A sequence of chains exhibits (total-variation) cutoff (resp., pre-cutoff) if for all , the ratio $t\mathrm{mix}(n)(ϵ)/t\mathrm{mix}(n)(1-ϵ)$ tends to 1 as $n \to \infty $ (resp., the $lim sup$ 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 $Gn$, such that the lazy simple random walks (LSRW) on $Gn$ satisfy the "product condition" $\mathrm{gap}(G_n) t_{\mathrm{mix}}^{(n)}(\epsilon) \to \infty $ as $n\to \infty$, where $\mathrm{gap}(Gn)$ is the spectral gap of the LSRW on $Gn$ (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 $Gn=(Vn,En)$ 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 $\Theta (\log |Vn|)$. Similarly, we also show that "lumping" states together may increase the order of the mixing-time by an optimal factor of $\Theta (\log |Vn|)$. This gives a negative answer to a question asked by Aldous and Fill.