A model $\mathfrak{M} = (M, E,...)$ of Zermelo-Fraenkel set theory ZF is said to be $\theta$-like, where E interprets $\in$ and $\theta$ is an uncountable cardinal, if $|M| = \theta$ but $|\{b \in M: bEa\}| < \theta$ for each $a \in M$. An immediate corollary of the classical theorem of Keisler and Morley on elementary end extensions of models of set theory is that every consistent extension of ZF has an $\aleph_1$-like model. Coupled with Chang's two cardinal theorem this implies that if $\theta$ is a regular cardinal $\theta$ such that $2^{<\theta} = \theta$ then every consistent extension of ZF also has a $\theta^+$-like model. In particular, in the presence of the continuum hypothesis every consistent extension of ZF has an $\aleph_2$-like model. Here we prove: THEOREM A. If $\theta$ has the tree property then the following are equivalent for any completion T of ZFC: (i) T has a $\theta$-like model. (ii) $\Phi \subseteq T$, where $\Phi$ is the recursive set of axioms $\{\exists \kappa(\kappa$ is n-Mahlo and "$V_{\kappa}$ is a $\Sigma_n$-elementary submodel of the universe"): $n \in \omega\}$. (iii) T has a $\lambda$-like model for every uncountable cardinal $\lambda$. THEOREM B. The following are equiconsistent over ZFC: (i) "There exists an $\omega$-Mahlo cardinal". (ii) "For every finite language $\mathscr{L}$, all $\aleph_2$-like models of ZFC($\mathscr{L}$) satisfy the scheme $\Phi(\mathscr{L})$.