7. As shown in Figure 2, MN is the diameter of $\odot O$. If
$$
\begin{array}{l}
\angle E=25^{\circ}, \angle P M Q \\
=35^{\circ}, \text { then } \angle M Q P \\
=(\quad) .
\end{array}
$$
(A) $30^{\circ}$
(B) $35^{\circ}$
Figure 2
(C) $40^{\circ}$
(D) $50^{\circ}$
7.C.

As shown in Figure 2, $\overparen{N Q}^{\circ}+\overparen{Q P}^{\circ}+\overparen{P M}^{\circ}=180^{\circ}$,
$$
\begin{array}{l}
\overparen{P M}^{\circ}-\overparen{N Q}^{\circ}=2 \angle E=2 \times 25^{\circ}=50^{\circ}, \\
\overparen{P Q}^{\circ}=2 \times 35^{\circ}=70^{\circ} .
\end{array}
$$

Therefore, $\overparen{P M}^{\circ}=80^{\circ}$, which means $\angle M Q P=40^{\circ}$.