FeedbackYou have not correctly calculated the concentration of added Fab at which half ofthe lysozyme will be bound to the Fab.The dissociation constant (K) is the equilibrium constant for the dissociation of areceptor and ligand: RL = R + L. In this case, lysozyme is the receptor and theFab fragment is the receptor.Ka is given by the equation:[R][L][RL]Kd ==Under these conditions, half of the ligand is bound and half is unbound, so[L] = [RL]. Making this substitution for [L] in the equation for Ka gives you:Ka=[R][RL][RL]= = [R]However, you must determine the total receptor concentration (Rtotal), which is thesum of free receptor and receptor bound to ligand.Rtotal = [R] + [RL]Because half of the ligand is bound and half is unbound in this case, [RL] is equalto one-half of Ltotal. In this case, Ltotal the concentration of lysozyme inthe solution.X An Fab fragment binds to lysozyme with a dissociation constant of Ka =10-11 M. A 1 nM (10-9 M) solution of lysozyme istreated with increasing concentrations of the Fab fragment. At what concentration of added Fab will half of the lysozyme bebound to the Fab?[Fab]9.9IncorrectnM

We are given the binding kinetics between lysozyme (receptor) and Fab fragment (ligand).

Answered: = An Fab fragment binds to lysozyme…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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A different Fab fragment binds to lysozyme with a dissociation constant of Kd=10-6 M. A 1 nM (10-9 M) solution of lysozyme is treated with increasing concentrations of this Fab fragment. At what concentration of added Fab will half of the lysozyme be bound to this Fab?
Bovine ribonuclease folds with ΔH° = -280 kJ mol-1 and ΔS° = -0.79 kJ mol-1 K -1. Assume ΔH° and ΔS° are independent of temperature. What fraction of bovine ribonuclease is unfolded at 42 °C?
Draw the structure of pentapeptide Trp-Asp-Cys-Lys-Gln that would predominate in aqueous solution at pH=5.5. (Relevant pKa values are 2.2, 3.8, 8.3, 9.6 and 10.5.)
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The octapeptide gly-cys-met-asn-lys-ala-tyr-gly was hydrolyzed consecutively by CNBr and then trypsin (3 fragments total). The mixture of fragments was buffered at a pH 8.5 and then chromatographed on an anion-exchanger (positive resin) with the same buffer. Draw sequence (abbreviated form) of all fragments at that pH (show charges) and predict elution order.
The enzymatic activity of lysozyme is optimal at pH 5.2 and decreases above and below this pH value. Lysozyme contains two amino acid residues in the active site essential for catalysis: Glu35 and Asp52. The pK value for the carboxyl side chains of these two residues are 5.9 and 4.5 respectively. What is the ionization state of each residue at the pH optimum of lysozyme? How can the ionization states of these 2 amino acid residues explain the pH-activity profile of lysozyme?
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The ESI-MS spectrum in positive ionization mode for lysozyme is obtained. a. What is the molecular weight of the protein to 5 significant figures based on the two highlighted ion species? b. What is the charge of the peaks at 1101.5 and 1789.2.
The active site of lysozyme contains two amino acid residues essential for catalysis: Glu35 and Asp52. The pKa values of the carboxyl side chains of these residues are 5.9 and 4.5, respectively. What is the ionization state(protonated or deprotonated) of each residue at pH 5.2, the pH optimum of lysozyme? How can the ionization states of these residues explain the pH-activity profile of lysozyme shown below?
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7. 1 B e here to search Required Study the two diagrams below. Diagram A 2 1 Diagram B 2 Based on the sequence of steps (1, 2, 3), which of the diagrams shows what would happen to proteins made at the ribosomes and transported to the outside of the cell? Diagram A Diagram B Both Diagram A and Diagram B Neither Diagram A nor Diagram B DELL 40)
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= An Fab fragment binds to lysozyme with a dissociation constant of Ka 10-11 M. A 1 nM (10-9 M) solution of lysozyme is treated with increasing concentrations of the Fab fragment. At what concentration of added Fab will half of the lysozyme be bound to the Fab?