Dissociation constants of leucine
Among the $4$ equilibria, only $3$ are linearly independent, because we have $(1)+(3)=(2)+(4)$. Let's then choose only three independent reactions

\begin{align}
\ce{NH3+-C5H10-COOH}&amp;\ce{&lt;=&gt; NH3+-C5H10-COO- + H+},\\
\ce{NH3+-C5H10-COO-}&amp;\ce{&lt;=&gt; NH2-C5H10-COO- + H+},\\
\ce{NH3+-C5H10-COO-}&amp;\ce{&lt;=&gt; NH2-C5H10-COOH},
\end{align}

as the basis equations, which are obtained from $(2)$, $(4)$, and $(1)-(2)$. The equilibrium constants of these equations are $K_{a1}$, $K_{a2}$, and $K_{a2}/K_{a1}=10^{-7.24}\ll 1$. Therefore most of the neutral amino acid molecules would exist in the zwitterionic form $\,\ce{NH3+-C5H10-COO-}$. Neglecting the last equilibrium, the equilibrium constants $K_{a1}$ and $K_{a2}$ are those of the $\ce{-COOH}\,$ and $\ce{-NH3+}$ groups in the zwitterions $\,\ce{NH3+-C5H10-COO-}$, which is much more abundant than the original molecules $\,\ce{NH2-C5H10-COOH}\,$ making the latter negligible.