2025 YIJC [H2] Prelim P1 (A) with Examiner Report

©YIJC 9744/01/PE/2025 [Turn over YISHUN INNOVA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CG INDEX NO BIOLOGY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet 9744/01 19 September 2025 1 hour READ THESE INSTRUCTIONS FIRST Do not use staples, paper clips, glue or correction fluid/tape. Write your name, index no. and CG on this cover page and OTAS provided unless this has been done for you. There are thirty questions on 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 Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 24 printed pages. ANSWERS

2 ©YIJC 9744/01/PE/2025 [Turn over 1 Which row correctly describes a cell organelle? organelle description A centriole a hollow cylinder that generates spindle microtubules and is enclosed by a single membrane B lysosome a spherical sac containing digestive enzymes that is formed by the smooth endoplasmic reticulum C mitochondrion a single membrane enclosing an inner membrane folded into cristae to house electron transport chains D ribosome two unequal subunits made of rRNA and proteins, attached permanently to the rough endoplasmic reticulum A – Incorrect: A centriole is a hollow cylinder made of nine triplets of microtubules and is not surrounded by any membrane. B – Incorrect: Lysosomes are spherical sacs containing digestive enzymes, but they are formed by the RER & Golgi apparatus, not the smooth endoplasmic reticulum. C – Correct: A mitochondrion has a double membrane — an outer membrane and a folded inner membrane (forming cristae) D – Incorrect: Ribosomal subunits are unequal in size (large and small subunit) and are found free in the cytoplasm or attached to the rough ER ER: Poorly done. Only 28% of candidates answered correctly. Most chose A (31%) or B (34%) instead of the correct answer C.

3 ©YIJC 9744/01/PE/2025 [Turn over 2 The table compares some features of three different proteins, P, Q and R. Of these proteins: • one has a fibrous quaternary structure • one has a globular tertiary structure • one has a globular quaternary structure protein number of genes coding for the protein % of protein composed of the commonest amino acid % of protein composed of small amino acids durability P 2 46 77 less sensitive to changes in pH and temperature Q 1 24 10 more sensitive to changes in pH and temperature R 2 21 8 more sensitive to changes in pH and temperature Which row correctly identifies each protein? fibrous quaternary structure globular tertiary structure globular quaternary structure A P R Q B P Q R C R P Q D R Q P A – Incorrect: R is not the globular tertiary structure — it has 2 genes coding for it, which is more typical of a quaternary protein. Its low % of small amino acids (8%) also does not fit fibrous proteins. B – Correct: P is the fibrous quaternary structure (2 genes, high % commonest amino acid, high % small amino acids, less sensitive to pH/temp), Q is the globular tertiary structure (1 gene, many amino acid types, more sensitive to pH/temp), and R is the globular quaternary structure (2 genes, more sensitive to pH/temp). C – Incorrect: P is not a globular quaternary structure — its high % of small amino acids (77%) and repetitive sequence (46% commonest amino acid) fit fibrous proteins, not globular. D – Incorrect: P is not a globular quaternary structure and R is not fibrous — R has many amino acid types and is more sensitive to pH/temp, which fits a globular protein. ` ER: Well done

4 ©YIJC 9744/01/PE/2025 [Turn over 3 Which of the following statements below are correct? 1. During endocytosis, vesicles bud inwards from the cell surface membrane, leading to a loss of phospholipids. 2. During endocytosis, vesicle membrane fuses with the cell surface membrane, leading to a gain of phospholipids. 3. During exocytosis, vesicles fuse with the cell surface membrane, leading to a gain of phospholipids. 4. During exocytosis, vesicles bud off from the cell surface membrane, leading to a loss of phospholipids. A 1 and 2 B 1 and 3 C 1 and 4 D 2 and 4 Statement 1 – Correct: In endocytosis, vesicles bud inwards from the plasma membrane. This removes a portion of the membrane, resulting in a net loss of phospholipids from the plasma membrane. Statement 2 – Incorrect: Endocytosis does not involve vesicle membrane fusion with the plasma membrane — that occurs in exocytosis. Endocytosis decreases, not increases, the phospholipid content of the plasma membrane. Statement 3 – Correct: In exocytosis, vesicles formed inside the cell (e.g., from the Golgi apparatus) fuse with the plasma membrane. This increases the plasma membrane surface area and results in a net gain of phospholipids. Statement 4 – Incorrect: Exocytosis does not involve vesicles budding off from the plasma membrane — instead, they fuse with it. Therefore, there is no loss of phospholipids during exocytosis. Take note! HIV Entry: Direct fusion of the viral envelope with the host cell membrane. Influenza Entry: Receptor- mediated endocytosis followed by fusion within an endosome. ER: Poorly done. Only 45% of candidates answered correctly. Most chose D (38%) instead of the correct answer C.

5 ©YIJC 9744/01/PE/2025 [Turn over 4 Carbon monoxide poisoning occurs when CO binds to haemoglobin with a much higher affinity than oxygen, reducing oxygen uptake. Which graph shows the effect of oxygen concentration on the rate of oxygen binding to haemoglobin, with and without carbon monoxide? Correct Answer: D • Haemoglobin shows sigmoidal (S-shaped) curves due to cooperative binding (binding of one O2 results in the easier binding of other O2 molecules). • CO acts as a competitive inhibitor, since it is structurally similar to oxygen and binds to the active site (haem group) of haemoglobin. • With CO present, the curve shifts to the right, showing that a higher oxygen concentration is needed to achieve the same saturation (reduced apparent affinity). • The maximum rate (Vmax) is also reduced, because some haem groups are permanently occupied by CO and cannot bind oxygen at all. • This matches graph D, where the “with inhibitor” curve is right-shifted and lower than the “without inhibitor” curve. ER: Poorly done. Only 39% of candidates answered correctly. Most chose C (48%) instead of the correct answer D.

6 ©YIJC 9744/01/PE/2025 [Turn over 5 Stem cells can be used to replace cells in damaged tissues, such as skin. The diagram shows two ways in which replacement skin cells can be produced. Which row correctly describes the stem cells P, Q and R? X Y Z A pluripotent multipotent totipotent B pluripotent induced pluripotent induced pluripotent C totipotent pluripotent induced pluripotent D totipotent totipotent totipotent X – Totipotent stem cell • X represents the zygote, formed immediately after fertilisation. • A zygote is totipotent, meaning it can give rise to all cell types in the body and extraembryonic tissues (e.g., placenta). Y – Pluripotent stem cell • Y represents embryonic stem cells derived from the early embryo. • These cells are pluripotent Z – Induced pluripotent stem cell (iPSC) • Z is generat