Find the isoelectric point (pI) of the given | vedclass

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(B) The isoelectric point $(pI)$ is calculated using the formula: $pI = \frac{pK_{a1} + pK_{a2}}{2}$.Given $pK_{a1} = 2.2$ (for the carboxyl group).Gi

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$5.9$

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(B) The isoelectric point $(pI)$ is calculated using the formula: $pI = \frac{pK_{a1} + pK_{a2}}{2}$.Given $pK_{a1} = 2.2$ (for the carboxyl group).Given $pK_b = 4.4$ for the amino group,we find $pK_{a2}$ using the relation $pK_{a2} = 14.0 - pK_b = 14.0 - 4.4 = 9.6$.Substituting these values into the formula: $pI = \frac{2.2 + 9.6}{2} = \frac{11.8}{2} = 5.9$.

Find the isoelectric point $(pI)$ of the given amino acid.
$CH_3-CH(NH_3^+)-CO_2H, \ pK_a = 2.2, \ pK_b = 4.4$

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Find the isoelectric point $(pI)$ of the given amino acid.$CH_3-CH(NH_3^+)-CO_2H, \ pK_a = 2.2, \ pK_b = 4.4$