Part VI: Enzyme Kinetics NAD +H +2e NADH E\", 0.32 V Eo 0.816 V 42 Provided The
ID: 1044581 • Letter: P
Question
Part VI: Enzyme Kinetics NAD +H +2e NADH E", 0.32 V Eo 0.816 V 42 Provided The overall reaction is NADH+ H 02 NAD + H20 What is the value of K/K' ratio? Show your detail work. 43 Calculate the value of pH when 40.0 mL of 0.0250M benzoic acid (HC,HsO2, Ka 6.3x10) is titrated with 5.0 mL of 0.050 M NaOH solution? 44 | The KM value of lysozyme is 6.0 x 10"M with hexa-N-acetylglucosamine as a substrate. The initial rate measured at substrate concentration [S]-0.061 M was 3.2 uM min1. Calculate the initial rates at [S) 1.5 x 102M 45 An enzyme has a KM value of [S] 5.00 x 10-M and a Vmax value of 53 HM min1. Calculate the value of vo if [S]-3.7x 104M and [I] 4.8 x 10 M for an uncompetitive inhibitor. (KI 1.7x 10 M.) An enzyme-catalyzed reaction (??-2.7 x 10 3M) is inhibited by a competitive inhibitor l (K- 3.1 x 10M). Suppose that the substrate concentration is 3.6 x 104M. How much of the inhibitor 46 is needed for 90% inhibition? 47 For the type of collision below: A + B-P Which formula can be used to calculate the rate constant of the reaction above? rate NANBdAB rate NANBdAB1e-/RT circle the parts representing the rate constant k in the correct formula. 48 | The values of KM and kcat for enzymes A and B are 9.5% ios M and 1.4 x 104s', o.44 M and 5.1 102 s, respectively. Which enzyme is more efficient? 49 Which quantity may be changed by an enzyme (circle all correction answers): b) Gibbs energy of reaction a) Gibbs energy of activation a) c) Enthalpy of reaction e) Equilibrium constant d) Enthalpy of activation tion of Lineweaver-Burk plot of an enzyme-catalyzed reaction is provided below Y 0.50134 x +6243.27 1) Calculate the value of Vmax of the reaction 2) Calculate the KM value of the reaction.Explanation / Answer
Inhibitor concentration = 1.02 x 10 -4 M
%inhibition = (?-1) x 100/(?+ ?),
? = 1 + I/Ki, l= inhibitor concentration
? = S/Km, S = substrate concentration
Km = 2.7 x 10-3 M (Michaelis-Menten constant)
Ki = 3.1 x 10-5 M
Substrate concentration = 3.6 x 10-4 M
l = x
% inhibition = 90
90 = ((1 + x/3.1 x 10-5 M)-1) x 100 / (1 + x/3.1 x 10-5 M) + (3.6 x 10-4 M/2.7 x 10-3 M)
Rearranging,
0.90 = ((1 + x/3.1 x 10-5)-1) / (1 + x/3.1 x 10-5) + 0.13333
Finding x = 0.000101859 = 1.02 x 10 -4 M
So the inhibitor concentration = x = 1.02 x 10 -4 M