XMSS 2025 4E Bio Prelim Paper 2 MS

Marking Scheme Biology (6093) Preliminary Examination 2025 Paper 2 – Section A Qn Marking Scheme Marks Marker’s Report 1a 1. Higher water potential in liquid/distilled water compared to (cytoplasm) of the red blood cells (ORA) 2. Water molecules enters the red blood cells by osmosis/ through partially permeable membrane 3. Cell expands and bursts, no cells seen/ contents released in solution 3 1b 1. crenate 1 1ci 2. nucleus 1 1cii 1. more space to carry more haemoglobin 2. ref to more efficient transport of oxygen to all parts of the body 2 2a 1. lipase with bile solution decreases in pH to 6.5 (+/- 0.1) faster at 45 minutes compared to lipase without bile at 60 minutes 2. bile emulsify large fat globules into smaller fat droplets 3. faster digestion of fats to fatty acid and glycerol by lipase + accumulation of fatty acid to reduce pH 3 2bi 1. straight line at 8.5 1 2bii 1. denatured + active site lost 2. lipase no longer able to digest (fats into fatty acid and glycerol) 2 2c 1. correct identification of E (rER /A. ribosomes) , F (mitochondrion) and G (nucleus) 2. E- ribosomes synthesise the lipase proteins + rER transport to the Golgi body for secretion 3. F – during aerobic respiration release energy required for synthesis of the proteins 4 Mark scheme will use these abbreviations / + R A Ig ref ecf AW AVP ORA OWTTE underline ( ) alternatives statements on both sides of the + are needed for that mark reject accept (for answers correctly cued by the question) ignore as irrelevant with reference to error carried forward alternative wording (where responses vary more than usual) alternative valid point or reverse argument or words to that effect actual word given must be used by candidate (grammatical variants excepted) the word / phrase in brackets is not required but sets the context

4. G – nucleus contains the genetic information/ gene that codes for the protein A: correct ref to transcription and translation process 3a 1. L- blind spot 2. no photoreceptors thus no light detected 2 3b 1. K (fovea), has no rods thus cannot detect at dim light 2. K (fovea) has the highest concentration of cones/photoreceptors thus higher chance to detect light at high intensity 3. J ( surrounding retina) contains rods that can detect light at dim light 2 3c 1. Ref to light as stimulus 2. Detected by photoreceptors + generate nerve impulses 3. Transmitted to sensory neurones to relay neurones (within brain/CNS) to motor neurone 4. Through synapse 5. Transmitted to effector 6. Circular muscles of iris contract + radial muscles of iris relax + pupil constrict R: if no mention of iris max 5 4a 1. A (8.8 - 6.9 = (+)1.9) B (8.95 – 6.9 = (+)2.05) C (6.1 – 6.9 = -0.8) all 3 correct- 1m 4b 1. Correct + and – axes labelled on y axis 2. Correct scale ,width and height of bar chart ( cover 2/3 grid) 3. Each bar labelled with (test-tube) A/B/C 3 4c 1. Ref. to increased light intensity in A compared to other test tube 2. Ref. to increased rate of photosynthesis thus higher oxygen produced 3. Test tube C no light present thus no photosynthesis occurs 4 + 1.9 + 2.05 -0.8 Test tube A Test tube B Test tube C

4. Ref. to higher respiration compared to photosynthesis in C thus oxygen decreases 5a 1. Mucus blocking the airway reduces oxygen to the alveoli 2. Lesser diffusion of oxygen into bloodstream and into the body cells thus lesser rate of respiration 2 5bi 1. Segment of DNA/ sequence of DNA nucleotide that codes for specific polypeptide/ protein 1 5bii 1. Change to nucleotide sequence causes a change in amino acid sequence 2. Amino acid sequence will change polypeptide chain structure 3. Changes polypeptide chain structure changes the specific 3D configuration/folding of proteins thus results in the changes in protein structure/function 3 5c 1. May not be equally accessible – only individual with sufficient financial means can afford gene technologies 2. Ref to animal welfare when conducting trials 3. AVP 1 6a 1. Fusion of (two) gamete to form a zygote + produce genetically dissimilar offspring 1 6bi 1. menstruation 1 6bii 1. pregnancy / fertilization + implantation of embryo 1 6c 1. thin fetal capillary + allows shorter distance for diffusion 2. branched/ numerous fetal capillary surrounded by maternal blood space + increases surface area 3. continuous blood flow + maintains steep concentration gradient by constantly supplying nutrients towards the fetal capillaries and constantly moving the waste substances away from fetal capillaries R: space filled with maternal blood increases surface area. 3 6d 1. Functions like lungs: allows oxygen to diffuse from maternal blood into fetal blood 2. Helps to remove carbon dioxide with diffuse from fetal blood into maternal blood 3. Functions like kindeys: helps remove excretory products (such as urea) which diffuses from fetal blood into maternal blood for removal 3 7ai 1. Sample A 1 7aii 1. The test with Biuret solution gave a violet colour, indicating the presence of proteins in the urine sample 2. Large molecules such as proteins will be able to diffuse into nephron 2 7b 1. When little or no ADH is produced, the cells of the walls of the collecting duct becomes less permeable for absorption of water into the bloodstream/ less water reabsorbed. 2. Hence, urine become less concentrated / volume of urine increases. 2 7ci 1. Urea 2. Excess water / mineral salts 2 7cii 1. Diffusion / osmosis (if answer in 7ci includes water) 1

7ciii 1. [Compulsory] Glucose / amino acid 2. Ensures no difference in concentration / No concentration gradient between blood and dialysing fluid / hence no net movement of glucose / amino acids molecules from blood to dialysing fluid 2 8ai 1. lichen → caribou → canis lupus 1 8aii 1. Ref. lichens had the lowest level of radioactivity while the canis lupus had the highest; 2. Ref. to bioaccumulation within one trophic level - As the radioactive caesium was non-biodegradable, it was stored and accumulated in the tissues 3. Ref to bioamplification/biomagnification – radioactive caesium passed along the food chain through feeding and increases in concentration / last trophic level should have the highest concentration R: if student incorrectly explain bioaccumulation as bioamplification 3 8b 1. Correct pyramid shape ( biggest width at the bottom followed) 2. Equal height + labels 2 8c 1. Ref to insufficient energy to sustain another trophic level 2. Ref to energy loss at each trophic level + at least one reason for energy loss ( heat during respiration / uneaten body part/ excretory products) 2 8d 1. Conservation of forests 2. Reduce use of fossil fuels 3. Reduce consumption of energy 4. AVP max 2 Paper 2 – Section B Qn Marking Scheme Marks Marker’s Report 9ai 1. Arterioles A: arteries 1 9aii 1. arterioles dilate/ vasodilation occurs 2. increases blood flow to muscles cells 2 9bi 1. General trend given → The volume of blood pumped by the left ventricle increases from rest to medium intensity but remains constant after at high intensity 2. At rest, the volume of blood pumped by left ventricle is smallest at ~53 cm³. 3. During medium-intensity exercise, the volume pumped increases to ~85 cm³. 4. During high-intensity exercise, the volume pumped is remains at ~85 cm³ but is pumped more frequently at a higher heat rate. max 3 9bii 1. [compulsory] Muscles require more oxygen and glucose during exercise for aerobic respiration to release energy for muscular contractions. max 2

2. The heart pumps at a higher volume of blood and higher heart rate per beat to meet this demand. 3. There is an increase in blood returning to the heart, causing more filling of the left ventricle before contraction. 4. The heart contracts