Question

What is the specific growth rate of a bacterial culture undergoing exponential growth, if the generation time is 27 min? Given the cell count at time zero is Zx10^7/L and the specific growth rate is 0.25/h, determine the cell count after The initial bacterial cell numbers in a batch reactor were measured to be 12,500 cells/L. After 4 hours, the cell numbers were measured to be 34,000/L. For the given reactor, determine: The specific growth rate b. Doubling time. Determine the specific growth rate of a bacterial culture using Monod growth kinetics. Given: the maximum growth rate (Umax) = 3.5/hr, substrate affinity constant (K) = 20 mg/L, substrate concentration of 200 mg/L, and maintenance constant (km) = 0.04/min. Determine the maximum growth rate of Saccharomyces organisms given the glucose concentration of 150 mg/L and half-saturation constant (K) for glucose as 25 mg/L. The cell count at time zero was 3x10^2/L and after 4.5 hours, the cell count was measured to be Zx10^9/L. The maintenance constant (km) is 0.05/min and the decay constant is 0.01/min.

          What is the specific growth rate of a bacterial culture undergoing exponential growth, if the generation time is 27 min?
Given the cell count at time zero is Zx10^7/L and the specific growth rate is 0.25/h, determine the cell count after
The initial bacterial cell numbers in a batch reactor were measured to be 12,500 cells/L. After 4 hours, the cell numbers were measured to be 34,000/L. For the given reactor, determine: The specific growth rate b. Doubling time.
Determine the specific growth rate of a bacterial culture using Monod growth kinetics. Given: the maximum growth rate (Umax) = 3.5/hr, substrate affinity constant (K) = 20 mg/L, substrate concentration of 200 mg/L, and maintenance constant (km) = 0.04/min.
Determine the maximum growth rate of Saccharomyces organisms given the glucose concentration of 150 mg/L and half-saturation constant (K) for glucose as 25 mg/L. The cell count at time zero was 3x10^2/L and after 4.5 hours, the cell count was measured to be Zx10^9/L. The maintenance constant (km) is 0.05/min and the decay constant is 0.01/min.

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Added by Robert D.

Biology for AP Courses

Julianne Zedalis, John Eggebrecht

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Solved by Expert Maitreya E

05/25/2022

Step 1

Given: Generation time (T) = 27 min Specific growth rate (μ) = ln(2) / T μ = ln(2) / 27 μ ≈ 0.0257 min^-1 Show more…

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What is the specific growth rate of a bacterial culture undergoing exponential growth, if the generation time is 27 min? Given the cell count at time zero is Zx10^7/L and the specific growth rate is 0.25/h, determine the cell count after The initial bacterial cell numbers in a batch reactor were measured to be 12,500 cells/L. After 4 hours, the cell numbers were measured to be 34,000/L. For the given reactor, determine: The specific growth rate b. Doubling time. Determine the specific growth rate of a bacterial culture using Monod growth kinetics. Given: the maximum growth rate (Umax) = 3.5/hr, substrate affinity constant (K) = 20 mg/L, substrate concentration of 200 mg/L, and maintenance constant (km) = 0.04/min. Determine the maximum growth rate of Saccharomyces organisms given the glucose concentration of 150 mg/L and half-saturation constant (K) for glucose as 25 mg/L. The cell count at time zero was 3x10^2/L and after 4.5 hours, the cell count was measured to be Zx10^9/L. The maintenance constant (km) is 0.05/min and the decay constant is 0.01/min.

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Transcript

00:01 Yeah, hi guys.

00:01 So, over here we have to find out the saturation constant.

00:05 Yes.

00:05 So if we will talk about the first part of the question, if we follow the formula that is equal, k is equal to ln2 divided by the cd.

00:19 Yes.

00:20 So here, this formula will be used and k is called as a specific growth rate and the td will be called as doubling or the generation rate.

00:31 Time, yes.

00:32 Here if we talk about the k, it will be equal to what? it will be equal to ln2, yes, divided by 27 minutes and it will be equal to the ln2 upon 0 .45, let is just right over here, the 4 5s and multiplied by h.

00:58 So it will be equal to or the k will come equal to 1 .5403 are raised to the power minus 1.

01:10 So this will be the answer of the fourth part.

01:13 Now if we talk about the second part portion, let us just solve that also.

01:19 Over here the n will be equal to the n knot multiplied by e.

01:26 Yes, raised to the power of k, please.

01:29 This will be equal to, or the n will become equal to two multiplied by cheng, which is of four.

01:37 Yes, and over here, it will be divided by l.

01:42 Yes, so 4 divided by l, let us just right over here.

01:48 Okay, and this will come into bracket, and then we have to multiply this by e.

01:53 It will be rate of power 0 .25 yes first r and in bracket it will be multiplied by 4 r yes so k and 2 okay now it will be equal to 2 multiplied by 10 risk of 4 divided by else and over here will put it in a bracket and multiply by e rate to power 1 and this will be equal to 2 multiplied by 5 10 x x x xx4 divided by l multiplied by 2 .71 phase...

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