IGCSE Biology Paper 4 (Theory / Extended)

Topic 2: Organisation of the Organism -- Challenge Exam 3

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

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 Cells at the Tip of a Root

Total: 12 marks

A gardener notices that a seedling wilts badly if its root tip is damaged when it is transplanted, even though the rest of the root is untouched. Under a microscope, the outer cells near the root tip are seen to have long narrow projections extending into the soil. These cells contain a nucleus, a large permanent vacuole and unusually large numbers of mitochondria, but no green structures.

(a) [4]

Name this type of cell and explain how its shape adapts it to its function. Explain also why damaging these cells causes the seedling to wilt.

Model Answer -- 1(a)

These are root hair cells [1]

The long narrow projection gives a very large surface area [1]

so water and mineral ions are absorbed from the soil more quickly [1]

If the root hair cells are damaged, much less water is absorbed, so the plant loses more water than it takes up and wilts [1]

⚠ If you missed marks here: “It has a hair so it absorbs more” earns the first mark only. The examiner pays for the intermediate step — large surface area — because that is the actual adaptation; the hair itself is only how the surface area is achieved. In the last part, link reduced absorption to the loss of support in the plant.

1 mark -- cells identified as root hair cells 1 mark -- long narrow projection gives a large surface area 1 mark -- so faster absorption of water and mineral ions 1 mark -- damage reduces water uptake so the seedling wilts

(b) [4]

Explain why these cells contain no green structures, and name the green structures found in other plant cells, giving their function.

Model Answer -- 1(b)

The green structures are chloroplasts [1]

They contain chlorophyll, which absorbs light energy [1]

They are the site of photosynthesis, which makes food for the plant [1]

Root cells are underground and receive no light, so photosynthesis could not take place and chloroplasts would be useless there [1]

⚠ If you missed marks here: The mark for the absence of chloroplasts is for the reasoning no light, therefore no photosynthesis, not for “the root does not need them”. Note also that this is why the comparison table says chloroplasts are found in the green parts of a plant, not simply in plant cells.

1 mark -- green structures named as chloroplasts 1 mark -- contain chlorophyll which absorbs light energy 1 mark -- site of photosynthesis which makes food 1 mark -- roots receive no light so photosynthesis cannot occur

Suggest why these cells contain unusually large numbers of mitochondria. Then explain why a student is wrong to say that a cell with no chloroplasts cannot be a plant cell.

Model Answer -- 1(c)

Mitochondria are the site of aerobic respiration, which releases energy [1]

Taking up mineral ions from the soil requires a large amount of energy, so many mitochondria are needed [1]

The student is wrong because chloroplasts are found only in the parts of a plant that receive light [1]

A plant cell is identified by its cell wall and large permanent vacuole, both of which these cells have [1]

⚠ If you missed marks here: Say that respiration releases energy; “produces energy” is refused. In the second half, the strongest answers supply the positive evidence as well as the correction — it is the wall and the vacuole that make a cell a plant cell, so quote them rather than simply saying the student is mistaken.

1 mark -- mitochondria are the site of aerobic respiration releasing energy 1 mark -- energy needed for uptake of mineral ions from the soil 1 mark -- chloroplasts occur only in parts of the plant receiving light 1 mark -- plant cells identified by cell wall and large permanent vacuole

Question 2 -- Down the Microscope

Total: 12 marks

A student uses a light microscope fitted with a ×10 eyepiece lens and a ×40 objective lens. With this combination the circular field of view she can see is 0.45 mm across in real terms. She observes a plant cell that occupies about one third of the width of the field of view.

(a) [3]

State the total magnification she is using, explain how it is obtained, and explain why a total of ×50 would be wrong.

Model Answer -- 2(a)

Total magnification = eyepiece magnification × objective magnification [1]

10 × 40 = ×400 [1]

×50 comes from adding the two values; the lenses enlarge one after the other, so their effects multiply rather than add [1]

⚠ If you missed marks here: This is a single mark that is lost by hundreds of candidates every year. Fix the idea rather than the number: the objective produces an enlarged image, and the eyepiece then enlarges that image again, so the two factors multiply.

1 mark -- rule that eyepiece and objective magnifications multiply 1 mark -- total magnification 400 1 mark -- explanation that adding is wrong because the lenses enlarge in succession

(b) [3]

Calculate the actual width of the plant cell. Give your answer in µm, and state whether the magnification of the microscope was needed for this calculation.

Model Answer -- 2(b)

0.45 ÷ 3 = 0.15 mm [1]

0.15 × 1000 = 150 µm [1]

The magnification was not needed, because the field of view was already given as an actual size [1]

⚠ If you missed marks here: The third mark separates candidates who understand the quantities from those who follow a recipe. Many will divide by 400 as well, producing 0.375 µm — smaller than a bacterium. Always ask whether a length you are given is an image length or an actual length before you calculate.

1 mark -- 0.45 divided by 3 = 0.15 mm 1 mark -- answer 150 micrometres 1 mark -- magnification not needed because the field of view was an actual size

The student photographs the cell through the microscope, and the photograph is printed at the same magnification as the microscope. Calculate the width of the cell on the printed photograph.

Model Answer -- 2(c)

image size = actual size × magnification [1]

0.15 × 400 [1]

= 60 mm [1]

⚠ If you missed marks here: Image size is the only quantity you ever multiply for. A quick reality check confirms the answer: 60 mm is about the width of a credit card, which is a sensible size for a printed cell.

1 mark -- rearrangement image = actual multiplied by magnification 1 mark -- substitution 0.15 multiplied by 400 1 mark -- answer 60 mm

(d) [3]

The photograph is later reprinted at exactly one quarter of its original width, but the caption still says ×400. State the width of the cell on the reprint, calculate the true magnification of the reprint, and explain why a scale bar printed on the photograph would not have been affected.

Model Answer -- 2(d)

New width = 60 ÷ 4 = 15 mm [1]

True magnification = 15 ÷ 0.15 = ×100 [1]

A scale bar is part of the image, so it is reduced by the same factor and continues to represent the same real length correctly [1]

⚠ If you missed marks here: The second mark can also be earned by reasoning rather than arithmetic: quartering every image length quarters the magnification, and 400 ÷ 4 = 100. The final mark is the reason serious micrographs carry scale bars instead of stated magnifications — a printed number does not survive resizing.

1 mark -- new printed width 15 mm 1 mark -- true magnification 100 1 mark -- a scale bar shrinks with the image so still represents the correct real length

Question 3 -- The Airways of a Smoker

Total: 12 marks

A pathologist compares tissue taken from the trachea of a lifelong non-smoker with tissue taken from the trachea of a person who has smoked for thirty years. In the non-smoker, the surface cells carry dense rows of short projections. In the smoker, most of these projections have been destroyed, although the cells themselves are still present and still produce mucus.

(a) [4]

Name the surface cells and the projections they carry, and describe fully the function these cells perform in a healthy trachea.

Model Answer -- 3(a)

The cells are ciliated cells [1]

The projections are cilia [1]

The cilia beat together in a rhythm [1]

moving mucus, which has trapped dust particles and bacteria, up the trachea and bronchi and away from the lungs [1]

⚠ If you missed marks here: Four marks for what feels like one idea, so keep writing: the cell, the structure, the movement and what is being moved are separate marking points. Be precise about the load — the cilia move mucus, not air and not oxygen.

1 mark -- cells named as ciliated cells 1 mark -- projections named as cilia 1 mark -- the cilia beat 1 mark -- moving mucus with trapped dust and bacteria away from the lungs

(b) [4]

Predict two consequences for the smoker of losing these projections, explaining each. Use the fact that mucus is still being produced in your answer.

Model Answer -- 3(b)

Mucus accumulates in the airways, because it is still produced but is no longer swept away [1]

The person coughs repeatedly to clear the mucus instead [1]

Dust particles and bacteria are not removed from the airways [1]

so infections of the lungs, such as bronchitis, become more likely [1]

⚠ If you missed marks here: The question hands you the key fact — mucus production continues — and expects you to use it, so build both consequences on the build-up rather than on a vague “the lungs get damaged”. Each consequence must be paired with its explanation to score both of its marks.

1 mark -- mucus accumulates because it is still produced but not swept away 1 mark -- the person coughs to clear the mucus 1 mark -- dust and bacteria are not removed from the airways 1 mark -- so lung infections become more likely

The lining of the trachea is described as a tissue, while the trachea is described as an organ and the lungs together with the trachea form part of a larger structure. Define tissue, organ and organ system, and state which level the trachea belongs to.

Model Answer -- 3(c)

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]

An organ system is a group of organs with related functions working together [1]

The trachea is an organ, because it contains more than one type of tissue [1]

⚠ If you missed marks here: The single words similar and different carry the first two marks; swap them and both definitions score zero. The final mark needs the justification as well as the classification, so do not simply write “organ” and move on.

1 mark -- tissue defined using cells of similar structure 1 mark -- organ defined using different tissues 1 mark -- organ system defined as a group of organs with related functions 1 mark -- trachea classified as an organ with a reason

Question 4 -- Three Cells, No Names

Total: 12 marks

A researcher records which structures are present in three unknown cells, P, Q and R.

structure	cell P	cell Q	cell R
cell wall	present	present	absent
nucleus	absent	present	present
chloroplasts	absent	absent	absent
large permanent vacuole	absent	present	absent
plasmids	present	absent	absent

(a) [4]

Identify cells P, Q and R, and for cell P give the one piece of evidence in the table that is conclusive on its own.

Model Answer -- 4(a)

P is a bacterial cell [1]

Q is a plant cell [1]

R is an animal cell [1]

The presence of plasmids is conclusive for P, because only bacterial cells contain them (the absence of a nucleus also identifies it) [1]

⚠ If you missed marks here: Do not identify from a single row of the table unless it really is decisive. A cell wall appears in both P and Q, so it separates neither from the other; plasmids and the missing nucleus are the features unique to bacteria. Q is still a plant cell despite having no chloroplasts.

1 mark -- P identified as bacterial 1 mark -- Q identified as plant 1 mark -- R identified as animal 1 mark -- plasmids or absent nucleus given as conclusive evidence for P

(b) [4]

Cell Q has no chloroplasts. Explain why this does not affect your identification, and name three structures that all three cells would contain even though the table does not list them.

Model Answer -- 4(b)

Chloroplasts occur only in the parts of a plant that receive light, so a plant cell from a root, bulb or tuber has none [1]

All three cells contain a cell membrane, which controls what enters and leaves [1]

All three contain cytoplasm, where most chemical reactions take place [1]

All three contain ribosomes, which are the site of protein synthesis [1]

⚠ If you missed marks here: The three universal structures — membrane, cytoplasm, ribosomes — are the backbone of every comparison question, and this part is a direct test of whether you know them. Naming mitochondria here would be wrong, because bacterial cells have none.

1 mark -- chloroplasts occur only in light-receiving parts so a plant cell may lack them 1 mark -- cell membrane named as present in all three 1 mark -- cytoplasm named as present in all three 1 mark -- ribosomes named as present in all three

Describe how the genetic material of cell P is arranged, and compare it with the arrangement in cell R. Explain why a student who says cell P has no genetic material is mistaken.

Model Answer -- 4(c)

In cell P the main genetic material is a single circular DNA molecule lying free in the cytoplasm [1]

together with plasmids, which are small separate rings of DNA carrying a few extra genes [1]

In cell R the DNA is contained within a nucleus, enclosed by a membrane [1]

The student has confused the absence of a nucleus with the absence of DNA; the DNA is present, it is simply not enclosed [1]

⚠ If you missed marks here: This distinction is examined deliberately and often. The absence of a nucleus describes the packaging of the DNA, not whether DNA exists. If cell P had no genetic material it could not make proteins or divide, so a growing bacterial culture disproves the claim by itself.

1 mark -- circular DNA free in the cytoplasm in cell P 1 mark -- plasmids as small separate rings of DNA 1 mark -- cell R keeps its DNA inside a nucleus 1 mark -- no nucleus does not mean no DNA

Question 5 -- Two Extremes

Total: 10 marks

A biologist compares two human cells. Cell M is a mature red blood cell, about 7 µm across, containing almost no organelles. Cell N is a neurone running from the base of the spine to a muscle in the foot, over one metre long.

(a) [4]

Explain how each of these two cells is adapted to its function, giving two adaptations for cell M and one for cell N.

Model Answer -- 5(a)

Cell M contains haemoglobin, which binds oxygen so the cell can transport it [1]

Cell M has no nucleus, so there is more room for haemoglobin and more oxygen can be carried [1]

Cell M is a biconcave disc, giving a large surface area so oxygen is absorbed and released more quickly [1]

Cell N is extremely long and thin, so it can conduct electrical impulses over long distances between distant parts of the body [1]

⚠ If you missed marks here: Each mark needs a feature and a consequence, so a list of three features about cell M will collect one mark at most. Note that only three of the four points above are strictly required for cell M, so choose the two you can explain most confidently and write them fully.

1 mark -- haemoglobin binds oxygen for transport 1 mark -- no nucleus so more room for haemoglobin and more oxygen carried 1 mark -- biconcave shape gives a large surface area for faster gas exchange 1 mark -- neurone is very long and thin so conducts impulses over long distances

(b) [3]

Cell M cannot divide and survives for only about four months. Explain why it cannot divide, and state where new red blood cells must therefore come from.

Model Answer -- 5(b)

Cell division is controlled by the nucleus, which contains the genetic material [1]

A mature red blood cell has lost its nucleus, so it cannot divide or repair itself [1]

New red blood cells are produced by the division of existing cells elsewhere in the body, in the bone marrow [1]

⚠ If you missed marks here: This is the trade-off behind the adaptation: losing the nucleus buys space for haemoglobin at the cost of the ability to divide. The final mark rewards the syllabus phrase division of existing cells; “the body makes new ones” is refused.

1 mark -- cell division is controlled by the nucleus 1 mark -- the mature red blood cell has no nucleus so cannot divide 1 mark -- new cells produced by division of existing cells in the bone marrow

A student says that because both cells are human cells they must contain the same organelles. Explain why the student is wrong, and state one structure that both cells do share, giving its function.

Model Answer -- 5(c)

Cells become specialised, so their structure is modified to suit the job they do [1]

Cell M has lost almost all its organelles, whereas cell N retains a nucleus and many mitochondria, so their contents are very different [1]

Both share a cell membrane, which controls what enters and leaves the cell (accept cytoplasm, or ribosomes for protein synthesis) [1]

⚠ If you missed marks here: The idea being tested is specialisation: cells from one organism start alike and are then modified, sometimes by losing structures. For the shared structure, avoid naming the nucleus — cell M has none — and always attach the function, since a bare name earns nothing.

1 mark -- cells become specialised so their structure is modified for their job 1 mark -- cell M has lost almost all organelles unlike cell N 1 mark -- a shared structure named with its function

Question 6 -- Chloroplasts Through the Depth of a Leaf

Total: 12 marks

A researcher measures the mean number of chloroplasts per cell at five depths below the upper surface of a leaf. The depths were measured on a micrograph printed at a magnification of ×250.

depth below upper surface / µm	20	60	100	140	180
mean number of chloroplasts per cell	42	38	26	14	9

(a) [4]

Describe the pattern shown by the data, quoting figures from the table, and calculate the percentage decrease in the mean number of chloroplasts between a depth of 20 µm and a depth of 180 µm.

Model Answer -- 6(a)

As the depth below the upper surface increases, the mean number of chloroplasts per cell decreases [1]

The count falls from 42 at 20 µm to 9 at 180 µm [1]

decrease = 42 − 9 = 33; percentage = (33 ÷ 42) × 100 [1]

= 78.6% (accept 79%) [1]

⚠ If you missed marks here: A description without figures rarely gets full credit, so always quote at least the two end values. In the calculation, the divisor is the original value of 42, not the final value of 9; dividing by 9 gives 367% and is a very common error.

1 mark -- chloroplast number decreases as depth increases 1 mark -- figures quoted from the table to support the description 1 mark -- working shown as 33 divided by 42 multiplied by 100 1 mark -- answer 78.6 percent or 79 percent

(b) [4]

Name the cell type found at a depth of 20 µm and explain the pattern shown by the data.

Model Answer -- 6(b)

At 20 µm the cells are palisade mesophyll cells [1]

They lie just below the upper surface, where the light intensity is greatest [1]

Light is absorbed by the chloroplasts in the upper layers, so less light reaches the cells deeper in the leaf [1]

Cells deeper in the leaf can carry out less photosynthesis, so fewer chloroplasts are needed there [1]

⚠ If you missed marks here: The explanation must involve light being absorbed on the way down, not simply “it is darker lower down”, which restates the observation. Naming the palisade mesophyll cell is a mark of its own, and it follows directly from the position and the high chloroplast count.

1 mark -- cells at 20 micrometres named as palisade mesophyll cells 1 mark -- these lie where light intensity is greatest 1 mark -- light is absorbed by upper layers so less reaches deeper cells 1 mark -- less photosynthesis deeper down so fewer chloroplasts needed

Calculate the distance on the printed micrograph that corresponds to a depth of 180 µm, and suggest two ways in which the reliability of the chloroplast counts could be improved.

Model Answer -- 6(c)

180 µm = 0.18 mm, and image = actual × magnification = 0.18 × 250 [1]

= 45 mm [1]

Improvement: count the chloroplasts in a larger number of cells at each depth and calculate a mean [1]

Improvement: repeat with leaves from several different plants, or grown in the same light conditions, so the results are not specific to one leaf [1]

⚠ If you missed marks here: In the calculation, this is the multiply case: you are finding an image length from an actual length. For the improvements, “be more careful” and “use a better microscope” earn nothing; name a change to the sampling that would reduce the effect of natural variation between cells or between leaves.

1 mark -- conversion and substitution 0.18 multiplied by 250 1 mark -- answer 45 mm 1 mark -- improvement of counting more cells at each depth and taking a mean 1 mark -- improvement of repeating with several leaves or plants under controlled conditions

Question 7 -- Neither of Them Has a Nucleus

Total: 10 marks

Two cells are photographed on the same electron micrograph, which is printed at a magnification of ×4000. Cell J is a flattened disc about 7 µm across with no cell wall and no nucleus. Cell K is a rod about 2 µm long with a cell wall, no nucleus, and several small rings of DNA in its cytoplasm.

(a) [3]

Identify cell J and cell K, giving one piece of evidence for each identification that does not rely on the absence of a nucleus.

Model Answer -- 7(a)

Cell J is a red blood cell [1]

because it is a flattened biconcave disc about 7 µm across with no cell wall [1]

Cell K is a bacterial cell, because it has a cell wall and plasmids, and is only 2 µm long [1]

⚠ If you missed marks here: The instruction to avoid the missing nucleus is the whole point of the question: both cells lack one, so it distinguishes nothing. Use the wall, the plasmids and above all the size, since a 2 µm cell is far too small to be a human cell of any kind.

1 mark -- cell J identified as a red blood cell 1 mark -- evidence of biconcave disc about 7 micrometres with no cell wall 1 mark -- cell K identified as a bacterium with evidence of cell wall or plasmids

(b) [3]

Calculate how long cell J and cell K each appear on the printed micrograph. Give both answers in millimetres.

Model Answer -- 7(b)

image size = actual size × magnification [1]

Cell J: 7 µm = 0.007 mm, so 0.007 × 4000 = 28 mm [1]

Cell K: 2 µm = 0.002 mm, so 0.002 × 4000 = 8 mm [1]

⚠ If you missed marks here: Both calculations multiply, because you are working from an actual size to an image size. Check the ratio afterwards: cell J is 3.5 times the length of cell K in real life, and 28 ÷ 8 = 3.5 on the print, which confirms both answers at once.

1 mark -- rearrangement image = actual multiplied by magnification 1 mark -- cell J appears 28 mm long 1 mark -- cell K appears 8 mm long

Explain why the two cells have no nucleus for completely different reasons, and state one advantage that the missing nucleus gives to cell J.

Model Answer -- 7(c)

Cell J lost its nucleus as it matured; it is an animal cell that began with one [1]

Cell K never had a nucleus, because bacterial cells do not have one at all [1]

Cell K still has DNA, present as a circular molecule free in the cytoplasm together with plasmids [1]

The advantage to cell J is that there is more space for haemoglobin, so it can transport more oxygen [1]

⚠ If you missed marks here: This is the trap the whole question is built around: “no nucleus” is an adaptation in one cell and a basic feature of the cell type in the other. Never identify a cell from a single missing structure, and never conclude that a cell without a nucleus has no DNA.

1 mark -- cell J lost its nucleus during development 1 mark -- cell K never had a nucleus as bacteria have none 1 mark -- cell K still contains DNA as a circular molecule with plasmids 1 mark -- advantage of more space for haemoglobin so more oxygen carried

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