oxygen] Fe2+ ion of the haem group binds reversibly to oxygen; 3. [Cooperative binding of oxygen] Weak intermolecular bonds (hydrogen bonds, ionic bonds, hydrophobic interactions) between the 4 subunits allow for cooperative binding of oxygen when binding of one oxygen molecule to one subunit induces a conformational change in the other 3 subunits, increasing their affinity for oxygen; [A: unloading of one oxygen molecule from one subunit induces conformational change in other 3 subunits, reducing their affinity for oxygen]

4 ©EJC 2020 9744/02/J1H2PROMO/2020 For Examiner’s Use 4. [Solubility] Polar/charged, hydrophilic R groups of amino acids project outwards while non-polar, hydrophobic R groups of amino acids are pointed towards the interior, shielded away from the aqueous environment. This makes haemoglobin soluble in the cytoplasm of red blood cells for efficient transport of oxygen; 5. [Compactness] Its globular structure makes it compact so that more haemoglobin can be packed into red blood cells (RBCs) for more efficient oxygen transport; Teacher’s comments:  Students who are clear about the structure of ha