Yes, this is possible. If $M_1^\#$ exists and is $\omega_1$-iterable then $x^\#$ exists for all reals $x$: By genericity iterations (either Woodin's or Neeman's version), for any real $x$ there is a countable iterate $N$ of $M_1^\#$ so that  $x$ is generic over $N$ for a forcing of size the Woodin cardinal of $N$. So $N[x]$ still "has a sharp" and $x^\#$ exists. 

If now, e.g. $M_1^\#$ exists, is $\omega_1$-iterable and $V=L[A]$ for a set $A$, then in the extension by $\mathrm{Col}(\omega, A)$, $(M_1^\#)^V$ cannot be $\omega_1$-iterable.