Cell Biology -- Diagnostic Tests

Cell Biology — Diagnostic Tests

Unit Tests

UT-1: Cell Structure and Comparison

Question: (a) Draw a labelled diagram of an animal cell and a plant cell, identifying at least 5 structures in each. (b) Create a comparison table of plant and animal cells, identifying structures unique to each and shared structures. (c) Explain the function of: mitochondria, ribosomes, cell membrane, and nucleus. (d) Describe two differences between prokaryotic and eukaryotic cells.

Solution:

(a) Animal cell: nucleus, cell membrane, cytoplasm, mitochondria, ribosomes, (rough and smooth) endoplasmic reticulum.

Plant cell (additional structures): cell wall, chloroplasts, permanent vacuole, all the above structures.

(b)	Feature	Animal Cell
Cell wall	Absent	Present (made of cellulose)
Chloroplasts	Absent	Present (for photosynthesis)
Permanent vacuole	Small, temporary (if any)	Large, central, filled with cell sap
Shape	Irregular, flexible	Fixed, rectangular
Nucleus	Present	Present
Mitochondria	Present	Present
Cell membrane	Present	Present (inside cell wall)

(c) Mitochondria: Site of aerobic respiration, where glucose and oxygen are converted to ATP (energy). They have a double membrane with folds (cristae) to increase surface area for the chemical reactions.

Ribosomes: Site of protein synthesis. They read mRNA and assemble amino acids into polypeptide chains (proteins). Found free in the cytoplasm or on the rough endoplasmic reticulum.

Cell membrane: Controls what enters and exits the cell (partially permeable). It is made of a phospholipid bilayer with embedded proteins, allowing selective transport via diffusion, osmosis, and active transport.

Nucleus: Contains genetic material (DNA) as chromosomes. It controls cell activities by regulating gene expression (which proteins are made). The nuclear envelope has pores for mRNA to exit.

(d) Two differences: (1) Prokaryotes have no membrane-bound organelles (no nucleus, mitochondria, or ER); their DNA is a single circular chromosome in the cytoplasm. Eukaryotes have membrane-bound organelles and linear chromosomes in a nucleus. (2) Prokaryotes are generally much smaller (1—5 $\mu$ M) than eukaryotes (10—100 $\mu$ M). Prokaryotes may also have plasmids (small extra DNA rings) and a capsule, which eukaryotes lack.

UT-2: Specialised Cells

Question: Describe how each of the following cells is specialised for its function: (a) red blood cell, (b) sperm cell, (c) root hair cell, (d) nerve cell (neuron).

Solution:

(a) Red blood cell: Biconcave disc shape maximises surface area for oxygen exchange. No nucleus maximises space for haemoglobin (which binds oxygen). Flexible membrane allows them to squeeze through narrow capillaries.

(b) Sperm cell: Has a tail (flagellum) for swimming to the egg. Contains many mitochondria to provide ATP for tail movement. The head contains the haploid nucleus with genetic material. The acrosome contains enzymes to penetrate the egg.

(c) Root hair cell: Has an elongated projection (root hair) that greatly increases the surface area for absorption of water and mineral ions from the soil. Has a large number of mitochondria for active transport of minerals against the concentration gradient.

(d) Nerve cell (neuron): Has long axons to transmit electrical impulses over long distances. The axon is insulated by a myelin sheath that speeds up impulse transmission. Has branched endings (dendrites) to connect to many other neurons, forming complex networks for processing information.

UT-3: Microscopy and Magnification

Question: (a) Calculate the total magnification of a microscope with a 10x eyepiece lens and a 40x objective lens. (b) A cell is observed to be 4.2 mm across the eyepiece scale. If the total magnification is 400x, calculate the actual size of the cell in $\mu$ M. (c) Explain the difference between resolution and magnification. (d) Describe two advantages of electron microscopy over light microscopy.

Solution:

(a) Total magnification $= 10 \times 40 = 400\times$ .

(b) Actual size $= \text{image size / \text{magnification = 4.2 \text{ mm / 400 = 0.0105 \text{ mm = 10.5 \mu$ M.

(c) Magnification is how much larger the image appears compared to the actual object (e.g., $400\times$ means 400 times larger). Resolution is the ability to distinguish between two closely spaced objects as separate — the minimum distance at which two points can be seen as distinct. Higher resolution means finer detail is visible.

(d) Two advantages: (1) Much higher resolution: Electron microscopes use electrons (much shorter wavelength than light) and can resolve structures down to about 0.2 nm, compared to about 200 nm for light microscopes. This reveals organelles like ribosomes, the structure of membranes, and internal details of mitochondria. (2) Much higher effective magnification: Electron microscopes can achieve magnifications of up to $2,000,000\times$ Compared to about $2000\times$ for light microscopes. Disadvantages: specimens must be in a vacuum (dead), preparation is complex, and images are black and white.

Integration Tests

IT-1: Osmosis and Cell Membranes (with Organisation)

Question: An experiment is set up with three potato cylinders placed in different solutions: distilled water, 0.5 mol/dm $^3$ sucrose, and 1.0 mol/dm $^3$ sucrose. The initial mass of each cylinder is 5.0 g. After 30 minutes, the masses are 5.8 g, 4.9 g, and 4.2 g respectively. (a) Explain the results using the concept of water potential. (b) Calculate the percentage change in mass for each cylinder. (c) Estimate the concentration of the potato cell sap. (d) Explain why this experiment is an example of osmosis, not diffusion.

Solution:

(a) Distilled water (5.8 g): Water moves into the potato cells by osmosis because the cell sap has a lower (more negative) water potential than the distilled water (which has the highest water potential of zero). The cells gain mass.

0.5 mol/dm $^3$ (4.9 g): The sucrose solution has a lower water potential than the potato cell sap, but the difference is small. Some water leaves the cells, causing a slight mass decrease.

1.0 mol/dm $^3$ (4.2 g): The concentrated sucrose solution has a much lower water potential than the cell sap. Water leaves the cells by osmosis, causing significant mass loss and the cells become flaccid.

(b) Distilled water: $(5.8 - 5.0)/5.0 \times 100\% = +16\%$ . 0.5 M: $(4.9 - 5.0)/5.0 \times 100\% = -2\%$ . 1.0 M: $(4.2 - 5.0)/5.0 \times 100\% = -16\%$ .

(c) The potato cell sap concentration is between 0 and 0.5 mol/dm $^3$ . At the cell sap concentration, there would be no net movement of water (no mass change). By interpolation, since the mass change crosses zero between 0 M (+16%) and 0.5 M (-2%), the concentration is approximately $0.5 \times 16/(16+2) = 0.44$ mol/dm $^3$ .

(d) Osmosis is the net movement of water molecules across a partially permeable membrane from a region of higher water potential to a region of lower water potential. Diffusion is the movement of any substance from high to low concentration. In this experiment, it is specifically water molecules moving through the cell membrane (partially permeable), not solute particles moving. The solute (sucrose) cannot cross the membrane, so only water moves.

IT-2: Cell Division and Growth (with Inheritance)

Question: (a) Describe the main stages of the cell cycle. (b) A cell in mitosis has 46 chromosomes. How many chromosomes will each daughter cell have? (c) Calculate the number of daughter cells produced from: (i) 5 cell divisions by mitosis, (ii) starting with a single cell that divides 10 times. (d) Explain why mitosis is important for growth and repair.

Solution:

(a) The cell cycle has two main phases: Interphase (cell growth and DNA replication — the cell copies its DNA so each chromosome consists of two identical sister chromatids) and Mitotic phase (cell division). The mitotic phase consists of: Prophase (chromosomes condense, nuclear envelope breaks down), Metaphase (chromosomes line up at the cell equator), Anaphase (sister chromatids are pulled to opposite poles by spindle fibres), Telophase (new nuclear envelopes form, chromosomes decondense), and Cytokinesis (the cytoplasm divides, forming two daughter cells).

(b) Each daughter cell has 46 chromosomes. Mitosis produces genetically identical daughter cells with the same chromosome number as the parent cell.

(c) (i) 1 cell $\to$ 2 (1 division) $\to$ 4 (2) $\to$ 8 (3) $\to$ 16 (4) $\to$ 32 (5). After 5 divisions: $2^5 = 32$ daughter cells. (ii) After 10 divisions: $2^{10} = 1024$ cells.

(d) Mitosis is important for: (1) Growth: A multicellular organism grows by producing more cells through mitosis, increasing the number of cells while maintaining the same chromosome number and genetic information. (2) Repair: Damaged or worn-out cells are replaced by new cells produced by mitosis (e.g., skin cells, blood cells, stomach lining cells). (3) Asexual reproduction: Some organisms reproduce by mitosis, producing offspring genetically identical to the parent.

IT-3: Microscopy and Cell Size (with Bioenergetics)

Question: (a) The diameter of a typical animal cell is 20 $\mu$ M. Calculate how many cells would fit along a line 1 mm long. (b) An electron micrograph shows a mitochondrion with a length of 5 mm on the image. The actual length is 1.5 $\mu$ M. Calculate the magnification. (c) A chloroplast is observed to be 3.8 mm long at a magnification of 15,000x. Calculate the actual length in $\mu$ M. (d) Explain why most cells are small ( 10—100 $\mu$ M) rather than large, relating your answer to the surface area to volume ratio.

Solution:

(a) $1 \text{ mm = 1000 \mu\text{m$ . Number of cells $= 1000/20 = 50$ cells.

(b) Magnification $= \text{image size / \text{actual size = 5 \text{ mm / 0.0015 \text{ mm = 3333\times$ .

(d) As a cell grows larger, its volume increases faster than its surface area ( $V \propto r^3$ vs $SA \propto r^2$ ). The surface area to volume ratio decreases, meaning: (1) Not enough cell membrane surface area for efficient exchange of oxygen, nutrients, and waste products by diffusion and osmosis. (2) The diffusion distance from the cell membrane to the centre of the cell increases, slowing down the rate of exchange. (3) The cell cannot sustain its metabolic demands. Small cells have a large SA:V ratio, ensuring efficient exchange. This is why large organisms are multicellular (many small cells rather than one giant cell) and why cells divide when they reach a certain size.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

Confusing terminology or concepts that appear similar but have distinct meanings.
Overlooking key assumptions or boundary conditions that limit applicability.

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