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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 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.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. ToconazoleThe 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.
- 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-)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?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)