What Will Be the pH After Incubating Carbonic Anhydrase with CO2?

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Discussion Overview

The discussion revolves around calculating the pH after incubating carbonic anhydrase with CO2, focusing on two scenarios: one without a buffer and one with a Tris buffer. Participants explore the implications of enzyme kinetics and the Henderson-Hasselbalch equation in determining pH changes over time.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a homework problem involving the calculation of pH after incubating carbonic anhydrase with CO2, noting the initial pH and the concentration of the enzyme.
  • Another participant claims to have solved the problem but does not provide details of their approach.
  • A request for the solution approach is made, highlighting the complexity of enzyme kinetics compared to general kinetics.
  • A detailed calculation is provided for the pH without buffer, resulting in a pH of 2.22 based on the amount of H+ produced.
  • The second part of the problem is addressed using the Henderson-Hasselbalch equation, leading to a calculated pH of approximately 6.5 when a buffer is included.
  • One participant discusses the rate of H+ production in relation to enzyme concentration and CO2, suggesting that under certain conditions, the reaction kinetics can simplify the calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solutions to the homework problem, as different approaches and calculations are presented without agreement on the final pH values.

Contextual Notes

There are assumptions made regarding the constancy of enzyme and substrate concentrations during the reaction, and the calculations depend on specific values for Kcat and buffer properties. The discussion reflects varying levels of understanding of enzyme kinetics and pH calculations.

Gannon
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Homework Statement



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)?


Homework 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.


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.
 
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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.
 
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
 
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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