Topic 2: Organisation of the Organism -- Challenge Exam 1
1 hour 15 minutes
80
7
75:00
0610
Instructions
Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.
Question Navigation
This paper covers the whole of Topic 2. Like a real Cambridge paper, the seven questions range across every sub-topic — cell structure and organisation, and the size of specimens — and they are deliberately mixed rather than grouped. All three Topic 2 papers do; they differ in the angle they come at it from, not in what they cover.
Question 1 -- The Cell That Was Not Labelled
Total: 12 marks
A student is shown a large electron micrograph of a single cell taken from an unfamiliar plant. She can see: structure A, a straight-edged layer forming the outer boundary of the cell; structure B, a very thin layer lying immediately inside A; structure C, a dense round body about 8 µm across containing a darker patch; and structure D, a single pale region that fills about 60% of the cell. Scattered through the remaining space are oval bodies 2 µm long with folded internal membranes. There are no green structures anywhere in the cell. A scale bar shows the whole cell is 95 µm long.
(a)[4]
Name structures A, B, C and D.
Model Answer -- 1(a)
A is the cell wall [1]
B is the cell membrane [1]
C is the nucleus [1]
D is the large permanent vacuole [1]
⚠ If you missed marks here: The wall and the membrane are two different structures lying one inside the other, and the order matters: the wall is always the outer layer and the membrane lies just inside it. Candidates who write “cell wall” for B lose the mark, and so do those who call D “cytoplasm” — cytoplasm is the jelly filling the rest of the cell, not one large pale sac.
(b)[4]
State one function of structure A and one function of structure B, and explain why these two functions must not be confused.
Model Answer -- 1(b)
A, the cell wall, gives the cell support and a fixed shape (and prevents it bursting when water enters) [1]
B, the cell membrane, controls what enters and leaves the cell [1]
The cell wall is fully permeable, so substances pass straight through it and it selects nothing [1]
Only the cell membrane is partially permeable, so only the membrane can control movement in and out [1]
⚠ If you missed marks here: This is the single most heavily penalised confusion in Topic 2. “The cell wall controls what enters the cell” scores zero every time, because a fully permeable layer cannot control anything. Learn the pair as one sentence: wall = support and shape, membrane = control.
(c)[4]
Name the oval bodies with folded internal membranes and state their function. Suggest which part of the plant this cell came from, and explain the evidence for your suggestion.
Model Answer -- 1(c)
The oval bodies are mitochondria [1]
They are the site of aerobic respiration, which releases energy for the cell [1]
The cell came from a part of the plant that receives no light, e.g. a root, a bulb or a tuber [1]
because there are no chloroplasts, and chloroplasts contain chlorophyll to absorb light for photosynthesis, which cannot happen in the dark [1]
⚠ If you missed marks here: “Mitochondria produce energy” is refused by mark schemes — energy is released from glucose by aerobic respiration, never created. In the second half, the mark is for linking the missing chloroplasts to the absence of light, not simply for naming a root: “it is a root cell” on its own earns one mark at most.
Question 2 -- Measuring What You Cannot See
Total: 12 marks
An electron micrograph of a cell is printed in a journal. It carries a scale bar labelled 20 µm, and when measured with a ruler this bar is 50 mm long. On the same micrograph the whole cell measures 150 mm long, and one mitochondrion inside it measures 6 mm long.
(a)[3]
Calculate the magnification of the micrograph. Show your working.
Model Answer -- 2(a)
magnification = image size ÷ actual size [1]
20 µm = 0.02 mm [1]
M = 50 ÷ 0.02 = ×2500 (no units) [1]
⚠ If you missed marks here: Dividing 50 by 20 without converting gives 2.5 and is wrong by exactly a factor of 1000 — the signature error of this topic. Write the conversion on its own line every time; it is worth a mark by itself. Also remember that a magnification carries no unit, so “2500 mm” loses the final mark.
(b)[3]
Calculate the actual length of the cell. Give your answer in µm.
Model Answer -- 2(b)
actual size = image size ÷ magnification [1]
150 ÷ 2500 = 0.06 mm [1]
0.06 × 1000 = 60 µm [1]
⚠ If you missed marks here: Two habits protect these marks. First, check the direction: the actual size must be smaller than the image size, so if you multiplied you have gone wrong. Second, finish in the unit the question demanded — 0.06 mm is correct arithmetic but the wrong unit, and it loses the accuracy mark.
(c)[3]
Calculate the actual length of the mitochondrion in µm, and state how many times longer the whole cell is than the mitochondrion.
Model Answer -- 2(c)
6 ÷ 2500 = 0.0024 mm [1]
= 2.4 µm [1]
the cell is 60 ÷ 2.4 = 25 times longer (or 150 ÷ 6 = 25) [1]
⚠ If you missed marks here: An answer of 2.4 mm rather than 2.4 µm should ring an alarm: a mitochondrion two millimetres long would be visible without a microscope. The ratio can be found directly from the image measurements because both structures were enlarged by the same factor, and quoting that shortcut is fully creditable.
(d)[3]
A second student measures the same cell and writes: “actual length = 150 × 2500 = 375 000 mm”. Identify her error, state why her answer is impossible, and explain why the magnification is not needed to compare the cell with the mitochondrion.
Model Answer -- 2(d)
She multiplied by the magnification instead of dividing by it [1]
Her answer is impossible because a microscope enlarges, so the actual size must be smaller than the image size (375 000 mm is 375 metres) [1]
Both structures were measured on the same image and so were enlarged by the same factor, which cancels when one is divided by the other [1]
⚠ If you missed marks here: The examiner wants the error named, not just corrected. “She should have divided” earns the first mark; the second needs the plausibility argument spelled out. The third mark is the discriminator on challenge papers — most candidates recalculate the ratio and never explain why the magnification cancels.
Question 3 -- Resistance in a Hospital Laboratory
Total: 12 marks
A hospital laboratory grows a bacterium on agar containing an antibiotic that prevents cells from building a cell wall. Almost every cell bursts and dies, but three colonies grow. DNA analysis shows that the cells in these colonies contain a small ring of DNA that the dead cells did not have.
(a)[4]
Name four structures found in a bacterial cell, and state which two of them are made of DNA.
Model Answer -- 3(a)
Any four of: cell wall, cell membrane, cytoplasm, ribosomes, circular DNA, plasmids — first two correct structures named [1]
third and fourth correct structures named [1]
The circular DNA is made of DNA [1]
The plasmids are also made of DNA [1]
⚠ If you missed marks here: The syllabus limits bacterial structures to exactly six, and a nucleus, mitochondria and chloroplasts are not among them — naming any of those scores nothing. The second half tests whether you realise that a bacterium contains two kinds of DNA, differing in size and importance rather than in chemistry.
(b)[4]
A student says the bacteria cannot be alive because they have no nucleus and therefore no genetic material. Explain why the student is wrong, and describe two ways in which the genetic material of this bacterium differs from that of a plant cell.
Model Answer -- 3(b)
The bacterium does contain genetic material [1]
Its DNA is a circular molecule lying free in the cytoplasm, not enclosed in a nucleus [1]
A plant cell keeps its DNA inside a nucleus, whereas the bacterial DNA is not enclosed by a membrane [1]
A bacterium also has plasmids, which a plant cell does not [1]
⚠ If you missed marks here: “No nucleus” describes the packaging, not the contents. The marks here are for the words circular and free in the cytoplasm, so an answer that only says “they do have DNA” stops after one mark. A growing colony is itself proof that the genetic material is present and working.
(c)[4]
Name the small ring of DNA found in the surviving cells and suggest how it allowed them to survive. Explain why this antibiotic does not harm the patient's own cells, and suggest why an antibiotic that attacks ribosomes would be far harder to use safely.
Model Answer -- 3(c)
The small ring of DNA is a plasmid [1]
It carries an extra gene giving resistance to the antibiotic, so those cells could still make a working cell wall or inactivate the drug, and they survived and divided [1]
Human (animal) cells have no cell wall, so a drug that blocks wall building has no target in them [1]
Ribosomes are present in bacterial and human cells alike, so a drug attacking ribosomes would also stop protein synthesis in the patient's own cells [1]
⚠ If you missed marks here: The last two marks are pure AO2: you are being asked to read the comparison table in your head. The winning idea is that a unique structure makes a safe drug target while a shared structure does not. Answers such as “antibiotics only affect bacteria” name no structure and earn nothing.
Question 4 -- Six Cells, Six Jobs
Total: 12 marks
The table below lists four specialised cells found in living organisms. A biologist is preparing revision cards and needs each card to show one visible feature of the cell and the function that feature makes possible.
cell
where it is found
ciliated cell
lining of the trachea and bronchi
root hair cell
surface of a young plant root
red blood cell
in the blood plasma
sperm cell
the male gamete
(a)[4]
For the ciliated cell and the root hair cell, describe one visible feature of each and explain how that feature adapts the cell to its function.
Model Answer -- 4(a)
The ciliated cell has many short hair-like cilia on its surface [1]
which beat to move mucus, with trapped dust and bacteria, along the trachea and bronchi away from the lungs [1]
The root hair cell has a long narrow extension [1]
which gives a large surface area, so water and mineral ions are absorbed from the soil more quickly [1]
⚠ If you missed marks here: Marks come in pairs here: feature, then consequence. A list such as “it has cilia and a root hair” earns the feature marks only. Use so that in every sentence, and be precise about the cilia — they move mucus, they do not increase surface area and they do not absorb oxygen.
(b)[4]
Explain how the red blood cell is adapted for the transport of oxygen. Give three separate adaptations, and state one disadvantage of one of them.
Model Answer -- 4(b)
It contains haemoglobin, which binds and carries oxygen [1]
It has no nucleus, so there is more room for haemoglobin and more oxygen can be carried [1]
It is a biconcave disc, giving a large surface area so oxygen is taken up and released faster [1]
A disadvantage: without a nucleus the cell cannot divide or repair itself, so it has a limited lifespan [1]
⚠ If you missed marks here: Three adaptations means three separate feature-plus-consequence statements; three ways of saying “it carries oxygen” is one mark. The final mark rewards genuine thinking rather than recall — losing the nucleus is a trade-off, and saying so shows you understand why the adaptation exists at all.
(c)[4]
A sperm cell contains a very large number of mitochondria, while a mature red blood cell contains very few organelles of any kind. Explain both observations.
Model Answer -- 4(c)
The sperm cell must swim to the egg, which requires energy [1]
Energy is released by aerobic respiration in the mitochondria, so many are needed [1]
The red blood cell is carried passively in the blood plasma and does little work of its own [1]
Having few organelles leaves the maximum space for haemoglobin, so more oxygen can be transported [1]
⚠ If you missed marks here: Write releases energy, never produces or makes — mark schemes reject the latter explicitly. For the red blood cell, the strongest answers give both halves: it does not need many organelles, and leaving them out is positively useful because it makes room for haemoglobin.
Question 5 -- Building an Organism
Total: 10 marks
A newly described marine worm has a feeding tube whose inner surface is lined with cells carrying hundreds of short beating projections. Around this lining is a layer of long thin cells that carry electrical signals to a nerve ring. The feeding tube joins a digestive sac, which in turn connects to a muscular pump.
(a)[4]
Define the terms tissue and organ, and use your definitions to classify the lining of the feeding tube and the feeding tube itself.
Model Answer -- 5(a)
A tissue is a group of cells with similar structures working together to perform a shared function [1]
An organ is a structure made of a group of different tissues working together to perform a specific function [1]
The lining of the feeding tube is a tissue, because it is made of one kind of cell [1]
The feeding tube is an organ, because it contains more than one type of tissue (the ciliated lining and the nervous layer) [1]
⚠ If you missed marks here: The words that carry the marks are similar for a tissue and different for an organ. Swap them and both definitions score zero. When classifying anything, apply one test only — count the number of tissue types — and ignore size, complexity or how important the structure looks.
(b)[3]
Name the two types of specialised cell described in the feeding tube, giving the function of each, and state the level of organisation formed by the feeding tube, the digestive sac and the muscular pump together.
Model Answer -- 5(b)
Ciliated cells, whose beating cilia move material along the tube [1]
Neurones, which conduct electrical impulses [1]
Together the three organs form an organ system [1]
⚠ If you missed marks here: The organism is unfamiliar but the cells are not — challenge papers routinely describe a known cell rather than naming it. A bare name without its function earns half, so always attach the job. The last mark needs the exact term organ system, not “a system” or “a body part”.
(c)[3]
The worm can regrow a damaged feeding tube. Explain, using the correct biological wording, where the new cells come from, and suggest why the beating cells would be expected to contain many mitochondria.
Model Answer -- 5(c)
New cells are produced by the division of existing cells [1]
Beating the cilia continuously requires energy [1]
Energy is released by aerobic respiration, which takes place in the mitochondria [1]
⚠ If you missed marks here: The first mark is for the exact syllabus phrase “division of existing cells” — “the worm grows new cells” or “the cells multiply” is refused. For the second half, give both steps: the work needs energy, and the energy is released by respiration in the mitochondria.
Question 6 -- Counting Mitochondria
Total: 12 marks
A researcher counts the mitochondria in samples of four cell types taken from the same mammal, using electron micrographs. Each cell was cut into thin sections, and the counts are the mean number of mitochondria seen in a single section.
cell type
mean number of mitochondria per section
number of cells examined
heart muscle cell
38
50
sperm cell
27
50
fat storage cell
4
50
mature red blood cell
0
6
(a)[4]
Describe the pattern shown by the data, and explain what the mitochondrion counts suggest about the activity of the heart muscle cell compared with the fat storage cell.
Model Answer -- 6(a)
The heart muscle cell has the most mitochondria and the mature red blood cell the fewest [1]
Heart muscle has about nine or ten times as many mitochondria per section as the fat storage cell [1]
Mitochondria are the site of aerobic respiration, which releases energy [1]
So the heart muscle cell releases much more energy and is far more active than the fat storage cell, which mainly stores material [1]
⚠ If you missed marks here: “Describe” wants the shape of the data, including at least one figure quoted or a comparison made — a bare list of the numbers already printed in the table adds nothing. The explanation mark is only awarded if you name aerobic respiration; “mitochondria give energy” is too loose and uses the wrong verb.
(b)[4]
Suggest why a sperm cell has a high count, and explain why the count for the mature red blood cell is zero rather than simply low.
Model Answer -- 6(b)
A sperm cell must swim to the egg, which requires a large amount of energy [1]
released by aerobic respiration in its many mitochondria [1]
A mature red blood cell has lost its nucleus and most of its organelles as it developed [1]
so that the maximum space is available for haemoglobin, allowing more oxygen to be transported [1]
⚠ If you missed marks here: The zero is not an anomaly to be explained away — it is a genuine adaptation. Candidates often write “the researcher made a mistake” or “too few cells were counted”, which misses the biology entirely. Link the missing organelles to the space gained and the oxygen carried.
(c)[4]
Evaluate the researcher's method. Give two weaknesses in the way the data were collected and suggest one improvement for each.
Model Answer -- 6(c)
Only six red blood cells were examined, far fewer than the 50 used for the other cell types [1]
Improvement: examine the same number of cells, at least 50, for every cell type so the comparison is fair [1]
Counting mitochondria in a single thin section does not give the number in the whole cell, and the number seen depends on where the cell was cut [1]
Improvement: count in several sections through each cell and calculate a mean, or state the count per unit volume rather than per section [1]
⚠ If you missed marks here: An evaluation must name a weakness and a matching improvement — “use more cells” without saying which sample was too small earns half. The sectioning point is the higher-level one: a thin slice through the edge of a cell will show fewer mitochondria than a slice through its middle, so single sections are not comparable.
Question 7 -- From Page to Reality
Total: 10 marks
A textbook prints a drawing of a plant cell. The drawing is 140 mm long and the caption states that the magnification of the drawing is ×2000. The same textbook prints a photomicrograph of a bacterium taken with a light microscope that has a ×10 eyepiece lens and a ×100 objective lens.
(a)[3]
Calculate the actual length of the plant cell shown in the drawing. Give your answer in µm and comment on whether your answer is a realistic size for a plant cell.
Model Answer -- 7(a)
actual = image ÷ magnification = 140 ÷ 2000 = 0.07 mm [1]
0.07 × 1000 = 70 µm [1]
This is realistic, because plant cells are typically around 100 µm long [1]
⚠ If you missed marks here: The third mark is given for the reality check, and it is free if you know a few typical sizes: bacterium 1–5 µm, mitochondrion about 2 µm, animal cell about 20 µm, plant cell about 100 µm. Candidates who answer “70 mm” and then call it realistic lose two marks for one mistake.
(b)[3]
State the total magnification of the light microscope, showing how it is obtained, and calculate how long a bacterium of actual length 3 µm would appear when viewed through it.
Model Answer -- 7(b)
Total magnification = eyepiece × objective (the two magnifications multiply) [1]
10 × 100 = ×1000 [1]
image = actual × magnification = 3 µm × 1000 = 3000 µm = 3 mm [1]
⚠ If you missed marks here: Adding the lens values instead of multiplying gives ×110 and is one of the most common errors on this topic. In the second half, image size is the only quantity you ever multiply for, and the answer must be converted to a sensible unit: 3000 µm is correct but 3 mm is how a scientist would write it.
(c)[4]
The textbook is reprinted in a smaller format in which every printed image is reduced to exactly half its original width, but the captions are not changed. Explain the effect of this on the stated magnification of the drawing, calculate the true magnification of the reprinted drawing, and explain why a scale bar would have avoided the problem.
Model Answer -- 7(c)
The stated magnification of ×2000 is now wrong, because the image is smaller while the cell is unchanged [1]
Magnification = image ÷ actual, so halving the image halves the magnification [1]
The true magnification of the reprinted drawing is ×1000 [1]
A scale bar is printed on the image, so it is reduced by the same amount as the picture and continues to represent the same real length correctly [1]
⚠ If you missed marks here: The mark for the reasoning is separate from the mark for the number, so show why halving the image halves the magnification rather than just asserting it. The final mark is the one that separates strong candidates: a scale bar survives resizing because it shrinks with the image, whereas a printed magnification does not.
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