5
$\begingroup$

I have been trying to confirm the Km of a substrate (which is 34 +/- 4 mM). This value was obtained in 50 mM MOPS, pH 6.3. I conducted my kinetics assay in a buffer of pH 7 and obtained a Km value in the 21.5. According to this paper, Fig. 2C, the normalized specific activity of the enzyme is about 70% at pH 6.3 and about 47% at pH 7. If I divide 34 mM at pH 6.3 by 0.7 (which should get me the optimum Km at the optimum pH of 5.5) and then multiply by 0.48, then I get 23 mM. However, the paper says Km is between 30 - 38 mM, so if I divide 30 and 38 separately by 0.7 and multiply by 0.47, I get 20 and 25.6 mM respectively. Because my value falls within this range, this then must mean that I have the right enzyme and the same result as the paper.

So my questions are:

  1. When the paper says Km is "34 +/- 4 mM", can I assume that means the Km can be anywhere between 30 - 38 mM? I'm surprised to see how wide the range is. I assumed Km is usually just one value with a deviation of at most 0.1.

  2. Do pH change Km values? I understand that pH changes the shape(s) of the enzyme and/or substrate. Therefore that must affect how much it wants to bind to the substrate. If the enzyme's desire to bind to a substrate decreases due to increase in pH, for example, that would mean more substrates are needed to surround the enzyme, thus increasing Km.

  3. If pH does change Km, is this how I determine the Km value of a different pH value if the Km value at another pH is already known? I know that specific activity and Michaelis constant are different, but how much product can be converted per minute depends on how much the enzyme likes to bind to a substrate, which is represented by the Michaelis constant. Did my reasoning and calculation arrive at the right conclusion? If not, how is the calculation done?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ A very brief answer is yes, pH levels almost always effect enzyme efficiency. An example is lysosomal enzymes which are optimally active in the acidic environment of the lysosome, but are ineffective in the near neutral pH of the cytosol. This protects the cell should a lysosome rupture; the enzymes will not digest the cell itself. From the wikipedia article "The value of K_\mathrm{M} is dependent on both the enzyme and the substrate, as well as conditions such as temperature and pH.". And try not to forget Maud Menten. $\endgroup$
    – AMR
    Commented Aug 21, 2015 at 18:46
  • $\begingroup$ Thanks for your reply. But is there an equation or method to calculate the relationship between Km and pH? If there isn't, was my calculation valid? I want to verify that if pH = 6.3 and Km =34, then Km must equal twenty-something at pH 7. $\endgroup$
    – wswr
    Commented Aug 21, 2015 at 19:44
  • 2
    $\begingroup$ @wswr No that assumption is not valid because the relationship between Km and pH can be non-linear and vary between different enzymes. $\endgroup$
    – WYSIWYG
    Commented Aug 21, 2015 at 19:45
  • $\begingroup$ Actually now that I think about it: I was confusing Michaelis-Menten constant with specific activity. My calculation was invalid. At pH 6.3, the specific activity is high, and the Km value is 34. The Km I got at pH is 21.5, but according to the paper, at pH 7, the specific activity is lower, which mean the Km should be higher. Can it can be assumed that the higher the Km, the lower the specific activity, and vice versa? $\endgroup$
    – wswr
    Commented Aug 25, 2015 at 15:28
  • $\begingroup$ This paper might help you: Current IUBMB recommendations on enzyme nomenclature and kinetics $\endgroup$
    – Zahra
    Commented May 28, 2017 at 13:31

1 Answer 1

Reset to default
5
$\begingroup$

From the derivation of Michaelis-Menten kinetics you can see that:

$$K_m=\frac{k_f + k_{cat}}{k_r}$$

Where $k_f$ and $k_r$ are binding and unbinding rate constants (for Enzyme-Substrate binding), respectively, and $k_{cat}$ is the turnover number. This is for the Quasi-Steady-State approximation (QSSA). For the equilibrium approximation:

$$K_m=\frac{k_f}{k_r}$$ which is same as the association constant.

In both the cases pH can affect the rate of binding and unbinding by affecting the affinity between the enzyme and the substrate. For example lets assume that the substrate binding site is negatively charged. Low pH would increase the electrostatic potential of the substrate binding site towards zero by affecting the ionization of the functional groups.

pH can also have indirect effects on the substrate binding site because it can modify the overall structure of the protein.

Many enzyme catalysed reactions involve acid-base catalysis i.e. there is a transfer of proton. In reactions like these, pH can affect $k_{cat}$.

If pH does change Km, is this how I determine the Km value of a different pH value if the Km value at another pH is already known?

You can do that by making a Lineweaver-Burk plot for the changed pH.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I did make a Lineweaver-burk plot to get the Km of 21.5 at pH 7. But because the only Km value I can compare to verify the enzyme activity is at the different pH value of 6.3, I want to know if there is another way to see if my result is correct. $\endgroup$
    – wswr
    Commented Aug 21, 2015 at 19:38
  • 1
    $\begingroup$ @wswr Make a LB plot for pH 6.3 and then compare. If your measurements are done accurately, your result should be correct. Result is what you observe; so there is no question of correct or incorrect. Unfortunately, there is no simple relationship between Km and pH. You can perform LB analysis at different pH and try to find a relationship by regression. But this would be applicable only for your enzyme. $\endgroup$
    – WYSIWYG
    Commented Aug 21, 2015 at 19:43
  • $\begingroup$ In a pedantic tone, "unbinding" would idiomatically be described as dissociation. $\endgroup$
    – RosieF
    Commented Aug 21, 2015 at 22:53
  • $\begingroup$ I just noticed, isn't Km = (kr + kcat)/kf = dissociation rate constants:association rate constant? $\endgroup$
    – wswr
    Commented Aug 25, 2015 at 17:59
  • $\begingroup$ @wswr kr is dissociation rate constant. kcat is the turnover number or the rate constant of product formation. Both these reduce the concentration of the [ES] complex. $\endgroup$
    – WYSIWYG
    Commented Aug 26, 2015 at 4:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .