## (solution) Can you please help me ? I saw that you had helped someone here

Can you please help me ? I saw that you had helped someone here before.

Problem 1.  Rate of dissociation, Rate of association and Kd(4 points total)

1 ml samples of tissue membranes were incubated overnight in a nearly saturating amount (100 nM) of radioligand L*.  At time 0, 1000 x [L*] of unlabeled ligand (100 mM) is added to each sample, and then bound ligand is separated from unbound (and non-specific binding is subtracted by using an excess competing ligand before adding the radioligand  in separate samples) at 30 minute intervals to give the following counts of specifically bound ligand as a function of time:

 min Specific CPM bound 0 662550 30 538158 60 437120 90 355052 120 288392 150 234247 180 190268 210 154545

a. Calculate K-1  (unit = min-1) if the specific activity of the ligand is 150 Ci/mMole, 1 Ci = 2.22 x 1012 disintegrations/min, and the efficiency of the counter is 0.5 counts/min per disintegration/min.  Calculate the number of CPM (Counts per Minute) expected for 1 pmole of bound ligand.  Assume that [LR]eq = binding at time 0.  Show your plot of the data and how you did your calculations.

b. Why does the concentration of radioligand not matter in this experiment, as long as almost all the binding sites are occupied at time 0, and that the nonspecific binding is not so high as to cause large errors in determining specific binding?

c. From k-1, calculate the T½ of L* unbinding from the receptor. T½ = -ln(2)/-k-1 for dissociation.

d. What technical problems might you have (if any) if your binding assay requires a 20 minute centrifugation step to separate bound from unbound? Calculate or estimate how much of a problem it would be

.

Problem 1, Part 2.  At time zero, 1 ml samples of the same tissue membranes used in problem 1 are mixed with 10 nM of radioligand L*, and then the counts bound to the membranes are determined at 2 minute intervals.  After subtracting nonspecific binding (determined in other samples), the following data are obtained:

 Time (min) Specific CPM bound 0 0 2 155836 4 272312 6 359369 8 424437 10 473071 12 509421 14 536590 16 556897 18 572075 20 583419

e. Use 10 nM [LR]eq  = 3.70563828 pMole (derived from a Scatchard plot) and the K-1 value from problem 1a to calculate K+1 (unit = L/mole-1 min-1), and then from both values, calculate Kd.   Show your graph and your work.  The values for the radioligand, the volume, and the counting efficiency are the same as above and throughout this entire problem set questions 1-3.  Hint: remember the difference between moles and molar.

f. Calculate the T½ of the ligand binding to the receptor at 10 nM ligand. T½ of association = ln(2)/Kon

g.  What would the graph look like for [L*] nM? Plot the data and calculate the T½ of binding under these conditions.

h.  If you were setting up an equilibrium binding assay at 1 nM radioligand would 1 hour of binding be sufficient?  Explain your answer.  Also give an estimate of how long you would expect to wait to have 87.5% of binding at equilibrium (remember T1/2 = time to reach 50% of binding at equilibrium, so 87.5% = 3 x T1/2, or Percent binding (1 - 1/(2x)), where x is the number of elapsed T1/2s as long as x>0). How long would it take to reach approximately 99% of the equilibrium binding?

Solution details:
STATUS
Answered
QUALITY
Approved
ANSWER RATING

STATUS

Answered

QUALITY

Approved

DATE ANSWERED

Sep 13, 2020

EXPERT

Tutor

ANSWER RATING

### Order New Solution. Quick Turnaround

Click on the button below in order to Order for a New, Original and High-Quality Essay Solutions. New orders are original solutions and precise to your writing instruction requirements. Place a New Order using the button below.

WE GUARANTEE, THAT YOUR PAPER WILL BE WRITTEN FROM SCRATCH AND WITHIN A DEADLINE. 