1. A creatinine clearance was performed on a 6-hour urine collection with a volu
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Question
1. A creatinine clearance was performed on a 6-hour urine collection with a volume of 400 ml. Urine creatinine = 50m/dl and the plasma creatinine = 2.5 mg/dl. What is the patient’s uncorrected creatinine clearance? What is the corrected clearance if patient surface area is 1.20 m2 ?
2. A creatinine clearance is performed on a 24-hour urine collection with a volume of 2200 ml. Urine creatinine = 150 mg/dl and the plasma creatinine = 1.5 mg/dl. Patient surface area is 2.68 m2 . What is the corrected clearance for this patient?
3. Given an absorbance of 1.125, what is the % T value?
4. Given the absorbance of a 150 mg/dl standard is 0.425, and the absorbance of QC level 1 is 0.205, calculate the concentration of QC level 1.
5. What is the final dilution on a twofold serial dilution made with four tubes?
6. A serum amylase is diluted 1/100 with a result of 45.0 IU/L. What is the patient’s actual amylase result?
7. If the mean for glucose in a quality control is 100 mg/dl, and 1 SD is 5 mg/dl, what is the probability that if the control sample is reanalyzed the result will fall between 95 and 105 mg/dl (+/- 1 SD)?
8. Given an SD of 1.5 mEq/L and a mean of 141 mEq/L for a sodium method, what is the CV? 9. What is the median for this set of numbers? 135, 128, 132, 134, 129, 131
Explanation / Answer
1) The uncorrected creatinine clearance is calculated by the formula,
Creatinine clearance = (volume x urinary creatinine concentration)/ plasma creatinine concentration
= (400 x 50)/2.5
= 8000 ml in six hours
= 1333 ml in 1 hour
= 22.22 ml/min
Corrected creatinine clearance is,
(Uncorrected creatinine clearance x 1.73)/ body surface area
= (22.22 x 1.73)/1.2
= 32.03 ml/min
Please post other questions separately.
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