An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V2) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.
Paper 12 Multiple Choice
There are forty questions on this paper. Answer all questions. Choose the one you consider correct.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
The Michaelis-Menten constant (\(K_m\)) and maximum velocity (\(V_{max}\)) of an enzyme-catalysed reaction were determined in the absence and presence of two different inhibitors, X and Y, at the same concentration. With inhibitor X, the \(V_{max}\) was decreased but the \(K_m\) remained the same. With inhibitor Y, the \(V_{max}\) remained the same but the \(K_m\) increased. Which statement correctly describes the modes of action of inhibitors X and Y?
A.Inhibitor X binds to the active site; inhibitor Y binds to an allosteric site.
B.Inhibitor X binds to an allosteric site; inhibitor Y binds to the active site.
C.Both inhibitors bind to the active site, but inhibitor X binds irreversibly.
D.Both inhibitors bind to an allosteric site, but inhibitor Y binds reversibly.
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Worked solution
Inhibitor X decreases \(V_{max}\) but leaves \(K_m\) unchanged, which is characteristic of non-competitive inhibition. Non-competitive inhibitors bind to an allosteric site (a site other than the active site) on the enzyme, which alters the enzyme shape and decreases catalytic activity. Inhibitor Y leaves \(V_{max}\) unchanged but increases \(K_m\) (reducing substrate affinity), which is characteristic of competitive inhibition. Competitive inhibitors bind directly to the active site of the enzyme, competing with the substrate.
Marking scheme
Correct option is B. 1 mark for identifying that inhibitor X is non-competitive (allosteric site) and inhibitor Y is competitive (active site).
Question 2 · multiple-choice
1 marks
In a biochemical reaction, the energy level of the substrate is \(+20\text{ kJ mol}^{-1}\) and the energy level of the products is \(-10\text{ kJ mol}^{-1}\). The activation energy of the uncatalysed reaction is \(+60\text{ kJ mol}^{-1}\). When an enzyme is added, the activation energy is reduced by \(35\text{ kJ mol}^{-1}\). What is the activation energy of the enzyme-catalysed reaction and the overall energy change (\(\Delta H\)) for the reaction?
A.Activation energy of catalysed reaction = \(25\text{ kJ mol}^{-1}\); \(\Delta H = -30\text{ kJ mol}^{-1}\)
B.Activation energy of catalysed reaction = \(25\text{ kJ mol}^{-1}\); \(\Delta H = -15\text{ kJ mol}^{-1}\)
C.Activation energy of catalysed reaction = \(35\text{ kJ mol}^{-1}\); \(\Delta H = -30\text{ kJ mol}^{-1}\)
D.Activation energy of catalysed reaction = \(35\text{ kJ mol}^{-1}\); \(\Delta H = -15\text{ kJ mol}^{-1}\)
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Worked solution
The activation energy of the catalysed reaction is calculated by subtracting the reduction from the uncatalysed activation energy: \(60\text{ kJ mol}^{-1} - 35\text{ kJ mol}^{-1} = 25\text{ kJ mol}^{-1}\). The overall energy change (\(\Delta H\)) of the reaction is the difference between the energy level of the products and that of the substrate: \(-10\text{ kJ mol}^{-1} - (+20\text{ kJ mol}^{-1}) = -30\text{ kJ mol}^{-1}\).
Marking scheme
Correct option is A. 1 mark for correct calculation of both catalysed activation energy (25 kJ mol^-1) and enthalpy change (-30 kJ mol^-1).
Question 3 · multiple-choice
1 marks
An ecologist sampled three different areas of a woodland to determine the species diversity of ground-nesting beetles. The data collected was used to calculate Simpson's Index of Diversity (\(D\)) for each area: Area 1: \(D = 0.85\); Area 2: \(D = 0.45\); Area 3: \(D = 0.12\). Which statement is a correct interpretation of these results?
A.Area 3 has the most stable ecosystem because a change in one species population is unlikely to affect other species.
B.A high value of \(D\) in Area 1 indicates that the community is dominated by a single highly successful species.
C.Area 2 has a higher species richness than Area 1 because its diversity index is closer to 0.5.
D.A small environmental change is most likely to lead to a significant change in the community structure of Area 3.
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Worked solution
A lower Simpson's Index of Diversity (\(D = 0.12\) in Area 3) indicates lower species diversity and evenness, meaning the community is dominated by very few species and is highly unstable and susceptible to environmental changes. A high value of \(D\) in Area 1 represents high diversity and evenness, meaning the community is stable and not dominated by a single species. Therefore, a small change in the environment is most likely to affect Area 3.
Marking scheme
Correct option is D. 1 mark for identifying the correct ecological interpretation of a low Simpson's Index of Diversity.
Question 4 · multiple-choice
1 marks
A student wants to investigate the change in distribution and abundance of plant species across a transition zone from a salt marsh to a sand dune. Which sampling method and tool should the student use to obtain the most valid data?
A.Random sampling using a frame quadrat placed at coordinates generated by a random number generator.
B.Systematic sampling using a frame quadrat placed at regular intervals along a line transect.
C.Random sampling using a line transect placed across the entire transition zone.
D.Systematic sampling using a mark-release-recapture method along a belt transect.
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Worked solution
When investigating a change in species distribution and abundance along an environmental gradient (such as from a salt marsh to a sand dune), systematic sampling is appropriate. This is carried out using a line or belt transect, with frame quadrats placed at regular intervals along the transect line.
Marking scheme
Correct option is B. 1 mark for selecting systematic sampling using a quadrat along a line transect for an environmental gradient.
Question 5 · multiple-choice
1 marks
A geneticist crosses two heterozygous fruit flies (\(AaBb \times AaBb\)) and obtains 160 offspring. The observed phenotypic ratio is compared to the expected Mendelian ratio of 9:3:3:1 using a chi-squared (\(\chi^2\)) test. The calculated \(\chi^2\) value is 5.82. The critical values of \(\chi^2\) at different degrees of freedom (df) and probability levels (p) are: df = 3 (p = 0.05 is 7.82; p = 0.01 is 11.34); df = 4 (p = 0.05 is 9.49; p = 0.01 is 13.28). What is the correct conclusion of this statistical test?
A.The difference between observed and expected results is significant because \(5.82 < 7.82\); the genes are linked.
B.The difference between observed and expected results is not significant because \(5.82 < 7.82\); the genes assort independently.
C.The difference between observed and expected results is significant because \(5.82 < 9.49\); the genes are on the same chromosome.
D.The difference between observed and expected results is not significant because \(5.82 > 3.84\); the genes are on different chromosomes.
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Worked solution
There are four phenotypic categories (9:3:3:1), meaning the degrees of freedom (df) is \(4 - 1 = 3\). The critical value at the standard significance level of \(p = 0.05\) for df = 3 is 7.82. Since the calculated \(\chi^2\) value of 5.82 is less than the critical value of 7.82, the difference between observed and expected results is not statistically significant (p > 0.05). Therefore, we fail to reject the null hypothesis, concluding that the genes assort independently.
Marking scheme
Correct option is B. 1 mark for correctly determining df = 3, comparing 5.82 to 7.82, and concluding that the differences are not significant, indicating independent assortment.
Question 6 · multiple-choice
1 marks
The lac operon in Escherichia coli regulates the expression of genes required for lactose metabolism. Under which combination of conditions will the transcription of the structural genes (lacZ and lacY) be at its highest level?
A.Lactose present, Glucose present
B.Lactose present, Glucose absent
C.Lactose absent, Glucose present
D.Lactose absent, Glucose absent
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Worked solution
For high-level transcription of the lac operon structural genes: 1. Lactose must be present so that allolactose binds to the repressor protein, altering its shape and preventing it from binding to the operator. This permits RNA polymerase to transcribe. 2. Glucose must be absent because low glucose levels lead to high cAMP concentrations. cAMP binds to the catabolite activator protein (CAP), which then binds to the promoter, facilitating a high rate of RNA polymerase binding.
Marking scheme
Correct option is B. 1 mark for identifying that high-level transcription requires lactose to be present (repressor off) and glucose to be absent (cAMP/CAP active).
Question 7 · multiple-choice
1 marks
The polymerase chain reaction (PCR) is used to amplify DNA. Which row correctly identifies the function of each component in a PCR mixture?
A.Taq polymerase: synthesises new DNA strands at high temperatures; Primers: provide a starting point for DNA polymerase to bind; dNTPs: act as building blocks and provide energy for DNA synthesis
B.Taq polymerase: unwinds the double-stranded DNA template; Primers: prevent the single strands from reannealing; dNTPs: act as catalysts to speed up the elongation step
C.Taq polymerase: synthesises new DNA strands at high temperatures; Primers: hydrolyse phosphodiester bonds to separate templates; dNTPs: selectively bind to the target sequence to initiate replication
D.Taq polymerase: denatures the DNA template at 95 degrees Celsius; Primers: provide a starting point for DNA polymerase to bind; dNTPs: act as coenzymes for the Taq polymerase
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Worked solution
Taq polymerase is a heat-stable DNA polymerase that synthesises new DNA strands during the elongation stage. Primers are short single-stranded DNA sequences that bind to target sequences and provide a starting 3'-OH group for Taq polymerase. Deoxynucleoside triphosphates (dNTPs) act as both the nucleotide building blocks for the new strand and the source of energy for phosphodiester bond synthesis when their outer phosphate groups are cleaved.
Marking scheme
Correct option is A. 1 mark for correctly matching the functions of Taq polymerase, primers, and dNTPs in PCR.
Question 8 · multiple-choice
1 marks
When the water potential of the blood decreases, antidiuretic hormone (ADH) is released. Which row correctly describes the molecular effects of ADH binding to receptors on the collecting duct cells?
A.Intracellular secondary messenger: cAMP; Position of aquaporins: Inserted into the luminal membrane; Water potential of urine: Decreased
B.Intracellular secondary messenger: cAMP; Position of aquaporins: Removed from the luminal membrane; Water potential of urine: Increased
C.Intracellular secondary messenger: Calcium ions; Position of aquaporins: Inserted into the basolateral membrane; Water potential of urine: Decreased
D.Intracellular secondary messenger: Calcium ions; Position of aquaporins: Removed from the basolateral membrane; Water potential of urine: Increased
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Worked solution
ADH binds to specific G-protein coupled receptors on the basolateral membrane of collecting duct cells. This activates adenylyl cyclase, converting ATP to cyclic AMP (cAMP), which acts as the intracellular secondary messenger. A phosphorylation cascade causes vesicles containing aquaporins to move to and fuse with the luminal (apical) membrane, increasing water reabsorption and producing a lower volume of highly concentrated urine (decreased water potential).
Marking scheme
Correct option is A. 1 mark for identifying cAMP as the second messenger, apical insertion of aquaporins, and decreased urine water potential.
Question 9 · multiple_choice
1 marks
An investigation was carried out into the effect of two different inhibitors, X and Y, on an enzyme-catalysed reaction. The maximum velocity (\(V_{\max}\)) and Michaelis-Menten constant (\(K_m\)) were determined for the reaction without inhibitors, and in the presence of inhibitor X and inhibitor Y.
Which statement correctly identifies the types of inhibition shown by inhibitor X and inhibitor Y?
A.X is a competitive inhibitor and Y is a non-competitive inhibitor.
B.X is a non-competitive inhibitor and Y is a competitive inhibitor.
C.Both X and Y are competitive inhibitors.
D.Both X and Y are non-competitive inhibitors.
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Worked solution
An increase in the Michaelis-Menten constant (\(K_m\)) with an unchanged maximum velocity (\(V_{\max}\)) is characteristic of a competitive inhibitor (Inhibitor X), which binds to the active site and can be overcome by high substrate concentrations. A decrease in \(V_{\max}\) with an unchanged \(K_m\) is characteristic of a non-competitive inhibitor (Inhibitor Y), which binds to an allosteric site and cannot be overcome by high substrate concentrations. Thus, option A is correct.
Marking scheme
1 mark for correctly identifying X as a competitive inhibitor and Y as a non-competitive inhibitor based on their respective effects on \(V_{\max}\) and \(K_m\).
Question 10 · multiple_choice
1 marks
The energy profile of a biochemical reaction is analysed. The energy levels of the key stages of the reaction are as follows:
- Initial energy level of reactants = \(150\text{ kJ mol}^{-1}\) - Energy of the transition state without enzyme = \(350\text{ kJ mol}^{-1}\) - Energy of the transition state with enzyme = \(220\text{ kJ mol}^{-1}\) - Final energy level of products = \(80\text{ kJ mol}^{-1}\)
Which row correctly shows the activation energy for the enzyme-catalysed reaction and the overall energy change (\(\Delta H\)) for this reaction?
| | Activation energy of catalysed reaction / \(\text{kJ mol}^{-1}\) | Overall energy change (\(\Delta H\)) / \(\text{kJ mol}^{-1}\) | |---|---|---| | A | +70 | -70 | | B | +70 | +70 | | C | +220 | -70 | | D | +120 | -150 |
A.Row A
B.Row B
C.Row C
D.Row D
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Worked solution
The activation energy (\(E_a\)) of the catalysed reaction is the energy difference between the reactants and the transition state in the presence of the enzyme: \(220\text{ kJ mol}^{-1} - 150\text{ kJ mol}^{-1} = +70\text{ kJ mol}^{-1}\). The overall energy change (\(\Delta H\)) is the difference between the final energy of the products and the initial energy of the reactants: \(80\text{ kJ mol}^{-1} - 150\text{ kJ mol}^{-1} = -70\text{ kJ mol}^{-1}\). This corresponds to row A.
Marking scheme
1 mark for correctly determining the activation energy (+70 kJ/mol) and the net energy change (-70 kJ/mol) for the catalysed reaction.
Question 11 · multiple_choice
1 marks
Which of the factors could explain a decrease in the rate of an enzyme-catalysed reaction over time when the reaction is carried out at the enzyme's optimum temperature of \(30^\circ\text{C}\)?
1. The activation energy of the reaction increases as product accumulates. 2. The frequency of successful collisions between enzyme and substrate molecules decreases. 3. Product molecules act as competitive inhibitors by binding to the active sites.
A.1, 2 and 3
B.1 and 2 only
C.2 and 3 only
D.3 only
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Worked solution
Statement 1 is incorrect because the activation energy barrier of the catalysed pathway is constant and does not change as products accumulate. Statement 2 is correct because substrate is consumed over time, reducing its concentration and the frequency of successful collisions with active sites. Statement 3 is correct because product molecules can bind to the active site, causing product/competitive inhibition. Therefore, statements 2 and 3 are correct.
Marking scheme
1 mark for selecting factors 2 and 3 as correct explanations for the decrease in reaction rate over time.
Question 12 · multiple_choice
1 marks
In a woodland habitat, there are five species of beetle. An environmental change causes the population of the most abundant beetle species to decrease significantly, while the populations of the other four species increase, such that all five species now have approximately equal numbers of individuals. The total number of individual beetles in the habitat remains unchanged.
Which row correctly shows the effect of this change on the species richness, species evenness, and Simpson's Index of Diversity (\(D\)) of this habitat?
| | Species richness | Species evenness | Simpson’s Index of Diversity (\(D\)) | |---|---|---|---| | A | remains the same | increases | increases | | B | remains the same | decreases | decreases | | C | decreases | increases | increases | | D | increases | increases | remains the same |
A.Row A
B.Row B
C.Row C
D.Row D
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Worked solution
Species richness is the total number of different species in a community; since all five species are still present, it remains the same. Species evenness measures how close in numbers each species in an environment is; since the populations are now more equal, species evenness increases. An increase in evenness increases the overall biodiversity, causing Simpson's Index of Diversity (\(D\)) to increase. This corresponds to row A.
Marking scheme
1 mark for identifying that species richness remains the same, species evenness increases, and Simpson's Index of Diversity increases.
Question 13 · multiple_choice
1 marks
Which features of the domain Archaea distinguish its members from those of the domain Bacteria?
1. They have membrane lipids containing branched hydrocarbon chains linked by ether bonds. 2. They have 70S ribosomes. 3. Their cell walls lack peptidoglycan. 4. Their DNA is associated with histone proteins in some species.
A.1, 3 and 4 only
B.2 and 3 only
C.1, 2 and 4
D.1 and 4 only
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Worked solution
Statement 1 is correct: Archaea possess ether-linked branched membrane lipids, whereas Bacteria have ester-linked unbranched lipids. Statement 2 is incorrect: both Archaea and Bacteria have 70S ribosomes, so this does not distinguish them. Statement 3 is correct: Archaea cell walls lack peptidoglycan, which is a major distinguishing feature from Bacteria. Statement 4 is correct: Archaea DNA can be associated with histones in some species, unlike Bacterial DNA, which is not associated with histones. Therefore, features 1, 3, and 4 distinguish Archaea from Bacteria.
Marking scheme
1 mark for identifying features 1, 3, and 4 as correct distinguishing criteria of Archaea from Bacteria.
Question 14 · multiple_choice
1 marks
In a species of sweet pea, the gene for flower colour (P/p) and the gene for pollen grain shape (L/l) are autosomal and linked. A heterozygous plant with the genotype PpLl was crossed with a double homozygous recessive plant (ppll).
The offspring phenotypes and their numbers were: - Purple flowers, long pollen grains: 382 - Red flowers, round pollen grains: 368 - Purple flowers, round pollen grains: 128 - Red flowers, long pollen grains: 122
What is the recombination frequency between these two gene loci?
A.12.5%
B.25.0%
C.33.3%
D.75.0%
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Worked solution
The parental gametes are PL and pl, so the non-recombinant (parental) offspring phenotypes are purple-long and red-round. The recombinant offspring phenotypes are purple-round (128) and red-long (122). Sum of recombinants = \(128 + 122 = 250\). Total number of offspring = \(382 + 368 + 128 + 122 = 1000\). Recombination frequency = \(\frac{250}{1000} \times 100\% = 25.0\%\).
Marking scheme
1 mark for identifying the recombinant phenotypes and correctly calculating the recombination frequency as 25.0%.
Question 15 · multiple_choice
1 marks
Which row correctly describes what occurs at each temperature phase during one cycle of the polymerase chain reaction (PCR)?
| | \(95^\circ\text{C}\) | \(55^\circ\text{C}\) | \(72^\circ\text{C}\) | |---|---|---|---| | A | Hydrogen bonds break to separate DNA strands | Primers bind to single-stranded DNA templates | Taq polymerase synthesises complementary strands | | B | Phosphodiester bonds break to separate DNA strands | Primers bind to single-stranded DNA templates | Taq polymerase synthesises complementary strands | | C | Hydrogen bonds break to separate DNA strands | Taq polymerase binds to single-stranded DNA | Primers synthesise complementary strands | | D | Phosphodiester bonds break to separate DNA strands | Taq polymerase binds to single-stranded DNA | Primers synthesise complementary strands |
A.Row A
B.Row B
C.Row C
D.Row D
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Worked solution
At \(95^\circ\text{C}\), hydrogen bonds between complementary strands break to separate the DNA double helix (phosphodiester bonds in the backbone are not broken). At \(55^\circ\text{C}\), primers anneal (bind) via complementary base pairing to the template strands. At \(72^\circ\text{C}\), the thermostable Taq polymerase synthesises complementary strands by adding free nucleotides (dNTPs) starting from the primers. This matches Row A.
Marking scheme
1 mark for correctly matching the molecular events to all three PCR temperatures.
Question 16 · multiple_choice
1 marks
The table shows the concentrations of four substances in the blood plasma, the glomerular filtrate in the Bowman's capsule, and the urine of a healthy human.
| Substance | Concentration in blood plasma / \(\text{g dm}^{-3}\) | Concentration in glomerular filtrate / \(\text{g dm}^{-3}\) | Concentration in urine / \(\text{g dm}^{-3}\) | |---|---|---|---| | W | 80.0 | 0.0 | 0.0 | | X | 1.0 | 1.0 | 0.0 | | Y | 0.3 | 0.3 | 20.0 | | Z | 3.0 | 3.0 | 6.0 |
Which row correctly identifies W, X, Y, and Z?
| | W | X | Y | Z | |---|---|---|---|---| | A | Protein | Glucose | Urea | Sodium ions | | B | Glucose | Protein | Sodium ions | Urea | | C | Protein | Sodium ions | Urea | Glucose | | D | Sodium ions | Glucose | Protein | Urea |
A.Row A
B.Row B
C.Row C
D.Row D
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Worked solution
W is protein because it is too large to pass through the basement membrane during ultrafiltration, leaving a concentration of 0.0 in the filtrate. X is glucose because it is filtered but completely reabsorbed in the proximal convoluted tubule, leaving a concentration of 0.0 in the urine. Y is urea because it is filtered and highly concentrated in the urine as water is reabsorbed. Z is sodium ions because they are filtered and partially reabsorbed, leaving a moderately higher concentration in the urine compared to plasma. This corresponds to Row A.
Marking scheme
1 mark for correctly identifying all four substances based on their physiological changes in concentration along the nephron.
Question 17 · multiple choice
1 marks
An experiment was carried out to investigate the effect of two different inhibitors, X and Y, on the rate of an enzyme-controlled reaction. The concentration of the enzyme was kept constant in all trials. The kinetic parameters of the reaction under different conditions were determined as follows:
Which statement is the correct conclusion about inhibitors X and Y?
A.Inhibitor X is a competitive inhibitor and binds to the active site; inhibitor Y is a non-competitive inhibitor and binds to an allosteric site.
B.Inhibitor X is a non-competitive inhibitor and binds to an allosteric site; inhibitor Y is a competitive inhibitor and binds to the active site.
C.Inhibitor X is a competitive inhibitor and increases the enzyme's affinity for the substrate; inhibitor Y is a non-competitive inhibitor and does not affect the affinity.
D.Inhibitor X is a non-competitive inhibitor and decreases the enzyme's affinity for the substrate; inhibitor Y is a competitive inhibitor and increases the affinity.
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Worked solution
Competitive inhibitors bind to the active site, increasing the value of \(K_m\) (apparent affinity decreases) but leaving \(V_{max}\) unchanged because high substrate concentrations can overcome the inhibition. Inhibitor X shows these exact properties (\(V_{max}\) remains 100, \(K_m\) increases from 2.0 to 6.0), meaning it is competitive and binds to the active site.
Non-competitive inhibitors bind to an allosteric site (a site other than the active site), decreasing the value of \(V_{max}\) but leaving \(K_m\) unchanged. Inhibitor Y shows these exact properties (\(V_{max}\) decreases from 100 to 50, while \(K_m\) remains 2.0), meaning it is non-competitive and binds to an allosteric site.
Marking scheme
Award 1 mark for the correct option (A). - Reject other options because they swap the inhibitors or misidentify the binding sites or affinity changes.
Question 18 · multiple choice
1 marks
Which statement correctly explains how an enzyme increases the rate of a metabolic reaction?
A.The active site of the enzyme alters its shape permanently to fit the transition state of the substrate, releasing energy that lowers the activation energy.
B.The enzyme provides an alternative pathway with a lower activation energy by orienting the substrates and putting physical strain on specific bonds.
C.The enzyme transfers thermal energy to the reactants, increasing their kinetic energy so that a higher proportion of collisions are successful.
D.The enzyme shifts the equilibrium position of the reaction towards the products, meaning that more product is formed per unit time.
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Worked solution
Enzymes increase the rate of chemical reactions by lowering the activation energy required. They achieve this by providing an alternative pathway, holding reactants in the correct orientation to react, and putting strain on chemical bonds to make them easier to break. They do not permanently alter their own structure, nor do they transfer thermal energy or shift the chemical equilibrium.
Marking scheme
Award 1 mark for the correct option (B). - Reject A: The active site changes shape temporarily (induced fit), not permanently. - Reject C: Enzymes do not transfer thermal energy to increase kinetic energy. - Reject D: Enzymes speed up the rate at which equilibrium is reached but do not change the position of equilibrium.
Question 19 · multiple choice
1 marks
A student measured the volume of oxygen released during the breakdown of hydrogen peroxide by catalase at regular intervals. The results are shown below:
* At \(0\text{ s}\): \(0.0\text{ cm}^3\) * At \(10\text{ s}\): \(15.0\text{ cm}^3\) * At \(20\text{ s}\): \(25.0\text{ cm}^3\) * At \(30\text{ s}\): \(31.0\text{ cm}^3\)
What is the average initial rate of reaction over the first 10 seconds, and what is the main reason for the subsequent decrease in the rate of reaction?
A.Rate: \(1.5\text{ cm}^3\text{ s}^{-1}\); Reason: Substrate concentration becomes limiting as substrate is converted to product.
B.Rate: \(1.5\text{ cm}^3\text{ s}^{-1}\); Reason: The active sites of the enzyme molecules become permanently denatured by the product.
C.Rate: \(0.7\text{ cm}^3\text{ s}^{-1}\); Reason: Substrate concentration becomes limiting as substrate is converted to product.
D.Rate: \(0.7\text{ cm}^3\text{ s}^{-1}\); Reason: High concentration of product competitively inhibits the active sites.
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Worked solution
The average initial rate of reaction over the first 10 seconds is calculated as: \text{Rate} = \frac{\text{Change in volume}}{\text{Change in time}} = \frac{15.0\text{ cm}^3 - 0.0\text{ cm}^3}{10\text{ s} - 0\text{ s}} = 1.5\text{ cm}^3\text{ s}^{-1}.
As the reaction proceeds, substrate is converted into product, meaning the concentration of substrate decreases. This makes the substrate concentration limiting, leading to fewer successful collisions per unit time between substrate molecules and the active sites of enzymes, thereby decreasing the rate.
Marking scheme
Award 1 mark for the correct option (A). - Reject options with incorrect rates (0.7). - Reject options suggesting that enzyme denatures (since temperature/pH remain constant and products do not denature enzymes).
Question 20 · multiple choice
1 marks
A student sampled a community of soil invertebrates in a woodland habitat and recorded the abundance of three species as shown in the table:
| Species | Number of individuals (\(n\)) | | :--- | :--- | | Woodlouse | 10 | | Earthworm | 6 | | Beetle | 4 |
Using the formula for Simpson's Index of Diversity: \(D = 1 - \Sigma \left(\frac{n}{N}\right)^2\)
What is the calculated value of Simpson's Index of Diversity (\(D\)) for this community?
A.0.38
B.0.50
C.0.62
D.0.80
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Worked solution
1. Calculate the total number of individuals (\(N\)): \(N = 10 + 6 + 4 = 20\).
3. Sum these squared values: \(\Sigma \left(\frac{n}{N}\right)^2 = 0.25 + 0.09 + 0.04 = 0.38\).
4. Subtract from 1: \(D = 1 - 0.38 = 0.62\).
Marking scheme
Award 1 mark for the correct option (C). - Option A (0.38) is the sum of \((n/N)^2\) before subtracting from 1. - Reject other options due to incorrect calculation steps.
Question 21 · multiple choice
1 marks
Which of the following describes features that are found in the domain Archaea but NOT in the domain Bacteria?
A.Ribosomes are of the 70S type and the cell wall contains peptidoglycan.
B.Ribosomes are of the 80S type and circular DNA is associated with histone proteins.
C.Cell membrane lipids contain branched hydrocarbon chains with ether linkages, and transcription features are similar to Eukarya.
D.Linear DNA is enclosed in a double membrane envelope and introns are present in all genes.
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Worked solution
Archaea and Bacteria are both prokaryotic, but they have distinct differences. Archaea have cell membrane lipids with branched hydrocarbon chains containing ether linkages (whereas Bacteria have unbranched ester-linked lipids). Furthermore, the machinery for transcription and translation in Archaea is much more similar to that of Eukarya than Bacteria.
Marking scheme
Award 1 mark for the correct option (C). - Reject A because peptidoglycan is unique to Bacteria, not Archaea. - Reject B because Archaea have 70S ribosomes (not 80S). - Reject D because Archaea have circular DNA, not linear DNA inside a nuclear envelope.
Question 22 · multiple choice
1 marks
In a plant species, two genes, A/a and B/b, control flower color. At least one dominant allele of each gene (genotype A_B_) is required to produce red flowers. All other genotypes result in white flowers.
A cross is carried out between two double heterozygous red-flowered plants (AaBb \(\times\) AaBb).
What is the expected ratio of phenotypes in the offspring?
A.9 red : 7 white
B.9 red : 3 pink : 4 white
C.12 red : 3 pink : 1 white
D.15 red : 1 white
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Worked solution
This is an example of complementary gene action (epistasis). When two double heterozygotes (AaBb) are crossed, the independent assortment of alleles yields a 9:3:3:1 genotypic ratio: - \(9/16\) A_B_ (have both dominant alleles, so they produce red flowers) - \(3/16\) A_bb (lack dominant B, so they produce white flowers) - \(3/16\) aaB_ (lack dominant A, so they produce white flowers) - \(1/16\) aabb (lack both, so they produce white flowers)
Grouping the white-flowered phenotypes: \(3 + 3 + 1 = 7\). Therefore, the expected phenotypic ratio is 9 red : 7 white.
Marking scheme
Award 1 mark for the correct option (A). - Reject B (9:3:4 is recessive epistasis). - Reject C (12:3:1 is dominant epistasis). - Reject D (15:1 is duplicate gene action).
Question 23 · multiple choice
1 marks
A student set up a PCR reaction mixture to amplify a specific sequence of DNA.
Which row correctly identifies the function of each component added to the reaction tube?
A.Primers: Provide a starting point with a free 3'-OH group for DNA polymerase to bind.
Taq polymerase: Synthesises the complementary DNA strands at high temperatures.
dNTPs: Act as substrates and provide energy for the synthesis of the new DNA strands.
Buffer with \(Mg^{2+}\): Acts as a cofactor to activate Taq polymerase and maintains pH.
B.Primers: Prevent the two single strands of template DNA from reannealing during cooling.
Taq polymerase: Unwinds the double-stranded DNA template at \(95\text{ }^\circ\text{C}\).
dNTPs: Act as fluorescent markers to detect the amplified DNA.
Buffer with \(Mg^{2+}\): Denatures the hydrogen bonds between DNA strands.
C.Primers: Provide a starting point with a free 3'-OH group for DNA polymerase to bind.
Taq polymerase: Unwinds the double-stranded DNA template at \(95\text{ }^\circ\text{C}\).
dNTPs: Synthesise the primer sequences during the annealing stage.
Buffer with \(Mg^{2+}\): Restores the double-helix structure of the DNA after elongation.
D.Primers: Act as restriction enzymes to cut the DNA at specific target sites.
Taq polymerase: Synthesises the complementary DNA strands at high temperatures.
dNTPs: Act as substrates and provide energy for the synthesis of the new DNA strands.
Buffer with \(Mg^{2+}\): Prevents contamination by other DNA polymerases.
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Worked solution
In a Polymerase Chain Reaction (PCR): 1. **Primers**: Single-stranded DNA sequences that attach to the target DNA. They provide the free 3'-OH group necessary for DNA polymerase to begin adding nucleotides. 2. **Taq polymerase**: A thermostable DNA polymerase that synthesises the complementary DNA strand at elevated temperatures (typically \(72\text{ }^\circ\text{C}\)). 3. **Deoxynucleoside triphosphates (dNTPs)**: Provide the free nucleotides (dATP, dTTP, dCTP, dGTP) that act as substrates and release energy when their phosphate bonds are cleaved to build the new strand. 4. **Buffer solution with \(Mg^{2+}\)**: Maintains optimal pH and acts as an essential cofactor for Taq polymerase activity.
Marking scheme
Award 1 mark for the correct option (A). - Reject B because Taq polymerase does not unwind DNA (high temperature of 95 degrees does), and dNTPs are not fluorescent markers. - Reject C because Taq polymerase does not unwind DNA, and dNTPs do not synthesise primers. - Reject D because primers do not act as restriction enzymes.
Question 24 · multiple choice
1 marks
When blood glucose concentration decreases below the normal set-point, glucagon is secreted by the alpha cells of the pancreas.
Which sequence of events describes the mechanism of action of glucagon on liver cells?
A.Glucagon binds to membrane receptors \(\rightarrow\) receptor activates G-protein \(\rightarrow\) G-protein activates adenylyl cyclase \(\rightarrow\) adenylyl cyclase converts ATP to cAMP \(\rightarrow\) cAMP activates protein kinases \(\rightarrow\) glycogen phosphorylase is activated to break down glycogen.
B.Glucagon enters the liver cell by active transport \(\rightarrow\) binds to intracellular receptors on the endoplasmic reticulum \(\rightarrow\) activates glycogen synthase \(\rightarrow\) glycogen synthase converts glycogen directly into glucose-1-phosphate.
C.Glucagon binds to membrane receptors \(\rightarrow\) receptor activates tyrosine kinase directly \(\rightarrow\) tyrosine kinase converts cAMP into ATP \(\rightarrow\) high ATP levels stimulate glycogen phosphorylase to break down glycogen.
D.Glucagon binds to membrane receptors \(\rightarrow\) receptor activates G-protein \(\rightarrow\) G-protein acts as a second messenger to directly hydrolyse glycogen into glucose \(\rightarrow\) glucose is released via facilitated diffusion through GLUT4 transporters.
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Worked solution
Glucagon is a peptide hormone, so it cannot cross the cell-surface membrane. It binds to specific receptors on the cell-surface membrane of hepatocytes. This binding activates a G-protein, which in turn activates the membrane-bound enzyme adenylyl cyclase. Adenylyl cyclase converts ATP into cyclic AMP (cAMP), which acts as a second messenger. cAMP activates a cascade of protein kinase enzymes, which eventually activates glycogen phosphorylase. This enzyme catalyses the hydrolysis of glycogen to glucose (glycogenolysis).
Marking scheme
Award 1 mark for the correct option (A). - Reject B because glucagon does not enter the cell and does not bind intracellular receptors. - Reject C because receptors activate G-proteins, not tyrosine kinases directly, and cAMP is made from ATP (not vice versa). - Reject D because G-proteins do not directly hydrolyse glycogen, and GLUT4 is associated with insulin's action in muscle/fat, not glucagon's action in liver cells.
Question 25 · Multiple Choice
1 marks
An enzyme-catalysed reaction is carried out at two different temperatures, \(25^\circ\text{C}\) and \(35^\circ\text{C}\), across a range of substrate concentrations. Assume that the enzyme is not denatured at either temperature.
Which parameter(s) of the reaction will change at \(35^\circ\text{C}\) compared to \(25^\circ\text{C}\)?
1. \(V_{\max}\) 2. \(K_m\) 3. Activation energy of the reaction (\(E_a\))
A.1 only
B.1 and 2 only
C.2 and 3 only
D.1, 2 and 3
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Worked solution
At a higher temperature (\(35^\circ\text{C}\) compared to \(25^\circ\text{C}\)), molecules have more kinetic energy. This increases the rate of catalysis, leading to an increase in the maximum velocity of the reaction (\(V_{\max}\)). The Michaelis-Menten constant (\(K_m\)) is also temperature-dependent because temperature alters the rates of enzyme-substrate binding and dissociation, thereby changing the affinity of the enzyme for its substrate. However, the activation energy (\(E_a\)) is a constant characteristic of the specific catalytic pathway of the enzyme-substrate complex and is independent of temperature. Therefore, only parameters 1 and 2 will change.
Marking scheme
Award 1 mark for identifying that both \(V_{\max}\) and \(K_m\) change with temperature but \(E_a\) remains unaffected.
Question 26 · Multiple Choice
1 marks
An experiment was conducted to investigate the effect of a specific inhibitor on the rate of an enzyme-catalysed reaction.
In the presence of a constant concentration of the inhibitor, the following observations were made: - At very high substrate concentrations, the rate of reaction approaches the maximum rate (\(V_{\max}\)) observed without the inhibitor. - The concentration of substrate required to reach half of this maximum rate is doubled.
Which statement about this inhibitor is correct?
A.It binds reversibly to an allosteric site on the enzyme, preventing the transition state from forming.
B.It binds covalently to the active site, permanently preventing substrate binding.
C.It has a molecular structure similar to the substrate and competes with it for the active site.
D.It binds only to the enzyme-substrate complex, preventing the release of products.
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Worked solution
The observations show that \(V_{\max}\) is unchanged (since the rate approaches the original \(V_{\max}\) at very high substrate concentrations) and the Michaelis-Menten constant (\(K_m\)) is doubled (since the substrate concentration required to reach \(\frac{1}{2} V_{\max}\) is doubled). These are the classic characteristics of a competitive inhibitor. A competitive inhibitor has a complementary shape/structure similar to the substrate, allowing it to compete with the substrate for binding to the active site.
Marking scheme
Award 1 mark for identifying that competitive inhibition matches the scenario and selecting the option describing competitive inhibitor characteristics.
Question 27 · Multiple Choice
1 marks
An ecologist sampled two different meadows, Meadow X and Meadow Y, to determine their plant diversity. The results of the sampling are shown in the table:
| Species | Number of individuals in Meadow X | Number of individuals in Meadow Y | | :--- | :---: | :---: | | Species A | 80 | 30 | | Species B | 10 | 25 | | Species C | 5 | 25 | | Species D | 5 | 20 |
Which statement correctly compares Meadow X and Meadow Y?
A.Meadow X has higher species richness and higher species evenness than Meadow Y.
B.Meadow X has the same species richness as Meadow Y, but Meadow Y has higher species evenness.
C.Meadow Y has higher species richness and higher species evenness than Meadow X.
D.Meadow Y has the same species richness as Meadow X, but Meadow X has higher species evenness.
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Worked solution
Species richness is the number of different species present in an area. Both Meadow X and Meadow Y contain exactly 4 species (Species A, B, C, and D), so they have the same species richness. Species evenness refers to how close in numbers each species in an environment is. Meadow X is heavily dominated by Species A (80%), indicating low species evenness. Meadow Y has a much more balanced distribution of individuals (30, 25, 25, 20), indicating higher species evenness. Therefore, Meadow Y has higher species evenness than Meadow X, while both have the same species richness.
Marking scheme
Award 1 mark for correctly comparing both species richness (equal in both) and species evenness (higher in Y).
Question 28 · Multiple Choice
1 marks
In a species of sweet pea, flower colour is controlled by two genes, \(A/a\) and \(B/b\), which assort independently.
- The presence of at least one dominant allele of both genes (\(A\_B\_\)) results in purple flowers. - All other genotypes (\(A\_bb\), \(aaB\_\), or \(aabb\)) result in white flowers.
A plant with purple flowers of genotype \(AaBb\) is self-pollinated.
What is the expected phenotypic ratio in the offspring?
A.9 purple : 7 white
B.12 purple : 4 white
C.15 purple : 1 white
D.9 purple : 3 white : 4 pink
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Worked solution
When a double heterozygote (\(AaBb\)) is self-pollinated, the expected Mendelian ratio of genotypes in the offspring is: 9 \(A\_B\_\) (possess at least one dominant allele of both genes), 3 \(A\_bb\) (possess dominant \(A\) but homozygous recessive \(b\)), 3 \(aaB\_\) (possess dominant \(B\) but homozygous recessive \(a\)), and 1 \(aabb\) (homozygous recessive for both genes). According to the rules provided: \(A\_B\_\) offspring (9/16) will have purple flowers, and the other classes (3/16 + 3/16 + 1/16 = 7/16) will have white flowers. Thus, the expected phenotypic ratio is 9 purple : 7 white.
Marking scheme
Award 1 mark for using the 9:3:3:1 ratio and combining the non-double-dominant classes to get a 9:7 ratio.
Question 29 · Multiple Choice
1 marks
During a single cycle of the Polymerase Chain Reaction (PCR), the mixture is subjected to three different temperatures in sequence: \(95^\circ\text{C}\), then \(55^\circ\text{C}\), and finally \(72^\circ\text{C}\).
Which row correctly describes the main event occurring at each temperature?
| | \(95^\circ\text{C}\) | \(55^\circ\text{C}\) | \(72^\circ\text{C}\) | | :--- | :--- | :--- | :--- | | **A** | Denaturation of DNA double helix | Annealing of primers to single strands | Optimum temperature for Taq polymerase extension | | **B** | Denaturation of DNA double helix | Activation of Taq polymerase | Annealing of primers to single strands | | **C** | Synthesis of complementary strands | Annealing of primers to single strands | Denaturation of DNA double helix | | **D** | Activation of Taq polymerase | Denaturation of DNA double helix | Synthesis of complementary strands |
A.Row A
B.Row B
C.Row C
D.Row D
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Worked solution
At \(95^\circ\text{C}\), high temperatures denature the DNA double helix into single strands. At \(55^\circ\text{C}\), the lower temperature allows short DNA primers to anneal to their complementary single-stranded targets. At \(72^\circ\text{C}\), Taq DNA polymerase is at its optimum temperature to extend the primers by synthesizing complementary strands.
Marking scheme
Award 1 mark for matching the temperatures with denaturation (95), annealing (55), and extension (72).
Question 30 · Multiple Choice
1 marks
What is the correct sequence of intracellular events that occurs immediately after antidiuretic hormone (ADH) binds to cell surface receptors on the collecting duct cells of a nephron?
1. Vesicles containing aquaporins fuse with the luminal (apical) membrane. 2. Adenylyl cyclase is activated, converting ATP to cyclic AMP (cAMP). 3. Cyclic AMP acts as a second messenger to activate a kinase cascade. 4. Active G-protein is produced following receptor conformational change. 5. Water permeability of the luminal membrane increases, allowing water to enter by osmosis.
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Worked solution
The correct biological sequence is: (4) ADH binding triggers G-protein activation; (2) active G-protein stimulates adenylyl cyclase to synthesize cAMP; (3) cAMP acts as a second messenger to activate a kinase cascade; (1) kinase activity leads to vesicle fusion bringing aquaporins to the membrane; (5) water moves by osmosis into the cells down the osmotic gradient.
Marking scheme
Award 1 mark for identifying the correct sequential path of G-protein coupled receptor signalling and cellular response.
Question 31 · Multiple Choice
1 marks
Which statement about the Michaelis-Menten constant (\(K_m\)) of an enzyme is correct?
A.A high \(K_m\) value indicates a high affinity of the enzyme for its substrate, meaning less substrate is needed to reach \(V_{\max}\).
B.A high \(K_m\) value indicates a low affinity of the enzyme for its substrate, meaning more substrate is needed to reach \(\frac{1}{2} V_{\max}\).
C.A low \(K_m\) value indicates a low affinity of the enzyme for its substrate, meaning less substrate is needed to reach \(V_{\max}\).
D.A low \(K_m\) value indicates a high affinity of the enzyme for its substrate, meaning more substrate is needed to reach \(\frac{1}{2} V_{\max}\).
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Worked solution
The Michaelis-Menten constant (\(K_m\)) is the substrate concentration at which the reaction rate is half of its maximum velocity (\(\frac{1}{2} V_{\max}\)). A high \(K_m\) indicates a low affinity of the enzyme for its substrate, because a larger amount of substrate is required to achieve half-maximal velocity. Conversely, a low \(K_m\) indicates a high affinity.
Marking scheme
Award 1 mark for selecting the option stating that a high \(K_m\) reflects low affinity and high substrate concentration needed for half \(V_{\max}\).
Question 32 · Multiple Choice
1 marks
Which features are characteristic of organisms in both the Domain Archaea and the Domain Bacteria, but are NOT characteristic of organisms in the Domain Eukarya?
1. Ribosomes that are 70S 2. Absence of a nucleus surrounded by a nuclear envelope 3. Cell walls containing peptidoglycan
A.1 and 2 only
B.1 and 3 only
C.2 and 3 only
D.1, 2 and 3
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Worked solution
Feature 1 (70S ribosomes) and Feature 2 (absence of a membrane-bound nucleus) are typical of both prokaryotic domains (Archaea and Bacteria) but not Eukarya (which have 80S cytoplasmic ribosomes and a true nucleus). Feature 3 is incorrect because only Bacteria contain peptidoglycan in their cell walls; Archaea cell walls do not contain peptidoglycan, meaning this feature is not shared by both.
Marking scheme
Award 1 mark for identifying that 70S ribosomes and absence of a nucleus are shared prokaryotic characteristics, while peptidoglycan cell walls are unique to Bacteria.
Question 33 · multiple_choice
1 marks
The Michaelis-Menten constant (\(K_m\)) and the maximum velocity (\(V_{\max}\)) of an enzyme-catalyzed reaction were measured in both the absence and presence of an inhibitor.
- In the absence of the inhibitor: \(K_m = 1.5\text{ mmol dm}^{-3}\) and \(V_{\max} = 50\text{ \mu mol min}^{-1}\) - In the presence of the inhibitor: \(K_m = 4.5\text{ mmol dm}^{-3}\) and \(V_{\max} = 50\text{ \mu mol min}^{-1}\)
Which statement describes the mechanism of this inhibitor?
A.It binds to an allosteric site, altering the shape of the active site so that the substrate can no longer bind.
B.It binds irreversibly to the active site, reducing the concentration of active enzyme molecules.
C.It competes with the substrate for the active site, and its effect is reduced at high substrate concentrations.
D.It covalently bonds to the enzyme-substrate complex, preventing the conversion of substrate to product.
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Worked solution
A competitive inhibitor competes with the substrate for binding at the active site. This increases the apparent Michaelis constant (\(K_m\)) because a higher substrate concentration is required to achieve half the maximum velocity. However, at sufficiently high substrate concentrations, the substrate virtually outcompetes the inhibitor, allowing the reaction to reach the same maximum velocity (\(V_{\max}\)). Therefore, a competitive inhibitor increases \(K_m\) while keeping \(V_{\max}\) unchanged. Non-competitive inhibitors decrease \(V_{\max}\) but typically do not change the affinity of the enzyme for its substrate in the same way (the substrate can still bind, but the complex is inactive).
Marking scheme
1 mark for identifying the correct option (C) by analyzing the effect of a competitive inhibitor on \(K_m\) (increased) and \(V_{\max}\) (unchanged).
Question 34 · multiple_choice
1 marks
Which of the following describe how an enzyme lowers the activation energy of a metabolic reaction?
1. Aligning substrate molecules in the correct orientation for reaction. 2. Changing the overall energy change (\(\Delta H\)) of the reaction. 3. Putting physical strain on specific bonds in the substrate. 4. Providing temporary alternative chemical pathways through active site residues.
A.1, 2, 3 and 4
B.1, 3 and 4 only
C.1 and 3 only
D.2 and 4 only
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Worked solution
Enzymes lower the activation energy of a reaction by: (1) orienting substrates correctly, (3) straining bonds within the substrate to make them easier to break, and (4) participating in the reaction chemistry (e.g., via acid/base catalysis or forming temporary covalent intermediates with active site residues). Enzymes do NOT alter the overall thermodynamics, such as the overall enthalpy change (\(\Delta H\)) or free energy change (\(\Delta G\)) of the reaction. Thus, statements 1, 3, and 4 are correct, while statement 2 is incorrect.
Marking scheme
1 mark for selecting B. Statement 2 is incorrect because enzymes do not alter the overall enthalpy change of a reaction.
Question 35 · multiple_choice
1 marks
The initial rate of an enzyme-controlled reaction is measured at \(20^\circ\text{C}\) and is found to be \(1.2\text{ mmol dm}^{-3}\text{ s}^{-1}\).
The temperature coefficient (\(Q_{10}\)) of this reaction is \(2.0\) between \(20^\circ\text{C}\) and \(40^\circ\text{C}\).
Assuming that the enzyme is not denatured at \(40^\circ\text{C}\), what is the expected rate of reaction at \(40^\circ\text{C}\)?
A.\(2.4\text{ mmol dm}^{-3}\text{ s}^{-1}\)
B.\(3.6\text{ mmol dm}^{-3}\text{ s}^{-1}\)
C.\(4.8\text{ mmol dm}^{-3}\text{ s}^{-1}\)
D.\(6.0\text{ mmol dm}^{-3}\text{ s}^{-1}\)
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Worked solution
The temperature coefficient (\(Q_{10}\) = 2.0) means that the rate of reaction doubles for every \(10^\circ\text{C}\) rise in temperature. The temperature increases from \(20^\circ\text{C}\) to \(40^\circ\text{C}\), which represents two intervals of \(10^\circ\text{C}\). - At \(20^\circ\text{C}\), rate = \(1.2\text{ mmol dm}^{-3}\text{ s}^{-1}\). - At \(30^\circ\text{C}\), rate = \(1.2 \times 2 = 2.4\text{ mmol dm}^{-3}\text{ s}^{-1}\). - At \(40^\circ\text{C}\), rate = \(2.4 \times 2 = 4.8\text{ mmol dm}^{-3}\text{ s}^{-1}\).
Marking scheme
1 mark for identifying the correct rate calculation (C), demonstrating understanding of the exponential nature of the temperature coefficient over multiple steps.
Question 36 · multiple_choice
1 marks
An ecological survey compared the plant biodiversity of two grassland communities, X and Y.
- Grassland X: Simpson's Index of Diversity (\(D\)) = 0.82 - Grassland Y: Simpson's Index of Diversity (\(D\)) = 0.24
Which statement is a correct conclusion based on these results?
A.Grassland X is dominated by a single species, making it highly stable.
B.Grassland Y has a higher species richness and evenness than Grassland X.
C.An outbreak of a disease affecting a dominant species in Grassland Y is likely to have a greater impact on community structure than a similar event in Grassland X.
D.A sudden change in abiotic factors in Grassland X will cause a rapid collapse of its food web because of low stability.
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Worked solution
Simpson's Index of Diversity (\(D\)) values range from 0 (no diversity) to 1 (infinite diversity). A high value, such as 0.82 in Grassland X, indicates a high species richness and evenness, resulting in a stable ecosystem with complex food webs that can resist environmental changes. A low value, such as 0.24 in Grassland Y, indicates low diversity where one or a few species dominate. In Grassland Y, because of the low diversity and dominance of certain species, any disease affecting a dominant species will have a profound, disproportionate impact on the whole community structure. Therefore, statement C is correct.
Marking scheme
1 mark for choosing C, which correctly interprets high vs low Simpson's Index of Diversity values and their relationship to community stability.
Question 37 · multiple_choice
1 marks
Which of the following conservation methods are correctly classified as *ex situ* conservation?
1. Captive breeding of endangered animals in a zoological park. 2. Controlled burning to maintain a specific successional stage in a nature reserve. 3. Storing seeds of wild crop relatives in a temperature-controlled seed bank. 4. Eradicating invasive alien species from an offshore island sanctuary.
A.1, 2 and 3
B.1 and 3 only
C.2 and 4 only
D.3 and 4 only
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Worked solution
*Ex situ* conservation involves the preservation of biological diversity outside of its natural habitats. Captive breeding in zoos (1) and seed storage in seed banks (3) are classic examples of *ex situ* conservation. Controlled burning (2) and invasive species eradication (4) are active management techniques performed within the natural habitat, which are examples of *in situ* conservation. Therefore, only statements 1 and 3 are correct.
Marking scheme
1 mark for identifying B as the correct answer based on distinguishing *ex situ* from *in situ* conservation methods.
Question 38 · multiple_choice
1 marks
In a species of sweet pea, flower color is controlled by two genes, \(A/a\) and \(B/b\), which assort independently.
The production of purple pigment requires the presence of at least one dominant allele of both genes (genotype \(A\_B\_\)). All other genotypes (such as \(A\_bb\), \(aaB\_\), or \(aabb\)) result in white flowers.
If two plants with the genotype \(AaBb\) are crossed, what is the expected ratio of phenotypes in the offspring?
A.9 purple : 7 white
B.9 purple : 3 pink : 4 white
C.15 purple : 1 white
D.12 purple : 3 white : 1 pink
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Worked solution
This cross is a dihybrid cross involving complementary epistasis. When two heterozygous plants (\(AaBb \times AaBb\)) are crossed, the standard Mendelian phenotypic ratio of \(9:3:3:1\) is modified: - \(9/16\) will have the genotype \(A\_B\_\) and will produce purple flowers. - \(3/16\) will be \(A\_bb\) (white flowers). - \(3/16\) will be \(aaB\_\) (white flowers). - \(1/16\) will be \(aabb\) (white flowers). Combining all the genotypes that result in white flowers: \(3 + 3 + 1 = 7\). Therefore, the phenotypic ratio is 9 purple : 7 white.
Marking scheme
1 mark for selecting A, showing understanding of epistatic gene interactions and modifying the dihybrid ratio accordingly.
Question 39 · multiple_choice
1 marks
The table shows steps in one cycle of the polymerase chain reaction (PCR).
Which row correctly identifies the temperature and molecular event occurring at each step of the cycle?
A.Step 1: \(94^\circ\text{C}\) to break phosphodiester bonds; Step 2: \(55^\circ\text{C}\) to allow primers to anneal; Step 3: \(72^\circ\text{C}\) for Taq polymerase to synthesize DNA strands
B.Step 1: \(94^\circ\text{C}\) to break hydrogen bonds; Step 2: \(55^\circ\text{C}\) to allow primers to anneal; Step 3: \(72^\circ\text{C}\) for Taq polymerase to synthesize DNA strands
C.Step 1: \(72^\circ\text{C}\) for Taq polymerase to synthesize DNA strands; Step 2: \(55^\circ\text{C}\) to allow primers to anneal; Step 3: \(94^\circ\text{C}\) to break hydrogen bonds
D.Step 1: \(94^\circ\text{C}\) to break hydrogen bonds; Step 2: \(72^\circ\text{C}\) for Taq polymerase to synthesize DNA strands; Step 3: \(55^\circ\text{C}\) to allow primers to anneal
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Worked solution
In a PCR cycle, the sequence of events is: 1. Denaturation (typically at \(90\text{--}95^\circ\text{C}\)): High temperature disrupts the hydrogen bonds holding the two complementary DNA strands together, separating them. It does not break the covalent phosphodiester bonds of the sugar-phosphate backbone. 2. Annealing (typically at \(50\text{--}65^\circ\text{C}\)): The temperature is lowered to allow primers to bind (anneal) to their complementary targets on the single-stranded DNA templates. 3. Extension (typically at \(70\text{--}75^\circ\text{C}\)): The temperature is raised to the optimum for Taq polymerase, which synthesizes new DNA strands by adding free nucleotides in the 5' to 3' direction. Therefore, option B is correct.
Marking scheme
1 mark for selecting B, which correctly identifies both the correct sequence of temperatures and the nature of molecular bonding disrupted/formed at each step.
Question 40 · multiple_choice
1 marks
The following events occur in the cells of the collecting duct when antidiuretic hormone (ADH) regulates water reabsorption in the mammalian kidney.
1. Aquaporins are inserted into the luminal membrane. 2. Active protein kinase A phosphorylates proteins that trigger vesicle movement. 3. Intracellular cyclic AMP (cAMP) levels increase. 4. ADH binds to G-protein coupled receptors on the cell surface membrane, activating adenylyl cyclase. 5. Water enters the cell by osmosis from the lumen of the collecting duct.
What is the correct sequence of these events?
A.4 \(\to\) 3 \(\to\) 2 \(\to\) 1 \(\to\) 5
B.4 \(\to\) 2 \(\to\) 3 \(\to\) 5 \(\to\) 1
C.3 \(\to\) 4 \(\to\) 2 \(\to\) 1 \(\to\) 5
D.3 \(\to\) 4 \(\to\) 1 \(\to\) 2 \(\to\) 5
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Worked solution
When ADH binds to G-protein coupled receptors on the cell surface membrane of the collecting duct cells, it activates the enzyme adenylyl cyclase (4). This enzyme catalyzes the conversion of ATP to cyclic AMP (cAMP), raising intracellular cAMP levels (3). cAMP acts as a second messenger to activate protein kinase A (PKA), which phosphorylates proteins that trigger the movement of vesicles containing aquaporins (2). These vesicles fuse with the luminal membrane, inserting the aquaporins into the membrane (1). This increases the permeability of the membrane to water, allowing water to enter the cell by osmosis down a water potential gradient (5). Hence, the correct order is 4 \to 3 \to 2 \to 1 \to 5.
Marking scheme
1 mark for identifying sequence A as the correct pathway of ADH signal transduction and subsequent water reabsorption.
Paper 22 AS Level Structured Questions
Answer all questions. Show your working and use appropriate units.
6 Question · 60 marks
Question 1 · Structured
10 marks
The enzyme urease catalyses the hydrolysis of urea into ammonia and carbon dioxide. (a) Explain the term activation energy and describe how enzymes such as urease lower it. [3] (b) An investigation was carried out to determine the effect of lead ions \(Pb^{2+}\) on the rate of urease activity at different substrate concentrations. The results showed that the maximum rate of reaction \(V_{max}\) remained unchanged, but the Michaelis-Menten constant \(K_m\) increased. (i) State the independent and dependent variables in this investigation. [2] (ii) State, with a reason, the type of inhibition caused by lead ions. [2] (c) Explain how an increase in temperature up to the optimum increases the rate of an enzyme-catalysed reaction, and why exceeding the optimum temperature decreases the rate. [3]
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Worked solution
Part (a): Activation energy is the minimum energy required to start a chemical reaction by bringing reacting molecules to a transition state. Enzymes lower activation energy by providing an alternative pathway with a lower activation energy, holding substrates in an orientation that facilitates bond-breaking and forming, or putting physical strain on bonds within the substrate. Part (b)(i): Independent variable: urea / substrate concentration. Dependent variable: rate of urease activity / rate of urea hydrolysis. Part (b)(ii): Competitive inhibition. This is because Vmax is unchanged (high substrate concentration overcomes the inhibitor's effect), but Km is increased (affinity of the enzyme for the substrate decreases because the inhibitor competes for the active site). Part (c): Up to the optimum temperature, increasing temperature increases the kinetic energy of enzymes and substrate molecules, leading to more frequent successful collisions and the formation of more enzyme-substrate complexes. Above the optimum temperature, the increased thermal energy disrupts hydrogen and ionic bonds stabilizing the tertiary structure of the enzyme, denaturing the active site so that the substrate can no longer bind.
Marking scheme
(a) Max 3 marks: 1. (activation energy is) minimum energy needed to start a reaction / reach transition state; 2. (enzymes) provide alternative pathway; 3. with lower activation energy; 4. by holding substrates in close proximity / correct orientation; 5. or putting strain on bonds in substrate. (b)(i) Max 2 marks: 1. Independent variable: urea / substrate concentration; 2. Dependent variable: rate of reaction / rate of ammonia or CO2 production. (b)(ii) Max 2 marks: 1. competitive inhibition; 2. Vmax remains unchanged because high substrate concentration can outcompete the inhibitor; 3. Km increases because substrate apparent affinity decreases / more substrate needed to reach half-Vmax. (c) Max 3 marks: 1. (up to optimum) increased kinetic energy increases successful collisions (between enzyme and substrate); 2. forming more enzyme-substrate complexes per unit time; 3. (above optimum) high temperature disrupts hydrogen/ionic bonds (and hydrophobic interactions); 4. alters tertiary structure / denatures active site, so substrate can no longer fit.
Question 2 · Structured
10 marks
Lactase is an enzyme used to produce lactose-free milk by hydrolysing lactose into glucose and galactose. The enzyme can be used in its free state or immobilised in alginate beads. (a) Describe how lactase is immobilised using alginate beads. [3] (b) Explain three advantages of using immobilised lactase compared to free lactase in the food industry. [3] (c) Suggest and explain how the process of immobilisation might affect the Michaelis-Menten constant \(K_m\) of lactase for its substrate, lactose. [4]
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Worked solution
Part (a): Lactase is mixed with a solution of sodium alginate. This mixture is then added dropwise using a syringe or pipette into a solution of calcium chloride. The sodium ions are replaced by calcium ions, causing cross-linking of the polymer chains and forming insoluble calcium alginate beads containing the entrapped enzyme. The beads are then rinsed with distilled water to remove any unbound enzyme. Part (b): Advantages include: 1. The enzyme can be easily recovered and reused, reducing production costs. 2. The final product (lactose-free milk) is not contaminated with the enzyme, avoiding the need for downstream purification. 3. Immobilisation often increases the thermal and chemical stability of the enzyme, allowing it to function over wider temperature/pH ranges. Part (c): Immobilisation typically increases the Km value of the enzyme. This is because: 1. The substrate (lactose) must diffuse through the alginate matrix to reach the active sites of the entrapped enzymes, which reduces the rate of diffusion. 2. Some active sites may become physically blocked or less accessible due to the gel matrix. 3. Consequently, a higher concentration of lactose is required to achieve half of the maximum velocity (Vmax), indicating a lower apparent affinity.
Marking scheme
(a) Max 3 marks: 1. mix lactase with sodium alginate solution; 2. add mixture dropwise into calcium chloride solution; 3. to form insoluble calcium alginate beads (trapping the enzyme); 4. rinse beads (with distilled water) to remove free enzyme. (b) Max 3 marks: 1. enzyme can be recovered/reused; 2. product (milk) is uncontaminated with enzyme / no downstream purification needed; 3. greater stability at higher temperatures / wider pH ranges; 4. allows continuous flow processes. (c) Max 4 marks: 1. Km increases / apparent affinity decreases; 2. substrate (lactose) must diffuse through the alginate bead matrix; 3. slows down the rate at which substrate reaches the active site (diffusion limitation); 4. active sites may be partially blocked / less accessible; 5. therefore, higher concentration of substrate is needed to reach half-Vmax.
Question 3 · Structured
10 marks
Ecologists studied the biodiversity of two marine habitats: a seagrass meadow and a coral reef. (a) Define the terms species richness and species evenness. [2] (b) The formula for Simpson's Index of Diversity is: \(D = 1 - \sum \left(\frac{n}{N}\right)^2\). In a sample from the seagrass meadow, four species (A, B, C, and D) were identified with the following individual counts: Species A = 45, Species B = 15, Species C = 30, Species D = 10. Calculate the Simpson's Index of Diversity (D) for the seagrass meadow. Show your working. [3] (c) The coral reef had a calculated D value of 0.88. Compare the biodiversity of the seagrass meadow and the coral reef, and explain what a higher D value indicates about the stability of the ecosystem. [3] (d) State two reasons why it is important to conserve marine ecosystems such as coral reefs. [2]
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Worked solution
Part (a): Species richness is the number of different species present in a particular habitat. Species evenness is a measure of the relative abundance of individuals in each of the species present in a habitat. Part (b): Total number of individuals (N) = 45 + 15 + 30 + 10 = 100. For each species, calculate (n/N)^2: Species A: (45/100)^2 = 0.2025. Species B: (15/100)^2 = 0.0225. Species C: (30/100)^2 = 0.0900. Species D: (10/100)^2 = 0.0100. Sum of (n/N)^2 = 0.2025 + 0.0225 + 0.0900 + 0.0100 = 0.325. D = 1 - 0.325 = 0.675 (or 0.68). Part (c): The coral reef has a higher biodiversity (D = 0.88) than the seagrass meadow (D = 0.68). A higher D value (closer to 1) indicates a highly stable ecosystem. In a highly stable ecosystem, there are complex food webs with many alternative food sources, making the community less vulnerable to changes or environmental disturbances (e.g., disease, temperature changes). Part (d): Reasons for conservation include: 1. Ethical/moral obligation to protect species. 2. Economic value (e.g., ecotourism, fisheries). 3. Ecological services (e.g., coastal protection from storms, carbon sequestration). 4. Potential source of new medicines/pharmaceuticals.
Marking scheme
(a) Max 2 marks: 1. Species richness: number of different species in a habitat; 2. Species evenness: relative abundance of each species in a habitat. (b) Max 3 marks: 1. N = 100 calculated; 2. Sum of (n/N)^2 = 0.325 (or shows correct calculation step for all species); 3. D = 0.675 or 0.68 (correct final answer). (c) Max 3 marks: 1. coral reef has higher biodiversity / higher D value than seagrass meadow; 2. higher D indicates greater ecosystem stability / less likely to collapse; 3. because food webs are more complex / alternative food sources exist; 4. environmental change / loss of one species has a smaller impact on the ecosystem as a whole. (d) Max 2 marks (any two from): 1. moral/ethical reasons to prevent extinction; 2. economic benefits (tourism / fisheries); 3. medicine / drug discovery; 4. ecosystem services / coastal protection / carbon storage.
Question 4 · Structured
10 marks
Coat color in a rodent species is determined by two independently assorting genes, Gene A and Gene B. Gene A determines pigment production; the dominant allele A allows pigment production, whereas the recessive allele a prevents pigment production, resulting in an albino coat. Gene B determines the distribution of pigment; the dominant allele B produces an agouti coat, while the recessive allele b produces a solid black coat. (a) Explain what is meant by the term epistasis. [2] (b) State the phenotypes of rodents with the following genotypes: (i) AaBb, (ii) Aabb, (iii) aaBB. [3] (c) Two heterozygous agouti rodents (AaBb) are crossed. Draw a genetic diagram to determine the expected phenotypic ratio of their offspring. [4] (d) State the statistical test that should be used to determine if the observed ratio of offspring phenotypes differs significantly from the expected ratio. [1]
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Worked solution
Part (a): Epistasis is a gene interaction where one gene masks or suppresses the expression of another gene at a different locus. In this case, the homozygous recessive condition (aa) of Gene A masks the phenotypic expression of Gene B. Part (b): (i) AaBb: Agouti (has pigment and is agouti distribution). (ii) Aabb: Black (has pigment but has the solid black recessive distribution). (iii) aaBB: Albino (has no pigment due to homozygous recessive 'aa', which masks 'BB'). Part (c): Gametes from both parents: AB, Ab, aB, ab. Punnett square results: - Agouti (A_B_): AABB (1), AABb (2), AaBB (2), AaBb (4) = 9. - Black (A_bb): Aabb (2), AAbb (1) = 3. - Albino (aa__): aaBB (1), aaBb (2), aabb (1) = 4. Expected phenotypic ratio: 9 Agouti : 3 Black : 4 Albino. Part (d): Chi-squared test (\(X^2\) test).
Marking scheme
(a) Max 2 marks: 1. gene interaction where one gene masks/suppresses/alters the expression of another gene; 2. at a different locus / position on a chromosome. (b) 3 marks (1 mark each): (i) AaBb: Agouti; (ii) Aabb: Black; (iii) aaBB: Albino. (c) Max 4 marks: 1. correct parental genotypes (AaBb x AaBb) and gametes (AB, Ab, aB, ab for both); 2. correct Punnett square containing 16 combinations; 3. correct assignment of phenotypes to genotypes (9 agouti, 3 black, 4 albino); 4. correct phenotypic ratio: 9 agouti : 3 black : 4 albino. (d) 1 mark: Chi-squared (test) / \(X^2\) test.
Question 5 · Structured
10 marks
Recombinant DNA technology is used to produce human proteins, such as human growth hormone (hGH), in bacterial cells. (a) Outline why complementary DNA (cDNA), rather than genomic DNA, is used when expressing eukaryotic genes in prokaryotic host cells such as Escherichia coli. [2] (b) Describe the roles of the following enzymes in the production of recombinant plasmids: (i) restriction endonuclease [2] (ii) DNA ligase [2] (c) Plasmids often contain marker genes, such as those coding for green fluorescent protein (GFP) or antibiotic resistance. Explain the purpose of including marker genes in recombinant plasmids and how they are used. [4]
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Worked solution
Part (a): Eukaryotic genomic DNA contains introns (non-coding regions) and exons (coding regions). Prokaryotes do not possess the splicing machinery required to remove introns from primary mRNA transcripts. Therefore, complementary DNA (cDNA), which is synthesized from mature mRNA (which has already had its introns removed) using reverse transcriptase, is used so that the bacterial cells can directly translate the continuous open reading frame into the correct protein. Part (b)(i): Restriction endonucleases recognize specific target sequences of DNA (restriction sites) and cut the phosphodiester bonds of the DNA backbone. This produces either blunt or sticky ends, allowing the gene of interest and the plasmid vector to be cut with the same enzyme to produce complementary sticky ends. Part (b)(ii): DNA ligase catalyzes the formation of phosphodiester bonds between the sugar-phosphate backbones of the cut plasmid and the gene of interest, annealing the complementary sticky ends together to form a stable recombinant DNA molecule. Part (c): Marker genes are used to identify which host bacterial cells have successfully taken up the recombinant plasmid (transformation). For example, antibiotic resistance genes allow only the transformed bacteria to survive and grow on a selective agar medium containing that antibiotic (non-transformed cells die). Alternatively, a GFP marker gene causes transformed cells to fluoresce green under UV light, allowing easy visual identification and selection without killing the host cells.
Marking scheme
(a) Max 2 marks: 1. eukaryotic genomic DNA contains introns; 2. prokaryotes/bacteria lack the mechanism to splice / remove introns (from primary mRNA); 3. cDNA contains only exons / coding sequences, allowing direct translation. (b)(i) Max 2 marks: 1. cuts DNA at specific recognition / restriction sites; 2. cuts phosphodiester bonds (of sugar-phosphate backbone); 3. produces complementary sticky ends (or blunt ends). (b)(ii) Max 2 marks: 1. seals nicks / joins gene of interest to plasmid vector; 2. forms phosphodiester bonds (between adjacent nucleotides); 3. joins sugar-phosphate backbones together. (c) Max 4 marks: 1. marker genes identify/select transformed bacteria / cells that have taken up the plasmid; 2. distinguishes them from non-transformed cells; 3. antibiotic resistance marker: cells grown on agar containing antibiotic, only those with plasmid survive / replicate; 4. GFP marker: transformed cells fluoresce under UV light; 5. screening allows selection for cloning/scaling up production.
Question 6 · Structured
10 marks
The regulation of blood glucose concentration is a key homeostatic mechanism in mammals. (a) Describe how a rise in blood glucose concentration is detected by the pancreas and how insulin is subsequently secreted by the beta cells of the islets of Langerhans. [4] (b) Explain how insulin acts on liver cells (hepatocytes) to decrease blood glucose concentration. [4] (c) Distinguish between the terms glycogenesis and gluconeogenesis. [2]
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Worked solution
Part (a): A rise in blood glucose concentration is detected by receptors on the cell surface membranes of beta (\(\beta\)) cells in the islets of Langerhans of the pancreas. Glucose enters the beta cells via GLUT2 facilitated diffusion transporters. Inside the cell, glucose is phosphorylated by glucokinase and respired to produce ATP. The resulting increase in the ATP to ADP ratio causes ATP-sensitive potassium (K+) channels to close. This depolarizes the cell surface membrane, opening voltage-gated calcium (Ca2+) channels. Calcium ions diffuse into the cell, triggering the exocytosis of vesicles containing insulin into the bloodstream. Part (b): Insulin binds to specific tyrosine kinase receptors on the cell surface membranes of hepatocytes. This triggers an intracellular signaling cascade that: 1. Causes vesicles containing GLUT4 glucose transporter proteins to fuse with the cell surface membrane, increasing its permeability to glucose. 2. Activates enzymes (e.g., glucokinase) to phosphorylate glucose, keeping intracellular glucose concentrations low to maintain a concentration gradient. 3. Activates glycogen synthase to convert glucose to glycogen (glycogenesis). 4. Increases the rate of glycolysis and conversion of glucose to fatty acids. Part (c): Glycogenesis is the synthesis of glycogen from glucose molecules. Gluconeogenesis is the synthesis of glucose from non-carbohydrate sources, such as lactate, glycerol, and amino acids.
Marking scheme
(a) Max 4 marks: 1. glucose enters beta cells via facilitated diffusion (GLUT transporters); 2. glucose is respired, increasing ATP concentration / ATP:ADP ratio; 3. ATP-sensitive potassium channels close, causing membrane depolarisation; 4. voltage-gated calcium channels open and calcium ions (Ca2+) enter the cell; 5. calcium ions trigger exocytosis of insulin-containing vesicles. (b) Max 4 marks: 1. insulin binds to specific receptors on hepatocyte cell membrane; 2. triggers cell signaling cascade / secondary messengers; 3. causes vesicle fusion with membrane to add more glucose transporter (GLUT) proteins / increases permeability to glucose; 4. activates enzymes for glycogenesis / conversion of glucose to glycogen; 5. activates enzymes for glycolysis / lipid synthesis; 6. maintains glucose concentration gradient into the cell. (c) Max 2 marks: 1. Glycogenesis: conversion/synthesis of glucose to glycogen; 2. Gluconeogenesis: production of glucose from non-carbohydrate precursors / amino acids / glycerol / lactate.
Paper 32 Advanced Practical Skills 2
Answer all questions. Use a sharp pencil for any diagrams or graphs.
2 Question · 40 marks
Question 1 · Practical Investigations
20 marks
Paper 32 Advanced Practical Skills 2
**Question 1**
You are required to investigate the effect of the concentration of an inhibitor, copper sulfate solution (\textbf{I}), on the activity of the enzyme amylase.
(a) (i) Complete Table 1.1 to show how you would dilute the stock \(2.0\%\) copper sulfate solution to prepare five different concentrations of copper sulfate solution: \(2.0\%\), \(1.5\%\), \(1.0\%\), \(0.5\%\), and \(0.0\%\), each with a final volume of \(10.0\text{ cm}^3\).
(ii) State the independent variable and the dependent variable in this investigation. [2]
(b) A student performed this investigation by mixing \(1\text{ cm}^3\) of each copper sulfate solution with \(2\text{ cm}^3\) of amylase solution, incubating the mixture for 2 minutes, and then adding \(2\text{ cm}^3\) of starch suspension. Samples were tested with iodine solution on a spotting tile every 10 seconds. The results are shown in Table 1.2.
Table 1.2 | Concentration of copper sulfate / % | Time taken for starch to be completely hydrolyzed / s | | :---: | :---: | | 0.0 | 40 | | 0.5 | 70 | | 1.0 | 120 | | 1.5 | 210 | | 2.0 | No change after 300 s |
Process these data to calculate the rate of reaction in \(\text{s}^{-1}\) using the formula: \[ \text{rate} = \frac{1000}{\text{time taken}} \] For the \(2.0\%\) concentration, assume the rate of reaction is \(0\text{ s}^{-1}\). Present the processed results in a suitable table, including both the concentration of copper sulfate and the calculated rate of reaction, to an appropriate number of significant figures. [4]
(c) Plot a graph of the rate of reaction (y-axis) against the concentration of copper sulfate (x-axis) on a grid. [4]
(d) Explain the effect of copper sulfate concentration on the activity of amylase, referencing the experimental data. [4]
(e) (i) The student identified that determining the "end point" (when the blue-black color of iodine no longer appears) is a major source of error. Suggest why this is a source of error. [1] (ii) Suggest an improvement to the method that would allow more accurate and objective measurement of the rate of starch hydrolysis. [2]
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### Step-by-Step Solution
**(a) (i) Dilution Calculations:** Using the formula \(C_1 V_1 = C_2 V_2\) where \(C_1 = 2.0\%\) and \(V_2 = 10.0\text{ cm}^3\): - For \(1.5\%\): \(V_1 = \frac{1.5 \times 10.0}{2.0} = 7.5\text{ cm}^3\) of I. \(10.0 - 7.5 = 2.5\text{ cm}^3\) of W. - For \(1.0\%\): \(V_1 = \frac{1.0 \times 10.0}{2.0} = 5.0\text{ cm}^3\) of I. \(10.0 - 5.0 = 5.0\text{ cm}^3\) of W. - For \(0.5\%\): \(V_1 = \frac{0.5 \times 10.0}{2.0} = 2.5\text{ cm}^3\) of I. \(10.0 - 2.5 = 7.5\text{ cm}^3\) of W.
**(a) (ii) Variables:** - Independent: Concentration of copper sulfate solution (inhibitor) / %. - Dependent: Time taken for starch to be completely hydrolyzed / s.
**(b) Processing Data:** Using \(\text{rate} = 1000 / \text{time}\): - At \(0.0\%\): \(1000 / 40 = 25.0\text{ s}^{-1}\) - At \(0.5\%\): \(1000 / 70 = 14.3\text{ s}^{-1}\) - At \(1.0\%\): \(1000 / 120 = 8.3\text{ s}^{-1}\) - At \(1.5\%\): \(1000 / 210 = 4.8\text{ s}^{-1}\) - At \(2.0\%\): Given as \(0.0\text{ s}^{-1}\)
**(c) Graph Plotting:** Ensure axes are fully labeled with units. Plot points with small 'x' or encircled dot. Join points with straight ruled lines or a smooth curve.
**(d) Explanation:** Copper ions (heavy metal ions) function as inhibitors. They bind to the enzyme amylase at a site other than the active site (allosteric site) or interact with disulfide bonds in the enzyme's tertiary structure. This alters the three-dimensional conformation of the enzyme, changing the shape of the active site. The substrate (starch) can no longer fit into the active site, preventing the formation of enzyme-substrate complexes, reducing the rate of reaction to zero at high concentrations (\(2.0\%\)).
**(e) Error & Improvement:** - Error: End point detection is highly subjective because the color changes gradually from blue-black to purple, reddish-brown, and finally yellow-brown. Identifying the exact moment starch is fully digested is inconsistent. - Improvement: Use a colorimeter calibrated with a blank. Measure absorbance of starch-iodine solution every 10 seconds. A stable low absorbance reading indicates the objective end point.
Marking scheme
**(a) (i) [3 marks total]** - 1 mark: All volumes of stock copper sulfate solution (I) calculated correctly (10.0, 7.5, 5.0, 2.5, 0.0). - 1 mark: All volumes of distilled water (W) calculated correctly (0.0, 2.5, 5.0, 7.5, 10.0). - 1 mark: All volumes in the table expressed to 1 decimal place (e.g. 10.0, 5.0).
**(a) (ii) [2 marks total]** - 1 mark: Independent variable identified as "concentration of copper sulfate / %". - 1 mark: Dependent variable identified as "time taken for starch to be hydrolyzed / s" or "rate of reaction / s⁻¹".
**(b) [4 marks total]** - 1 mark: Table compiled with clear headings including units for both columns ("Concentration of copper sulfate / %" and "Rate of reaction / s⁻¹"). - 1 mark: Calculated rates for 0.0%, 0.5%, 1.0%, and 1.5% correct (25.0, 14.3, 8.3, 4.8). - 1 mark: Rate for 2.0% listed as 0.0. - 1 mark: All rates rounded consistently to 1 decimal place or 2-3 significant figures.
**(c) [4 marks total]** - 1 mark: Axes correctly orientated (y-axis = Rate, x-axis = Concentration) and labeled with units. - 1 mark: Scale chosen so that the plotted points occupy more than 50% of both horizontal and vertical grids. - 1 mark: All 5 points plotted accurately using a small cross (x) or dot-in-circle. - 1 mark: Points connected with thin, clean, straight ruled lines or a smooth curve of best fit, with no extrapolation beyond the plotted points.
**(d) [4 marks total]** - 1 mark: States that increasing copper sulfate concentration decreases the rate of reaction. - 1 mark: Refers to data from the experiment (e.g., rate decreases from 25.0 s⁻¹ to 0.0 s⁻¹ as concentration increases from 0.0% to 2.0%). - 1 mark: Identifies copper ions as non-competitive/heavy metal inhibitors that bind to the enzyme (allosteric site/disulfide bridges). - 1 mark: Explains that this changes the tertiary structure/shape of the active site, meaning substrate can no longer bind / no enzyme-substrate complexes can form.
**(e) (i) [1 mark total]** - 1 mark: Explains that the visual determination of the color change end point is subjective / difficult to judge consistently.
**(e) (ii) [2 marks total]** - 1 mark: Suggests using a colorimeter. - 1 mark: To measure absorbance or transmission of the samples mixed with iodine at regular time intervals, giving a quantitative/objective measurement.
Question 2 · Practical Investigations
20 marks
Paper 32 Advanced Practical Skills 2
**Question 2**
Figure 2.1 is a photomicrograph of a transverse section through the stem of a typical herbaceous dicotyledonous plant, *Ranunculus*.
(a) Draw a large, low-power plan diagram of the segment of the stem shown in Figure 2.1, representing approximately one-quarter of the stem section. Your drawing should show the distribution of tissues (such as epidermis, cortex, vascular bundles, and pith) but should not show individual cells. Use a sharp pencil. [5]
(b) Figure 2.2 shows a high-power view of a single vascular bundle from the same stem section.
Make a large, detailed drawing of a group of three adjacent xylem vessel elements and two adjacent sclerenchyma (cap) fibres. Label one xylem vessel and one sclerenchyma cell on your drawing. [5]
(c) The actual diameter of a xylem vessel in the stem is \(60\ \mu\text{m}\). A student uses a light microscope with an eyepiece graticule to measure the same xylem vessel.
Under the high-power objective (\(\times 400\)), the stage micrometer calibration shows that \(1\) eyepiece graticule unit (\text{epu}) is equal to \(2.5\ \mu\text{m}\).
(i) Calculate the expected diameter of this xylem vessel in eyepiece graticule units (\text{epu}). Show your working. [2]
(ii) The student then measures the vessel in a drawing they made. The diameter of the vessel in the drawing is \(36\text{ mm}\). Calculate the magnification of the student's drawing compared to the actual size of the vessel (\(60\ \mu\text{m}\)). Show your working and state your answer to an appropriate number of significant figures. [3]
(d) Describe how the structure of a xylem vessel element is adapted to its function of transporting water and mineral ions under tension. [5]
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### Step-by-Step Solution
**(a) Plan Drawing:** - Draw a sector representing one-quarter of the stem. - Use a sharp pencil to draw clean, single, continuous lines. Do not shade. - Represent the tissue layers proportionally: narrow outer epidermis, cortex layer, a ring of discrete vascular bundles, and large central pith.
**(b) High-Power Drawing:** - Draw only the requested cells (three xylem vessels and two sclerenchyma fibres). - Draw double-lines for cell walls to show thickness. - Sclerenchyma cells must have much thicker walls and smaller lumens than xylem vessels. - Label one xylem vessel and one sclerenchyma cell with straight, uncrossed label lines.
**(c) (i) Eyepiece Graticule Unit Calculation:** - Actual size = \(60\ \mu\text{m}\) - Calibration: \(1\text{ epu} = 2.5\ \mu\text{m}\) - \(\text{Diameter in epu} = \frac{60\ \mu\text{m}}{2.5\ \mu\text{m/epu}} = 24\text{ epu}\)
**(d) Xylem Adaptations:** - Thick walls containing lignin: prevents collapse of vessels under high tension (negative pressure) and provides structural support. - Absence of living contents (no cytoplasm, nucleus, or organelles): reduces resistance to water flow. - Loss of end-walls: creates a continuous open column for uninterrupted water movement. - Pits (non-lignified areas in walls): allow lateral water transport to adjacent cells.
Marking scheme
**(a) [5 marks total]** - 1 mark: Clean, single, continuous lines drawn with a sharp pencil. No shading or sketching lines. - 1 mark: Plan shows the correct sector of the stem (approximately one-quarter / quadrant) with outer and inner boundary lines. - 1 mark: Draws at least 3 separate vascular bundles arranged in a distinct arc. - 1 mark: No individual cells are drawn anywhere in the plan diagram. - 1 mark: Correct proportion of tissues shown (epidermis is thin, vascular bundles are correctly located between cortex and pith).
**(b) [5 marks total]** - 1 mark: Draws exactly three xylem vessels and two sclerenchyma cells, all adjacent to each other. - 1 mark: Cells drawn with double-line walls to represent thickness. - 1 mark: Sclerenchyma cells drawn with significantly thicker walls and smaller lumens compared to the larger, thinner-walled xylem vessel elements. - 1 mark: Xylem vessels drawn with characteristic angular or rounded, open lumens. - 1 mark: Correct labeling of one "xylem vessel" and one "sclerenchyma cell" with straight, ruled lines touching the respective structures.
**(c) (ii) [3 marks total]** - 1 mark: Converts drawing diameter correctly from mm to \(\mu\text{m}\) (\(36\text{ mm} = 36,000\ \mu\text{m}\)). - 1 mark: Shows division of drawing size by actual size (\(\frac{36,000}{60}\)). - 1 mark: Correct final magnification of \(\times 600\) (accept "600" or "x600", must be a whole number or to an appropriate number of significant figures).
**(d) [5 marks total]** - 1 mark: Walls are thickened with lignin to withstand tension / negative pressure without collapsing. - 1 mark: Xylem vessels are dead / contain no cytoplasm / no organelles, creating a hollow lumen with low resistance to water flow. - 1 mark: End-walls are broken down / absent, allowing a continuous, uninterrupted column of water. - 1 mark: Pits in the lateral walls allow the movement of water sideways / bypass blockages. - 1 mark: Narrow lumen allows adhesion of water molecules to hydrophilic wall surface, supporting the water column.
Paper 42 A Level Structured Questions
Answer all questions. Show your working and use appropriate units.
10 Question · 100 marks
Question 1 · Structured
10 marks
Pyruvate kinase is a key enzyme in glycolysis that catalyses the final step of the pathway, converting phosphoenolpyruvate (PEP) to pyruvate with the concurrent production of ATP. It is allosterically regulated by fructose 1,6-bisphosphate (FBP), which acts as an activator, and by ATP, which acts as an inhibitor.
(a) Describe how pyruvate kinase lowers the activation energy of the reaction. [3] (b) Explain how the binding of FBP (an allosteric activator) affects the activity of pyruvate kinase. [3] (c) Distinguish between competitive and non-competitive inhibition of enzymes such as pyruvate kinase. [4]
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(a) Pyruvate kinase lowers activation energy by holding the substrate (PEP) in a specific orientation within the active site, putting strain on the chemical bonds of the substrate to make them break more easily, providing an alternative reaction pathway with a lower activation energy, and forming temporary bonds (e.g., hydrogen or ionic) between the enzyme's active site R-groups and the substrate.
(b) FBP binds to an allosteric site (a site other than the active site) of pyruvate kinase. This binding induces a conformational change (change in 3D shape) of the enzyme, which alters the shape of the active site. The active site becomes more complementary to the substrate (PEP), increasing the affinity of the enzyme for the substrate, thus increasing the rate of reaction.
(c) Competitive inhibitors have a molecular shape similar to the substrate and bind directly to the active site, blocking substrate entry. Their inhibitory effect can be overcome by increasing substrate concentration. Non-competitive inhibitors bind to an allosteric site on the enzyme, causing a conformational change that alters the shape of the active site so the substrate can no longer bind or react. Their effect cannot be overcome by increasing substrate concentration, and they decrease the maximum velocity \(V_{\max}\) of the enzyme.
Marking scheme
(a) [Max 3 marks]: 1. Holds substrate / PEP in a specific orientation; 2. Puts strain on bonds within the substrate to facilitate breakage; 3. Provides an alternative reaction pathway with lower activation energy; 4. R-groups of active site amino acids form temporary / transition state bonds with substrate.
(b) [Max 3 marks]: 1. FBP binds to allosteric site / site other than active site; 2. Causes conformational change / change in tertiary structure / 3D shape of enzyme; 3. Active site shape changes to become (more) complementary to substrate / PEP; 4. Increases affinity of enzyme for substrate / increases rate of reaction.
(c) [Max 4 marks]: 1. Competitive: inhibitor has similar shape to substrate AND binds to active site; 2. Non-competitive: inhibitor binds to allosteric site / site other than active site; 3. Competitive: blocks substrate from entering active site / competes with substrate; 4. Non-competitive: causes conformational change that alters active site shape (preventing substrate binding / reaction); 5. Competitive: can be overcome by increasing substrate concentration (\(V_{\max}\) unchanged); 6. Non-competitive: cannot be overcome by increasing substrate concentration (\(V_{\max}\) decreased).
Question 2 · Structured
10 marks
An investigation was conducted to determine the effect of copper ions (\(Cu^{2+}\)), which act as non-competitive inhibitors, on the rate of starch hydrolysis by amylase.
(a) State three variable factors, other than the concentration of copper ions, that must be controlled in this investigation and suggest how each would be controlled. [3] (b) Describe and explain the expected results as copper ion concentration increases from \(0\) to \(0.5\text{ mol dm}^{-3}\). [4] (c) Explain why using immobilized amylase instead of free amylase is advantageous in industrial processes. [3]
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Worked solution
(a) Temperature should be controlled using a thermostatically-controlled water bath. pH should be controlled using a buffer solution. Amylase concentration and starch concentration should be controlled by using the same volume and concentration of stock solutions.
(b) As copper ion concentration increases, the rate of starch hydrolysis decreases. This is because copper ions act as non-competitive inhibitors, binding to the allosteric site of amylase. This causes a change in the tertiary structure of the enzyme, altering the shape of the active site so that starch can no longer bind. Fewer enzyme-substrate complexes form. At very high concentrations of copper ions, the rate will reach a low plateau as all enzymes are fully inhibited.
(c) Immobilized enzymes can be easily recovered and reused, reducing production costs. They are more stable at wider ranges of temperature and pH due to the stabilizing effect of the support matrix. The final product is not contaminated with the enzyme, eliminating the need for expensive downstream purification steps.
Marking scheme
(a) [Max 3 marks - 1 mark for each factor and control method]: 1. Temperature - using a thermostatically-controlled water bath; 2. pH - using a buffer solution; 3. Enzyme / substrate concentration - using a constant volume and concentration of enzyme/starch stock solutions.
(b) [Max 4 marks]: 1. As copper ion concentration increases, the rate of starch hydrolysis decreases; 2. Copper ions bind to the allosteric site of amylase; 3. Causes conformational change / change in tertiary structure; 4. Active site shape changes so it is no longer complementary to starch / substrate; 5. Fewer enzyme-substrate complexes (ESCs) form per unit time / rate reaches a low plateau.
(c) [Max 3 marks]: 1. Enzyme can be easily recovered / reused; 2. Product is free from enzyme contamination / no downstream purification needed; 3. Greater thermal stability / stability over a wider range of pH (due to protective support matrix).
Question 3 · Structured
10 marks
Succinate dehydrogenase is an enzyme of both the citric acid (Krebs) cycle and the electron transport chain, located in the inner mitochondrial membrane. It catalyses the oxidation of succinate to fumarate. Malonate is a structural analogue of succinate and acts as a competitive inhibitor of this enzyme.
(a) Outline the role of succinate dehydrogenase in the Krebs cycle. [2] (b) Suggest and explain the effect of adding an excess of succinate to a mitochondrial suspension containing a constant concentration of malonate. [3] (c) Explain the role of NAD and FAD in the electron transport chain during aerobic respiration. [5]
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Worked solution
(a) Succinate dehydrogenase catalyses the dehydrogenation (oxidation) of succinate to fumarate. The hydrogens removed from succinate are transferred to FAD, reducing it to \(FADH_2\).
(b) Adding an excess of succinate will overcome the competitive inhibition of malonate. This increases the rate of succinate oxidation back to near-normal levels, because the substrate molecules outcompete the malonate molecules for binding to the active site of succinate dehydrogenase.
(c) Reduced NAD and reduced FAD transfer hydrogen atoms (protons and electrons) to the electron transport chain on the inner mitochondrial membrane (cristae). The electrons are passed along a series of electron carriers in a series of redox reactions, releasing energy. This energy is used to pump protons (\(H^+\)) from the matrix into the intermembrane space, creating a proton gradient. Protons diffuse back into the matrix through ATP synthase (chemiosmosis), driving the synthesis of ATP from ADP and inorganic phosphate. Oxygen acts as the final electron acceptor, combining with protons and electrons to form water.
Marking scheme
(a) [Max 2 marks]: 1. Catalyses the dehydrogenation / oxidation of succinate to fumarate; 2. Transfers hydrogen / electrons to FAD to form reduced FAD / \(FADH_2\).
(b) [Max 3 marks]: 1. Rate of reaction / succinate oxidation increases; 2. Succinate outcompetes malonate for the active site of the enzyme; 3. High substrate concentration reduces the probability of inhibitor binding (to active site); 4. Maximum rate of reaction (\(V_{\max}\)) can still be achieved.
(c) [Max 5 marks]: 1. Reduced NAD / FAD donate electrons and protons / H atoms to the ETC; 2. Electrons pass along a series of electron carriers / cytochromes via redox reactions; 3. Energy is released as electrons pass along the chain; 4. This energy is used to pump protons (\(H^+\)) into the intermembrane space; 5. Creates a proton / electrochemical / pH gradient; 6. Protons diffuse down their gradient through ATP synthase (chemiosmosis) to generate ATP; 7. Oxygen acts as the final electron (and proton) acceptor to form water.
Question 4 · Structured
10 marks
An ecologist investigated the plant biodiversity of two grassland areas, Area X and Area Y, by counting the individuals of three species. The results are shown below: Area X: Species A = 50, Species B = 30, Species C = 20 (Total \(N = 100\)) Area Y: Species A = 90, Species B = 8, Species C = 2 (Total \(N = 100\))
(a) Calculate the Simpson's Index of Diversity (\(D\)) for Area X and Area Y using the formula: \(D = 1 - \sum \left(\frac{n}{N}\right)^2\) Show your working. [4] (b) Explain what the calculated values of \(D\) indicate about the biodiversity and stability of Area X compared to Area Y. [3] (c) Describe how random sampling using quadrats can be used to estimate the abundance of plant species in a grassland. [3]
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Worked solution
(a) For Area X: \(n/N\) values: Species A = 50/100 = 0.5; Species B = 30/100 = 0.3; Species C = 20/100 = 0.2. \((n/N)^2\) values: 0.25, 0.09, 0.04. \(\sum (n/N)^2 = 0.25 + 0.09 + 0.04 = 0.38\). \(D = 1 - 0.38 = 0.62\).
For Area Y: \(n/N\) values: Species A = 90/100 = 0.9; Species B = 8/100 = 0.08; Species C = 2/100 = 0.02. \((n/N)^2\) values: 0.81, 0.0064, 0.0004. \(\sum (n/N)^2 = 0.81 + 0.0064 + 0.0004 = 0.8168\). \(D = 1 - 0.8168 = 0.1832\) (or 0.18).
(b) Area X has a higher \(D\) value (0.62) than Area Y (0.18), indicating higher biodiversity and greater species evenness. Area X is more stable than Area Y because a change in one species (e.g., due to disease or environmental change) is less likely to disrupt the whole community, as there are alternative species to maintain food webs and community structure.
(c) Lay out two tape measures at right angles to create a coordinate grid. Use a random number generator to obtain pairs of coordinates to place the quadrat. Count the number of individuals of each species or estimate percentage cover within the quadrat. Repeat with a large number of quadrats (minimum 10-15) and calculate the mean per quadrat, then scale up to the total area of the grassland.
Marking scheme
(a) [Max 4 marks]: 1. Area X: Correct calculation of \((n/N)^2\) values (0.25, 0.09, 0.04) and sum (0.38); 2. Area X: Correct final value of \(D = 0.62\); 3. Area Y: Correct calculation of \((n/N)^2\) values (0.81, 0.0064, 0.0004) and sum (0.8168); 4. Area Y: Correct final value of \(D = 0.18\) or \(0.183\) / \(0.1832\).
(b) [Max 3 marks]: 1. Area X has higher biodiversity / species evenness than Area Y; 2. Area X is more stable than Area Y; 3. In Area X, an environmental change / disease is less likely to affect the ecosystem / food web because alternative species exist; 4. Area Y is dominated by a single species (Species A), making it vulnerable.
(c) [Max 3 marks]: 1. Use tape measures to set up a grid system / coordinate axes; 2. Use a random number generator / table to select coordinates for quadrat placement (to avoid bias); 3. Record species presence / count individuals / estimate percentage cover in each quadrat; 4. Repeat many times (e.g. minimum 10 times) to obtain a representative sample and calculate a mean.
Question 5 · Structured
10 marks
The black rhinoceros (*Diceros bicornis*) is critically endangered due to habitat loss and poaching. Conservation efforts involve both *in-situ* and *ex-situ* strategies.
(a) Distinguish between *in-situ* and *ex-situ* conservation, providing one example of each method used for the black rhinoceros. [3] (b) Explain the role of captive breeding programmes in zoos in conserving endangered species such as the black rhinoceros, and discuss the difficulties associated with these programmes. [5] (c) State two reasons why maintaining biodiversity is of economic importance to local human populations. [2]
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Worked solution
(a) In-situ conservation involves protecting species in their natural habitat (e.g., establishing national parks, game reserves, or anti-poaching patrols for black rhinos). Ex-situ conservation involves protecting species outside their natural habitat (e.g., captive breeding in zoos or keeping rhinos in wildlife sanctuaries/safari parks outside their native range).
(b) Captive breeding programmes aim to increase the population size of endangered species, maintain genetic diversity through managed breeding (using studbooks to avoid inbreeding), and eventually reintroduce individuals back into the wild. Difficulties include: animals refusing to breed in captivity due to stress or lack of natural stimuli; small gene pools leading to inbreeding depression; loss of wild behaviors/learned survival skills; and high financial costs of maintaining large mammals in zoos.
(c) Economic benefits include ecotourism, which generates income and employment for local communities, and the preservation of ecosystem services (such as clean water, soil stability, and pollination) which support local agriculture and prevent natural disasters.
Marking scheme
(a) [Max 3 marks]: 1. In-situ: conserving species in their natural habitat AND Ex-situ: conserving species outside their natural habitat; 2. In-situ example: national parks / wildlife reserves / anti-poaching patrols / habitat restoration; 3. Ex-situ example: zoos / captive breeding facilities / safari parks.
(b) [Max 5 marks]: 1. Role: to increase population size of endangered species; 2. Role: to maintain genetic diversity / prevent inbreeding by using studbooks / genetic testing / exchanging animals between zoos; 3. Role: to reintroduce animals back into the wild / native habitat; 4. Difficulty: low reproductive success / stress of captivity prevents mating; 5. Difficulty: small gene pool / risk of inbreeding depression / loss of genetic variation; 6. Difficulty: habituation to humans / loss of natural behaviors / survival skills (e.g., foraging, avoiding predators); 7. Difficulty: high cost / space requirements.
(c) [Max 2 marks]: 1. Ecotourism generates revenue / provides jobs for local populations; 2. Sources of medicine / genetic resources for crop breeding / agriculture; 3. Ecosystem services (e.g., erosion control, water purification, flood prevention) protect local infrastructure / agricultural land.
Question 6 · Structured
10 marks
In sweet peas (*Lathyrus odoratus*), the gene for flower colour has two alleles: purple (\(B\)) is dominant to red (\(b\)). The gene for pollen grain shape also has two alleles: long (\(L\)) is dominant to round (\(l\)). These two gene loci are autosomally linked. A cross was carried out between a dihybrid plant with purple flowers and long pollen (\(BbLl\)) and a homozygous recessive plant with red flowers and round pollen (\(bbll\)).
(a) Explain what is meant by the terms *autosomal linkage* and *crossing over*. [4] (b) The offspring from the cross described above were: Purple flowers, long pollen: 242 Purple flowers, round pollen: 28 Red flowers, long pollen: 32 Red flowers, round pollen: 238 Calculate the crossover value (recombination frequency) for these two genes. Show your working. [3] (c) Explain how crossing over during meiosis produces recombinant phenotypes in the offspring. [3]
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Worked solution
(a) Autosomal linkage refers to two or more genes being located on the same autosome (non-sex chromosome) and therefore tending to be inherited together because they do not assort independently during meiosis. Crossing over is the exchange of alleles/segments of non-sister chromatids between homologous chromosomes during prophase I of meiosis.
(b) The recombinant phenotypes are the non-parental combinations: purple-round (28) and red-long (32). Total recombinants = \(28 + 32 = 60\). Total offspring = \(242 + 28 + 32 + 238 = 540\). Recombination frequency (crossover value) = \(\frac{\text{Total recombinants}}{\text{Total offspring}} \times 100 = \frac{60}{540} \times 100 = 11.1\%\) (or 11.11%).
(c) During prophase I of meiosis, homologous chromosomes pair up to form bivalents. Chiasmata form where non-sister chromatids break and rejoin, exchanging alleles between paternal and maternal chromosomes. This breaks the linkage between the genes on the same chromosome, producing recombinant gametes (\(Bl\) and \(bL\)) in the heterozygous parent. When these recombinant gametes fuse with the gametes of the homozygous recessive parent (\(bl\)), recombinant phenotypes (purple/round and red/long) are produced.
Marking scheme
(a) [Max 4 marks]: 1. Autosomal linkage: two or more genes located on the same autosome / non-sex chromosome; 2. Inherited together / do not undergo independent assortment (unless crossing over occurs); 3. Crossing over: exchange of genetic material / alleles between non-sister chromatids; 4. Occurs between homologous chromosomes during prophase I of meiosis.
(b) [Max 3 marks]: 1. Identifies recombinants as purple, round (28) AND red, long (32) [Total = 60]; 2. Identifies total offspring as 540; 3. Correct calculation: \(\frac{60}{540} \times 100 = 11.1\%\) (accept 11% or 11.11%).
(c) [Max 3 marks]: 1. Homologous chromosomes pair up / bivalent forms (during prophase I); 2. Chiasmata form where non-sister chromatids break and rejoin, exchanging alleles / DNA segments; 3. This separates linked alleles / creates new combinations of alleles on chromatids / produces recombinant gametes (\(Bl\) and \(bL\)); 4. Fusion of recombinant gametes with gametes from homozygous recessive parent results in recombinant phenotypes.
Question 7 · Structured
10 marks
Recombinant DNA technology is used to produce human insulin in genetically modified *Escherichia coli* bacteria.
(a) Explain why complementary DNA (cDNA), synthesized using reverse transcriptase, is used to obtain the human insulin gene for expression in bacteria, rather than using genomic DNA. [3] (b) Describe the role of restriction endonucleases and DNA ligase in the production of a recombinant plasmid. [4] (c) Explain how marker genes, such as those coding for fluorescent proteins, are used to identify bacteria that have successfully taken up the recombinant plasmid. [3]
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Worked solution
(a) Human genomic DNA contains introns (non-coding sequences) as well as exons (coding sequences). Bacteria do not possess the splicing machinery (spliceosomes) required to remove introns from pre-mRNA. If genomic DNA were used, the bacteria would translate the introns, resulting in a non-functional, oversized protein. Using cDNA, which is synthesized from mature mRNA that has already had its introns removed, ensures the bacteria can translate the continuous coding sequence into functional human insulin.
(b) Restriction endonucleases cut both the human DNA (containing the insulin gene) and the plasmid vector at specific, palindromic recognition sites. This produces complementary single-stranded overhanging ends known as sticky ends. DNA ligase is then used to catalyze the formation of phosphodiester bonds between the sugar-phosphate backbones of the plasmid DNA and the insert DNA, sealing the nick and permanently joining them into a single recombinant plasmid.
(c) Marker genes are incorporated into the plasmid alongside the insulin gene. If green fluorescent protein (GFP) is used, the bacteria that have successfully taken up the plasmid will express the GFP gene and glow green under ultraviolet (UV) light. This allows researchers to easily identify and select transformed colonies without killing them, unlike antibiotic selection which requires replica plating.
Marking scheme
(a) [Max 3 marks]: 1. Human genomic DNA contains introns (and exons) / non-coding regions; 2. Bacteria lack the splicing machinery / spliceosomes to remove introns; 3. cDNA is synthesized from mature mRNA (which has no introns / only exons); 4. Ensures bacteria can translate the correct sequence into functional / active insulin protein.
(b) [Max 4 marks]: 1. Restriction endonuclease cuts DNA / plasmid at specific recognition site / sequence; 2. Cuts to produce complementary sticky ends (single-stranded overhangs); 3. Same restriction enzyme is used on both human DNA and plasmid; 4. DNA ligase joins the human DNA insert and plasmid vector; 5. Forms phosphodiester bonds in the sugar-phosphate backbone.
(c) [Max 3 marks]: 1. Marker gene is integrated into plasmid alongside target gene; 2. Transformed bacteria (that have taken up plasmid) express the marker gene; 3. If GFP is used, transformed colonies fluoresce / glow under UV light; 4. Allows easy identification / selection of transformed cells / distinguishes them from non-transformed cells.
Question 8 · Structured
10 marks
Osmoregulation is an essential homeostatic mechanism in mammals that maintains the water potential of the blood within narrow limits. When water potential decreases, antidiuretic hormone (ADH) is released.
(a) Explain how osmoreceptors in the hypothalamus detect a decrease in the water potential of the blood and stimulate the release of ADH. [3] (b) Describe the mechanism of action of ADH on the cells of the collecting duct of the kidney nephron. [5] (c) Suggest the effect of a drug that blocks G-protein-coupled receptors in the collecting duct cells on urine volume and concentration. [2]
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Worked solution
(a) When blood water potential decreases, water leaves the osmoreceptor cells in the hypothalamus by osmosis down a water potential gradient. This causes the osmoreceptor cells to shrink (lose volume). This physical shrinkage stimulates sensory neurones in the hypothalamus, which generate action potentials that travel down the axons of neurosecretory cells into the posterior pituitary gland. This triggers the release of ADH into the blood capillaries via exocytosis.
(b) ADH binds to specific G-protein-coupled receptors on the cell surface membrane of the collecting duct cells. This binding activates a G-protein, which activates the enzyme adenylyl cyclase. Adenylyl cyclase catalyses the conversion of ATP to cyclic AMP (cAMP), which acts as a second messenger. cAMP activates a protein kinase cascade, leading to the fusion of intracellular vesicles containing aquaporins (water channel proteins) with the luminal (apical) cell surface membrane. This increases the permeability of the collecting duct to water, allowing water to leave the collecting duct lumen by osmosis into the hypertonic medulla.
(c) If G-protein-coupled receptors are blocked, ADH cannot bind and activate the signaling pathway. Consequently, aquaporins will not fuse with the luminal membrane. Water cannot be reabsorbed from the collecting duct, leading to the production of a large volume of dilute urine.
Marking scheme
(a) [Max 3 marks]: 1. Decrease in blood water potential causes water to leave osmoreceptors (in hypothalamus) by osmosis; 2. Osmoreceptor cells shrink / decrease in volume; 3. This stimulates / triggers action potentials in neurosecretory cells / sensory neurones; 4. Action potentials travel to the posterior pituitary gland, triggering exocytosis of ADH into the blood.
(b) [Max 5 marks]: 1. ADH binds to (specific) G-protein-coupled receptors on cell surface membrane (of collecting duct cells); 2. Activates G-protein which activates adenylyl cyclase; 3. Adenylyl cyclase converts ATP to cyclic AMP / cAMP (second messenger); 4. cAMP activates protein kinase (cascade); 5. Vesicles containing aquaporins / water channel proteins move to and fuse with the luminal / apical membrane; 6. Increases permeability of membrane to water; 7. Water moves out of collecting duct lumen into the tissue fluid / blood by osmosis (down water potential gradient).
(c) [Max 2 marks]: 1. Large / increased volume of urine; 2. Low / decreased concentration / dilute urine; 3. Explain: no aquaporins inserted into luminal membrane, so water cannot be reabsorbed (despite presence of ADH).
Question 9 · Structured
10 marks
**(a)** Homeostasis relies on self-regulating mechanisms. Define the term *negative feedback* as used in homeostasis. [2]
**(b)** Osmoregulation is a key homeostatic process in mammals. When the water potential of the blood decreases below a set point, antidiuretic hormone (ADH) is released from the posterior pituitary gland. ADH travels in the blood and binds to specific receptors on the basolateral membranes of collecting duct cells in the kidney. Describe the intracellular cell-signalling pathway that occurs within these cells after ADH binds to its receptor, leading to an increase in the permeability of the luminal membrane to water. [5]
**(c)** Nephrogenic diabetes insipidus (NDI) is a rare disorder that can be caused by an inherited mutation in the *AVPR2* gene, which codes for the ADH receptor (V2 receptor). Explain how this mutation results in the production of large volumes of dilute urine, even when the individual is dehydrated. [3]
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Worked solution
**(a)** Negative feedback is a control mechanism where a change in a parameter/factor (such as blood water potential or temperature) is detected by receptors. This triggers a corrective mechanism that reverses the direction of change to return the factor back to its normal/set level.
**(b)** The binding of ADH to the V2 receptor on the basolateral membrane activates a G-protein. This active G-protein stimulates the enzyme adenylyl cyclase, which catalyses the conversion of ATP into cyclic AMP (cAMP), acting as a second messenger. cAMP activates protein kinase A (PKA). This phosphorylation cascade triggers the movement of intracellular vesicles containing aquaporins (water channel proteins) towards the luminal/apical membrane. The vesicle membranes fuse with the luminal membrane, increasing the density of aquaporin channels, which allows water molecules to rapidly enter the cell down a water potential gradient.
**(c)** A mutation in the *AVPR2* gene changes the primary sequence of amino acids in the V2 receptor, altering its tertiary structure. Consequently, ADH can no longer bind to the receptor, or the receptor cannot activate the G-protein. Because the intracellular signalling pathway is blocked, vesicles containing aquaporins do not fuse with the luminal membrane of the collecting duct cells. The luminal membrane remains impermeable to water, so water cannot be reabsorbed by osmosis down the water potential gradient. The unabsorbed water remains in the filtrate and is excreted as a large volume of dilute urine.
Marking scheme
**(a)** 1. change / deviation from a set point / norm (in an internal environment) is detected (by a receptor); [1] 2. corrective mechanism(s) are activated to reverse the change / return the factor to the set point / norm; [1]
**(b)** 1. ADH binding (to receptor) activates a G-protein; [1] 2. active G-protein activates the enzyme adenylyl cyclase; [1] 3. adenylyl cyclase catalyses conversion of ATP to cyclic AMP / cAMP / second messenger; [1] 4. cAMP activates a kinase enzyme / protein kinase A (PKA); [1] 5. vesicle (membranes) containing aquaporins / water channel proteins move towards / fuse with the luminal / apical membrane; [1] [Max 5]
**(c)** 1. mutation alters the primary structure / amino acid sequence of the V2 receptor, changing its 3D tertiary structure; [1] 2. ADH cannot bind to the receptor (or G-protein is not activated); [1] 3. aquaporins are not inserted into the luminal membrane (which remains impermeable to water); [1] 4. water cannot be reabsorbed by osmosis (down the water potential gradient into the tissue fluid/blood); [1] [Max 3]
Question 10 · Structured
10 marks
**(a)** Enzyme tools are essential in genetic engineering and research. Contrast the functions of *restriction endonucleases* and *reverse transcriptase*. [3]
**(b)** Microarray technology allows researchers to detect and measure the expression of thousands of genes simultaneously. Describe how a microarray is used to compare the gene expression profiles of a cancerous lung cell and a normal, healthy lung cell. [5]
**(c)** Explain how the results of a microarray analysis can be used to assist in the development of personalized medical treatment for a patient with cancer. [2]
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Worked solution
**(a)** Restriction endonucleases cut double-stranded DNA molecules at specific palindromic target sequences (recognition sites) by cleaving phosphodiester bonds, which can leave either sticky or blunt ends. In contrast, reverse transcriptase synthesises a complementary DNA (cDNA) strand using a single-stranded mRNA molecule as a template, forming new phosphodiester bonds.
**(b)** To compare gene expression, mRNA is first extracted from both cancerous and healthy lung cells. Reverse transcriptase is used to transcribe the mRNA molecules into cDNA. The cDNA from the cancer cells is labelled with a specific fluorescent dye (e.g., red), and the cDNA from the healthy cells is labelled with a different coloured fluorescent dye (e.g., green). The two sets of labelled cDNA are mixed together and applied to the microarray, where they hybridize with matching single-stranded DNA probes anchored to the spots. Unbound cDNA is washed away, and the microarray is scanned with lasers to excite the fluorescent dyes. The intensity of red, green, or yellow (where both hybridize) signals at each spot indicates the relative levels of transcription (expression) of each gene.
**(c)** Microarray analysis identifies the exact genes that are overexpressed (such as oncogenes) or underexpressed (such as tumour suppressor genes) in a patient's specific tumour. This allows clinicians to select targeted therapies, such as specific monoclonal antibodies or small-molecule inhibitors, that block the products of these active genes. It prevents the use of ineffective general treatments, reduces side effects, and allows monitoring of how the tumor responds to treatment over time.
Marking scheme
**(a)** 1. restriction endonucleases cut/cleave double-stranded DNA, whereas reverse transcriptase synthesises (single-stranded) cDNA from an mRNA/RNA template; [1] 2. restriction endonucleases break phosphodiester bonds, whereas reverse transcriptase forms phosphodiester bonds; [1] 3. restriction endonucleases target specific (palindromic) recognition sites, whereas reverse transcriptase copies any mRNA molecule present; [1] 4. restriction endonucleases can produce sticky or blunt ends, whereas reverse transcriptase produces a continuous cDNA strand; [1] [Max 3]
**(b)** 1. mRNA is extracted from both cancerous and healthy cells; [1] 2. reverse transcriptase is used to make cDNA (from the mRNA templates); [1] 3. cDNA from cancerous cells is labelled with one fluorescent dye AND cDNA from healthy cells is labelled with a different fluorescent dye; [1] 4. labelled cDNA samples are mixed and hybridize with single-stranded DNA probes fixed on the microarray; [1] 5. unbound cDNA is washed off; [1] 6. microarray is scanned with laser/UV light to detect fluorescence; [1] 7. the intensity/colour of fluorescence at each spot indicates the level of expression of that gene; [1] [Max 5]
**(c)** 1. identifies which specific genes are overexpressed / underexpressed (e.g., oncogenes / tumour suppressor genes); [1] 2. allows selection of targeted drugs / therapies that specifically inhibit/target the products of those overexpressed genes; [1] 3. prevents prescribing drugs that will be ineffective or have severe side effects; [1] [Max 2]
Paper 52 Planning, Analysis and Evaluation
Answer all questions. You should show all your working.
2 Question · 30 marks
Question 1 · Experimental Planning and Evaluation
15 marks
Alkaline phosphatase is an enzyme that catalyzes the hydrolysis of phosphate esters. In an investigation, the artificial substrate p-nitrophenyl phosphate (pNPP) is used. Alkaline phosphatase converts colorless pNPP into yellow p-nitrophenyl (pNP), which absorbs light at 405 nm.
A student wants to investigate the effect of different concentrations of a competitive inhibitor, phenylalanine, on the initial rate of this reaction.
The student is provided with: - 10.0% (w/v) stock solution of phenylalanine (inhibitor) - 1.0% (w/v) alkaline phosphatase enzyme solution - 0.05 mol dm\(^{-3}\) pNPP substrate solution - Buffer solution at pH 9.0 - Standard laboratory glassware and access to a colorimeter with a 405 nm filter.
(a) Describe a method the student could use to carry out this investigation.
Your description should be detailed enough to allow another person to follow it, and should include: - the preparation of five different concentrations of phenylalanine using a simple or serial dilution of the stock 10.0% solution - how the dependent variable will be measured - how key confounding variables will be controlled - a suitable control experiment - the safety precautions that should be taken. [9]
(b) (i) The student plotted a graph of the initial rate of reaction against substrate concentration at two different concentrations of phenylalanine. State how the student could use this graph to confirm that phenylalanine is a competitive inhibitor and not a non-competitive inhibitor. [2]
(ii) The student carried out 10 replicates at each inhibitor concentration and calculated the mean initial rate and standard deviation. Explain why calculating the standard deviation of the replicate data is more useful than finding the range. [2]
(iii) State the statistical test the student should use to determine whether there is a significant difference between the mean initial rates of reaction at two different inhibitor concentrations, and justify your choice. [2]
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Worked solution
(a) Method: 1. Prepare five concentrations of phenylalanine (e.g., 10.0%, 8.0%, 6.0%, 4.0%, and 2.0%) using the 10.0% stock solution. For example, to prepare 10 cm\(^{3}\) of 2.0% phenylalanine, mix 2.0 cm\(^{3}\) of 10.0% stock solution with 8.0 cm\(^{3}\) of distilled water/buffer in a test-tube. Do this for each required concentration. 2. Set up a colorimeter with a 405 nm filter and calibrate (zero) it using a blank consisting of buffer, substrate, and distilled water (without enzyme). 3. In a reaction tube, mix 2.0 cm\(^{3}\) of the substrate (pNPP), 1.0 cm\(^{3}\) of the pH 9.0 buffer, and 1.0 cm\(^{3}\) of the selected concentration of phenylalanine. 4. Place this tube and a tube of the enzyme solution in a water bath at a constant temperature (e.g., 37 \(^{\circ}\)C) for 5 minutes to equilibrate. 5. Add 1.0 cm\(^{3}\) of the enzyme solution to the reaction mixture, quickly transfer it to a cuvette, and place it in the colorimeter. 6. Measure and record the absorbance at 405 nm every 10 seconds for 2 minutes. 7. Plot a graph of absorbance against time and determine the gradient of the initial linear portion of the curve to find the initial rate of reaction. 8. Repeat the procedure with the remaining concentrations of phenylalanine. 9. Set up a control experiment where the enzyme solution is replaced with boiled, denatured enzyme (or distilled water) to prove that any absorbance change is enzymatically catalyzed. 10. Safety precautions: Wear safety goggles and lab coat. Phenylalanine and pNPP should be handled with gloves as they may be skin irritants.
(b) (i) If phenylalanine is a competitive inhibitor, increasing the substrate concentration will overcome the inhibition. The graph will show that at high substrate concentrations, the initial rate of reaction reaches the same maximum rate (\(V_{\max}\)) in both concentrations of phenylalanine. If it were non-competitive, the maximum rate would be significantly lower at the higher inhibitor concentration.
(ii) Standard deviation shows the dispersion or spread of all data points around the mean, whereas the range only shows the difference between the maximum and minimum values, which is heavily influenced by single extreme values or anomalies. Standard deviation allows further statistical testing and gives a better indication of data reliability.
(iii) The student should use the Student's t-test (unpaired t-test) because they are comparing the means of two independent, continuous, and normally distributed sets of replicate data.
Marking scheme
(a) Planning Method (Max 9 marks): 1. Dilution: Describes a clear method to prepare five different concentrations of phenylalanine. 2. Dilution calculation: Shows a correct volume calculation for making at least one concentration (e.g., 2.0 cm\(^{3}\) of 10% stock + 8.0 cm\(^{3}\) of water to make 10 cm\(^{3}\) of 2% phenylalanine). 3. Use of colorimeter: Mentions using a colorimeter with a 405 nm filter and zeroing with a suitable blank. 4. Time-course: Measures absorbance at regular, frequent time intervals (e.g., every 10 to 30 seconds) for the first 1 to 2 minutes. 5. Rate calculation: Plots absorbance against time and calculates the initial gradient to find the initial rate of reaction. 6. Control of temperature: Uses a thermostatically controlled water bath (e.g., 37 \(^{\circ}\)C) and equilibrates solutions before mixing. 7. Control of pH: Adds an equal volume of pH 9.0 buffer to each reaction mixture. 8. Standardization: Keeps enzyme volume/concentration and substrate volume/concentration constant across all runs. 9. Control experiment: Describes replacing the enzyme with water or boiled enzyme to show no reaction occurs. 10. Safety: Wear eye protection and/or gloves due to chemicals being potential irritants.
(b) (i) Graph interpretation (Max 2 marks): 1. At high substrate concentrations, the initial rate of reaction reaches the same maximum velocity (\(V_{\max}\)) for both inhibitor concentrations [1]. 2. This shows that competitive inhibition can be overcome by high substrate concentrations (unlike non-competitive inhibition) [1].
(b) (ii) Standard deviation vs range (Max 2 marks): 1. Standard deviation takes all data points into account to show the spread of data around the mean, whereas range only uses the two extreme values [1]. 2. Standard deviation is less affected by anomalies or extreme outliers than the range / allows statistical analysis [1].
(b) (iii) Statistical test (Max 2 marks): 1. Student's t-test / unpaired t-test [1]. 2. Because it compares the means of two distinct/independent sets of continuous data [1].
Question 2 · Experimental Planning and Evaluation
15 marks
An ecological researcher wants to compare the species diversity of ground-dwelling beetles (Coleoptera) in two areas of a temperate deciduous forest: - Area A: heavily grazed by wild deer. - Area B: enclosed by a deer-exclusion fence for 10 years to prevent grazing.
(a) Describe a method the researcher could use to sample the beetle populations and collect the data required to calculate Simpson's Index of Diversity (\(D\)) for each area.
Your description should be detailed enough to allow another person to follow it, and should include: - the sampling technique used to capture ground-dwelling beetles - how sampling locations within each area are determined to prevent bias - how standardisation is maintained to ensure the comparison between the two areas is valid - how the beetles are identified and the data recorded - safety and ethical considerations. [8]
(b) The researcher calculated Simpson's Index of Diversity (\(D\)) for both areas using the formula:
\(D = 1 - \sum \left(\frac{n}{N}\right)^2\)
where \(n\) is the number of individuals of a particular species and \(N\) is the total number of individuals of all species.
(i) Explain what the values of Simpson's Index of Diversity (\(D\)) close to 1 and close to 0 indicate about the stability of the communities in these areas. [2]
(ii) In Area A, the total number of beetles caught (\(N\)) was 240. The sum of the squares of the proportions, \(\sum \left(\frac{n}{N}\right)^2\), was calculated as 0.382. In Area B, the total number of beetles caught (\(N\)) was 310, and the sum of the squares of the proportions, \(\sum \left(\frac{n}{N}\right)^2\), was 0.145.
Calculate the Simpson's Index of Diversity (\(D\)) for Area A and Area B, and state which area has the higher species diversity. Show your working. [3]
(iii) Suggest two biotic factors, other than grazing by deer, that could influence the species diversity of beetles in these woodland areas. [2]
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Worked solution
(a) Method: 1. Use pitfall traps to capture ground-dwelling beetles. 2. In each area, lay out two long tape measures at right angles to form a grid system (e.g., 20 m by 20 m). 3. Generate random pairs of coordinates using a random number generator or computer program to determine where to place the traps, thus avoiding observer bias. 4. At each selected coordinate, dig a hole and insert a pitfall trap (e.g., a plastic cup) so that the rim is exactly level with the ground surface. 5. Standardize the traps by using cups of identical diameter and depth, and leave them open for the same period (e.g., exactly 48 hours) in both Area A and Area B. 6. Place a raised cover (e.g., a piece of wood supported by small stones) over each trap to prevent rain from flooding it and to hide the catch from predators. 7. To adhere to ethical standards and minimize unnecessary killing, use dry traps with some leaf litter inside to provide shelter, and check them daily to release non-target animals. 8. Retrieve the traps after 48 hours, identify each captured beetle species using an identification key/field guide, and count the number of individuals of each species. 9. Record the count of individuals of each species (\(n\)) and calculate the total count of all beetles (\(N\)) for each area. 10. Safety precautions: Wear stout boots to protect feet on rough terrain, use gloves when handling soil, and wash hands thoroughly after collecting traps to prevent infections from soil-borne pathogens.
(b) (i) - A value of \(D\) close to 1 indicates high species diversity and high community stability. If a disturbance or disease occurs, the community is resilient because there are many alternative species and pathways in the food web. - A value of \(D\) close to 0 indicates low species diversity and low stability (a fragile ecosystem). The community is dominated by one or two species, making it vulnerable to environmental changes.
(b) (ii) For Area A: \(D = 1 - 0.382 = 0.618\) (or 0.62)
For Area B: \(D = 1 - 0.145 = 0.855\) (or 0.86)
Area B has a higher Simpson's Index of Diversity (0.855 vs 0.618), which indicates that the ungrazed area (Area B) has higher species diversity than the grazed area (Area A).
(b) (iii) Two biotic factors (other than deer grazing) could include: 1. Predation by other species (e.g., insectivorous birds, shrews, spiders). 2. Interspecific competition between beetle species or with other insects for resources like food (e.g., organic matter, smaller invertebrates) or breeding/sheltering sites.
Marking scheme
(a) Planning Method (Max 8 marks): 1. Sampling device: Use of pitfall traps placed in the ground with the rim flush/level with the soil surface. 2. Randomisation: Setting up a grid with tape measures and using a random number generator to obtain sampling coordinates. 3. Standardisation of equipment: Using the same type/size/dimensions of trap in both Area A and Area B. 4. Standardisation of time: Leaving the traps in place for the same duration (e.g., 48 hours) in both areas. 5. Replication: Placing a large number of traps (e.g., at least 15-20) in each area to ensure a representative sample. 6. Ethical consideration: Dry traps with leaf litter/cover used AND checked frequently to minimize harm to trapped animals/non-target species. 7. Identification: Use of an identification key/field guide to determine beetle species. 8. Recording: Explains recording the number of individuals of each beetle species (\(n\)) and the total number of all beetles (\(N\)). 9. Safety: Wear sturdy boots/footwear to prevent injury on uneven forest floors OR wash hands/use gloves when handling traps/soil.
(b) (i) Simpson's Index Interpretation (Max 2 marks): 1. Value close to 1: High species diversity, meaning the community is stable/resilient to environmental change [1]. 2. Value close to 0: Low species diversity, meaning the community is unstable/vulnerable/dominated by few species [1].
(b) (ii) Calculation (Max 3 marks): 1. Correct calculation for Area A: \(D = 0.618\) (accept 0.62) [1]. 2. Correct calculation for Area B: \(D = 0.855\) (accept 0.86) [1]. 3. Correct statement that Area B has higher species diversity [1].
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