a) What is the Steady State assumption; how does steady state differ from equilibrium? b) Transition state; what are two ways that enzymes can decrease the transition state energy?

A) Competitive inhibition: Where do these inhibitors bind? To what mechanistic form of the enzyme do these inhibitors bind? What are changes to Km and Vmax?

Would you expect an “enzyme” designed to bind its target substrate as tightly as it binds the reaction transition state to show a rate enhancement over the uncatalyzed reaction? In other words, would such protein be a catalyst? Use a reaction energy diagram to explain why or why not.

Which of the following aspects of catalysis by enzymes can NOT be explained by the Fischer Lock and Key Hypothesis? Enzymes will lower the activation energy barrier for reaction. Enzymes will specifically recognize their substrates. Appropriate substrates will bind to the enzyme to form a Michaelis complex. Enzymes have an "active site" where appropriate substrates will be bound. Two of the above cannot be explained by the Fischer "Lock and Key" Hypothesis.

Which of the following is TRUE under the following conditions: the enzyme concentration is 2.5 nM, substrate concentration is 75 nM, the KM = 150 nM, and the Vmax = 20 nmol/min a) The rate of the reaction is 20 nmol/min! b) The rate of the reaction is between 10 nmol/min and 20 nmol/min. c) The rate of the reaction is 10 nmol/min. d) The rate of the reaction is below 10 nmol/min. e) The rate cannot be determined from the above information.

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

Ben is studying an enzyme-catalyzed reaction. In the reaction, H+ ions are formed. He found that after a specific time, the reaction stopped, even though there was still a lot of substrate available in the mixture. a) Propose a hypothesis to explain why the reaction stopped.

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?

Soy sauce is prepared in a CLOSED tank by fermenting a salted mixture of soybeans and wheat with yeast, over a period of 8 to 12 months. The resulting sauce (after solids are removed) is rich in ethanol.  (a) How is ethanol produced in the resulting soy sauce? What is the importance of this process? Include the enzymes involved.  (b) Could ethanol be produced when the reaction mixture is incubated in an OPEN tank?

What type of bonds in the tertiary structure of the enzyme break at high temperatures? Which ones will not break?