ds S dt k+S where S= substrate concentration (moles/L), Vm = maximum uptake rate (moles/L/d), and k, = the half- saturation constant, which is the substrate level at which uptake is half of the maximum [moles/L]. If the initial substrate level at t = 0 is So, this differential equation can be solved for S = S, – v„t +k, In(S, / S) f(S) = S, – V, t+ k In (S, /S)– S Determining the root of f(S) where So = 8 moles/L, Vm= 0.7 moles/L/d, and k, = 2.5 moles/L. Iterate until ɛ< 0.1 or the number of iterations reaches 3, whichever comes first.
ds S dt k+S where S= substrate concentration (moles/L), Vm = maximum uptake rate (moles/L/d), and k, = the half- saturation constant, which is the substrate level at which uptake is half of the maximum [moles/L]. If the initial substrate level at t = 0 is So, this differential equation can be solved for S = S, – v„t +k, In(S, / S) f(S) = S, – V, t+ k In (S, /S)– S Determining the root of f(S) where So = 8 moles/L, Vm= 0.7 moles/L/d, and k, = 2.5 moles/L. Iterate until ɛ< 0.1 or the number of iterations reaches 3, whichever comes first.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![ds
S
V.
dt
k+S
where S = substrate concentration (moles/L), Vm= maximum uptake rate (moles/L/d), and k, = the half-
saturation constant, which is the substrate level at which uptake is half of the maximum [moles/L]. If the
initial substrate level at t = 0 is So, this differential equation can be solved for
S = S, - v„t +k, In(S, / S)
m
f(S)= S, – V t+ k̟ n (S, /S)– S
Determining the root of f(S) where So = 8 moles/L, Vm= 0.7 moles/L/d, and k, = 2.5 moles/L.
Iterate until ɛ, < 0.1 or the number of iterations reaches 3, whichever comes first.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7076d277-031b-44bc-a665-cd0b1f8be22a%2F1792292b-1522-40c2-8962-eac2110e510c%2Fsw7qnhp_processed.png&w=3840&q=75)
Transcribed Image Text:ds
S
V.
dt
k+S
where S = substrate concentration (moles/L), Vm= maximum uptake rate (moles/L/d), and k, = the half-
saturation constant, which is the substrate level at which uptake is half of the maximum [moles/L]. If the
initial substrate level at t = 0 is So, this differential equation can be solved for
S = S, - v„t +k, In(S, / S)
m
f(S)= S, – V t+ k̟ n (S, /S)– S
Determining the root of f(S) where So = 8 moles/L, Vm= 0.7 moles/L/d, and k, = 2.5 moles/L.
Iterate until ɛ, < 0.1 or the number of iterations reaches 3, whichever comes first.
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