When [I] is 10-7 M, 99% of P's activity is inhibited. What is the Kd of this Protein- Inhibitor interaction?
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- 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 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.)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)
- 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 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?Calculate the concentration of the ligand required to fill 25% of the binding site. X+A ⇌ XA , ka = 1.5 ×10-6 Y+A ⇌ YA , ka = 4.7 × 10-4 (i.e, q = 0.25) for both proteins.
- 12 mM of protein A is combined with 6 mM of ligand X in water. After the protein-ligand complex binding reaches equilibrium, you measure that the free ligand concentration is 3 mM and the concentration of protein-ligand complex is 3 mM. What is the Kd for protein A? Although they would be in mM, do not include units in your answer, only the number as a whole integer.A family of proteins known as cupredoxins contain a single redoxactive Cu ion coordinated by a Cys, a Met, and two His residues. The reduction potentials of cupredoxins range from about 0.15 V to 0.68 V. What does this information reveal about the role of the protein component of the cupredoxins?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?
- Intramitochondrial ATP concentrations are about 5 mM, and phosphate concentration is about 10 mM. If ADP is five times more abundant than AMP, calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. Calculate AG for ATP hydrolysis at 37 °C under these condi- tions. The energy charge is the concentration of ATP plus half the concen- tration of ADP divided by the total adenine nucleotide concentration: [ATP] + 1/2[ADP] [ATP] + [ADP] + [AMP]Consider a protein with two surface-exposed histidine residues: HisA is a “typical” histidine residue with a pKa = 6.2 HisB is involved in a stabilizing interaction (His-NH+ ..... -O2C-Glu) with a neighboring glutamic acid residue. For HisB, the Gibbs free energy of deprotonation at pH = 7.0 and T = 293K is ΔG'o = +15 kj mol-1. If you had a solution, at pH = 7.0 and T = 293K, containing this protein: a) What fraction of HisA residues are protonated? b) What fraction of HisB residues are protonated? c) What is the pKa of HisB?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).