Thủy phân hoàn toàn a mol Gly-Ala-Val-Glu trong dung dịch NaOH dư, có b

$$\begin{gathered} Gly-Ala-Val-Glu+3H_{2}O \\ \to Gly+Ala+Val+Glu \\ Gly-H_{2}N-C{H}_2-COOH+NaOH \\ \to H_{2}N-C{H}_2-COONa+H_{2}O \\ Ala H_{2}N-CH\left(C{H}_3\right)-COOH+NaOH \\ \to H_{2}N-CH\left(C{H}_3\right)-COONa+H_{2}O \\ Val H_{2}N-CH\left(C_3{H}_7\right)-COOH+NaOH \\ \to H_{2}N-CH\left(C_3{H}_7\right)-COONa+H_{2}O \\ Glu H_{2}N-C_3{H}_5{\left(COOH\right)}_2+2NaOH \\ \to H_{2}N-C_3{H}_5{\left(COOH\right)}_2+2H_{2}O \end{gathered}$$

$$\begin{gathered} Gly-Ala-Val-Glu+5NaOH \\ \to Gly-Na+Ala-Na+Val-Na+Glu-Na+2H_{2}O \end{gathered}$$