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Tricky Logical Q
Started by SonhosDaVida
I was wondering if anyone can shed some light on this this problem? Thank you
The isoelectric point (pI) is the pH at which a protein has no net charge.
Higher pH means the solution becomes more basic --> this means carboxyl groups lose protons (R-COOH --> R-COO-) and ammonium groups lose protons (R-NH3+ --> R-NH2)
So when pH > pI, the protein has more R-COO- and R-NH2 groups and becomes more negatively charged.
Higher pH means the solution becomes more basic --> this means carboxyl groups lose protons (R-COOH --> R-COO-) and ammonium groups lose protons (R-NH3+ --> R-NH2)
So when pH > pI, the protein has more R-COO- and R-NH2 groups and becomes more negatively charged.
Lower pH means the solution becomes more acidic --> this means carboxylate groups gain protons (R-COO- --> R-COOH) and amine groups gain protons (R-NH2 --> R-NH3+)
So when pH < pI, the protein has more R-COOH and R-NH3+ groups and becomes more positively charged.
Another way to think about this is acidic solutions have lots of H+ available that can be added to proteins to neutralize negative charges and make them more positive. Basic solutions have more OH- that neutralize positive charges and make proteins more negative.
Charge on a ProteinSo when pH < pI, the protein has more R-COOH and R-NH3+ groups and becomes more positively charged.
Another way to think about this is acidic solutions have lots of H+ available that can be added to proteins to neutralize negative charges and make them more positive. Basic solutions have more OH- that neutralize positive charges and make proteins more negative.
