robin082006
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A diabetic teenager is found to have a pH of 7.1 and normal electrolyte levels (Na+ = 140 mM, K+ = 4 mM, Cl– = 103 mM) except for a bicarbonate of 11 mM (normal 22 to 28 mM). The urine tests positive for ketone bodies, mostly due to acetoacetic acid and acetoacetate (CH3C=OCH2COOH and CH3C=OCH2COO–), which have a pK of 4.8. In this case, it is assumed that acetoacetate is the only significant anion in the blood besides chloride, and that each acetoacetate anion binds and removes one sodium cation during excretion by the kidney. Given that the patient has a normal glomerular filtration rate of about 7 L of blood per hour without any retention of acetoacetate/acetoacetic acid, the rates of sodium, acetoacetate, and acetoacetic acid loss will be
A. 10 mmol/h of each species
B. 50 mmol/h of sodium and acetoacetate, virtually no acetoacetic acid excretion
C. 100 mmol/h of sodium and acetoacetic acid, virtually no acetoacetate excretion
D. 200 mmol/h of sodium and acetoacetate, virtually no acetoacetic acid excretion
E. 300 mmol/h of each species
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| Andrew000
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where can i get the answere
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| drk1980
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i'll try with E
(btw, is it significant that acetoacetic acid is in a non-ionised state, hence easily crosses membranes?)
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| msyamp
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how do you work out this case? plzzzzzzzzzzzzzzz
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| sturge_weber
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whats the answer
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| anjushree
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AS PH of 7.1 is more than pk of 4.8 the medium is bosic ,the ketone will be inacetoacetate form andacidic sub in basic medium will be in ionised form -thus will be excreated
now if we consider diff bet ph &pk -7.1-4.8 near to 2 ,99 I &1 UI -99ACETOACETATE wILLBE EXCREted
Na is excreted along with acetoacetate
with this background ans CAN BE BET -B or D
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| anjushree
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robin please post the ans
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| Doc2378
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Well...whats the answer Robin??
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| robin082006
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D is correct.
Here is the explanation:
it is a state of met acidosis
Anion gap =(Na +K) -( Cl + HCO3) = (140+4) - (103 + 11) =30
this large AG is due to acetoacetate
pH = pK + log acetoacetate/acetic acid
7.1 = 4.8 + log acetoacetate/ acetic acid
log acetoacetate/acetic acid = 2.3
=>acetoacetate/acetic acid > 100
=>99% of acetic acid is in ionized form i.e acetoacetate and is excreted. (ketone bodies are freely filtered)
filtered load = GFR* plasma conc= 7L/h*30mmol/L=210mmol/L
bcoz acetoacetate anion binds and removes one sodium cation during excretion so, Na+ loss wud also be same
99% is in acetoacetate form so, acetoacetic acid excretion wud almost be nil
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| msyamp
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GREAT GOD. A VERY GOOD EXCELLENT QUESTION. INVOLVES EVERY THING OF RENAL AND ACID BASE
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