An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V1) Cambridge International A Level Biology (0610) paper. Not affiliated with or reproduced from Cambridge.
Paper 21: Multiple Choice (Extended)
Answer all 40 multiple-choice questions. Choose the one you consider correct.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
Which row correctly identifies the products of anaerobic respiration in yeast cells and in human muscle cells?
A.Yeast: ethanol and carbon dioxide | Muscle: lactic acid only
B.Yeast: lactic acid and carbon dioxide | Muscle: ethanol only
C.Yeast: ethanol only | Muscle: lactic acid and carbon dioxide
D.Yeast: carbon dioxide only | Muscle: ethanol and lactic acid
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Worked solution
In yeast cells, anaerobic respiration (fermentation) produces ethanol and carbon dioxide. In human muscle cells during vigorous exercise, anaerobic respiration produces lactic acid only. Therefore, row A is correct.
Marking scheme
1 mark for identifying the correct products for both yeast and muscle cells. Reject options where lactic acid is produced in yeast or ethanol in human muscle.
Question 2 · multiple-choice
1 marks
Four cylinders of potato tissue, each of equal initial mass of 5.0 g, were placed in sucrose solutions of different concentrations. After one hour, the final mass of each cylinder was measured:
- Cylinder W: final mass 5.8 g - Cylinder X: final mass 5.2 g - Cylinder Y: final mass 4.8 g - Cylinder Z: final mass 4.1 g
In which solutions did water leave the potato cells by osmosis?
A.W and X
B.Y and Z
C.W only
D.Z only
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Worked solution
Water leaves cells by osmosis when the external solution has a lower water potential than the cell sap, resulting in a loss of mass. Since the initial mass was 5.0 g, a final mass of less than 5.0 g (as seen in Y and Z) indicates water loss.
Marking scheme
1 mark for identifying that cylinders Y and Z lost mass, showing that water left the cells by osmosis.
Question 3 · multiple-choice
1 marks
What is the correct pathway of an oxygen molecule as it diffuses from the lumen of an alveolus into the blood?
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Worked solution
An oxygen molecule starts in the alveolus. It must first diffuse through the single-celled alveolar wall, then across the single-celled capillary wall into the blood plasma, and finally across the red blood cell membrane to bind with hemoglobin.
Marking scheme
1 mark for the correct sequence of barriers crossed by oxygen during diffusion into the blood.
Question 4 · multiple-choice
1 marks
Which row correctly matches a region of the alimentary canal with its function?
A.colon — absorption of remaining water
B.stomach — mechanical digestion only
C.mouth — chemical digestion of proteins
D.ileum — secretion of amylase and lipase
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Worked solution
The colon is responsible for the absorption of remaining water from undigested food material. The stomach performs both mechanical and chemical digestion; the mouth digests starch (not proteins); the ileum absorbs digested food molecules but does not secrete amylase or lipase (which are produced by the pancreas).
Marking scheme
1 mark for identifying the correct function of the colon as specified by the Cambridge syllabus. Reject options with incorrect digestion sites or organ secretions.
Question 5 · multiple-choice
1 marks
Which statement describes how DNA controls the production of a specific protein in a cell?
A.The sequence of bases in a gene determines the sequence of amino acids in the protein.
B.The sequence of amino acids in DNA determines the sequence of bases in the protein.
C.DNA molecules leave the nucleus and act as templates for protein assembly at the ribosome.
D.Each gene contains the code to produce all types of proteins required by a cell.
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Worked solution
A gene is a length of DNA. The sequence of bases within this gene acts as a code that determines the specific order of amino acids assembled to form a protein. DNA itself does not contain amino acids, does not leave the nucleus (mRNA does), and each gene only codes for one specific protein.
Marking scheme
1 mark for identifying that the sequence of bases in a gene determines the sequence of amino acids in a protein.
Question 6 · multiple-choice
1 marks
A student draws a plant cell. The actual length of the cell is 0.08 mm. The student's drawing of the cell has a length of 40 mm. What is the magnification of the student's drawing?
A.\(\times 0.002\)
B.\(\times 50\)
C.\(\times 500\)
D.\(\times 5000\)
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Worked solution
Using the formula: \(M = \frac{I}{A}\), where \(I\) is the image (drawing) size and \(A\) is the actual size. \(M = \frac{40\text{ mm}}{0.08\text{ mm}} = 500\). Thus, the magnification is \(\times 500\).
Marking scheme
1 mark for calculating the magnification correctly as \(\times 500\). Reject options with inverted calculation ratios or decimal errors.
Question 7 · multiple-choice
1 marks
Which row correctly compares natural selection and artificial selection?
A.Selection pressure: exerted by the environment in natural selection, and by humans in artificial selection.
B.Rate of evolution: slower in natural selection, and even slower in artificial selection.
C.Outcome: beneficial to humans only in natural selection, and beneficial to the species in artificial selection.
D.Genetic variation: always eliminated in natural selection, and always increased in artificial selection.
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Worked solution
In natural selection, environmental factors (like predation, competition, or climate) exert the selection pressure. In artificial selection, humans select individuals with desirable features for breeding. Therefore, option A is correct.
Marking scheme
1 mark for identifying the correct agent of selection pressure for both types of selection. Reject incorrect descriptions of rate, outcome, or genetic variation.
Question 8 · multiple-choice
1 marks
Which statement describes the effect of a very high temperature (above \(60^\circ\text{C}\)) on an enzyme-controlled reaction?
A.The kinetic energy of the substrate and enzyme molecules decreases, reducing the rate of collisions.
B.The shape of the active site changes permanently, preventing the substrate from binding.
C.The substrate molecules are denatured, so they can no longer fit into the active site.
D.The activation energy of the reaction is lowered, causing the reaction rate to increase.
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Worked solution
At high temperatures, the thermal energy breaks the bonds maintaining the three-dimensional shape of the enzyme. This permanently alters the shape of the active site (denaturation), meaning the substrate can no longer fit into it, and the reaction stops.
Marking scheme
1 mark for identifying that high temperatures cause permanent changes to the shape of the enzyme's active site (denaturation), preventing substrate binding.
Question 9 · Multiple Choice
1 marks
A respirometer is used to measure the rate of respiration in germinating seeds. It contains potassium hydroxide solution to absorb carbon dioxide. During the experiment, the level of liquid in the capillary tube moves towards the test-tube containing the seeds. Which process causes this movement of liquid?
A.Absorption of carbon dioxide by the seeds
B.Release of carbon dioxide by the seeds
C.Consumption of oxygen by the seeds
D.Production of water vapour by the seeds
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Worked solution
Potassium hydroxide absorbs carbon dioxide. As the germinating seeds respire aerobically, they consume oxygen gas from the tube. Since the carbon dioxide produced is continuously absorbed, the total volume and pressure of the gas inside the tube decreases. This pressure drop draws the liquid in the capillary tube towards the respirometer.
Marking scheme
1 mark: Correctly identifies oxygen consumption as the process that decreases gas volume, pulling the liquid index.
Question 10 · Multiple Choice
1 marks
Four cylinders of potato tissue were weighed and placed in sucrose solutions of different concentrations: \(0.0\text{ mol/dm}^3\), \(0.2\text{ mol/dm}^3\), \(0.4\text{ mol/dm}^3\), and \(0.6\text{ mol/dm}^3\). After two hours, they were reweighed and the percentage change in mass was calculated. In which solution would the potato cylinders show the greatest increase in mass, and in which solution would they show the greatest decrease in mass?
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Worked solution
In \(0.0\text{ mol/dm}^3\) sucrose solution (pure water), the water potential is higher outside the potato cells than inside, causing water to enter by osmosis and giving the greatest mass increase. In \(0.6\text{ mol/dm}^3\) sucrose solution, the water potential is lowest outside, causing the maximum net movement of water out of the cells by osmosis, resulting in the greatest mass decrease.
Marking scheme
1 mark: Correctly identifies \(0.0\text{ mol/dm}^3\) for greatest increase and \(0.6\text{ mol/dm}^3\) for greatest decrease in mass.
Question 11 · Multiple Choice
1 marks
Which row correctly shows the approximate percentages of oxygen, carbon dioxide, and nitrogen in expired air under normal resting conditions?
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Worked solution
During gas exchange, the body absorbs oxygen and releases carbon dioxide. Therefore, expired air contains less oxygen (around 16%) and more carbon dioxide (around 4.0%) than inspired air (which is 21% oxygen and 0.04% carbon dioxide). The percentage of nitrogen remains relatively unchanged at around 78%.
Marking scheme
1 mark: Selects the row showing 16% oxygen, 4.0% carbon dioxide, and 78% nitrogen as correct expired air values.
Question 12 · Multiple Choice
1 marks
An aquatic plant was placed in a beaker of water under a light source. The rate of photosynthesis was determined by counting the number of oxygen bubbles released per minute. Which modification would increase the rate of bubble production?
A.Moving the light source further away from the beaker
B.Adding sodium hydrogencarbonate to the water
C.Lowering the temperature of the water to \(5^\circ\text{C}\)
D.Covering the beaker with a black cardboard box
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Worked solution
Sodium hydrogencarbonate releases dissolved carbon dioxide when added to water. Since carbon dioxide is a reactant and often a limiting factor for photosynthesis under these conditions, increasing its concentration increases the rate of photosynthesis, leading to a higher rate of oxygen bubble production.
Marking scheme
1 mark: Recognizes that adding sodium hydrogencarbonate increases the carbon dioxide concentration, which increases the photosynthetic rate.
Question 13 · Multiple Choice
1 marks
Where is amylase produced, and where does it carry out its function in the human digestive system?
A.Produced by pancreas only; active in duodenum only
B.Produced by salivary glands only; active in mouth only
C.Produced by salivary glands and pancreas; active in mouth and duodenum
D.Produced by stomach and pancreas; active in stomach and duodenum
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Worked solution
Amylase is produced in both the salivary glands (salivary amylase) and the pancreas (pancreatic amylase). Salivary amylase is active in the mouth, while pancreatic amylase is secreted into the duodenum (small intestine) where it breaks down starch into maltose.
Marking scheme
1 mark: Identifies both salivary glands and pancreas as sites of production, and mouth and duodenum as sites of action.
Question 14 · Multiple Choice
1 marks
Which statement correctly describes the relationship between DNA, genes, and chromosomes?
A.A chromosome contains DNA, and a gene is a length of DNA that codes for a specific protein.
B.A gene is a chromosome, and DNA is a length of protein that codes for a gene.
C.A chromosome is a protein, and a gene is a length of DNA that codes for a chromosome.
D.A gene contains chromosomes, and a protein is a length of DNA that codes for a gene.
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Worked solution
A chromosome is a thread-like structure consisting of DNA. A gene is defined as a specific length of DNA that codes for a protein. Thus, chromosomes contain DNA, and a gene is a length of that DNA.
Marking scheme
1 mark: Selects the correct relationship showing that a chromosome contains DNA and a gene is a coding length of DNA.
Question 15 · Multiple Choice
1 marks
A student observes a palisade mesophyll cell under a microscope. The actual length of the cell is \(0.08\text{ mm}\). In the student's drawing, the cell is \(40\text{ mm}\) long. What is the magnification of the drawing?
A.\(\times 0.002\)
B.\(\times 5\)
C.\(\times 50\)
D.\(\times 500\)
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Worked solution
Magnification is calculated using the formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Here, the image size (length of drawing) is \(40\text{ mm}\) and the actual size is \(0.08\text{ mm}\). Substituting these values: \(\text{Magnification} = \frac{40}{0.08} = 500\), which is written as \(\times 500\).
Marking scheme
1 mark: Correctly calculates magnification using the formula, showing a magnification of \(\times 500\).
Question 16 · Multiple Choice
1 marks
Which statement about natural selection is correct?
A.It results in the extinction of all species within an ecosystem over a short period.
B.It occurs when humans selectively breed organisms with desirable features.
C.It leads to the transmission of advantageous alleles to the next generation.
D.It is driven by the choice of individual organisms to mutate their genes to survive.
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Worked solution
Natural selection occurs when environmental factors select for individuals with advantageous phenotypes. These individuals are more likely to survive, reproduce, and pass on their advantageous alleles to their offspring, leading to evolutionary change over generations.
Marking scheme
1 mark: Identifies that natural selection results in the transmission of advantageous alleles to subsequent generations.
Question 17 · Multiple Choice
1 marks
During vigorous exercise, human muscle cells respire anaerobically. Yeast cells also undergo anaerobic respiration when oxygen is absent. Which row correctly compares the products of anaerobic respiration in human muscle cells and in yeast cells?
A.Muscle cells: Lactic acid and carbon dioxide | Yeast: Ethanol and carbon dioxide
B.Muscle cells: Lactic acid only | Yeast: Ethanol and carbon dioxide
C.Muscle cells: Lactic acid and carbon dioxide | Yeast: Ethanol only
D.Muscle cells: Lactic acid only | Yeast: Ethanol only
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Worked solution
In humans, anaerobic respiration in muscle cells produces lactic acid only (no carbon dioxide is produced). In yeast cells, anaerobic respiration (fermentation) produces ethanol and carbon dioxide.
Marking scheme
Award 1 mark for the correct option (B) which correctly identifies the products for both muscle cells and yeast.
Question 18 · Multiple Choice
1 marks
Four similar cylinders of potato tissue were immersed in sucrose solutions of different concentrations: \(0.0\text{ mol/dm}^3\), \(0.2\text{ mol/dm}^3\), \(0.6\text{ mol/dm}^3\), and \(0.8\text{ mol/dm}^3\) for two hours. Which solution concentration would cause the potato cells to become flaccid and show the greatest percentage decrease in mass?
A.\(0.0\text{ mol/dm}^3\)
B.\(0.2\text{ mol/dm}^3\)
C.\(0.6\text{ mol/dm}^3\)
D.\(0.8\text{ mol/dm}^3\)
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Worked solution
The highest concentration of sucrose (\(0.8\text{ mol/dm}^3\)) has the lowest water potential. Water moves out of the potato cells by osmosis down the water potential gradient, causing the cells to become flaccid and leading to the greatest percentage decrease in mass.
Marking scheme
Award 1 mark for identifying the highest concentration (D) as causing the greatest water loss and flaccidity.
Question 19 · Multiple Choice
1 marks
What are the changes in the volume of the thorax and the pressure in the lungs when a person inhales?
A.Volume of thorax: decreases | Pressure in lungs: decreases
B.Volume of thorax: decreases | Pressure in lungs: increases
C.Volume of thorax: increases | Pressure in lungs: decreases
D.Volume of thorax: increases | Pressure in lungs: increases
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Worked solution
During inhalation, the diaphragm and external intercostal muscles contract, which increases the volume of the thorax. This increase in volume leads to a decrease in the air pressure inside the lungs below atmospheric pressure, causing air to enter the lungs.
Marking scheme
Award 1 mark for selecting the correct row (C) showing increased volume and decreased pressure.
Question 20 · Multiple Choice
1 marks
At a certain light intensity, a plant has a photosynthetic rate of 25 units at \(15\text{ }^\circ\text{C}\) and \(0.04\%\) \(\text{CO}_2\). When the temperature is increased to \(25\text{ }^\circ\text{C}\) (with \(\text{CO}_2\) kept at \(0.04\%\)), the rate remains 25 units. When the \(\text{CO}_2\) concentration is increased to \(0.1\%\) (at \(15\text{ }^\circ\text{C}\)), the rate increases to 45 units. Which factor is limiting the rate of photosynthesis at \(15\text{ }^\circ\text{C}\) and \(0.04\%\) \(\text{CO}_2\)?
A.Light intensity
B.Carbon dioxide concentration
C.Temperature
D.Water availability
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Worked solution
Because increasing the concentration of carbon dioxide increases the rate of photosynthesis (from 25 to 45 units), while increasing the temperature does not, carbon dioxide is the limiting factor under the initial conditions.
Marking scheme
Award 1 mark for selecting the correct limiting factor (B) based on the experimental data provided.
Question 21 · Multiple Choice
1 marks
Which row correctly describes the production, storage, and site of action of bile in the human digestive system?
A.Produced by: gall bladder | Stored in: liver | Site of action: duodenum
B.Produced by: liver | Stored in: gall bladder | Site of action: duodenum
C.Produced by: liver | Stored in: gall bladder | Site of action: stomach
D.Produced by: pancreas | Stored in: gall bladder | Site of action: ileum
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Worked solution
Bile is produced by cells in the liver, stored in the gall bladder, and released into the duodenum (the first part of the small intestine) where it emulsifies fats and neutralises acidic stomach contents.
Marking scheme
Award 1 mark for identifying liver as the producer, gall bladder as the storage organ, and duodenum as the site of action (B).
Question 22 · Multiple Choice
1 marks
A student measures a drawing of a plant cell. The measured width of the drawing is \(48\text{ mm}\). The actual width of the plant cell is \(80\text{ }\mu\text{m}\). What is the magnification of the drawing?
A.\(\times 0.6\)
B.\(\times 6\)
C.\(\times 600\)
D.\(\times 6000\)
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Worked solution
First, convert the image width to the same units as the actual width: \(48\text{ mm} = 48\text{ }000\text{ }\mu\text{m}\). Use the formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{48\text{ }000\text{ }\mu\text{m}}{80\text{ }\mu\text{m}} = 600\). Therefore, the magnification is \(\times 600\).
Marking scheme
Award 1 mark for calculating the correct magnification (C) by converting units correctly and using the magnification formula.
Question 23 · Multiple Choice
1 marks
The sequence of bases in a short section of a DNA template strand is: \(\text{A-T-G-C-C-G-T-A-A}\). What is the sequence of bases in the mRNA strand transcribed from this DNA template?
A.\(\text{U-A-C-G-G-C-A-U-U}\)
B.\(\text{T-A-C-G-G-C-A-T-T}\)
C.\(\text{U-T-C-G-G-C-T-U-U}\)
D.\(\text{A-U-G-C-C-G-U-A-A}\)
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Worked solution
During transcription, complementary base pairing occurs. Adenine (A) in DNA pairs with Uracil (U) in RNA, Thymine (T) pairs with Adenine (A), Guanine (G) pairs with Cytosine (C), and Cytosine (C) pairs with Guanine (G). Therefore, \(\text{A-T-G-C-C-G-T-A-A}\) transcribes to \(\text{U-A-C-G-G-C-A-U-U}\).
Marking scheme
Award 1 mark for the correct complementary RNA sequence (A), noting that RNA contains Uracil (U) instead of Thymine (T).
Question 24 · Multiple Choice
1 marks
Which statement describes a feature that is characteristic of natural selection but NOT of artificial selection?
A.Humans choose which individuals breed based on desirable traits.
B.New alleles are created directly by the selection process to adapt to the environment.
C.Environmental factors determine which individuals are more likely to survive and reproduce.
D.It leads to a decrease in the genetic variation of a population over many generations.
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Worked solution
Natural selection is driven by environmental factors (such as predators, climate, and disease) which determine survival and reproductive success. In contrast, artificial selection is driven by human choice to propagate desirable features.
Marking scheme
Award 1 mark for identifying environmental selection (C) as the distinguishing factor of natural selection.
Question 25 · Multiple Choice
1 marks
A student places epidermal cells of a red onion in a highly concentrated sucrose solution. Which row correctly describes the net movement of water and the appearance of the cells after 30 minutes?
A.Net movement of water: out of the cells | Appearance: cytoplasm pulls away from the cell wall (plasmolysed)
B.Net movement of water: out of the cells | Appearance: cells swell and burst (lysed)
C.Net movement of water: into the cells | Appearance: cytoplasm presses firmly against the cell wall (turgid)
D.Net movement of water: into the cells | Appearance: cells shrink and shrivel (crenated)
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Worked solution
The concentrated sucrose solution has a lower water potential than the vacuole and cytoplasm of the onion cells. Water moves out of the cells by osmosis down a water potential gradient, causing the vacuole and cytoplasm to shrink and pull away from the cell wall. This state is known as plasmolysis (the cell is plasmolysed). Plant cells do not burst (lyse) due to the presence of a rigid cell wall, nor do they become turgid when water leaves.
Marking scheme
Award 1 mark for the correct option A. - Reject other options: B (plant cells do not burst due to the cell wall), C and D (water potential gradient causes water loss, not uptake).
Question 26 · Multiple Choice
1 marks
Red blood cells were placed into two different solutions: distilled water and a \(5.0\%\) concentrated salt solution. Which row correctly describes the changes observed in the appearance of these cells?
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Worked solution
Distilled water has a higher water potential than the red blood cell cytoplasm, so water enters the cells by osmosis. Because animal cells lack a cell wall, they swell and burst (lyse). A \(5.0\%\) salt solution has a lower water potential than the cytoplasm, so water leaves the cells by osmosis, causing them to shrink and shrivel (crenate).
Marking scheme
Award 1 mark for the correct option A. - B is incorrect because red blood cells will shrink, not remain unchanged in a highly concentrated solution. - C is incorrect because distilled water causes lysis, not remaining unchanged. - D is incorrect because the effects of the solutions are reversed.
Question 27 · Multiple Choice
1 marks
Which statement correctly describes aerobic respiration?
A.It is the release of a relatively small amount of energy by the breakdown of glucose in the cytoplasm.
B.It is the release of a relatively large amount of energy by the breakdown of food molecules in the presence of oxygen.
C.It is the chemical breakdown of glucose to produce ethanol and carbon dioxide in the absence of oxygen.
D.It is the conversion of lactic acid back into glucose in the liver using oxygen.
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Worked solution
Aerobic respiration is the release of a relatively large amount of energy by the chemical breakdown of glucose (and other nutrient molecules) in the presence of oxygen. Carbon dioxide and water are produced as waste products.
Marking scheme
Award 1 mark for the correct option B. - A is incorrect as it describes anaerobic respiration's energy release and wrong cellular location (aerobic occurs primarily in mitochondria). - C describes anaerobic respiration in yeast. - D describes the repayment of an oxygen debt, which is not the primary definition of aerobic respiration.
Question 28 · Multiple Choice
1 marks
Yeast cells and human muscle cells can both perform anaerobic respiration when oxygen is absent. Which row correctly identifies the chemical products of anaerobic respiration in each of these cell types?
A.Yeast: ethanol and carbon dioxide | Muscle cells: lactic acid only
B.Yeast: lactic acid only | Muscle cells: ethanol and carbon dioxide
C.Yeast: lactic acid and carbon dioxide | Muscle cells: lactic acid and water
D.Yeast: ethanol only | Muscle cells: lactic acid and carbon dioxide
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Worked solution
In yeast, anaerobic respiration (fermentation) produces ethanol (alcohol) and carbon dioxide. In human muscle cells, anaerobic respiration during vigorous exercise produces lactic acid only. No carbon dioxide is produced during anaerobic respiration in mammalian muscles.
Marking scheme
Award 1 mark for the correct option A. - B is incorrect because it reverses the products. - C is incorrect because anaerobic respiration in human muscles does not produce water or carbon dioxide. - D is incorrect because yeast fermentation also produces carbon dioxide.
Question 29 · Multiple Choice
1 marks
How do the structural features of an alveolus and its associated capillary network maximize the rate of diffusion of oxygen into the blood?
A.Thick alveolar walls provide a short diffusion pathway, and a slow rate of blood flow maintains a steep gradient.
B.Thin alveolar walls decrease the diffusion distance, and a high concentration of oxygen in the capillary blood maintains a steep gradient.
C.A large surface area increases contact, and a steep concentration gradient is maintained by continuous blood flow and ventilation.
D.A dry surface area prevents water loss, and a narrow capillary diameter slows down red blood cells to allow more time for diffusion.
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Worked solution
The rate of diffusion is maximized by a large total surface area of many alveoli, a very thin diffusion barrier (both alveolar and capillary walls are only one cell thick), and the maintenance of a steep concentration gradient by constant ventilation (breathing) and continuous blood flow carrying oxygen away.
Marking scheme
Award 1 mark for the correct option C. - A is incorrect because thick walls decrease the rate of diffusion. - B is incorrect because a high concentration of oxygen in the capillary blood would reduce the diffusion gradient. - D is incorrect because the gas exchange surface must be moist to dissolve oxygen, not dry.
Question 30 · Multiple Choice
1 marks
Which sequence of events correctly describes the synthesis of a protein from the genetic code in DNA?
A.mRNA is manufactured in the cytoplasm \(\rightarrow\) mRNA passes into the nucleus \(\rightarrow\) DNA copies the code \(\rightarrow\) amino acids are assembled at the cell membrane
B.DNA template is transcribed to make mRNA in the nucleus \(\rightarrow\) mRNA moves to a ribosome \(\rightarrow\) ribosome assembles amino acids in the sequence determined by the codons
C.Ribosomes make mRNA in the nucleus \(\rightarrow\) mRNA moves to the cytoplasm \(\rightarrow\) DNA matches amino acids to the mRNA sequence
D.Amino acids are linked in the nucleus to form mRNA \(\rightarrow\) mRNA moves to a ribosome \(\rightarrow\) the ribosome copies the DNA to make a protein
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Worked solution
Protein synthesis begins in the nucleus where the DNA sequence is transcribed to form mRNA. This mRNA molecule then leaves the nucleus and travels to a ribosome in the cytoplasm. At the ribosome, transfer molecules assemble amino acids in the specific sequence dictated by the codons on the mRNA to construct the protein.
Marking scheme
Award 1 mark for the correct option B. - A is incorrect because transcription (copying the DNA code to mRNA) occurs in the nucleus, not the cytoplasm first. - C is incorrect because the ribosome does not make mRNA, nor does it reside in the nucleus. - D is incorrect because amino acids are not linked in the nucleus to form mRNA.
Question 31 · Multiple Choice
1 marks
A micrograph shows a mitochondrion. The actual length of the mitochondrion is \(2.5\ \mu\text{m}\). The image length of the mitochondrion in the micrograph is \(5.0\text{ cm}\). What is the magnification of this micrograph?
A.\(\times 2\)
B.\(\times 200\)
C.\(\times 2,000\)
D.\(\times 20,000\)
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Worked solution
To calculate magnification, use the formula: \(M = \frac{I}{A}\). 1. Convert the image length (I) to micrometers (\(\mu\text{m}\)) so that units are consistent: \(5.0\text{ cm} = 50\text{ mm}\) \(50\text{ mm} \times 1000 = 50,000\ \mu\text{m}\). 2. Apply the formula: \(M = \frac{50,000\ \mu\text{m}}{2.5\ \mu\text{m}} = 20,000\). Therefore, the magnification is \(\times 20,000\).
Marking scheme
Award 1 mark for the correct calculation leading to option D. - Reject A, B, and C due to incorrect unit conversions (e.g., failing to multiply by 1000 or dividing directly by different metric units).
Question 32 · Multiple Choice
1 marks
The rate of an enzyme-controlled reaction was monitored at various temperatures. Which statement correctly explains why the rate of reaction drops rapidly to zero at temperatures significantly above the optimum temperature?
A.The kinetic energy of the enzyme molecules decreases, resulting in fewer successful collisions.
B.The active site of the enzyme changes shape permanently, preventing substrate molecules from binding.
C.The substrate molecules gain excessive kinetic energy and break apart before they can reach the active site.
D.The activation energy of the reaction decreases, making the catalytic activity of the enzyme redundant.
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Worked solution
At high temperatures, the thermal energy breaks the bonds that maintain the specific three-dimensional shape of the enzyme's active site. Once this shape is permanently altered, the enzyme is denatured. The substrate molecule can no longer fit into the active site, preventing any further reaction.
Marking scheme
Award 1 mark for the correct option B. - A is incorrect because kinetic energy increases (not decreases) at high temperatures, though denaturation prevents successful collisions. - C is incorrect because substrates do not spontaneously break down before binding. - D is incorrect because denaturation does not lower the activation energy of a reaction; it stops the enzyme from lowering it.
Question 33 · multiple-choice
1 marks
Which row correctly identifies the products of anaerobic respiration in yeast cells and human muscle cells, and the relative amount of energy released per glucose molecule compared to aerobic respiration?
A.Yeast: carbon dioxide and ethanol; Muscle: lactic acid; Energy released: much less
B.Yeast: carbon dioxide and water; Muscle: lactic acid and carbon dioxide; Energy released: same amount
C.Yeast: ethanol only; Muscle: lactic acid only; Energy released: much less
D.Yeast: carbon dioxide and ethanol; Muscle: lactic acid and carbon dioxide; Energy released: much more
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Worked solution
During anaerobic respiration, yeast (fungus) undergoes alcoholic fermentation, producing carbon dioxide and ethanol. Human muscle cells undergo lactic acid fermentation, producing lactic acid only (no carbon dioxide). Both pathways release a much smaller amount of energy per glucose molecule compared to aerobic respiration because glucose is only partially broken down.
Marking scheme
1 mark for the correct option. Correct choice: A.
Question 34 · multiple-choice
1 marks
Four cylinders of potato tissue, each with an initial length of 40 mm, are placed into four different test-tubes containing sucrose solutions of different concentrations. After two hours, the lengths of the potato cylinders are measured. Tube 1 (0.0 mol/dm3) has a final length of 43 mm. Tube 2 (0.2 mol/dm3) has a final length of 41 mm. Tube 3 (0.5 mol/dm3) has a final length of 39 mm. Tube 4 (0.8 mol/dm3) has a final length of 37 mm. Which concentration is closest to the sucrose concentration inside the potato cells?
A.0.10 mol/dm3
B.0.35 mol/dm3
C.0.65 mol/dm3
D.0.90 mol/dm3
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Worked solution
The concentration inside the cells is closest to the point where there is no net movement of water, resulting in no change in length (final length remains 40 mm). Since the potato cylinder increased in length at 0.2 mol/dm3 (gained water) and decreased in length at 0.5 mol/dm3 (lost water), the internal concentration must lie between these two values. 0.35 mol/dm3 is the only value within this range.
Marking scheme
1 mark for the correct option. Correct choice: B.
Question 35 · multiple-choice
1 marks
The list shows some of the structures through which a molecule of oxygen passes as it travels from the trachea to a red blood cell in a capillary: 1. Alveolar wall, 2. Bronchiole, 3. Bronchus, 4. Capillary wall. Which sequence represents the correct path taken by the oxygen molecule?
A.2 -> 3 -> 1 -> 4
B.3 -> 2 -> 1 -> 4
C.3 -> 2 -> 4 -> 1
D.2 -> 3 -> 4 -> 1
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Worked solution
Oxygen travels from the trachea down the bronchus, then into the smaller bronchioles. From the bronchioles, it reaches the alveoli, where it diffuses across the alveolar wall and then across the capillary wall to enter the blood.
Marking scheme
1 mark for the correct option. Correct choice: B.
Question 36 · multiple-choice
1 marks
An experiment is set up to investigate the effect of light intensity on the rate of photosynthesis of an aquatic plant under different conditions of temperature and carbon dioxide concentration. Condition X: 15 degrees Celsius and 0.04% carbon dioxide. Condition Y: 25 degrees Celsius and 0.04% carbon dioxide. Condition Z: 25 degrees Celsius and 0.15% carbon dioxide. At high light intensities, the rate of photosynthesis is higher in Condition Z than in Condition Y, but it is the same in Condition X and Condition Y. Which factor is limiting the rate of photosynthesis at high light intensities in Condition Y?
A.Carbon dioxide concentration
B.Light intensity
C.Temperature
D.Water availability
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Worked solution
Comparing Condition Y and Condition Z, the temperature is kept at 25 degrees Celsius but the carbon dioxide concentration is increased from 0.04% to 0.15%, which causes an increase in the rate of photosynthesis. This shows that carbon dioxide concentration was the factor limiting the rate under Condition Y. Comparing X and Y shows that increasing the temperature while keeping CO2 at 0.04% does not increase the rate, confirming temperature is not the limiting factor at that CO2 concentration.
Marking scheme
1 mark for the correct option. Correct choice: A.
Question 37 · multiple-choice
1 marks
Which row correctly identifies where bile is produced, where bile is stored, and where bile acts in the human alimentary canal?
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Worked solution
Bile is produced by cells in the liver, stored in the gallbladder, and released via the bile duct into the duodenum (the first part of the small intestine) where it emulsifies fats and neutralizes stomach acid.
Marking scheme
1 mark for the correct option. Correct choice: A.
Question 38 · multiple-choice
1 marks
A structural protein consists of a chain of 240 amino acids. What is the minimum number of bases in the coding region of the mRNA molecule, and the minimum number of base pairs in the DNA gene required to code for this protein?
A.mRNA bases: 80; DNA base pairs: 80
B.mRNA bases: 240; DNA base pairs: 240
C.mRNA bases: 720; DNA base pairs: 720
D.mRNA bases: 720; DNA base pairs: 1440
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Worked solution
Each amino acid is coded for by a sequence of three bases (a codon) on the mRNA. Therefore, 240 amino acids require a minimum of 240 multiplied by 3 = 720 bases on the mRNA. Each base on the single-stranded mRNA is complementary to one base on the template DNA strand, which corresponds to one base pair in the double-stranded DNA gene. Thus, the gene must contain at least 720 base pairs.
Marking scheme
1 mark for the correct option. Correct choice: C.
Question 39 · multiple-choice
1 marks
A student uses a light microscope to view a plant palisade cell. The image of the cell under the microscope has a length of 7.5 cm. The actual length of the palisade cell is 150 \(\mu\text{m}\). What is the magnification of the image?
A.x 50
B.x 200
C.x 500
D.x 2000
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Worked solution
To find magnification, use the formula Magnification = Image size / Actual size. Convert 7.5 cm to micrometers: 7.5 cm = 75 mm = 75,000 micrometers. Magnification = 75,000 micrometers / 150 micrometers = 500. Therefore, the magnification is x 500.
Marking scheme
1 mark for the correct option. Correct choice: C.
Question 40 · multiple-choice
1 marks
Which statement is a feature of artificial selection but not of natural selection?
A.It involves the struggle for survival in a changing environment.
B.It results in organisms that are better adapted to survive in their wild habitat.
C.Humans choose which individuals with desirable features are allowed to reproduce.
D.New alleles arise in the population through random mutations.
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Worked solution
Artificial selection is driven by human choice to select individuals with features that humans find useful or desirable, whereas natural selection is driven by environmental factors where individuals with advantageous adaptations are more likely to survive and reproduce.
Marking scheme
1 mark for the correct option. Correct choice: C.
Paper 41: Theory (Extended)
Answer all questions in the spaces provided. Show all calculations where required.
6 Question · 79.98 marks
Question 1 · Structured
13.33 marks
Fig. 1.1 shows a setup used to investigate anaerobic respiration in yeast.
(a) Define the term *anaerobic respiration*. [2]
(b) Compare anaerobic respiration in yeast cells with anaerobic respiration in human muscle cells during vigorous exercise. [4]
(c) Yeast is widely used in bread-making. Explain how the products of anaerobic respiration in yeast contribute to the process of making bread and explain why bread is not an alcoholic food product. [4]
(d) During an investigation, yeast produced \( 12.0\text{ cm}^3 \) of carbon dioxide in 10 minutes at 20°C. When the temperature was increased to 30°C, the volume of carbon dioxide produced in 10 minutes was \( 27.6\text{ cm}^3 \). Calculate the percentage increase in the rate of carbon dioxide production. Show your working. [3]
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Worked solution
(a) Anaerobic respiration is the chemical reactions in cells that break down nutrient molecules to release energy without using oxygen. (b) Yeast cells produce ethanol and carbon dioxide, whereas muscle cells produce lactic acid. Both processes release much less energy per glucose molecule compared to aerobic respiration, but they do not require oxygen. (c) Yeast produces carbon dioxide, which forms gas bubbles trapped in the dough, causing it to rise. The ethanol produced is volatile and evaporates during the baking process due to the high oven temperatures. (d) Change in volume = \( 27.6 - 12.0 = 15.6\text{ cm}^3 \). Percentage increase = \( \frac{15.6}{12.0} \times 100 = 130\% \).
Marking scheme
(a) - Release of energy from nutrient molecules / glucose [1] - Without using oxygen [1]
(b) - Yeast produces ethanol AND carbon dioxide [1] - Muscle cells produce lactic acid only [1] - Both produce less energy than aerobic respiration [1] - Both do not use oxygen [1]
(c) - Carbon dioxide is trapped in dough / creates bubbles [1] - Causes the dough to expand / rise [1] - Ethanol / alcohol is produced [1] - Ethanol evaporates during baking / heating [1]
(d) - Method: Calculates change in volume: \( 27.6 - 12.0 = 15.6\text{ cm}^3 \) [1] - Method: Divides change by original and multiplies by 100: \( \frac{15.6}{12.0} \times 100 \) [1] - Accuracy: 130% [1]
Question 2 · Structured
13.33 marks
A student investigated the effect of different sucrose concentrations on the mass of sweet potato cylinders.
(a) Define *osmosis* with reference to water potential. [3]
(b) A sweet potato cylinder had an initial mass of 4.50 g. After being placed in a 0.8 mol/dm³ sucrose solution for 2 hours, its final mass was 3.78 g. Calculate the percentage change in mass of this cylinder. Show your working. [3]
(c) Explain, in terms of water potential, why the sweet potato cylinder decreased in mass when placed in the 0.8 mol/dm³ sucrose solution. [4]
(d) Describe how the student could use a graph of percentage change in mass against sucrose concentration to determine the concentration of sucrose that is equal to the internal water potential of the sweet potato cells. [3]
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Worked solution
(a) Osmosis is the net movement of water molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution), through a partially permeable membrane. (b) Change in mass = \( 3.78\text{ g} - 4.50\text{ g} = -0.72\text{ g} \). Percentage change = \( \frac{-0.72}{4.50} \times 100 = -16.0\% \). (c) The water potential of the 0.8 mol/dm³ sucrose solution is lower than the water potential inside the sweet potato cells. Water moved out of the cells by osmosis, down a water potential gradient, through the partially permeable cell membranes, causing the cells to become flaccid. (d) Plot percentage change in mass on the y-axis and sucrose concentration on the x-axis. Find the point where the line of best fit crosses the x-axis (where the percentage change in mass is zero). At this point, there is no net movement of water because the water potential inside and outside the cells is equal.
Marking scheme
(a) - Net movement of water molecules [1] - From higher water potential to lower water potential [1] - Through a partially permeable membrane [1]
(b) - Method: Correct calculation of mass change: \( 3.78 - 4.50 = -0.72\text{ g} \) [1] - Method: Correct fraction set up: \( \frac{-0.72}{4.50} \times 100 \) [1] - Accuracy: -16.0% or 16.0% decrease (must indicate negative or decrease) [1]
(c) - Sucrose solution has a lower water potential than the sweet potato cells [1] - Water moves out of the sweet potato cells [1] - By osmosis down the water potential gradient [1] - Through the partially permeable membrane [1]
(d) - Plot percentage change on y-axis and concentration on x-axis [1] - Identify the concentration where the line crosses the x-axis / zero mass change [1] - At this point, there is no net water movement / equal water potential inside and outside [1]
Question 3 · Structured
13.33 marks
The gas exchange system in humans is adapted to maximize the rate of diffusion of respiratory gases.
(a) State three structural features of the alveoli that adapt them for efficient gas exchange. [3]
(b) Explain how goblet cells and ciliated cells work together to protect the lungs from pathogens and dust. [4]
(c) Emphysema is a chronic lung disease often caused by long-term tobacco smoking. Explain how emphysema affects gas exchange in the lungs. [4]
(d) Table 3.1 shows the percentage of oxygen and carbon dioxide in inspired and expired air.
| Gas | Inspired air (%) | Expired air (%) | |---|---|---| | Oxygen | 21.0 | 16.0 | | Carbon dioxide | 0.04 | 4.00 |
Explain the differences in the percentages of oxygen and carbon dioxide between inspired and expired air. [2]
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Worked solution
(a) Three features of alveoli are: one-cell thick walls (short diffusion path), large surface area (millions of alveoli), and a dense network of blood capillaries maintaining a steep concentration gradient. (b) Goblet cells produce sticky mucus that traps inhaled pathogens, dust, and dirt particles. Ciliated cells have tiny hair-like structures called cilia that beat in a coordinated way to sweep the mucus upwards away from the lungs towards the throat to be swallowed. (c) Chronic exposure to cigarette smoke damages and breaks down the alveolar walls, causing them to fuse together. This significantly reduces the surface area available for gas exchange. As a result, less oxygen diffuses into the blood, leading to shortness of breath and fatigue. (d) Oxygen is lower in expired air because cells use oxygen for aerobic respiration. Carbon dioxide is higher in expired air because it is produced as a waste product during aerobic respiration and is excreted via the alveoli.
Marking scheme
(a) Any three from: - Large surface area [1] - Thin walls / one cell thick epithelium (short diffusion distance) [1] - Good / rich blood capillary supply [1] - Moist lining (gases dissolve) [1]
(b) - Goblet cells secrete/produce mucus [1] - Mucus traps pathogens / dust [1] - Cilia / ciliated cells beat/move [1] - To sweep mucus up / out of the trachea/airway [1]
(c) - Breakdown / destruction of alveolar walls [1] - Alveoli fuse / merge to form larger air spaces [1] - Decreased surface area for diffusion [1] - Less oxygen enters blood / slower rate of gas exchange [1]
(d) - Oxygen is lower in expired air because it is used in aerobic respiration [1] - Carbon dioxide is higher in expired air because it is produced in aerobic respiration [1]
Question 4 · Structured
13.33 marks
Photosynthesis is the process by which plants manufacture carbohydrates.
(a) Write the balanced chemical equation for photosynthesis. [3]
(b) Describe how a student could investigate the effect of light intensity on the rate of photosynthesis in an aquatic plant, such as *Elodea*. [4]
(c) Explain why, at high light intensities, the rate of photosynthesis in a plant may reach a plateau (level off) and no longer increase. [3]
(d) Outline the role of chlorophyll in photosynthesis and name the mineral ion required by plants to synthesize it. [3]
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Worked solution
(a) The balanced chemical equation is: \( 6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \) (in the presence of light and chlorophyll). (b) Place the *Elodea* in a beaker of water. Position a light source (lamp) at a measured distance from the beaker. Allow the plant to adjust, then count the number of oxygen bubbles produced per minute. Change the light intensity by moving the lamp to different distances (e.g., 10 cm, 20 cm, 30 cm) and repeat the process, ensuring other variables like temperature and carbon dioxide concentration are controlled. (c) At high light intensities, light is no longer the limiting factor. Some other factor, such as carbon dioxide concentration or temperature, is in short supply and limits the rate of photosynthesis. (d) Chlorophyll absorbs light energy and transfers this energy into chemical energy for the synthesis of carbohydrates. Magnesium ions are required by plants to produce chlorophyll.
(b) - Move light source to different distances (to vary light intensity) [1] - Count the number of bubbles / volume of oxygen gas produced in a set time period [1] - Allow time for plant to adapt/equilibriate at each distance [1] - Control variable: temperature (using water bath) OR carbon dioxide concentration (using sodium hydrogencarbonate) [1]
(c) - Light is no longer the limiting factor [1] - Another factor is limiting the reaction rate [1] - Named factor: Carbon dioxide concentration OR temperature [1]
(d) - Absorbs light / solar energy [1] - Converts light energy into chemical energy [1] - Magnesium (ion) is required for synthesis [1]
Question 5 · Structured
13.33 marks
The human digestive system is specialized for physical and chemical digestion.
(a) Identify the specific region of the human alimentary canal where: (i) Amylase, lipase, and trypsin are all secreted. [1] (ii) Bile is stored before being released into the duodenum. [1] (iii) The majority of water is absorbed. [1]
(b) Describe the role of bile in the digestion of fats. [4]
(c) Explain how the structure of a villus is adapted to maximize the absorption of digested food molecules. [5]
(d) State the function of lacteals located inside the villi. [1]
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Worked solution
(a) (i) Duodenum (or small intestine), as these pancreatic enzymes are delivered there. (ii) Gallbladder stores bile. (iii) Small intestine absorbs the largest total volume of water, though the colon is also involved in water absorption from undigested material. (b) Bile emulsifies fats, breaking large fat droplets into many smaller droplets. This increases the surface area for the enzyme lipase to act on, speeding up chemical digestion. Bile is also alkaline, so it neutralizes the acidic mixture coming from the stomach, providing the optimum pH for lipase. (c) A villus is adapted in several ways: its wall (epithelium) is only one cell thick, which minimizes the diffusion distance; it contains microvilli on its surface, greatly increasing the surface area for absorption; it has an extensive blood capillary network to rapidly transport absorbed sugars and amino acids away, maintaining a steep concentration gradient; and a central lacteal to transport digested lipids. (d) Lacteals absorb and transport fatty acids and glycerol (lipids) into the lymphatic system.
Marking scheme
(a) - (i) Duodenum / small intestine [1] - (ii) Gallbladder [1] - (iii) Small intestine (accept Colon/large intestine) [1]
(b) - Emulsifies fats / lipids [1] - Breaks large fat droplets into smaller droplets [1] - Increases surface area for lipase action [1] - Neutralizes stomach acid / provides alkaline conditions / optimum pH [1]
(c) Max 5 marks: - Villi / microvilli increase surface area [1] - Epithelium is one cell thick / thin wall [1] - Decreases diffusion distance [1] - Good capillary network / rich blood supply [1] - Lacteal absorbs fatty acids and glycerol [1] - Blood flow / transport of nutrients maintains concentration gradient [1]
(d) - Absorption / transport of fats / lipids / fatty acids and glycerol [1]
Question 6 · Structured
13.33 marks
DNA is the genetic material that carries instructions for the synthesis of proteins.
(a) Describe the structure of a DNA molecule. [4]
(b) Define the term *gene*. [2]
(c) Explain how the sequence of bases in a gene determines the specific structure of a protein, including the roles of mRNA and ribosomes. [5]
(d) Explain why a change in the DNA base sequence of a gene can result in an enzyme that does not function. [2]
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Worked solution
(a) DNA is a double-stranded helix. The two strands are held together by complementary base pairs: adenine (A) always pairs with thymine (T), and cytosine (C) always pairs with guanine (G). These base pairs are linked by hydrogen bonds. (b) A gene is a length of DNA that codes for a specific protein. (c) The DNA base sequence determines the sequence of bases in the mRNA molecule during transcription. The mRNA molecule leaves the nucleus and binds to a ribosome in the cytoplasm. The ribosome reads the mRNA sequence in triplets (codons). Each codon codes for a specific amino acid. This determines the exact sequence of amino acids joined together to form a protein, which then folds into a specific three-dimensional shape. (d) A change in the DNA base sequence (mutation) alters the codon sequence of the mRNA. This results in different amino acids being incorporated, which changes the shape of the enzyme's active site. The substrate will no longer fit into the active site, meaning no enzyme-substrate complexes can form.
Marking scheme
(a) - Two strands [1] - Double helix shape [1] - Nitrogenous bases: adenine, thymine, cytosine, guanine (A, T, C, G) [1] - Complementary base pairing / A pairs with T and C pairs with G [1]
(b) - Length of DNA [1] - That codes for a protein [1]
(c) - DNA sequence is transcribed into mRNA [1] - mRNA travels from the nucleus to the cytoplasm / to a ribosome [1] - Ribosome reads mRNA base sequence / codons [1] - Base sequence determines the sequence of amino acids [1] - Protein folds into a specific 3D shape based on the amino acid chain [1]
(d) - Changes the amino acid sequence [1] - Changes the shape of the active site so substrate cannot bind / fit [1]
Paper 51: Practical Test
Complete all practical tasks, record all experimental observations in the spaces provided, and complete the design task.
2 Question · 40 marks
Question 1 · practical
20 marks
Complete all practical tasks, record all experimental observations in the spaces provided, and complete the design task. ### Question 1 Yeast cells respire glucose or sucrose, releasing carbon dioxide and transferring hydrogen to hydrogen-acceptor molecules. Methylene blue is a blue dye that acts as an artificial hydrogen acceptor; it becomes colourless when reduced by respiring yeast cells. You are going to investigate the effect of different concentrations of sucrose solution on the rate of respiration in yeast. You are provided with: - Active yeast suspension, labelled **Yeast** - 10% sucrose solution, labelled **S1** - Distilled water, labelled **Water** - 0.1% methylene blue solution, labelled **Indicator** - Test-tubes, syringes, a water bath maintained at 35 °C, and a stopwatch. **(a)** Describe how you would prepare 10 cm³ of a 5% sucrose solution using the 10% sucrose solution (**S1**) and distilled water (**Water**). [2] **(b)** A student performed the investigation using 10% (**S1**), 5%, 2% and 0% (distilled water) sucrose solutions. The student mixed 5 cm³ of yeast suspension, 5 cm³ of the respective sucrose solution, and 1 cm³ of methylene blue solution in a test-tube. The tube was placed in a water bath maintained at 35 °C. The time taken for the methylene blue to lose its blue colour and turn colourless was recorded using a stopwatch. Three trials were carried out for each concentration. **(i)** Prepare a table to record the results of this investigation. Your table must show: - the concentration of sucrose solution (%) - the volume of 10% sucrose solution (**S1**) used to prepare 10 cm³ of each solution (cm³) - the volume of distilled water used to prepare 10 cm³ of each solution (cm³) - the time taken for the blue colour to disappear for each of the three trials (seconds) - the mean time taken (seconds). [4] **(ii)** The student recorded the following raw times: - For 10% sucrose: 145 s, 151 s, 148 s - For 5% sucrose: 280 s, 292 s, 286 s - For 2% sucrose: 580 s, 610 s, 595 s - For 0% sucrose: Did not decolourise after 15 minutes (900 s) in all three trials. Calculate the mean time taken for the 10% sucrose solution to decolourise. Show your working. [2] **(iii)** Calculate the rate of respiration for the 10% and 5% sucrose solutions using the formula: \(\text{Rate} = \frac{1000}{\text{mean time in seconds}}\). State the unit of rate as \(\text{s}^{-1}\). Show your working. [3] **(c)** State three variables that must be controlled in this investigation to ensure valid results. [3] **(d) (i)** Identify one potential source of error in this method. [1] **(ii)** Suggest an improvement to overcome the error identified in **(d)(i)**. [1] **(e)** Yeast respiration is affected by temperature. Design an investigation to find the effect of temperature on the rate of anaerobic respiration in yeast. [4]
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Worked solution
Part (a): 5 cm³ of 10% sucrose (S1) and 5 cm³ of distilled water. Part (b)(i): A correctly constructed table with appropriate headings and units. Part (b)(ii): Calculation: (145 + 151 + 148) / 3 = 148 s. Part (b)(iii): Rate for 10% = 1000 / 148 = 6.76 s⁻¹; Rate for 5% = 1000 / 286 = 3.50 s⁻¹. Part (c): Control variables: volume of yeast, temperature, pH, methylene blue concentration. Part (d): Endpoint judgment is subjective; solve by using a colorimeter or a color comparison tube. Part (e): Setup tubes at different temperatures, collect gas in a gas syringe over a fixed time, control yeast concentration, repeat trials.
Marking scheme
Part (a) [2 marks]: 1 mark for measuring 5 cm³ of 10% sucrose solution; 1 mark for adding 5 cm³ of distilled water. Part (b)(i) [4 marks]: 1 mark for structured table layout with clear borders; 1 mark for column headings with correct units (%, cm³, s); 1 mark for showing correct volumes (10% sucrose: 10, 5, 2, 0 cm³ and water: 0, 5, 8, 10 cm³ respectively); 1 mark for including columns for 3 trials and mean. Part (b)(ii) [2 marks]: 1 mark for correct working shown; 1 mark for final value of 148 s. Part (b)(iii) [3 marks]: 1 mark for calculating 10% rate as 6.76 (or 6.8); 1 mark for calculating 5% rate as 3.50 (or 3.5); 1 mark for correct unit s⁻¹. Part (c) [3 marks]: 1 mark for each valid controlled variable up to 3 (e.g., yeast volume, methylene blue volume, water bath temperature). Part (d)(i) [1 mark]: Subjective judgment of when blue colour has completely disappeared. Part (d)(ii) [1 mark]: Use a colorimeter to measure absorbance or keep a control tube with decolourised yeast as a color comparison. Part (e) [4 marks]: 1 mark for varying temperature using water baths (at least 5 temperatures); 1 mark for maintaining other variables constant (yeast concentration, volume, substrate concentration); 1 mark for measuring the volume of gas produced in a set time (or time taken for decolourisation); 1 mark for repeating the experiment at each temperature.
Question 2 · practical
20 marks
Complete all practical tasks, record all experimental observations in the spaces provided, and complete the design task. ### Question 2 Potato tissue is composed of cells containing cell walls, cell membranes, cytoplasm, a vacuole, and starch grains. An investigation was carried out to study the effect of placing potato cylinders in different concentrations of sodium chloride (salt) solution. Potato cylinders of equal initial length (50.0 mm) were prepared and placed in test-tubes containing five different concentrations of sodium chloride solution (0.0, 0.2, 0.4, 0.6 and 0.8 mol dm⁻³). The cylinders were left in the solutions for 60 minutes. They were then removed, blotted dry with paper towels, and their final lengths were measured. The results are shown in Table 2.1. Table 2.1 - Concentration 0.0 mol dm⁻³: Initial length 50.0 mm, Final length 54.0 mm, Change in length +4.0 mm, Percentage change +8.0% - Concentration 0.2 mol dm⁻³: Initial length 50.0 mm, Final length 52.0 mm, Change in length +2.0 mm, Percentage change +4.0% - Concentration 0.4 mol dm⁻³: Initial length 50.0 mm, Final length 49.5 mm, Change in length -0.5 mm, Percentage change -1.0% - Concentration 0.6 mol dm⁻³: Initial length 50.0 mm, Final length 47.5 mm, Change in length -2.5 mm, Percentage change -5.0% - Concentration 0.8 mol dm⁻³: Initial length 50.0 mm, Final length 45.5 mm, Change in length -4.5 mm, Percentage change -9.0% **(a) (i)** Plot a line graph on a grid of the percentage change in length (y-axis) against the concentration of sodium chloride solution (x-axis). [4] **(ii)** Use your graph to estimate the concentration of sodium chloride solution at which there is no change in length of the potato cylinders. [1] **(iii)** Explain, in terms of osmosis, why the potato cylinder in the 0.0 mol dm⁻³ solution increased in length. [3] **(iv)** Explain why calculating the percentage change in length is better than using the raw change in length when comparing samples of potato tissue. [1] **(b)** Describe how you would carry out a food test to show that potato cells contain starch. State the reagent used and the positive result. [2] **(c)** Fig. 2.1 represents a photomicrograph of a potato cell containing starch grains. The length of one large starch grain in the photomicrograph is measured as 42 mm. The actual length of this starch grain is 0.07 mm. Calculate the magnification of the starch grain in the photomicrograph. Show your working using the formula \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). [2] **(d)** Fig. 2.2 shows a diagram of a potato cell. The cell has a thick cell wall drawn with a double line, a thin inner cell membrane, a large central vacuole, and five oval starch grains of different sizes inside. In the space below, make a large, clear line drawing of this potato cell. Do not label the drawing. [4] **(e)** Design an experiment using model agar blocks to investigate the effect of surface area to volume ratio on the rate of diffusion. [3]
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Worked solution
Part (a)(i): Graph with correct axes, scales, plotted points, and line. Part (a)(ii): Value read where percentage change is 0% (around 0.36 mol dm⁻³). Part (a)(iii): Water moves into the cell from high water potential to low water potential across the partially permeable cell membrane by osmosis. Part (a)(iv): Normalises different initial lengths for fair comparison. Part (b): Iodine solution turns blue-black in the presence of starch. Part (c): Image / Actual = 42 mm / 0.07 mm = x600. Part (d): Large drawing of the cell with clean lines, thick double cell wall, central vacuole, and 5 distinct starch grains. Part (e): Use agar blocks of different SA:Vol ratios, immerse in acid, and measure diffusion time or depth of penetration.
Marking scheme
Part (a)(i) [4 marks]: 1 mark for correct labels with units on axes (x: Concentration of sodium chloride solution / mol dm⁻³, y: Percentage change in length / %); 1 mark for linear scale using more than 50% of the grid; 1 mark for all 5 points plotted accurately; 1 mark for drawing a clean line of best fit or connecting points with neat straight lines. Part (a)(ii) [1 mark]: Correct reading of x-intercept from their plotted graph (expected 0.36 mol dm⁻³, accept 0.34 to 0.38 mol dm⁻³). Part (a)(iii) [3 marks]: 1 mark for stating distilled water has a higher water potential than the potato cells; 1 mark for water moving in by osmosis down a water potential gradient; 1 mark for water passing across the partially permeable cell membrane (or cell becomes turgid/expands). Part (a)(iv) [1 mark]: To account for differences in initial lengths of the potato cylinders / allows comparison. Part (b) [2 marks]: 1 mark for using iodine solution; 1 mark for colour change from yellow-brown/orange to blue-black. Part (c) [2 marks]: 1 mark for showing correct formula or division (42 / 0.07); 1 mark for correct magnification of x600 (or 600). Part (d) [4 marks]: 1 mark for large drawing (occupying at least half the space); 1 mark for clear single lines without shading/sketching; 1 mark for drawing the cell wall as a double line (thickness); 1 mark for including a vacuole and exactly 5 oval starch grains of different sizes. Part (e) [3 marks]: 1 mark for preparing agar blocks of at least three different sizes (varying SA:V ratio); 1 mark for measuring the time taken for the indicator to change colour completely (or measuring distance acid diffused in a fixed time); 1 mark for keeping temperature or acid concentration constant.
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