Chemistry - Biomolecules Question with Solution | TestHub

ChemistryBiomoleculesAMINO ACIDEasy2 min

ChemistryEasyinteger

If pKa values for protonated Lysine molecule are $pKa_{1} = 2.18; pKa_{2} = 8.95; pKa_{3} = 10.79$ respectively then, Calculate the value of $(x - 1)$ where $x$ is iso-electric point of lysine.

Answer:

Solution:

Iso electric point of an amino acid with an acidic side chain is average of pKa values of two most acidic groups.

Hence pI = (10.79 +8.95)/2 = 9.87 = x

Hence x-1 is 8.87≈ 9

Stream:JEESubject:ChemistryTopic:BiomoleculesSubtopic:AMINO ACID

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