Answer
Please see the image. In the case that Q and S are present in high concentrations, the reaction L->N would predominate.
Work Step by Step
In the case that Q and S are present in high concentrations, that would (via inhibition) cause O to build up (see image). In turn, that would inhibit the conversion of L to M, making the conversion of L to N the only remaining reaction uninhibited and, thus, the predominant activity.
