Secondary 3 Biology Practice Paper 5

TuitionGoWhere Practice Paper Answers - Biology Secondary 3

Version: 5 of 5 Total Marks: 80

Section A: Multiple Choice Answers

Section A Total: [10]

Section B: Structured Response Answers

11.

(a) [2 marks]

• Membrane-bound nucleus / nuclear envelope [1]
• Mitochondria [1]
• Endoplasmic reticulum (rough or smooth) [1]
• Golgi body / Golgi apparatus [1]
• Lysosomes [1]

(Any two correct; accept also: nuclear pores, 70S vs 80S ribosomes if visible, though typical EM distinction is membrane-bound organelles)

(b) [3 marks]

• Ribosomes: protein synthesis using mRNA template [1]
• Rough endoplasmic reticulum: folding and initial modification of polypeptide; transport in vesicles [1]
• Golgi body / apparatus: further modification, sorting, and packaging into secretory vesicles [1]
• Secretory vesicles fuse with cell membrane to release hormone by exocytosis [1]

(Maximum 3 marks; must have correct sequence)

(c) [2 marks]

• Protein synthesis and secretion are energy-requiring processes (ATP needed) [1]
• Mitochondria are sites of aerobic respiration, producing large amounts of ATP [1]
• More mitochondria provide sufficient ATP to power ribosome function, vesicle transport, and exocytosis [1]

(Any 2 points)

12.

(a) [1 mark]

• C [1]

(b) [3 marks]

• In solution C (concentrated solution / hypertonic), water potential is lower outside the cell than inside [1]
• Water leaves the cell by osmosis (down water potential gradient, from high to low water potential) [1]
• Cytoplasm and vacuole lose water, shrink, and pull away from cell wall (plasmolysis) [1]
• Cell membrane no longer pushed against cell wall; chloroplasts cluster in center [1]

(Maximum 3 marks)

(c) [3 marks]

• Red blood cells would burst / undergo haemolysis / cytolysis [1]
• Animal cells lack cell wall, so no protective structure against water entry [1]
• Distilled water is hypotonic compared to cytoplasm, so water enters continuously by osmosis [1]
• Plant cell wall is rigid and provides resistance to expansion, preventing bursting [1]

(First 3 valid points; must include contrast with plant cell)

13.

(a) [1 mark]

• The region of the enzyme where the substrate binds / a specific pocket or cleft in the tertiary structure [1]

(b) [3 marks]

• High temperature increases kinetic energy of molecules, increasing collision frequency [1]
• Above optimum temperature, bonds holding enzyme in shape (hydrogen bonds, ionic bonds) vibrate and break [1]
• Tertiary structure / 3D shape of enzyme is altered / denatured [1]
• Active site shape changes permanently, substrate can no longer fit / enzyme-substrate complexes cannot form [1]

(Maximum 3 marks; must include denaturation and active site change)

(c)(i) [1 mark]

• pH 7 [1]

(c)(ii) [2 marks]

• Change in oxygen: 28 → 3.1 = decrease of 25.3 cm³ [1]
• Percentage decrease = (25.3 / 28) × 100 = 90.4% (accept 90–91%) [1]

(c)(iii) [2 marks]

• Concentration / volume of hydrogen peroxide (substrate) [1]
• Concentration / volume of liver extract (enzyme source) [1]
• Temperature of incubation [1]
• Time of collection / same apparatus [1]
• Same batch of liver / same pH of buffer used [1]

(Any two with brief explanation of why control is needed)

14.

(a) [2 marks]

• Phospholipids have hydrophilic (water-loving) phosphate heads and hydrophobic (water-fearing) fatty acid tails [1]
• In aqueous environment, heads face water on both sides while tails are shielded in the membrane interior, minimising contact with water / creating energetically favourable arrangement [1]

(b)(i) [1 mark]

• Cell recognition / antigen markers / cell adhesion / receptors for hormones and neurotransmitters [1]

(b)(ii) [1 mark]

• Stabilises membrane / maintains fluidity / prevents phospholipids packing too closely at low temperatures / prevents excessive fluidity at high temperatures [1]

(c) [3 marks]

• Selectively permeable means some substances can cross while others are restricted / controlled entry and exit of substances [1]
• Small non-polar molecules (O₂, CO₂) and small polar uncharged molecules (water, urea) diffuse through lipid bilayer [1]
• Large, polar, or charged molecules need protein channels or carriers [1]
• Specific proteins allow selectivity — only molecules matching channel size or carrier specificity can pass / active transport moves substances against concentration gradients [1]

(Maximum 3 marks; must explain both lipid barrier and protein-mediated transport)

15.

(a) [2 marks]

• Large central lipid droplet pushes cytoplasm and nucleus to periphery, maximising storage volume [1]
• Thin cell membrane / absence of large cytoplasmic organelles reduces metabolic demands, allowing more space for fat [1]

(b) [6 marks]

(Maximum 6 marks; need 3 valid differences with structures and functions linked)

16.

(a) [2 marks]

• X: Granum / stack of thylakoids [1]
• Y: Stroma [1] (Accept: thylakoid for X if clearly pointing to single disc; stroma must be ground substance)

(b) [3 marks]

• Thylakoid membranes / grana contain chlorophyll and electron transport chains for light harvesting [1]
• Large surface area of thylakoid membranes provides extensive area for photosystems and electron carriers [1]
• Stacked grana allow interconnected membrane system for efficient electron and proton transfer [1]
• Surrounding stroma contains enzymes for ATP production via chemiosmosis and the Calvin cycle [1]

(Maximum 3 marks; must link structure to light-dependent reactions specifically)

(c) [2 marks]

• Mitochondria evolved from free-living prokaryotes (endosymbiotic theory) and retained their own DNA for some proteins [1]
• Chloroplasts also evolved from endosymbionts but have transferred most genes to nucleus / lost most original DNA over evolution [1]
• Chloroplasts are more dependent on nuclear-encoded proteins imported from cytoplasm / different evolutionary reduction of genome [1]

(Any valid evolutionary explanation; accept endosymbiotic theory discussion)

17.

(a) [2 marks]

• Amylase diffuses from well into surrounding agar [1]
• Amylase hydrolyses / digests starch to maltose; starch absent so iodine no longer gives blue-black colour [1]
• Cleared zone indicates region where starch has been broken down [1]

(Maximum 2 marks)

(b) [3 marks]

• 37°C is near optimum temperature for human salivary / pancreatic amylase [1]
• Kinetic energy of enzyme and substrate molecules is higher at 37°C than 0°C [1]
• More frequent successful collisions / more enzyme-substrate complexes form per unit time [1]
• Rate of starch digestion is faster, so cleared zone is larger [1]

(Maximum 3 marks; must compare temperatures and explain rate difference)

(c) [2 marks]

• 60°C is above optimum temperature for amylase [1]
• Enzyme begins to denature / tertiary structure disrupted / active site shape altered [1]
• Fewer effective collisions / enzyme-substrate complexes cannot form efficiently [1]
• Some remaining activity as not all molecules denatured immediately, giving small zone [1]

(Maximum 2 marks)

(d) [1 mark]

• Measure the diameter / area of cleared zones at regular time intervals [1]
• Use a colorimeter to measure starch disappearance / reducing sugar appearance [1]
• Maintain constant temperature with water bath for each test [1]

(Any valid improvement for precision)

18.

(a) [2 marks]

• Fibrous structure with cross-linked polypeptides (disulfide bonds) provides strength and insolubility [1]
• Coiled arrangement allows some elasticity while resisting stretching / mechanical stress [1]
• Keratin is insoluble in water, making it durable for protective functions (hair, nails, skin) [1]

(Maximum 2 marks)

(b) [3 marks]

• Preproinsulin contains a signal sequence directing it to RER [1]
• Signal sequence removed to form proinsulin; C-peptide removed during processing in Golgi [1]
• Final active insulin (two chains) is smaller, allowing packaging into secretory vesicles [1]
• Folding and disulfide bond formation in ER/Golgi ensures correct tertiary structure for binding to receptors [1]

(Maximum 3 marks; must explain why modification is necessary for transport and secretion)

(c) [3 marks]

• Stomach pH is approximately 1.5–2 (highly acidic) due to hydrochloric acid [1]
• Amylase has optimum pH 7; at stomach pH the enzyme is denatured / active site charges altered [1]
• Amylase cannot bind starch / enzyme-substrate complexes do not form in stomach [1]
• In small intestine, pH rises to 7–8 due to bicarbonate from pancreas; if amylase secreted from pancreas, it can function again / salivary amylase may have some residual activity if pH neutralises [1]

(Must explain inactivation in stomach and reactivation/continuation in small intestine)

19.

(a)(i) [1 mark]

• Prevents osmotic water gain or loss that would burst or shrink organelles / maintains organelle integrity and shape [1]

(a)(ii) [1 mark]

• Reduces enzyme activity / metabolic reactions that could degrade organelles / prevents denaturation of proteins [1]

(b) [2 marks]

• Different organelles have different sizes and therefore different sedimentation rates [1]
• Larger / denser organelles sediment at lower centrifugal force; smaller ones need higher force [1]
• Sequential centrifugation at increasing speeds pellets organelles from largest to smallest [1]

(Maximum 2 marks)

(c) [3 marks]

• Cytosol contains many soluble enzymes involved in glycolysis and other metabolic pathways [1]
• No membrane to pellet during centrifugation, so remains in supernatant [1]
• Expected proteins: enzymes of glycolysis (phosphofructokinase, pyruvate kinase), transaminases, dehydrogenases [1]
• High protein concentration because cytosol is major metabolic compartment / many reactions occur there [1]

(Maximum 3 marks; must explain why high proportion and name examples)

20.

(a) [4 marks]

• Membrane proteins are embedded within the lipid bilayer, not just on the surface [1]
  • Earlier Davson-Danielli model had proteins as separate layers on either side [1]
• Phospholipid bilayer is fluid — lipids and proteins can move laterally [1]
  • Earlier models depicted rigid sandwich structure [1]
• Proteins are a mosaic — varied types with different functions, distributed asymmetrically [1]
• Hydrophobic and hydrophilic regions of proteins match their positions in/through the membrane [1]

(Maximum 4 marks; need two features with contrast to earlier model)

(b) [2 marks]

• Davson-Danielli proposed continuous protein layers on membrane surfaces [1]
• Integral (transmembrane) proteins span the membrane, with regions exposed on both sides and within the lipid core [1]
• This is incompatible with surface-only protein layers — proteins must penetrate through the hydrophobic core [1]

(Maximum 2 marks)

(c) [3 marks]

• Freeze-fracture splits the membrane along the hydrophobic core between phospholipidleaflets [1]
• Reveals internal view showing proteins embedded throughout the lipid matrix as bumps/particles [1]
• Distribution of proteins is irregular mosaic pattern, not continuous layers [1]
• Particle movement at different temperatures demonstrates membrane fluidity [1]

(Maximum 3 marks)

Section B Total: [50]

Section C: Data Response and Analysis Answers

21.

(a) [2 marks]

• Low absorbance at 0–40°C with slight gradual increase [1]
• Sharp / steep increase in absorbance between 40°C and 60°C [1]
• Highest absorbance / near-maximum leakage at 80°C [1]

(Maximum 2 marks; need description of overall trend with key temperature points)

(b) [4 marks]

• Phospholipid bilayer has hydrophobic fatty acid tails in membrane interior [1]
• Above 40°C, thermal energy increases phospholipid movement / membrane becomes more fluid [1]
• At 60°C and above, increased kinetic energy disrupts bonds holding membrane structure [1]
• Membrane proteins denature / channel proteins change shape, creating gaps [1]
• Betacyanin leaks through damaged membrane; extensive leakage at 80°C indicates complete loss of membrane integrity [1]

(Maximum 4 marks; must explain membrane structure changes causing increased permeability)

(c) [4 marks]

Improvement 1: [2 marks]

• Use a water bath or thermostatically controlled heater to maintain precise temperatures [1]
• Reason: Reduces temperature fluctuation, ensures all samples at exact target temperature, improves reliability / reproducibility [1]

Improvement 2: [2 marks]

• Use replicates at each temperature (e.g., 5 cylinders per temperature) and calculate mean absorbance [1]
• Reason: Identifies and reduces effect of anomalous results / biological variation between beetroot samples; statistical validity improves [1]

(Other valid improvements: standardise cylinder size with cork borer; use buffer instead of distilled water to control pH; standardise rinsing time; use colorimeter calibration; blind measurement to reduce bias)

22.

(a) [4 marks]

(Maximum 4 marks; need clear distinction using diagram information)

(b) [3 marks]

• HMG-CoA reductase catalyses formation of mevalonate, a precursor for cholesterol synthesis [1]
• Statins occupy / compete with substrate for the active site of this enzyme [1]
• Reduced enzyme activity means less mevalonate produced, therefore less cholesterol synthesised [1]
• Blood cholesterol levels decrease as liver takes in more LDL from blood to compensate / de novo synthesis is blocked [1]

(Maximum 3 marks)

(c) [3 marks]

• Irreversible inhibition permanently modifies the active site / forms covalent bonds with enzyme [1]
• Inhibited enzyme molecules cannot function again; they must be replaced by new enzyme synthesis [1]
• Enzymes are continually broken down and resynthesised in cells / have finite lifespan [1]
• Regular aspirin administration needed to inhibit newly synthesised COX enzymes / maintain therapeutic enzyme inhibition as old enzymes degrade [1]

(Maximum 3 marks; must explain irreversibility and need for repeated dosing)

Section C Total: [20]

Grand Total: [80]

END OF ANSWER KEY