Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at a certain ratio of [A]inside to [A]outside, the AG for the transport of substance A from outside the cell to the inside, Aoutside Ainside, is -11.3 kJ/mol at 25°C.
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- Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at a certain ratio of [A]inside to [A]outside, the AG for the transport of substance A from outside the cell to the inside, Aoutside → Ainside, is -12.1 kJ/mol at 25°C. What is the ratio of the concentration of substance A inside the cell to the concentration outside? [A]inside [A]outside || 656275.63 Incorrect Choose the true statement about the transport of A under the conditions described. Increasing [A]outside will cause AG for movement of Aoutside to Ainside to become a smaller negative number. Decreasing the concentration of the uniport protein in the membrane will cause AG to become a larger negative number.A one-to-one protein (P)-ligand (L) complexation (P + L PL) has a dissociation equilibrium constant (Kd) value of 100 nM at 25°C, and the Kd remains the same at 37°C. 1) What is AS of binding at 25°C? Assume ACp of the binding is 0 over the temperature range. AS = 1.34E2 kJ/(mol*K) (note the unit!!) (sig. fig =3) 2) What is the concentration of the PL complex formed at equilibrium when you mix 0.20 uM (microM) of Protein and 1.0 uM of Ligand together at 37°C? PL at equilibrium = 8.1E-1 uM (note the unit!!) (sig. fig =2)The effect of temperature on the hydrolysis of lactose by a ß-galactosidase is shown below in Table 1. The temperature coefficient, Q10 is the factor by which the rate increases by raising the temperature 10°C. The universal gas constant, R is 8.314 J/mol.K. (a) (b) Table 1: Data of Vmax over temperature T (°C) 20 30 35 40 45 Vmax (umoles/min.mg protein) 4.50 8.65 11.80 15.96 21.36 Plot the graph of In Vm vs 1/T using any spreadsheet software (include all appropriate labels and equation). Calculate the activation energy Ea and temperature coefficient Q10.
- The acidity of the stomach is maintained by the H*/K* ATPase in parietal cells of the gastric mucosa. These cells have an internal pH = 7.4. The H*/K* ATPase transports H* across the cellular membrane into the stomach where pH = 0.8. (a) Given a membrane potential of 65 mV (inside negative), calculate AG in kJ/mol for the transport of H* into the stomach at 37 °C. Show your work. (b) The hydrolysis of ATP is directly coupled to H* transport and the stoichiometry of the transport reaction is such that 1 mol of ATP is hydrolyzed for every 1 mol of H* transported into the stomach. ATP is hydrolyzed by the ATPase on the inside of the cell. Under the conditions given in part (a) and assuming the steady-state concentrations of ATP, ADP and P₁ given below, is the transport of H* in to the stomach thermodynamically favorable at 37 °C? Show the calculation that supports your answer. Steady-state intracellular concentrations: ATP = 4.2 mM ADP = 360 μM Pi = 5.4 mM For ATP hydrolysis at 37 °C: ATP…Neutral sphingomyelinase 2 converts sphingomyelin into ceramide and phosphorcholine. What kind of enzyme is it? Assume Vmax is 35 µM min-1. When you provide 3.0 x 10-5 M of sphingomyelin, you observe an initial velocity of 6.0 µM min-1. Calculate the KM.A 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 m mol/min. What is the reaction velocity when the concentration of maltose is 1.0 mM?
- 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, compareAn antibody binds to another protein with anequilibrium constant, K, of 5 × 109 M–1. When it binds toa second, related protein, it forms three fewer hydrogenbonds, reducing its binding affinity by 11.9 kJ/mole. Whatis the K for its binding to the second protein? (Free-energychange is related to the equilibrium constant by the equa-tion ΔG° = –2.3 RT log K, where R is 8.3 × 10–3 kJ/(mole K)and T is 310 K.)A 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)
- Consider a suspension of particles (isoelectric point is at pH 6) in water at pH 2 and a NaCl concentration of 0.001 M. Describe how the strength of repulsion varies with the following changes, assuming all other conditions remain constant. Give a description (more than just increase or decrease) in terms of the effect on the double layer thickness and the zeta potential. (a) Change from 0.001 M NaCl to 0.1 M NaCl, (b) Change from pH = 2 to pH = 5.An enzyme has the following values: vmax's; 0.028 M s-' and 0.021 M s-' (I), and km's; 0.00198 M and 0.00197 M (I), when the total concentration of enzyme is 10-º M. The (I) values were determined with 1 microM inhibitor. Select the value closest to the k, for this enzyme. (hint 1 microM = 10-6 M).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)