1. Human hemoglobin is a tetrameric protein, under normal conditions. This is to

Question

1. Human hemoglobin is a tetrameric protein, under normal conditions. This is to say that it is loosely bound complex made of four monomeric hemoglobin molecules. Suppose one mole of tetrameric hemoglobin has entropy 10R, where R is the universal gas constant. Imagine now this tetrameric hemoglobin has fully dissociated into monomeric hemoglobin. Ignoring the rotational and vibrational entropy and pretending hemoglobin is an ideal gas, what is the entropy change during this process, in units of R?

2. In the preceding problem: What is energy change during the process, assuming the bond energy is, numerically, 10kBT per tetramer?

Ignore vibrational energy and give answer in terms of RT.

Explanation / Answer

Question 1

According to the question,

we have to unfold following equation i.e -RTInKF = -TS = -T (S2 -S1) equation a

where R is gas constant ; T is constant temperature ; S2 is entropy of polypetides ; S1 is entropy of unfolded protein

Step 1: Find KF

For a molecule with two non-identical subunits, at any temperature, T, the constant of folding, KF, is:

KF = molar concentration of unfolded protein / (molar concentration of subunit 1)n (moloar concentration of subunit 2)n

where n = no. of each subunit

In our case, hemoglobulin has 4 polypeptides of which two are identical,

As protein gets fully denatured into 4 polypeptides, so molar concentration of each polypeptide will be 2 molar   

Hence, KF = 1 / (2)2 (2)2 = 1 / 16 = 0.0625

Step 2: Find S2

S1 = 10R

putting values in equation a, we get

S2 = 8.8R

Hence, change in entropy = 8.8R - 10R = - 1.2   

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