Cysteine has three groups that ionise: the α-COOH group (pK = 1.96), the thiol group of the side…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Cysteine has three groups that ionise: the α-COOH group (), the thiol group of the side chain (, neutral when protonated) and the α-NH3+ group (). Use these to find the isoelectric point. Then find the net charge of cysteine at , to two decimal places. Take the carboxyl group as fully ionised, and the ammonium group as protonated.
Need a reminder? The key ideaShow
The isoelectric point is the pH at which the average charge is zero. For a neutral side chain, . For other amino acids, average the two pK values on either side of the neutral form. The fraction of a group that is still protonated is for a group that is positive when protonated.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. List the charges from the acid side to the base side.
Step 1. The charges. Below 1.96: (COOH, NH3+, SH). Between 1.96 and 8.18: (COO-, NH3+, SH). Between 8.18 and 10.28: (COO-, NH3+, S-). Above 10.28: .
Step 2. pI. The neutral form lies between 1.96 and 8.18. So .
Step 3. Charge at pH 7.0. The ammonium group is . The thiol is deprotonated by , so it gives . The carboxylate gives .
Step 4. Net. .
Answer: pI and the net charge at pH 7.0 is .
Check. The pH of 7.0 is above the pI of 5.07, so a small negative charge is expected.
Watch out. The side-chain pK is used here, not the α-NH3+ pK, because the neutral form is bounded by the first two ionisations.
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