The reaction: A + B <-> P + Q is catalysed by enzyme E. Draw a Cleland diagram for a Non-sequential (Ping-Pong) mechanism for this reaction.

The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V% for an enzyme-catalyzed, single-substrate reaction E + S=ES → E + P. The model can be more readily understood when comparing three conditions: [S] > Km- Match each statement with the condition that it describes. Note that "rate" refers to initial velocity Vo where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km Not true for any of these conditions [ES] is much lower than [Efree]. Reaction rate is independent of Increasing [Etotal] will lower Almost all active sites will Km- be filled. [S). [Efree] is about equal to [Etotal]. Show All W- 5179933 (3).docx 5179933 (4).docx PCR-MINI RES....docx MacBook Pro

Consider an enzyme that catalyzes the reaction S2 P, by the following simple reaction mechanism: k, E + S 2 E•S →E kcat + P Suppose the enzyme acquires a mutation that causes k1 to be 10-times smaller than for the wild-type (non-mutant) enzyme. Suppose you measure the initial reaction rate (vo) at several different [S] for the mutant and the wild-type enzymes. Under what conditions would the mutation have a greater effect on the reaction rate (vo) of the mutant enzyme compared to the wild-type enzyme - at very low [S], or at very high [S]? Explain briefly how you decided.

In the scheme below which represents the mechanism of action for a large number of enzymes: A+B⟺AB⟶C The steady state approximation is reached when: d[AB]/dt≈0 k2≫k1 k−1≫k1 k−1=k1

The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate ?0V0 for an enzyme‑catalyzed, single‑substrate reaction E+S↽−−⇀ES⟶E+PE+S↽−−⇀ES⟶E+P. The model can be more readily understood when comparing three conditions: [S]<<?m[S]<<Km, [S]=?m[S]=Km, and [S]>>?m[S]>>Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity ?0V0 where steady state conditions are assumed. [Etotal][Etotal] refers to the total enzyme concentration and [Efree][Efree] refers to the concentration of free enzyme.

The rate constants of an enzyme-catalyzed reaction, obeying the Michaelis-Menten kinetics, have been determined : E + S K₁ = 2 x 108 M-¹ S-¹ -1 -1 -1 K-₁= 1 x 10³ S K₂ = 5 x 10³ S-1 K₁ 1 K-1 ES K₂ E +P 1- Determine the Michaelis constant Km of the enzyme. 2- Determine the catalytic constant (kcat) of the enzyme. 3- Determine the catalytic efficiency of the enzyme.

Consider this intermediate in the derivation of the Michaelis-Menten equation. [E] [S] [ES| k-1 + kz km Assume that k is negligible compared to the other rate constants. If the k is very small, it suggests that the enzyme has a Select an option affinity for its substrate, while if the if the km is very large, it suggests that the enzyme has a Select an option. affinity for its substrate. Select an option Submit You have used 0 of high Sav low moderate

Sketch the complete reaction free energy diagram for an enzyme-catalyzed conversion of a single substrate (S) into product (P), where the reaction is spontaneous in the forward direction Overlay the free energy diagram for the uncatalyzed reaction and indicate  delta delta G〒 on your sketch:  Chemical step is rate limiting

Given the following reaction and equation for the initial velocity of the reaction: k₁ k3 E+S ES E + P V=Keat [ES] = k3 [ES] k₂ where keat is the rate constant for the reaction which forms the product from the ES complex. Explain in words why the velocity is directly proportional to theamount of enzyme added in the presence of saturating substrate levels.

Which of these statements about enzyme-catalyzed reactions is false? The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction. The Michaelis-Menten constant Km equals the [S] at which V = 1/2 V, max: At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration. The rate of a reaction decreases steadily with time as substrate is depleted. If enough substrate is added, the normal V, of max a reaction can be attained even in the presence of a competitive inhibitor.