50% Decreased Flow in the Coronary Artery - USMLE Forums
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#1
01-19-2011
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50% Decreased Flow in the Coronary Artery

A 49 YO woman with CAD undergoes coronary arteriography that shows a 50% decrease in the lumen of the LAD. For any arteriovenous pressure gradient, the flow thru this artery (compared with normal) wall decrease by a factor of which of the following?

a) 2
b) 4
c) 8
d) 16
e) 32

#2
01-19-2011
 USMLE Forums Guru Steps History: Step 1 Only Posts: 487 Threads: 95 Thanked 1,500 Times in 361 Posts Reputation: 1510

its very simple... lumen is reduced by half (50%) which mean that the radius is 1/2 the size, therefore

1/(0.5)^4 = 16

The flow thru LAD artery wall decrease by a factor of 16.
 The above post was thanked by: Mondoshawan (01-27-2011), rigbbm (06-11-2018)
#3
01-19-2011
 USMLE Forums Guru Steps History: 1 + CS Posts: 304 Threads: 9 Thanked 324 Times in 139 Posts Reputation: 334

I thought Poiseulle Law is talking about resistance not flow, Can we apply it to flow?

#4
01-19-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,270 Times in 416 Posts Reputation: 1294

Quote:
 Originally Posted by laithbv I thought Poiseulle Law is talking about resistance not flow, Can we apply it to flow?
Poiseulle's law is

and Flow = ΔP/resistance

so the factor of change in ΔP (1/r^4 in this case) is the same as that in Flow, by the transitivity of multiplication.
 The above post was thanked by: aktorque (01-27-2011), Sabio (01-19-2011)
#5
01-27-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,270 Times in 416 Posts Reputation: 1294

Quote:
 Originally Posted by Mondoshawan Poiseulle's law is and Flow = ΔP/resistance so the factor of change in ΔP (1/r^4 in this case) is the same as that in Flow, by the transitivity of multiplication.
The italicized part should say "Flow is Q in the equation, so the factor of change in Flow is the inverse of that in ΔP (1/r^4 in this case) = r^4 = 1/16. So, as aktorque said, the flow would decrease by a factor of 16."
Hope that's clearer, and sorry for the bad explanation before!
 The above post was thanked by: aktorque (01-27-2011)
#6
01-27-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,270 Times in 416 Posts Reputation: 1294

Quote:
 Originally Posted by Mondoshawan Rearranging for Q, you can see that Q = (blahblahblah) r^4.
This is really the gist, and all we really need to know about Poiseuille's equation. Flow is directly proportional to radius^4, and resistance is inversely proportional to radius^4.
 The above post was thanked by: Jorabim (06-28-2014)
#7
01-27-2011
 USMLE Forums Scout Steps History: 1 + CK Posts: 88 Threads: 12 Thanked 178 Times in 49 Posts Reputation: 188

Mondoshawan is right. this is another way to look at it;

Resistance = (8 x viscosity x length)/ radius^4
Resistance = Pressure change/Flow

If radius reduces by 1/2, then resistance increases by 16. (inverse proportnl)
If resistance increases by 16,then Flow reduces by 16.(inversely proportional)

for questions like this, Goljan said, just remember '16', the boards wont go beyond that.
hope this helps.
 The above post was thanked by: Jorabim (06-28-2014), Mondoshawan (01-27-2011)

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