The tripeptide glycylarginylglutamate contains four ionizable groups with pKas of 2.1, 4.1, 9.8, and 12.5. Calculate the pI for this molecule.

The isolectric point for this molecule is 7.1

Explanation:

First, list the pka states that the tripeptide glycylarginylglutamate which can be found

pKa_1 = 2.1

pKa_2 = 4.1

pKa_3 = 9.8

pKa_4 = 12.5

Now it is necessary to find the isoelectric point (pI)

pl = SUM(pKa_1 + ... + pka_n)/n

pl = (2.1 + 4.1 + 9.8 + 12.5)/4

pl = 7.1

The isolectric point for this molecule is 7.1

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error