Consider two neurons, A and B, which reciprocally inhibit each other. If both of the neurons receive input at the same time, it seems to me that no oscillation will occur between the two. Such would occur only if one of the two is initially "favored" over the other. So what I'm wondering is, how does that happen? Is it essentially random?

  • 2
    $\begingroup$ Honi's got most of it, but it also involves refractory periods and post-inhibitory characteristics of the membrane ("rebound"). $\endgroup$ Apr 30, 2013 at 23:07

1 Answer 1


Noise will ensure that there will not be perfect symmetry and therefore that one has an advantage to fire first. Whichever one fires first will inhibit the other and ensure that the first one to fire will stay on for however long the dynamics of the pattern generator allow.

  • $\begingroup$ I had a feeling that that's the case. So randomly (e.g. Gaussian) distributed noise is the sole or main deciding factor here? $\endgroup$ Apr 30, 2013 at 21:13
  • 1
    $\begingroup$ noise comes in so many forms. you can have noise in the synapses, you can have noise in somatic conductances, the cells can be different sizes, the cells can have different time constants, etc. $\endgroup$
    – honi
    May 2, 2013 at 1:55
  • $\begingroup$ I down-voted this answer. Noise does inject assymetry into the system. However, a much stronger, system-regulated mechanism of assymetey is connection strength. This can come in the form of 1) number of interconnections, 2) regulation of the amount of neurotransmitter released 3) autoregulation of a specific cell 4) regulation of neurotransmitter receptor density 5) other connections or astrocytes or even local environment regulating the excitability of individual cells. $\endgroup$ May 3, 2013 at 11:15
  • $\begingroup$ as i described in my explanatory note just above your comment. $\endgroup$
    – honi
    May 3, 2013 at 13:09
  • $\begingroup$ I would call it signal, not noise. But I think you could anyway improve your answer a lot... $\endgroup$ May 4, 2013 at 3:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.