Question

8. Enzymes often exhibit pH-dependent activity due to protonation/deprotonation of key amino acids. A general schematic is shown in i) for a pH-dependent enzyme, in which the "active" pathway requires EH and ESH. The constants ka, kz, kcat are the usual rate constants for the Michaelis-Menten reaction scheme, and the dissociation constants KEI, KESI, KE2, Kesz are defined as shown in ii). a. Derive a M-M rate equation for the pH-dependent scheme, below. To do so: 1) express total enzyme concentration (E.) as a function of all six species containing enzyme and solve for (EH) as a function of [E.], [ESH], [H], and dissociation constants; 2) substitute (EH) into our usual M- Mequation [ESH]=[EH][S]/km and solve for (ESH]. b. Use a computer to create a plot of rate vs pH over the range of pH 0 to 7. Use Ke1=KES1=0.079 M, Kez=Kesz=3.2x104 M, and assume that kcat[E][S]/[KM+[S])=1 mm/sec. Recall that [H+=1 M x 10 PM Your curve should resemble the one in iii) for pepsin, a key digestive enzyme in the stomach. C. Healthy humans have a stomach pH of 3.0 after a meal. The treatment of some stomach disorders includes use of drugs like Prilosec that reduce the amount of acid secreted into the stomach. Using the values stated in part b, above, calculate the factor by which the enzymatic activity of pepsin would decrease if you raised the pH from 3.0 to 5.0. What concern would you have for this treatment? EH ESH; [EH]H] EH; (x-1 sec.-1) LLLLLLL EH+V ESH_>EH+P Kisi [ESH] H [ESH;] ko/K PK, 1.1 EES [EH] ESTH] [ESH] KES2 =

Answer 1

You have asked three questions here. As per Chegg Q&A Guidelines I am answering the first question since you haven't specified which question to answer.

We are given that:

KE1= EHH+11 EH,

[ESH][H+] Kesi = [ES]

KE2 = E- H+] [ΕΗ]

KES2 = [ES-H+1 ESH

Now, the reaction involves protonation and deprotonation steps. These reactions are very fast and hence all these acid-base reactions are considered to be in equilibrium.

The total enzyme concentration in the protonated/deprotonated state is given by:

EHT = [EH,] + [EH] + [E-

From the equilibrium constants mentioned above, we have,

(EH)= [EH][Η]. 1 ΚΕΙ + (EH) +- ΚΕ2[ΕΗ) [H+]

H+1 KE2 :: [EH]T = [EH] +1+ H+1

:EHT = Eha

where, $\alpha = (\frac{[H^+]}{K_{E1}} + 1 + \frac{K_{E2}}{[H^+]})$

Similarly, the total enzyme-substrate concentration in the protonated/deprotonated state is given by:

ESHT = [ESH,1 + ESH] + ES-

[ESH][H+] KES2[ESH] :: [ESHT = + ESH+ KESI H+1

[H+1 KES2 :: [ESHT = [ESH KESI+1+ (H+1

: ESHT = ESH

where, $\beta = (\frac{[H^+]}{K_{ES1}} + 1 + \frac{K_{ES2}}{[H^+]})$

And the total enzyme concentration in any form is given by:

ET = EHT + ESHT

Now, assuming steadu-state condition, we have, the rate of change of [ESH] is 0.

DESH

DESH T = k1[EH][S] – (k-1 +km)[ESH] = 0 at

.. ki[EH] S1 = (k_1 + k) [ESH

::EH - (k-1 + k2) ESH ki[S]

Now, let

KM= k_1 tkr ki

KMESH :: EH] =-

Therefore, we have,

KMESHA LEHT = ST

KMα :. [E] = [ESH] - + β)

Now, the initial rate is given by

Vo = k ESH

2 ET rivo =

k2 E ::00 = KM + [S]

Let

Kn= KM

and

Vimax = Vmax

where,

Vmar = k) ET

Thus, substituting back into the above equation gives the modified Michaelis-Menten equation as follows:

Vma: [S] 00 KM + (S)