RayBerg
Forum Senior
Topics: 32
Posts: 150
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I'll play
1 c
2 e (I was debating b/t b and e..because central chemorec are under control of CO2. So I guess it indirectly controls it?)
3 b
4 f
5 b
About #3, from the info given 3mg/dL was filtered out of 12mg/dl.
Therefore, FF = 3/12 = 0.25. HOpe it helps
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| doc179
Forum Guru
Topics: 67
Posts: 1,217
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1.c
2.b
3.b
4.f
5.b
#3= I guess, 9 is 75% of 12 and the 12-9= 3 which is 25% of 12.
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| RayBerg
Forum Senior
Topics: 32
Posts: 150
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Doc, can you go over #2 for me? I had it b/t b and e anyways. Thanks!
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| drduck
Forum Guru
Topics: 82
Posts: 529
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1.c
2.b
3.b
4.f
5.b
those were fine...
in Q 2....due to hypoxia there is less pulmonary flow....bcos there is vasoconstriction....specific to pulmonary circulation.
and normally it is pCO2 that regulates....but if pO2 goes considerably low it can also stimulate chemoreceptors..
hope that helps
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| venureddy82
Forum Newbie
Topics: 0
Posts: 3
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3. A freely filterable substance that is neither reabsorbed nor secreted has a renal artery concentration of 12 mg/ml and a renal vein concentration of 9 mg/ml. Calculate the filtration fraction.
Ans is B. Can u explain
a. 0.15
b. 0.25
c. 0.75
d. 1.00
e. not enough information given to make this calculation
FF=GFR/RPF
GFR= 12-9=3
RPF= 12
So, FF= 3/12=1/4=0.25
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