A one-to-one protein (P)-ligand (L) complexation (P + L <-> PL) has a equilibrium constant (Kd) value of 100 nM at 25°C, and the Kd remains the same at 37°C.
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- An 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 protein-ligand binding reaction is run. At equilibrium, half the protein is ligand bound, the unboundligand concentration is 0.657 nM. Calculate the koff value for this reaction. Assume the kon value is typical ofprotein-ligand interactions.Consider the dissociation reaction for a protein-ligand complex: P•L P + L A. Sketch a binding curve (fractional saturation θ vs. ligand concentration [L]) for this protein-ligand complex (ligand A). Show where on that curve you could obtain the dissociation equilibrium constant Kd for the reaction. B. Now sketch on the same axes a θ vs. [L] plot for a different ligand (B) that binds more weakly than the first ligand. C. Does the weaker binding ligand have a higher, or lower, Kd than the tighter binding ligand? D. Sketch a binding curve for a cooperatively bound ligand with K0.5 higher than that of Kd for A or B. (Note: for cooperative binding, each protein molecule would have to have more than 1 binding site for the ligand; K0.5 is the experimentally determined ligand concentration that gives θ = 0.5.)
- Consider the binding reaction L + R → LR, where L is a ligand and R is its receptor. When 1 × 10−3 M of L is added to a solution containing 5 × 10−2 M of R, 90 percent of the L binds to form LR. What is the Keq of this reaction? How will the Keq be affected by the addition of a protein that facilitates (catalyzes) this binding reaction? What is the dissociation equilibrium constant Kd?Two proteins, A and B, bind to the same ligand, L, with the binding curvesshown below. What is the dissociation constant, Kd , for each protein? Which protein (A or B) has a greater affinity for ligand L?Molecular diameters of proteins IgG and TNFα are 11.2 nm and 4.4 nm, respectively. The diffusion coefficient of IgG in a dilute aqueous solution at 37oC is 2.8 ⤬ 10-7 cm2/s. Estimate the diffusion coefficient of TNFα in a dilute aqueous solution at the same temperature.
- Calculate θ for a certain protein-ligand pair when the ligand concentration = 1 M and the Kd = 1 X 10-15 M.Three different ligands, Ligand Q, Ligand T, and Ligand W, bind to the same protein but with different affinity: The association constant (Ka) for the binding of Ligand Q to the protein is 0.033 nM-1. The fractional saturation (Y) of the protein is 0.20 when the concentration of Ligand T is 1.25 nM. The fractional saturation (Y) of the protein is 0.80 when the concentration of Ligand W is 72 nM. Given this information, Calculate Kd for the binding of each ligand to this protein. Which ligand binds with greatest affinity? Which ligand binds with the lowest affinity?Give the general Adiar equation for the binding of a ligand to a dimeric protein. Explain further what your understanding is of the terms "no-, positive-, and negative cooperativity” and graphically present the relationship between Ȳ and [S] for each of these cases. Also, give the relationship between the constants Kb1 and Kb2 in each case.
- 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. What is the ratio of the concentration of substance A inside the cell to the concentration outside? [A]inside [A]outside = 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. Movement of Aoutside to Ainside will be spontaneous. Because AG is negative, the ratio [A]inside/[A]outside must be greater than one.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.affinity of a protein-protein or protein-ligand interaction can be described by the Dissociation Constant, Kd (written below). Consider a protein P and its inhibitor, I. I inhibits P's activity when bound to it: koff _ [A][B] Dissociation Constant: Ka = koN [AB] Question When [I] is 10-7 M, 99% of P's activity is inhibited. What is the Kd of this Protein- Inhibitor interaction?