Biologi
Actual concentration
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
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:
Where:
μ = is the biomass growth rate,
μmax =? is the maximum specific growth rate,
S = is the substrate concentration (acetate in this case),
Ks =? is the half-saturation constant,
D = is the dilution rate.
In this case, μmax=1 hr−1, Ks=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 and the endogenous biomass consumption rate:
Substituting the known values:
Now, solve for S, the acetate concentration:
Multiply both sides by 5+S:
Distribute the terms:
Combine like terms and divide with 0,1S on both sides:
Divide by 0.1:
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
Skriv et svar til: Actual concentration
Du skal være logget ind, for at skrive et svar til dette spørgsmål. Klik her for at logge ind.
Har du ikke en bruger på Studieportalen.dk?
Klik her for at oprette en bruger.