How does a supersymmetry transformation relate bosons to fermions? A superfield is defined as a function of of a spacetime coordinate and two spinor coordinates, $\Phi (x, \theta, \theta^*)$. A supersymmetric transformation is then generated by the operator $\partial_{a}+i\sigma^{\mu}_{a\dot{c}}\theta^{\dot{c}}\partial_{\mu}$.
What does it mean when we say this transformation relates fermionic to bosonic fields? Does the parameter $\theta$ measure the 'bosonic' content of a superfield?
 A: A superfield is a function of the bosonic coordinates $x^\mu$ as well as the supercoordinates $\theta^a$ and $\theta^{\dot c}$. It may be Taylor-expanded in the latter. The Taylor expansion in terms of anticommuting, Grassmann variables is inevitably finite because the square of each Grassmann variable is already zero, e.g. $\theta^1\theta^1=0$.
Consequently, the expansions of the superfields have a finite number of terms such as 
$$ \phi(x,\theta) = h(x) + \theta^a \psi_a + \theta^{\dot c}\psi_{\dot c} + \theta^a\theta^b \epsilon_{ab} H +\dots $$
Note that both sides of the equation above are Grassmann-even, bosonic fields, which implies that the coefficients in the terms on the right hand side with an even number of $\theta$-like factors are bosonic, i.e. Grassmann-even, for example $h,H$, while those with an odd number such as $\psi_a$ are Grassmann-odd i.e. fermionic.
If you act on the expression above by $\partial / \partial \theta^a$, it erases the $\theta^a$ factor (and if there's none, the derivative of the term vanishes). Consequently, the $\theta$-derivative of e.g. $\phi(x,\theta)$ produces terms like $\psi_a$ without any $\theta$ factor.
So if the original superfield is varied by a multiple of the $\theta$-derivative of itself,
$$\delta\phi = \epsilon^a \psi_a+\dots $$
you see that the term proportional to $\psi_a$ has the same $\theta$-structure (namely no $\theta$ factors) as the $h$ fields had at the beginning. So the variation of $h$ (a bosonic field) is proportional to $\psi_a$ (a fermionic field) – $\delta h \sim \epsilon_a \psi^a $ – which must of course be multiplied by another fermionic object, namely $\epsilon^a$, a Grassmann parameter of the supertransformation (you should view this as a pure number, an angle, even though it's anticommuting: it's not a field). This "variation of boson is proportional to a fermion and vice versa" is what we mean by "SUSY transforms bosons to fermions and vice versa".
Of course, all the other terms in $Q^a$ (and when acting on all the other terms in $\phi$ or any other superfield) have the same effect of exchanging bosons and fermions. It boils down to the statistics or grading.
