Given the curves below, what are the dissociation constants for proteins A and B? 100 80- 60 40 20 0 0 10 20 30 Concentration (M) A=28, B=3 A=3, B=28 Cannot determine Kp from this plot O A=10, B=2 B A=2, B=10 A 40
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- 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)Based on the Hill Plot below, this protein-ligand interaction displays log Y 1-Y 5 2.5 0 -2.5 -2.5 nH1 2.5 Binding that obeys the Law of Mass Action Negative cooperativity 5Three 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 number of low affinity binding sites occupied by a ligand at 100% saturation is greater than the number of high affinity binding sites occupied by a ligand at 100% saturation true or false?From the Hill Plot below, the of the first binding event for the receptor-ligand system under study is: Q1 4 nM 10 μΜ -2 nM 2 nM Calculate the Hill Coefficient from the receptor-ligand binding data below: Q2 4 100 2 3 (077) log 10 8 6 4 2 0 -2 -4 -6 -6 -4 -2 0 2 4 log [L] (nM) 6 8 10From the Hill Plot below, the of the first binding event for the receptor-ligand system under study is: Q1 4nM Ο 10 μΜ -2 nM 2 nM Calculate the Hill Coefficient from the receptor-ligand binding data below: Q2 4 100 2 3 (0-1) 60 log 10 8 6 4 2 0 -2 -4 -6 -6 -4 -2 0 2 4 log [L] (nM) 6 8 10
- The dissociation equilibrium constant (Kd) is:_____ (select all that apply) Group of answer choices: Is a measure of ligand efficacy Typically equivalent to 1- Ka The concentration of ligand that gives maximal receptor occupancy The concentration of ligand that gives half-maximal receptor occupancy Measured in 1/M or M-1 Typically measured in mol/L or M A measure of the strength of a ligand - receptor binding reactionYou are a technician in a biochemistry lab running receptor binding experiments. The target membrane-bound receptor has been partially purified from mouse, rat, and human cell lines. Using the same radioactive ligand in a saturation binding assay for each species' receptor, you generate the binding data in the table. The dependent variable, Y, is the fraction of binding sites occupied by the ligand. Ligand Y for mouse Y for rat Y for human concentration receptor receptor receptor (nM) 0.20 0.048 0.29 0.17 0.50 0.11 0.50 0.33 1.0 0.20 0.67 0.50 4.0 0.50 0.89 0.80 10 0.71 0.95 0.91 20 0.83 0.97 0.95 50 0.93 0.99 0.98 Determine the Ka for the human receptor in this binding experiment. human receptor Kd nMAn engineered ligand binds its target with 5 nM affinity at high pH. However, protonation of a histidine residue in the binding site of the ligand renders it unable to bind. Plot the ligand:target complex concentration versus the initial ligand concentration (in protonated or unprotonated form) at pH 4, 5, 6, 7, or 8 (five lines on one plot). Use a total target concentration of 10 nM.
- An experiment was carried out to measure the reaction rate of hydrolysis of acetylcholme (substrate) with serum enzymes (Eadie, 1949). In the experiment, two experiments were conducted, namely experiment 1 without using a prostigmine inhibitor and experiment 2 using a prostigmine inhibitor at 1.5 x 10^-7 mol/l. the data obtained are: a. Is prostigmine competitive or noncompetitive inhibitor? b. determine the value of km and rmax for the two experiments, compareBiomedical Engineering: An equilibrium dialysis experiment was performed to characterize the binding affinity of a mouse IgG or human IgG antibody for an antigen, GAD65. (The association constant for the monoclonal mouse IgG is 4.75 x 10^8 M-1 and for a human IgG is 1.3 x 10^10 M-1) a- calculate the dissociation constant for each antibody and explain which antibody has a higher binding affinity for GAD65? b-calculate the fraction of free Ab sites for each antibody for a For a GAD65 concentration of 0.7 nMTwo co-op students at your start-up company have been asked to evaluate the rate of reaction occurring in a transparent gel particle containing immobilized mouse melanoma cells. The equation for the reaction rate ris is: Vmax T'As Km+CAs where R=3.2 mm, Vmax= 0.12 gmol s' m3, CAS = 41 gmol m3, and Km=0.8 gmol m3. One student reports a reaction rate of 1.6x10 gmol st; the other reports 1.6x10 1º gmol s1. You left your calculator on the bus this morning, but must know quickly which student is correct. Use an order-of- magnitude calculation to identify the right answer. Which student is correct?