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

answer; i believe the correct answer is (solid gold);

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.