In Zee Chapter VII.5 QFT book:

He showed that the standard model fermions form $5^*$ and $10$ representations of $SU(5)$:
$$
5^* \oplus 10.
$$
In particular, the $10$ is the anti-symmetric matrix representation from the tensor product of $5 \otimes 5 $:
$$5 \otimes 5 = 10 \oplus 15$$
or 
$$(5 \otimes 5)_A = 10  $$

Question 

1. If the fermion representation $10$ is from the anti-symmetric  part of $5 \otimes 5$, can the fermion $10$ be the composite of fermion $5^*$?

In Zee's notation, can we regard the fermion $10$ (the left-handed up and down quark doublet, the right handed $u$ quark, the right-handed electron) as the bound state of some sort of two $5$ (each $5$ forms the left-handed neutrino and electron doublet, the right handed $d$ quark):
$$
\psi^{\mu\nu} = (\psi^{*\mu}\psi^{*\nu}-\psi^{*\nu}\psi^{*\mu})^*?
$$

2. How does the fermion statistics match on the left and right handed sides?