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A-Level · Biology

AB2 Cells

AQA 7402 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A student used a light microscope fitted with an eyepiece graticule to examine a section of onion epidermis. Before taking measurements, the student calibrated the eyepiece graticule against a stage micrometre scale.

(a)At a particular magnification, 20 eyepiece graticule units aligned exactly with 0.40 mm on the stage micrometre scale. Calculate the value of 1 eyepiece unit in micrometres (um) at this magnification.(2)

(b)The student then measured the width of one epidermal cell as 15 eyepiece units. Calculate the actual width of this cell in micrometres.(2)

(c)A photomicrograph of the same cell measured 45 mm across the image. Calculate the magnification of this image. Give your answer as 'x' followed by a whole number.(3)

(d)State two differences between the magnification and the resolution of a microscope.(2)

(Total for Question 1 is 9 marks)

Table 1 compares some features of a typical bacterial (prokaryotic) cell and a typical animal (eukaryotic) cell.

Table 1

Feature	Bacterial cell	Animal cell
Membrane-bound nucleus	...	...
Circular DNA (plasmid)	...	...
80S ribosomes	...	...
Mitochondria	...	...

(a)Complete Table 1 by writing 'present' or 'absent' for each feature in each cell type.(4)

(b)Mitochondria have a folded inner membrane called cristae. Explain how this feature of mitochondrial structure relates to its function.(3)

(c)Describe the pathway taken by a protein from its synthesis to its secretion from a eukaryotic cell.(4)

(Total for Question 2 is 11 marks)

This question is about the resolution and magnification of microscopes.

(a)Define the term resolution as applied to microscopy.(1)

(b)Explain why an electron microscope has a higher resolution than a light microscope.(2)

(c)A transmission electron micrograph of a mitochondrion measured 500 mm across the printed image. The actual length of the mitochondrion was 4.2 um. Calculate the magnification of this image. Give your answer to 2 significant figures.(3)

(d)Suggest one limitation of using a transmission electron microscope (TEM) to study a living cell such as this mitochondrion.(2)

(Total for Question 3 is 8 marks)

Cell fractionation is used to separate organelles so they can be studied individually.

(a)Describe the process of homogenisation and filtration in the preparation of a cell fraction.(2)

(b)Explain why the solution used in cell fractionation must be: (i) ice-cold (ii) isotonic (iii) buffered.(3)

(c)During ultracentrifugation, the homogenate is spun at progressively increasing speeds. State the order in which nuclei, mitochondria and ribosomes would appear in the pellet.(2)

(d)A student wants to check that a cell fraction is pure mitochondria and not contaminated with other organelles. Suggest how this could be checked using microscopy.(2)

(Total for Question 4 is 9 marks)

This question is about mitosis and the cell cycle.

(a)State what happens during interphase, mitosis and cytokinesis in the cell cycle.(3)

(b)Table 2 gives brief descriptions of stages of mitosis, labelled W, X, Y and Z, in a random order.
W: Chromosomes line up individually along the equator of the cell, attached to spindle fibres.
X: Chromatids are pulled to opposite poles of the cell by the spindle fibres.
Y: Chromosomes condense and become visible; the nuclear envelope breaks down.
Z: Chromatids arrive at the poles and uncoil; nuclear envelopes reform.
Identify which stage of mitosis (prophase, metaphase, anaphase or telophase) matches each description.(4)

(c)A microscope field of view of a root tip squash showed 240 cells in total, of which 18 cells were undergoing mitosis. Calculate the mitotic index for this sample.(2)

(d)A tumour biopsy was found to have a much higher mitotic index than the surrounding healthy tissue. Explain why this might indicate that the tissue is cancerous.(2)

(Total for Question 5 is 11 marks)

This question is about stem cells and their use in medicine.

(a)Define the terms totipotent, pluripotent and multipotent, as applied to stem cells.(3)

(b)Discuss the potential medical uses of stem cells and the ethical issues associated with obtaining stem cells from human embryos.(6)

(Total for Question 6 is 9 marks)

This question is about the structure of the cell-surface membrane and the effect of temperature on membrane permeability.

(a)Describe the fluid mosaic model of the cell-surface membrane.(4)

(b)A group of students investigated the effect of temperature on the permeability of beetroot cell membranes as part of a required practical. Beetroot cylinders of equal size were placed in water baths at 10, 20, 30, 40, 50 and 60 degrees C for 10 minutes. The absorbance of the surrounding water (due to leaked red pigment, betalain) was then measured using a colorimeter. Absorbance increased only slightly between 10 and 40 degrees C, then increased sharply between 40 and 60 degrees C. Explain these results.(3)

(c)State two variables, other than temperature, that should be controlled in this investigation.(2)

(d)In the investigation at 50 degrees C, three repeat absorbance readings were 0.62, 0.65 and 0.70. Calculate the mean absorbance at this temperature.(1)

(Total for Question 7 is 10 marks)

This question is about diffusion and the surface area to volume ratio of cells and organisms.

(a)Define the term diffusion.(1)

(b)State three factors that affect the rate of diffusion across a cell-surface membrane.(3)

(c)A cell is cube-shaped with sides of length 3 mm. Calculate the surface area, the volume, and the surface area to volume ratio of this cell. Show your working.(4)

(d)A larger cube-shaped organism has sides of length 9 mm. Calculate its surface area to volume ratio, and explain why larger organisms like this often need specialised exchange surfaces rather than relying on diffusion over their whole body surface.(4)

(Total for Question 8 is 12 marks)

A group of students carried out a required practical investigation into water potential using potato tissue. Cylinders of potato of equal mass were placed into sucrose solutions of different concentrations for 30 minutes, then blotted and reweighed. Table 3 shows the percentage change in mass of the potato cylinders at each sucrose concentration.

Table 3
Sucrose concentration (mol/dm3): 0.0, 0.2, 0.4, 0.6, 0.8, 1.0
Percentage change in mass: +8, +3, -2, -7, -12, -15

(a)Define water potential.(1)

(b)Using the data in Table 3, estimate the sucrose concentration of a solution that has the same water potential as the potato tissue (i.e. where there would be no change in mass). Show your working.(3)

(c)Explain, in terms of water potential, why the potato cylinders in 1.0 mol/dm3 sucrose solution decreased in mass.(3)

(d)Suggest one improvement to the method that would increase the validity of the results.(2)

(Total for Question 9 is 9 marks)

This question is about the absorption of glucose in the ileum by active transport and co-transport.

(a)Describe how glucose is absorbed from the lumen of the ileum into epithelial cells by co-transport with sodium ions.(4)

(b)Explain why epithelial cells involved in this process contain a large number of mitochondria.(2)

(c)Compare active transport and facilitated diffusion in terms of the use of ATP and the direction of movement relative to the concentration gradient.(3)

(Total for Question 10 is 9 marks)

This question contains short recall and application items on cell structure and specialisation.

(a)Which organelle is the main site of aerobic respiration in a eukaryotic cell?(1)

A) Golgi apparatus
B) Mitochondrion
C) Ribosome
D) Lysosome

(b)Which of the following is NOT typically found in a prokaryotic cell?(1)

A) Plasmid
B) Cell wall
C) Membrane-bound nucleus
D) Ribosomes

(c)State the function of the nucleolus.(1)

(d)A mature sperm cell contains a large number of mitochondria concentrated in its mid-piece. Explain how this adaptation helps the sperm cell carry out its function.(2)

(e)State two structural adaptations of a root hair cell that suit it to the absorption of water and mineral ions.(2)

(Total for Question 11 is 7 marks)

A group of students used agar cubes containing phenolphthalein indicator (which turns from colourless to pink in alkaline conditions) to model diffusion into cells. Cubes with sides of 1 cm, 2 cm and 3 cm were prepared and placed into a beaker of sodium hydroxide solution. The time taken for each cube to turn completely pink (indicating the NaOH had diffused fully to the centre) was recorded.

(a)Predict, with a reason, which cube would turn completely pink in the shortest time.(2)

(b)The 1 cm cube took 4 minutes 30 seconds to turn completely pink. Assuming diffusion occurs from the outer surface towards the centre of the cube (a distance of 5 mm), calculate the mean rate of diffusion in mm per minute.(3)

(c)Evaluate the use of agar cubes as a model for diffusion into living cells in this investigation.(3)

(Total for Question 12 is 8 marks)

Mark scheme · AB2 Cells

Question 1

(a) M1 converts 0.40 mm to 400 um (or equivalent correct method)
(a) A1 20 um cao
(a) Answer: 20 um
(b) M1 15 x 20 (or correct method using their value from (a), ft)
(b) A1 300 um cao (ft from (a))
(b) Answer: 300 um
(c) M1 converts 45 mm to 45000 um
(c) dM1 magnification = image size / actual size = 45000 / 300 (dependent on M1)
(c) A1 x150 cao
(c) Answer: x150
(d) B1 magnification is the ratio of image size to actual (real) object size, i.e. how much bigger the image is (oe)
(d) B1 resolution is the minimum distance between two points at which they can still be distinguished as separate (oe)
(d) Answer: Magnification = how much bigger the image is than the real object (ratio of image size to actual size); Resolution = the minimum distance between two points at which they can still be seen as separate objects

Question 2

(a) B1 membrane-bound nucleus: absent (bacterial); present (animal)
(a) B1 plasmid: present (bacterial); absent (animal)
(a) B1 80S ribosomes: absent, bacteria have 70S (bacterial); present (animal)
(a) B1 mitochondria: absent (bacterial); present (animal)
(a) Answer: Bacterial: nucleus absent, plasmid present, 80S ribosomes absent (has 70S), mitochondria absent. Animal: nucleus present, plasmid absent, 80S ribosomes present, mitochondria present
(b) B1 cristae increase the surface area of the inner membrane (oe)
(b) B1 providing more space for electron transport chain proteins / ATP synthase enzymes
(b) B1 increasing the rate of ATP production by aerobic respiration
(b) Answer: Cristae increase the surface area of the inner mitochondrial membrane, providing more space for electron transport chain proteins and ATP synthase, increasing the rate of ATP production by aerobic respiration
(c) B1 protein synthesised by ribosomes on the rough endoplasmic reticulum (RER)
(c) B1 protein enters the RER and is folded/processed, then a vesicle buds off carrying the protein
(c) B1 vesicle fuses with the Golgi apparatus, where the protein is modified and packaged
(c) B1 a secretory vesicle buds off the Golgi, moves to and fuses with the cell-surface membrane, releasing the protein by exocytosis
(c) Answer: Ribosomes on the RER synthesise the protein; it enters the RER and is packaged into a vesicle; the vesicle fuses with the Golgi apparatus where the protein is modified and packaged; a secretory vesicle buds off the Golgi and fuses with the cell-surface membrane, releasing the protein by exocytosis

Question 3

(a) B1 the minimum distance between two points/objects at which they can still be distinguished as separate (oe)
(a) Answer: The minimum distance between two points that can still be seen/distinguished as two separate points
(b) B1 electrons have a much shorter wavelength than light
(b) B1 a shorter wavelength gives a higher resolution (allows smaller/closer structures to be distinguished)
(b) Answer: Electrons have a much shorter wavelength than light, and a shorter wavelength gives a higher resolution
(c) M1 converts 500 mm to 500000 um
(c) dM1 magnification = image size / actual size = 500000 / 4.2 (dependent on M1)
(c) A1 x120000 (1.2 x 10⁵) awrt 2 s.f.
(c) Answer: x120000 (1.2 x 10⁵) to 2 s.f.
(d) B1 specimens must be dead, dehydrated, chemically fixed and stained, and viewed in a vacuum, so cannot be living (oe)
(d) B1 only a thin 2D section is seen, so the whole 3D living structure / dynamic processes cannot be observed (oe, any second valid point)
(d) Answer: TEM specimens must be dead, dehydrated and stained (viewed in a vacuum), so living processes cannot be observed; only a thin 2D section is seen, not the whole 3D living structure

Question 4

(a) B1 tissue is homogenised (e.g. in a blender) in a cold, buffered, isotonic solution to break open cells and release organelles
(a) B1 the homogenate is filtered to remove debris/whole cells/connective tissue
(a) Answer: The tissue is homogenised (blended) in a cold, buffered, isotonic solution to break open the cells and release the organelles; the homogenate is then filtered to remove any large debris or unbroken cells/tissue
(b) B1 ice-cold: reduces enzyme activity/respiration, preventing autolysis/digestion of organelles
(b) B1 isotonic: prevents organelles gaining or losing water by osmosis and bursting or shrinking
(b) B1 buffered: keeps pH constant, preventing denaturation of proteins/enzymes
(b) Answer: Ice-cold: reduces enzyme activity, preventing autolysis of organelles; Isotonic: prevents organelles gaining or losing water by osmosis and bursting or shrinking; Buffered: keeps pH constant, preventing denaturation of proteins/enzymes
(c) B1 nuclei first, at the lowest speed (densest/largest organelle)
(c) B1 then mitochondria, then ribosomes last, at the highest speed (both needed for the mark, correct order)
(c) Answer: Nuclei first (lowest speed), then mitochondria, then ribosomes last (highest speed) - i.e. order of decreasing density/size
(d) B1 view the fraction using an electron microscope (high enough resolution to see organelle detail)
(d) B1 check that only structures with the features of mitochondria (e.g. double membrane, cristae) are present, with no other organelle types visible
(d) Answer: View the fraction under an electron microscope and check that only structures with the features of mitochondria (double membrane, cristae) are present, with no other organelle types visible

Question 5

(a) B1 interphase: cell grows and DNA/organelles are replicated (cell prepares for division)
(a) B1 mitosis: the nucleus divides, producing two genetically identical nuclei
(a) B1 cytokinesis: the cytoplasm (and cell membrane) divides, producing two separate daughter cells
(a) Answer: Interphase: cell growth and DNA replication/organelle duplication; Mitosis: nuclear division producing two genetically identical nuclei; Cytokinesis: division of the cytoplasm to form two separate daughter cells
(b) B1 W = metaphase
(b) B1 X = anaphase
(b) B1 Y = prophase
(b) B1 Z = telophase
(b) Answer: W = metaphase; X = anaphase; Y = prophase; Z = telophase
(c) M1 18 / 240
(c) A1 0.075 (or 7.5%) cao
(c) Answer: 0.075 (7.5%)
(d) B1 a higher mitotic index means a greater proportion of cells are dividing at any one time
(d) B1 this suggests mitosis/the cell cycle is no longer controlled properly (uncontrolled cell division), characteristic of a tumour
(d) Answer: A high mitotic index shows that a large proportion of cells are dividing at any given time, indicating uncontrolled cell division (loss of control of the cell cycle), which is characteristic of a tumour

Question 6

(a) B1 totipotent: can differentiate into any cell type, including extra-embryonic/placental tissue (found only in the early embryo)
(a) B1 pluripotent: can differentiate into almost any cell type but not extra-embryonic tissue (e.g. embryonic stem cells)
(a) B1 multipotent: can differentiate into a limited/restricted range of related cell types (e.g. adult bone marrow stem cells)
(a) Answer: Totipotent: can become any cell type including extra-embryonic tissue; Pluripotent: can become almost any cell type (not extra-embryonic tissue); Multipotent: can become a limited range of related cell types
(b) L3 (5-6): Detailed and accurate discussion of both medical uses and ethical issues, with a balanced argument and clear scientific reasoning linking stem cell type to therapeutic potential. Uses a range of relevant, correctly applied biological terminology throughout.
(b) L2 (3-4): Reasonable discussion of medical uses and/or ethical issues, with some scientific reasoning, though the argument may be one-sided or contain minor inaccuracies or omissions. Mostly correct use of terminology.
(b) L1 (1-2): Basic or generalised statements about stem cells, medical uses or ethical issues with little linking or scientific detail. Limited use of appropriate terminology.
(b) L0 (0): No relevant content.
(b) Answer: Level 3 response balancing therapeutic potential (embryonic vs adult vs iPS cells) against the ethical concerns of embryo destruction, consent and safety

Question 7

(a) B1 phospholipid bilayer forms the basic structure, hydrophilic heads facing out (towards water) and hydrophobic tails facing inwards
(a) B1 phospholipids/proteins can move (laterally) within the layer, giving a fluid structure
(a) B1 proteins are scattered throughout the bilayer in a mosaic pattern; some span the whole membrane (integral), others sit within one layer (peripheral)
(a) B1 cholesterol is present between phospholipids, adding stability and regulating fluidity (accept glycoproteins/glycolipids for cell signalling/recognition)
(a) Answer: A bilayer of phospholipids (hydrophilic heads out, hydrophobic tails in) that can move, giving a fluid structure; proteins are scattered through the bilayer in a mosaic pattern, some spanning the whole membrane; cholesterol sits between phospholipids adding stability
(b) B1 at higher temperatures, phospholipids/proteins gain more kinetic energy and move more, making the bilayer more fluid (oe)
(b) B1 above about 40 degrees C, membrane (channel/carrier) proteins begin to denature and the phospholipid bilayer structure begins to break down
(b) B1 this makes the membrane much more permeable, so more pigment (betalain) leaks out, increasing absorbance sharply
(b) Answer: Above about 40 degC, increased kinetic energy causes phospholipids to move more and membrane proteins to denature, breaking down the membrane structure; this makes the membrane much more permeable, so more pigment leaks out and absorbance increases sharply
(c) B1 size (surface area/volume) of the beetroot cylinders
(c) B1 any second valid control variable, e.g. time in the water bath, volume of water surrounding each cylinder, source/species of beetroot
(c) Answer: Any two of: size (surface area/volume) of the beetroot cylinders; time left in the water bath; volume of water surrounding each cylinder; species/source of beetroot used
(d) A1 0.66 awrt (0.657 to 3 s.f.) cao
(d) Answer: 0.66 (0.657 to 3 s.f.)

Question 8

(a) B1 net movement of molecules/particles from a region of higher concentration to a region of lower concentration (down a concentration gradient), due to random motion; requires no metabolic energy (oe)
(a) Answer: Net movement of molecules/particles from a region of higher concentration to a region of lower concentration (down a concentration gradient), as a result of their random motion; requires no input of metabolic energy
(b) B1 surface area available for diffusion (larger SA = faster rate)
(b) B1 concentration gradient/difference in concentration (steeper gradient = faster rate)
(b) B1 thickness of membrane/diffusion distance, or temperature (thinner/warmer = faster rate)
(b) Answer: Any three of: surface area available for diffusion; the concentration gradient (difference in concentration); the thickness of the membrane/diffusion distance; temperature
(c) M1 surface area = 6 x 3² = 54 mm²
(c) M1 volume = 3³ = 27 mm³
(c) dM1 ratio = 54 / 27 (dependent on both previous marks)
(c) A1 2 : 1 (or 2.0) cao
(c) Answer: SA = 54 mm²; V = 27 mm³; SA:V ratio = 2 : 1
(d) M1 surface area = 6 x 9² = 486 mm² and volume = 9³ = 729 mm³
(d) A1 ratio = 0.67 : 1 awrt (486/729 = 0.667) cao
(d) B1 as size/volume increases, the SA:V ratio decreases, so the diffusion distance/pathway to the centre increases
(d) B1 surface area is too small relative to volume to supply oxygen/nutrients or remove waste/heat fast enough by diffusion alone, so specialised exchange surfaces are needed
(d) Answer: SA:V ratio = 0.67:1 (486 mm² : 729 mm³). Larger organisms have a smaller SA:V ratio, so diffusion alone across the outer body surface cannot supply oxygen/nutrients or remove wastes fast enough, so specialised exchange surfaces (with adaptations to increase SA and reduce diffusion distance) are needed

Question 9

(a) B1 a measure of the tendency of water molecules to diffuse out of a solution; pure water has a water potential of zero and all solutions have negative water potentials (oe)
(a) Answer: A measure of the tendency of water molecules to diffuse out of a solution; pure water has a water potential of 0 kPa, and solutions have negative water potentials (the more concentrated the solution, the more negative the water potential)
(b) M1 identifies the zero crossing point lies between 0.2 mol/dm3 (+3%) and 0.4 mol/dm3 (-2%)
(b) dM1 correct linear interpolation, e.g. 3/(3+2) x 0.2 (dependent on M1)
(b) A1 awrt 0.30 to 0.35 mol/dm3 (e.g. 0.32)
(b) Answer: Approximately 0.32 mol/dm3 (accept 0.30 to 0.35 mol/dm3)
(c) B1 the sucrose solution has a lower (more negative) water potential than the cells/cytoplasm of the potato tissue
(c) B1 water moves out of the potato cells across the partially permeable cell-surface membrane by osmosis
(c) B1 down (from a higher to a lower) water potential gradient, so the tissue loses mass
(c) Answer: The 1.0 mol/dm3 sucrose solution has a lower (more negative) water potential than the potato cells. Water therefore moves out of the cells, across the partially permeable membrane, by osmosis, down the water potential gradient, causing the tissue to lose mass
(d) B1 any suitable improvement, e.g. blot each cylinder dry with paper towel before weighing (removes surface solution)
(d) B1 reasoning: so only the change in mass of the tissue itself is measured (or second valid improvement with reasoning, e.g. use more repeats/cylinders per concentration and take a mean)
(d) Answer: Any suitable improvement, e.g. gently blot each cylinder dry with paper towel before weighing (to remove surface solution) so only the change in mass of the tissue itself is measured

Question 10

(a) B1 sodium ions are actively transported (using ATP) out of the epithelial cell across the basal membrane into the blood, by the sodium-potassium pump
(a) B1 this maintains a low concentration of sodium ions inside the epithelial cell, creating a concentration gradient for sodium ions between the lumen and the cytoplasm
(a) B1 sodium ions diffuse from the lumen into the epithelial cell down this concentration gradient, through a co-transporter protein in the cell-surface (apical) membrane
(a) B1 glucose is carried into the cell at the same time (against its own concentration gradient) via the co-transporter protein, then diffuses into the blood by facilitated diffusion
(a) Answer: Sodium ions are actively pumped out of the epithelial cell into the blood by the sodium-potassium pump, maintaining a low Na+ concentration inside the cell; Na+ then diffuses from the lumen into the cell down its concentration gradient via a co-transporter protein, carrying glucose in with it (against its own gradient); glucose then diffuses into the blood via facilitated diffusion
(b) B1 active transport (of sodium ions by the sodium-potassium pump) requires ATP
(b) B1 mitochondria carry out aerobic respiration to produce this ATP, so many are needed to supply enough energy for active transport
(b) Answer: Active transport of sodium ions by the sodium-potassium pump requires ATP; mitochondria produce ATP by aerobic respiration, so many are needed to supply enough energy for this active transport
(c) B1 active transport requires ATP (uses metabolic energy), whereas facilitated diffusion does not (it is passive)
(c) B1 active transport can move substances against (up) a concentration gradient, whereas facilitated diffusion only moves substances down a concentration gradient
(c) B1 both use specific membrane (carrier or channel) proteins to move substances that cannot diffuse directly through the phospholipid bilayer
(c) Answer: Active transport requires ATP and can move substances against a concentration gradient; facilitated diffusion needs no ATP and only moves substances down a concentration gradient. Both use specific membrane proteins to transport substances that cannot cross the phospholipid bilayer directly

Question 11

(a) B1 B - Mitochondrion cao
(a) Answer: B - Mitochondrion
(b) B1 C - Membrane-bound nucleus cao
(b) Answer: C - Membrane-bound nucleus
(c) B1 produces/assembles ribosomes (makes ribosomal RNA, rRNA) (oe)
(c) Answer: It produces ribosomes / makes ribosomal RNA (rRNA) and assembles ribosomes
(d) B1 the mitochondria carry out aerobic respiration to release ATP
(d) B1 ATP provides the energy needed for the tail (flagellum) to move/beat, propelling the sperm towards the egg
(d) Answer: The mitochondria respire aerobically to produce ATP, which provides the energy needed for the tail (flagellum) to move, allowing the sperm to swim towards the egg
(e) B1 long hair-like extension of the cell, increasing surface area for absorption
(e) B1 large number of mitochondria to provide ATP for active transport of mineral ions (or thin cell wall/large permanent vacuole maintaining a steep water potential gradient)
(e) Answer: Any two of: a long root hair/extension of the cell, increasing surface area for absorption; many mitochondria, providing ATP for active transport of mineral ions; thin cell wall/large permanent vacuole to maintain a steep water potential/concentration gradient

Question 12

(a) B1 the 1 cm cube
(a) B1 because it has the largest surface area to volume ratio, so the diffusion pathway/distance to the centre is shortest relative to its volume, so NaOH reaches the centre fastest
(a) Answer: The 1 cm cube, because it has the largest surface area to volume ratio, meaning NaOH has the shortest distance to diffuse to reach the centre of the cube
(b) M1 converts time to minutes: 4 min 30 s = 4.5 min
(b) dM1 rate = distance / time = 5 / 4.5 (dependent on M1)
(b) A1 1.1 mm/min awrt (1.11 mm/min to 3 s.f.)
(b) Answer: 1.1 mm per minute (1.11 mm/min to 3 s.f.)
(c) B1 strength: agar cubes are cheap, safe and easy to prepare in precise sizes/shapes, allowing SA:V ratio to be varied systematically and quantitatively
(c) B1 limitation: agar is non-living and has no cell-surface membrane, so cannot model active transport or membrane permeability
(c) B1 limitation: real organisms/cells are not simple cubes and have specialised exchange surfaces, so the model may not accurately reflect real diffusion distances
(c) Answer: Strength: agar cubes are cheap, safe and easy to prepare in precise sizes, allowing SA:V ratio to be varied systematically. Limitation: agar is non-living and has no cell membrane, so cannot model active transport or membrane permeability; real organisms are not simple cubes, so the model may not accurately reflect real diffusion distances/exchange surfaces

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