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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:
- Answer all questions in the spaces provided.
- Read each question carefully, referring to figures where indicated.
- Show all working for calculations.
- The marks for each question or part-question are shown in brackets.
Section A: Multiple Choice (5 marks)
Choose the most appropriate answer for each question. Write the letter in the box provided.
1. Which organelle is the primary site of aerobic respiration in a eukaryotic cell?
A. Nucleus
B. Golgi apparatus
C. Mitochondrion
D. Rough endoplasmic reticulum
[1 mark]
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2. The fluid mosaic model of the cell membrane describes membrane proteins as:
A. fixed in a rigid lipid bilayer
B. freely moving within a phospholipid bilayer
C. absent from the membrane
D. forming a continuous sheet on the membrane surface
[1 mark]
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3. Which of the following correctly pairs the monomer with its polymer?
A. Amino acid – starch
B. Glucose – protein
C. Nucleotide – DNA
D. Glycerol – glycogen
[1 mark]
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4. Competitive inhibition of an enzyme can be overcome by:
A. increasing substrate concentration
B. lowering temperature
C. adding a non-competitive inhibitor
D. removing the cofactor
[1 mark]
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5. The semi-conservative replication of DNA means that each new double helix contains:
A. two new strands
B. two original strands
C. one original strand and one new strand
D. a mixture of broken fragments from both old strands
[1 mark]
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Section B: Structured Questions (30 marks)
6. Figure 6.1 shows a diagram of a typical animal cell as seen under an electron microscope.
(a) Identify the structures labelled P, Q, and R.
[3 marks]
P: ________________________
Q: ________________________
R: ________________________
(b) Explain the role of the Golgi apparatus in processing and transporting proteins synthesised on the rough endoplasmic reticulum.
[3 marks]
7. Figure 7.1 illustrates the energy profile of an enzyme-catalysed reaction in the absence and presence of an inhibitor.
(a) With reference to Fig. 7.1, state the effect of the inhibitor on the activation energy of the reaction.
[1 mark]
(b) Using your knowledge of enzyme action, explain how a non-competitive inhibitor reduces the rate of reaction.
[3 marks]
8. Figure 8.1 represents a small region of the cell surface membrane.
(a) Name the components labelled W, X, and Y.
[3 marks]
W: ________________________
X: ________________________
Y: ________________________
(b) Describe one function of component X in the membrane.
[2 marks]
9. A student carried out Benedict’s test on three unknown solutions. The observations are shown in Table 9.1.
| Solution | Colour after heating |
|---|---|
| A | Blue |
| B | Green |
| C | Brick-red |
(a) Identify which solution contains the highest concentration of reducing sugar. Give a reason for your answer.
[2 marks]
(b) Explain why solution A remained blue.
[1 mark]
10. Figure 10.1 shows the structure of a polysaccharide.
(a) State whether the polysaccharide is amylose or amylopectin. Give one visible feature that supports your choice.
[2 marks]
(b) Describe how the structure of this polysaccharide makes it suitable as a storage molecule in plants.
[2 marks]
11. Figure 11.1 shows the rate of glucose uptake into erythrocytes as the external glucose concentration increases.
(a) With reference to Fig. 11.1, describe the shape of the curve.
[2 marks]
(b) Explain the pattern observed, and identify the transport mechanism responsible for this glucose uptake.
[3 marks]
12. Figure 12.1 shows a simplified diagram of part of a DNA double helix.
(a) Name the type of bond indicated by the arrow Z.
[1 mark]
(b) Explain the significance of complementary base pairing in ensuring accurate replication of the genetic information.
[3 marks]
13. Water performs many essential functions in living organisms.
With reference to its molecular structure, explain the role of water as a solvent for polar and ionic substances.
[3 marks]
14. Figure 14.1 shows the structural formula of a lipid molecule.
(a) Name the type of lipid shown.
[1 mark]
(b) Describe how the structure of this lipid relates to its function as an energy reserve in animals.
[2 marks]
15. Table 15.1 shows the first ten amino acids of the β-globin polypeptide in three different species.
| Species | Amino acid sequence (positions 1–10) |
|---|---|
| Human | Val-His-Leu-Thr-Pro-Glu-Glu-Lys-Ser-Ala |
| Gorilla | Val-His-Leu-Thr-Pro-Glu-Glu-Lys-Ser-Ala |
| Mouse | Val-His-Leu-Thr-Asp-Ala-Glu-Lys-Ser-Ala |
(a) Compare the sequences of human and mouse β-globin.
[2 marks]
(b) Using the data, discuss what the amino acid sequences suggest about the evolutionary relationship between humans, gorillas, and mice.
[2 marks]
Section C: Data-Based and Extended Response Questions (15 marks)
16. A student investigated the activity of catalase in liver extract at different pH values. The volume of oxygen produced in 60 seconds was measured. The results are shown in Figure 16.1 and Table 16.1.
| pH | Volume of O₂ produced / cm³ |
|---|---|
| 2.0 | 0.2 |
| 4.0 | 2.5 |
| 6.0 | 8.0 |
| 7.0 | 9.8 |
| 8.0 | 9.6 |
| 10.0 | 3.1 |
(a) State the optimum pH for catalase activity in this experiment.
[1 mark]
(b) Explain why the rate of oxygen production decreases sharply at pH 10.0.
[2 marks]
(c) Suggest one practical improvement the student could make to increase the reliability of the results.
[1 mark]
17. Figure 17.1 shows a diagram of the four levels of protein structure.
Describe each level of protein structure and explain how a single amino acid substitution in the primary structure can alter the tertiary structure and function of a protein. Use the example of sickle‑cell haemoglobin to support your answer.
[6 marks]
18. Figure 18.1 shows the results of gel electrophoresis of haemoglobin proteins from three individuals: a person with sickle‑cell anaemia, a healthy individual, and a carrier (heterozygote).
(a) With reference to Fig. 18.1, explain why the carrier shows two bands whereas the healthy individual shows only one band.
[3 marks]
(b) Name the molecular difference between normal HbA protein and sickle‑cell HbS protein that accounts for their different positions on the gel.
[1 mark]
19. Figure 19.1 shows the arrangement of genes in the lac operon of E. coli.
(a) Explain how the lac operon is regulated when lactose is absent from the growth medium.
[2 marks]
(b) Explain how the presence of lactose leads to transcription of the structural genes.
[2 marks]
20. An experiment was performed to study the effect of an inhibitor on enzyme X. The initial reaction rate was measured at different substrate concentrations, both in the absence of inhibitor and in the presence of inhibitor. The data are shown in Table 20.1.
| [Substrate] / mmol dm⁻³ | Rate without inhibitor / μmol min⁻¹ | Rate with inhibitor / μmol min⁻¹ |
|---|---|---|
| 1.0 | 12 | 4 |
| 2.0 | 20 | 10 |
| 5.0 | 35 | 28 |
| 10.0 | 40 | 38 |
(a) For a substrate concentration of 5.0 mmol dm⁻³, calculate the percentage inhibition caused by the inhibitor.
[2 marks]
(b) Based on the data, what type of inhibition is most likely? Justify your answer.
[2 marks]
End of Quiz
Answers
A-Level Biology H2 Quiz - Cells Biomolecules – ANSWERS
Total marks: 50
Marking points are shown in brackets; accept equivalent scientific wording.
Section A: Multiple Choice
1. C – Mitochondrion (1)
2. B – freely moving within a phospholipid bilayer (1)
3. C – Nucleotide – DNA (1)
4. A – increasing substrate concentration (1)
5. C – one original strand and one new strand (1)
Section B: Structured Questions
6.
(a)
P – Nucleus/nuclear envelope (1)
Q – Mitochondrion (1)
R – Rough endoplasmic reticulum (1)
(b)
Proteins from the RER are transported to the Golgi in vesicles (1). Within the Golgi, the proteins are modified (e.g. by addition of carbohydrate groups – glycosylation) (1). The modified proteins are then sorted and packaged into secretory vesicles for transport to the cell membrane or to other organelles (1).
7.
(a) The inhibitor does not change the activation energy; the peak height (activation energy) remains the same in the presence of inhibitor (1).
(Accept: activation energy is unchanged.)
(b) A non‑competitive inhibitor binds to an allosteric site (a site other than the active site) (1). This changes the shape (conformation) of the enzyme, including the active site, so the substrate can no longer bind (1). Since the inhibitor does not compete with the substrate, increasing substrate concentration cannot overcome the inhibition, and Vmax decreases (1).
8.
(a)
W – Phospholipid bilayer / phospholipid molecule (1)
X – (Integral/intrinsic) protein / channel protein / carrier protein (1)
Y – Glycoprotein / glycocalyx / carbohydrate chain (1)
(b) Component X (protein) can act as a channel or carrier for facilitated diffusion of ions/polar molecules across the membrane (1), or as a pump for active transport (1).
9.
(a) Solution C contains the highest concentration of reducing sugar (1). A brick‑red precipitate indicates a large amount of reducing sugar has reduced the Cu²⁺ ions in Benedict’s reagent to Cu⁺ (1).
(b) Solution A remained blue because it contained no (or a very low concentration of) reducing sugar, so no reduction of Cu²⁺ occurred (1).
10.
(a) Amylopectin (1). The molecule shows a branched structure / 1→6 glycosidic bonds at branch points (1).
(b) Amylopectin has many branches, which provides many free ends for rapid hydrolysis by enzymes (1). This allows glucose to be released quickly when energy is needed. The molecule is compact and insoluble, so it does not affect the water potential of the cell (1).
11.
(a) The rate of uptake increases rapidly at low glucose concentrations and then levels off (plateaus) at high concentrations, giving a hyperbolic curve (1). The curve does not increase linearly; it approaches a maximum rate (Vmax) (1).
(b) The pattern is characteristic of facilitated diffusion (1). The rate increases as more carrier/channel proteins are occupied, but once all carriers are saturated (occupied), the rate cannot increase further, hence the plateau (1). Since no metabolic energy is required and glucose moves down its concentration gradient, the mechanism is facilitated diffusion (1).
12.
(a) Hydrogen bond (1).
(b) Complementary base pairing (A with T, G with C) ensures that the sequence of the new strand is exactly determined by the template strand (1). When the double helix unwinds, each parental strand acts as a template; the specificity of hydrogen bonding ensures that only the correct nucleotide is added by DNA polymerase (1). This results in two identical daughter DNA molecules, preserving the genetic information accurately (1).
13.
Water is a polar molecule because the oxygen atom has a slight negative charge (δ⁻) and the hydrogen atoms have a slight positive charge (δ⁺) (1). This polarity allows water molecules to surround and separate charged ions (e.g. Na⁺, Cl⁻) and polar molecules (e.g. glucose) (1). The formation of hydration shells keeps the solutes in solution, enabling them to dissolve, react, and be transported in biological systems (1).
14.
(a) Triglyceride (or triacylglycerol) (1).
(b) Triglycerides are composed of a glycerol backbone esterified to three fatty acid chains (1). The long hydrocarbon tails contain a high proportion of carbon–hydrogen bonds and are highly reduced; they yield a large amount of energy per gram when oxidised in respiration. Being hydrophobic, triglycerides are stored as anhydrous fat droplets, which do not increase water content and therefore do not affect osmosis (1).
15.
(a) The human and mouse sequences differ at two positions: positions 5 and 6 (Pro/Asp and Glu/Ala) (1). The remaining eight amino acids are identical (1).
(b) The identical sequences between human and gorilla suggest a very close evolutionary relationship (recent common ancestor) (1). The greater number of differences between human and mouse indicates that humans and mice diverged earlier in evolutionary history (1).
Section C: Data-Based and Extended Response Questions
16.
(a) pH 7.0 (1)
(b) At pH 10.0, the pH is far from the enzyme’s optimum, causing disruption of hydrogen and ionic bonds that maintain the active site’s specific shape (denaturation) (1). The substrate can no longer bind effectively, so the rate of product formation falls sharply (1).
(c) Repeat the experiment at each pH at least twice more and calculate a mean value (1). (Accept: use a more precise buffer, control temperature more carefully, etc.)
17.
Model answer:
- Primary structure: Linear sequence of amino acids in a polypeptide chain, held together by peptide bonds (1).
- Secondary structure: Regular local folding patterns – α‑helices and β‑pleated sheets – stabilised by hydrogen bonds between the –NH and –C=O groups of the backbone (1).
- Tertiary structure: Overall three‑dimensional shape of a single polypeptide, maintained by interactions between R‑groups: disulfide bonds, ionic bonds, hydrophobic interactions, hydrogen bonds (1).
- Quaternary structure: Assembly of two or more polypeptide subunits (e.g. haemoglobin has four subunits) (1).
A single amino acid substitution in the primary structure changes the sequence of R‑groups. In sickle‑cell haemoglobin, the substitution of glutamic acid (hydrophilic, charged) by valine (hydrophobic) at position 6 of the β‑globin chain (1) leads to a change in tertiary structure: the hydrophobic valine creates a “sticky” patch that causes haemoglobin molecules to aggregate into long rods when deoxygenated (1). This alters the quaternary structure and distorts the red blood cell into a sickle shape, impairing oxygen transport (1).
18.
(a) The carrier is heterozygous: they possess one allele for normal HbA and one allele for sickle‑cell HbS (1). Each allele produces a slightly different protein (HbA and HbS). Gel electrophoresis separates proteins mainly by charge and/or size; HbS has a different net charge (or molecular size) and therefore migrates to a different position from HbA (1). The presence of both proteins results in two distinct bands (1).
(b) In HbS, the amino acid glutamic acid (negatively charged) is replaced by valine (neutral/hydrophobic), changing the overall charge of the protein (1). (Accept: substitution of glutamic acid by valine.)
19.
(a) When lactose is absent, the regulatory gene produces the active repressor protein, which binds to the operator region (1). This blocks RNA polymerase from binding to the promoter, so transcription of the structural genes is prevented (1).
(b) When lactose is present, it is converted to allolactose, which acts as an inducer and binds to the repressor protein (1). The repressor changes shape and can no longer bind to the operator, allowing RNA polymerase to bind to the promoter and transcribe the structural genes (1).
20.
(a) At [S] = 5.0 mmol dm⁻³:
Rate without inhibitor = 35 μmol min⁻¹; rate with inhibitor = 28 μmol min⁻¹.
Percentage inhibition = [(35 – 28) / 35] × 100 = (7/35) × 100 = 20% (2 marks; 1 for working, 1 for correct answer).
(b) The data suggest competitive inhibition (1). In competitive inhibition, the inhibitor competes with the substrate for the active site. At low substrate concentration, the inhibitor has a large effect (rates reduced significantly), but at high substrate concentration, the substrate out‑competes the inhibitor, and the rates approach the uninhibited rate (Vmax almost reached). This is reflected in the data: at 10 mmol dm⁻³, the rates are nearly equal (40 vs 38 μmol min⁻¹) (1).
(If the student explains why non‑competitive inhibition is unlikely—e.g. Vmax would remain reduced even at high [S]—award credit.)
End of Answer Key