QUESTION 3 In the presence of 0.000005 M competitive inhibitor, an enzyme exhibits an apparent Km of 0.002M. If the Kt of the inhibitor is 0.0000003 M, what is the true Km of the enzyme? Give your answer with 3 significant figures an scientific notation. For example, 0.000023434 is 2.34e

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You run an experiment using 0.5 µM of an enzyme. The kinetic parameters for your enzyme are kcat = 1800 s-1 and KM = 25 µM. (a) If k-1 is 240 s -1, what is the value of k1? (Report your answer in units of M-1 s -1) (b) What is Vmax for the enzyme (Don't forget to include units in your answer)? (c) You set up an experiment with the following initial concentrations of substrate. Calculate the rate of the reaction at each substrate concentration (report your answer in units of µM sec-1 [S](μM) 1 100 1000 [S][μM) 1 100 1000 v (μMsec ¹) (d) You now add 10 nM of a competitive inhibitor with a KI = 2 nM. What is the rate of the reaction (v) for the following substrate concentrations and how does it compare to the rate without inhibitor (VO)? Vinhib (uMsec-¹) Vinhih/V

You have an initial solution in which you added quantities of “A” and “B” such that there is 4.5 M “A” and 2.5 M “B” and no complex (“AB”) at time 0.  After equilibrium, you are able to isolate and quantitate the “AB” complex, and find its concentration is 1.5 M.  Given that RT is 0.59 kcal/mol, what is the delta Go’ for the association reaction?

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?

How can you find Kcat if you are only given Vmax, Km and [E] ? I tried using Kcat=Vmax/[E] but that didn't work. How do you know if [E] is the same as [E]total, and if it isn't how do you find it from this information: The Vmax for a particular enzyme is 10 nmols/L/s. The Km for its substrate is 5 microM. If the enzyme concentration is 10 nM, what is the kcat? a.80 nmoles/L/s b.8000 nmoles/L/s c.2 nmoles/L/s d.50 nmoles/L/s

Please don't provide handwritten solution ....

Only 15-14

Could you explain step by step how to solve this? I'm not sure how to get all the numbers that are shown in the solved answer.