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Q: question
A: 79ac4c41-0543-4987-bd8c-f7f9d1d1b7c1
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- 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?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 tetrapeptide, glutamate-glycine-alanine-lysine, is prepared at at concentration of 1 mM (0.001 M) and is measured in the standard setup (pathlength of 1 cm). What is the approximate absorbance of this peptide at 280 nm? Hint: if the peptide contained a single tryptophan, the answer would be about 10. 10 280 1 0
- Which of these heterocyclic drugs is likely to be the least soluble in water? Use the Fsp³ parameter to decide. OH Tramadol Chemical Formula: C16H25NO2 YOUR OW Pantoprazole Torasemide Chemical Formula: C16H15F2N3O4S Chemical Formula: C16H20N4O3S Temazepam -OH Chemical Formula: C16H13CIN₂O2 Tioconazole Chemical Formula: C16H13C3N₂OS A. Tramadol B. Pantoprazole C. Torasemide D. Temazepam E. ToconazoleWhich of the following situations would produce a Hill plot with nH < 1.0? Explain your reasoning in each case.(a) The protein has multiple subunits, each with a single ligand-binding site. Binding of ligand to one site decreases the binding affinity of other sites for the ligand.(b) The protein is a single polypeptide with two ligand-binding sites, each having a different affinity for the ligand.(c) The protein is a single polypeptide with a single ligand-binding site. As purified, the protein preparation is heterogeneous, containing some protein molecules that are partially denatured and thus have a lower binding affinity for the ligand.The KM for the reaction of chymotrypsin with Substrate A is 8.8 x 10-4 M, while the KM for the reaction of chymotrypsin with Substrate B is 8.7 x 10-3 M. Which of the following statements are likely true? Chymotrypsin has a higher apparent affinity for Substrate A. The Vmax would be higher in the presence of Substrate A. The kcat would be higher with Substrate A. The V max would be higher in the presence of Substrate B. Two of the above are true.
- 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?The 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-)Calculate θ for a certain protein-ligand pair when the ligand concentration = 1 M and the Kd = 1 X 10-15 M.
- 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.)If instead of using 3.5 µM myoglobin (receptor) you used half of this (that is, 1.75 µM myoglobin), what would be that value of the Kd, that you calculated ( how would it change)? Please explain so I can solve on my own :) (How does changing concentration of the receptor in a ligand-receptor binding experiment affect the dissociation constant?)The figure shows binding curves for two proteins that bind the same ligand. Which binding curve represents simple equilibrium binding of a ligand? 1.0- Protein 1 0.8- Protein 2 0.6 0.4 0.2 0.0 25 50 75 100 [Ligand] (mM) 1) neither curve 2) the curve for protein 1 3) the curve for protein 2 4) both curves Y (fractional saturation)