a) Calculate the enzyme and specific activity of a reaction with 3 pM Hsp90 using the following information: The rate is measured in a spectrophotometer as 0.028 OD units/min in a 1 ml reaction volume. The absorbance was detected at 340nm and the extinction coefficient for NADH at this wavelength is 6200L M- 1 min-1 and the molecular mass of Hsp90 is 82.7 kDa. The rate of NADH utilisation is equivalent to the rate of ATP utilised by Hsp90. Show all your calculations and the units for your answers. b) Calculate the turnover number for the reaction described in (a) above
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- a) Calculate the enzyme and specific activity of a reaction with 3 μM Hsp90 using the following information: The rate is measured in a spectrophotometer as 0.028 OD units/min in a 1 ml reaction volume. The absorbance was detected at 340nm and the extinction coefficient for NADH at this wavelength is 6200 L M-1 min-1 and the molecular mass of Hsp90 is 82.7 kDa. The rate of NADH utilisation is equivalent to the rate of ATP utilised by Hsp90. Show all your calculations and the units for your answers. b) Calculate the turnover number for the reaction described in (a) aboveA bacterial enzyme catalyzes the hydrolysis of maltose as shown in the reaction given below: Maltose + H2O -> 2 glucose If the reaction has a Km of 0.135 mM and a V max of 65 umol/min. What is the reaction velocity when the concentration of maltose is 1.0 mM? (Please take note of the units)Calculate the catalytic efficiency (in mM) for the uninhibited reaction based on the red line in the graph below. Note the concentration of the enzyme in the reaction is 1.55 micromolar. Velocity is in units (U), and substrate concentration is in mM. The equation of the red line is: y = 0.167x + 0.133 The equation of the blue line is: y = 0.500x + 0.133 -2 -4 -2 4 10 12 1/[S]
- An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :Consider the two half-reactions below and their standard reduction potentials. NAD+ + H+ + 2e → NADH Elo= -0.32 V a-Ketoglutarate + CO₂ + 2H+ + 2e → Isocitrate E' = -0.38 V (a) What is AE" for the spontaneous redox reaction that is, the reaction that actually occurs under standard biochemical conditions (pH 7)? (b) Which of the following statements are correct under standard biochemical conditions? i. The concentration of H+ is 1.0 M. ii. The reaction NAD+ + Isocitrate → NADH + H+ + a-Ketoglutarate + CO₂ is favor- able. iii. NAD+ accepts electrons from isocitrate. iv. The NAD → NADH reaction actually occurs in reverse. The a-ketoglutarate → isocitrate reaction occurs as written. (c) Calculate AG" for the reaction in (a). (pH 7, 25°C, pressure, 1 atm.) (d) Suppose that the actual conditions are T = 25°C, pH = 7, CO₂ = 1 atm, [a-Ketoglutarate] 10 mM, [NAD+] = 2.5 mM, and [NADH] = 0.5 mM. PAR = 2 mM, [Isocitrate] DE What is the value of AG under those conditions? (Hints: pH 7 is already…Carbonic anhydrase catalyzes the hydration of CO. CO2 + H2O ¬ H½CO3 The Km of this enzyme for CO, is 1.20×104 µ.M. When [CO,] = 3.60×104 µM, the rate of reaction was 4.50 umol·mL! sec-1 a What is Vmax for this enzyme? umol·mL-!sec-!
- The standard reduction potential for ubiquione (A or coenzyme Q) is .045 V, and the standard reduciton potential (E) for FAD is -0.219 V. Using these values, show that the oxidation for FADH2 by ubiquinone theoretically liberates enough energy to drive the synthesis of ATP. Faraday constant =96.48KJ/Vol delta G' standard for ATP Synthesis is +30.5 KJ/mol R=8.314 J/mol K=1.987 cal/mol KAn enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?What is the Vmax of an enzyme at 1 µM with an activation free energy (AG#) of 55000 J/mol at 298K? -AG # RT k cat = 6.2× 10¹² (Ideal gas constant R = 8.314 J K¯1 mol-¹1) 0.14 μM s-1 -1 S-1 1.4 μM s 14 μM s-1 0.14 mM s-1 1.4 mM s-1
- An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?You make reaction progress curve by plotting absorbance vs time (seconds) and find the equation of the line to be y = -0.00235x + 0.7129. Calculate the U/µL and U/mL of lactate dehydrogenase activity in this fraction. The LDH activity is done identical to what is indicated in the lab manual. Show each step of the calculation from AU/time to M/min, to mol/min, to µmol/min to µmol/min/µL (=U/µL).A particular enzyme has a ΔΔG‡ of -22.1 kJ mol-1 at 37.0 °C. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)