3. The succinate dehydrogenase reaction in the direction it proceeds within the CAC is shown below (left to right): FADH, FAD H-C-H H-C-H coo succinate dehydrogenase coo- Succinate Fumarate AG" - O kl/mol a) Calculate the standard reduction potential for the overall reaction above. b) Based on the reduction potential, calculate and compare your standard transformed free energy change to the textbook value (above), and venture to explain any discrepancies. ( c) Assuming the temperature of the cell is 37 °C, calculate the equilibrium ratio of oxidized to reduced cofactor when the ratio of oxidized product to reduced substrate is 1.2. .

You are working on an enzyme that obeys standard Michaelis-Menten kinetics. What variable is the V, dependant on if the concentration of the substrate is substantially higher than the concentration of the enzyme? [S] [E] [ES] O [P] O not enough information provided

An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s

at what substrate concentration is v= 5.0 mM s^-1 for the enzyme catalyzed hydrolysis of trehalose?

. The optimal conditions for salivary lysozyme (hydrolyzing glycoproteins of bacterial wall) are 37 C - temperature and pH is 5.2. Explain the decrease in this enzyme activity if the temperature will rise up to 60 °C and pH will be changed to 8.0. To answer the question: a) draw the graph of the velocity dependency on temperature and pH; b) calculate the relative enzyme activity if 10 mg of lysozyme catalyzes the formation of 5 uM of the product per 2 minutes. Concidor NH3: 5.

(b) You are investigating the effects of several agents on the activity of alcohol dehydrogenase. The enzyme activity data are shown in the table below. Construct a [substrate] vs. activity plot and a double-reciprocal plot for this enzyme. Be sure to label all axes. Determine the Vmax and KM for AD from the graphs in each type of plot. AD activity (nM/min) AD activity + agent A (nM/min) AD activity + agent B (nM/min) [Alcohol] (nM) 0.1 14 2 0.5 50 7 8. 1.0 65 10 30 2.0 72 12 45 4.0 80 14 62 8.0 85 15 75 32.0 90 16 90

The Lineweaver-Burke plot was originally developed in order to "linearize" the data obtained from enzyme kinetics experiments, in order to facilitate the determination of kinetic parameters. Why is it not considered to be an accurate method for this purpose? It is very difficult to draw a straight line on a computer. It is very difficult to calculate the variables required for the "x" and "y" axis. It is more accurate to use the standard "V versus [S]" plot to determine Vmax and KM- The plot weights the least accurate data points the most heavily. It is no longer considered to be acceptable to extrapolate from known data.

An enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.

1. In a catabolic pathway, metabolite X gets converted into metabolite Y, metabolite X 0 J mol-¹ and AG' is <0 J mol-¹ for this reaction. a) At standard states, is this reaction favorable? b) Is the reaction favorable or unfavorable at cellular condition? c) Is this an endergonic or exergonic reaction in the cell? d) Would you describe this as a thermodynamically downhill or uphill reaction in the cell?

Two enzyme catalyzed reactions, RA (red) and Rp (blue), have kinetic profiles shown below. Report which reaction has the higher maximal rate and which reaction has the smaller Km. в A 1/ [S] Higher Vm [ Select ] Lower Km [ Select ] A/T