Is a family's Turán number governed by one bipartite member? (Erdős #575)

For a finite family $\mathcal{F}$ of finite graphs, let $\mathrm{ex}(n;\mathcal{F})$ be the maximum number of edges in an $n$-vertex graph containing no member of $\mathcal{F}$ as a subgraph. Trivially $\mathrm{ex}(n;\mathcal{F})\le\mathrm{ex}(n;G)$ for every $G\in\mathcal{F}$. Conjecture: for every such $\mathcal{F}$, if $\mathcal{F}$ contains at least one bipartite graph, then there exists a bipartite $G\in\mathcal{F}$ with $$\mathrm{ex}(n;G)\ll_{\mathcal{F}}\mathrm{ex}(n;\mathcal{F}),$$ where $\ll_{\mathcal{F}}$ hides a constant depending only on $\mathcal{F}$. Combined with the trivial inequality this says $\mathrm{ex}(n;\mathcal{F})\asymp_{\mathcal{F}}\mathrm{ex}(n;G)$: the family's Turán number is, up to a constant factor, already determined by a single bipartite member.

Acceptance. FULLY RESOLVES: a complete proof that for every finite family $\mathcal{F}$ containing a bipartite graph there is a bipartite $G\in\mathcal{F}$ with $\mathrm{ex}(n;\mathcal{F})\gg_{\mathcal{F}}\mathrm{ex}(n;G)$ (Lean/Coq preferred, else a full written proof); OR a counterexample — an explicit finite family $\mathcal{F}$ containing a bipartite graph, with a proof that $\mathrm{ex}(n;\mathcal{F})/\mathrm{ex}(n;G)\to 0$ for every bipartite $G\in\mathcal{F}$ (a genuine asymptotic separation, not merely a constant-factor one). ADVANCES: prove the conjecture for a structured subclass — e.g. two-graph families $\mathcal{F}=\{G_1,G_2\}$, families whose bipartite member is a complete bipartite graph $K_{s,t}$, or families of trees — with complete proof; or reduce it to a known statement about Turán exponents of bipartite graphs. Deliver the proof file, or the counterexample family together with the separation proof.