Glucose-1-phosphate is converted to Glucose-6 phosphate by an enzyme. The equilibrium constant in biochemical standard conditions is 19. Answer the following four questions regarding this enzyme-catalyzed reaction. K! = 19 R 8.314 AG-RTln K' AG' = AG' + RTlnQ' Ran J mol K Glucose 1- PO²- →→ Glucose - 6 - PO

The M and H subunits of lactate dehydrogenase have very similar sizes and shapes but differ in amino acidcomposition. If the only difference between the two were that theH subunit had a glutamic acid in a position where the M subunithad a serine, how would the five isozymes of LDH separate on electrophoresis using a gel at pH 8.6?

For an enzyme catalyzed reaction of the form: S + E → P + E, the rate of product formation, [P], is given by: d[P]/dt = k2[E}total [S] /(Km + [S]) = For the enzymatically catalyzed hydrolysis of ATP at 25 °C and pH 7.0, the Michaelis Menten constant, Km was found to be equal to 16.8 μmol L 1 and the value of k2[E]total was found to be 0.220 μmol L-¹s¹. Find the initial rate at an initial ATP concentration of 30.0 μmol L-1

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.

When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added to a solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of 65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in units of min-1)?

The initial velocities of two different enzyme-catalyzed reactions were measured over a series of substrate concentrations. The following results were obtained: Enyme A: KM = 1.5 mM, Vmax = 10 μM s-1 Enyme B: KM = 5.0 mM, Vmax = 85 µM s-1 (a) Which enzyme binds to its substrate more tightly (assume k.1 >> k₂ in the Michaelis-Menten model)? (b) Calculate the initial velocities of each reaction when the substrate concentration is 2.5 mM. (c) Calculate the Kcat of each enzyme if the total enzyme concentration is 100 nM. (d) Which enzyme is the more efficient catalyst? Explain your answer. The enzyme carbonic anhydrase is strongly inhibited by the drug acetazolamide. A plot of the initial reaction velocity (as a percentage of Vmax) in the absence and presence of the inhibitor is shown below. What type of inhibition is taking place? Explain your reasoning. V (% of Vmax) 100 50 0.2 0.4 No inhibitor Acetazolamide [S] (MM) 0.6 0.8 1

A particular reaction has a ΔG‡ of 30.0 kJ mol-1 at 25.0 °C. In the presence of an enzyme, the same reaction has a ΔG‡ of 1.50 kJ mol-1 at the same temperature. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)

We want to measure the activity of alanine aminotransferase (ALAT) present in a serum. The reaction catalyzed by the enzyme is: Reaction 1: I- *H₂N- glutamate H - C-COO CH₂ CH₂ COO 0.1 M phosphate buffer pH 7.4 : 550 μL 1.2 M alanine: 100 μL CH3 pyruvate CH3 C time (min) A340 CIO O COO The enzyme reaction is alized in the following conditions: In a 1 cm-cuvette are added: COO™ pyruvate lactate dehydrogenase* (LDH, 300 µμg.mL-¹): 50 μL 1.5 mM NADH : 200 μL 0.04 M a-ketoglutarate: 500 µL serum containing ALAT: 600 μµL ALAT NADH + H+ 0 0.915 a-cétoglutarate COO LDH * Lactate dehydrogenase (LDH) reduces pyruvate into lactate, with the concomitant oxydation of NADH. This allows to indirectly measure the amount of product formed. с=0 CH₂ 1 0.741 Reaction 2: NAD+ CH₂ COO™ H-C CH3 OH COO™ lactate alanine H + *H3N-C The reaction is performed at 25 °C and the absorbance at 340 nm is monitored every minute, for 5 min. The absorbance values are given in the table below: Data: ENADH at 340 nm =…

For a Michaelis-Menten reaction, k₁=5 x 107/M-s, k-1-2 x 104/s, and k2=4 x 10²/s. Calculate the Ks and KM for this reaction. Does substrate binding achieve equilibrium or steady state?

The KM values for the reaction of chymotrypsin with two different substrates are given in the table below. Considering this information, which substrate has the lower apparent affinity for the enzyme? Which substrate is likely to give a lower value for Vmax? Substrate N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester KM (M) 8.8 X 10-² 6.6 X 10-4 N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give a lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give the lower V₁ max. N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; N- acetyltyrosine ethyl ester is likely to give the lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; N- acetylvaline will likely to give the lower Vmax. None of the above statements are correct.