 Solution

Take Euclidean gamma matrices with $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\delta }_{\mu \nu }$, and independent Grassmann fields $\psi ,\bar{\psi }$. The minimally coupled Dirac action isIts mass term is the Dirac mass term, and the interaction is $-ig{\bar{\psi }}^a{\gamma }_{\mu }{A}_{\mu }^b(T_{\mathrm{ad}}^b{)}_{ac}{\psi }^c$. With Fourier convention $e^{ip\cdot x}$, the free kinetic operator is $i/p+m$. The Euclidean adjoint Dirac Feynman rules are thereforeThe vertex sign follows by expanding $e^{-S_{\psi }}$, not $e^{i{S}_{\psi }}$. Momentum conservation supplies the usual Dirac delta function at each vertex. There is only a one-gluon fermion vertex, with no two-gluon seagull because the fermionic covariant derivative is first order. A closed fermion loop has the usual fermionic sign $-1$. The identity $(i/p+m)(-i/p+m)=(p^2+m^2)I$ checks the Dirac propagator.