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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.

Answer :

gabrielcardoso

Answer:

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

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