4. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B$ $=2, A A_{1}=A D=1$, points $E$, $F$, and $G$ are the midpoints of edges $A A_{1}$, $C_{1} D_{1}$, and $B C$ respectively. Then, the volume of the tetrahedron $B_{1} E F G$ is ( ).
(A) $\frac{1}{4}$
(B) $\frac{3}{8}$
(C) $\frac{1}{2}$
(D) $\frac{5}{8}$
4.B.

Take a point $H$ on the extension of $D_{1} A_{1}$ such that $A_{1} H=\frac{1}{4}$. It is easy to prove that $H E / / B_{1} G, H E / /$ plane $B_{1} F G$. Therefore,
$$
V_{B_{1}-E F C}=V_{E-B_{1} F C}=V_{H-B_{1} F G}=V_{G-B_{1} F H} .
$$

And $S_{\triangle B_{1} F H}=\frac{9}{8}$, the distance from point $G$ to plane $B_{1} F H$ is 1. Hence $V_{B_{1}-E F C}=\frac{3}{8}$.