Let $(X,x)$ be an affine variety with a normal isolated singularity and $Y$ be a smoothing of $X$ (for this we mean $Y$ can be realized as a smooth fibre of a deformation of $X$).
Question. Can we say that the topology of $Y$ is nontrivial, i.e. $\pi_i(Y)\neq \{1\}$ for some $i\geq 1$? Here the topology is induced from the analytic structure instead of the Zariski topology.
Edit. This does not answer the question, but it just provides examples of homotopically non-trivial smoothings $Y$ in every dimension $n \geq 1$. Perhaps someone might find it useful, hence I will not delete it.
Consider the nodal quadric hypersurface $$X=\{x_1^2+x_2^2+ \cdots + x_{n+1}^2=0\} \subset \mathbb{C}^{n+1},$$ admitting the smoothing family $$X_t=\{x_1^2+x_2^2+ \cdots + x_{n+1}^2-t=0\}.$$ Set $Y=X_1$, which is a smooth quadric in $\mathbb{C}^{n+1}$.
By [1], $Y$ is diffeomorphic to the tangent bundle $TS^{n}$ which is, in turn, homotopically equivalent to $S^n$, see [2]. So we get $$\pi_n(Y)=\pi_n(S^n)=\mathbf{Z}.$$
References.
[1] https://math.stackexchange.com/questions/1784898/tangent-bundle-of-sphere-as-a-complex-manifold
[2] https://math.stackexchange.com/questions/2034443/tangent-bundle-and-manifold-are-homotopy-equivalent
