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Old 04-05-2013
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Drug Decreasing the slope of phase 4?

Propanalol, esmolol, metoprolol, atenolol, and timolol slow abnormal pacemakers by "decreasing the slope of phase 4"
What does "decreasing the slope of phase 4" do exactly?
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Old 04-05-2013
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thats the phase of Inward Na funny current which brings membrane potential to threshold in pacemakers
this drugs decrease If Na current and increases time required to reach threshold=decreased slope or decreased heart rate
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Old 04-05-2013
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I've reached a conclusion: phase 4 = SA node =heart rate. Phase 0 = AV node = PR interval /conduction

Correct me if any doesn't agree
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Old 04-05-2013
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Quote:
Originally Posted by ` Faith ` View Post
I've reached a conclusion: phase 4 = SA node =heart rate. Phase 0 = AV node = PR interval /conduction

Correct me if any doesn't agree
I guess u xactly need the meaning of slope.
In any graph,slope =y-intercept/x-inetrcept (The higher the y-axis value,the higher the slope).This implies increased slope is closer to y-axis(more standing) and decreased slope is closer to x-axis (more sleeping). Try to recollect ur school mathematics..

When plotting on graph, phase 4 of the specialized heart cells with automaticity(SA,AV nodes) is caused by pacemaker potential(It is special and is not stable unlike other stable muscle potential).This pacemaker potential is a slow,gradual depolarisation towards thershold.

Beta blockers decrease heart rate by decreasing slope of phase 4.This means they increase the time of automatic cells to reach threshold thus decrease their rate of firing.
So wat I meant to say is phase 4 can be for SA node,AV node,Purkinje fibers or anyother abnormal pacemakers.
Phase 0 in nodal cells is bcoz of influx of calcium thru L-type Calcium channels seen in SA and AV node.
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