
Km represents the.... formation of EP the back reaction of product to ES formation of ES kcat Km represents the......
Km represents the.... (click all that apply) the back reaction of product to ES kcat formation of EP formation of ES
)* (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
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where [S]=6mM, it was found that Vo=480nM/min. What was the enzyme concentration? Give your answer in nM. Using the same kcat and Km as the previous question, if [Et]=0.5uM gives a Vo=5 uM/min, what wat the [S]? Give your answer in uM. reaction is run with the kcat=20/min and the Km=4uM. Use the enzyme concentration from question 1 above. A very strong inhibitor is added creating...
D-Lactose is the substrate for B-galactosidase. Given Vo = kcat [Et] [S]/km + [S], calculate [S], when Km = 4.0 nM, V. = 10.5 M s', kcat = 500 s, and [Et] = 40 uM Calculate the catalytic efficiency. Below is a double-reciprocal plot for an enzyme reaction in the absence and presence of of inhibitor. Give the equation for the line. Calculate Vmax and Km for the enzyme and enzyme plus inhibitor. Which type of inhibition is apparent. 0.10...
2 (10 marks)The KM and kcat of a carboxypeptidase isoform were found to be 2.00mM and 150s-1 respectively for substrate A. What is the initial rate of the reaction if [A] = 4.00mM and [E]total = 0.010 mM? The presence of a competitive inhibitor [I] = 5.00mM decreases the initial rate by a factor of 2, what is KI?