X là tetrapeptit Gly-Val-Ala-Val, Y là tripeptit Val-Ala-Val

* Hướng dẫn giải

Đáp án A

+Bản chất phản ứng: $\begin{array}{l} Gly-Val-Ala-Val + 4NaOH\to muối + H_{2}O \left(1\right)\\ mol: x \leftarrow 4x \to x\\ Val-Ala-Val + 3NaOH\to muóai + H_{2}O \left(2\right)\\ mol: y \leftarrow 3y \to y\\ + \left\{\begin{array}{l}{m}_{(X, Y)}=344x+287y=14,055\\ BTKL: \underset{m_{(X, Y)}}{\underbrace{14,055}}+\underset{m_{NaOH}}{\underbrace{40(4x+3y)}}=\underset{muối}{\underbrace{19,445}}+\underset{m_{H_{2}O}}{\underbrace{18(x+y)}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,02\\ y=0,025\end{array}\right.\\ \Rightarrow \%m_X=\frac{0,02.344}{14,055}.100\%=48,95\%\end{array}$