Suppose I take a short exact sequence of filtered chain complexes: $$0\to A\xrightarrow{p} B\xrightarrow{q} C\to 0$$ We assume that $p$ and $q$ are filtration-preserving, so that $p(F_rA)\subseteq F_rB$ and $q(F_rB)\subseteq F_rC$ for all $r$. We also assume that $p$ and $q$ are strict, so that $p^{-1}(F_rB) = F_rA$ and $q(F_rB) = F_rC$ for all $r$.
Question: Let $r\geq 0$. In what ways, if any, is the above situation reflected on the $r$-th pages of the spectral sequences associated to $A$, $B$ and $C$?
Our assumption that $p$ and $q$ are strict means that we have a short exact sequence of filtration quotients: $$ 0 \to F_iA/F_{i-1}A\to F_iB/F_{i-1}B\to F_iC/F_{i-1}C\to 0$$ But this precisely means that we have a short exact sequence of $E^0$-pages: $$ 0 \to E^0_{i,j}A\xrightarrow{p^0} E^0_{i,j}B\xrightarrow{q^0} E^0_{i,j}C\to 0$$ (I'm writing $E^r_{i,j}A$ for the terms of the $r$-th page of the spectral sequence associated to $A$, and similarly for $B$ and $C$. I'm writing $p^r$ and $q^r$ for the maps of $r$-th pages induced by $p$ and $q$. And I'm using the convention that $E^0_{i,j} = (F_i/ F_{i-1})_{i+j}$ so that $d^0\colon E^0_{i,j}\to E^0_{i,j-1}$ and more generally $d^r\colon E^r_{i,j}\to E^r_{i-r,j+r-1}$.)
Since $p^0$ and $q^0$ respect the differentials of the $E^0$ pages, the short exact sequences above turn into long exact sequences of homology groups, or in other words, long exact sequences of $E^1$ pages: $$ \cdots \to E^1_{i,j+1}C\to E^1_{i,j}A\xrightarrow{p^1} E^1_{i,j}B\xrightarrow{q^1} E^1_{i,j}C\to E^1_{i,j-1}A\xrightarrow{p^1}\cdots$$ But what happens next?
Refined sub-question: For $r\geq 2$, is there any structure at all that relates the image of $p^r$ with the kernel of $q^r$?
I wonder if there is some kind of obstruction-theory flavoured result involving with primary, secondary, tertiary, ... connecting homomorphisms that all have a different degree shift.
