- A.Two separate light chains and one dimer of two heavy chains
- B.Two separate light chains and two separate heavy chains
- C.Two fragments, each consisting of one light chain bound to one heavy chain
- D.Four separate polypeptide chains and no intact disulfide bonds
Cambridge IAS-Level · Thinka-original Practice Paper
2023 Cambridge IAS-Level Biology (9700) Practice Paper with Answers
Thinka Nov 2023 (V3) Cambridge International A Level-Style Mock — Biology (9700)
Paper 1 (Multiple Choice)
* Control: \(K_m = 2.0 \text{ mmol dm}^{-3}\), \(V_{\max} = 100 \mu\text{mol dm}^{-3} \text{ s}^{-1}\)
* With substance X: \(K_m = 5.5 \text{ mmol dm}^{-3}\), \(V_{\max} = 100 \mu\text{mol dm}^{-3} \text{ s}^{-1}\)
* With substance Y: \(K_m = 2.0 \text{ mmol dm}^{-3}\), \(V_{\max} = 45 \mu\text{mol dm}^{-3} \text{ s}^{-1}\)
Which statement correctly describes the binding sites of substances X and Y on the enzyme?
- A.X binds to the active site; Y binds to a site other than the active site.
- B.X binds to a site other than the active site; Y binds to the active site.
- C.X and Y both bind to the active site of the enzyme.
- D.X and Y both bind to sites other than the active site of the enzyme.
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2. Substance Y decreases the \(V_{\max}\) from \(100\) to \(45 \mu\text{mol dm}^{-3} \text{ s}^{-1}\) but leaves the \(K_m\) unchanged at \(2.0 \text{ mmol dm}^{-3}\). This is characteristic of a non-competitive inhibitor, which binds to an allosteric site (a site other than the active site) and reduces catalytic activity without affecting substrate binding affinity.
Therefore, X binds to the active site, and Y binds to a site other than the active site.
Marking scheme
- A.Collagen: Every third amino acid is glycine, which allows the three polypeptide chains to pack closely together. Haemoglobin: Hydrophobic R-groups are oriented towards the center of the molecule, leaving hydrophilic R-groups on the outside.
- B.Collagen: Three alpha-helices are held together by disulfide bonds to form a strong, stretchable fibre. Haemoglobin: Four polypeptide chains are linked by covalent peptide bonds between their R-groups.
- C.Collagen: It is a globular protein that is highly soluble in water because of its outer hydrophilic R-groups. Haemoglobin: It is a fibrous protein with high tensile strength due to the presence of hydrophobic prosthetic groups.
- D.Collagen: Individual triple helices are joined by hydrogen bonds to form myofibrils. Haemoglobin: The four polypeptide chains are identical, and each contains an iron ion (\(Fe^{3+}\)) in its haem group.
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- A.3.75 \(\mu\text{m}\)
- B.15.0 \(\mu\text{m}\)
- C.60.0 \(\mu\text{m}\)
- D.240.0 \(\mu\text{m}\)
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2. These 4 divisions coincide with 20 eyepiece graticule divisions (epd). Therefore: \(1 \text{ epd} = 400 \mu\text{m} / 20 = 20 \mu\text{m}\) at \(\times 100\).
3. When the magnification is increased from \(\times 100\) to \(\times 400\) (an increase by a factor of 4), the physical size represented by each eyepiece division decreases by the same factor: \(1 \text{ epd at } \times 400 = 20 \mu\text{m} / 4 = 5 \mu\text{m}\).
4. The chloroplast measures 3 epd under \(\times 400\) magnification. Its actual length is: \(3 \times 5 \mu\text{m} = 15 \mu\text{m}\).
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- A.It acts as a competitive inhibitor of the enzyme transpeptidase, preventing the cross-linking of peptidoglycan chains in growing cell walls.
- B.It binds to the small ribosomal subunit (30S), preventing the binding of tRNA and halting bacterial protein synthesis.
- C.It disrupts the phospholipid bilayer of the outer membrane of Gram-negative bacteria, causing leakage of essential ions.
- D.It inhibits bacterial DNA gyrase, preventing the replication and transcription of the bacterial circular chromosome.
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- A.Water moves through the non-living parts of the cells, including cell walls and intercellular spaces, until it reaches the endodermis where it is forced into the symplast pathway by the Casparian strip.
- B.Water moves through the cytoplasm and plasmodesmata of adjacent cells, driven by a concentration gradient of mineral ions established by active transport.
- C.Water and solutes are actively pumped across the cell surface membrane into the vacuole and then pass from vacuole to vacuole via specialized transport proteins.
- D.Water moves through the xylem vessel elements against gravity, driven entirely by positive hydrostatic pressure generated by the Casparian strip.
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- A.Carbon dioxide reacts with water, catalysed by carbonic anhydrase, to form carbonic acid, which dissociates into hydrogen ions and hydrogencarbonate ions. The hydrogen ions bind to oxyhaemoglobin, releasing oxygen, and hydrogencarbonate ions diffuse out into the plasma in exchange for chloride ions.
- B.Carbon dioxide binds directly to the haem group of haemoglobin to form carboxyhaemoglobin, displacing oxygen. Hydrogencarbonate ions are actively pumped into the red blood cells, causing chloride ions to leave.
- C.Carbon dioxide is converted to carbonic acid by carbonic anhydrase in the blood plasma. Carbonic acid then enters the red blood cell to dissociate into hydrogen ions and hydrogencarbonate ions.
- D.Carbon dioxide reacts with water slowly without an enzyme to form carbonic acid. This dissociates, and the resulting hydrogen ions increase the affinity of haemoglobin for oxygen, holding it tighter.
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1. Carbon dioxide diffuses into the red blood cell, where the enzyme carbonic anhydrase rapidly catalyses its reaction with water to form carbonic acid (\(H_2CO_3\)).
2. Carbonic acid spontaneously dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)).
3. The hydrogen ions bind to oxyhaemoglobin, promoting the release of oxygen to the respiring tissues due to a conformational change that lowers haemoglobin's oxygen affinity (the Bohr effect).
4. Hydrogencarbonate ions diffuse out of the red blood cell down their concentration gradient into the plasma, while chloride ions enter the cell to maintain electrical neutrality (the chloride shift).
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- A.16 chromosomes, 32 sister chromatids, 64 telomeres
- B.16 chromosomes, 16 sister chromatids, 32 telomeres
- C.32 chromosomes, 32 sister chromatids, 64 telomeres
- D.32 chromosomes, 64 sister chromatids, 128 telomeres
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2. Sister chromatids: Since DNA replication occurred during S phase of interphase, each of the 16 chromosomes consists of 2 sister chromatids. Therefore, there are \(16 \times 2 = 32\) sister chromatids.
3. Telomeres: A telomere is a region of repetitive nucleotide sequences at each end of a chromatid. Each sister chromatid is a linear DNA molecule and thus has 2 telomeres (one at each end). Since there are 32 chromatids in the cell, the total number of telomeres is \(32 \times 2 = 64\) telomeres.
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- A.\(36\ \mu\text{m}\)
- B.\(180\ \mu\text{m}\)
- C.\(360\ \mu\text{m}\)
- D.\(1500\ \mu\text{m}\)
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- A.Disulfide bonds are formed between the carboxyl groups of two adjacent cysteine residues.
- B.Hydrogen bonds are formed only between the oxygen of a carbonyl group and the hydrogen of an amine group in the peptide backbone.
- C.Hydrophobic interactions occur between non-polar R-groups which tend to cluster together in the interior of the protein.
- D.Ionic bonds are weak electrostatic attractions between positively charged R-groups and hydrophobic R-groups.
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- A.The inhibitor binds reversibly to an allosteric site on the enzyme, altering the shape of the active site.
- B.The inhibitor has a similar molecular shape to the substrate and competes with the substrate for the active site.
- C.The inhibitor forms strong, covalent bonds with the active site, irreversibly inactivating the enzyme.
- D.Increasing the substrate concentration has no effect on the degree of inhibition by this inhibitor.
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- A.act as receptors: glycoproteins and glycolipids; help to stabilize membrane: cholesterol
- B.act as receptors: phospholipids and cholesterol; help to stabilize membrane: glycoproteins
- C.act as receptors: glycoproteins only; help to stabilize membrane: glycolipids and cholesterol
- D.act as receptors: glycolipids and cholesterol; help to stabilize membrane: phospholipids
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- A.\(3'\text{-ATCGGTACG-5'}\)
- B.\(3'\text{-TAGCCATGC-5'}\)
- C.\(5'\text{-TAGCCATGC-3'}\)
- D.\(5'\text{-UAGCCAUGC-3'}\)
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- A.sieve tube: cytoplasm present, non-lignified walls, companion cells present; xylem: cytoplasm absent, lignified walls, companion cells absent
- B.sieve tube: cytoplasm absent, lignified walls, companion cells present; xylem: cytoplasm present, non-lignified walls, companion cells absent
- C.sieve tube: cytoplasm present, non-lignified walls, companion cells absent; xylem: cytoplasm present, non-lignified walls, companion cells present
- D.sieve tube: cytoplasm absent, lignified walls, companion cells absent; xylem: cytoplasm absent, lignified walls, companion cells absent
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- A.W: carbonic anhydrase, X: chloride ions, Y: oxygen, Z: carbaminohaemoglobin
- B.W: carbonic anhydrase, X: potassium ions, Y: carbon dioxide, Z: carboxyhaemoglobin
- C.W: catalase, X: chloride ions, Y: oxygen, Z: carboxyhaemoglobin
- D.W: hydrogenase, X: hydrogen carbonate ions, Y: oxygen, Z: carbaminohaemoglobin
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- A.The secondary response has a shorter delay because memory B-lymphocytes immediately start producing large amounts of antibodies.
- B.The secondary response is faster and produces a higher concentration of antibodies because memory B-lymphocytes rapidly divide and differentiate into plasma cells.
- C.The primary response is slower because T-helper cells must mutate to produce the correct antibodies.
- D.The secondary response lasts longer because the antibodies produced are more stable and resist breakdown for a longer period of time.
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- A.1, 2 and 3 only
- B.1, 2 and 4 only
- C.2 and 4 only
- D.1, 3 and 4 only
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- A.\(V_{\max}\) decreases and \(K_m\) remains the same
- B.\(V_{\max}\) remains the same and \(K_m\) increases
- C.\(V_{\max}\) decreases and \(K_m\) increases
- D.\(V_{\max}\) remains the same and \(K_m\) remains the same
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- A.Physical length is \(6.0\ \mu\text{m}\); it will appear \(10\) graticule units long.
- B.Physical length is \(6.0\ \mu\text{m}\); it will appear \(1.6\) graticule units long.
- C.Physical length is \(1.5\ \mu\text{m}\); it will appear \(4\) graticule units long.
- D.Physical length is \(2.4\ \mu\text{m}\); it will appear \(10\) graticule units long.
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- A.Prophase: chromatin condenses, spindle fibres begin to form, and centrioles migrate to opposite poles of the cell.
- B.Metaphase: chromosomes align along the equator of the cell and attach to spindle fibres via their centromeres.
- C.Anaphase: sister chromatids are pulled apart by the shortening of spindle microtubules towards the centrioles.
- D.Telophase: chromosomes decondense, nuclear envelope reforms, and a cleavage furrow pinches the cell membrane.
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- A.Type of immunity: Artificial passive | Reason: No memory cells were produced from the first injection because no antigen was introduced.
- B.Type of immunity: Natural passive | Reason: The injected antibodies were rapidly broken down by the child's immune system.
- C.Type of immunity: Artificial active | Reason: The memory cells produced had a short lifespan of less than three months.
- D.Type of immunity: Artificial passive | Reason: The second bite contained a different antigen that could not be recognized.
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- A.Fluidity: cholesterol | Receptor: glycoprotein
- B.Fluidity: glycolipid | Receptor: protein channel
- C.Fluidity: cholesterol | Receptor: phospholipid
- D.Fluidity: glycoprotein | Receptor: glycolipid
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- A.\(5'\text{-AUGCCGUAG}-3'\)
- B.\(5'\text{-UACGGCAUC}-3'\)
- C.\(3'\text{-AUGCCGUAG}-5'\)
- D.\(5'\text{-ATGCCGTAG}-3'\)
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- A.Largest percentage: hydrogencarbonate ions (\(\text{HCO}_3^-\)) | Enzyme: carbonic anhydrase
- B.Largest percentage: carbaminohaemoglobin | Enzyme: carbonic anhydrase
- C.Largest percentage: dissolved carbon dioxide in plasma | Enzyme: haemoglobinase
- D.Largest percentage: hydrogencarbonate ions (\(\text{HCO}_3^-\)) | Enzyme: carbonic decarboxylase
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- A.cholesterol and glycoproteins
- B.glycolipids and phospholipids
- C.protein channels and cholesterol
- D.glycoproteins and intrinsic proteins mohs scale value close to standard definition of channel activity but wrong context in recognition context.
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- A.\(\times 8\)
- B.\(\times 80\)
- C.\(\times 800\)
- D.\(\times 8000\)
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- A.It binds to an allosteric site and changes the shape of the active site.
- B.It binds reversibly to the active site of the enzyme.
- C.It forms permanent covalent bonds with the active site.
- D.It decreases the \(V_{max}\) of the enzyme-controlled reaction.
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- A.5'- AUG CCG AAU GAU -3'
- B.3'- AUG CCG AAU GAU -5'
- C.3'- UAC GGC UUA CUA -5'
- D.5'- UAC GGC UUA CUA -3'
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- A.The curve shifts to the left, decreasing the affinity of haemoglobin for oxygen, making oxygen more readily available to tissues.
- B.The curve shifts to the right, decreasing the affinity of haemoglobin for oxygen, making oxygen more readily available to tissues.
- C.The curve shifts to the right, increasing the affinity of haemoglobin for oxygen, reducing the release of oxygen to tissues.
- D.The curve shifts to the left, increasing the affinity of haemoglobin for oxygen, reducing the release of oxygen to tissues.
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- A.1, 2, 3 and 4
- B.1, 2 and 4 only
- C.1 and 3 only
- D.2 and 4 only
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- A.Artificial active immunity, because no memory cells are produced.
- B.Artificial passive immunity, because the injected antibodies are broken down and no memory cells are produced.
- C.Natural passive immunity, because the antibodies are biological molecules.
- D.Artificial passive immunity, because the immune system produces its own antibodies but they decay rapidly.
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- A.Anaphase followed by Telophase
- B.Metaphase followed by Anaphase
- C.Anaphase followed by Metaphase
- D.Telophase followed by Cytokinesis
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- A.\(7\ \mu\text{m}\)
- B.\(35\ \mu\text{m}\)
- C.\(70\ \mu\text{m}\)
- D.\(700\ \mu\text{m}\)
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- A.\(V_{max}\) is unchanged; \(K_m\) is increased.
- B.\(V_{max}\) is decreased; \(K_m\) is increased.
- C.\(V_{max}\) is unchanged; \(K_m\) is decreased.
- D.\(V_{max}\) is decreased; \(K_m\) is unchanged.
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- A.1, 2 and 3
- B.1 and 2 only
- C.1 and 3 only
- D.2 and 3 only
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- A.The antibodies from the first injection were recognized as foreign and destroyed by the child's immune system.
- B.The first injection triggered a primary immune response, but the memory cells produced had died.
- C.The injection of antivenom provided artificial passive immunity, which does not result in the production of memory cells.
- D.The snake venom acts as an antigen but is too small to stimulate B-lymphocytes to undergo clonal expansion.
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- A.Plasma pH: Decreases | Affinity of haemoglobin: Decreases | Shift of curve: To the right
- B.Plasma pH: Decreases | Affinity of haemoglobin: Increases | Shift of curve: To the left
- C.Plasma pH: Increases | Affinity of haemoglobin: Decreases | Shift of curve: To the right
- D.Plasma pH: Increases | Affinity of haemoglobin: Increases | Shift of curve: To the left
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- A.cell walls of cortex \(\rightarrow\) cytoplasm of endodermis \(\rightarrow\) cell walls of pericycle \(\rightarrow\) xylem lumen
- B.cell walls of cortex \(\rightarrow\) cell walls of endodermis \(\rightarrow\) cell walls of pericycle \(\rightarrow\) xylem lumen
- C.plasmodesmata of cortex \(\rightarrow\) cytoplasm of endodermis \(\rightarrow\) plasmodesmata of pericycle \(\rightarrow\) xylem lumen
- D.cytoplasm of cortex \(\rightarrow\) Casparian strip \(\rightarrow\) cytoplasm of endodermis \(\rightarrow\) xylem lumen
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- A.Row A
- B.Row B
- C.Row C
- D.Row D
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- A.UAC, GGC, UUU, ACG
- B.TAC, GGC, TTT, ACG
- C.AUG, CCG, AAA, UGC
- D.ATG, CCG, AAA, TGC
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Paper 2 (AS Structured)
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Paper 3 (Advanced Practical Skills)
You are provided with:
- \(10\text{ cm}^3\) of \(1.0\%\) solution of the inhibitor, I
- \(50\text{ cm}^3\) of distilled water, W
- \(20\text{ cm}^3\) of \(1.0\%\) amylase solution, A
- \(20\text{ cm}^3\) of \(1.0\%\) starch solution, S
- Iodine in potassium iodide solution (in a dropper bottle)
(a) You are required to prepare a range of concentrations of the inhibitor solution, I, using simple dilution of the stock \(1.0\%\) solution I with distilled water, W.
Prepare five concentrations of I: \(1.0\%\), \(0.8\%\), \(0.6\%\), \(0.4\%\), and \(0.2\%\). The total volume of each solution should be \(10.0\text{ cm}^3\).
Complete Table 1.1 to show the volumes of \(1.0\%\) solution I and distilled water, W, required to prepare each concentration. [4]
Table 1.1
| Concentration of inhibitor I / % | Volume of 1.0% inhibitor solution I / \(\text{cm}^3\) | Volume of distilled water, W / \(\text{cm}^3\) |
|---|---|---|
| 1.0 | | |
| 0.8 | | |
| 0.6 | | |
| 0.4 | | |
| 0.2 | | |
(b) Carry out the investigation using the following instructions:
1. Label five test-tubes with the concentrations of I prepared in (a).
2. Pipette \(2.0\text{ cm}^3\) of each concentration of I into its respective test-tube.
3. Add \(2.0\text{ cm}^3\) of amylase solution A to each test-tube and mix gently.
4. Leave the mixtures for 2 minutes to allow interaction.
5. Put one drop of iodine solution into each well of a spotting tile.
6. Add \(2.0\text{ cm}^3\) of starch solution S to the first test-tube (containing \(1.0\%\) I and A), mix quickly, and start a timer.
7. Every 30 seconds, use a clean dropper pipette to remove a drop of the mixture and add it to a well containing iodine solution on the spotting tile.
8. Record the time when the starch is completely hydrolysed (when the blue-black colour no longer appears and the iodine remains yellow-brown). If the starch is not hydrolysed after 300 seconds, stop timing and record 'more than 300'.
9. Repeat steps 6 to 8 for the remaining four concentrations of I.
Record your results in an appropriate table. [6]
(c) In another investigation, a student measured the rate of amylase activity at different pH values, with no inhibitor present. The results are shown in Table 1.2.
Table 1.2
| pH | Rate of reaction / arbitrary units (a.u.) |
|---|---|
| 4.0 | 1.2 |
| 5.0 | 3.5 |
| 6.0 | 6.8 |
| 7.0 | 8.5 |
| 8.0 | 4.2 |
| 9.0 | 0.8 |
(i) Plot a line graph of the data in Table 1.2 on a grid. [4]
(ii) State and explain the effect of pH on the activity of amylase shown in the graph. [2]
(d) (i) Identify two sources of error in the method you used in (b). [2]
(ii) Suggest how you would modify the method in (b) to investigate the effect of temperature on the rate of starch hydrolysis. [4]
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To prepare \(10.0\text{ cm}^3\) of each concentration using \(1.0\%\) inhibitor solution:
- \(1.0\%\): \(10.0\text{ cm}^3\) of I, \(0.0\text{ cm}^3\) of W
- \(0.8\%\): \(8.0\text{ cm}^3\) of I, \(2.0\text{ cm}^3\) of W
- \(0.6\%\): \(6.0\text{ cm}^3\) of I, \(4.0\text{ cm}^3\) of W
- \(0.4\%\): \(4.0\text{ cm}^3\) of I, \(6.0\text{ cm}^3\) of W
- \(0.2\%\): \(2.0\text{ cm}^3\) of I, \(8.0\text{ cm}^3\) of W
(b) Table of Results (Specimen Data)
- Heading of 1st column: 'concentration of inhibitor I / %'
- Heading of 2nd column: 'time taken for starch hydrolysis / s'
- All times must be recorded in whole seconds (multiples of 30 seconds).
- Higher concentration of inhibitor results in a longer time taken for hydrolysis.
- Example data:
- 1.0% -> 240 s (or 'more than 300')
- 0.8% -> 180 s
- 0.6% -> 120 s
- 0.4% -> 90 s
- 0.2% -> 60 s
(c) (i) Graph Plotting
- Axes: x-axis labeled 'pH', y-axis labeled 'Rate of reaction / arbitrary units' or 'Rate of reaction / a.u.'.
- Scale: x-axis 1.0 unit per 2 cm; y-axis 1.0 or 2.0 units per 2 cm. Must be linear and use more than half the grid.
- Plotting: All 6 points plotted precisely to within half a small square with a small, sharp cross or circled dot.
- Line: Points connected with straight, ruled lines from point to point, or a smooth, clean curve passing exactly through all points with no double lines.
(c) (ii) Explain the effect of pH
- As pH increases from 4.0 to 7.0, the rate of reaction increases to the optimum at pH 7.0 (8.5 a.u.), and then decreases at higher pH values.
- Explanation: Extreme pH values (both low and high) alter the ionic charge on the amino acid R-groups in the active site of amylase. This disrupts hydrogen and ionic bonds, changing the tertiary structure and denaturing the enzyme. Consequently, the active site is no longer complementary to starch, and fewer enzyme-substrate complexes form.
(d) (i) Sources of Error
1. End-point color change is subjective / difficult to judge exactly when the blue-black color completely disappears.
2. The sampling interval of 30 seconds is too wide, meaning the exact end-point could occur between sampling times.
3. Lack of temperature control / temperature fluctuated during the experiment.
(d) (ii) Modifications for Temperature Investigation
1. Maintain the concentration of the inhibitor I constant (e.g., at 0.5%) or omit it entirely.
2. Use a range of at least five different temperatures (e.g., \(20^\circ\text{C}\), \(30^\circ\text{C}\), \(40^\circ\text{C}\), \(50^\circ\text{C}\), \(60^\circ\text{C}\)) using thermostatically-controlled water baths.
3. Pre-incubate the enzyme solution A and substrate starch solution S separately in the water bath at the target temperature for 5 minutes before mixing.
4. Ensure all other variables remain controlled (e.g., same volumes and concentrations of amylase and starch, same pH using a buffer).
Marking scheme
- 1 mark for all five concentrations listed in order with correct unit in the header.
- 1 mark for correct volumes of stock 1.0% inhibitor I (10.0, 8.0, 6.0, 4.0, 2.0).
- 1 mark for correct volumes of distilled water W (0.0, 2.0, 4.0, 6.0, 8.0).
- 1 mark for all volumes written consistently to 1 decimal place (e.g., 10.0, 8.0, 0.0) or whole numbers.
(b) Results Table [6 marks]
- 1 mark for table with fully drawn borders containing raw data.
- 1 mark for clear column headings with appropriate units: 'concentration of inhibitor I / %' and 'time / s' (do not accept 'time / sec' or 'time / mins').
- 1 mark for independent variable (concentration) in the first column.
- 1 mark for all raw times recorded as whole numbers (multiples of 30 seconds).
- 1 mark for showing the correct trend: time increases as the concentration of inhibitor increases.
- 1 mark for carrying out the experiment at all 5 concentrations.
(c) (i) Graph [4 marks]
- 1 mark (O): Orientation of axes (pH on x-axis, Rate on y-axis) and complete labels with units: 'Rate of reaction / arbitrary units' or 'Rate of reaction / a.u.'.
- 1 mark (S): Scale on both axes must be linear, easy to read, and allow plotted points to occupy more than 50% of the grid area in both directions.
- 1 mark (P): All points plotted accurately within half a small square using small, thin crosses (x or +).
- 1 mark (L): Sharp, thin line connecting all points cleanly, either by straight ruled lines or a smooth curve.
(c) (ii) Explanation of pH [2 marks]
- 1 mark for describing the trend: optimum activity is at pH 7.0; rate decreases as pH deviates below or above pH 7.0.
- 1 mark for explaining that changes in pH alter the charges on amino acid R-groups, breaking hydrogen/ionic bonds, changing active site shape (denaturation) so substrate no longer binds.
(d) (i) Sources of Error [2 marks]
- 1 mark each for any two of:
- End-point color change is subjective / difficult to determine when yellow-brown is reached.
- 30-second sampling interval is too wide.
- Temperature was not controlled / temperature fluctuated.
- Only one trial was completed / no replicates to identify anomalies.
(d) (ii) Modifications [4 marks]
- 1 mark for using at least 5 different temperatures (e.g., \(20, 30, 40, 50, 60^\circ\text{C}\)).
- 1 mark for using thermostatically controlled water baths to maintain constant temperatures.
- 1 mark for pre-incubating solutions A and S separately in the water baths before mixing.
- 1 mark for controlling other key variables (e.g., pH using buffer, keeping volume/concentration of enzyme and substrate constant).
Slide J1 is a prepared slide of a transverse section through a herbaceous plant stem.
(a) Draw a large, low-power plan diagram of a sector of the stem shown on slide J1 to show the distribution of tissues. Your drawing should show one complete vascular bundle and the tissues on either side of it, from the outer epidermis to the central pith. Label the xylem and the sclerenchyma. [6]
(b) Use the high-power objective lens of your microscope to locate the cortex of the stem on slide J1. Select three adjacent parenchyma cells. Make a large drawing of these three cells to show their cell walls and arrangement. [5]
(c) (i) A student calibrated an eyepiece graticule using a stage micrometer. Using a high-power objective lens, 1 eyepiece graticule unit (epu) was found to be equivalent to 2.5 micrometres (\(\mu\)m). On slide J1, the width of a single vascular bundle was measured as 68 epu. Calculate the actual width of this vascular bundle in micrometres (\(\mu\)m). Show your working. [3]
(ii) Prepare a table to compare three observable differences between the xylem vessels and the phloem sieve tube elements visible on slide J1. [4]
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Worked solution
(a) Plan Diagram guidelines:
- Clear, single-line sector drawing representing the epidermis, cortex, vascular bundle (with xylem, phloem, and sclerenchyma cap), and pith boundary.
- No cell details drawn.
- Label lines drawn with a ruler, ending precisely on the target tissues (xylem vessels and sclerenchyma cap).
(b) High-power drawing guidelines:
- Drawing shows exactly three adjacent parenchyma cells.
- Cell walls shown as double lines to represent thickness.
- Small intercellular air spaces shown at the junctions between cells.
(c) (i) Calculation:
\(\text{Actual width} = 68 \text{ epu} \times 2.5 \mu\text{m/epu}\)
\(\text{Actual width} = 170 \mu\text{m}\).
(c) (ii) Comparison Table:
A table comparing xylem vessels and phloem sieve tube elements should look like this:
Marking scheme
Part (a) [6 marks]:
1. Sharp, continuous, thin lines used; no shading or sketching.
2. Sector drawn is large (occupies at least 50% of the available area).
3. Correct proportion: cortex width is less than half the radial width of the vascular bundle.
4. Vascular bundle drawn as a wedge/oval shape with distinct layers representing xylem, phloem, and sclerenchyma.
5. Label 'xylem' points precisely to the inner region of the vascular bundle with large open vessels.
6. Label 'sclerenchyma' points precisely to the cap of cells on the outer edge of the phloem.
Part (b) [5 marks]:
1. Only three adjacent cells drawn, each at least 40 mm in size.
2. Cell walls shown as double lines with middle lamella visible or represented.
3. Cells are rounded or polygonal, showing parenchyma characteristics.
4. Intercellular spaces drawn correctly at the cell junctions (corners).
5. No cell contents (like nuclei or cytoplasm) drawn or shaded.
Part (c) (i) [3 marks]:
1. Shows correct working: \(68 \times 2.5\). [1]
2. Correct numerical value: \(170\). [1]
3. Correct unit given as \(\mu\)m (or micrometres) AND expressed to appropriate significant figures (2 or 3 s.f.). [1]
Part (c) (ii) [4 marks]:
1. Complete table format with fully ruled lines, columns, and clear headings: 'Feature', 'Xylem vessel', and 'Phloem sieve tube element'. [1]
2. Thickness difference: Xylem walls are thick/lignified AND phloem walls are thin. [1]
3. Lumen diameter difference: Xylem vessels have a larger lumen/diameter AND phloem sieve tubes have a smaller lumen/diameter. [1]
4. End wall structures: Xylem has no end walls/perforated ends AND phloem sieve tubes have sieve plates/end walls. [1]
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