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Engineering Life: Miracles in Biotechnology - Fermenters (Bioreactors)-advanced

Grade 9CBSE

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

🔑Concepts

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A bioreactor (or fermenter) is a sophisticated vessel designed to provide an optimal environment for the growth of microorganisms, plant cells, or animal cells to produce specific biological products like enzymes, antibiotics, or vaccines.

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The Stirred-Tank Bioreactor is the most common type, featuring an impeller for mixing and a sparger at the bottom to provide air (oxygen). It ensures a uniform distribution of nutrients and oxygen throughout the culture medium.

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Critical control parameters include temperature (TT), pHpH, dissolved oxygen (DODO), agitation speed, and nutrient concentration. Deviations in these can significantly reduce yield or lead to cell death.

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Growth Kinetics: Microorganisms in a bioreactor typically follow a growth curve consisting of the lag phase, log (exponential) phase, stationary phase, and decline phase. The log phase is where the specific growth rate (μ\mu) is calculated.

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Aseptic Conditions: It is vital to maintain a sterile environment within the fermenter to prevent the growth of contaminating microbes. This is achieved using pressurized steam and high-efficiency particulate air (HEPA) filters.

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Scale-up: The process of increasing the volume of production from a laboratory scale (few liters) to an industrial scale (thousands of liters) while maintaining identical environmental conditions for the cells.

📐Formulae

dXdt=μX\frac{dX}{dt} = \mu X

Xt=X0eμtX_t = X_0 e^{\mu t}

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

td=ln⁡(2)μt_d = \frac{\ln(2)}{\mu}

μ=ln⁡Xt−ln⁡X0t\mu = \frac{\ln X_t - \ln X_0}{t}

💡Examples

Problem 1:

In a batch fermenter, a bacterial culture with an initial concentration of X0=2×104X_0 = 2 \times 10^4 cells/mL grows exponentially with a specific growth rate of μ=0.4 h−1\mu = 0.4 \text{ h}^{-1}. Calculate the cell concentration after t=5t = 5 hours.

Solution:

Given: X0=2×104X_0 = 2 \times 10^4 cells/mL μ=0.4 h−1\mu = 0.4 \text{ h}^{-1} t=5 ht = 5 \text{ h}

Using the exponential growth formula: Xt=X0eμtX_t = X_0 e^{\mu t} Xt=(2×104)×e(0.4×5)X_t = (2 \times 10^4) \times e^{(0.4 \times 5)} Xt=(2×104)×e2X_t = (2 \times 10^4) \times e^{2}

Since e2≈7.389e^2 \approx 7.389: Xt=2×104×7.389X_t = 2 \times 10^4 \times 7.389 Xt=1.4778×105 cells/mLX_t = 1.4778 \times 10^5 \text{ cells/mL}

Explanation:

We apply the integrated form of the rate equation for exponential growth to find the final cell density after a specific duration.

Problem 2:

If a yeast population in a bioreactor doubles every 22 hours, what is its specific growth rate (μ\mu)?

Solution:

Given: Doubling time td=2 hourst_d = 2 \text{ hours}

The formula for specific growth rate is: μ=ln⁡2td\mu = \frac{\ln 2}{t_d} μ=0.6932\mu = \frac{0.693}{2} μ=0.3465 h−1\mu = 0.3465 \text{ h}^{-1}

Explanation:

The specific growth rate is inversely proportional to the doubling time. Using the constant ln⁡2≈0.693\ln 2 \approx 0.693, we can determine how fast the biomass increases per unit time.