pI Isoelectric Pt. Calculation

First, i use "my way" to give u the answer and help u understand this question. And u can do it in your own way, if u got the key point.


1.At the isoelectric point, which means its net charge is zero.

2. The groups that' ll affect Alanyl Histidine' s net charge is Ala' s -NH3+ , His' R group, and His' -COOH (Cause the peptide is formed)

Now start to calculate, try to find when it' s net charge is zero.

(a)Assume that its ionic groups are all protonated(e.g. pH <1.8)
which means its net charge is +2
(b) when pH arise over 1.8(but < 6), the H+ leaves the -COOH of His (C- ter of Alanyl Histidine) which makes its net charge becomes + 1
(c) when pH arise over 6 , the H+ leaves the R group of His, which makes
the net charge become very close to zero.
(d)When pH arise over 9.9 , the H+ leaves the -NH3+ of Ala(N- ter of Alanyl Histidine), which makes its net charge becomes -1

See the pic below,

+2 ----1.8---> +1 ----6---> 0 ----9.9---> -1

So Alanyl Histidine' s isoelectric point is about : (9.9 + 6)/2 = 7.95

I just gave u the way of thinking this question, next time when u' r doing this kind of question, maybe u can get the ans within a minute. But of course, you have to know the amino acids' structure very well. (Oh yeh and of course the basic calculating of pH, pKa... )


Any question ?

2011-09-25 22:15:05 補充：
oops ... some typing mistakes...
"Now start to calculate, try to find out when its net charge is zero."