Cell Biology · Unit 1: Seeing and testing cells
A Measured Eukaryotic Cell
What is inside a eukaryotic cell, how much room does each part take, and why are cells the size they are? This lesson compares animal, plant and fungal cells using real measurements, each tied to the cell it came from, then works through surface-to-volume arithmetic and its limits. Afterwards you can read a volume table, set up a supply-and-demand size estimate, and say what yeast, cultured cells, flies, worms and plants can and cannot tell you.
Watch first The Surface Feeds the Volume · 2:54
A log-phase Chinese hamster cell in culture has a mean volume of about 710 cubic micrometres, and a little more than a quarter of that is nucleus.
That figure comes from a 1972 study of V79-S171 cells by D. W. Ross and H. C. Mel, who used electron microscopy to divide the cell’s volume among its parts. A textbook drawing of “the animal cell” gives you the parts but not the shares. This lesson asks four questions in turn. What is inside a eukaryotic cell? How much room does each part take, in which cell? Why are cells small? And which organism should you study to answer a given question?
The idea: a cell is a set of measured compartments
A eukaryotic cell keeps its DNA inside a membrane-bounded nucleus. Its interior is divided into compartments, regions enclosed by a membrane. The space inside an organelle is its lumen; the fluid outside the organelles but inside the plasma membrane is the cytosol. Think of a floor plan. A sketch tells you there is a kitchen and a bedroom; a floor plan with square metres tells you how the house is used.
Animal, plant and fungal cells share a plasma membrane, a nucleus and a cytoskeleton, the network of protein filaments that shapes the cell and moves things within it. All three have an endoplasmic reticulum (ER) and Golgi apparatus, which sort and transport proteins headed for secretion, and mitochondria, which carry out oxidative phosphorylation. The bilayer that bounds all of these is covered in membrane structure and function, and the broad picture of the cell in the introduction to biochemistry and cell biology.
The differences are where students slip:
| Part | Animal cell | Plant cell | Fungal cell (yeast) |
|---|---|---|---|
| Wall | none | principally cellulose | β-glucans, chitin and mannoproteins (S. cerevisiae) |
| Plastids | none | chloroplasts in photosynthetic cells; leucoplasts, without pigment, in nonphotosynthetic tissues | none |
| Vacuole | small or absent | often large, can fill most of the cell | acidic, with degradative and storage roles |
Chloroplasts carry out the light reactions, but only in cells that photosynthesize. A root cell deep in soil is a plant cell without them. Yeasts are unicellular fungi, and budding yeast has the same core organelles as your cells, which is why it is so useful.
Numbers with their cells: volumes, counts and fractions
Each number below belongs to one cell type. Carry the label with the number.
- Chinese hamster V79-S171 cells, log phase, in culture: mean volume 710 µm³, nucleus 190 µm³, total mitochondria 37.5 µm³, about 90 mitochondria per cell. Ross and Mel took whole-cell volume from electronic cell sizing and apportioned it among nucleus, cytoplasm and mitochondria by electron microscopy.
- Wild-type fission yeast, growing: nuclear-to-cell volume ratio 0.080 ± 0.013, measured by Frank Neumann and Paul Nurse with a fluorescent nuclear-envelope marker.
- Arabidopsis root epidermis: the vacuole fills about 30 to 40% of young cells and 80 to 90% after elongation. Across that comparison the cells grew about 14-fold in volume, while the space between vacuole and cell boundary grew only about two- to threefold.
- Arabidopsis leaf mesophyll, wild type (Wassilewskija): a mean of 83 chloroplasts per cell. That is one tissue of one plant, not every plant cell.
Now read the hamster data as you would a results table.
Worked problem, part 1: reading a volume table. Using the V79 means:
nuclear fraction = 190 / 710 = 0.27; mitochondrial fraction = 37.5 / 710 = 0.053
So the nucleus is about 27% of the cell and mitochondria about 5%. If you treat the cell as a sphere of the same volume, r³ = 3 × 710 / (4π) = 169.5 µm³, so r = 5.5 µm and the sphere-equivalent diameter is about 11.1 µm. That is a calculation, not a measured diameter; cells spread on a dish are flatter than that.
Compare 0.27 with fission yeast’s 0.08. The nuclear fraction is not one universal constant. Each value is real for its own cell.
Counts need a definition too. The 90 mitochondria in a V79 cell were counted from EM images. In budding yeast, image analysis shows that mitochondria can form one interconnected network or many separate tubules, so the count depends on the rule: counting each connected structure as one object makes a single network one mitochondrion. This lesson gives no census of organelle numbers across animals, plants and fungi. How these volumes are measured, and what limits each method, comes in microscopy and quantitative images.
Why cells are small: surface, volume and supply
Everything a cell takes in crosses its surface, while everything it uses is spread through its volume. Picture a city fed by trucks entering at its border: double the city’s width and the border doubles, but the area to feed goes up fourfold.
Worked problem, part 2: geometry. Take a hypothetical smooth spherical cell that grows from radius 5 µm to 10 µm.
A = 4πr² V = 4πr³/3 A/V = 3/r
At r = 5 µm: A = 314.2 µm², V = 523.6 µm³, A/V = 0.6 µm⁻¹. At r = 10 µm: A = 1256.6 µm², V = 4188.8 µm³, A/V = 0.3 µm⁻¹.
The radius doubles, the area rises fourfold, the volume eightfold, and the ratio halves.
Worked problem, part 3: supply against demand. Now invent two rates; they are made-up inputs, not measurements. Suppose the membrane can take up a nutrient at no more than J = 20 molecules µm⁻² s⁻¹, and the cell uses q = 6 molecules µm⁻³ s⁻¹ evenly through its volume. Assume the whole surface takes up nutrient and nothing else limits the cell.
supply meets demand when J × A ≥ q × V, that is, 3/r ≥ q/J, so r_max = 3J/q
Here r_max = 3 × 20 / 6 = 10 µm. The ratio of capacity to demand is (J/q) × (3/r) = 10/r: at 5 µm it is 2 (supply 6,283 molecules per second against demand of 3,142), and at 10 µm it is exactly 1. Change J, q or the shape and the answer changes.
- rmax = 3J/q
- A/V = 3/r
- Area A = 4πr²
- Volume V = 4πr³/3
- Supply J × A
- Demand q × V
- Capacity to demand = (J/q) × (3/r)
In this sketch, a smooth sphere whose whole surface takes up nutrient and nothing else limits it, supply meets demand up to rmax = 3J/q. That is a scaling penalty, not a universal maximum. The rates are invented, not measurements; shape, vacuoles and biosynthetic scaling are left out.
So the arithmetic gives a scaling penalty, not a universal maximum. It leaves out four things.
Shape. A flat or long thin cell keeps much more surface per volume than a sphere. Diffusion time grows with the square of distance, which also favours cells where no point is far from the surface.
Vacuoles. A plant root epidermal cell adds volume mostly as vacuole, so its cytoplasm grows far less than the cell does.
Biosynthetic scaling. In 2019 Gabriel Neurohr, Angelika Amon and colleagues held budding-yeast cells in G1 of the cell cycle with a temperature-sensitive cdc28-13 allele. The cells kept growing, from about 65 fL to about 600 fL in six hours. When macromolecule synthesis failed to keep pace with volume, the cytoplasm became diluted; the same paper reports this in primary human fibroblasts (IMR90) enlarged by senescence induction. A cell can be too big for its own synthesis machinery, not only for its membrane.
The size guide itself. “Eukaryotic cells are 10 to 100 µm across” is a guide, not a bound. A 65 fL yeast cell has a sphere-equivalent diameter of about 5 µm.
Model systems and the limits of transfer
Choose the organism for the question, and remember what a result in it does and does not show.
Budding yeast (Saccharomyces cerevisiae) is fast and easy to manipulate genetically; how genes are deleted is covered in CRISPR and functional genomics. In 2015 Aashiq Kachroo, Edward Marcotte and colleagues tested 414 essential yeast genes and found that nearly 47% could be replaced by their human orthologs, judged by rescue of growth. That shows a conserved function. It does not show that the other half are unconserved, since failure can have several causes, and growth rescue says nothing about tissue roles.
Fission yeast (Schizosaccharomyces pombe) is a rod that grows at its ends, which makes size easy to measure; its nuclear-size results follow in the design problem.
Cultured mammalian lines give precise numbers, like the V79 volumes, but a line is not an animal, and a line’s name is not a guarantee. When Uri Ben-David and colleagues characterized 27 strains of the breast cancer line MCF7 in 2018, they found rapid genetic diversification, and at least 75% of the compounds that strongly inhibited some strains were completely inactive in others. Report the strain and conditions.
Flies (Drosophila): a 2000 survey by Fortini and colleagues found apparent fly counterparts for 178 of 287 human disease genes, but none for α-synuclein. Conservation is gene by gene.
Worms (C. elegans): Sydney Brenner’s 1974 paper linked about 300 EMS-induced mutants to about 100 genes, establishing gene-to-phenotype links in a whole animal. Human effects need more evidence.
Plants (Arabidopsis): the arc6 mutant, with one to three chloroplasts per mesophyll cell against 83 in wild type, and the root vacuole measurements show control of chloroplast number and vacuole size that animal cells do not offer. Leaf and root data are not interchangeable.
Design problem
Does DNA content or cell size set the size of the nucleus? You work with fission yeast and have size mutants that divide at very different sizes, a way to raise DNA content without division (switching off the mitotic cyclin Cdc13 so the genome is copied repeatedly), a fluorescent nuclear-envelope marker, and images that outline each cell. Design the experiment and say what each model predicts before you look.
One way to do it
First, measure nuclear and cell volume in each cell, across the widest size range you can get from the mutants. If cell size sets nuclear size, the nuclear-to-cell ratio stays constant as cells get bigger. If DNA alone sets it, nuclear volume stays roughly fixed in cells with the same DNA content, so the ratio falls as cells grow.
Second, change DNA content without changing the rule for cell size. Switch off Cdc13 so the genome is copied several times, and compare cells with 2C and with much higher DNA content. The DNA model predicts a relatively larger nucleus in the high-DNA cells; the cell-size model predicts the same ratio.
Controls: measure wild-type cells in the same conditions, use the same marker and thresholds for every strain, and measure enough cells per strain to see the spread.
Neumann and Nurse did this in 2007. The ratio stayed roughly constant, near 0.08, across a 35-fold range of average cell sizes, and a 16-fold difference in DNA content (2C against 32C) left relative nuclear size unchanged. In cells with several nuclei, a nucleus surrounded by more cytoplasm grew faster. The answer in fission yeast is cell size, probably through something in the cytoplasm. It does not give one nuclear fraction for every cell: V79 hamster cells sit near 0.27.
How we measure it
Electron-microscope stereology
Thin sections of fixed cells are photographed in the electron microscope, and the area each compartment covers in many random sections is measured. Area fractions in sections estimate volume fractions in the whole cell, the kind of measurement used to apportion the volume of V79 hamster cells among nucleus and mitochondria.
Fluorescent envelope marker with Nomarski optics
A fluorescent protein fused to a nuclear-envelope protein outlines the nucleus, while Nomarski (differential interference contrast) imaging outlines the unstained cell. Measuring both in the same cell gives the nuclear-to-cell volume ratio, as in fission yeast with Cut11-GFP.
Electronic cell sizing (Coulter counter)
Cells in salt solution pass one at a time through a small aperture, and each one changes the electrical resistance in proportion to its volume. Thousands of cells give a volume distribution in minutes, which is how yeast volumes of 65 to 600 femtolitres were followed.
Complementation by a human ortholog
An essential yeast gene is removed or shut off, and a human ortholog is supplied on a plasmid. If the cells grow, the human gene supplies the function yeast needs for growth; if not, the result alone does not say why.
Finish line
Test yourself
8 questions
Practice and MCAT-style, with explained answers.
Next · Lesson 2
Microscopy and Quantitative Images
Two GFP-filled vesicles 150 nm apart look like one blob in the best ordinary light microscope. Zooming in will not split them. Why not?
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