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IB Biology B2.3: Cell specialization worked solutions

Worked solutions to every practice question in B2.3: Cell specialization of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
A pancreatic beta cell and a skeletal muscle cell from one person both contain the insulin gene, but only the beta cell makes insulin. What is the best explanation? (A) The muscle cell lost the insulin gene. (B) The insulin gene is expressed in the beta cell only. (C) The two cells have different genomes. (D) Only dividing cells make insulin.
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Both cells keep the whole genome. Differentiation changes which genes are expressed, so the beta cell makes insulin and the muscle cell does not.

Final answer

B

Question B1

[1 mark]
Which pair of properties defines a stem cell? (A) divides repeatedly; can differentiate (B) divides once; then specialises (C) cannot divide; can differentiate (D) divides repeatedly; never specialises
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A stem cell renews itself by repeated division and also produces daughters that differentiate. A cell lacking either property is not a stem cell.

Final answer

A

Question C1

[1 mark]
A cell from the inner cell mass of a blastocyst can form every cell type of the body but not the placenta. Which term describes it? (A) totipotent (B) pluripotent (C) multipotent (D) fully specialised
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Every body cell type but not the placenta is the range of a pluripotent cell. A totipotent cell would also form placental cells; a multipotent cell is limited to one lineage.

Final answer

B

Question D1

[1 mark]
A signalling protein spreads from one side of an early embryo. Cells near the source switch on one set of genes and cells far away switch on another set. Which conclusion is supported? (A) The DNA of cells near the source mutates. (B) The concentration of the signal affects which genes are expressed. (C) Cells far from the source cannot divide. (D) Only cells near the source differentiate.
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A concentration gradient gives each cell positional information. Different concentrations switch on different genes, so the cells differentiate differently without any change in DNA.

Final answer

B

Question E1

[1 mark]
Which cell has the greatest length? (A) egg: diameter 100 \mum (B) sperm: length 50 \mum (C) muscle fibre: length 3 cm (D) red cell: diameter 7 \mum
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3 cm is 30 000 \mum, far longer than the other values. Compare the same dimension: the egg and the red cell are given as diameters.

Final answer

C

Question F1

[2 marks]
A cuboid cell keeps its shape and every edge doubles in length. (a) State the factor by which the surface area changes. [1] (b) State the factor by which the surface area to volume ratio changes. [1]
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(a) Area scales with the square of the linear factor, 2^2=4 (1). (b) Volume rises by 2^3=8, so the ratio changes by \frac48=\frac12 and halves (1).

Final answer

(a) \times4. (b) It halves (\times\frac12).

Question G1

HL[1 mark]
Which feature of a proximal convoluted tubule cell increases the membrane area facing the tubule lumen? (A) a thick basement membrane (B) microvilli (C) many mitochondria (D) a flagellum
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Microvilli are finger-like projections that multiply the membrane area on the lumen side. Mitochondria supply ATP; they do not add surface.

Final answer

B

Question H1

HL[1 mark]
Which row correctly matches each type of pneumocyte to its feature? (A) type I: lamellar bodies; type II: very thin (B) type I: very thin; type II: lamellar bodies (C) type I: secretes surfactant; type II: very thin (D) type I: lamellar bodies; type II: secretes surfactant
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Type I cells are very thin for gas exchange. Type II cells hold lamellar bodies and secrete surfactant. The other rows swap the roles.

Final answer

B

Question I1

HL[1 mark]
Which feature is found in both cardiac muscle cells and skeletal muscle fibres? (A) branched cells (B) striations (C) many nuclei per cell (D) voluntary control
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Both are striated, with myofibrils of sarcomeres. Branching is cardiac, many nuclei is skeletal, and voluntary control applies only to skeletal muscle.

Final answer

B

Tutorial practice

Question K2

[5 marks]
Adult humans keep stem cells in niches inside several tissues. (a) Define a stem cell niche. [1] (b) Name two niches in adult humans and the cells that each one supplies. [2] (c) Explain how a niche can supply replacement cells without using up its stem cells. [2]
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(a) A niche is the small region of tissue, with its supporting cells and signals, in which stem cells are maintained and regulated (1). (b) Any two named niches with their output: bone marrow giving red cells, white cells and platelets (1); the base of a gut crypt giving lining cells, or a hair follicle giving follicle and skin cells (1). (c) A stem cell divides and one daughter stays in the niche as a stem cell, which keeps the reserve (1); the other daughter leaves and differentiates to replace lost cells, and signals from the niche regulate which happens (1).

Final answer

(a) The region of tissue, with supporting cells and signals, where stem cells are kept and regulated. (b) Bone marrow: blood cells; hair follicle: follicle and skin cells (or gut crypt: lining cells). (c) Some daughters stay as stem cells; others leave and differentiate; niche signals control which.

Question L2

[4 marks]
Cells can be modelled as cubes. A cube of side 2 \mum has a surface area of 24 \mum^2 and a volume of 8 \mum^3; a cube of side 4 \mum has 96 \mum^2 and 64 \mum^3. (a) Calculate the surface area and the volume of a cube of side 6 \mum. [2] (b) Deduce the surface area to volume ratio of a cube of side 12 \mum. [1] (c) State one limitation of using cubes as a model of cells. [1]
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(a) Area =6\times6^2=216\ \mum^2 (1); volume =6^3=216\ \mum^3 (1). (b) The ratio is 6/L, which for L=12 gives 0.5 per \mum (1). (c) Real cells are not cubes: they have folds, flattened regions and organelles, so the model gives a trend, not exact values (1).

Final answer

(a) 216\ \mum^2 and 216\ \mum^3. (b) 0.5 per \mum. (c) Real cells are folded, flattened or irregular, and the model ignores diffusion distance.

Question M2

[4 marks]
Spherical cell A has a radius of 5 \mum and cell B a radius of 15 \mum. Both use oxygen at the same rate per unit volume and take it up at the same rate per unit of membrane area. For a sphere the area to volume ratio is 3/r. (a) Calculate the surface area to volume ratio of each cell. [2] (b) Explain why cell B is less able to meet its oxygen demand, although its membrane area is larger. [2]
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(a) A: 3/5=0.60\ \mum^{-1} (1); B: 3/15=0.20\ \mum^{-1} (1). (b) Cell B has 3^2=9 times the membrane area but 3^3=27 times the volume (1), so each unit of cytoplasm has less membrane to supply oxygen and the supply falls behind the demand (1).

Final answer

(a) A: 0.60; B: 0.20 per \mum. (b) B has 9 times the area but 27 times the volume, so each unit of volume has less membrane and supply falls behind demand.

Question N2

[6 marks]
mRNA from the actin gene was detected in a red blood cell precursor, a pancreatic beta cell and a neuron of one person. mRNA from the globin gene was detected only in the precursor, and mRNA from the insulin gene only in the beta cell. (a) State why all three cell types contain the insulin gene. [1] (b) Suggest why the actin gene is expressed in all three cell types. [2] (c) Explain how cells with the same genes can have different structures and functions. [3]
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(a) All three descend from the zygote by mitosis, which passes on the same genome (1). (b) Actin filaments form part of the cytoskeleton (1), needed for the shape and movement of every cell type (1). (c) Different sets of genes are expressed in each cell type (1), so different proteins and mRNA are made (1), and these proteins give each cell its own structure and function (1).

Final answer

(a) They descend from one zygote by mitosis, so they have the same genome. (b) Actin is needed for the cytoskeleton and cell shape in every cell. (c) Different genes are expressed; so different proteins are made; the proteins determine structure and function.

Question O2

[4 marks]
In culture a zygote (P) gave all body cell types and placental cells. The inner cell mass of a blastocyst (Q) gave all body cell types but no placental cells. Adult bone marrow (R) gave only red cells, white cells and platelets. (a) Classify each of P, Q and R as totipotent, pluripotent or multipotent. [3] (b) Explain why the cells of R could not be used to regenerate a damaged liver. [1]
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(a) P forms body cells and placental cells: totipotent (1); Q forms every body cell type but no placenta: pluripotent (1); R forms only the cells of one lineage, blood: multipotent (1). (b) Cells of R are restricted to blood cell types, so they cannot form liver cells (1).

Final answer

(a) P totipotent; Q pluripotent; R multipotent. (b) R makes blood cells only, so it cannot give liver cells.

Question P2

HL[5 marks]
A proximal convoluted tubule cell has a brush border facing the tubule lumen and deep infoldings, with many mitochondria beside them, on the side facing the blood. (a) Name the structures that form the brush border. [1] (b) Explain how the brush border and the basal infoldings with their mitochondria increase the rate of reabsorption. [4]
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(a) Microvilli (1). (b) Microvilli greatly increase the membrane area for the cell volume (1), so more transport proteins can reabsorb glucose, ions and water per second (1). The basal infoldings add membrane that carries pumps and channels (1); the mitochondria beside them supply the ATP for active transport (1).

Final answer

(a) Microvilli. (b) Microvilli: more membrane per volume, so more transport proteins. Infoldings: more membrane for pumps; mitochondria: ATP for active transport; result: faster uptake of glucose, ions and water.

Question Q2

HL[5 marks]
A baby born several weeks early has difficulty inflating its lungs because its alveoli make too little surfactant. (a) Identify the cell type that secretes surfactant and the structures in which it is stored. [2] (b) Explain the effect of too little surfactant on the alveoli and on gas exchange. [3]
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(a) Type II pneumocytes (1); their lamellar bodies (1). (b) Without surfactant the surface tension of the water film lining the alveolus is high (1), so the walls stick together and the alveoli collapse as air leaves (1); the area for gas exchange falls and the lungs are hard to inflate (1).

Final answer

(a) Type II pneumocytes; lamellar bodies. (b) High surface tension in the water film; walls stick together and the alveoli collapse as air leaves; less area for gas exchange and harder to inflate.

Question R2

HL[6 marks]
A human sperm cell is adapted to reach and enter an egg. (a) Draw a labelled diagram of a human sperm cell showing the nucleus, the acrosome, the midpiece and the tail. [4] (b) State the function of two of the structures you labelled. [2]
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(a) One mark each for a drawn structure that is correctly placed and labelled with a line ending on it: nucleus in the head (1); acrosome as a cap at the front of the head (1); midpiece behind the head with mitochondria (1); tail extending from the midpiece, with a continuous cell outline (1). (b) Any two: the acrosome releases enzymes that digest a path through the outer layers of the egg (1); the mitochondria in the midpiece supply ATP (1); the tail beats to propel the sperm (1); the nucleus carries the haploid set of chromosomes (1).

Final answer

(a) Head with the nucleus and, in front of it, the acrosome; midpiece with mitochondria behind the head; long tail. (b) Acrosome: enzymes digest a path through the egg’s outer layers; midpiece: mitochondria supply ATP for the tail.

Advanced practice

Question S3

[7 marks]
Factor N is a signalling protein made by niche cells in mouse hair follicles. Four groups of mice were compared after three weeks. Group 1 normal niche: 12 stem cells per follicle, 94% of follicles regrow; group 2 niche cells lack factor N: 3, 21%; group 3 niche cells lack factor N, factor N added: 11, 88%; group 4 stem cells lack the receptor, factor N added: 2, 18%. (a) Deduce, using the data, the role of factor N. [2] (b) Explain why group 3 strengthens the conclusion beyond the result for group 2 alone. [2] (c) Deduce what group 4 shows about how factor N acts. [2] (d) State one limitation of applying these results to human hair loss. [1]
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(a) In group 2 the number of stem cells per follicle falls from 12 to 3 (1) and regrowth falls from 94% to 21%, so factor N maintains the stem cell reserve and hair regrowth (1). (b) Adding factor N to group 3 restores values near normal, 11 stem cells and 88% regrowth (1), so the fall in group 2 was caused by the missing factor and not by other damage to the niche cells (1). (c) Without the receptor the stem cells stay low at 2 and regrowth at 18% even though factor N is present (1), so the factor acts directly on the stem cells through that receptor (1). (d) Any one: mice may use different niche signals from humans; the groups are small; follicle counts alone do not show hair quality (1).

Final answer

(a) Without N, stem cells fall from 12 to 3 and regrowth from 94% to 21%, so N maintains the stem cell reserve and regrowth. (b) Adding N back restores near-normal values, so the loss was caused by missing N and not by other damage. (c) Stem cells need the receptor to respond, so N acts directly on them. (d) Mouse niche signals may differ from human ones; small sample.

Question T3

[7 marks]
A spherical alga takes up nutrients at 4.0 units per hour per \mum^2 of membrane and uses them at 1.0 unit per hour per \mum^3 of cytoplasm. For a sphere, area =4\pi r^2 and volume =\frac43\pi r^3. (a) Show that the supply to demand ratio is 12/r, with r in \mum. [2] (b) Calculate the ratio at r=6\ \mum and the radius at which supply equals demand. [2] (c) Deduce what happens to a sphere of radius 18 \mum. [1] (d) Explain how a flat disc of the same volume can exceed that radius limit. [2]
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(a) Supply =4.0\times4\pi r^2=16\pi r^2 and demand =1.0\times\frac43\pi r^3 (1); dividing gives \frac{16\pi r^2}{\frac43\pi r^3}=\frac{12}{r} (1). (b) 12/6=2.0 (1); supply equals demand when 12/r=1, so r=12\ \mum (1). (c) At 18 \mum the ratio is 12/18=0.67, so supply meets only two thirds of demand and the cell starves unless it divides (1). (d) For the same volume a flat disc has a larger membrane area, so supply per unit of demand is higher (1) and the path from membrane to centre is shorter, so it can be larger before supply falls short (1).

Final answer

(a) 16\pi r^2\div\frac43\pi r^3=12/r. (b) 2.0; 12 \mum. (c) Supply is 0.67 of demand, so it starves or divides. (d) The disc has more area for the same volume, so supply per unit demand rises, and the diffusion path is shorter.

Question U3

HL[6 marks]
A tissue section contains two kinds of striated muscle cell, X and Y. X: length 120 \mum, width 20 \mum, 1 central nucleus, branched, striated. Y: length 3.0 cm, width 100 \mum, more than 100 nuclei, unbranched, striated. (a) Identify X and Y, giving one piece of evidence from the table for each. [2] (b) Explain how the structure of X suits the function of heart muscle. [2] (c) Suggest why Y has so many nuclei. [2]
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(a) X is cardiac muscle: branched, with one central nucleus (1); Y is skeletal muscle: long, unbranched and with many nuclei (1). (b) The cells are joined end to end and branch (1), so the stimulus is propagated through the wall and many cells contract together (1). (c) Y is one very large cell, formed by the fusion of many precursor cells, so it keeps many nuclei (1); many nuclei supply the mRNA for the very large number of proteins and myofibrils along its length (1).

Final answer

(a) X cardiac (branched, one nucleus); Y skeletal (long, unbranched, many nuclei). (b) End-to-end joins and branches pass the stimulus along the heart wall, so many cells contract together. (c) It is a very large cell, formed by fusion of precursor cells, so many nuclei supply the mRNA for its many proteins.

Question V3

[6 marks]
A signal protein is released at the edge of a developing limb bud and its concentration falls with distance. Cells exposed to 80 units or more become type R, cells at 40 units up to but not including 80 become type S, and cells below 40 units become type T. Distance (mm) 0.2, 0.4, 0.6, 0.8, 1.0; concentration (units) 82, 67, 55, 40, 37. (a) Deduce the cell type formed at 0.2 mm, 0.8 mm and 1.0 mm. [3] (b) Explain how cells with the same genome become three different types. [2] (c) A cell from 1.0 mm is moved to 0.2 mm before it differentiates. Predict its type and state what this shows. [1]
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(a) 82 units is at least 80, so type R (1); 40 units meets the lower limit of type S (1); 37 units is below 40, so type T (1). (b) Each concentration range switches on a different set of genes in the cells (1), so different proteins are made even though every cell has the same genome (1). (c) It becomes type R, because the local concentration, not the origin of the cell, decides its fate (1).

Final answer

(a) 0.2 mm: R; 0.8 mm: S (exactly 40); 1.0 mm: T. (b) Different concentrations switch on different genes, so different proteins are made. (c) Type R: position, through concentration, decides fate.

Connected questions

Question X1

Connected questions[6 marks]
B2.3 + D2.2 + D1.2. A transcription factor is present in pancreatic beta cells but not in red blood cell precursors. It binds to the promoter of the insulin gene. In the precursors the DNA of the insulin gene carries many methyl groups. (a) Outline how transcription of the insulin gene leads to the production of insulin. [2] (b) Explain how the transcription factor restricts insulin production to beta cells. [2] (c) Explain how methylation can keep the insulin gene silent in all the descendants of a precursor. [2]
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(a) RNA polymerase transcribes the gene into mRNA (1); the mRNA is translated at ribosomes to make the insulin polypeptide (1). (b) The factor binds the promoter and allows RNA polymerase to start transcription (1); it is present only in beta cells, so the gene is expressed only there (1). (c) Methylation blocks transcription of the gene (1), and the methyl pattern is copied during DNA replication, so each daughter cell inherits the silent gene and the identity is kept through mitosis (1).

Final answer

(a) RNA polymerase makes mRNA from the gene; ribosomes translate it into insulin. (b) Only beta cells have the factor, so only there can RNA polymerase be recruited to the promoter. (c) The methyl marks are copied when DNA is replicated, so daughter cells keep the silent pattern.

Question X2

Connected questions · HL[6 marks]
B2.3 + B3.1 + B2.1. The gas exchange barrier in an alveolus consists of a type I pneumocyte 0.15 \mum thick, a fused basement membrane 0.10 \mum thick and a capillary endothelial cell 0.15 \mum thick. (a) Calculate the total thickness of the barrier. [1] (b) Explain how oxygen crosses the membrane of a type I pneumocyte. [2] (c) Explain how features of the alveolus maintain a rapid rate of diffusion of oxygen. [3]
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(a) 0.15+0.10+0.15=0.40\ \mum (1). (b) Oxygen is a small non-polar molecule and passes through the phospholipid bilayer by simple diffusion (1), moving from high concentration in the alveolus to low concentration in the blood (1). (c) Any three: the barrier is only 0.40 \mum thick, so the path is short (1); the flat type I cells and many alveoli give a very large area (1); blood flow removes oxygen, keeping a steep gradient (1); breathing renews the air, keeping alveolar oxygen high (1).

Final answer

(a) 0.40 \mum. (b) Oxygen is small and non-polar, so it diffuses through the phospholipid bilayer down its concentration gradient. (c) Thin barrier; large area; ventilation and blood flow keep the gradient steep.

Question X3

Connected questions[6 marks]
B2.3 + D2.1 + D3.1. Stem cells at the base of a gut crypt divide by mitosis about once a day. A mutation can arise in a crypt stem cell or in a differentiated lining cell nearby. Sperm cells are haploid. (a) Explain why the daughter cells of a stem cell division are genetically identical. [2] (b) Explain why a mutation in a crypt stem cell is more serious than one in a lining cell. [2] (c) Explain why sperm cannot be produced by mitosis alone. [2]
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(a) The DNA is replicated before division, giving two identical chromatids per chromosome (1), and mitosis shares one copy of each chromosome to each daughter (1). (b) A stem cell persists and keeps dividing, so every descendant inherits the mutation (1); a lining cell is short-lived and is shed, so its mutation does not last (1). (c) Mitosis keeps the diploid number of chromosomes (1); sperm must be haploid, which needs meiosis, so that fertilisation restores the diploid number and not double it (1).

Final answer

(a) DNA is replicated, then sister chromatids separate equally. (b) The stem cell persists and divides, so all its descendants carry the mutation; a lining cell is short-lived. (c) Mitosis keeps the diploid number; meiosis halves it so fertilisation restores the diploid number.

Question X4

Connected questions · HL[6 marks]
B2.3 + B3.3 + C1.2. Cardiac muscle cells contract about once a second throughout life and contain many mitochondria. In a skeletal muscle fibre, calcium ions are stored in specialised endoplasmic reticulum and released when the fibre is stimulated. (a) Explain why cardiac muscle cells contain many mitochondria. [2] (b) Outline the role of the released calcium ions in contraction. [2] (c) Explain how a sarcomere shortens. [2]
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(a) Contraction needs ATP (1), which a cell that contracts continuously makes by aerobic respiration in many mitochondria (1). (b) Calcium ions bind to troponin (1), which moves tropomyosin so that the binding sites on actin are exposed and myosin heads can attach (1). (c) Myosin heads attach to actin and pivot, pulling the actin filaments towards the centre of the sarcomere, using ATP (1); the filaments slide past each other so the sarcomere shortens although neither filament shortens (1).

Final answer

(a) Continuous contraction needs a steady supply of ATP, made by aerobic respiration in mitochondria. (b) Calcium binds troponin, which moves tropomyosin and exposes the binding sites on actin. (c) Myosin heads pull actin filaments towards the centre, using ATP, so the filaments slide and the sarcomere shortens.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Cell specialization from the core ideas to full exam questions, with question types, worked examples and graded practice.

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