Following is a Dixon plot for PNPP inhibition by inorganic phosphates. What answer choice shows the correct combination of the mode of inhibition and Ki of the El complex? [S] = 0.2 mM "[S] = 0.4 mM I/Vav 8 7 y=0.01357x+3.6976 6 10 5 4 3 2 1 y=0.0795x+2.7693 -30 -20 -10 0 10 20 30 [I](M) Mixed inhibition, Ki= -21.2 UM Mixed inhibition, Ki= 1/16.5 uM Competitive inhibition, Ki= 16.5 UM
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Following is a Dixon plot for PNPP inhibition by inorganic phosphates. What answer choice shows the correct combination of the mode of inhibition and Ki of the EI complex?
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- Calculate, or make a best estimate of, the unknown factors in the situations listed in Table 1. Table 1. Data for problem 1. To find Data (i) Ek [K+]out = 4 mM; [K+]in = 130 mM (ii) [Cl-]in [Cl-]out = 570 mM; ECl = -65 mV (iii) ECl [Cl-]out = 150 mM; [Cl-]in = 8 mM; in a mammal (iv) [Na+]in Overshoot of action potential = +35 mV; saline [Na+] = 112 mM (v) [K+]in Blood [K+] = 3.2 mM; undershoot of action potential = -87 mV (vi) ECa [Ca2+]out = 5.6 mM; free [Ca2+]in = 0.8 mM (vii) [K+]out [K+]in = 350 mM; Ek = -82 mVWhat is the Keq for the conversion of Glucose 6-Phosphate to Glucose 1-Phosphate if the phosphate transfer potential for Glucose 1-Phosphate and Glucose 6-Phosphate are -20.9 kJ/mol and -13.8 kJ/mol respectively? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a 0.27 b 0.45 c 10.5 d 0.019 e 0.057 f 24.2 g 7.10 h 0.24 i 2477The enzyme serine hydroxymethy ltransferase (SHMT) catalyses the conversion of serine into glycine. The fo llowing table gives the initial rates, vo, for the SHMT-catalysed reaction of the substrate serine at var ious concentratio ns of serine, lSI.[S]/(mmol dm-3) 10 20 30 40vo(μmol dm-3 s-1) 1.63 2.94 4.10 4.95Use the data to determine the values of the MichaelisMenten constant, the maximum velocity of the reaction, and the maximum turnover number of the enzyme.
- The value of AH° for HBr(g) was first evaluated using the following standard enthalpy values obtained experimentally. Use these data to calculate the value of AHf° for HBr(g). Cl,(g)+2KB1(ag) – Br, (ag)+2KCI(ag) AH° =-96.2 kJ H,(g)+Cl, (g)- HCl(aq)+ KOH(aq) KC(aq)+H,O(!) 2HCI(g) AH° =-184 kJ AH° =-57.3 kJ HBr(aq)+ KOH(aq) KBr(aq)+ H,O(!) AH° =-57.3 kJ AH° =-77.0 kJ HC(g) Br,(g) HBr(g)- HCl(aq) disdlving in water Br, (aq) AH° =-4.2 kJ disolving in water HBr(aq) AH° =-79.9 kJ dissolving in water ΔΗΡ- i kJHexokinase in red blood cells has a Km of approximately 50 µM. What concentration of blood glucose would yield a velocity equal to 90 % Vmax? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a 500 mM 0.0450 mM 4500 μΜ 5000 µM e 5.0 mM 0.50 mM 450 μΜ 4.5 mM 50 μΜ.Use the concept of Hess' Law to determine AH for the process V 2X +Z given the following numbered processes: V 2 W AH1 W X + Y AH2 Z- 2Y AH3 ΔΗ+ ΔH +ΔΗ3 ΟΔΗ+ 2ΔΗ-ΔΗ3 ΟΔΗ+ 2ΔΗ2+ ΔΗ3 ΔΗ + ΔH,-ΔΗ3
- The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)An experiment was carried out to measure the reaction rate of hydrolysis of acetylcholme (substrate) with serum enzymes (Eadie, 1949). In the experiment, two experiments were conducted, namely experiment 1 without using a prostigmine inhibitor and experiment 2 using a prostigmine inhibitor at 1.5 x 10^-7 mol/l. the data obtained are: a. Is prostigmine competitive or noncompetitive inhibitor? b. determine the value of km and rmax for the two experiments, compareThe plasma profiles of codeine (COD) and metabolites for 2 individuals (labeled A and B) are shown below. The X-axis is time in hours after an oral dose of codeine. [M=morphine; C6G=COD-6-glucuronide; M3G = morphine-3-glucuronide; NM (ignore)]. Note the data is shown on a log scale on the Y-axis. (A) Which individual is the poor metabolizer? Explain how you know this from the profiles? (B) Is this a problem for cough suppression? Explain. -CH HO Codeine COD 10 000 1000 C6G COD 100 M3G M6G NM 10 M 10 20 30 0 10 20 30 Plasma concentration (nmol I-)
- The hexapeptide Ala-Met-Leu-Lys-Phe-Asp is digested in the same tube with both Cyanogen Bromide and Trypsin at the same time at pH=7. Draw the structure of the product(s) that would bind to a cation exchange column. (Relevant pKa values are 2.2, 3.9, 9.5 and 10.5. Assume pKa values for any newly generated a-amino and a-carboxyl groups are 9.5 and 2.2 respectively)What is the Keg for the conversion of Glucose 1-Phosphate to Glucose 6-Phosphate if the phosphate transfer potential for Glucose 1-Phosphate and Glucose 6-Phosphate are 20.9 kJ/mol and 13.8 kJ/mol respectively? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a 0.27 b 0.45 10.5 d 0.019 e 0.057 f 24.2 7.10 h 0.24 i 2477The E1% at 280 nm of the protein myosin is 0.842. You take 40µl of a myosin solution of unknown concentration and add water to a final volume of 2.0 ml. The absorbance of this solution at 280 nm is 0.771. What is the concentration of the undiluted myosin solution? (units = mg/ml).