(iv) Consider the molecule 1 being derivatized to yield molecule 2. How would you expect AG for the binding process (to their target) of 2 to compare to 1? Justify your answer.
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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?) To better understand how the lacR works, let's start let's approximate the volume of a bacterial cell by treating it as a cylinder with radius that is 0.5 um and length that is 2 um. What is the volume of the cylinder in liters, L)? Next, let's calculate the concentrations of the lacR (ligand) and the operator site (Op) (primary receptor) in this volume. There are 10 lacR molecules in the cell and 1 operator site. Using this information calculate [lacR] and [Op] in molar units in one cell? (see eq. 13.17 and 13.18).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)
- Calculate the net charge on the following tripeptides at pH 5.0: (a) Leu-His-Asp [0] (b) Ala-Ile-Val [0] (c) Met-Lys-Arg [+2] (d) Which tripeptide will be retained the shortest on a cation-exchange chromatographic column in a pH 5.0 buffer? Why?The extinction coefficient or absorptivity (ɛ) of protein A at 340 nm is 6440 M-1 cm-1, whereas protein B does not absorb at 340 nm. What absorbance will be observed when light at 340 nm passes through a 5 mm cuvette containing 10 µM of protein A and 10 µM of protein B? Beer-Lambert-law; A = ɛ x C x1; A = absorbance, C= concentration, 1= pathlength).The equilibrium constant for the hydrolysis of the peptide alanylglycine (Gly-Ala in the reaction from Part B) by a peptidase is K = 9.0 × 10² at 310 K. Calculate AG for this reaction. Express the Gibbs free energy to three significant figures. AG = Submit ΠΑΠΙ ΑΣΦ Request Answer ? kJ/mol Keq [Gly] [Ala] [Gly-Ala]
- The following data describe binding of ligand A to a protein, using both Scatchard plots and Hills plots, tell as much as you can about the binding reaction. [A](M) 1X10-6 5 X10-6 1 X10-5 5 x 10-5 1 X 10-4 5 x 10-4 1x 10-3 5X10-3 ṽ 0.101 0.381 0.591 1.116 1.409 1.813 1.899 1.955..Calculate the solubility of Ag,CO3 (in mol/L) at 25 °C in a 0.02 M Na2CO3 solution. Hint: Ag,CO3(s) → 2Ag*(aq) + co3?(ag) Ksp (Ag,CO3) = 8.1 x 10 12 NażCO3(aq) → 2Na*(aq) + CO3 (aq)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.)
- 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.Asp residue (both of which are essential for catalysis) with pK, values of 5.9 and 4.5, respectively. If the enzyme is found in the lysosome (pH = 5.2), which residue will act as the general acid and which will act as the general base during the initial steps of the reaction? Explain your reasoning. (Protein concentration can readily be determined using the Beer-Lambert law: A = e l c where A = absorbance e = molar absorption coefficient (M-1cm-1) l = light path length (cm) c = concentration (M) If the molar absorption coefficient at 280 nm for yeast ADH is 48860 M-1cm-1 and a 10 mL solution of the protein has an absorbance at 280 nm of 0.4 (as measured by a spectrometer with pathlength 1 cm), then what is the concentration of the protein solution (in μM)? i.e. concentration = ______ μM If the molecular weight of the protein is 36849, what is its concentration in mg/mL? i.e. concentration = _______ mg/mL For each part of the question, show your calculations to arrive at your answers.