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A Level H2 Biology Cells Biomolecules Quiz

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A-Level Biology H2 Quiz - Cells Biomolecules

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a scientific calculator.

Section A: Cell Structure and Membrane Transport (Questions 1–5)

1. Fig. 1.1 shows a simplified diagram of a plasma membrane.

(Imagine a diagram showing a phospholipid bilayer with embedded proteins, cholesterol, and glycoproteins. Label A points to a phospholipid head, Label B points to a channel protein, Label C points to cholesterol.)

(a) Identify the structures labelled A, B, and C. [3] A: _______________________________________________________ B: _______________________________________________________ C: _______________________________________________________

(b) Explain how the structure of structure A contributes to the barrier function of the membrane. [2]

(c) Structure B allows for facilitated diffusion. State two features of facilitated diffusion that distinguish it from simple diffusion. [2]

2. A student investigated the effect of temperature on the permeability of beetroot cell membranes. Beetroot cylinders were placed in water at different temperatures for 30 minutes. The intensity of the red pigment (betacyanin) released into the water was measured using a colorimeter.

(a) Explain why betacyanin is normally retained within the vacuole of intact beetroot cells. [2]

(b) At 60°C, the absorbance reading was significantly higher than at 20°C. Explain this result in terms membrane structure. [3]

3. Which of the following statements correctly describes the movement of water molecules during osmosis? [1]

A. Water moves from a region of lower water potential to a region of higher water potential through a partially permeable membrane. B. Water moves from a region of higher solute concentration to a region of lower solute concentration through a fully permeable membrane. C. Water moves from a region of higher water potential to a region of lower water potential through a partially permeable membrane. D. Water moves against its concentration gradient using ATP.

4. Sodium ions are pumped out of animal cells by the Na⁺/K⁺ ATPase pump.

(a) Explain why this process is classified as active transport. [2]

(b) Cyanide is a metabolic poison that inhibits aerobic respiration. Predict and explain the effect of adding cyanide to the cells on the uptake of potassium ions. [3]

5. Fig. 5.1 shows the ultrastructure of a prokaryotic cell and a eukaryotic cell.

(Imagine two diagrams: one prokaryote with nucleoid, plasmid, 70S ribosomes; one eukaryote with nucleus, mitochondria, 80S ribosomes.)

(a) Complete the table below by placing a tick (✓) if the feature is present and a cross (✗) if it is absent. [3]

Feature	Prokaryotic Cell	Eukaryotic Cell
Membrane-bound nucleus
80S ribosomes
Circular DNA associated with histones

(b) Mitochondria are thought to have originated from prokaryotic cells via endosymbiosis. State two structural features of mitochondria that support this theory. [2]

Section B: Biological Molecules and Enzymes (Questions 6–10)

6. Hemoglobin is a globular protein with a quaternary structure.

(a) Define the term quaternary structure. [2]

(b) Explain how the specific shape of the hemoglobin molecule is maintained. [2]

7. Fig. 7.1 shows the structure of a triglyceride.

(Imagine a glycerol molecule esterified to three fatty acid chains.)

(a) Name the type of reaction that joins glycerol to the fatty acids. [1]

(b) Triglycerides are insoluble in water. Explain how this property is advantageous for their role as storage molecules in organisms. [2]

8. An enzyme-catalysed reaction was carried out at varying substrate concentrations. The results are shown in Fig. 8.1.

(Imagine a hyperbolic curve: Rate of reaction vs Substrate Concentration. The curve plateaus at Vmax.)

(a) Explain why the rate of reaction increases between point X (low substrate) and point Y (medium substrate). [2]

(b) Explain why the rate of reaction remains constant after point Z (high substrate). [2]

9. Competitive and non-competitive inhibitors affect enzyme activity differently.

(a) Describe how a competitive inhibitor affects the $K_m$ and $V_{max}$ of an enzyme. [2] $K_m$ : __________________________________________________________________ $V_{max}$ : _______________________________________________________________

(b) Suggest why increasing the substrate concentration can overcome the effect of a competitive inhibitor but not a non-competitive inhibitor. [3]

10. Which of the following bonds is responsible for maintaining the primary structure of a protein? [1]

A. Hydrogen bonds B. Peptide bonds C. Disulfide bridges D. Ionic bonds

Section C: DNA, RNA, and Protein Synthesis (Questions 11–15)

11. Fig. 11.1 shows a short segment of a DNA molecule.

(Imagine a double helix segment with base pairs A-T and G-C.)

(a) Name the bond that holds the two strands of DNA together. [1]

(b) If the percentage of Adenine in a DNA sample is 22%, calculate the percentage of Cytosine. Show your working. [2] Working:

Answer: _______________ %

12. Describe the process of DNA replication. In your answer, include the roles of:

DNA helicase
DNA polymerase
DNA ligase [5]

13. Table 13.1 compares DNA and mRNA.

Feature	DNA	mRNA
Sugar	Deoxyribose	(a) _______________
Bases	A, T, C, G	(b) _______________
Strands	Double	(c) _______________

Complete the table. [3]

14. During translation, tRNA molecules bring amino acids to the ribosome.

(a) Explain the significance of the anticodon on the tRNA molecule. [2]

(b) State the site within the ribosome where peptide bonds are formed. [1]

15. A mutation occurs in a gene, changing the DNA triplet from GAA to GTA.

Original mRNA codon: CUU (codes for Leucine)
Mutated mRNA codon: CAU (codes for Histidine)

(a) Name the type of gene mutation described. [1]

(b) Explain why this mutation might have a significant effect on the protein's function. [2]

Section D: Application and Data Analysis (Questions 16–20)

16. Fig. 16.1 shows the results of gel electrophoresis performed on DNA samples from four individuals (1–4) for a specific gene.

(Imagine a gel image. Lane 1: One band at top. Lane 2: One band at bottom. Lane 3: Two bands (top and bottom). Lane 4: One band at top.)

(a) Explain the principle behind the separation of DNA fragments in gel electrophoresis. [2]

(b) Identify which individual is heterozygous for this gene. Explain your answer. [2] Individual: _______ Explanation: ______________________________________________________________

17. Sickle cell anaemia is caused by a mutation in the $\beta$ -globin gene. The normal allele is $Hb^A$ and the sickle cell allele is $Hb^S$ .

(a) Explain how the substitution of valine for glutamic acid in the $\beta$ -globin chain leads to the sickling of red blood cells under low oxygen conditions. [3]

(b) Individuals who are heterozygous ( $Hb^A Hb^S$ ) often show no symptoms of sickle cell anaemia. Explain why. [2]

18. The lac operon in E. coli controls the production of enzymes required for lactose metabolism.

(a) State the function of the lacI gene. [1]

(b) Explain what happens to the lac operon when lactose is absent from the environment. [3]

19. ATP is the universal energy currency of the cell.

(a) Describe the structural difference between an ATP molecule and a DNA nucleotide containing adenine. [2]

(b) Explain why ATP is a suitable immediate energy source for cellular reactions. [2]

20. A student isolated mitochondria from liver cells and suspended them in a buffer solution. Oxygen consumption was measured over time.

(a) Explain why the mitochondria consume oxygen. [2]

(b) If an uncoupling agent (which makes the inner mitochondrial membrane permeable to protons) is added, oxygen consumption increases but ATP production stops. Explain this observation. [3]

End of Quiz

Answers

A-Level Biology H2 Quiz - Cells Biomolecules (Answer Key)

1. (a) A: Phospholipid (head) [1] B: Channel protein / Transmembrane protein [1] C: Cholesterol [1] (b) The hydrophilic phosphate heads face the aqueous environment (outside/inside cell), while the hydrophobic fatty acid tails face inward. [1] This creates a barrier to water-soluble (polar/charged) molecules/ions, preventing them from passing freely. [1] (c)

Requires specific carrier/channel proteins. [1]
Does not require energy (ATP) / Moves down concentration gradient. [1] (Note: Must distinguish from simple diffusion which doesn't need proteins.)

2. (a) The tonoplast (vacuolar membrane) and plasma membrane are selectively/permeable. [1] Betacyanin is a large/polar molecule that cannot pass through the phospholipid bilayer or transport proteins under normal conditions. [1] (b) High temperature causes the phospholipids to gain kinetic energy and move more, increasing membrane fluidity. [1] Proteins in the membrane denature/change shape at high temperatures. [1] This creates gaps/holes in the membrane, allowing the pigment to leak out. [1]

3. C [1]

4. (a) It moves ions against their concentration gradient. [1] It requires energy from the hydrolysis of ATP. [1] (b) Uptake of potassium ions will decrease/stop. [1] Cyanide inhibits aerobic respiration, preventing the production of ATP. [1] The Na⁺/K⁺ pump requires ATP to function; without ATP, active transport cannot occur. [1]

5. (a) Membrane-bound nucleus: Prokaryotic (✗), Eukaryotic (✓) [1] 80S ribosomes: Prokaryotic (✗), Eukaryotic (✓) [1] Circular DNA associated with histones: Prokaryotic (✗), Eukaryotic (✓) [1] (Note: Prokaryotic DNA is circular but NOT associated with histones.) (b)

Mitochondria have their own circular DNA (similar to prokaryotes). [1]
Mitochondria have 70S ribosomes (similar to prokaryotes). [1] (Alternative: Double membrane / Binary fission replication.)

6. (a) The arrangement/interaction of two or more polypeptide chains/subunits. [1] To form a functional protein. [1] (b) By bonds/interactions between the R-groups of the different polypeptide chains. [1] These include hydrogen bonds, ionic bonds, disulfide bridges, and hydrophobic interactions. [1]

7. (a) Condensation [1] (b) Being insoluble, they do not affect the water potential of the cell/cytoplasm. [1] This prevents water from entering the cell by osmosis, which would cause swelling/lysis. [1]

8. (a) As substrate concentration increases, there are more frequent collisions between substrate and enzyme active sites. [1] More enzyme-substrate complexes are formed per unit time. [1] (b) All active sites are saturated/occupied. [1] The enzyme is working at its maximum rate ( $V_{max}$ ); adding more substrate cannot increase the rate further. [1]

9. (a) $K_m$ : Increases [1] $V_{max}$ : Unchanged / Remains the same [1] (b) Competitive inhibitors bind to the active site; high substrate concentration outcompetes the inhibitor for the active site. [1] Non-competitive inhibitors bind to an allosteric site, changing the shape of the active site. [1] Substrate cannot bind regardless of concentration because the active site is no longer complementary. [1]

10. B [1]

11. (a) Hydrogen bonds [1] (b) If A = 22%, then T = 22% (Chargaff's rule). [1] A + T = 44%. Therefore G + C = 100% - 44% = 56%. Since G = C, C = 56% / 2 = 28%. [1]

12.

DNA helicase breaks hydrogen bonds between base pairs, unzipping the double helix. [1]
This creates two template strands. [1]
Free nucleotides pair with complementary bases on the template strands (A-T, C-G). [1]
DNA polymerase joins nucleotides together by forming phosphodiester bonds (in the 5' to 3' direction). [1]
On the lagging strand, DNA ligase joins Okazaki fragments together. [1]

13. (a) Ribose [1] (b) A, U, C, G (Uracil replaces Thymine) [1] (c) Single [1]

14. (a) The anticodon is complementary to the mRNA codon. [1] This ensures the correct amino acid is brought to the ribosome in the correct sequence. [1] (b) Peptidyl transferase center / Between the A site and P site. [1] (Accept: Ribosome catalyses bond formation.)

15. (a) Substitution (point mutation) [1] (b) The amino acid sequence (primary structure) is changed. [1] This may change the folding/tertiary structure of the protein, affecting the shape of the active site/function. [1]

16. (a) DNA is negatively charged and moves towards the positive electrode (anode). [1] Smaller fragments move faster/further through the gel matrix than larger fragments. [1] (b) Individual 3 [1] Heterozygotes have two different alleles, producing two different sized DNA fragments (two bands). [1]

17. (a) Valine is hydrophobic, whereas glutamic acid is hydrophilic. [1] Under low oxygen, the hydrophobic valine interacts with other hydrophobic regions on adjacent hemoglobin molecules. [1] This causes hemoglobin to polymerize/aggregate, distorting the red blood cell into a sickle shape. [1] (b) They produce both normal ( $Hb^A$ ) and sickle ( $Hb^S$ ) hemoglobin. [1] The presence of sufficient normal hemoglobin prevents significant sickling under normal oxygen conditions. [1]

18. (a) Codes for the repressor protein. [1] (b) The repressor protein is active and binds to the operator region. [1] This blocks RNA polymerase from binding to the promoter. [1] Transcription of the structural genes (lacZ, lacY, lacA) is prevented/inhibited. [1]

19. (a) ATP has three phosphate groups; a DNA nucleotide has one phosphate group. [1] ATP contains ribose sugar; DNA nucleotide contains deoxyribose sugar. [1] (b) The hydrolysis of the terminal phosphate bond releases a manageable/small amount of energy suitable for cellular work. [1] It can be rapidly regenerated/recycled from ADP and Pi. [1]

20. (a) Oxygen acts as the final electron acceptor in the electron transport chain. [1] It combines with electrons and protons to form water. [1] (b) The uncoupling agent allows protons to leak back into the matrix without passing through ATP synthase. [1] The electron transport chain continues to pump protons (consuming oxygen) to try to maintain the gradient. [1] However, because the proton gradient is dissipated, ATP synthase cannot generate ATP. [1]