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Bacterial growth and metabolism

Microbiology (B.Sc) · General microbiology · free study notes for exams, placements and interviews

In short

Bacterial growth means an increase in the number of cells, not in the size of one cell: a bacterium grows by binary fission, so a population doubles at a fixed interval called the generation time. Metabolism is the set of reactions that supply that growth with energy (ATP), reducing power (NADH, NADPH) and carbon skeletons — catabolism breaking substrates down, anabolism building cell material up. The topic is examined in B.Sc General Microbiology, M.Sc entrance tests (CUET-PG, JAM, GAT-B, ICMR, CSIR-NET life sciences) and in every fermentation or quality-control interview, because yield, batch time and sterility all follow from growth kinetics. It is used daily in Indian industry — Serum Institute, Biocon, Dr Reddy's, Hindustan Unilever, Amul and Mother Dairy, Nestlé India, water-testing labs under BIS and FSSAI — wherever cells are counted, grown in a fermenter or killed to a specification.

Explained

How bacteria grow, and how growth is measured

The growth curve in batch culture

A closed system with fixed nutrients gives four phases on a plot of log₁₀(cell number) against time.

  1. Lag phase: no increase in number; cells synthesise enzymes, ribosomes, ATP and cofactors and repair damage. Length depends on inoculum age, size and whether the medium is the same as before — a large, young, log-phase inoculum into identical medium gives almost no lag.
  2. Log (exponential) phase: balanced growth at a constant maximum specific growth rate μ; cells are most uniform, physiologically active and most sensitive to antibiotics and disinfectants — so this is the phase used for inocula, transformation (competent cells), enzyme production and Gram staining.
  3. Stationary phase: growth rate = death rate, number constant (typically 10⁹-10¹⁰ cells/mL for *E. coli*) because a nutrient is exhausted, oxygen is limiting or toxic products (acids, alcohols) accumulate. Cells become smaller, tougher, more resistant; secondary metabolites (most antibiotics), endospores and many exotoxins appear here. The stringent response (ppGpp) and σˢ (RpoS)-controlled genes switch on.
  4. Death (decline) phase: exponential loss of viability; some cells persist as viable but non-culturable (VBNC) forms.
  5. Diauxic growth: two log phases separated by a second lag, as in glucose + lactose for *E. coli* — glucose is used first (catabolite repression, low cAMP), then the *lac* operon is induced.
  6. Cryptic growth / synchronous growth (all cells dividing together, obtained by filtration or temperature shift) are examined as special cases.

Continuous culture and the chemostat

Physical and chemical requirements

Nutritional and energy classification

Catabolism: central pathways and ATP

Anabolism and regulation

Key points

Formulas and examples

N = N₀ × 2ⁿ — N = final cell number, N₀ = initial number, n = number of generations (dimensionless)
n = (log₁₀N − log₁₀N₀) / log₁₀2 = (log₁₀N − log₁₀N₀) / 0.301
g = t / n — g = generation (doubling) time, t = time of exponential growth (min or h)
μ = 0.693 / g — μ = specific growth rate (h⁻¹); from ln2 = 0.693
dX/dt = μX, so ln(X/X₀) = μt — X = biomass concentration (g/L)
Growth rate constant k = n / t — generations per hour (divisions h⁻¹); μ = k × 0.693
CFU/mL = (colonies counted) / (volume plated in mL × dilution factor) — count plates with 30-300 colonies
Dilution rate D = F / V — F = medium flow rate (L/h), V = culture volume (L); units h⁻¹; at steady state μ = D
Monod: μ = μ_max × S / (K_s + S) — S = limiting substrate (g/L), K_s = half-saturation constant (g/L)
Growth yield Y_X/S = ΔX / ΔS — g dry biomass formed per g substrate consumed
Specific substrate uptake q_s = μ / Y_X/S — g substrate per g cells per hour

Example 1 — generations, generation time and specific growth rate. A broth inoculated with 5 × 10³ cells/mL reaches 3.2 × 10⁸ cells/mL in 6 h of exponential growth.

Example 2 — predicting a count (food-safety style). *Staphylococcus aureus* at 10² cells/g in a mithai sample, g = 30 min at 37 °C, left out 5 h.

Example 3 — viable count from a dilution series. 0.1 mL of a 10⁻⁵ dilution spread-plated gives 148 colonies.

Example 4 — chemostat. A 2 L chemostat is fed at 0.6 L/h; the organism's μ_max = 0.5 h⁻¹, K_s = 0.2 g/L.

Example 5 — ATP and yield. Per glucose, aerobic respiration via EMP + TCA + ETC yields (theoretical maxima): 10 NADH × 3 + 2 FADH₂ × 2 + 4 substrate-level ATP = 30 + 4 + 4 = 38 ATP; homolactic fermentation yields 2 ATP.

FeatureAerobic respirationAnaerobic respirationFermentation
Terminal electron acceptorO₂NO₃⁻, SO₄²⁻, CO₂, Fe³⁺, fumarateOrganic intermediate (pyruvate, acetaldehyde)
ETC usedYesYesNo
ATP mechanismOxidative phosphorylation + SLPOxidative phosphorylation + SLPSubstrate-level only
ATP per glucose (textbook)up to 38~5-36 depending on acceptor2 (EMP), 1 (ED)
Example*Pseudomonas* on O₂*Desulfovibrio*, denitrifying *Pseudomonas**Lactobacillus*, *Clostridium*
PathwayNet ATPNAD(P)HDiagnostic organism
EMP (glycolysis)22 NADH*E. coli*, most bacteria
Entner-Doudoroff11 NADH + 1 NADPH*Pseudomonas*, *Zymomonas*
Pentose phosphate (oxidative)0 (net)2 NADPH per G6P*Bacillus*, many aerobes
Phosphoketolase11 NADH*Leuconostoc*

Where it is used

Interview questions

Common mistakes

Related topics

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