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A Level H1 Biology Practice Paper 3
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Questions
TuitionGoWhere Exam Practice (AI) - Biology H1 A-Level
TuitionGoWhere Secondary School (AI)
Subject: Biology
Level: A-Level H1
Paper: Practice Paper - Version 3 of 5
Topic: Cells and Biomolecules
Duration: 1 hour 15 minutes
Total Marks: 60
Name: ________________________
Class: ________________________
Date: ________________________
Instructions to Candidates
- Write your name, class, and date in the spaces provided.
- Answer all questions.
- Write your answers in the spaces provided in this question paper.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- You are advised to spend approximately 15 minutes on Section A and 60 minutes on Section B.
Section A: Structured Questions
Answer all questions in this section.
1. Fig. 1.1 shows a diagram of a phospholipid molecule.
(a) Identify the parts labelled A and B. [2] A: _______________________________________________________ B: _______________________________________________________
(b) Describe how phospholipids arrange themselves when placed in an aqueous environment to form a membrane. [2]
(c) Explain why this arrangement is essential for the function of the cell membrane as a barrier. [2]
2. A student investigated the effect of temperature on the activity of the enzyme amylase. The results are shown in Table 2.1.
| Temperature / °C | Rate of Reaction / arbitrary units |
|---|---|
| 10 | 12 |
| 20 | 28 |
| 30 | 45 |
| 40 | 58 |
| 50 | 35 |
| 60 | 5 |
| 70 | 0 |
(a) State the optimum temperature for this enzyme based on the data. [1]
(b) Explain the decrease in the rate of reaction between 40°C and 70°C with reference to the structure of enzymes. [3]
(c) Suggest why the rate of reaction at 10°C is low, but not zero. [2]
3. Fig. 3.1 shows a simplified diagram of a cell membrane illustrating two methods of transport, Method X and Method Y.
(a) Identify Method X and Method Y. [2] Method X: ____________________________________________________ Method Y: ____________________________________________________
(b) State one difference between Method X and Method Y regarding the use of energy. [1]
(c) Glucose enters red blood cells via facilitated diffusion. Explain why glucose cannot enter cells by simple diffusion through the phospholipid bilayer. [2]
4. Mitochondria were isolated from liver cells and incubated in a solution containing pyruvate. Oxygen consumption was measured. In a second experiment, mitochondria were incubated with glucose.
(a) Explain why oxygen was consumed in the first experiment but not in the second. [3]
(b) Name the stage of aerobic respiration that occurs in the mitochondrial matrix. [1]
5. Water is essential for life.
(a) Explain how the polarity of water molecules contributes to its effectiveness as a solvent. [2]
(b) State two properties of water that help organisms maintain a stable internal temperature. [2]
Section B: Data Response and Extended Structured Questions
Answer all questions in this section.
6. Fig. 6.1 shows the structure of a triglyceride and a phospholipid.
(a) Compare the structure of a triglyceride and a phospholipid. [3]
(b) Triglycerides are used for energy storage in animals, while phospholipids are used for membrane structure. Explain how the structure of triglycerides makes them suitable for energy storage. [3]
(c) Unsaturated fatty acids have kinks in their hydrocarbon chains. Explain how the presence of unsaturated fatty acids affects the fluidity of cell membranes at low temperatures. [3]
7. Haemoglobin is a globular protein found in red blood cells. Fig. 7.1 shows the四级 structure of haemoglobin.
(a) Describe the levels of protein structure present in haemoglobin. [4]
(b) Explain how a change in a single amino acid in the primary structure of haemoglobin (as seen in sickle cell anaemia) can affect its function. [3]
(c) Distinguish between globular and fibrous proteins, giving one example of each. [3]
8. DNA and RNA are nucleic acids involved in protein synthesis.
(a) Complete Table 8.1 to show three differences between DNA and RNA. [3]
| Feature | DNA | RNA |
|---|---|---|
| Sugar | ||
| Bases | ||
| Structure |
(b) Explain the significance of complementary base pairing in DNA replication. [3]
(c) A molecule of mRNA contains 1200 nucleotides. Calculate the maximum number of amino acids in the polypeptide chain coded for by this mRNA. Show your working. [2]
Working:
Answer: ________________________ amino acids
9. Fig. 9.1 shows the results of an experiment investigating the uptake of potassium ions by root hair cells at different oxygen concentrations.
(a) Describe the relationship between oxygen concentration and potassium ion uptake shown in Fig. 9.1. [2]
(b) Explain the relationship described in (a) with reference to the mechanism of transport. [3]
(c) At very high oxygen concentrations, the rate of uptake levels off. Suggest a reason for this. [2]
10. Enzymes are biological catalysts.
(a) Define the term 'active site'. [1]
(b) Explain the 'induced fit' model of enzyme action. [3]
(c) Competitive inhibitors and non-competitive inhibitors affect enzyme activity differently. Explain how a competitive inhibitor reduces the rate of an enzyme-catalysed reaction. [3]
11. Cell fractionation is a technique used to separate organelles.
(a) State why the homogenate is kept ice-cold during cell fractionation. [1]
(b) State why the homogenate is buffered. [1]
(c) Explain why the homogenate is filtered before centrifugation. [1]
(d) Describe how differential centrifugation separates organelles. [3]
12. Fig. 12.1 shows a prokaryotic cell and a eukaryotic cell.
(a) State two structural features present in the eukaryotic cell but absent in the prokaryotic cell. [2]
(b) Prokaryotic cells have ribosomes. State the sedimentation coefficient (S value) of prokaryotic ribosomes. [1]
(c) Explain why antibiotics that target prokaryotic ribosomes do not usually harm human cells. [2]
13. Glycogen and starch are polysaccharides.
(a) State the monomer unit of glycogen and starch. [1]
(b) Explain why glycogen is a suitable storage molecule in animals. [3]
(c) Cellulose is also a polysaccharide made of glucose. Explain why humans cannot digest cellulose. [2]
14. Fig. 14.1 shows the fluid mosaic model of the cell membrane.
(a) Explain the term 'fluid mosaic'. [2]
(b) Cholesterol is present in animal cell membranes. State two functions of cholesterol in the membrane. [2]
(c) Glycoproteins are found on the surface of the membrane. State one function of glycoproteins. [1]
15. Water potential is a key concept in understanding water movement.
(a) Define water potential. [1]
(b) Pure water has a water potential of _________ kPa. [1]
(c) A plant cell is placed in a solution with a lower water potential than the cell sap. Describe what happens to the cell and name the condition. [3]
16. ATP is the universal energy currency.
(a) Describe the structure of an ATP molecule. [3]
(b) Explain why ATP is a suitable immediate energy source for cellular processes. [2]
(c) State one cellular process that uses ATP. [1]
17. Fig. 17.1 shows the effect of substrate concentration on the rate of an enzyme-catalysed reaction.
(a) Explain the shape of the curve between point A and point B. [2]
(b) Explain why the rate of reaction remains constant after point B. [2]
(c) On Fig. 17.1, sketch the curve you would expect if the enzyme concentration was doubled. [1]
18. Collagen is a fibrous protein.
(a) Describe the structure of a collagen molecule. [3]
(b) Explain how the structure of collagen relates to its function in tendons. [2]
19. The cell cycle consists of interphase and mitosis.
(a) State what happens to the DNA during the S phase of interphase. [1]
(b) Explain why the G1 phase is important for cell growth. [2]
(c) Distinguish between chromatin and chromosomes. [2]
20. Viruses are acellular particles.
(a) State two components found in all viruses. [2]
(b) Explain why viruses are not considered living organisms. [2]
(c) Some viruses have an envelope. State the origin of this envelope. [1]
End of Paper
Answers
TuitionGoWhere Exam Practice (AI) - Biology H1 A-Level
Answer Key and Marking Scheme
Topic: Cells and Biomolecules
Version: 3 of 5
Section A: Structured Questions
1. (a) A: Phosphate head / Hydrophilic head [1] B: Fatty acid tails / Hydrophobic tails [1]
(b) Phospholipids form a bilayer [1]. The hydrophilic heads face outward towards the aqueous environment (cytoplasm/tissue fluid) and the hydrophobic tails face inward, away from water [1].
(c) The hydrophobic interior prevents the passage of water-soluble (polar/charged) substances [1]. This allows the cell to control what enters and leaves (selective permeability) / maintains distinct internal environment [1].
2. (a) 40°C [1]
(b) High temperature causes the enzyme to denature [1]. The hydrogen bonds (and other bonds) maintaining the tertiary structure break [1]. The shape of the active site changes, so the substrate no longer fits / enzyme-substrate complexes cannot form [1].
(c) At low temperatures, molecules have low kinetic energy [1]. There are fewer successful collisions between enzyme and substrate per unit time [1].
3. (a) Method X: Active Transport [1] Method Y: Facilitated Diffusion [1] (Note: Assuming Fig 3.1 shows X moving against gradient with ATP, and Y moving down gradient via channel/carrier)
(b) Method X requires ATP / energy; Method Y does not [1].
(c) Glucose is a large / polar molecule [1]. It cannot pass through the hydrophobic fatty acid tails of the phospholipid bilayer [1].
4. (a) Pyruvate can enter the mitochondria and undergo the Link Reaction and Krebs cycle, which produce reduced NAD/FAD for the Electron Transport Chain (ETC) where oxygen is used [1]. Glucose cannot enter the mitochondria directly [1]. Glucose must first be broken down into pyruvate via glycolysis, which occurs in the cytoplasm, not in isolated mitochondria [1].
(b) Krebs Cycle / Citric Acid Cycle [1]
5. (a) Water molecules are polar (dipole) [1]. They form hydrogen bonds with ions/polar molecules, surrounding them and keeping them in solution [1].
(b) Any two of:
- High specific heat capacity [1]
- High latent heat of vaporisation [1]
- High thermal conductivity [1]
Section B: Data Response and Extended Structured Questions
6. (a) Both contain glycerol and fatty acids [1]. Triglycerides have 3 fatty acids; Phospholipids have 2 fatty acids and 1 phosphate group [1]. Triglycerides are non-polar/hydrophobic; Phospholipids are amphipathic (have hydrophilic head and hydrophobic tails) [1].
(b) The fatty acid tails contain many C-H bonds which store high energy [1]. They are insoluble in water, so they do not affect the water potential of cells/osmosis [1]. They can be packed densely without water, providing more energy per gram than carbohydrates [1].
(c) The kinks prevent the fatty acid tails from packing closely together [1]. This increases the fluidity of the membrane [1]. At low temperatures, this prevents the membrane from becoming too rigid/solidifying [1].
7. (a) Primary: Sequence of amino acids [1]. Secondary: Alpha-helix or beta-pleated sheet formed by hydrogen bonds [1]. Tertiary: 3D folding of polypeptide chains held by ionic, hydrogen, disulphide bonds [1]. Quaternary: Association of multiple polypeptide subunits (4 in haemoglobin) [1].
(b) Change in primary structure changes the sequence of amino acids [1]. This alters the R-group interactions, changing the tertiary structure/shape of the protein [1]. This changes the shape of the heme group or oxygen binding site, reducing oxygen carrying capacity / causing polymerization (sickling) [1].
(c) Globular: Compact, spherical, soluble, metabolic function (e.g., Haemoglobin/Insulin) [1.5 for description + example]. Fibrous: Long, insoluble, structural function (e.g., Collagen/Keratin) [1.5 for description + example].
8. (a) Sugar: Deoxyribose (DNA) vs Ribose (RNA) [1]. Bases: Thymine (DNA) vs Uracil (RNA) [1]. Structure: Double helix / Double stranded (DNA) vs Single stranded (RNA) [1].
(b) Ensures accurate replication / fidelity [1]. Each strand acts as a template for the new strand [1]. Complementary base pairing ensures the new DNA molecule is identical to the original [1].
(c) 1200 nucleotides / 3 (bases per codon) = 400 codons [1]. One codon codes for one amino acid (stop codon does not code for amino acid, so max is 399 or 400 depending on interpretation, but usually 1200/3 = 400 is accepted for 'maximum' unless stop is specified. Accept 399-400). Answer: 400 (or 399) [1].
9. (a) As oxygen concentration increases, the rate of potassium uptake increases [1]. The rate levels off / reaches a maximum at higher oxygen concentrations [1].
(b) Uptake is by active transport [1]. Active transport requires ATP / energy [1]. ATP is produced by aerobic respiration, which requires oxygen [1].
(c) All carrier proteins are saturated / working at maximum rate [1]. Limited number of carrier proteins available [1].
10. (a) The region on the enzyme where the substrate binds [1].
(b) The active site is not rigid/complementary to the substrate initially [1]. The substrate induces a change in the shape of the active site [1]. This creates a precise fit for the substrate (enzyme-substrate complex) [1].
(c) Competitive inhibitor has a similar shape to the substrate [1]. It binds to the active site [1]. It blocks the substrate from binding / reduces the frequency of successful collisions between enzyme and substrate [1].
11. (a) To reduce enzyme activity / prevent digestion of organelles by lysosomal enzymes [1].
(b) To maintain pH / prevent denaturation of enzymes/proteins [1].
(c) To remove unbroken cells / large tissue debris [1].
(d) Spin at low speed to pellet heaviest organelles (nuclei) [1]. Remove supernatant and spin at higher speed to pellet next heaviest (mitochondria/chloroplasts) [1]. Repeat at increasing speeds to separate lighter organelles (ribosomes) [1].
12. (a) Any two: Nucleus, Membrane-bound organelles (mitochondria, Golgi, ER), Linear DNA, Histones, 80S ribosomes [1 each].
(b) 70S [1]
(c) Human cells have 80S ribosomes which have a different structure [1]. The antibiotic specifically targets the 70S ribosome structure / binding site [1].
13. (a) Alpha-glucose / Glucose [1]
(b) It is insoluble, so it does not affect water potential [1]. It is compact/coiled, allowing storage of many units in small space [1]. It is highly branched, allowing rapid hydrolysis/release of glucose when needed [1].
(c) Cellulose contains beta-glucose [1]. Humans lack the enzyme (cellulase) to break the beta-1,4-glycosidic bonds [1].
14. (a) Fluid: Phospholipids and proteins can move laterally [1]. Mosaic: Proteins are embedded in the bilayer in a scattered pattern [1].
(b) Any two:
- Regulates membrane fluidity (prevents packing at low temp / stabilizes at high temp) [1].
- Reduces permeability to small water-soluble molecules [1].
- Mechanical stability [1].
(c) Cell recognition / Cell signalling / Antigen receptor [1].
15. (a) The tendency of water molecules to move from one region to another / The pressure exerted by water molecules [1].
(b) 0 [1]
(c) Water leaves the cell by osmosis [1]. The vacuole shrinks and the cytoplasm pulls away from the cell wall [1]. The cell becomes plasmolysed [1].
16. (a) Adenine (nitrogenous base) [1]. Ribose (pentose sugar) [1]. Three phosphate groups [1].
(b) Releases a small, manageable amount of energy suitable for cellular reactions [1]. Hydrolysis is a single-step reaction / releases energy quickly [1].
(c) Any one: Active transport, Muscle contraction, Protein synthesis, DNA replication, Nerve impulse transmission [1].
17. (a) Rate increases as substrate concentration increases [1]. More enzyme-substrate complexes form per unit time / More frequent collisions [1].
(b) All active sites are saturated / occupied [1]. Enzyme concentration is the limiting factor [1].
(c) Curve rises more steeply initially and plateaus at a higher rate (Vmax doubles) [1].
18. (a) Three polypeptide chains wound together to form a triple helix [1]. Held together by hydrogen bonds [1]. Many triple helices bundle together to form fibrils/fibres [1].
(b) High tensile strength [1]. Resists pulling forces / stretches without breaking [1].
19. (a) DNA replication / DNA synthesis [1].
(b) Cell grows in size [1]. Synthesis of organelles and proteins required for division [1].
(c) Chromatin: Uncondensed, long thin threads, present during interphase [1]. Chromosomes: Condensed, thick, visible structures, present during mitosis [1].
20. (a) Genetic material (DNA or RNA) [1]. Protein coat / Capsid [1].
(b) They cannot reproduce independently / require a host cell [1]. They do not carry out metabolism / do not have cellular structure [1].
(c) Derived from the host cell membrane (during budding) [1].