Let $Y\subset \mathbb{C}P^4$ be the quintic threefold given by the equation $$X^5_0+X^5_1+X^5_2+X^5_3+X^5_4+5X_0X_1X_2X_3X_4=0$$ it has 125 singular points whose links are homeomorphic to $S^2\times S^3$. This threefold is a deformation of a smooth Calabi-Yau manifold $Y_{\epsilon}$ $$X^5_0+X^5_1+X^5_2+X^5_3+X^5_4+5(1+\epsilon)X_0X_1X_2X_3X_4=0$$ The Betti numbers of $Y$ are: $b_0=b_6=1$, $b_1=b_5=0$, $b_2=1$, $b_3=103$, $b_4=25$, those of $Y_{\epsilon}$ are: $b_0=b_6=1$, $b_1=b_5=0$, $b_2=b_4=1$, $b_3=204$.

For a small value of $\epsilon$ one can perform a conifold transition:

one can contract 125 3-spheres in $Y_{\epsilon}$ and get $Y$,

each singular point of $Y$ is the vertex of a cone whose base is $S^2\times S^3$, we can replace each of this singular point by a copy of $\mathbb{C}P^1$, we get a Calabi-Yau manifold $Y_s$ wich is a small resolution of $Y$.

I have some very naive questions on conifold transitions:

In this case conifold transitions depend on choices of 125 3-cycles which are 125 embedded 3-spheres in $Y_{\epsilon}$ whose contractions kill a copy of $\mathbb{Z}^{\oplus 101}$ in $H_3(Y_{\epsilon},\mathbb{Z})\cong \mathbb{Z}^{\oplus 204}$.

1) How do we choose these 125 cycles to get $Y$? More precisely, we need at least 101 non-trivial homological 3-spheres, a naive choice would be to pick 101 copies of $S^3$ representing the 101 generators of $\mathbb{Z}^{101}\subset H_3(Y_{\epsilon},\mathbb{Z})$ and 24 $S^3$ which are homologicaly trivial, but I guess this is too naive. I suppose that the 3-cycles need to satisfy some relations what are these relations? Are these homological conditions if they exist sufficient?

2) I suppose that the small resolution also depends on these choices, in particular do we know the cohomology algebra of the small resolution. I know the Hodge Diamond of the cohomology of $Y_s$ but I would like to know its algebra structure.