BCMB/CHEM 8190 ANSWERS
PROBLEM SET 7

1)

a) dV = 1/(pi T2*) + 1/(pi tau) = 0.5 + 1/(pi tau) = 1.0; tau = 0.64s

b) dV = 1/(pi T2*) + (tau/(8 pi))(wa-wb)**2 = 0.5 + (tau/(8 pi))(2pi*100)**2 = 5.0;

tau = 2.9x10**(-4)s

2) Dissociation constant = 2X10^-4 X 5X10^-6 = 1X10^-9

3) The relative rate of build-up of NOEs will be 300,000/2000. A 20% contribution from the free NOE would occur with a free to bound ratio of X/(X-1) where X(2)/((1-X)300) = 0.2. Assuming the ligand binding constant is relatively high, 20% would be bound at a 30:1 ligand to protein ratio.

4) 1/T1(Mn) - !/T1(Mg) is proportional to 1/r**6 where r is the metal site to proton distance.  Using two distances to eliminate the proportionality constant,  r(A8)**6/R(a9)**6 = 1.5/0.167.  This implies r(A8)/r(A9) = 1.442.  Since G& and A9 are on a straight line and see equal effects, the A8-metal vector is perpendicular to the A8-A9 vector.  Relating the lengths of the sides of this right triangle we get a metal to A8 distance of 4.81 and a metal to A9 distance of 6.94.