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A Level H2 Biology Cells Biomolecules Quiz
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Questions
A-Level Biology H2 Quiz - Cells Biomolecules
Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 50
Duration: 1 hour
Total Marks: 50
Instructions:
- This quiz contains 20 questions on Cells and Biomolecules.
- Answer ALL questions in the spaces provided.
- Marks for each question are indicated in brackets.
- Show all working for calculation questions.
- Use diagrams where appropriate to support your answers.
Section A: Short Answer (15 marks)
Answer all questions in this section.
1. State the primary function of the smooth endoplasmic reticulum in liver cells. [1]
2. Distinguish between the terms "hydrophilic" and "hydrophobic" with reference to the structure of a phospholipid molecule. [2]
3. Name the type of bond that stabilises the tertiary structure of a protein between the side chains of two cysteine amino acids. [1]
4. A student prepared a 0.5 mol dm⁻³ solution of sucrose (C₁₂H₂₂O₁₁). Calculate the mass of sucrose required to prepare 250 cm³ of this solution. (Relative molecular mass of sucrose = 342) [2]
5. State the role of cholesterol in animal cell membranes. [1]
6. Explain why the cell surface membrane is described as having a "fluid mosaic" structure. [3]
7. A red blood cell is placed in a hypotonic solution. Describe and explain the changes that occur to the cell. [3]
8. State two structural differences between prokaryotic and eukaryotic cells. [2]
Section B: Structured Questions (20 marks)
Answer all questions in this section.
9. Figure 1 shows the effect of increasing substrate concentration on the rate of an enzyme-catalysed reaction in the presence and absence of an inhibitor.
| Substrate concentration (mmol dm⁻³) | Rate without inhibitor (arbitrary units) | Rate with inhibitor (arbitrary units) |
|---|---|---|
| 0.5 | 12 | 4 |
| 1.0 | 20 | 8 |
| 2.0 | 30 | 15 |
| 4.0 | 38 | 25 |
| 8.0 | 40 | 35 |
| 16.0 | 40 | 40 |
(a) Using the data in Figure 1, identify the type of inhibition shown. Explain your answer. [2]
(b) Explain, at the molecular level, how this type of inhibitor reduces enzyme activity. [2]
(c) Predict the effect on the rate of reaction if a large excess of substrate is added in the presence of the inhibitor. Justify your answer. [1]
10. Figure 2 shows a diagram of the fluid mosaic model of the cell surface membrane.
(a) Name the structures labelled A, B, and C in Figure 2. [3]
A: ________________________
B: ________________________
C: ________________________
(b) Explain how the structure of component A enables it to perform its function in the membrane. [2]
(c) Describe the role of component C in cell signalling. [2]
11. A student investigated the effect of temperature on the rate of an enzyme-catalysed reaction. The results are shown in Figure 3.
| Temperature (°C) | Rate of reaction (arbitrary units) |
|---|---|
| 10 | 5 |
| 20 | 12 |
| 30 | 24 |
| 40 | 35 |
| 50 | 20 |
| 60 | 2 |
(a) Describe the trend shown by the data. [2]
(b) Explain why the rate of reaction decreases at temperatures above 40°C. [3]
(c) Suggest why the rate of reaction at 10°C is low. [1]
12. A plant cell with a water potential (Ψ) of −0.8 MPa is placed in a sucrose solution with a water potential of −0.3 MPa. The solute potential (Ψₛ) of the cell is −1.2 MPa.
(a) Calculate the pressure potential (Ψₚ) of the plant cell. Show your working. [2]
(b) State the direction of net water movement. Explain your answer. [2]
Section C: Data-Based and Extended Response (15 marks)
Answer all questions in this section.
13. Figure 4 shows the molecular structures of two monosaccharides, α-glucose and β-glucose.
(a) Describe one structural difference between α-glucose and β-glucose. [1]
(b) Name the polysaccharide formed by the condensation of α-glucose molecules in plant cells. [1]
(c) Explain how the structure of the polysaccharide named in (b) is related to its function as a storage molecule. [3]
14. A suspension of mitochondria was prepared in a buffer containing ADP and inorganic phosphate (Pi). The oxygen concentration in the buffer was monitored over time. At point X, sodium azide (an inhibitor of cytochrome c oxidase) was added. The results are shown in Figure 5.
(a) Describe the change in oxygen concentration before point X. [1]
(b) Explain why oxygen is consumed by the mitochondrial suspension. [2]
(c) Predict and explain the change in oxygen concentration after the addition of sodium azide at point X. [2]
15. Figure 6 shows the primary structure of a short polypeptide.
Gly – Ala – Val – Cys – Ser – Lys – Pro
(a) State the number of peptide bonds in this polypeptide. [1]
(b) Explain how the primary structure of a protein determines its tertiary structure. [3]
(c) A mutation causes the valine (Val) at position 3 to be replaced by glutamic acid (Glu). Suggest how this change could affect the function of the protein. [1]
16. Describe the process by which a glycoprotein is synthesised and secreted from a eukaryotic cell. [5]
17. Compare and contrast the structure and function of DNA and RNA. [4]
18. Explain how the properties of water make it an essential molecule for living organisms. [4]
19. A student carried out the Benedict's test on a solution of an unknown carbohydrate. A brick-red precipitate was formed. The student concluded that the solution contained glucose. Evaluate the student's conclusion. [3]
20. Discuss the importance of hydrogen bonding in the structure and function of biological molecules. Use named examples to support your answer. [5]
END OF QUIZ
Answers
A-Level Biology H2 Quiz - Cells Biomolecules: Answer Key
Total Marks: 50
Section A: Short Answer (15 marks)
1. State the primary function of the smooth endoplasmic reticulum in liver cells. [1]
Answer: Synthesis of lipids / detoxification of drugs and poisons / carbohydrate metabolism. (Any one correct function = 1 mark)
2. Distinguish between the terms "hydrophilic" and "hydrophobic" with reference to the structure of a phospholipid molecule. [2]
Answer:
- Hydrophilic means "water-loving" / attracted to water (1). The phosphate head of a phospholipid is hydrophilic because it is polar / charged and can form hydrogen bonds with water (1).
- Hydrophobic means "water-fearing" / repelled by water (1). The fatty acid tails are hydrophobic because they are non-polar / uncharged and cannot form hydrogen bonds with water (1).
Marking note: Award 1 mark for correct definition of each term with reference to phospholipid structure. Max 2 marks.
3. Name the type of bond that stabilises the tertiary structure of a protein between the side chains of two cysteine amino acids. [1]
Answer: Disulfide bond / disulfide bridge.
4. A student prepared a 0.5 mol dm⁻³ solution of sucrose (C₁₂H₂₂O₁₁). Calculate the mass of sucrose required to prepare 250 cm³ of this solution. (Relative molecular mass of sucrose = 342) [2]
Answer:
- Moles of sucrose = concentration × volume = 0.5 × (250/1000) = 0.125 mol (1)
- Mass = moles × Mᵣ = 0.125 × 342 = 42.75 g (1)
Marking note: Award 1 mark for correct calculation of moles, 1 mark for correct final answer with units. Allow ecf if moles calculation is incorrect but mass calculation is correctly applied.
5. State the role of cholesterol in animal cell membranes. [1]
Answer: Regulates membrane fluidity / maintains membrane stability / prevents the membrane from becoming too fluid at high temperatures or too rigid at low temperatures.
6. Explain why the cell surface membrane is described as having a "fluid mosaic" structure. [3]
Answer:
- "Fluid" refers to the phospholipid bilayer, where phospholipid molecules can move laterally within the layer (1).
- "Mosaic" refers to the proteins that are embedded in the phospholipid bilayer, forming a scattered / patchy pattern (1).
- The proteins and phospholipids are not fixed in position; they can move within the membrane, giving it a dynamic structure (1).
Marking note: Award 1 mark for each correct explanation of "fluid" and "mosaic", and 1 mark for linking the two concepts to describe the overall structure.
7. A red blood cell is placed in a hypotonic solution. Describe and explain the changes that occur to the cell. [3]
Answer:
- The solution has a higher water potential than the cell cytoplasm (1).
- Water enters the cell by osmosis down the water potential gradient (1).
- The cell swells and eventually bursts (lysis / haemolysis) because the cell membrane cannot withstand the increased pressure (1).
Marking note: Award 1 mark for identifying the direction of water movement, 1 mark for explaining osmosis, and 1 mark for describing the outcome (lysis).
8. State two structural differences between prokaryotic and eukaryotic cells. [2]
Answer: Any two from:
- Prokaryotic cells have no nucleus / have a circular chromosome free in the cytoplasm; eukaryotic cells have a membrane-bound nucleus.
- Prokaryotic cells have no membrane-bound organelles (e.g., mitochondria, ER); eukaryotic cells have membrane-bound organelles.
- Prokaryotic cells have smaller ribosomes (70S); eukaryotic cells have larger ribosomes (80S).
- Prokaryotic cells have a cell wall made of peptidoglycan; eukaryotic plant cells have a cell wall made of cellulose (animal cells have no cell wall).
- Prokaryotic cells may have plasmids; eukaryotic cells do not.
Marking note: Award 1 mark for each correct structural difference. Must be a clear comparison.
Section B: Structured Questions (20 marks)
9. Figure 1 data on enzyme inhibition.
(a) Using the data in Figure 1, identify the type of inhibition shown. Explain your answer. [2]
Answer: Competitive inhibition (1). At high substrate concentrations (16.0 mmol dm⁻³), the rate with inhibitor reaches the same maximum rate (40 units) as without inhibitor, indicating that the inhibitor can be overcome by excess substrate (1).
Marking note: Award 1 mark for correct identification, 1 mark for explanation referencing the data (V_max unchanged).
(b) Explain, at the molecular level, how this type of inhibitor reduces enzyme activity. [2]
Answer:
- The inhibitor has a shape similar to the substrate and competes for the active site of the enzyme (1).
- When the inhibitor binds to the active site, it blocks the substrate from binding, preventing the formation of enzyme-substrate complexes and reducing the rate of reaction (1).
Marking note: Award 1 mark for describing structural similarity/competition, 1 mark for explaining the consequence (blocked active site).
(c) Predict the effect on the rate of reaction if a large excess of substrate is added in the presence of the inhibitor. Justify your answer. [1]
Answer: The rate of reaction will reach the maximum rate (V_max) / be the same as without inhibitor. The excess substrate out-competes the inhibitor for the active site.
10. Figure 2 diagram of fluid mosaic model.
(a) Name the structures labelled A, B, and C in Figure 2. [3]
Answer: A: Phospholipid / phospholipid bilayer (1) B: Channel protein / carrier protein / transport protein (accept integral protein) (1) C: Glycoprotein / glycolipid (accept carbohydrate chain) (1)
(b) Explain how the structure of component A enables it to perform its function in the membrane. [2]
Answer:
- Phospholipids have a hydrophilic phosphate head and hydrophobic fatty acid tails (1).
- This amphipathic nature allows them to form a bilayer in an aqueous environment, with the hydrophilic heads facing outwards towards the water and the hydrophobic tails facing inwards, creating a selectively permeable barrier (1).
Marking note: Award 1 mark for describing the amphipathic structure, 1 mark for explaining how this forms a barrier.
(c) Describe the role of component C in cell signalling. [2]
Answer:
- Glycoproteins/glycolipids act as receptors for signalling molecules such as hormones or neurotransmitters (1).
- The carbohydrate chain on the extracellular surface allows specific binding of signalling molecules, triggering a cellular response (1).
Marking note: Award 1 mark for identifying the receptor role, 1 mark for explaining specific binding/signalling.
11. Figure 3 data on temperature and enzyme activity.
(a) Describe the trend shown by the data. [2]
Answer:
- The rate of reaction increases as temperature increases from 10°C to 40°C (1).
- Above 40°C, the rate of reaction decreases sharply, falling to near zero at 60°C (1).
Marking note: Award 1 mark for describing the increase, 1 mark for describing the decrease.
(b) Explain why the rate of reaction decreases at temperatures above 40°C. [3]
Answer:
- At high temperatures, the increased kinetic energy causes the hydrogen bonds and other weak bonds maintaining the enzyme's tertiary structure to break (1).
- This causes the enzyme to lose its specific three-dimensional shape / denature (1).
- The active site is no longer complementary to the substrate, so enzyme-substrate complexes cannot form, and the rate of reaction decreases (1).
Marking note: Award 1 mark for identifying bond breakage, 1 mark for denaturation, 1 mark for linking to active site function.
(c) Suggest why the rate of reaction at 10°C is low. [1]
Answer: At low temperatures, the enzyme and substrate molecules have low kinetic energy, so they move more slowly and collide less frequently, resulting in fewer successful enzyme-substrate collisions.
12. Plant cell water potential calculation.
(a) Calculate the pressure potential (Ψₚ) of the plant cell. Show your working. [2]
Answer: Ψ = Ψₛ + Ψₚ −0.8 = −1.2 + Ψₚ (1) Ψₚ = −0.8 − (−1.2) = +0.4 MPa (1)
Marking note: Award 1 mark for correct substitution into equation, 1 mark for correct answer with units.
(b) State the direction of net water movement. Explain your answer. [2]
Answer: Water will move from the sucrose solution into the cell (1). The solution has a higher water potential (−0.3 MPa) than the cell (−0.8 MPa), so water moves down the water potential gradient by osmosis (1).
Marking note: Award 1 mark for correct direction, 1 mark for explanation referencing water potential values.
Section C: Data-Based and Extended Response (15 marks)
13. Figure 4 monosaccharide structures.
(a) Describe one structural difference between α-glucose and β-glucose. [1]
Answer: In α-glucose, the hydroxyl (−OH) group on carbon-1 is below the plane of the ring; in β-glucose, the −OH group on carbon-1 is above the plane of the ring.
(b) Name the polysaccharide formed by the condensation of α-glucose molecules in plant cells. [1]
Answer: Starch (accept amylose or amylopectin).
(c) Explain how the structure of the polysaccharide named in (b) is related to its function as a storage molecule. [3]
Answer:
- Starch is a large, insoluble molecule that does not affect the water potential of the cell, so it can be stored without causing osmotic water uptake (1).
- Amylose is a helical, coiled structure that makes starch compact, allowing a large amount of glucose to be stored in a small space (1).
- Amylopectin is branched, providing many free ends for rapid hydrolysis by enzymes to release glucose when energy is needed (1).
Marking note: Award 1 mark for each correct structure-function relationship. Max 3 marks.
14. Figure 5 mitochondrial oxygen consumption.
(a) Describe the change in oxygen concentration before point X. [1]
Answer: The oxygen concentration decreases steadily / at a constant rate over time.
(b) Explain why oxygen is consumed by the mitochondrial suspension. [2]
Answer:
- Oxygen acts as the final electron acceptor in the electron transport chain during oxidative phosphorylation (1).
- It accepts electrons and combines with hydrogen ions to form water, allowing the electron transport chain to continue functioning and ATP to be synthesised (1).
Marking note: Award 1 mark for identifying oxygen as the final electron acceptor, 1 mark for explaining its role in the ETC/ATP synthesis.
(c) Predict and explain the change in oxygen concentration after the addition of sodium azide at point X. [2]
Answer:
- The rate of oxygen consumption will decrease / stop (1).
- Sodium azide inhibits cytochrome c oxidase, blocking electron transfer to oxygen, so oxygen is no longer reduced to water and the electron transport chain stops (1).
Marking note: Award 1 mark for correct prediction, 1 mark for explanation linking inhibitor to ETC.
15. Figure 6 polypeptide primary structure.
(a) State the number of peptide bonds in this polypeptide. [1]
Answer: 6 peptide bonds. (A polypeptide of 7 amino acids has n−1 = 6 peptide bonds.)
(b) Explain how the primary structure of a protein determines its tertiary structure. [3]
Answer:
- The primary structure is the specific sequence of amino acids in the polypeptide chain (1).
- The sequence determines which R groups are present and their positions, which influences the types of bonds that form (hydrogen bonds, ionic bonds, disulfide bridges, hydrophobic interactions) (1).
- These bonds cause the polypeptide to fold into a specific three-dimensional tertiary structure, which is unique to each protein (1).
Marking note: Award 1 mark for defining primary structure, 1 mark for linking to bond formation, 1 mark for explaining folding into tertiary structure.
(c) A mutation causes the valine (Val) at position 3 to be replaced by glutamic acid (Glu). Suggest how this change could affect the function of the protein. [1]
Answer: Valine is non-polar/hydrophobic while glutamic acid is polar/charged/hydrophilic. This change in R group properties could alter the folding of the protein / disrupt bonds in the tertiary structure, potentially changing the shape of the active site (if an enzyme) or binding site, affecting function.
Marking note: Accept any reasonable suggestion linking amino acid property change to altered protein structure/function.
16. Describe the process by which a glycoprotein is synthesised and secreted from a eukaryotic cell. [5]
Answer:
- The polypeptide chain is synthesised by ribosomes attached to the rough endoplasmic reticulum (RER) (1).
- The polypeptide enters the lumen of the RER, where it folds into its tertiary structure (1).
- The protein is transported in vesicles from the RER to the Golgi apparatus (1).
- In the Golgi apparatus, the protein is modified by the addition of carbohydrate groups to form a glycoprotein (1).
- The glycoprotein is packaged into secretory vesicles, which move to and fuse with the cell surface membrane, releasing the glycoprotein by exocytosis (1).
Marking note: Award 1 mark for each correct step in sequence. Max 5 marks.
17. Compare and contrast the structure and function of DNA and RNA. [4]
Answer: Similarities:
- Both are nucleic acids composed of nucleotides containing a sugar, phosphate group, and nitrogenous base (1).
- Both are involved in protein synthesis (DNA stores genetic information; RNA transfers and translates it) (1).
Differences:
- DNA contains deoxyribose sugar; RNA contains ribose sugar (1).
- DNA is double-stranded; RNA is single-stranded (1).
- DNA contains thymine (T); RNA contains uracil (U) (1).
- DNA stores genetic information; RNA has various roles (mRNA carries genetic code, tRNA transfers amino acids, rRNA forms ribosomes) (1).
Marking note: Award 1 mark for each valid comparison point. Must include at least one similarity and two differences for full marks. Max 4 marks.
18. Explain how the properties of water make it an essential molecule for living organisms. [4]
Answer:
- Water is a polar molecule that forms hydrogen bonds, making it an excellent solvent for ionic and polar substances, allowing metabolic reactions to occur in solution (1).
- Water has a high specific heat capacity, meaning it resists temperature changes, providing a stable environment for aquatic organisms and maintaining constant body temperature (1).
- Water has a high latent heat of vaporisation, so evaporation of sweat provides an effective cooling mechanism for organisms (1).
- Water is cohesive (due to hydrogen bonding), which supports the transpiration stream in plants and surface tension for some organisms (1).
- Ice is less dense than liquid water, so it floats and insulates bodies of water, allowing aquatic life to survive in cold conditions (1).
Marking note: Award 1 mark for each correctly explained property and its biological significance. Max 4 marks.
19. A student carried out the Benedict's test on a solution of an unknown carbohydrate. A brick-red precipitate was formed. The student concluded that the solution contained glucose. Evaluate the student's conclusion. [3]
Answer:
- The conclusion is not fully valid because the Benedict's test is not specific to glucose (1).
- A positive result (brick-red precipitate) indicates the presence of a reducing sugar, which includes all monosaccharides (e.g., glucose, fructose, galactose) and some disaccharides (e.g., maltose, lactose) (1).
- To confirm the presence of glucose specifically, additional tests such as a glucose oxidase test or chromatography would be required (1).
Marking note: Award 1 mark for identifying the limitation of the test, 1 mark for explaining what a positive result actually indicates, 1 mark for suggesting further testing.
20. Discuss the importance of hydrogen bonding in the structure and function of biological molecules. Use named examples to support your answer. [5]
Answer:
- Hydrogen bonds form between a hydrogen atom covalently bonded to an electronegative atom (e.g., O, N) and another electronegative atom (1).
- In water, hydrogen bonds give water its cohesive properties, high specific heat capacity, and solvent properties, which are essential for life (1).
- In proteins, hydrogen bonds between the −NH and −C=O groups of the polypeptide backbone stabilise secondary structures such as α-helices and β-pleated sheets (1).
- In DNA, hydrogen bonds between complementary base pairs (A−T and C−G) hold the two strands of the double helix together, allowing replication and transcription (1).
- In cellulose, hydrogen bonds between parallel cellulose chains form microfibrils, providing high tensile strength to plant cell walls (1).
- The specificity of enzyme-substrate binding often involves hydrogen bonding between the active site and the substrate, enabling catalysis (1).
Marking note: Award 1 mark for each valid point with a named example. Must include at least three different biological molecules for full marks. Max 5 marks.
END OF ANSWER KEY