A sequence of chains exhibits (total-variation) cutoff (resp., pre-cutoff) if for all $0<ε< 1/2$, the ratio $t_{\mathrm{mix}}^{(n)}(ε)/t_{\mathrm{mix}}^{(n)}(1-ε)$ tends to 1 as $n \to \infty $ (resp., the $\limsup$ of this ratio is bounded uniformly in $ε$), where $t_{\mathrm{mix}}^{(n)}(ε)$ is the $ε$-total-variation mixing-time of the $n$th chain in the sequence. We construct a sequence of bounded degree graphs $G_n$, such that the lazy simple random walks (LSRW) on $G_n$ satisfy the "product condition" $\mathrm{gap}(G_n) t_{\mathrm{mix}}^{(n)}(ε) \to \infty $ as $n \to \infty$, where $\mathrm{gap}(G_n)$ is the spectral gap of the LSRW on $G_n$ (a known necessary condition for pre-cutoff that is often sufficient for cutoff), yet this sequence does not exhibit pre-cutoff. Recently, Chen and Saloff-Coste showed that total-variation cutoff is equivalent for the sequences of continuous-time and lazy versions of some given sequence of chains. Surprisingly, we show that this is false when considering separation cutoff. We also construct a sequence of bounded degree graphs $G_n=(V_{n},E_{n})$ that does not exhibit cutoff, for which a certain bounded perturbation of the edge weights leads to cutoff and increases the order of the mixing-time by an optimal factor of $Θ(\log |V_n|)$. Similarly, we also show that "lumping" states together may increase the order of the mixing-time by an optimal factor of $Θ(\log |V_n|)$. This gives a negative answer to a question asked by Aldous and Fill.