-Inhibitor +Inhibitor [S] (mM) Vη&νβσπ:(μmol/sec). V0&νβσπ&ν βσπ (μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Vmax of this enzyme WITHOUT iinhibitor?

For the following problem, indicate whether enzyme is repressed (-) or produced (+). Please show all possible outcomes.

The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (uM) 1.3 - Inhibitor + Inhibitor 2.50 0.62 2.6 4.00 1.42 6.5 6.30 2.65 13.0 7.60 3.12 26.0 9.00 3.58 What is the Km of the inhibited enzyme reaction?

you are trying to come up with a drug to inhibit the activity of an enzyme thuogth to have a role in a liver disease. in the lab the enzyme was shown to have a Km of 1x10-6 M and Vmax of 0.1 micromoles/min.mg measruing at room temperature. you developed a mixed non-competitve inhibitor with a ki=0.4x10-6M and a Ki` pf 0/2 x 10-5 What will be the apparent Km in the presence of 1.0x10-6 M?

Using the appropriate graph and table above, explain what the R48C mutation appears to be doing to the enzyme’s function.  Discuss the kinetic parameter changes and their meaning in this context, not the structure of the enzyme, which was not given to you.

8L.3.4

The table contains data about how different ASFV mutants affect enzyme activity-- for example the D29A mutant, where the D at position 29 was mutated to an A.  Select three mutants at random, and for each mutant explain how that mutation could affect intermolecular force interactions (an example: hydrogen bonding could be disrupted by this mutation because…

The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity. K is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (ro) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for Vmax and Km- Vmax= µM/min v (µM/min) 300 275 250 225 200 175 150 125 100 75 50 Km = 25 K 0 10 20 30 V max 40 50 [S] (M) 1/2 V max Michaelis-Menten curve 60 70 80 90 100 HM

5.50 1/V, min/umol 5.00 4.50 4.00 y = 0.9474x + 2.6649 y = 0.9997x + 2.032 0.00 1.00 2.00 2.50 3.00 1/[S], uM -1 Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in the presence of two concentrations of inhibitor, what would be the Vmax for the uninhibited enzyme? (bottom graph) Equation is given. Choose the one best answer. 3.50 3.00 2.50 2.00

PTP1B Substrate kcat Km kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEPYLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 ± 0.22 0.53 + 0.02 34.7 + 0.25 52.7 ± 0.7 0.066 + 0.001 (d) (. ) The units for kcat/KM in the above are given according to standard scientific notation. On this basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?