Prokaryotic cells and growth for the MCAT

MCAT content library / Biology & biochemistry / Prokaryotic cells and growth

Combine structural clues with controlled culture data, calculate generations, and distinguish a growing population from a moving one.

About 17 minutes, including the worked examples and checks.

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Separate cellular organization from a growth pattern
Use a combination of structural features
Growth means more cells; the rate depends on conditions
Worked application: compare matched cultures
Count the quantity that answers the question
Movement can redistribute cells without producing them
Check your reasoning
Continue learning
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Separate cellular organization from a growth pattern

A small cell that multiplies rapidly is not identified completely by either observation. Structure tells you about its cellular organization; a time course tells you what it does under particular conditions. The AAMC places prokaryotic classification, fission, environmental physiology, and chemotaxis together because these kinds of evidence can be combined without treating every microorganism as biologically identical.¹

Cell theory connects organisms through two central ideas: cells are fundamental units of living organization, and new cells arise from existing cells. Microscopy of plant and animal tissues helped establish the first idea; evidence of cell division supported the second. Hooke’s observations of cork, Schleiden and Schwann’s work on tissues, and Remak’s evidence of division contributed different pieces. Virchow helped popularize the existing-cell principle. Seeing a culture become cloudy is not, by itself, evidence that cells appeared spontaneously; contamination, division, and the measurement method must be considered.²

Use a combination of structural features

Bacteria and Archaea both have prokaryotic organization: their main genetic material occupies a nucleoid without a surrounding nuclear envelope. They have membranes, ribosomes, and organized molecular machinery, but not the typical nucleus and mitotic apparatus of eukaryotic cells. Many have a circular chromosome, although chromosome number and shape have exceptions. Prokaryotic does not mean an unstructured bag of molecules.³

Peptidoglycan is a strong clue to a bacterial wall. Archaeal walls lack peptidoglycan, and some bacteria lack walls entirely, so absence of a wall does not by itself identify an archaeon. Archaeal membrane lipids also commonly use ether-linked isoprenoid chains, unlike the ester-linked fatty-acid lipids typical of bacteria. Cell-wall support and membrane selectivity are different functions; removing one need not immediately eliminate the other.⁴

Shape names describe morphology: cocci are approximately spherical, bacilli are rods, and spirilla are spiral forms. Shape can narrow a description without establishing a species, oxygen requirement, or capacity to cause disease. Some bacteria move using rotating flagella; many bacterial flagellar motors obtain energy from an ion electrochemical gradient rather than by directly spending ATP on each rotation.³⁸

Growth means more cells; the rate depends on conditions

During binary fission, a cell duplicates and partitions its genetic material and cellular components, then divides. In a specified exponential interval with negligible death and a constant doubling time d, N = N₀ × 2^t/d. The exponent counts generations, not cells added. Equal time intervals produce equal fold changes rather than equal absolute increases.⁵

A closed batch culture can pass through lag, exponential growth, stationary behavior, and decline. Lag can include active adaptation without much increase in cell number. A stationary viable count may reflect balanced division and death or reduced division; the flat count alone cannot distinguish them. Nutrient depletion, accumulated waste, and changing environmental conditions prevent indefinite extrapolation of the earlier doubling time.⁵

Oxygen physiology also varies. Obligate aerobes require oxygen for growth; obligate anaerobes are harmed by oxygen under conditions they cannot tolerate. Facultative anaerobes can grow with or without it, often more effectively with it. Aerotolerant anaerobes tolerate oxygen without using aerobic respiration. Growth without oxygen does not uniquely establish fermentation: some organisms can use anaerobic respiration with another terminal electron acceptor.⁶

Worked application: compare matched cultures

The following original illustrative data track viable cells in equal, constant volumes of the same strain. Temperature, pH, inoculum, and all unmentioned conditions are matched. A and B have abundant nutrient and remain exponential over 0–80 minutes, with negligible death. Culture C lacks a required nutrient. The counts are direct viable-cell measurements, not inferred from cloudiness.

Time (min)	A: oxygen, complete nutrient	B: no oxygen, complete nutrient	C: oxygen, required nutrient absent
0	1.00	1.00	1.00
20	2.00	1.41	1.00
40	4.00	2.00	1.00
60	8.00	2.83	1.00
80	16.00	4.00	1.00

Viable-cell counts under three conditions, in units of 10⁵ cells

Original illustrative cultures. Equal vertical intervals represent equal fold changes. A and B follow the stated exponential models; C has no net population increase.

For A, the final-to-initial ratio is 16/1 = 16 = 2⁴. Four generations occurred in 80 minutes, so d = 80/4 = 20 minutes. For B, the ratio is 4 = 2², giving d = 40 minutes. The displayed intermediate values for B are rounded; its supplied model is an exact doubling every 40 minutes.

This strain grows without oxygen and grows faster with it in the tested medium. That supports a facultative pattern here. The table does not identify its oxygen-free pathway or establish that every nutrient formulation would support the same rates. Culture C remains at its starting viable count. Its missing nutrient prevents the population expansion seen in A, but the flat viable count does not show that oxygen killed the cells.

Notice the distinction between fold change and added cells. A adds 15 × 10⁵ cells, whereas B adds 3 × 10⁵. A therefore has four times B’s final population but five times B’s increase from the common starting count. Specify which comparison the question requests. On the graph’s logarithmic vertical scale, equal distances represent equal ratios; the two exponential trajectories become straight lines with different slopes.

Count the quantity that answers the question

A direct microscopic count can include dead cells unless viability is assessed separately. A colony-forming-unit count measures units that produce colonies under the supplied culture conditions; a cluster can form one colony, and a viable cell may fail to grow on a particular medium. Turbidity is an indirect signal requiring suitable calibration. These measurements cannot be substituted for one another without checking their assumptions.⁵

Movement can redistribute cells without producing them

Chemotaxis is a biased movement response to a chemical gradient. In a run-and-tumble bacterium such as E. coli, changing the probability of reorientation can extend travel when conditions improve. Individual trajectories remain variable; population accumulation does not require every cell to steer directly toward the source. Motility without directional bias is different from chemotaxis, and not all bacteria use the same swimming pattern.³⁸

A useful experiment compares a gradient with a spatially uniform concentration, reverses the gradient, and measures the total cell number over an interval too short for appreciable division. If an enrichment follows the high-concentration side while the total stays fixed, redistribution is the supported explanation. Without those observations, local density alone could reflect movement, differential growth, adhesion, or loss elsewhere.

Interactions with other organisms also depend on their consequences. Mutualism benefits both partners; commensalism benefits one without a detectable effect on the other; parasitism benefits one while harming the other. These are relationships, not permanent labels attached to every member of a domain. A bacterial culture result should therefore be interpreted at the level actually tested: a particular population, environment, measurement, and interval.⁷

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A replicating microorganism grows in a medium containing no other cells. It contains ribosomes and a DNA-rich nucleoid without a nuclear envelope. Chemical analysis identifies peptidoglycan in its cell wall. Which classification is best supported by the combined evidence?

A. Archaeal cell.

Review this choice. Archaea have prokaryotic organization, but their walls do not contain bacterial peptidoglycan. The wall chemistry distinguishes the two prokaryotic domains here.

B. Bacterial cell.

Correct. The nucleoid and independent cellular growth support prokaryotic organization; peptidoglycan specifically supports a bacterial cell.

C. Eukaryotic cell.

Review this choice. The supplied organization lacks a nuclear envelope, and the wall contains peptidoglycan rather than a typical eukaryotic wall material.

D. Viral particle.

Review this choice. A viral particle does not have its own ribosomes and does not replicate as an independent cellular organism in cell-free medium.

The principle to carry forward

Combine organization, wall chemistry, and replication evidence; size or shape alone would be much less diagnostic.

Check 2

In a 90-minute interval established to be exponential, a bacterial culture increases from 4.0 × 10⁴ to 3.2 × 10⁵ viable cells. Volume is constant, death is negligible, and the doubling time is constant. What is the doubling time?

A. 10 minutes

Review this choice. This would require nine doublings and a 512-fold increase. The observed ratio is only eightfold.

B. 15 minutes

Review this choice. This would require six doublings and a 64-fold increase. An eightfold increase corresponds to three doublings.

C. 30 minutes

Correct. The ratio is (3.2 × 10⁵)/(4.0 × 10⁴) = 8 = 2³. Three generations in 90 minutes gives 30 minutes per doubling.

D. 45 minutes

Review this choice. A 45-minute doubling time would give two doublings in 90 minutes, producing a fourfold increase rather than eightfold.

The principle to carry forward

Divide final count by initial count before taking the number of doublings; do not treat the count difference as the growth exponent.

Check 3

A sealed chamber begins with a uniform distribution of 10,000 motile bacteria. Over two minutes, division, death, attachment, and fluid flow are negligible. A nutrient gradient produces 70% of cells in its high-concentration half; reversing the gradient reverses the enriched half. A uniform nutrient concentration leaves 50% in each half. The total remains 10,000 in every condition. Which explanation best fits?

A. Biased movement along the chemical gradient.

Correct. The enrichment changes sides with the gradient while total cell number is conserved. That pattern supports chemotactic redistribution.

B. Faster reproduction in the nutrient-rich half.

Review this choice. Division is negligible during the stated interval, and the total does not increase. Reproduction cannot account for the observed redistribution.

C. Loss of cells from the nutrient-poor half.

Review this choice. Death and loss are excluded, and all 10,000 cells remain. The cells redistribute rather than disappear.

D. Uniform movement with no gradient-dependent bias.

Review this choice. Random motility without a gradient-dependent bias does not explain reproducible enrichment that follows reversal of the chemical gradient.

The principle to carry forward

Chemotaxis changes where cells are; population growth changes how many cells exist. The controls distinguish those outcomes.