[qz465]

Hi Everyone,
Here is a question I am stuck on.

Reduced NAD emits fluorescent light at 460 nm when it is excited at 340 nm, and this fluorescence has been found to be enhanced in the presence of lactate dehydrogenase. This effect was used to determine the stoichiometry of binding.

To 2ml of a 0.473 mg/ml solution of beef-heart lactate dehydrogenase in a 0.05M Tris-acetate buffer, pH 7.2, were added small amounts of 10-3 M NADH solution in the same buffer.
The fluorescence emission at 460 nm was measured (in arbitrary units) after each addition and, as a control, readings were also made of the fluorescence changes when the NADH was added to 2ml of the buffer

http://imgur.com/qrWhDqU

The concentrations of protein in the effluent were estimated by measurements of the absorbance at 280 nm, and the volume of buffer necessary to elute each protein form the column was thus determined.

http://imgur.com/5tEaq9o

I started by using the gel-filtration data to estimate the MW of LDH at 150000 g/mol. I did this just by inspection and making comparisons.

After determining this, I found the number of moles of LDH present in the solution:

(0.473mg/mL)(2mL) = 0.946 mg LDH

0.946 mg / 150000 g/mol = 6.31 x 10^-9 mol LDH.

Then I thought that if I could find the point at which adding additional NADH produces the same effect as not having LDH in the solution at all, at this point all of the LDH would be bound to NADH. This occurs when 35 μL of NADH have been added (For example, adding a total of 40 μL of NADH in the LDH solution produces the same flourescense as 35 μL + 5 μL of NADH in a solution containing no LDH) so:

6.31 x 10^-9 mol LDH/((1x10^-3 mol/L)(35x10^-6 L)) = 0.180.

This ratio seems to make no sense. How can LDH bind to NADH in a ratio that is less than 1:1?

I would appreciate any help

[Yggdrasil]

Proteins can exist as multimeric complexes that contain multiple binding sites per molecule.  For example, hemoglobin has four oxygen binding sites.

As for the math, I haven't checked the numbers, but your approach is roughly correct.  Your estimate of the total amount of NADH bound by LDH is slightly higher than it should.  Your method assumes that all NADH is immediately bound by LDH until LDH is saturated.  However, binding reactions are equilibrium reactions, so there will always be some free NADH in addition to the bound NADH.

[Babcock_Hall]

Your MW estimate for LDH is in the right ballpark.  It might be possible to analyze the binding data in a more rigorous way, but I am not sure how to go about that process.  I am also not certain that doing so was expected of you.