In a patient that has two pathways connected proximally and distally, with different conduction velocities but the same refractory period, would it be easy to induce reentrant tachycardia?

• A. True
• B. False
• C. Not enough information
• D. Depends on the direction of conduction

Answer: (B)

Reentrant tachycardia needs a closed loop where an impulse can travel around and re-excite tissue after it has recovered. Here there are two parallel pathways between the same two points, with different speeds but the same refractory period. The same refractory period means the tissue on both pathways recovers at essentially the same time after being activated. That makes unidirectional block less likely: for reentry to start, one direction must conduct while the opposite direction is still refractory. If both paths recover together, that timing advantage is lost, so a premature impulse is unlikely to set up a stable reentrant circuit. So, despite the velocity difference, it would not be easy to induce reentrant tachycardia.