Biochemistry question? I really need a DETAILED explanation on how to find the n
ID: 845543 • Letter: B
Question
Biochemistry question? I really need a DETAILED explanation on how to find the net charge of Asp amino acid at pH of 1.0, 3.0, 6.0 and 11.0? I know the pKa values are 1.99 (carboxyl group), 9.90 (amine group) and 3.90 (side chain). I know that you look at the groups and if the pH < pKa they're protonated, if pH > pKa they're deprotonated but I just can't understand how to assign the charges for the different groups and add them up. Example, for Asp at pH 3, I'm getting (+1 plus +1 plus -1) = +1 which is incorrect because the overall net charge is really equal to 0. PLEASE help me clarify and understand this.
Explanation / Answer
Henderson Hasselbach equation is expressed as follows :
pH = pKa + log10(a/1-a)
Here, 'a' is degree of dissociation
In the given amino acid, we have 3 different parts which undergo dissociation ( deprotonation ) :
Carboxyl group: -COOH with pKa 1.99
Amine group: NH3+ with pKa 9.9
Side chain containig OH: with pKa 3.9
At pH = 3
For the carboxyl group : complete dissociation ( COOH to COO- ) gives -1 charge
Putting the values of pH and pKa in above eqn and solving for 'a' , we get :
a = 0.91
This means at this pH the carboxyl group is about 90% dissociated, thus it will give a charge of -0.9 , and not -1 because it is not 100% dissociated
For the amine group : complete dissociation ( NH3+ to NH2 ) gives 0 charge
Putting the values of pH and pKa in above eqn and solving for 'a' , we get :
a = 1.25*10-7
This means at this pH the carboxyl group is almost undissociated, thus it will give a charge of +1
For the side chain OH group : complete dissociation ( OH to O- ) gives -1 charge
Putting the values of pH and pKa in above eqn and solving for 'a' , we get :
a = 0.11
This means at this pH the side chain group is about 10% dissociated, thus it will give a charge of -0.1 , and not -1 because it is not 100% dissociated
Thus, adding all the charges, we have : net charge = -0.9 + 1 - 0.1 = 0
In this way, you can calculate for other pH values as well. Just remember to calculate 'a' correctly
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