4) The simplified mechanism below is often used to describe an enzyme, E, cataly

Question

4) The simplified mechanism below is often used to describe an enzyme, E, catalyzing conversion of a substrate, S, to product P k2 The ATP hydrolysis activity of a helicase is often examined in this way. Where E would be the helicase, S would be the ATP substrate, and P would be ADP and inorganic phosphate. The experiment would be carried out by holding the substrate in large excess over the enzyme so that the total substrate concentration can be considered to be constant. The initial velocity would be determined at multiple substrate concentrations at fixed enzyme concentrations 1) Write down the rate equation for ES and P 2) Assume that the [ES] is in the steady-state and solve the equation algebraically for [ES]. You will need the conservation of mass equation that says the total enzyme, [Elo, is given by the sum of [E] and [ES], i.e. E E+ ES Plug your solution for ES into your rate equation for d[Pl/dt Show that the velocity has the form v=-= 3) 4) dt 1+R2 [S]k

Explanation / Answer

ES is formed from the 1st reaction, and decomposed to E and S and P as well

d(ES)/dt= K1[E][S]-K2[ES]- K3[ES]

at steady state

d[ES]/dt= 0

or K1[E] [S]- K2[ES]= K3[ES]=0

[ES]= K1[E][S]/(K2+K3)

[E]= (K2+K3)[ES]/K1[S]

But Eo= [E]+[ES]

Eo= (K2+K3) [ES]/K1[S]+ [ES]

= [ES] ( K2+K3)/K1S+1]

=[ES] ( K2+K3+K1S]/K1S

[ES]= Eo*K1S/ (K2+K3+K1S)

[ES]= [EO] /[(K2+K3)/K1S+ 1]

dP/dt= K3[ES]= K3[ Eo] /[1+ (K2+K3)/K1S]

V= dP/dt= rate of formati of P= K3[ Eo] /[1+ (K2+K3)/K1S]

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.