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A Level H1 Biology Practice Paper 5
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Questions
TuitionGoWhere Practice Paper - Biology H1 A-Level
TuitionGoWhere Practice Paper (AI) – Version 5
Subject: Biology H1
Level: A-Level
Paper: Topic Test: Cells & Biomolecules
Duration: 1 hour
Total Marks: 47
Name: ____________________
Class: ____________________
Date: ____________________
Instructions:
- This paper consists of three sections (A, B, and C) covering the topic Cells & Biomolecules.
- Answer all questions in the spaces provided.
- Marks are indicated in brackets [ ].
- You may use a scientific calculator where necessary.
Section A: Short Answer (15 marks)
Answer all questions in this section.
-
Fig. 1 shows a transmission electron micrograph of a eukaryotic cell.
Fig. 1 (not shown): organelle X has a double membrane and contains DNA; organelle Y consists of flattened membrane-bound sacs stacked together.
(a) Identify organelle X. [1]
(b) Identify organelle Y. [1]
(c) State one function of organelle Y in a secretory cell. [1]
-
Describe the arrangement of phospholipid molecules in the cell membrane. [2]
-
A plant cell from a potato tuber is placed in a concentrated sucrose solution (0.5 mol dm⁻³). After 10 minutes, the vacuole has shrunk and the cytoplasm has pulled away from the cell wall.
(a) Name the condition observed in the cell. [1]
(b) Explain why the vacuole shrinks. [2]
-
An enzyme (amylase) is incubated with starch at different temperatures. The following results are obtained:
Temperature / °C 20 30 40 50 60 70 Rate of reaction / mg starch hydrolysed min⁻¹ 12 28 45 30 5 0 Explain the results at: (a) 20 °C to 40 °C. [2]
(b) 50 °C to 70 °C. [2]
-
Distinguish between the secondary and tertiary structure of a protein. [3]
Section B: Structured Questions (18 marks)
Answer all questions in this section.
-
Radioactive thymine is added to a culture of dividing animal cells.
(a) Explain why thymine is used as a marker for DNA synthesis. [1]
(b) With reference to the cell cycle, name the phase in which the radioactive thymine would first be detected in the nucleus. [1]
(c) Explain your answer to part (b). [2]
-
Compare and contrast simple diffusion and active transport. [3]
-
Skeletal muscle cells require large amounts of ATP for contraction.
(a) Name the organelle that is present in very high numbers in these cells. [1]
(b) Explain why a large amount of this organelle is essential. [2]
-
Isolated mitochondria were incubated in a medium containing either glucose or pyruvate. Carbon dioxide was produced only in the medium with pyruvate.
(a) Name the metabolic pathway that produces carbon dioxide in the mitochondria. [1]
(b) Explain why carbon dioxide is produced when pyruvate is added but not when glucose is added. [3]
-
Malonate is a competitive inhibitor of the enzyme succinate dehydrogenase.
(a) Explain how malonate reduces the activity of succinate dehydrogenase. [2]
(b) Suggest how the inhibition could be overcome. [1]
Section C: Data-Based and Extended Response (14 marks)
Answer all questions in this section.
-
The table below shows some structural features of two biomolecules.
Feature Molecule A (glycogen) Molecule B (triglyceride) Monomer(s) Glucose Glycerol and fatty acids Type of bonds formed Glycosidic bonds Ester bonds Solubility in water Soluble Insoluble Using the information in the table, suggest two reasons why glycogen is a more useful energy source for rapidly respiring muscle cells than a triglyceride. [2]
-
Describe the structure of a DNA nucleotide. [2]
-
Water is often described as a ‘universal solvent’ for biological systems.
Explain why water can dissolve ionic compounds such as sodium chloride. [2]
-
State two different functions of proteins embedded in the cell membrane. [2]
-
Describe the process of exocytosis and explain the role of the cell membrane in this process. [2]
-
A student measures the effect of pH on the activity of catalase from potato tissue.
Suggest a suitable method to measure the rate of the reaction. [2]
-
Explain why a large change in pH can cause a permanent loss of enzyme activity. [2]
-
Describe the role of the Golgi apparatus in the processing and secretion of a protein hormone. [2]
-
A plant cell has a solute potential (Ψₛ) of –500 kPa. The surrounding solution has a solute potential of –300 kPa. Assume the pressure potential (Ψₚ) in the cell is zero.
(a) Calculate the water potential (Ψ) of the cell and of the solution. [1]
(b) State the direction of water movement. [1]
-
The fluid mosaic model describes the structure and behaviour of cell membranes.
Discuss how the arrangement of phospholipids and membrane proteins enables the selective movement of substances into and out of cells. [6]
END OF PAPER
Answers
TuitionGoWhere Practice Paper – Biology H1 A-Level – Version 5
Answer Key and Marking Scheme
Total Marks: 47
Section A
-
(a) Mitochondrion / mitochondria [1]; reject ‘mitochondria’ if spelling incorrect but if consistent with plural accept.
(b) Golgi body / Golgi apparatus / Golgi complex [1].
(c) Modification / packaging / sorting of proteins (for secretion) [1]; accept formation of secretory vesicles / lysosomes. -
Phospholipids form a bilayer [1]; hydrophilic phosphate‑containing heads face outward toward aqueous environment on both sides of membrane; hydrophobic fatty‑acid tails face inward away from water [1]. Award [1] for bilayer concept and [1] for orientation of heads and tails.
-
(a) Plasmolysis [1].
(b) Sucrose solution has lower water potential than cell sap [1]; water moves out of vacuole by osmosis down water potential gradient / from higher to lower water potential [1]. -
(a) As temperature rises from 20 °C to 40 °C, kinetic energy of enzyme and substrate molecules increases [1]; more frequent successful collisions / more enzyme‑substrate complexes formed per unit time [1].
(b) At 50 °C, the enzyme begins to denature; hydrogen / ionic / hydrophobic bonds that maintain tertiary structure break [1]; by 70 °C denaturation is complete – active site has lost its specific shape and no longer complementary to substrate; no more enzyme‑substrate complexes form / rate drops to zero [1]. -
Secondary structure refers to local folding of the polypeptide chain into α‑helices or β‑pleated sheets, held by hydrogen bonds between backbone –NH and –C=O groups [1]; tertiary structure is the overall 3‑D folding of the whole polypeptide [1], stabilized by hydrogen bonds, ionic bonds, hydrophobic interactions and disulfide bridges between R‑groups [1].
Section B
-
(a) Thymine is a base found only in DNA, not in RNA [1].
(b) S phase (synthesis phase) [1].
(c) During S phase, DNA replication occurs; new DNA molecules are synthesised using existing strands as templates [1]; thus radioactive thymine is incorporated into newly made DNA / nuclear radioactivity increases [1]. -
Simple diffusion involves movement of molecules down a concentration gradient, from high to low concentration, without the expenditure of metabolic energy (passive); substances move directly through the phospholipid bilayer or through channel proteins [1]. Active transport moves molecules against a concentration gradient, from low to high concentration, requiring energy (ATP) and specific carrier proteins [1]. Both processes transport substances across the cell membrane, but they differ in energy requirement, direction of movement and protein involvement [1].
-
(a) Mitochondria [1].
(b) Mitochondria are the site of aerobic respiration [1]; they produce large quantities of ATP via oxidative phosphorylation, which is required for muscle contraction [1]. -
(a) Krebs cycle (citric acid cycle) [1].
(b) Pyruvate can enter the matrix of the mitochondrion, where it is converted to acetyl‑CoA and then enters the Krebs cycle, producing carbon dioxide as a waste product [1]. Glucose cannot enter the Krebs cycle directly; glycolysis occurs in the cytoplasm and breaks down glucose to pyruvate, but isolated mitochondria lack the enzymes for glycolysis [1]; therefore, incubation with glucose alone does not yield pyruvate inside the mitochondrion and no CO₂ is produced [1]. -
(a) Malonate has a shape similar to the substrate (succinate) [1]; it competes for the active site of succinate dehydrogenase, occupying it temporarily and preventing the substrate from binding, thus reducing enzyme activity [1].
(b) Increasing the concentration of the substrate (succinate) will out‑compete the inhibitor [1].
Section C
-
Glycogen is soluble in water / cell sap, so it can be readily mobilised / transported within the cell, while triglycerides are insoluble [1]; glycogen is broken down by enzymes that recognise glycosidic bonds, releasing glucose rapidly for respiration, while triglycerides require more extensive processing (e.g., lipolysis, then β‑oxidation) before use [1].
-
A DNA nucleotide is composed of a deoxyribose sugar [1], a phosphate group, and a nitrogenous base (adenine, thymine, cytosine, or guanine) [1].
-
Water is a polar molecule with a partial negative charge on the oxygen atom and partial positive charges on the hydrogen atoms [1]; the charged regions surround and separate the Na⁺ and Cl⁻ ions, forming hydration shells and keeping them in solution [1].
-
Any two from, e.g.:
- channel proteins allow facilitated diffusion of ions / small polar molecules [1];
- carrier proteins mediate active transport / facilitated diffusion [1];
- receptors for hormones / neurotransmitters (signal transduction) [1];
- cell‑cell recognition (glycoproteins) [1];
- enzymatic activity [1];
- attachment to cytoskeleton / extracellular matrix [1]. Award [1] for each valid, distinct function.
-
Exocytosis is the process by which vesicles containing materials fuse with the cell membrane [1]; the membrane of the vesicle becomes part of the cell membrane, releasing the vesicle contents to the outside of the cell [1].
-
Use catalase extracted from potato tissue [1]; add to hydrogen peroxide at different pH buffers (e.g., pH 4–9); measure the volume of oxygen gas produced per unit time using a gas syringe / measuring cylinder over water, or measure the time taken for a filter‑paper disc soaked in enzyme to rise in a test tube of peroxide [1]. Accept any appropriate method that quantifies rate.
-
Extreme pH changes disrupt the ionic and hydrogen bonds that maintain the enzyme’s tertiary structure [1]; the active site becomes permanently altered and no longer complementary to the substrate; the enzyme is denatured [1].
-
Protein from ribosomes on rough ER is transported to Golgi in vesicles [1]; in the Golgi, the protein is modified (e.g., glycosylation) and packaged into secretory vesicles that bud off and move to the cell membrane for exocytosis [1].
-
(a) Ψ of cell = Ψₛ = –500 kPa (since Ψₚ = 0); solution Ψ = –300 kPa [1].
(b) Water moves from the solution (higher water potential) into the cell (lower water potential) [1]. -
(award up to 6 marks for a well‑structured discussion)
- Phospholipid bilayer: hydrophobic core prevents free passage of large polar molecules and ions; only small, non‑polar molecules (O₂, CO₂) and water (via osmosis) diffuse freely [1].
- Channel proteins: provide hydrophilic pores for facilitated diffusion of specific ions / small polar molecules down their concentration gradient [1].
- Carrier proteins: undergo conformational change to transport molecules (e.g., glucose) by facilitated diffusion or active transport against gradient using ATP [1].
- Active transport driven by ATP hydrolysis (e.g., Na⁺/K⁺ pump) [1].
- Endocytosis / exocytosis: large particles or macromolecules are engulfed or secreted via vesicle formation requiring membrane flexibility [1].
- Cholesterol in membrane stabilizes fluidity, affecting permeability; unsaturated fatty acids maintain fluidity for protein movement and transport [1].
Mark according to quality of response linking structural features to selective permeability.