Km represents the.... (click all that apply)
| the back reaction of product to ES |
| kcat |
| formation of EP |
| formation of ES |
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Km represents the.... (click all that apply) the back reaction of product to ES kcat formation...
Km represents the.... formation of EP the back reaction of product to ES formation of ES kcat
Km represents the.... formation of EP the back reaction of product to ES formation of ES kcat
)* (Vmax). 4. When [S] = KM, Vo=( A) 0.5 B) KM C) 0.75 D) kcat E) [S] 5. The overall transformation shown in the following reaction: E s. Es p + E For the reaction, the steady state assumption A) ES breakdown occurs at the same rate as ES formation B) [P]>>[E] C) implies that ki-k-1 D) implies that k., and k2 are such that the [ES] = k1[ES] E) [S] = [P]
For enzymes in which the slowest (rate-limiting) step is the reaction: K2 ES → E+P Km becomes equivalent to: A) kcat. B) the [S] where V0 = Vmax. C) the dissociation constant, Kd, for the ES complex. D) the maximal velocity. E) the turnover number. The answer is C), could you please explain?
Glucokinase is a liver enzyme with a kcat of 33/second and a Km for Glucose of 7 mM. What is the initial reaction rate (in mM product / second) of 0.1 mM glucokinase in the presence of 50 mM glucose? Please show your work clearly.
2. In a single substrate enzyme-catalyzed reaction, the forward rate constant (formation of ES) is 2.1x105 M-1 s-1 , the reverse rate constant (dissociation of ES to E +S) is 9.4x103 s-1 , and the catalytic rate constant (turnover of ES to P) is 7.2x102 s-1 . From this data, KM is:
Consider a description of an enzymatic reaction pathway that begins with the binding of substrate S to enzyme E and ends with the release of product P from the enzyme. E+S →ES → EP E+P Under many circumstances, KM = [E] [S] / [ES] What proportion of enzyme molecules are bound to substrate when [S] = KM? Why? Recall that when [S] = KM, the reaction rate is Vmax/2. Does your answer to Part A make sense in light of...
54. What is the catalyze reaction rate? Km=2 mM Kcat= 3 s-1 At the enzyme concentration=10 nM and substrate concentration= 3 mM
Q1: Please derive rate of product formation (rp) of the following enzymatic reaction. Please apply Pseudo- Steady-State Hypothesis (PSSH) on the two enzyme-substrate complexes. E+S S E S,+S,ESS E SS, P+E
01: Please derive rate of product formation (rp) of the following enzymatic reaction. Please apply Pseudo- Steady-State Hypothesis (PSSH) on the two enzyme-substrate complexes. E+ SE. E•S,+8, E•Ss E•S,S, “P+E
01: Please derive rate of product formation (rp) of the following enzymatic reaction. Please apply Pseudo- Steady-State Hypothesis (PSSH) on the two enzyme-substrate complexes. E+ SE. E•S,+8, E•Ss E•S,S, “P+E