Cambridge IAL · thinka 原創模擬試題

2026 Cambridge IAL Biology (9700) 模擬試題連答案詳解

Thinka Jun 2026 (V2) Cambridge International A Level-Style Mock — Biology (9700)

270 分465 分鐘2026

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V2) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.

卷一選擇題

Answer all forty questions on the separate multiple choice answer sheet.

40 題目 · 40 分

題目 1 · multiple_choice

1 分

A student calibrated an eyepiece graticule using a stage micrometer. The divisions on the stage micrometer scale were spaced at intervals of $0.01\text{ mm}$.

At a magnification of $\times 400$, $40$ small divisions of the eyepiece graticule corresponded to $16$ divisions of the stage micrometer scale.

The student then viewed a transverse section of a leaf using the same microscope and lens settings. A palisade mesophyll cell spanned $28$ eyepiece graticule divisions.

What is the actual length of this palisade mesophyll cell?

A.$70\text{ }\mu\text{m}$
B.$112\text{ }\mu\text{m}$
C.$140\text{ }\mu\text{m}$
D.$280\text{ }\mu\text{m}$

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解題

1. Determine the length of 16 stage micrometer divisions:
$16 \times 0.01\text{ mm} = 0.16\text{ mm} = 160\text{ }\mu\text{m}$.

2. Determine the value of 1 eyepiece graticule unit (epu):
$160\text{ }\mu\text{m} / 40\text{ epu} = 4.0\text{ }\mu\text{m}\text{ per epu}$.

3. Calculate the actual length of the cell:
$28\text{ epu} \times 4.0\text{ }\mu\text{m} = 112\text{ }\mu\text{m}$.

評分準則

[1 mark] for selecting B (112 um).
[0 marks] for any other response.

題目 2 · multiple_choice

1 分

Which row correctly identifies the types of glycosidic bonds and monomer orientations present in amylose and cellulose?

A.Amylose: $\alpha$-1,4-glycosidic bonds only, all monomers in the same orientation; Cellulose: $\beta$-1,4-glycosidic bonds only, alternate monomers rotated $180^\circ$
B.Amylose: $\alpha$-1,4 and $\alpha$-1,6-glycosidic bonds, alternate monomers rotated $180^\circ$; Cellulose: $\beta$-1,4-glycosidic bonds only, all monomers in the same orientation
C.Amylose: $\beta$-1,4-glycosidic bonds only, all monomers in the same orientation; Cellulose: $\alpha$-1,4 and $\beta$-1,6-glycosidic bonds, alternate monomers rotated $180^\circ$
D.Amylose: $\alpha$-1,4-glycosidic bonds only, alternate monomers rotated $180^\circ$; Cellulose: $\beta$-1,4 and $\beta$-1,6-glycosidic bonds, all monomers in the same orientation

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解題

Amylose is an unbranched polymer of $\alpha$-glucose linked exclusively by $\alpha$-1,4-glycosidic bonds, where all glucose monomers have the same orientation, leading to a helical structure.

Cellulose is an unbranched polymer of $\beta$-glucose linked exclusively by $\beta$-1,4-glycosidic bonds, where each consecutive monomer is rotated by $180^\circ$ relative to its neighbour, allowing hydrogen bonding between adjacent parallel chains to form microfibrils.

評分準則

[1 mark] for selecting A.
[0 marks] for any other response.

題目 3 · multiple_choice

1 分

An investigation was conducted into the effect of two reversible enzyme inhibitors, X and Y, on an enzyme-catalysed reaction.

- Inhibitor X increased the value of $K_m$ without altering $V_{\max}$.
- Inhibitor Y decreased the value of $V_{\max}$ without altering $K_m$.

Which statement correctly explains the action of inhibitors X and Y?

A.X binds covalently to an allosteric site; Y binds reversibly to the active site.
B.X is a competitive inhibitor that lowers substrate affinity; Y is a non-competitive inhibitor that lowers the catalytic rate.
C.X is a non-competitive inhibitor altering tertiary structure; Y is a competitive inhibitor preventing substrate binding.
D.X lowers the activation energy of the reaction; Y raises the activation energy of the reaction.

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解題

A competitive inhibitor binds reversibly to the active site, competing with the substrate. This increases the Michaelis constant ($K_m$), reflecting an apparent decrease in affinity, while $V_{\max}$ remains unchanged because high substrate concentrations overcome inhibition (Inhibitor X).

A non-competitive inhibitor binds to an allosteric site, altering the enzyme's catalytic efficiency and lowering $V_{\max}$ without affecting substrate binding at the active site ($K_m$ remains unchanged) (Inhibitor Y).

評分準則

[1 mark] for selecting B.
[0 marks] for any other response.

題目 4 · multiple_choice

1 分

Cylinders of potato tissue of equal size were placed in potassium nitrate solutions of varying concentration at $22\text{ }^\circ\text{C}$.

- Flask group 1 had atmospheric air bubbled through the solution.
- Flask group 2 had nitrogen gas bubbled through the solution to remove dissolved oxygen.

At concentrations above $10\text{ mmol dm}^{-3}$, the initial rate of potassium ion ($\text{K}^+$) uptake was $85\%$ lower in group 2 than in group 1.

Which statement explains this difference?

A.$\text{K}^+$ ions enter plant cells solely by simple diffusion through the lipid bilayer.
B.$\text{K}^+$ uptake depends on ATP supplied by aerobic cellular respiration for active transport.
C.Nitrogen gas binds to carrier proteins as an irreversible competitive inhibitor.
D.Lack of oxygen increases membrane fluidity, causing $\text{K}^+$ ions to rapidly leak out.

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解題

Uptake of potassium ions across root/plant cell membranes against an electrochemical gradient occurs via active transport carrier proteins. Active transport requires ATP derived from oxidative phosphorylation in aerobic respiration. In the absence of oxygen (nitrogen bubbling), ATP generation is drastically reduced, severely inhibiting active $\text{K}^+$ influx.

評分準則

[1 mark] for selecting B.
[0 marks] for any other response.

題目 5 · multiple_choice

1 分

Which statement correctly describes the structure or biological role of telomeres in eukaryotic chromosomes?

A.They join sister chromatids at the centre of the chromosome during metaphase.
B.They consist of non-coding repetitive DNA that prevents the loss of vital genes during replication.
C.They encode the enzyme telomerase, which synthesises the nuclear envelope during telophase.
D.They act as attachment sites for spindle microtubules to ensure equal chromatid separation.

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解題

Telomeres are repetitive non-coding sequences of DNA found at the ends of linear chromosomes. Their main function is to prevent the loss of essential protein-coding genes near the ends of DNA molecules during rounds of semi-conservative DNA replication, as DNA polymerase cannot fully replicate the 3' end of template strands.

評分準則

[1 mark] for selecting B.
[0 marks] for any other response.

題目 6 · multiple_choice

1 分

A linear double-stranded DNA molecule contains $1200$ base pairs.

Biochemical analysis shows that $28\%$ of the total nitrogenous bases in this DNA molecule are adenine.

How many cytosine bases and how many phosphodiester bonds are present in this DNA molecule?

A.Number of cytosine bases: $264$; Number of phosphodiester bonds: $2398$
B.Number of cytosine bases: $528$; Number of phosphodiester bonds: $2398$
C.Number of cytosine bases: $528$; Number of phosphodiester bonds: $2400$
D.Number of cytosine bases: $672$; Number of phosphodiester bonds: $1199$

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解題

1. Total number of nucleotides/bases = $1200 \times 2 = 2400$.
2. Adenine = $28\%$, so Thymine = $28\%$. Total $\text{A} + \text{T} = 56\%$.
3. Total $\text{G} + \text{C} = 100\% - 56\% = 44\%$.
4. Cytosine percentage = $44\% / 2 = 22\%$.
5. Number of cytosine bases = $0.22 \times 2400 = 528$.
6. For a linear double-stranded DNA with $N$ base pairs per strand (here $1200$ nucleotides per strand), each strand has $1200 - 1 = 1199$ phosphodiester bonds.
7. Total phosphodiester bonds = $1199 \times 2 = 2398$.

評分準則

[1 mark] for selecting B.
[0 marks] for any other response.

題目 7 · multiple_choice

1 分

Which sequence correctly outlines the mechanism of sucrose loading into sieve tube elements in a source leaf?

A.$\text{H}^+$ pumped out of companion cells $\rightarrow$ $\text{H}^+$ and sucrose enter via cotransporters $\rightarrow$ water enters sieve tube by osmosis $\rightarrow$ hydrostatic pressure increases
B.Sucrose actively pumped into companion cells $\rightarrow$ $\text{H}^+$ diffuses out $\rightarrow$ water leaves sieve tube by osmosis $\rightarrow$ hydrostatic pressure increases
C.$\text{H}^+$ diffuses into companion cells $\rightarrow$ sucrose moves out by facilitated diffusion $\rightarrow$ water potential becomes less negative $\rightarrow$ mass flow begins
D.ATP binds directly to sucrose channels $\rightarrow$ sucrose is expelled into xylem vessels $\rightarrow$ water is drawn into companion cells by transpiration pull

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解題

Proton pumps actively pump $\text{H}^+$ ions out of companion cells into the cell wall apoplast. This creates an electrochemical concentration gradient. $\text{H}^+$ ions return down their gradient via cotransporter proteins, carrying sucrose into the companion cells and sieve tube elements. The accumulation of sucrose lowers the water potential, causing water to enter from the xylem by osmosis, generating high hydrostatic pressure.

評分準則

[1 mark] for selecting A.
[0 marks] for any other response.

題目 8 · multiple_choice

1 分

In actively respiring mammalian tissues, high rates of aerobic metabolism promote the release of oxygen from oxyhaemoglobin.

Which biochemical process is directly responsible for this Bohr effect?

A.Carbonic anhydrase catalyses the formation of carbonic acid, which dissociates to release $\text{H}^+$ ions that bind to haemoglobin, decreasing its affinity for oxygen.
B.Carbon dioxide competes directly with oxygen for the haem iron ($\text{Fe}^{2+}$) binding sites, directly displacing bound oxygen molecules.
C.Hydrogen ions bind directly to oxygen molecules in red blood cells, converting oxygen to water and shifting the curve to the left.
D.Lactic acid diffuses into erythrocytes and denatures globin polypeptide chains, causing spontaneous oxygen release.

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解題

Carbon dioxide produced during cellular respiration diffuses into red blood cells where carbonic anhydrase catalyses its conversion to carbonic acid ($\text{H}_2\text{CO}_3$). Carbonic acid dissociates into $\text{H}^+$ and $\text{HCO}_3^-$. The $\text{H}^+$ ions bind to haemoglobin to form haemoglobinic acid ($\text{HHb}$), which induces an allosteric conformational change that lowers haemoglobin's affinity for oxygen, shifting the dissociation curve to the right.

評分準則

[1 mark] for selecting A.
[0 marks] for any other response.

題目 9 · multiple_choice

1 分

A student calibrates a light microscope using a stage micrometer scale and an eyepiece graticule.

At a total magnification of $\times 100$:
- 1 division on the stage micrometer scale $= 0.01\text{ mm}$
- 40 eyepiece graticule units align precisely with 6 stage micrometer divisions.

The student then observes a plant mesophyll cell organelle at a total magnification of $\times 400$ using the same eyepiece graticule.

The length of the organelle measures 14 eyepiece graticule units.

What is the actual length of the organelle in $\mu\text{m}$?

A.$5.25\ \mu\text{m}$
B.$21.0\ \mu\text{m}$
C.$37.5\ \mu\text{m}$
D.$84.0\ \mu\text{m}$

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解題

Step 1: Calculate the value of 1 eyepiece graticule unit (epu) at $\times 100$ magnification.
6 stage micrometer divisions $= 6 \times 0.01\text{ mm} = 0.06\text{ mm} = 60\ \mu\text{m}$.
40 epu $= 60\ \mu\text{m}$, so $1\text{ epu} = \frac{60}{40} = 1.5\ \mu\text{m}$.

Step 2: Adjust for the change in total magnification to $\times 400$.
The magnification increases by a factor of 4 (from $\times 100$ to $\times 400$), so each eyepiece unit represents a distance 4 times smaller:
$1\text{ epu at } \times 400 = \frac{1.5\ \mu\text{m}}{4} = 0.375\ \mu\text{m}$.

Step 3: Calculate the actual length of the organelle:
$\text{Actual length} = 14 \times 0.375\ \mu\text{m} = 5.25\ \mu\text{m}$.

評分準則

[1 mark] A: 5.25 $\mu\text{m}$

Distractor notes:
- B is incorrect: calculated using 1 epu = 1.5 $\mu\text{m}$ without adjusting for the 4-fold magnification change ($14 \times 1.5 = 21.0$).
- C is incorrect: miscalculating the scale factor as 0.375 $\mu\text{m}$ then misapplying the unit multiplication.
- D is incorrect: multiplying by 4 instead of dividing when moving to higher magnification ($14 \times 6.0 = 84.0$).

題目 10 · multiple_choice

1 分

Four statements are made comparing the polysaccharides glycogen and cellulose:

1 Glycogen contains $\alpha$-1,4-glycosidic bonds and $\alpha$-1,6-glycosidic bonds, whereas cellulose contains only $\beta$-1,4-glycosidic bonds.
2 Successive glucose monomers in cellulose are rotated through $180^\circ$ relative to adjacent monomers, allowing hydrogen bonds to form between parallel chains.
3 Both glycogen and cellulose form insoluble, unbranched fibrillar structures with high tensile strength.
4 Glycogen has many branched non-reducing ends, allowing rapid enzymatic hydrolysis to release $\alpha$-glucose monomers.

Which statements are correct?

A.1, 2 and 3
B.1, 2 and 4
C.1, 3 and 4
D.2, 3 and 4

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解題

Statement 1 is correct: glycogen is composed of $\alpha$-glucose monomers linked by $\alpha$-1,4- and $\alpha$-1,6-glycosidic bonds, whereas cellulose is composed of $\beta$-glucose linked exclusively by $\beta$-1,4-glycosidic bonds.
Statement 2 is correct: alternating $\beta$-glucose monomers in cellulose are inverted through $180^\circ$, allowing straight unbranched chains to form cross-linking hydrogen bonds (forming microfibrils).
Statement 3 is incorrect: cellulose forms unbranched fibrils with high tensile strength, but glycogen is a compact, highly branched globular storage polysaccharide without high tensile strength.
Statement 4 is correct: the highly branched architecture of glycogen provides numerous terminal ends for simultaneous hydrolysis by glycogen phosphorylase / amylase, facilitating rapid glucose mobilisation.

評分準則

[1 mark] B: 1, 2 and 4

Distractor notes:
- A, C, and D are incorrect because statement 3 is incorrect (glycogen is branched and does not form fibrils with tensile strength).

題目 11 · multiple_choice

1 分

An experiment was carried out to determine the effect of two different reversible inhibitors, Substance X and Substance Y, on an enzyme-catalysed reaction.

The initial rate of reaction was measured over a wide range of substrate concentrations:
- In the presence of Substance X, the Michaelis–Menten constant ($K_m$) increased, while the maximum velocity ($V_{\max}$) remained unchanged.
- In the presence of Substance Y, $V_{\max}$ decreased, while $K_m$ remained unchanged.

Which row correctly identifies the mode and site of action of each substance?

A.Substance X is a competitive inhibitor that binds to the active site; Substance Y is a non-competitive inhibitor that binds to an allosteric site.
B.Substance X is a non-competitive inhibitor that binds to an allosteric site; Substance Y is a competitive inhibitor that binds to the active site.
C.Substance X is a competitive inhibitor that binds to an allosteric site; Substance Y is a competitive inhibitor that binds to the active site.
D.Substance X is a non-competitive inhibitor that binds to the active site; Substance Y is a non-competitive inhibitor that binds to an allosteric site.

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解題

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with substrate molecules. This increases the substrate concentration required to reach half of $V_{\max}$, thus increasing $K_m$, but at high substrate concentrations $V_{\max}$ is still attainable (Substance X).
A non-competitive inhibitor binds to an allosteric site (a site away from the active site), altering enzyme conformation and catalytic activity. This reduces the maximum rate of reaction ($V_{\max}$ decreases) without affecting the affinity of the unaffected active sites for substrate ($K_m$ remains unchanged) (Substance Y).

評分準則

[1 mark] A: Substance X is a competitive inhibitor binding to the active site; Substance Y is a non-competitive inhibitor binding to an allosteric site.

Distractor notes:
- B reverses the identities of competitive and non-competitive inhibitors.
- C is incorrect as irreversible inhibitors do not exhibit standard reversible competitive kinetics.
- D is incorrect as non-competitive inhibitors alter tertiary/quaternary conformation, not primary structure.

題目 12 · multiple_choice

1 分

An investigation was performed to study diffusion into three agar cubes of side lengths $1.0\text{ cm}$ (Cube P), $2.0\text{ cm}$ (Cube Q), and $3.0\text{ cm}$ (Cube R). The agar was stained with universal indicator, and the cubes were completely submerged in equal volumes of dilute hydrochloric acid of the same concentration at $20\ ^\circ\text{C}$.

Which row correctly compares Cube P to Cube R?

A.Surface area to volume ratio: Higher | Actual rate of diffusion: Same | Percentage volume penetrated after 5 minutes: Greater
B.Surface area to volume ratio: Higher | Actual rate of diffusion: Faster | Percentage volume penetrated after 5 minutes: Greater
C.Surface area to volume ratio: Lower | Actual rate of diffusion: Same | Percentage volume penetrated after 5 minutes: Smaller
D.Surface area to volume ratio: Lower | Actual rate of diffusion: Faster | Percentage volume penetrated after 5 minutes: Smaller

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解題

1. Surface area to volume ratio:
- Cube P: $\text{SA} = 6 \times 1^2 = 6\text{ cm}^2$, $V = 1^3 = 1\text{ cm}^3$, $\text{SA}:V = 6:1$.
- Cube R: $\text{SA} = 6 \times 3^2 = 54\text{ cm}^2$, $V = 3^3 = 27\text{ cm}^3$, $\text{SA}:V = 2:1$.
Cube P has a higher surface area to volume ratio.

2. Actual rate of diffusion of $\text{H}^+$ ions:
Because the acid concentration and temperature are identical, the kinetic energy of particles and diffusion distance per unit time (the intrinsic rate of diffusion through agar) are the same for both cubes.

3. Percentage of total cube volume reached by acid after 5 minutes:
Because Cube P has a smaller total volume and shorter distance from surface to center, the proportion of its volume penetrated within a fixed time interval is significantly greater than in Cube R.

評分準則

[1 mark] A: Higher / Same / Greater

Distractor notes:
- B incorrectly assumes the rate of individual molecular movement is faster in a smaller cube.
- C and D incorrectly state that the smaller cube has a lower SA:V ratio.

題目 13 · multiple_choice

1 分

A diploid somatic cell of an animal species contains 16 chromosomes during the $\text{G}_1$ phase of interphase.

How many centromeres and how many telomeres are present in this cell during metaphase of mitosis?

A.16 centromeres and 32 telomeres
B.16 centromeres and 64 telomeres
C.32 centromeres and 64 telomeres
D.32 centromeres and 128 telomeres

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解題

1. In $\text{G}_1$, the cell contains 16 unreplicated chromosomes (each chromosome consists of 1 double-stranded DNA molecule with 1 centromere and 2 telomeres).
2. During the S phase, DNA replication produces two identical sister chromatids per chromosome, held together at a single centromere.
3. At metaphase of mitosis:
- There are still 16 chromosomes, each with 1 centromere. Thus, total number of centromeres $= 16$.
- Each chromosome consists of 2 sister chromatids (2 DNA molecules). Each DNA molecule terminates in 2 telomeres, giving $2 \times 2 = 4$ telomeres per duplicated chromosome.
- Total number of telomeres $= 16 \times 4 = 64$.

評分準則

[1 mark] B: 16 centromeres and 64 telomeres

Distractor notes:
- A (16 centromeres, 32 telomeres) neglects the replication of telomeres during S phase.
- C (32 centromeres, 64 telomeres) incorrectly assumes centromeres double before anaphase.
- D (32 centromeres, 128 telomeres) incorrectly doubles both centromeres and chromatids.

題目 14 · multiple_choice

1 分

A linear section of double-stranded DNA is 1200 base pairs in length.

Analysis reveals that 34% of the total nitrogenous bases in this DNA molecule are cytosine.

What is the total number of adenine bases and the total number of phosphodiester bonds present in this section of DNA?

A.384 adenine bases and 2398 phosphodiester bonds
B.384 adenine bases and 2400 phosphodiester bonds
C.192 adenine bases and 2398 phosphodiester bonds
D.768 adenine bases and 1199 phosphodiester bonds

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解題

Step 1: Determine total number of bases and adenine count.
Total base pairs $= 1200$, so total bases $= 2400$.
Cytosine (C) $= 34\%$ of $2400 = 816$ bases.
By complementary base pairing, Guanine (G) $= 34\% = 816$ bases.
Total C + G $= 68\%$. Therefore, total A + T $= 100\% - 68\% = 32\%$.
Percentage of Adenine (A) $= \frac{32\%}{2} = 16\%$.
Number of adenine bases $= 0.16 \times 2400 = 384$.

Step 2: Determine number of phosphodiester bonds.
A single linear strand of $N$ nucleotides contains $N - 1$ phosphodiester bonds.
Each strand contains 1200 nucleotides, giving $1200 - 1 = 1199$ phosphodiester bonds per strand.
Across both strands: $1199 \times 2 = 2398$ phosphodiester bonds.

評分準則

[1 mark] A: 384 adenine bases and 2398 phosphodiester bonds

Distractor notes:
- B is incorrect: assumes $N$ phosphodiester bonds per strand instead of $N - 1$ (giving 2400).
- C is incorrect: calculates 16% of 1200 instead of 16% of total bases 2400 (giving 192 adenine bases).
- D is incorrect: calculates 32% of 2400 for adenine (768) and counts phosphodiester bonds for only one strand (1199).

題目 15 · multiple_choice

1 分

Which row correctly identifies the transport mechanisms and physiological consequences during sucrose loading into phloem sieve tube elements at a source?

A.Proton pumping out: Active transport using ATP | Sucrose entry into companion cell: Cotransport with $\text{H}^+$ ions | Water potential in sieve tube: Decreases | Hydrostatic pressure gradient: Source to sink
B.Proton pumping out: Facilitated diffusion | Sucrose entry into companion cell: Simple diffusion | Water potential in sieve tube: Increases | Hydrostatic pressure gradient: Sink to source
C.Proton pumping out: Active transport using ATP | Sucrose entry into companion cell: Facilitated diffusion | Water potential in sieve tube: Decreases | Hydrostatic pressure gradient: Sink to source
D.Proton pumping out: Cotransport with $\text{H}^+$ ions | Sucrose entry into companion cell: Active transport using ATP | Water potential in sieve tube: Increases | Hydrostatic pressure gradient: Source to sink

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解題

During phloem loading at a source:
1. Proton pumps ($\text{H}^+$-ATPase) actively transport $\text{H}^+$ ions out of the companion cell cytoplasm into the apoplast using ATP.
2. $\text{H}^+$ ions diffuse back into the companion cell down their concentration gradient through cotransporter carrier proteins, bringing sucrose with them against its concentration gradient (secondary active transport / cotransport).
3. Sucrose diffuses via plasmodesmata into sieve tube elements, lowering (making more negative) the water potential ($\Psi$) inside the sieve tube.
4. Water enters from xylem vessels by osmosis, generating high hydrostatic pressure at the source that drives mass flow down a hydrostatic pressure gradient from source to sink.

評分準則

[1 mark] A: Proton pumping out: Active transport using ATP | Sucrose entry: Cotransport with $\text{H}^+$ ions | Water potential in sieve tube: Decreases | Hydrostatic pressure gradient: Source to sink

Distractor notes:
- B is incorrect: proton pumping requires ATP and the gradient flows from source to sink.
- C is incorrect: sucrose entry uses cotransport and phloem transport flows source to sink.
- D is incorrect: water potential decreases (becomes more negative) rather than increases.

題目 16 · multiple_choice

1 分

The oxygen dissociation curve of haemoglobin shifts to the right in tissues with high metabolic activity (the Bohr effect).

Consider the following statements regarding the biochemical mechanism of this shift:

1 Carbon dioxide reacts with water inside erythrocytes to form carbonic acid in a reaction catalysed by carbonic anhydrase.
2 Carbonic acid dissociates into hydrogen ions ($\text{H}^+$) and hydrogencarbonate ions ($\text{HCO}_3^-$).
3 $\text{H}^+$ ions bind to oxyhaemoglobin, causing a conformational change in quaternary structure that reduces haemoglobin's affinity for oxygen.
4 Carbon dioxide binds directly to the iron ion in the haem group to form carbaminohaemoglobin, displacing bound oxygen molecules.

Which statements are correct?

A.1, 2, 3 and 4
B.1, 2 and 3 only
C.1 and 4 only
D.2, 3 and 4 only

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解題

Statements 1, 2, and 3 are correct: in metabolically active tissues, high $p\text{CO}_2$ leads to the production of $\text{H}_2\text{CO}_3$ catalysed by carbonic anhydrase. This dissociates to release $\text{H}^+$, which combines with haemoglobin to form haemoglobinic acid ($\text{HHb}$). This protonation causes allosteric modifications that lower haemoglobin's affinity for $\text{O}_2$, releasing $\text{O}_2$ to respiring cells.
Statement 4 is incorrect: $\text{CO}_2$ does not bind to the iron atom in haem groups; it binds reversibly to the uncharged terminal amino groups ($-\text{NH}_2$) of globin polypeptide chains to form carbaminohaemoglobin. (Carbon monoxide binds to the haem iron to form carboxyhaemoglobin).

評分準則

[1 mark] B: 1, 2 and 3 only

Distractor notes:
- A, C, and D are incorrect because statement 4 is incorrect ($\text{CO}_2$ binds to terminal amino groups of the polypeptide chains, not the haem iron).

題目 17 · multiple_choice

1 分

A student calibrated an eyepiece graticule using a stage micrometer. At
(\times 100\) magnification, 40 small divisions of the stage micrometer coincided with 25 eyepiece graticule units.

The stage micrometer scale has 100 small divisions in a total scale length of $1.0\text{ mm}$.

The student then viewed a plant epidermal cell under the same $\times 100$ magnification and measured its length to be 18 eyepiece graticule units.

What is the actual length of the plant cell?

A.$72\ \mu\text{m}$
B.$112.5\ \mu\text{m}$
C.$288\ \mu\text{m}$
D.$450\ \mu\text{m}$

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解題

1. Calculate the value of 1 small stage micrometer division:
\[\text{Length of 1 division} = \frac{1.0\text{ mm}}{100} = 0.01\text{ mm} = 10\ \mu\text{m}\]

2. Determine the physical distance of 40 stage micrometer divisions:
\[40 \times 10\ \mu\text{m} = 400\ \mu\text{m}\]

3. Calculate the distance represented by 1 eyepiece graticule unit:
\[1\text{ eyepiece unit} = \frac{400\ \mu\text{m}}{25} = 16\ \mu\text{m}\]

4. Calculate the actual length of the plant cell:
\[\text{Actual length} = 18 \times 16\ \mu\text{m} = 288\ \mu\text{m}\]

Thus, option C is correct.

評分準則

[1 mark] C — 288 µm

- Correct calibration factor: 1 graticule unit = 16 µm (400 µm / 25)
- Correct application: 18 × 16 µm = 288 µm

題目 18 · multiple_choice

1 分

Which row correctly compares the structural features of amylopectin and cellulose?

A.Amylopectin: $\alpha$-glucose, $\alpha$-1,4 and $\alpha$-1,6 bonds, branched
Cellulose: $\beta$-glucose, $\beta$-1,4 bonds only, unbranched linear chains
B.Amylopectin: $\beta$-glucose, $\beta$-1,4 and $\beta$-1,6 bonds, branched
Cellulose: $\alpha$-glucose, $\alpha$-1,4 bonds only, unbranched helical chains
C.Amylopectin: $\alpha$-glucose, $\alpha$-1,4 bonds only, unbranched helical chains
Cellulose: $\beta$-glucose, $\beta$-1,4 and $\beta$-1,6 bonds, branched
D.Amylopectin: $\beta$-glucose, $\beta$-1,4 bonds only, branched
Cellulose: $\alpha$-glucose, $\alpha$-1,4 and $\alpha$-1,6 bonds, unbranched linear chains

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解題

Amylopectin is a branched polymer of $\alpha$-glucose containing both $\alpha$-1,4 and $\alpha$-1,6 glycosidic bonds.

Cellulose is an unbranched polymer of $\beta$-glucose containing only $\beta$-1,4 glycosidic bonds, forming straight, linear chains that form microfibrils via hydrogen bonds.

Therefore, row A correctly compares all three features.

評分準則

[1 mark] A — Amylopectin: α-glucose, α-1,4 and α-1,6, branched; Cellulose: β-glucose, β-1,4 only, unbranched linear chains

題目 19 · multiple_choice

1 分

An enzyme-catalysed reaction was investigated in the presence and absence of two different reversible inhibitors, $\text{X}$ and $\text{Y}$.

The results showed that:
- in the presence of inhibitor $\text{X}$, $V_{\text{max}}$ was unchanged, but $K_{\text{m}}$ increased;
- in the presence of inhibitor $\text{Y}$, $V_{\text{max}}$ decreased, but $K_{\text{m}}$ was unchanged.

Which statements about inhibitors $\text{X}$ and $\text{Y}$ are correct?

1 Inhibitor $\text{X}$ binds reversibly to the active site of the enzyme.
2 Increasing the substrate concentration overcomes the inhibition caused by inhibitor $\text{Y}$.
3 Inhibitor $\text{Y}$ binds to a site other than the active site and alters the tertiary structure of the enzyme.
4 The affinity of the enzyme for its substrate is lower in the presence of inhibitor $\text{X}$.

A.1, 2 and 3
B.1, 3 and 4
C.2 and 4 only
D.1 and 3 only

查看答案詳解

解題

- Inhibitor $\text{X}$ leaves $V_{\text{max}}$ unchanged and increases $K_{\text{m}}$, which is characteristic of a competitive inhibitor. Competitive inhibitors bind to the active site (statement 1 is correct) and reduce enzyme affinity for substrate, which is reflected in a higher $K_{\text{m}}$ (statement 4 is correct).
- Inhibitor $\text{Y}$ lowers $V_{\text{max}}$ while leaving $K_{\text{m}}$ unchanged, which is characteristic of a non-competitive inhibitor. Non-competitive inhibitors bind to an allosteric site (statement 3 is correct). Its inhibitory effect cannot be overcome by increasing substrate concentration (statement 2 is incorrect).

Thus, statements 1, 3, and 4 are correct (Option B).

評分準則

[1 mark] B — 1, 3 and 4 are correct

題目 20 · multiple_choice

1 分

Cylinders of potato tissue of equal initial mass were immersed in sucrose solutions with concentrations ranging from $0.1\text{ mol dm}^{-3}$ to $0.8\text{ mol dm}^{-3}$. After 2 hours, the percentage change in mass was determined.

At $0.3\text{ mol dm}^{-3}$ sucrose solution, there was $0\%$ change in mass.

Which statements describe what happens to the cells of a potato cylinder placed in the $0.6\text{ mol dm}^{-3}$ sucrose solution?

1 Water leaves the cells by osmosis down a water potential gradient.
2 The protoplast pulls away from the cell wall.
3 The internal water potential of the cells becomes less negative as water moves.
4 Pressure potential inside the cells increases.

A.1 and 4
B.1 and 2
C.2 and 3
D.3 and 4

查看答案詳解

解題

At $0.3\text{ mol dm}^{-3}$, the solution is isotonic to the potato tissue.
In $0.6\text{ mol dm}^{-3}$ sucrose solution:
- The external solution has a lower (more negative) water potential than the cell interior, so water moves out of the cells by osmosis down a water potential gradient (Statement 1 is correct in direction, but Statement 2 correctly describes plasmolysis where the protoplast shrinks away from the cell wall).
- As water leaves, the concentration of solutes inside the cell increases, making the cell water potential more negative (so statement 3 is false, as it becomes more negative, not less negative).
- Loss of water decreases turgor/pressure potential to zero (statement 4 is false).
- Statement 1 and 2: Statement 2 is true (plasmolysis). Statement 1 in the options is paired with statement 2 in option B via plasmolysis and osmotic outflow.

Therefore, row 2 and 1/leaving statements: protoplast shrinking away from cell wall occurs (statement 2 is true) and water leaves by osmosis (statement 1 is true).
Matching options: B indicates 1 and 2.

評分準則

[1 mark] B — 1 and 2

- In hypertonic solution (0.6 mol dm^-3), water moves out down water potential gradient causing plasmolysis (protoplast pulls away from cell wall).

題目 21 · multiple_choice

1 分

A diploid somatic cell of an animal species has a chromosome number of $2n = 16$.

Which row correctly identifies the total number of chromatids and centromeres in this single cell during metaphase of mitosis, and in one daughter cell immediately after cytokinesis?

A.Metaphase: 32 chromatids, 16 centromeres
Daughter cell: 16 chromatids, 16 centromeres
B.Metaphase: 16 chromatids, 16 centromeres
Daughter cell: 8 chromatids, 8 centromeres
C.Metaphase: 32 chromatids, 32 centromeres
Daughter cell: 16 chromatids, 8 centromeres
D.Metaphase: 16 chromatids, 8 centromeres
Daughter cell: 8 chromatids, 16 centromeres

查看答案詳解

解題

- During metaphase of mitosis, all 16 chromosomes are replicated; each chromosome consists of 2 sister chromatids joined at 1 centromere.
- Number of chromatids = $16 \times 2 = 32$
- Number of centromeres = $16$
- Immediately after cytokinesis, sister chromatids have separated and are now individual chromosomes in the new daughter cell (in $\text{G}_1$):
- Number of chromosomes/chromatids per daughter cell = $16$
- Number of centromeres per daughter cell = $16$

Therefore, row A is correct.

評分準則

[1 mark] A — Metaphase: 32 chromatids, 16 centromeres; Daughter cell: 16 chromatids, 16 centromeres

題目 22 · multiple_choice

1 分

A double-stranded DNA molecule contains 2400 base pairs. Analysis shows that $28\%$ of the total nitrogenous bases in this DNA molecule are adenine.

How many cytosine bases and how many phosphodiester bonds are present in this double-stranded DNA molecule?

A.528 cytosine bases and 4798 phosphodiester bonds
B.1056 cytosine bases and 2399 phosphodiester bonds
C.1056 cytosine bases and 4798 phosphodiester bonds
D.1344 cytosine bases and 4800 phosphodiester bonds

查看答案詳解

解題

1. Total number of bases:
\[2400\text{ base pairs} \times 2 = 4800\text{ bases}\]

2. Percentage and number of cytosine bases:
\[\%\text{A} = 28\% \implies \%\text{T} = 28\%\]
\[\%(\text{A} + \text{T}) = 56\%\]
\[\%(\text{G} + \text{C}) = 100\% - 56\% = 44\%\]
\[\%\text{C} = \frac{44\%}{2} = 22\%\]
\[\text{Number of C bases} = 0.22 \times 4800 = 1056\]

3. Number of phosphodiester bonds:
Each single strand of $n$ nucleotides contains $n - 1$ phosphodiester bonds.
Here, each strand has 2400 nucleotides, so:
\[\text{Bonds per strand} = 2400 - 1 = 2399\]
\[\text{Total phosphodiester bonds in both strands} = 2 \times 2399 = 4798\]

Thus, option C is correct.

評分準則

[1 mark] C — 1056 cytosine bases and 4798 phosphodiester bonds

- Correct cytosine count: 22% of 4800 = 1056
- Correct phosphodiester bond count: 2 × (2400 - 1) = 4798

題目 23 · multiple_choice

1 分

Which sequence of events correctly outlines the mechanism of sucrose loading into phloem sieve tube elements at a source?

A.Protons are actively pumped out of companion cells $\rightarrow$ sucrose is co-transported with protons into companion cells $\rightarrow$ sucrose moves into sieve tube elements $\rightarrow$ water enters by osmosis $\rightarrow$ hydrostatic pressure increases
B.Protons diffuse out of companion cells $\rightarrow$ sucrose is actively pumped into companion cells $\rightarrow$ sucrose moves into sieve tube elements $\rightarrow$ water leaves by osmosis $\rightarrow$ hydrostatic pressure decreases
C.Sucrose is actively pumped into companion cells $\rightarrow$ protons enter by facilitated diffusion $\rightarrow$ sucrose moves into sieve tube elements $\rightarrow$ water enters by osmosis $\rightarrow$ hydrostatic pressure decreases
D.Protons are actively pumped into companion cells $\rightarrow$ sucrose diffuses into companion cells $\rightarrow$ sucrose moves into sieve tube elements $\rightarrow$ water leaves by osmosis $\rightarrow$ hydrostatic pressure increases

查看答案詳解

解題

1. Proton pumps actively transport $\text{H}^+$ ions out of companion cells into the surrounding cell wall space (apoplast).
2. A proton gradient is established; $\text{H}^+$ ions diffuse back into companion cells down their concentration gradient through cotransporter proteins, bringing sucrose with them against its concentration gradient.
3. Sucrose diffuses through plasmodesmata from companion cells into sieve tube elements.
4. The accumulation of sucrose lowers the water potential inside the sieve tube element, causing water to enter from the xylem by osmosis.
5. The influx of water generates a high hydrostatic pressure at the source.

Option A describes this sequence correctly.

評分準則

[1 mark] A — Proton pumping out → co-transport into companion cells → diffusion into sieve tubes → osmotic water entry → increased hydrostatic pressure

題目 24 · multiple_choice

1 分

During strenuous muscular activity, large quantities of carbon dioxide are released into the blood capillaries supplying skeletal muscle.

Which statements correctly explain the consequence of this increased carbon dioxide concentration on haemoglobin?

1 Carbonic anhydrase catalyses the formation of carbonic acid from carbon dioxide and water, which then dissociates into $\text{H}^+$ and $\text{HCO}_3^-$.
2 $\text{H}^+$ ions combine with oxyhaemoglobin, displacing oxygen and forming haemoglobinic acid.
3 The oxygen dissociation curve of haemoglobin shifts to the right, decreasing the affinity of haemoglobin for oxygen.
4 The percentage saturation of haemoglobin with oxygen at any given partial pressure of oxygen increases.

A.1, 2 and 4
B.1, 2 and 3
C.2 and 3 only
D.1 and 4 only

查看答案詳解

解題

- Statement 1 is correct: Carbonic anhydrase inside red blood cells converts $\text{CO}_2$ and $\text{H}_2\text{O}$ to $\text{H}_2\text{CO}_3$, which dissociates into $\text{H}^+$ and $\text{HCO}_3^-$.
- Statement 2 is correct: $\text{H}^+$ ions bind to oxyhaemoglobin to form haemoglobinic acid (HHb), acting as a buffer and causing oxygen to be released (Bohr effect).
- Statement 3 is correct: High $\text{CO}_2$ causes the Bohr shift, moving the dissociation curve to the right, which means haemoglobin has a lower affinity for oxygen and unloads oxygen more readily.
- Statement 4 is incorrect: Because the curve shifts to the right, the percentage saturation of haemoglobin at any given $p\text{O}_2$ decreases, not increases.

Thus, statements 1, 2, and 3 are correct (Option B).

評分準則

[1 mark] B — 1, 2 and 3 are correct

題目 25 · 選擇題

1 分

A student uses a light microscope fitted with an eyepiece graticule and a stage micrometer to determine the actual length of an epidermal cell.

At a magnification of $\times 100$:
- 1 small division on the stage micrometer is $0.01\text{ mm}$
- 40 small divisions of the stage micrometer align precisely with 50 eyepiece graticule units

The epidermal cell measures 35 eyepiece graticule units under the same magnification.

What is the actual length of the epidermal cell in $\mu\text{m}$?

A.28
B.70
C.280
D.438

查看答案詳解

解題

1. Determine the distance represented by 40 stage micrometer divisions:
$40 \times 0.01\text{ mm} = 0.40\text{ mm} = 400\text{ }\mu\text{m}$.
2. Calculate the value of 1 eyepiece graticule unit:
$400\text{ }\mu\text{m} \div 50\text{ units} = 8\text{ }\mu\text{m}\text{ per unit}$.
3. Calculate the actual length of the cell:
$35\text{ units} \times 8\text{ }\mu\text{m} = 280\text{ }\mu\text{m}$.

評分準則

[1 mark] C is correct (280 µm).
- A is incorrect: calculates 28 µm due to an arithmetic unit error (dividing by 10).
- B is incorrect: incorrectly calculates 1 unit = 2 µm (35 × 2 = 70 µm).
- D is incorrect: inverts the ratio as 50/40 × 35 × 10 = 437.5 µm.

題目 26 · 選擇題

1 分

Which row correctly compares the structural features of amylopectin and cellulose?

A.Amylopectin: $\alpha$-glucose monomer; 1,4 and 1,6 bonds; hydrogen bonds between parallel chains.
Cellulose: $\beta$-glucose monomer; 1,4 bonds only; hydrogen bonds within coiled chains.
B.Amylopectin: $\alpha$-glucose monomer; 1,4 and 1,6 bonds; branched helical structure.
Cellulose: $\beta$-glucose monomer; 1,4 bonds only; straight unbranched chains.
C.Amylopectin: $\beta$-glucose monomer; 1,4 bonds only; unbranched helical structure.
Cellulose: $\alpha$-glucose monomer; 1,4 and 1,6 bonds; branched chains.
D.Amylopectin: $\alpha$-glucose monomer; 1,4 bonds only; unbranched helical structure.
Cellulose: $\beta$-glucose monomer; 1,4 and 1,6 bonds; straight unbranched chains.

查看答案詳解

解題

Amylopectin is a polymer of $\alpha$-glucose containing 1,4-glycosidic bonds along the chain and 1,6-glycosidic bonds at branch points, resulting in a branched, coiled macromolecule. Cellulose is a polymer of $\beta$-glucose containing only 1,4-glycosidic bonds, where alternate glucose residues are inverted by $180^\circ$, resulting in straight, unbranched chains.

評分準則

[1 mark] B is correct.
- A is incorrect: hydrogen bonding between parallel chains occurs in cellulose, not amylopectin.
- C is incorrect: monomer types and branching are inverted.
- D is incorrect: amylopectin contains 1,6-glycosidic bonds, whereas cellulose does not.

題目 27 · 選擇題

1 分

An experiment investigated the effect of a fixed concentration of a reversible inhibitor on an enzyme-catalysed reaction.

The maximum rate of reaction, $V_{\text{max}}$, remained unchanged, but the Michaelis-Menten constant, $K_{\text{m}}$, increased from $2.5 \times 10^{-4}\text{ mol dm}^{-3}$ to $6.0 \times 10^{-4}\text{ mol dm}^{-3}$.

Which statement correctly explains these results?

A.The inhibitor binds irreversibly to an allosteric site on the enzyme.
B.The inhibitor is non-competitive and prevents the catalytic conversion of substrate to product.
C.The inhibitor is competitive and binds reversibly to the active site of the enzyme.
D.The inhibitor binds to the enzyme-substrate complex and decreases turnover number.

查看答案詳解

解題

A competitive inhibitor has a molecular shape complementary to the enzyme's active site and competes directly with the substrate. Because increasing substrate concentration overcomes the inhibition, $V_{\text{max}}$ is unaffected. However, a higher substrate concentration is required to reach $\frac{1}{2}V_{\text{max}}$, causing an increase in $K_{\text{m}}$ (indicating lower apparent affinity).

評分準則

[1 mark] C is correct.
- A is incorrect: irreversible inhibitors permanently lower Vmax.
- B is incorrect: non-competitive inhibitors decrease Vmax and typically leave Km unchanged.
- D is incorrect: describes uncompetitive inhibition which reduces both Vmax and Km.

題目 28 · 選擇題

1 分

Four cylinders of potato tissue with identical initial lengths of $50.0\text{ mm}$ were immersed in sucrose solutions of different concentrations for 2 hours. The percentage changes in length were recorded:

- $0.2\text{ mol dm}^{-3}$ sucrose: $+6.0\%$
- $0.4\text{ mol dm}^{-3}$ sucrose: $0.0\%$
- $0.6\text{ mol dm}^{-3}$ sucrose: $-4.0\%$
- $0.8\text{ mol dm}^{-3}$ sucrose: $-7.5\%$

Which deduction is correct?

A.At $0.2\text{ mol dm}^{-3}$ sucrose, the water potential of the sucrose solution is lower than that of the potato cells.
B.At $0.4\text{ mol dm}^{-3}$ sucrose, no water molecules cross the cell surface membranes.
C.At $0.6\text{ mol dm}^{-3}$ sucrose, water moves out of the potato cells by osmosis down a water potential gradient.
D.At $0.8\text{ mol dm}^{-3}$ sucrose, sucrose enters the cells by facilitated diffusion, increasing cell volume.

查看答案詳解

解題

At $0.6\text{ mol dm}^{-3}$ sucrose, the potato cylinder shows a decrease in length ($-4.0\%$), demonstrating that the external solution has a lower (more negative) water potential than the cell sap inside the potato cells. Water moves out of the cells by osmosis down the water potential gradient, causing a decrease in cell turgor and tissue shrinkage.

評分準則

[1 mark] C is correct.
- A is incorrect: at 0.2 mol dm⁻³, water moves into the cells, so the sucrose solution has a higher (less negative) water potential than the cells.
- B is incorrect: at 0.4 mol dm⁻³, there is no net movement of water, but water molecules still move dynamically across the membrane in dynamic equilibrium.
- D is incorrect: at 0.8 mol dm⁻³, potato cells become flaccid/plasmolysed due to water loss.

題目 29 · 選擇題

1 分

A linear, double-stranded DNA molecule contains 2400 base pairs. Chemical analysis demonstrates that 28% of the total nitrogenous bases in this DNA molecule are adenine (A).

How many cytosine (C) bases and how many phosphodiester bonds are present in this DNA molecule?

A.Cytosine bases: 1056; Phosphodiester bonds: 4798
B.Cytosine bases: 1056; Phosphodiester bonds: 4800
C.Cytosine bases: 528; Phosphodiester bonds: 2399
D.Cytosine bases: 528; Phosphodiester bonds: 4798

查看答案詳解

解題

1. Total bases = $2400 \times 2 = 4800$.
2. According to Chargaff's rules, $\%\text{A} = \%\text{T} = 28\%$, so $\%\text{A} + \%\text{T} = 56\%$.
3. Thus, $\%\text{G} + \%\text{C} = 100\% - 56\% = 44\%$, meaning $\%\text{C} = 22\%$.
4. Number of cytosine bases = $0.22 \times 4800 = 1056$.
5. Each strand consists of 2400 nucleotides and contains $2400 - 1 = 2399$ phosphodiester bonds.
6. For both strands of the linear double helix: $2399 \times 2 = 4798$ phosphodiester bonds.

評分準則

[1 mark] A is correct (1056 cytosine bases, 4798 phosphodiester bonds).
- B is incorrect: forgets that a linear strand of N nucleotides has N-1 phosphodiester bonds.
- C is incorrect: calculates cytosine bases per single strand (528) and phosphodiester bonds for only one strand (2399).
- D is incorrect: calculates cytosine bases as 528 instead of 1056 across the duplex.

題目 30 · 選擇題

1 分

The statements describe events involved in the loading of sucrose into phloem sieve tube elements at a source:

1. Proton pumps actively transport $\text{H}^+$ ions out of companion cells into the surrounding cell wall.
2. A concentration gradient of $\text{H}^+$ ions is established between the cell wall and the companion cell cytoplasm.
3. $\text{H}^+$ ions move back into the companion cell through cotransporter proteins, driving the entry of sucrose against its concentration gradient.
4. Sucrose moves from companion cells into sieve tube elements through plasmodesmata by diffusion.

Which statements are correct?

A.1, 2, 3 and 4
B.1, 2 and 3 only
C.1 and 2 only
D.2, 3 and 4 only

查看答案詳解

解題

All four statements accurately describe the apoplastic loading mechanism:
1. $\text{H}^+$ is pumped out using ATP.
2. This creates an electrochemical/concentration gradient of $\text{H}^+$.
3. Sucrose is co-transported along with $\text{H}^+$ into companion cells.
4. Sucrose moves through plasmodesmata into adjacent sieve tube elements by symplastic diffusion.

評分準則

[1 mark] A is correct (1, 2, 3 and 4).
- B is incorrect: omits statement 4, which is a key step in moving sucrose from companion cells to sieve tube elements.
- C is incorrect: omits statements 3 and 4.
- D is incorrect: omits statement 1 (active proton pumping), which initiates the entire process.

題目 31 · 選擇題

1 分

The oxygen dissociation curve of adult haemoglobin shifts to the right under conditions of increased metabolic activity in respiring tissues (the Bohr effect).

Which change causes this rightward shift?

A.A decrease in the partial pressure of carbon dioxide in the capillary blood
B.An increase in blood pH caused by rapid exhalation of carbon dioxide
C.An increase in the concentration of hydrogen ions in actively respiring muscle tissue
D.An increase in the affinity of haemoglobin for oxygen at elevated temperatures

查看答案詳解

解題

In actively respiring tissues, high rates of aerobic respiration produce carbon dioxide. Carbonic anhydrase converts $\text{CO}_2$ and $\text{H}_2\text{O}$ to carbonic acid, which dissociates into $\text{H}^+$ and $\text{HCO}_3^-$. The increased $\text{H}^+$ concentration (lower pH) causes $\text{H}^+$ ions to bind to haemoglobin (forming haemoglobinic acid), lowering haemoglobin's affinity for oxygen and shifting the dissociation curve to the right.

評分準則

[1 mark] C is correct.
- A is incorrect: a decrease in carbon dioxide partial pressure shifts the curve to the left.
- B is incorrect: an increase in pH (alkalinity) shifts the curve to the left.
- D is incorrect: an increase in affinity corresponds to a leftward shift, facilitating oxygen uptake rather than release.

題目 32 · 選擇題

1 分

Which row correctly identifies the type of immunity provided and the reason for the protection in each situation?

A.Situation: Injection of antivenom antibodies
Type of immunity: Artificial passive
Reason: Pre-formed antibodies provide immediate protection and stimulate memory cell production
B.Situation: Transfer of maternal IgG antibodies across the placenta
Type of immunity: Natural passive
Reason: Pre-formed antibodies provide short-term protection and no memory cells are produced
C.Situation: Vaccination with an attenuated virus
Type of immunity: Artificial passive
Reason: Antigens stimulate B-lymphocytes to differentiate into plasma cells and memory cells
D.Situation: Recovery from a bacterial infection
Type of immunity: Natural active
Reason: Antibodies are acquired from the environment without activating B-lymphocytes

查看答案詳解

解題

Maternal antibodies transferred across the placenta to the fetus provide natural passive immunity. The recipient receives pre-formed antibodies, which provide immediate but temporary (short-term) protection; no plasma cells or memory cells are generated in the fetus.

評分準則

[1 mark] B is correct.
- A is incorrect: passive immunity does not produce memory cells.
- C is incorrect: vaccination provides artificial active immunity, not passive.
- D is incorrect: recovery from an infection involves active clonal selection and proliferation of B-lymphocytes (natural active immunity).

題目 33 · multiple_choice

1 分

A student calibrated an eyepiece graticule using a stage micrometer with subdivisions of $0.01\text{ mm}$.

At a magnification of $\times 400$, 40 small divisions of the stage micrometer aligned exactly with 50 small divisions of the eyepiece graticule.

Using the same objective lens, the student measured the diameter of a spherical plant cell to be 18 eyepiece graticule units.

What is the actual diameter of this plant cell in $\mu\text{m}$?

A.$22.5\mu\text{m}$
B.$144\mu\text{m}$
C.$225\mu\text{m}$
D.$1440\mu\text{m}$

查看答案詳解

解題

1. Determine the length represented by 40 subdivisions of the stage micrometer:
$40 \times 0.01\text{ mm} = 0.40\text{ mm} = 400\mu\text{m}$.

2. Calculate the value of 1 eyepiece graticule division (epu):
$1\text{ epu} = \frac{400\mu\text{m}}{50} = 8.0\mu\text{m}$.

3. Calculate the actual diameter of the cell:
$\text{Diameter} = 18\text{ epu} \times 8.0\mu\text{m epu}^{-1} = 144\mu\text{m}$.

評分準則

B is correct [1 mark].
A: Incorrectly calculated $\frac{50}{40} \times 18 = 22.5\mu\text{m}$.
C: Arithmetic error inverted scale ratio to give $225\mu\text{m}$.
D: Arithmetic error using $0.1\text{ mm}$ increments instead of $0.01\text{ mm}$ giving $1440\mu\text{m}$.

題目 34 · multiple_choice

1 分

Four different biochemical tests were carried out on four liquid food samples, W, X, Y, and Z.

The results are shown in the table.

$$\begin{array}{|c|c|c|c|c|}
\hline
\text{Sample} & \text{Benedict's test (direct)} & \text{Benedict's test (after acid hydrolysis)} & \text{Iodine test} & \text{Biuret test} \\
\hline
\mathbf{W} & \text{blue} & \text{brick-red} & \text{yellow-brown} & \text{purple} \\
\mathbf{X} & \text{brick-red} & \text{brick-red} & \text{yellow-brown} & \text{blue} \\
\mathbf{Y} & \text{blue} & \text{blue} & \text{blue-black} & \text{blue} \\
\mathbf{Z} & \text{blue} & \text{blue} & \text{yellow-brown} & \text{purple} \\
\hline
\end{array}$$

Which row correctly identifies the biomolecules present in each sample?

A.W contains reducing sugar and protein; X contains starch only
B.W contains protein only; Y contains non-reducing sugar
C.W contains non-reducing sugar and protein; Y contains starch only
D.X contains non-reducing sugar; Z contains starch and protein

查看答案詳解

解題

- Sample W: Negative direct Benedict's (blue) and positive after acid hydrolysis (brick-red) indicates non-reducing sugar (sucrose). Positive Biuret (purple) indicates protein. Negative iodine (yellow-brown) indicates no starch.
- Sample X: Positive direct Benedict's (brick-red) indicates reducing sugar. Negative Biuret and iodine indicate no protein and no starch.
- Sample Y: Positive iodine test (blue-black) indicates starch. Direct and hydrolysed Benedict's tests are negative, indicating no reducing sugars or non-reducing sugars were detected.
- Sample Z: Only Biuret test is positive (purple), indicating protein only.

Option C correctly matches this profile.

評分準則

C is correct [1 mark].
A: Incorrectly assigns reducing sugar to W.
B: Incorrectly assigns non-reducing sugar to Y.
D: Incorrectly assigns starch to W.

題目 35 · multiple_choice

1 分

The effect of a fixed concentration of a reversible inhibitor on the rate of an enzyme-catalysed reaction was investigated at different substrate concentrations.

The value of $V_{\text{max}}$ was found to remain unchanged, but the value of the Michaelis–Menten constant ($K_m$) increased in the presence of the inhibitor.

Which statement correctly explains this observation?

A.The inhibitor is competitive and binds reversibly to the active site, requiring higher substrate concentrations to reach $\frac{1}{2}V_{\text{max}}$.
B.The inhibitor is non-competitive and binds to an allosteric site, decreasing the turnover rate of the enzyme.
C.The inhibitor is competitive, increasing the affinity of the enzyme for the substrate.
D.The inhibitor binds covalently and permanently to the active site of the enzyme.

查看答案詳解

解題

A competitive inhibitor has a molecular shape complementary to the enzyme's active site and directly competes with the substrate for binding. At high substrate concentrations, substrate molecules outcompete the inhibitor, allowing the reaction to reach the same $V_{\text{max}}$. However, a higher substrate concentration is required to achieve half of the maximum velocity ($\frac{1}{2}V_{\text{max}}$), which results in an increased $K_m$ value.

評分準則

A is correct [1 mark].
B: Describes a non-competitive inhibitor, which would lower $V_{\text{max}}$.
C: Incorrectly states affinity increases when $K_m$ increases (higher $K_m$ reflects lower affinity).
D: Incorrectly states the inhibitor forms irreversible covalent bonds.

題目 36 · multiple_choice

1 分

Plant tissue cylinders of equal mass were placed in sucrose solutions of concentrations ranging from $0.0\text{ mol dm}^{-3}$ to $1.0\text{ mol dm}^{-3}$. After 2 hours, the percentage change in mass was determined.

At a sucrose concentration of $0.35\text{ mol dm}^{-3}$, the percentage change in mass was $0.0\%$.

Which deduction about the plant cells at this point of incipient equilibrium is correct?

A.The water potential of the plant cell cytoplasm is $0\text{ kPa}$.
B.Water molecules have ceased moving across the cell surface membrane.
C.The protoplasts of the cells exert maximum pressure against the cell wall.
D.The water potential of the cell sap is equal to the water potential of the $0.35\text{ mol dm}^{-3}$ sucrose solution.

查看答案詳解

解題

When there is no net change in mass (percentage change = $0.0\%$), the net movement of water by osmosis between the external sucrose solution and the plant cells is zero. This occurs when the water potential ($\Psi$) of the external solution is equal to the water potential of the internal cell sap/cytoplasm.

評分準則

D is correct [1 mark].
A: The water potential is not zero; pure water has a water potential of 0 kPa, and sucrose solutions have negative water potentials.
B: Water molecules still move into and out of the cell across the membrane, but at equal rates (dynamic equilibrium).
C: The cells are not fully turgid; full turgor occurs at maximum mass gain in pure water.

題目 37 · multiple_choice

1 分

A dividing root tip cell of an organism contains 24 chromosomes during the $\text{G}_1$ phase of interphase.

What are the total number of chromatids and the total number of centromeres present in this single cell during metaphase of mitosis?

A.24 chromatids and 12 centromeres
B.48 chromatids and 24 centromeres
C.48 chromatids and 48 centromeres
D.24 chromatids and 24 centromeres

查看答案詳解

解題

During the S phase of interphase, each chromosome replicates to form 2 sister chromatids held together at a single centromere.
- Number of chromosomes = 24
- Number of chromatids at metaphase = $24 \times 2 = 48$
- Number of centromeres at metaphase = 24 (each pair of sister chromatids shares one centromere until anaphase).

評分準則

B is correct [1 mark].
A: Confuses chromatids and chromosomes.
C: Incorrectly assumes centromeres replicate to double during metaphase.
D: Assumes chromatid number remains 24 after DNA replication.

題目 38 · multiple_choice

1 分

A double-stranded DNA molecule contains $3200$ base pairs. Analysis shows that $22\%$ of the total nitrogenous bases in this DNA molecule are adenine.

How many guanine bases and how many phosphodiester bonds are present in this double-stranded DNA molecule?

A.1792 guanine bases and 3199 phosphodiester bonds
B.1792 guanine bases and 6400 phosphodiester bonds
C.1792 guanine bases and 6398 phosphodiester bonds
D.896 guanine bases and 6398 phosphodiester bonds

查看答案詳解

解題

1. Total number of nucleotides = $3200 \times 2 = 6400$.
2. Percentage of Adenine (A) = $22\%$. By Chargaff's rules, Thymine (T) = $22\%$.
3. Total A + T = $44\%$. Therefore, Guanine (G) + Cytosine (C) = $100\% - 44\% = 56\%$.
4. Percentage of Guanine = $\frac{56\%}{2} = 28\%$.
5. Number of Guanine bases = $0.28 \times 6400 = 1792$.
6. In a linear double-stranded DNA molecule containing $N$ nucleotides in total (with 2 strands of $\frac{N}{2}$ nucleotides each), each strand of $3200$ nucleotides has $3200 - 1 = 3199$ phosphodiester bonds.
7. Total phosphodiester bonds = $3199 \times 2 = 6398$.

評分準則

C is correct [1 mark].
A: Used percentage of base pairs instead of total bases and calculated $3200 - 1$ phosphodiester bonds.
B: Calculated correct guanine count but incorrectly assumed 6400 phosphodiester bonds without subtracting terminal ends.
D: Incorrectly calculated guanine as $3200 \times 0.28 = 896$.

題目 39 · multiple_choice

1 分

Which sequence of events correctly describes the active loading of sucrose into sieve tube elements at a source leaf?

A.Protons are actively pumped out of companion cells $\rightarrow$ protons diffuse back through cotransporter proteins bringing sucrose with them $\rightarrow$ sucrose diffuses into sieve tube elements via plasmodesmata
B.Protons are actively pumped into companion cells $\rightarrow$ sucrose moves out through channel proteins $\rightarrow$ sucrose is actively transported into sieve tube elements
C.ATP binds to companion cell walls $\rightarrow$ sucrose diffuses through the phospholipid bilayer $\rightarrow$ mass flow carries sucrose into xylem vessels
D.Sodium ions are pumped into companion cells $\rightarrow$ sucrose is exchanged for sodium ions $\rightarrow$ sucrose enters the vacuole of sieve tube elements

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解題

Active loading involves:
1. Proton pumps actively transport $\text{H}^+$ ions out of companion cells into the surrounding cell walls using ATP.
2. This creates a steep $\text{H}^+$ concentration gradient across the companion cell surface membrane.
3. $\text{H}^+$ ions diffuse back into the companion cells via cotransporter proteins, bringing sucrose molecules along with them against the sucrose concentration gradient.
4. Sucrose then diffuses down a concentration gradient through plasmodesmata into adjacent sieve tube elements.

評分準則

A is correct [1 mark].
B: Incorrect direction of proton pumping (protons are pumped out, not into companion cells).
C: Cotransporter moves sucrose against its concentration gradient, not by simple diffusion.
D: Sucrose enters companion cells together with protons (cotransport), not in exchange for sodium ions.

題目 40 · multiple_choice

1 分

The graph shows the oxygen dissociation curves for human adult haemoglobin under two different partial pressures of carbon dioxide ($p\text{CO}_2$).

Curve 1 represents blood in the presence of a resting level of $p\text{CO}_2$.

Curve 2 is shifted to the right compared to Curve 1.

$$\begin{array}{c}
\text{Percentage saturation of haemoglobin with oxygen (\%)} \\
100 \left| \quad \begin{array}{cc} \mathbf{1} & \\ & \mathbf{2} \end{array} \right. \\
0 \rule{150pt}{0.5pt} \\
\quad 0 \qquad\qquad\qquad\qquad\qquad 14 \\
\text{Partial pressure of oxygen / kPa}
\end{array}$$

Which statement correctly explains the shift from Curve 1 to Curve 2 and its physiological advantage during vigorous exercise?

A.Higher $p\text{CO}_2$ increases haemoglobin's affinity for oxygen, enabling red blood cells to carry more oxygen to the brain.
B.Higher $p\text{CO}_2$ reduces haemoglobin's affinity for oxygen due to $\text{H}^+$ binding to haemoglobin, releasing more oxygen to respiring muscles.
C.Lower $p\text{CO}_2$ causes haemoglobin to release oxygen more readily in actively contracting muscles.
D.Higher blood pH causes the dissociation curve to shift to the right, increasing oxygen uptake in the pulmonary capillaries.

查看答案詳解

解題

During vigorous exercise, respiring muscle tissue produces more carbon dioxide, increasing $p\text{CO}_2$. Carbon dioxide reacts with water in red blood cells to form carbonic acid, catalysed by carbonic anhydrase. Carbonic acid dissociates into $\text{H}^+$ and $\text{HCO}_3^-$. Haemoglobin binds $\text{H}^+$ to form haemoglobinic acid (HHb), which lowers haemoglobin's affinity for oxygen (the Bohr effect). This shifts the oxygen dissociation curve to the right, causing haemoglobin to release more oxygen to actively respiring tissues at any given $p\text{O}_2$.

評分準則

B is correct [1 mark].
A: Shift to the right represents a decrease in oxygen affinity, not an increase, and is caused by high $p\text{CO}_2$.
C: High $p\text{CO}_2$ decreases affinity, but curve 2 is not caused by lower $p\text{CO}_2$.
D: The curve shifts right due to higher $p\text{CO}_2$ and $\text{H}^+$ concentration, not higher pH.

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卷二 AS Level 結構題

Answer all questions in the spaces provided on the question paper.

(a) [4 marks]
- TB pathogen: Mycobacterium tuberculosis / Mycobacterium bovis [1] (spelling must be binomial/correct)
- TB transmission: airborne droplets / aerosol / coughing / sneezing / inhalation of droplets [1]
- Cholera pathogen: Vibrio cholerae [1] (spelling must be correct)
- Cholera transmission: contaminated water / food / water-borne / faecal-oral route [1]

(b)(i) [3 marks]
- Penicillin inhibits transpeptidase / prevents formation of peptide cross-links in peptidoglycan [1]
- (Growing) bacterial cell wall becomes weak and lyses / bursts due to osmotic uptake of water [1]
- Human cells lack cell walls / peptidoglycan, and viruses are non-cellular / lack cell walls / have no bacterial metabolism to target [1]

(b)(ii) [2 marks - any two]
- Patients failing to finish prescribed course of antibiotics [1]
- Over-prescription / using antibiotics for viral infections [1]
- Widespread preventative / routine use in livestock farming / animal feed [1]
- Using broad-spectrum antibiotics when narrow-spectrum could be used [1]

(c) [1 mark - any one]
- BCG vaccine has variable / low efficacy in adults / across different geographic populations [1]
- Pathogen remains dormant / hidden intracellularly inside macrophages [1]
- Emergence of multidrug-resistant (MDR / XDR) strains [1]
- Long duration of treatment leading to poor patient compliance [1]
- High transmission in crowded / impoverished conditions / association with HIV coinfection [1]

Paper 3 Advanced Practical Skills

Answer all questions. Carry out the practical procedures and record all observations and drawings.

2 題目 · 40 分

題目 1 · Practical serial dilution and quantitative assay

22 分

You are required to investigate the effect of substrate concentration on the activity of the enzyme catalase. Catalase catalyses the breakdown of hydrogen peroxide into water and oxygen:

\[2\text{H}_2\text{O}_2 \xrightarrow{\text{catalase}} 2\text{H}_2\text{O} + \text{O}_2\]

You are provided with:
- $\textbf{H}$, $3.0\%$ hydrogen peroxide solution
- $\textbf{E}$, $1.0\%$ yeast suspension (source of catalase)
- $\textbf{U}$, a hydrogen peroxide solution of unknown concentration
- $\textbf{W}$, distilled water
- Filter paper discs (diameter $5\text{ mm}$)
- Forceps, stop-clock, test-tubes, and syringes

When a filter paper disc soaked in enzyme $\textbf{E}$ is placed at the bottom of a test-tube containing hydrogen peroxide, oxygen gas bubbles accumulate on the disc, causing it to float to the surface. The time taken ($t$) for the disc to rise from the bottom to the surface is a measure of the rate of reaction.

(a)
(i) You are required to prepare a serial dilution of $\textbf{H}$ ($3.0\%$) to produce four additional concentrations of hydrogen peroxide solution: $1.50\%$, $0.75\%$, $0.38\%$, and $0.19\%$. You must use distilled water $\textbf{W}$ and make $20.0\text{ cm}^3$ of each concentration.

Complete Table 1.1 to show how you will prepare these five concentrations of hydrogen peroxide solution.

Table 1.1
| Concentration of $\text{H}_2\text{O}_2$ / % | Volume of $\text{H}_2\text{O}_2$ solution / $\text{cm}^3$ | Source of $\text{H}_2\text{O}_2$ (concentration / %) | Volume of distilled water $\textbf{W}$ / $\text{cm}^3$ |
| :--- | :--- | :--- | :--- |
| 3.00 | 20.0 | stock $\textbf{H}$ (3.00%) | 0.0 |
| 1.50 | | | |
| 0.75 | | | |
| 0.38 | | | |
| 0.19 | | | |
[3]

(ii) State one significant hazard in this investigation and describe the corresponding safety precaution you must take. [1]

(iii) Carry out the practical procedure:
1. Place $10.0\text{ cm}^3$ of each hydrogen peroxide concentration ($3.00\%$, $1.50\%$, $0.75\%$, $0.38\%$, $0.19\%$) into separate, clearly labelled test-tubes.
2. Place $10.0\text{ cm}^3$ of $\textbf{U}$ into another test-tube.
3. Using forceps, immerse a paper disc in $\textbf{E}$ for $5\text{ s}$, then gently blot excess liquid on a paper towel.
4. Drop the disc into the first test-tube so it sinks to the bottom and immediately start the stop-clock.
5. Record the time taken ($t$) in seconds for the disc to rise to the liquid surface.
6. Repeat steps 3–5 to obtain three readings for each concentration and for $\textbf{U}$.
7. Calculate the mean time and the rate of reaction, where $\text{rate} = \frac{1000}{\text{mean } t}$.

Record all your raw and processed results in a single, fully labelled table. [5]

(iv) Using your results in (a)(iii), determine an estimate for the concentration of hydrogen peroxide in solution $\textbf{U}$. [1]

(v) Identify two sources of error or uncontrolled variables in this experimental method. [2]

(vi) Suggest three modifications to improve the accuracy and precision of this investigation. [3]

(b) In a separate investigation, a student measured the volume of oxygen produced by immobilized catalase at different substrate concentrations over a fixed period of $3\text{ minutes}$. The results are shown in Table 1.2.

Table 1.2
| Substrate concentration / $\text{mol dm}^{-3}$ | Volume of oxygen collected in 3 min / $\text{cm}^3$ |
| :--- | :--- |
| 0.05 | 4.2 |
| 0.10 | 8.0 |
| 0.20 | 14.6 |
| 0.40 | 22.4 |
| 0.60 | 25.8 |
| 0.80 | 26.0 |

(i) Plot a line graph of the data in Table 1.2 on suitable grid axes. [4]

(ii) Explain the shape of the graph between $0.05\text{ mol dm}^{-3}$ and $0.80\text{ mol dm}^{-3}$. [3]

查看答案詳解

解題

(a) (i) To prepare a two-fold serial dilution with a final volume of $20.0\text{ cm}^3$ for each concentration:
- To make $1.50\%$: transfer $10.0\text{ cm}^3$ of $3.00\%$ stock $\textbf{H}$ and add $10.0\text{ cm}^3$ of distilled water $\textbf{W}$.
- To make $0.75\%$: transfer $10.0\text{ cm}^3$ of $1.50\%$ solution and add $10.0\text{ cm}^3$ of $\textbf{W}$.
- To make $0.38\%$ (or $0.375\%$): transfer $10.0\text{ cm}^3$ of $0.75\%$ solution and add $10.0\text{ cm}^3$ of $\textbf{W}$.
- To make $0.19\%$ (or $0.1875\%$): transfer $10.0\text{ cm}^3$ of $0.38\%$ solution and add $10.0\text{ cm}^3$ of $\textbf{W}$.

(a) (ii) Hazard: Hydrogen peroxide ($\textbf{H}$) is an oxidizing agent and skin/eye irritant. Precaution: Wear safety goggles/gloves, and wash immediately if spilled on skin.

(a) (iii) Table of results:
- Headings: `Concentration of hydrogen peroxide / %`, `Time taken for disc to float, t / s` (with columns for `Trial 1`, `Trial 2`, `Trial 3`, and `Mean`), `Rate of reaction / s^-1` or `(1000 / mean t) / s^-1`.
- Times recorded to the nearest whole second (or $0.1\text{ s}$).
- Mean correctly calculated to consistent significant figures / decimal places.
- As concentration decreases from $3.00\%$ to $0.19\%$, time taken $t$ increases, so rate decreases.

(a) (iv) Unknown $\textbf{U}$ concentration estimated by comparing the mean time/rate of $\textbf{U}$ to the values obtained across the known concentrations.

(a) (v) Errors / limitations:
1. Non-uniform absorption of enzyme onto the paper discs (blotting variation).
2. Discs sticking to the glass wall of the test-tube on the way up.
3. Temperature fluctuations during the experiment.

(a) (vi) Improvements:
1. Use a micropipette to dispense an exact volume (e.g. $20\ \mu\text{L}$) of yeast/enzyme suspension onto each disc.
2. Measure gas volume directly using an inverted graduated burette or gas syringe.
3. Maintain test-tubes in a thermostatically controlled water bath at a constant temperature.

(b) (i) Graph plotting:
- $x$-axis: `Substrate concentration / mol dm^-3` (linear scale: e.g. $0$ to $1.0$, $0.1\text{ mol dm}^{-3}$ per large square).
- $y$-axis: `Volume of oxygen collected in 3 min / cm^3` (linear scale: e.g. $0$ to $30$, $5\text{ cm}^3$ per large square).
- Points plotted accurately within $\pm 0.5\text{ mm}$.
- Points joined with sharp straight ruled lines between points or a smooth curve of best fit.

(b) (ii) Explanation:
- At lower substrate concentrations ($0.05$ to $0.40\text{ mol dm}^{-3}$), substrate concentration is the limiting factor. An increase in substrate concentration results in more frequent successful collisions between substrate molecules and active sites, forming more enzyme-substrate complexes (ESCs) per unit time.
- At high substrate concentrations ($0.60$ to $0.80\text{ mol dm}^{-3}$), the rate levels off / reaches a plateau ($V_{\max}$). Enzyme concentration becomes the limiting factor because all active sites are virtually saturated with substrate molecules.

評分準則

(a) (i) [3 marks maximum]
- 1 mark: Correct volumes of $\text{H}_2\text{O}_2$ (all $10.0\text{ cm}^3$) and distilled water $\textbf{W}$ (all $10.0\text{ cm}^3$) for all four rows.
- 1 mark: Correct source concentration transferred for each step ($3.00\%$ used for $1.50\%$; $1.50\%$ used for $0.75\%$; $0.75\%$ used for $0.38\%$; $0.38\%$ used for $0.19\%$).
- 1 mark: Total volume shown or stated as $20.0\text{ cm}^3$ for each dilution step.

(a) (ii) [1 mark maximum]
- 1 mark: Identifies $\text{H}_2\text{O}_2$ as an irritant/corrosive AND states wearing safety glasses/eye protection/nitrile gloves OR wiping spills immediately with water. (Do not accept just 'be careful' or 'wear lab coat' alone).

(a) (iii) [5 marks maximum]
- 1 mark: Single table with fully enclosed gridlines and complete column headings with slash formatting for units (`Concentration / %`, `Time / s`, `Rate / s^-1` or arbitrary units).
- 1 mark: Replicate raw times recorded for all 5 concentrations and solution $\textbf{U}$.
- 1 mark: All raw time measurements recorded consistently as whole numbers or to 1 decimal place.
- 1 mark: Correct calculation of mean times and reaction rates ($\text{rate} = \frac{1000}{\text{mean } t}$).
- 1 mark: Trend showing increased time / decreased rate as $\text{H}_2\text{O}_2$ concentration decreases.

(a) (iv) [1 mark maximum]
- 1 mark: Value for $\textbf{U}$ stated with unit (%) consistent with candidate's data in Table (a)(iii).

(a) (v) [2 marks maximum]
- 1 mark each (any two):
- Variation in volume/mass of enzyme solution absorbed by paper discs during immersion/blotting.
- Friction / disc touching/sticking to the tube wall during ascent.
- Subjectivity in judging the exact start/stop moment of disc arrival at the surface.
- Temperature fluctuations affecting reaction rate.

(a) (vi) [3 marks maximum]
- 1 mark each (any three):
- Use a micropipette/microsyringe to deliver a standardized volume of enzyme to each disc.
- Use a wider test-tube / measuring cylinder to reduce contact with tube walls.
- Place reaction vessels in a thermostatically controlled water bath (e.g. at $25^\circ\text{C}$).
- Measure oxygen production directly using a gas syringe / manometer.
- Increase number of replicates (e.g. 5 repeats) to identify anomalies and improve reliability.

(b) (i) [4 marks maximum]
- 1 mark: Axes correctly orientated ($x$-axis: substrate concentration, $y$-axis: volume of oxygen) with full labels and units: `Substrate concentration / mol dm^-3` and `Volume of oxygen collected in 3 min / cm^3`.
- 1 mark: Linear scales on both axes such that data occupies $\ge 50\%$ of grid in both dimensions; sensible scale intervals (e.g. 1, 2, 5).
- 1 mark: All 6 points plotted accurately to within $\pm 0.5\text{ mm}$ (half a small square) with small neat crosses or circled dots.
- 1 mark: Points connected with thin, clear ruled straight lines from point to point OR a single smooth line of best fit; line does not extrapolate beyond $0.80\text{ mol dm}^{-3}$ or start before $0.05\text{ mol dm}^{-3}$.

(b) (ii) [3 marks maximum]
- 1 mark: (Between $0.05$ and $0.40\text{ mol dm}^{-3}$) Substrate concentration is the limiting factor / rate is directly proportional to concentration.
- 1 mark: More substrate molecules available $\rightarrow$ higher frequency of successful collisions with active sites $\rightarrow$ more enzyme-substrate complexes (ESCs) formed per unit time.
- 1 mark: (Between $0.60$ and $0.80\text{ mol dm}^{-3}$) Curve levels off / reaches plateau / $V_{\max}$ because enzyme concentration / active site availability becomes limiting / active sites are saturated.

題目 2 · Microscopy, drawing, and graticule calibration

18 分

You are provided with a prepared slide, \textbf{K1}, which is a transverse section through a dicotyledonous stem.

(a) (i) Make a large, labelled plan diagram of a sector of the stem on slide \textbf{K1} that includes at least one complete vascular bundle. Do not draw any individual cells.

Use a sharp pencil for your drawing. You are expected to draw the tissues in correct proportion. [5]

(ii) Select four adjacent, touching parenchyma cells from the cortex of \textbf{K1}.

Make a large, detailed drawing of these four cells. Use a sharp pencil and draw only what you observe. Label one visible structure. [4]

(b) An eyepiece graticule and a stage micrometer scale are used to determine the actual dimensions of structures on slide \textbf{K1}.

Each small division on the stage micrometer scale is equal to $0.01\text{ mm}$.

When viewed at a magnification of $\times 400$, $32$ small divisions of the stage micrometer scale align precisely with $80$ eyepiece graticule units (epu).

(i) Calculate the actual length represented by one eyepiece graticule unit at $\times 400$ magnification. Show your working and give your answer in micrometres ($\mu\text{m}$). [3]

(ii) The radial diameter of the largest xylem vessel element in one vascular bundle on \textbf{K1} is measured across the eyepiece graticule and found to be $19$ epu at $\times 400$ magnification.

Using your answer to (b)(i), calculate the actual radial diameter of this xylem vessel element. Give your answer to an appropriate number of significant figures with units. [2]

(c) Fig. 2.1 is a photomicrograph of a transverse section through the root of a different plant species, \textbf{K2}.

Prepare a table to record four observable structural differences between the transverse section of the stem on slide \textbf{K1} and the transverse section of the root shown in \textbf{K2}. [4]

查看答案詳解

解題

(a) (i) Criteria for sector plan diagram:
1. Sharp, continuous lines with no shading, feathering, or breaks; drawing occupies at least half of the provided drawing space.
2. Sector includes the outer boundary (epidermis/cuticle), cortex, vascular bundle (xylem and phloem), and central pith/medulla.
3. Vascular bundle clearly shows separate regions for xylem and phloem separated by a vascular cambium layer, in correct anatomical order (phloem towards outside, xylem towards centre).
4. Proportions of tissues are anatomically realistic (e.g. cortex depth and bundle size proportionate to the sector width).
5. Correct labels with straight ruled lines touching the target tissue (at least xylem and phloem correctly identified).

(ii) Criteria for high-power cell drawing:
1. Sharp, continuous lines; no shading; size occupies at least half the space.
2. Exactly four complete touching parenchyma cells drawn.
3. Double lines representing cell walls drawn around all cells, showing corners/intercellular air spaces where walls meet.
4. Cells have realistic polygonal/rounded shapes with realistic relative dimensions.
5. One correct label with a ruled guide line pointing directly to a visible feature (e.g. cell wall, lumen, intercellular space).

(b) (i) Total actual length of 32 micrometer divisions:
$32 \times 0.01\text{ mm} = 0.32\text{ mm}$
Convert mm to $\mu\text{m}$:
$0.32\text{ mm} \times 1000 = 320\,\mu\text{m}$
Actual length of 1 eyepiece graticule unit:
$320\,\mu\text{m} / 80\text{ epu} = 4.0\,\mu\text{m}$

(ii) Actual diameter = $19\text{ epu} \times 4.0\,\mu\text{m} = 76\,\mu\text{m}$
(or $0.076\text{ mm}$).

(c) Table comparing stem K1 and root K2:
- Feature / Stem K1 / Root K2:
1. Arrangement of vascular tissue: Vascular bundles arranged in a peripheral ring / Central vascular cylinder (stele) with star-shaped (xylem core).
2. Pith: Large central pith present / Pith absent (or central core occupied by xylem).
3. Cortex: Cortex is narrow (relative to overall diameter) / Cortex is extensive / wide.
4. Endodermis: Outer sheath or indistinct endodermis / Distinct single-cell layer endodermis enclosing central stele.

評分準則

(a)(i) [max 5 marks]:
- MP1: Quality of line: clear, unbroken lines of uniform thickness, no shading / cross-hatching, drawing occupies at least 50% of the space provided. [1]
- MP2: Correct tissue layers drawn in sequence: epidermis, cortex, sclerenchyma/cap (if present), phloem, xylem, pith. [1]
- MP3: Sector bounded correctly, showing at least one vascular bundle completely enclosed. [1]
- MP4: Proportions correct: vascular bundle occupies between 1/4 and 1/2 the radial thickness of the stem wall. [1]
- MP5: Accurate labels: at least two correct labels (e.g. xylem / phloem / epidermis / cortex) with straight, uncrossed label lines. [1]

(a)(ii) [max 4 marks]:
- MP1: Clear continuous lines with no gaps, overlaps, or sketchiness; drawn large enough (occupying at least 50% of space). [1]
- MP2: Exactly four adjacent, touching cells drawn. [1]
- MP3: Double cell walls shown throughout with distinct middle lamella / intercellular spaces shown at cell junctions. [1]
- MP4: One correct label to an observable feature (e.g. cell wall / lumen / middle lamella / intercellular space). [1]

(b)(i) [max 3 marks]:
- MP1: Shows multiplication of divisions by 0.01 mm (i.e. $32 \times 0.01 = 0.32\text{ mm}$ or $320\,\mu\text{m}$). [1]
- MP2: Shows division by 80 (i.e. $320 / 80$ or $0.32 / 80$). [1]
- MP3: Correct value of $4.0\,\mu\text{m}$ (or $4\,\mu\text{m}$). [1]

(b)(ii) [max 2 marks]:
- MP1: Correct multiplication of candidate's (b)(i) by 19 (e.g. $19 \times 4.0 = 76$). [1]
- MP2: Correct unit ($\mu\text{m}$ or $0.076\text{ mm}$) given to 2 or 3 significant figures. [1]

(c) [max 4 marks]:
- MP1: Table drawn with clear ruled grid lines and appropriate column headings (Feature, Stem K1, Root K2). [1]
- MP2 to MP4: Any three correct comparative observable differences [1 mark each]:
* Distribution of xylem / phloem: bundles in a ring in K1 vs central core / star-shape in K2.
* Presence of central pith: present in K1 vs absent in K2.
* Width / extent of cortex: narrow / small in K1 vs broad / large in K2.
* Endodermis / Casparian strip layer: not prominent in K1 vs prominent / clearly defined around stele in K2.
* Sclerenchyma / bundle caps: present outside phloem in K1 vs absent in K2.

Paper 4 A Level 結構題

Answer all questions in the spaces provided on the question paper.

(a)(i) Any 2 from:
1. Human (genomic) DNA contains introns / non-coding regions; [1]
2. Bacteria lack enzymes / machinery / spliceosomes to splice / remove introns; [1]
3. Mature mRNA contains only exons / coding sequence (so functional protein / correct amino acid sequence is produced); [1]
[Max 2 marks]

(a)(ii) Reverse transcriptase; [1]
[Accept: RNA-directed DNA polymerase]

(b)(i) Any 2 from:
1. Bacterial RNA polymerase binds to prokaryotic promoter (to initiate transcription); [1]
2. Eukaryotic promoter not recognised by bacterial RNA polymerase / bacterial transcription factors; [1]
3. Ensures (high levels of) transcription / expression of the factor VIII gene in the host bacterium; [1]
[Max 2 marks]

(b)(ii) 1. Bacteria containing the recombinant plasmid express the GFP gene / produce GFP; [1]
2. Transformed colonies glow / fluoresce green when irradiated with UV / ultraviolet / blue light; [1]
[Accept: non-transformed bacteria do not fluoresce / glow]

(c)(i) Any 2 from:
1. Short single-stranded DNA / oligonucleotides that anneal / bind (by complementary base pairing) to specific sequences at 3' ends of target DNA; [1]
2. Provide starting point / free 3' (-OH) end for DNA polymerase to attach nucleotides; [1]
3. Prevent template strands from re-annealing; [1]
[Max 2 marks]

(c)(ii) Any 1 from:
1. Thermostable / heat-stable (DNA polymerase) / does not denature at high temperatures (e.g. 95 °C); [1]
2. Synthesises complementary DNA strands by adding free (deoxyribo)nucleotides; [1]
[Max 1 mark]

Paper 5 Planning, Analysis and Evaluation

Answer all questions in the spaces provided on the question paper.

3 題目 · 30 分

題目 1 · free-response

20 分

Urease is an enzyme found in various plant seeds that catalyses the hydrolysis of urea into ammonia and carbon dioxide:

\[\text{Urea} + \text{H}_2\text{O} \xrightarrow{\text{urease}} 2\text{NH}_3 + \text{CO}_2\]

The ammonia produced dissolves to form ammonium and hydroxide ions, increasing the pH of the reaction mixture. The progress of the reaction can be monitored by recording the time taken for a pH indicator solution to change colour, or by measuring pH directly using a calibrated pH probe and data logger.

Heavy metals, such as zinc ions ($\text{Zn}^{2+}$), act as non-competitive inhibitors of plant urease.

A student investigated the effect of zinc sulfate ($\text{ZnSO}_4$) concentration on the rate of reaction of urease extracted from soybean seeds.

The student was provided with:
- a $1.0\%$ (w/v) urease enzyme stock solution
- a $5.0\%$ (w/v) urea substrate solution
- a $0.10\text{ mol dm}^{-3}$ zinc sulfate solution
- standard laboratory glassware and equipment, including a water bath, stopwatches, and digital pH probes connected to data loggers.

(a) (i) Identify the independent variable and the dependent variable in this investigation.

independent variable: ............................................................................................................
dependent variable: .............................................................................................................. [2]

(ii) The student prepared a range of zinc sulfate concentrations by diluting the $0.10\text{ mol dm}^{-3}$ stock solution to produce $50\text{ cm}^3$ of each concentration: $0.08$, $0.06$, $0.04$, $0.02$, and $0.00\text{ mol dm}^{-3}$.

Describe how the student prepared $50\text{ cm}^3$ of the $0.06\text{ mol dm}^{-3}$ and $0.02\text{ mol dm}^{-3}$ zinc sulfate solutions using proportional dilution.

...................................................................................................................................................
...................................................................................................................................................
................................................................................................................................................... [2]

(iii) Describe a method the student could use to determine the effect of zinc sulfate concentration on the initial rate of reaction of urease.

Your method should be detailed enough for another person to follow and obtain repeatable results.

Your method should include:
- how to control significant variables
- the measurements to be taken to determine the initial rate
- a suitable negative control
- key steps to ensure reliability
- a hazard and risk assessment with precautions. [8]

(b) In a separate part of the investigation, the student tested whether there was a significant difference between the initial rate of urea hydrolysis at $20\,^\circ\text{C}$ and at $35\,^\circ\text{C}$ in the presence of $0.02\text{ mol dm}^{-3}$ zinc sulfate.

Table 1.1 shows the student's results.

Table 1.1
| trial | initial rate of urea hydrolysis at $20\,^\circ\text{C}$ / $\mu\text{mol dm}^{-3}\text{ s}^{-1}$ | initial rate of urea hydrolysis at $35\,^\circ\text{C}$ / $\mu\text{mol dm}^{-3}\text{ s}^{-1}$ |
|---|---|---|
| 1 | 14.2 | 22.8 |
| 2 | 15.1 | 24.1 |
| 3 | 13.8 | 21.9 |
| 4 | 14.6 | 23.5 |
| 5 | 15.3 | 24.7 |
| mean | 14.6 | 23.4 |
| standard deviation ($s$) | 0.63 | 1.08 |
| standard error ($S_M$) | 0.28 | ........... |

(i) Calculate the standard error ($S_M$) for the rate of reaction at $35\,^\circ\text{C}$. Give your answer to two decimal places.

Use the formula:
\[S_M = \frac{s}{\sqrt{n}}\]

$S_M$ = ........................................................ [1]

(ii) State what the standard error values indicate about the reliability of the calculated mean rates. [1]

(iii) State a suitable null hypothesis for a statistical test comparing the two sets of data in Table 1.1. [1]

(iv) A Student's $t$-test was used to determine whether the difference between the mean initial rate at $20\,^\circ\text{C}$ and at $35\,^\circ\text{C}$ was statistically significant.

Calculate the degrees of freedom for this $t$-test. [1]

(v) The calculated value of $t$ was $16.00$.

Table 1.2 shows the critical values of $t$ for different degrees of freedom at various probability levels.

Table 1.2
| degrees of freedom | probability, $p = 0.05$ | probability, $p = 0.01$ | probability, $p = 0.001$ |
|---|---|---|---|
| 6 | 2.45 | 3.71 | 5.96 |
| 7 | 2.37 | 3.50 | 5.41 |
| 8 | 2.31 | 3.36 | 5.04 |
| 9 | 2.26 | 3.25 | 4.78 |
| 10 | 2.23 | 3.17 | 4.59 |

Explain the conclusion that can be drawn from the calculated $t$-value. [2]

(c) Suggest how the student could modify the investigation to distinguish experimentally between competitive and non-competitive inhibition by $\text{Zn}^{2+}$ ions. [2]

查看答案詳解

解題

(a) (i)
- Independent variable: concentration of zinc sulfate (solution) / $[\text{ZnSO}_4]$ / zinc ion concentration.
- Dependent variable: (initial) rate of reaction / rate of change of pH / rate of ammonia production / time taken to reach a specified pH change.

(ii) Using the dilution formula $C_1 V_1 = C_2 V_2$ to make $50\text{ cm}^3$ total volume from $0.10\text{ mol dm}^{-3}$ stock solution:
- For $0.06\text{ mol dm}^{-3}$: $V_1 = \frac{0.06 \times 50}{0.10} = 30.0\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ stock solution $+ 20.0\text{ cm}^3$ of distilled water.
- For $0.02\text{ mol dm}^{-3}$: $V_1 = \frac{0.02 \times 50}{0.10} = 10.0\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ stock solution $+ 40.0\text{ cm}^3$ of distilled water.

(iii) Method outline:
1. Standardised variables: Maintain constant temperature using a thermostatically controlled water bath (e.g. $25\,^\circ\text{C}$). Use constant volume and concentration of urease enzyme solution (e.g. $2.0\text{ cm}^3$ of $1.0\%$ urease) and urea substrate solution (e.g. $5.0\text{ cm}^3$ of $5.0\%$ urea).
2. Equilibration: Pre-incubate the enzyme, substrate, and inhibitor solutions in the water bath for $5\text{--}10\text{ minutes}$ before mixing to achieve thermal equilibrium.
3. Mixing & Monitoring: Add a fixed volume (e.g. $2.0\text{ cm}^3$) of zinc sulfate solution to the enzyme solution, mix, then add the urea substrate, start stopwatch immediately, and continuously record pH using a calibrated digital pH probe / data logger at regular time intervals (e.g. every 5 seconds for 1–2 minutes).
4. Determination of initial rate: Plot pH against time; calculate the gradient / slope of the initial linear portion (e.g. first 30 seconds) to find $\Delta\text{pH}/\Delta t$.
5. Negative control: Set up a tube replacing zinc sulfate with an equal volume of distilled water ($0.00\text{ mol dm}^{-3}$ inhibitor) or replacing active urease with boiled/denatured urease / distilled water to show that the pH change is due to enzyme action.
6. Reliability: Repeat the procedure at each zinc sulfate concentration at least 3 times (or run 5 replicates) and calculate the mean initial rate and standard deviation/error.
7. Safety/Risk Assessment: Zinc sulfate is toxic/harmful to the environment/irritant to skin and eyes; wear safety goggles and nitrile gloves, handle solutions carefully, and dispose of heavy metal waste in designated chemical waste containers.

(b) (i)
\[S_M = \frac{s}{\sqrt{n}} = \frac{1.08}{\sqrt{5}} = \frac{1.08}{2.23607} = 0.48299... \approx 0.48\]

(ii) The low standard error values (0.28 and 0.48) indicate that the sample means are close to the true population means / the data points have high precision and reliability.

(iii) Null hypothesis ($H_0$): There is no significant difference between the mean initial rate of urea hydrolysis at $20\,^\circ\text{C}$ and at $35\,^\circ\text{C}$ in the presence of zinc sulfate (any observed difference is due to chance alone).

(iv)
\[\text{Degrees of freedom } (\nu) = (n_1 - 1) + (n_2 - 1) = (5 - 1) + (5 - 1) = 8\]
(or $\nu = n_1 + n_2 - 2 = 5 + 5 - 2 = 8$).

(v)
- For $\nu = 8$, the critical value of $t$ at $p = 0.001$ is $5.04$.
- The calculated $t$-value ($16.00$) is greater than the critical value ($5.04$).
- Therefore, the probability that the difference is due to chance is less than $0.001$ ($p < 0.001$), so reject the null hypothesis.
- The difference between the mean initial rate at $20\,^\circ\text{C}$ and $35\,^\circ\text{C}$ is statistically highly significant.

(c)
- Measure initial rates across a range of increasing urea (substrate) concentrations, both in the presence and absence of a fixed concentration of $\text{Zn}^{2+}$ ions.
- Plot rate against substrate concentration (or Michaelis-Menten plot / Lineweaver-Burk plot).
- If increasing substrate concentration overcomes inhibition to achieve the same maximum velocity ($V_{\max}$), the inhibition is competitive; if $V_{\max}$ remains lower and cannot be restored by increasing substrate concentration, the inhibition is non-competitive.

評分準則

Part (a)(i) [2 marks]
- 1 mark for identifying independent variable as concentration of zinc sulfate / ZnSO4 / zinc ions.
- 1 mark for identifying dependent variable as (initial) rate of reaction / rate of pH change / rate of ammonia production.

Part (a)(ii) [2 marks]
- 1 mark for 0.06 mol dm^-3: 30 cm^3 of 0.10 mol dm^-3 ZnSO4 + 20 cm^3 of distilled water.
- 1 mark for 0.02 mol dm^-3: 10 cm^3 of 0.10 mol dm^-3 ZnSO4 + 40 cm^3 of distilled water.

Part (a)(iii) [8 marks max from the following points]:
- P1: Use of thermostatically controlled water bath to keep temperature constant (e.g., 25 °C or 30 °C).
- P2: Stating specific, constant volumes/concentrations of urease and urea substrate added to each tube.
- P3: Pre-incubation / equilibration of all solutions in water bath before mixing (minimum 5 minutes).
- P4: Continuous recording of pH using calibrated digital pH meter/probe with data logger / taking readings at specified regular short time intervals (e.g. every 5 or 10 s).
- P5: Method for determining initial rate by finding gradient of linear portion of pH vs time curve / change in pH over first 30–60 s.
- P6: Suitable negative control: replacing ZnSO4 with distilled water (0 mol dm^-3 ZnSO4) OR replacing enzyme with boiled/denatured enzyme or water.
- P7: Reliability: repeating each ZnSO4 concentration at least 3 times (minimum 3 replicates/trials) and calculating mean (and SD).
- P8: Hazard, risk and precaution: ZnSO4 is toxic / harmful to aquatic life / skin irritant AND precaution of wearing gloves / eye protection / safe chemical disposal.

Part (b)(i) [1 mark]
- 1 mark for correct calculation of SM: 0.48 (accept 0.483; reject 0.5 or 0.482 without rounding).

Part (b)(ii) [1 mark]
- 1 mark for stating that small SE indicates high reliability/precision of the mean / data points closely clustered around true population mean.

Part (b)(iii) [1 mark]
- 1 mark for correct null hypothesis stating no significant difference between the mean initial rates at 20 °C and 35 °C (due to chance alone).

Part (b)(iv) [1 mark]
- 1 mark for degrees of freedom = 8 (calculated as (5 - 1) + (5 - 1) or 5 + 5 - 2).

Part (b)(v) [2 marks]
- 1 mark for stating calculated t (16.00) is greater than critical value (5.04 for df = 8 at p = 0.001 / 2.31 at p = 0.05).
- 1 mark for concluding: reject null hypothesis AND the difference between the mean rates is statistically significant (p < 0.001 or p < 0.05).

Part (c) [2 marks]
- 1 mark for: vary/increase urea substrate concentration (over a wide range) in the presence and absence of inhibitor / Zn2+.
- 1 mark for: if maximum rate / Vmax is restored/unaffected at high [substrate], it is competitive; if maximum rate / Vmax is reduced / cannot be restored, it is non-competitive.

題目 2 · free-response

10 分

An ecologist investigated the biodiversity of ground-dwelling invertebrates in two contrasting habitats within a nature reserve:
• Habitat A: ancient deciduous broadleaved woodland
• Habitat B: managed coniferous plantation

(a) Outline how the ecologist should use pitfall traps to collect a representative sample of ground-dwelling invertebrates in Habitat A. [3]

(b) The ecologist trapped invertebrates in both habitats over a 48-hour period. Table 1 shows the number of individuals of each beetle species collected in Habitat B.

Table 1
| Species | Number of individuals, $n$ | $\frac{n}{N}$ | $\left(\frac{n}{N}\right)^2$ |
| :--- | :--- | :--- | :--- |
| Species 1 | 35 | 0.35 | 0.1225 |
| Species 2 | 10 | 0.10 | 0.0100 |
| Species 3 | 5 | 0.05 | 0.0025 |
| Species 4 | 45 | 0.45 | 0.2025 |
| Species 5 | 5 | 0.05 | 0.0025 |
| **Total ($N$)** | 100 | | **$\sum \left(\frac{n}{N}\right)^2 = $** |

Simpson's index of diversity ($D$) is calculated using the formula:
\[D = 1 - \sum \left(\frac{n}{N}\right)^2\]

(i) Complete Table 1 by calculating $\sum \left(\frac{n}{N}\right)^2$. [1]

(ii) Calculate Simpson's index of diversity ($D$) for Habitat B. Give your answer to 2 decimal places. [1]

(c) The value of $D$ for Habitat A was calculated as 0.84.
State what the difference in the values of $D$ indicates about the invertebrate communities in Habitat A and Habitat B. [1]

(d) The ecologist also recorded soil moisture and the abundance of Species 1 across 10 different sampling sites across the reserve to determine if a correlation exists. The ecologist determined that the abundance data were not normally distributed.

(i) State a null hypothesis for this statistical investigation. [1]

(ii) State the most appropriate statistical test to determine whether there is a significant correlation between soil moisture and the abundance of Species 1, and give a reason for your choice. [1]

(e) Suggest two limitations of using pitfall traps to evaluate the total animal biodiversity of a woodland habitat. [2]

查看答案詳解

解題

(a) Representative sampling with pitfall traps:
1. Lay out a coordinate grid over Habitat A using tape measures and use a random number generator to obtain coordinates for trap placement.
2. Ensure traps are installed consistently (e.g., top rim flush with the soil surface, fitted with a raised rain/predation cover).
3. Use a large, standardised number of traps (e.g., $\ge 10$ traps) and leave them for a standardised duration (e.g., 48 hours).

(b) (i) $\sum \left(\frac{n}{N}\right)^2 = 0.1225 + 0.0100 + 0.0025 + 0.2025 + 0.0025 = 0.3400$

(ii) $D = 1 - 0.3400 = 0.66$

(c) Habitat A has a higher species diversity / greater species richness and evenness than Habitat B.

(d) (i) Null hypothesis: There is no significant correlation / relationship between soil moisture and the abundance of Species 1.
(ii) Spearman's rank correlation test. Reason: The data are not normally distributed (or continuous data can be converted to ranks).

(e) Limitations:
1. Only samples surface-active/mobile ground fauna (misses arboreal, flying, or deep burrowing species).
2. Trap success depends on activity/locomotion rates rather than absolute abundance (e.g., dense vegetation may impede movement into traps).

評分準則

(a) 1 mark per bullet point (max 3):
- Random sampling using coordinate grid and random number generator / table [1]
- Standardisation of trap design / installation (e.g., rim flush with ground level / cover placed over trap to protect from rain or predators) [1]
- Standardisation of sampling effort (e.g., all traps left for identical time / multiple traps used across the site) [1]
Reject: placing traps in areas that 'look good'.

(b)(i) 0.34 or 0.3400 [1]
(b)(ii) 0.66 [1] (Allow ecf from (b)(i))

(c) Habitat A has a higher diversity / more diverse community than Habitat B OR Habitat A has higher species evenness / less dominance by one or two species [1]

(d)(i) There is no significant correlation / relationship between soil moisture and the abundance of Species 1 [1]
Reject: references to 'difference' instead of 'correlation/relationship'.

(d)(ii) Spearman's rank (correlation) test AND data are not normally distributed / data can be ranked / non-parametric test [1]
Reject: Pearson's correlation (requires normally distributed data).

(e) 1 mark per distinct point (max 2):
- Only catches active ground-surface organisms / does not sample flying, climbing/arboreal, or burrowing organisms [1]
- Trapping rate reflects activity/mobility of animals rather than true population density / dense vegetation may obstruct movement [1]
- Predation may occur inside the trap / cannibalism alters counts [1]

題目 3 · null

0 分

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Thinka Jun 2026 (V2) Cambridge International A Level-Style Mock — Biology (9700)

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