Biologi

Actual concentration

27. november 2023 af alma00 - Niveau: Universitet/Videregående

Hej

Jeg sidder lidt fast med denne opgave:

A bacterium is grown in a chemostat with acetate as its sole carbon source and its rate limiting substrate. Its growth on acetate obeys Monod kinetics, with values for the maximum specific growth rate, umax = 1 hr-1 and half-saturation coefficient Ks = 5 mg/L acetate. If the chemostat is operated at a dilution rate of 0.2 hr-1 and the endogenous biomass consumption rate is 0.1 hr-1, what is the actual acetate concentration in the reactor?

- 1.25 mg/L

- Less than 1.25 mg/L

- More than 1.25 mg/L

Håber i kan hjælpe

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Svar #1
28. november 2023 af AngelOneOne

In a chemostat, the biomass growth rate (μ) is related to the dilution rate (D) and Monod kinetics as follows:

$\mu =\frac{\mu_{max}\cdot S}{K_{s}+S}-D$

Where:

μ = is the biomass growth rate,
μ_max =? is the maximum specific growth rate,
S = is the substrate concentration (acetate in this case),
K_s =? is the half-saturation constant,
D = is the dilution rate.

In this case, μ_max=1 hr⁻¹, K_s=5 mg/L, and D=0.2 hr⁻¹.

Given that the endogenous biomass consumption rate is 0.1 hr⁻¹, the actual growth rate (μ) is the sum of the specific growth rate $\left (\frac{\mu_{max}\cdot S}{K_{s}+S} \right )$ and the endogenous biomass consumption rate:

$\mu = \frac{\mu_{max}\cdot S}{K_{s}+S}+\textup{endogenous rate}-D$

Substituting the known values:

$\mu = \frac{1\cdot S}{5+S}+0,1-0,2$

Now, solve for S, the acetate concentration:

$0=\frac{S}{5+S}-0,1$

Multiply both sides by 5+S:

$0=S-0,1\cdot (5+S)$

Distribute the terms:

$0=S-0,5-0,1S$

Combine like terms and divide with 0,1S on both sides:

$0,1S=0,5$

Divide by 0.1:

$S=5\; \textup{mg/L}$

Therefore, the actual acetate concentration in the reactor is 5 mg/L.

- - -

/Angel

"The Universe is under no obligation to make sense to you" - Niel deGrasse Tyson
“Look deep into nature, and then you will understand everything better” - Albert Einstein

Svar #2
28. november 2023 af alma00

Thank you :)))

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