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Engineering Life: Miracles in Biotechnology - Growth of Microorganisms in a Fermenter-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Fermenter (or Bioreactor) is a large vessel used to grow microorganisms, plant, or animal cells in a controlled environment to produce substances like antibiotics, enzymes, or vitamins.

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The Growth Curve of microorganisms in a batch culture consists of four distinct phases: 1. Lag Phase (adaptation), 2. Log/Exponential Phase (rapid division), 3. Stationary Phase (growth rate equals death rate), and 4. Decline/Death Phase (depletion of nutrients).

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Aeration and Agitation: Microorganisms require oxygen for aerobic respiration. A Sparger bubbles air into the medium, while an Impeller (stirrer) ensures uniform distribution of nutrients and oxygen.

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Specific Growth Rate (mu\\mu): This represents the increase in cell mass per unit time per unit cell mass, typically expressed in units of h−1h^{-1}.

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Sterilization: To prevent contamination, fermenters are sterilized using high-pressure steam. The standard condition is 121∘C121^\circ\text{C} at 15 psi15 \text{ psi} for 1515-2020 minutes.

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Batch vs. Continuous Culture: In batch culture, nutrients are added once at the start. In Continuous Culture, fresh medium is added and waste is removed at a constant rate to keep cells in the log phase.

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pH and Temperature Control: Sensors monitor these parameters. Cooling jackets or coils remove excess heat generated by microbial metabolism to maintain an optimal temperature (e.g., 37∘C37^\circ\text{C}).

📐Formulae

Nt=N0×2nN_t = N_0 \times 2^n

n=log⁡10Nt−log⁡10N0log⁡102n = \frac{\log_{10} N_t - \log_{10} N_0}{\log_{10} 2}

g=tng = \frac{t}{n}

μ=ln⁡2g≈0.693g\mu = \frac{\ln 2}{g} \approx \frac{0.693}{g}

μ=2.303log⁡10(NtN0)t\mu = \frac{2.303 \log_{10}(\frac{N_t}{N_0})}{t}

💡Examples

Problem 1:

A bacterial culture starts with N0=103N_0 = 10^3 cells. If the generation time (gg) is 3030 minutes, calculate the total number of cells (NtN_t) after 33 hours of exponential growth.

Solution:

First, calculate the number of generations (nn) in 33 hours: t=3 hours=180 minutest = 3 \text{ hours} = 180 \text{ minutes} n=tg=18030=6n = \frac{t}{g} = \frac{180}{30} = 6 Now, use the population growth formula: Nt=N0×2nN_t = N_0 \times 2^n Nt=103×26N_t = 10^3 \times 2^6 Nt=1000×64N_t = 1000 \times 64 Nt=64,000 cellsN_t = 64,000 \text{ cells}

Explanation:

Since the bacteria double every 3030 minutes, they undergo 66 doubling cycles in 180180 minutes. The final population is the initial population multiplied by 22 raised to the power of the number of generations.

Problem 2:

If the specific growth rate (μ\mu) of a yeast culture in a fermenter is 0.231 h−10.231 \text{ h}^{-1}, calculate the generation time (gg) in hours.

Solution:

Using the relationship between specific growth rate and generation time: g=ln⁡2μg = \frac{\ln 2}{\mu} g=0.6930.231g = \frac{0.693}{0.231} g=3 hoursg = 3 \text{ hours}

Explanation:

The specific growth rate is inversely proportional to the generation time. By dividing the natural log of 22 (approx 0.6930.693) by the growth rate, we find that the population doubles every 33 hours.