- A.The rate of the reaction is at its maximum value (Vmax) because all enzyme active sites are fully saturated.
- B.Half of the enzyme active sites are occupied by substrate molecules, and the reaction rate is half of Vmax.
- C.The rate of reaction is negligible because the substrate concentration is too low to overcome the activation energy.
- D.The enzyme is completely denatured because the substrate concentration has reached a critical inhibitory threshold.
Cambridge IAS-Level · Thinka-original Practice Paper
2024 Cambridge IAS-Level Biology (9700) Practice Paper with Answers
Thinka Jun 2024 (V3) Cambridge International A Level-Style Mock — Biology (9700)
Paper 1 (Multiple Choice)
- A.x50
- B.x500
- C.x5000
- D.x50000
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- A.Osmosis across selectively permeable cell surface membranes and through plasmodesmata.
- B.Active transport of mineral ions followed by the movement of water by facilitated diffusion.
- C.Movement of water through cell walls and intercellular spaces driven by cohesive forces and transpiration pull.
- D.Bulk flow of water through the cytoplasm of adjacent cells linked by plasmodesmata.
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- A.It is a globular protein containing prosthetic haem groups and is soluble in water.
- B.It is composed of three polypeptide chains wound around each other to form a triple helix.
- C.It contains hydrophobic amino acids on its outer surface, making it highly soluble.
- D.Its quaternary structure is stabilized solely by disulfide bonds between adjacent alpha-helices.
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- A.Prophase
- B.Metaphase
- C.Anaphase
- D.Telophase
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- A.The bicuspid (mitral) valve opens, and blood flows into the left ventricle.
- B.The semi-lunar valve in the aorta opens, and blood is ejected from the left ventricle.
- C.The semi-lunar valve in the aorta closes, preventing backflow of blood.
- D.The left atrium contracts to force the remaining blood into the left ventricle.
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- A.22%
- B.28%
- C.44%
- D.56%
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- A.Active artificial immunity
- B.Active natural immunity
- C.Passive artificial immunity
- D.Passive natural immunity
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- A.At high temperatures, it decreases fluidity by stabilizing the phospholipid tails; at low temperatures, it increases fluidity by preventing tight packing of tails.
- B.At high temperatures, it increases fluidity by disrupting the hydrophobic core; at low temperatures, it decreases fluidity by binding tightly to fatty acid chains.
- C.It always decreases membrane fluidity across all physiological temperatures by forming covalent bonds with glycolipids.
- D.It always increases membrane fluidity by promoting the formation of unsaturated fatty acid chains in phospholipids.
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- A.It increases the rate of reaction by providing alternative pathways with a lower activation energy, without changing the net energy change of the reaction.
- B.It decreases the activation energy of the reaction by raising the free energy of the products.
- C.It increases the equilibrium constant of the reaction, ensuring more products are formed at equilibrium.
- D.It increases the kinetic energy of the substrate molecules, making collisions more frequent and successful.
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- A.Active transport of protons out of the companion cells into the cell wall space using ATP.
- B.Facilitated diffusion of protons from the sieve tube elements into the companion cells.
- C.Co-transport of protons and sucrose into the companion cells via symporter proteins.
- D.Active transport of sucrose into the phloem sap using a sucrose-proton antiporter.
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- A.CO2 reacts with water to form carbonic acid; carbonic acid dissociates into hydrogen ions and hydrogencarbonate ions; hydrogencarbonate ions diffuse out of the cell; chloride ions diffuse in.
- B.CO2 reacts with water to form carbonic acid; carbonic acid dissociates into hydrogen ions and hydrogencarbonate ions; chloride ions diffuse out of the cell; hydrogencarbonate ions diffuse in.
- C.CO2 binds directly to hemoglobin forming carbaminohemoglobin; carbonic acid is formed by carbonic anhydrase in the plasma; chloride ions enter the red blood cell.
- D.CO2 reacts with water to form carbaminohemoglobin; carbonic acid dissociates into hydrogen ions and hydrogencarbonate ions; hydrogencarbonate ions diffuse into the cell.
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- A.24 chromatids and 12 centromeres
- B.12 chromatids and 12 centromeres
- C.24 chromatids and 24 centromeres
- D.12 chromatids and 6 centromeres
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- A.5'-AUG CGU AGC UGA-3'
- B.5'-UAC GCA UCG ACU-3'
- C.5'-UCG ACU GCA UAC-3'
- D.5'-UCA GCU ACG CAU-3'
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- A.breast milk = natural passive immunity; tetanus toxoid = artificial active immunity
- B.breast milk = natural passive immunity; tetanus toxoid = artificial passive immunity
- C.breast milk = natural active immunity; tetanus toxoid = artificial active immunity
- D.breast milk = natural active immunity; tetanus toxoid = artificial passive immunity
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- A.ribosome on rough endoplasmic reticulum -> lumen of rough endoplasmic reticulum -> transport vesicle -> Golgi apparatus -> lysosome
- B.free ribosome -> lumen of smooth endoplasmic reticulum -> transport vesicle -> Golgi apparatus -> lysosome
- C.ribosome on rough endoplasmic reticulum -> transport vesicle -> Golgi apparatus -> secretory vesicle -> cell surface membrane
- D.nucleolus -> cytoplasm -> Golgi apparatus -> transport vesicle -> lysosome
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- A.1, 2 and 3 only
- B.1 and 3 only
- C.2 and 4 only
- D.1, 2, 3 and 4
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- A.1, 2, 3 and 4
- B.1 and 2 only
- C.1, 2 and 3 only
- D.3 and 4 only
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- A.Water will enter the cell because the water potential of the solution is higher than that of the cell.
- B.Water will leave the cell because the water potential of the cell is higher than that of the solution.
- C.No net movement of water will occur because the solute potentials are balanced.
- D.Water will leave the cell until the solute potential of the cell becomes \(+0.3\text{ MPa}\).
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- A.Amylose has only \(\alpha\)-1,4-glycosidic bonds, while glycogen is more highly branched than amylopectin due to more frequent \(\alpha\)-1,6-glycosidic bonds.
- B.Amylopectin has only \(\alpha\)-1,6-glycosidic bonds, while amylose has both \(\alpha\)-1,4 and \(\alpha\)-1,6-glycosidic bonds.
- C.Glycogen has only \(\alpha\)-1,4-glycosidic bonds, while amylopectin has only \(\alpha\)-1,6-glycosidic bonds.
- D.Amylose and glycogen are unbranched polymers, while amylopectin is a highly branched polymer.
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- A.Prophase: 24 chromatids, 12 chromosomes; Daughter cells: 12 chromosomes
- B.Prophase: 24 chromatids, 24 chromosomes; Daughter cells: 12 chromosomes
- C.Prophase: 12 chromatids, 12 chromosomes; Daughter cells: 6 chromosomes
- D.Prophase: 12 chromatids, 24 chromosomes; Daughter cells: 12 chromosomes
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- A.1, 2 and 3
- B.1 and 2 only
- C.2 and 3 only
- D.1 only
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- A.mRNA codons: 5'-AUG CCA GUG-3'; tRNA anticodons: 3'-UAC GGU CAC-5'
- B.mRNA codons: 5'-ATG CCT GTG-3'; tRNA anticodons: 3'-UAC GGU CAC-5'
- C.mRNA codons: 5'-AUG CCA GUG-3'; tRNA anticodons: 3'-ATG CCT GTG-5'
- D.mRNA codons: 5'-UAC GGU CAC-3'; tRNA anticodons: 3'-AUG CCA GUG-5'
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- A.Trachea: cartilage, smooth muscle, goblet cells; Bronchi: cartilage, smooth muscle, goblet cells; Bronchioles: smooth muscle only (no cartilage, no goblet cells)
- B.Trachea: cartilage and goblet cells only; Bronchi: cartilage, smooth muscle, goblet cells; Bronchioles: cartilage and smooth muscle only
- C.Trachea: cartilage, smooth muscle, goblet cells; Bronchi: smooth muscle and goblet cells only; Bronchioles: goblet cells only
- D.Trachea: smooth muscle and goblet cells only; Bronchi: cartilage and goblet cells only; Bronchioles: smooth muscle and cartilage only
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- A.\(V_{max}\) is decreased and \(K_m\) is unchanged
- B.\(V_{max}\) is unchanged and \(K_m\) is increased
- C.\(V_{max}\) is decreased and \(K_m\) is increased
- D.\(V_{max}\) is unchanged and \(K_m\) is decreased
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1. Ribosomes
2. Nucleolus
3. Centrioles
4. Rough endoplasmic reticulum
- A.1 and 2 only
- B.1, 2 and 4 only
- C.1, 3 and 4 only
- D.2, 3 and 4 only
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- A.Protons are pumped out of the companion cell by active transport; sucrose and protons enter the companion cell by co-transport down the proton gradient.
- B.Protons are pumped into the companion cell by active transport; sucrose and protons leave the companion cell by co-transport down the sucrose gradient.
- C.Protons diffuse out of the companion cell; sucrose enters the companion cell by active transport using ATP.
- D.Protons and sucrose enter the companion cell together via facilitated diffusion down their respective concentration gradients.
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- A.Haemoglobin contains hydrophobic amino acid residues on its outer surface, whereas collagen has hydrophobic residues strictly in its core.
- B.Haemoglobin is made of four identical polypeptide chains, whereas collagen consists of three different polypeptide chains.
- C.Haemoglobin has a quaternary structure containing prosthetic groups, whereas collagen has a quaternary structure with no prosthetic groups.
- D.Haemoglobin polypeptides are held together by peptide bonds to form a triple helix, whereas collagen polypeptides form a globular shape held by disulfide bonds.
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- A.8 chromatids in metaphase; 4 chromosomes in telophase
- B.16 chromatids in metaphase; 8 chromosomes in telophase
- C.16 chromatids in metaphase; 16 chromosomes in telophase
- D.8 chromatids in metaphase; 8 chromosomes in telophase
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- A.Carbonic anhydrase catalyses the formation of carbonic acid; hydrogen carbonate ions diffuse out of the cell; chloride ions diffuse into the cell.
- B.Carbonic anhydrase catalyses the formation of carbonic acid; hydrogen carbonate ions diffuse into the cell; chloride ions diffuse out of the cell.
- C.Carbonic anhydrase catalyses the breakdown of carbonic acid; hydrogen carbonate ions diffuse out of the cell; chloride ions diffuse into the cell.
- D.Carbonic anhydrase catalyses the breakdown of carbonic acid; hydrogen carbonate ions diffuse into the cell; chloride ions diffuse out of the cell.
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- A.\(3'\text{- UAC -}5'\), \(3'\text{- GGU -}5'\), \(3'\text{- CAC -}5'\)
- B.\(3'\text{- AUG -}5'\), \(3'\text{- CCA -}5'\), \(3'\text{- GUG -}5'\)
- C.\(5'\text{- UAC -}3'\), \(5'\text{- GGU -}3'\), \(5'\text{- CAC -}3'\)
- D.\(3'\text{- ATG -}5'\), \(3'\text{- GGT -}5'\), \(3'\text{- CAC -}5'\)
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1. For mRNA codon \(5'\text{- AUG -}3'\), the anticodon is \(3'\text{- UAC -}5'\).
2. For mRNA codon \(5'\text{- CCA -}3'\), the anticodon is \(3'\text{- GGU -}5'\).
3. For mRNA codon \(5'\text{- GUG -}3'\), the anticodon is \(3'\text{- CAC -}5'\).
Thus, the anticodons in the \(3'\) to \(5'\) direction are \(3'\text{- UAC -}5'\), \(3'\text{- GGU -}5'\), and \(3'\text{- CAC -}5'\).
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- A.It binds to bacterial 70S ribosomes and prevents the translation of essential proteins.
- B.It inhibits the enzyme transpeptidase, preventing the cross-linking of peptidoglycan chains in the cell wall.
- C.It breaks the phosphodiester bonds in circular DNA, preventing bacterial DNA replication.
- D.It disrupts the fluid mosaic structure of the bacterial cell surface membrane, causing cell lysis.
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- A.At high temperatures, cholesterol increases fluidity by preventing phospholipids from packing closely together.
- B.At high temperatures, cholesterol decreases fluidity by stabilizing the hydrocarbon tails of phospholipids.
- C.At low temperatures, cholesterol decreases fluidity by promoting hydrophobic interactions between fatty acids.
- D.At low temperatures, cholesterol increases fluidity by increasing the kinetic energy of polar head groups.
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- A.\(V_{max}\) decreases, \(K_m\) increases
- B.\(V_{max}\) decreases, \(K_m\) remains unchanged
- C.\(V_{max}\) remains unchanged, \(K_m\) increases
- D.\(V_{max}\) remains unchanged, \(K_m\) decreases
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- A.Bonds broken during denaturation: Hydrogen and ionic. Bonds broken during hydrolysis: Peptide.
- B.Bonds broken during denaturation: Peptide. Bonds broken during hydrolysis: Hydrogen and ionic.
- C.Bonds broken during denaturation: Peptide and hydrogen. Bonds broken during hydrolysis: Peptide.
- D.Bonds broken during denaturation: Ionic only. Bonds broken during hydrolysis: Hydrogen and peptide.
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- A.Prophase: 16 chromosomes, 16 pg DNA. Daughter cell: 16 chromosomes, 8 pg DNA.
- B.Prophase: 32 chromosomes, 16 pg DNA. Daughter cell: 16 chromosomes, 8 pg DNA.
- C.Prophase: 16 chromosomes, 16 pg DNA. Daughter cell: 8 chromosomes, 4 pg DNA.
- D.Prophase: 32 chromosomes, 8 pg DNA. Daughter cell: 16 chromosomes, 4 pg DNA.
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- A.100 amino acids and 99 peptide bonds
- B.99 amino acids and 98 peptide bonds
- C.99 amino acids and 99 peptide bonds
- D.98 amino acids and 97 peptide bonds
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- A.Xylem: lignified wall present, cytoplasm absent. Phloem: lignified wall absent, cytoplasm present.
- B.Xylem: lignified wall present, cytoplasm present. Phloem: lignified wall absent, cytoplasm absent.
- C.Xylem: lignified wall absent, cytoplasm absent. Phloem: lignified wall present, cytoplasm present.
- D.Xylem: lignified wall present, cytoplasm absent. Phloem: lignified wall present, cytoplasm absent.
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- A.Cartilage: absent. Smooth muscle: present. Ciliated epithelium: present.
- B.Cartilage: present. Smooth muscle: absent. Ciliated epithelium: present.
- C.Cartilage: absent. Smooth muscle: present. Ciliated epithelium: absent.
- D.Cartilage: present. Smooth muscle: present. Ciliated epithelium: absent.
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- A.Antivenom: passive artificial. Tetanus vaccine: active artificial.
- B.Antivenom: active artificial. Tetanus vaccine: passive artificial.
- C.Antivenom: passive natural. Tetanus vaccine: active artificial.
- D.Antivenom: passive artificial. Tetanus vaccine: active natural.
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Paper 2 (AS Level Structured Questions)
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Paper 3 (Advanced Practical Skills)
Investigation of Osmosis in Onion Epidermal Cells
Red onion epidermal cells contain a pigmented cell sap inside their vacuoles, which allows them to be easily observed under a light microscope. When these cells are placed in hypertonic solutions, water leaves the vacuole, causing the protoplast to shrink away from the cell wall in a process known as plasmolysis. In this investigation, you will determine the effect of sucrose concentration on the percentage of plasmolysed cells.
You are provided with a stock solution of 1.0 mol dm-3 sucrose, labeled S, and distilled water, labeled W.
(a) (i) Complete a plan to show how you would prepare 10.0 cm3 of each of the following sucrose concentrations: 0.8, 0.6, 0.4, and 0.2 mol dm-3, using stock solution S and distilled water W. Show all of your calculations and record the volumes in a suitable table. [3 marks]
(a) (ii) You are to mount small, single-layer strips of red onion epidermis on separate microscope slides, each covered with a different sucrose concentration. After leaving the slides for 10 minutes to equilibrate, observe them under the high-power objective lens of a microscope. Count a representative sample of cells in each treatment. In the space below, construct a single table to record all your raw data, including the number of plasmolysed cells, the total number of cells counted, and the calculated percentage of plasmolysis for each concentration. [6 marks]
(a) (iii) Explain, in terms of water potential, the relationship between sucrose concentration and the percentage of plasmolysed onion cells. [4 marks]
(a) (iv) Identify two sources of error in this practical procedure and suggest an improvement for each. [3 marks]
(a) (v) Describe how you would modify this procedure to estimate the solute potential of the onion epidermal cells more accurately. [4 marks]
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Solution for Onion Plasmolysis Investigation
(a) (i) Dilution Calculations and Table:
The volume of stock sucrose solution S required is calculated using the formula: \( V_1 = \frac{C_2 \times V_2}{C_1} \), where \( C_1 = 1.0 \text{ mol dm}^{-3} \) and \( V_2 = 10.0 \text{ cm}^3 \).
- For 0.8 mol dm-3: \( 8.0 \text{ cm}^3 \) of S + \( 2.0 \text{ cm}^3 \) of W
- For 0.6 mol dm-3: \( 6.0 \text{ cm}^3 \) of S + \( 4.0 \text{ cm}^3 \) of W
- For 0.4 mol dm-3: \( 4.0 \text{ cm}^3 \) of S + \( 6.0 \text{ cm}^3 \) of W
- For 0.2 mol dm-3: \( 2.0 \text{ cm}^3 \) of S + \( 8.0 \text{ cm}^3 \) of W
(a) (ii) Results Table:
A well-constructed results table must have complete grid lines and clear headers with appropriate slash-separated units. Example of a standard data layout with typical biological trends:
- 0.2 mol dm-3: 0 plasmolysed / 20 total cells counted = 0% plasmolysis
- 0.4 mol dm-3: 3 plasmolysed / 20 total cells counted = 15% plasmolysis
- 0.6 mol dm-3: 12 plasmolysed / 20 total cells counted = 60% plasmolysis
- 0.8 mol dm-3: 19 plasmolysed / 20 total cells counted = 95% plasmolysis
- 1.0 mol dm-3: 20 plasmolysed / 20 total cells counted = 100% plasmolysis
(a) (iii) Explanation:
As sucrose concentration increases, the solute potential of the external solution decreases, which lowers its water potential. This creates a water potential gradient where the water potential inside the vacuole is higher than the external water potential. Consequently, water moves out of the vacuole and cytoplasm by osmosis down the water potential gradient across the selectively permeable cell membrane. This causes the protoplast to shrink away from the cell wall, increasing the percentage of plasmolysed cells observed.
(a) (iv) Errors and Improvements:
- Error: Subjectivity in deciding whether a cell is 'just' starting to plasmolyse. Improvement: Establish a standard reference set of photomicrographs showing clear criteria for a plasmolysed cell.
- Error: Epidermal cells obtained from different parts of the onion bulb scale may have different pre-existing water potentials. Improvement: Take all tissue strips from the exact same scale and layer of a single onion bulb.
(a) (v) Modification:
Prepare a much narrower range of sucrose concentrations around the concentration that yielded approximately 50% plasmolysis (e.g., 0.45, 0.50, 0.55, 0.65 mol dm-3). Count a larger sample size of cells (at least 100 cells per slide) to accurately determine the concentration that causes 50% plasmolysis (incipient plasmolysis), at which point the internal solute potential of the cells is equal to the external osmotic potential. Ensure temperature is controlled using a water bath.
Marking scheme
(a) (i) [3 marks]
- 1 mark: Correctly calculates volumes of stock S and distilled water W to make all 4 concentrations.
- 1 mark: Formulates volumes to total exactly 10.0 cm3 for each concentration.
- 1 mark: Presents all calculated volumes consistently to 1 decimal place (e.g., 8.0 cm3, not 8 cm3).
(a) (ii) [6 marks]
- 1 mark: Table drawn with clear ruled borders and distinct columns.
- 1 mark: Headers with correct units presented as: 'Sucrose concentration / mol dm-3'.
- 1 mark: Records raw count data for at least 5 different sucrose concentrations.
- 1 mark: Raw counts of cells are whole numbers.
- 1 mark: Correct calculation of percentage plasmolysis for all rows.
- 1 mark: Percentage values expressed consistently to either the nearest whole number or 1 decimal place.
(a) (iii) [4 marks]
- 1 mark: Higher sucrose concentration leads to a lower/more negative external water potential.
- 1 mark: Water moves out of the cell vacuole down a water potential gradient.
- 1 mark: Movement occurs by osmosis across a selectively permeable cell membrane.
- 1 mark: Protoplast shrinks away from the cell wall, resulting in a higher proportion of plasmolysed cells.
(a) (iv) [3 marks]
- 1 mark: Identifies one valid systematic or random error (e.g., subjective boundary of plasmolysis, cell variation).
- 1 mark: Identifies a second valid error.
- 1 mark: Proposes an appropriate, feasible improvement corresponding to at least one of the identified errors.
(a) (v) [4 marks]
- 1 mark: Uses a narrower interval/range of sucrose concentrations.
- 1 mark: Replicates and counts a larger number of cells (minimum 50-100 cells per concentration) to reduce random error.
- 1 mark: Identifies the precise point of incipient plasmolysis (where 50% of the cells are plasmolysed).
- 1 mark: Mentions controlling temperature of the solutions using a thermostatic water bath.
Investigation of Catalase Activity using the Floating Disc Method
Catalase is an enzyme found inside living tissues that catalyzes the breakdown of toxic hydrogen peroxide into water and oxygen gas. In this investigation, you will use a 'floating disc' assay to measure the activity of yeast catalase. Paper discs soaked in yeast suspension are placed at the bottom of a test-tube of hydrogen peroxide. Oxygen bubbles produced by the reaction get trapped in the paper fibers, making the disc buoyant and causing it to rise. The time taken (t) for the disc to rise to the surface is recorded.
You are provided with: 3.0% hydrogen peroxide solution, labeled H; 10% yeast suspension, labeled Y; and distilled water, labeled W.
(a) (i) Complete the table below to show how you would prepare 20.0 cm3 of each of the following hydrogen peroxide concentrations: 2.5%, 2.0%, 1.5%, 1.0%, and 0.5% by diluting the 3.0% stock solution H with distilled water W. Show all working. [4 marks]
(a) (ii) Carry out the procedure for each concentration. Dip a uniform paper disc into yeast suspension Y with forceps, blot excess liquid onto a paper towel, and place it at the bottom of a test-tube containing 20.0 cm3 of the hydrogen peroxide concentration. Record the time taken (t) in seconds for the disc to rise to the surface. Perform two trials for each concentration and construct a single table to record all raw times, calculated mean times, and rates of reaction calculated as 1/mean time (s-1). [6 marks]
(a) (iii) Other than hydrogen peroxide concentration, identify two variables that must be controlled in this experiment and describe how you would maintain them constant. [4 marks]
(a) (iv) Outline a control experiment that should be carried out to confirm that active catalase inside the yeast suspension is the agent responsible for the floating of the discs. [3 marks]
(a) (v) Explain how you could modify this experimental setup to investigate the effect of pH on catalase activity. [3 marks]
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Solution for Catalase Floating Disc Investigation
(a) (i) Dilution Calculations:
To find the volume of stock H (3.0%) needed to prepare 20 cm3 of a desired concentration: \( V_1 = \frac{C_2 \times 20.0}{3.0} \).
- To make 2.5%: \( (2.5 / 3.0) \times 20.0 = 16.7 \text{ cm}^3 \) of H + \( 3.3 \text{ cm}^3 \) of W
- To make 2.0%: \( (2.0 / 3.0) \times 20.0 = 13.3 \text{ cm}^3 \) of H + \( 6.7 \text{ cm}^3 \) of W
- To make 1.5%: \( (1.5 / 3.0) \times 20.0 = 10.0 \text{ cm}^3 \) of H + \( 10.0 \text{ cm}^3 \) of W
- To make 1.0%: \( (1.0 / 3.0) \times 20.0 = 6.7 \text{ cm}^3 \) of H + \( 13.3 \text{ cm}^3 \) of W
- To make 0.5%: \( (0.5 / 3.0) \times 20.0 = 3.3 \text{ cm}^3 \) of H + \( 16.7 \text{ cm}^3 \) of W
(a) (ii) Results Table:
The table must have complete grid lines and headers with units, e.g., 'Hydrogen peroxide concentration / %', 'Time, t / s', 'Mean time / s', and 'Rate / s-1'. Raw times should be whole numbers. Typical trends show that as hydrogen peroxide concentration increases, time decreases and rate increases:
- 2.5%: Trial 1 = 9s, Trial 2 = 11s, Mean = 10.0s, Rate = 0.100 s-1
- 1.5%: Trial 1 = 18s, Trial 2 = 22s, Mean = 20.0s, Rate = 0.050 s-1
- 0.5%: Trial 1 = 58s, Trial 2 = 62s, Mean = 60.0s, Rate = 0.017 s-1
(a) (iii) Controlled Variables:
- Yeast volume / concentration on the disc: Keep the disc size identical by using the same paper hole-punch, and use a standardized blotting duration (e.g., 3 seconds on each side on a paper towel) to ensure uniform yeast volume.
- Temperature: Place all solution tubes and yeast suspension in a water bath maintained at a constant temperature (e.g., 25°C) and check with a thermometer.
(a) (iv) Control Experiment:
Boil the yeast suspension for 10 minutes to denature the catalase enzyme, cool it to room temperature, and repeat the procedure. The discs should remain at the bottom of the test-tubes and not rise, confirming that active enzyme is required for oxygen production and disc flotation.
(a) (v) Modification for pH:
Keep the concentration of hydrogen peroxide constant (e.g., 1.5%). Prepare a range of buffer solutions (e.g., pH 4, 5, 6, 7, 8). Mix a fixed volume of each buffer with the hydrogen peroxide before placing the yeast-soaked discs. Keep other factors like temperature and yeast concentration constant.
Marking scheme
(a) (i) [4 marks]
- 1 mark: Correctly calculates volume of stock H for at least three dilution values.
- 1 mark: Correctly calculates volume of distilled water W for all dilutions.
- 1 mark: Total volume of each dilution sums to exactly 20.0 cm3.
- 1 mark: Records all calculated volumes consistently to 1 decimal place.
(a) (ii) [6 marks]
- 1 mark: Table constructed with clean ruled grid lines and clear headers.
- 1 mark: Appropriate units listed in column headers, e.g., 't / s', 'Rate / s-1'.
- 1 mark: Records raw data for 2 trials across 5 different concentrations.
- 1 mark: Raw stopwatch times recorded as whole numbers (no decimal places for manual timing).
- 1 mark: Correct calculation of mean times from the raw values.
- 1 mark: Rate values calculated as 1/mean time and expressed to 3 significant figures.
(a) (iii) [4 marks]
- 1 mark: Identifies size/material of the filter paper disc as a variable to control.
- 1 mark: Explains control method (e.g., use the same paper grade, same punch tool).
- 1 mark: Identifies temperature of reaction as a variable to control.
- 1 mark: Explains control method (e.g., use of a water bath monitored with a thermometer).
(a) (iv) [3 marks]
- 1 mark: Uses denatured / boiled and cooled yeast suspension (or distilled water instead of yeast).
- 1 mark: Keeps all other physical conditions and volumes identical.
- 1 mark: States the expected result: the paper disc does not rise to the surface.
(a) (v) [3 marks]
- 1 mark: Keeps the hydrogen peroxide concentration constant.
- 1 mark: Uses at least 5 different pH buffer solutions.
- 1 mark: Describes mixing the buffer solution with the reaction mixture before starting timing.
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