Microequilibrium constants for binding of metal to a protein. The iron-transport protein, transferrin, has two distinguishable metal-binding

Question:

Microequilibrium constants for binding of metal to a protein. The iron-transport protein, transferrin, has two distinguishable metal-binding sites, designated a and b. The microequilibrium formation constants for each site are defined as follow
Microequilibrium constants for binding of metal to a protein. The

For example, the formation constant k1a refers to the reaction Fe3+ + transferring Featransferrin, in which the metal ion binds to site a:

Microequilibrium constants for binding of metal to a protein. The

(a) Write the chemical reactions corresponding to the conventional macroscopic formation constants, K1 and K2.
(b) Show that K1 = k1a = k1b and K2 -1 = k2a-1 + k2b-1.
(c) Show that k1ak2b = k1bk2a. This expression means that, if you know any three of the microequilibrium constants, you automatically know the fourth one.
(d) A challenge to your sanity. From the equilibrium constants below, find the equilibrium fraction of each of the four species in the diagram above in circulating blood that is 40% saturated with iron (that is, Fe/transferrin = 0.80, because each protein can bind 2 Fe).
Effective formation constants for blood plasma at pH 7.4

Microequilibrium constants for binding of metal to a protein. The

The binding constants are so large that you may assume that there is negligible free Fe3+. To get started, let's use the abbreviations [T] = [transferrin] + [FeT] = [FeaT] = [FebT], and [Fe2T = [Fe2transferrin]. Now we can write
Mass balance for protein: [T] + [FeT] + [Fe2T] = 1 (A)
Mass balance for iron:

Microequilibrium constants for binding of metal to a protein. The
Microequilibrium constants for binding of metal to a protein. The

Now you have three equations with three unknowns and should be able to tackle this problem.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: