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Enyme Kinetics: Carbonic Anhydrase

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A carbonic anhydrase can generate H+ at a rate constant (Kcat) of 106 s-1.

    1) What will be the pH in the solution after 0.1 nM of this carbonic anhydrase was incubated with unlimited substrate (CO2) for 1 min (assume the initial pH = 7.0)?

    2) What will be the pH in the above reaction solution (0.1 nM of the carbonic anhydrase incubated with unlimited substrate (CO2) for 1 min) if 100 mM Tris (pKa = 8.0) was included as buffer (assume the initial pH = 7.0)?


    2. Relevant equations
    The Henderson-Hasselbach equation, and maybe something to do with Kcat? I have an idea of how to do part 2, but I think it requires the answer to part 1, which I have no idea where to begin with.


    3. The attempt at a solution
    My instructor is very bad about giving us problems after barely explaining or not explaining the method to solve them. Any help is greatly appreciated.
     
  2. jcsd
  3. Oct 3, 2008 #2
    Solved.
     
  4. Oct 3, 2008 #3

    Borek

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    Staff: Mentor

    Can you share approach used? I have started answering your post twice and aborted it twice. It is obvious to me how to solve it using very general approach to kinetics, but enzyme kinetics has its own laws and methods so I preferred not to confuse you with my approach.
     
  5. Oct 3, 2008 #4
    Sure:

    1) H+ is formed at a rate of Kcat is 106 /mol /sec.

    In one min, 60(106) = 60000000 H+ /mol /min.

    60000000(1E-9M Carbonic Anhydrase) = .006M H+ formed.

    pH = -log[H+] = -log[.006] = 2.22




    2) [Bt] = + [BH+], where Bt stands for total buffer, [Bt] = 100mM

    pH = 7, pKa = 8

    pH = pKa + log(/[BH+])

    7 = 8 + log(/[BH+])

    -1 = log(/[BH+])

    1/10 = /[BH+], (1/10)[BH+] =

    (make substitution)
    at pH = 7, [Bt] = 1/10[BH+] + [BH+]

    [BH+] = 90.9mM, = 9.09mM

    pH = pKa + log(/[BH+])

    ([H+] changes buffer ratio)
    pH = 8 + log[(9.09E-3 - .006)/(90.9 +.006)]

    pH = 8 + log(.0031/.969)

    pH = 8 + (-1.5)

    pH = 6.5
     
  6. Oct 4, 2008 #5

    Borek

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    Staff: Mentor

    Yes. That follows from

    [tex]\frac {d[H^+]} {dt} = k [Anhydrase] [CO_2] [/tex]

    just - which is unusual for the classic kinetics calculations - you can assume that both [Anhydrase] and [CO2] are constant, and if there is enough [CO2] reaction kinetics doesn't depend on it. That leads to the conclusion that amount of H+ produced is just proportional to the product of anhydrase concentration and time.
     
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