CLINICAL CHEMISTRY 2
NAME: ___________________________________________________YEAR/SEC: __________
DATE:__________________________________________
I. MULTIPLE CHOICE ( I PT. EACH)
SHADE THE BOX THAT CORRESPONDS TO THE LETTER OF YOUR CHOICE (Best answer) IN THE ANSWER SHEET PROVIDED.
1. The most common specimen for blood gas analysis is:
a. Plasma c. whole blood e. arterial blood
b. Serum d. buffy coat
CASE ANALYSIS
The following lab results were obtained from a 50-year old male patient, complaining of persistent diarrhea for 3 days and rapid respiration : Laboratory results were:
pH = 7.21
pCO2 = 19 mm Hg
PO2 = 96 mm Hg
Total Bilirubin – 25 mg/dL
HCO3 = 7 mmol/L
SO2 = 96 %
K Injection:
2. What is the patient’s acid-base status?
a. Respiratory acidosis
b. Respiratory alkalosis
c. Metabolic acidosis
d. Metabolic alkalosis
e. all of the above
f. none of the above
3. Based on the laboratory results given in question no.2. Why is the HCO3 level so low?
a. Because of rapid respiration
b. Because of persistent diarrhea
c. Because it compensates the respiratory aspect
d. A & C
e. C & D
f. None of the above
g. All of the above
4. Why does the patient have rapid respiration?
a. To restore normal pH
b. To decrease pCO2
c. To restore 20:1 HCO3 to H2CO3 ratio
d. A & C only
e. All of the above
f. none of the above
CHOICES FOR NOS. 5-onwards
EXISTING CONDITION:
A. UNCOMPENSATED RESPIRATORY ACIDOSIS
B. UNCOMPENSATED RESPIRTATORY ALKALOSIS
C. UNCOMPENSATED METABOLIC ACIDOSIS
D. UNCOMPENSATED METABOLIC ALKALOSIS
E. PARTIALLY COMPENSATED RESPIRATORY ACIDOSIS
F. PARTIALLY COMPENSATED RESPIRATORY ALKALOSIS
G. PARTIALLY COMPENSATED METABOLIC ACIDOSIS
H. PARTIALLY COMPENSATED METABOLIC ALKALOSIS
I. FULLY COMPENSATED RESPIRATORY ACIDOSIS
J. FULLY COMPENSATED RESPIRATORY ALKALOSIS
K. FULLY COMPENSATED METABOLIC ACIDOSIS
L. FULLY COMPENSATED METABOLIC ALKALOSIS
M. NONE OF THE ABOVE
5. pH= 7.36, HCO3= 23 mmol/L, pCO2= 45 mmHg=M
6. pH= 7.55, HCO3 = 28 mmol/L, pCO2 = 59 mmHg=H
7. pH= 7.15, HCO3 = 9 mmol/L, pCO2 = 25 mmHg=G
8. pH = 7.8, HCO3 = 18 mmol/L, pCO2 = 30 mmHg=F
9. pH= 7.48, TCO2= 39 mmol/L, pCO2 = 28 mmHg=H
10. pH= 7.50, HCO3= 25 mmol/L, pCO2 = 18 mmHg=B
CHOICES FOR 66 TO 70
COMPENSATORY MECHANISM:
INDICATE WITH AN ARROW HOW THE SPECIFIC SUBSTANCES WOULD COMPENSATE IN THE FOLLOWING EXISTING CONDITIONS:
11. FEVER –RESULTS TO METABOLIC ACIDOSIS – LUNGS WOULD COM[PENSATE- PCO2 INC.EXCRETION, H+ INCREASED EXCRETION.
12. SWEATING –SAME AS NO. 11
13. ANXIETY –RESULTS TO HYPERVENTILATION CAUSING RESPIRATORY ALKALOSIS- KIDNEYS WOULD COMPENSATE BY DECREASING HCO3, H+ RETENTION AND INCREASING EXCRETION,
14. PHYSICAL EXERTION –SAME AS NO. 13
15. COPD – MOST CAUSE HYPOVENTILATION CAUSING RESPIRATORY ACIDOSIS, KIDNEYS WOULD COMPENSATE BY INCREASING RETENTION OF HCO3 AND DECREASING ITS EXCRETION, H+ INCREASED EXCRETION.
16. RESULTS OF LAB TESTS:
• pCO2= 53 mm Hg
• O2 Saturation: 79%
• HCO3= 29 mmol/L
QUESTIONS:
• 1. WHAT IS THE PH?-7.36
• 2. IDENTIFY THE CONDITION. DEFEND YOUR ANSWER. -REFER TO NOTES
• 3. WHAT IS THE BODY’S COMPENSATORY MECHANISM? -REFER TO NOTES
• 4. WHAT ADDITIONAL TEST COULD YOU PERFORM? - DELTA CHECK, ELECTROLYTES, 5 COMMON PARAMETERS (BUA,TAG,CREA, FBS, CHOLE), CBC.
ANSWER: FULLY COMPENSATED RESPIRATORY ACIDOSIS.
17. RESULTS OF LAB TESTS:
TCO2= 27 mmol/L
PCO2= 45 mmHg
QUESTIONS:
• 1. WHAT IS THE PH?=7.38
• 2. IDENTIFY THE CONDITION. DEFEND YOUR ANSWER. -ALL VALUES ARE NORMAL
• 3. WHAT IS THE BODY’S COMPENSATORY MECHANISM? - N/A
• 4. WHAT ADDITIONAL TEST COULD YOU PERFORM? - IF PATIENT FEELS UNWELL, RECOMMEND EXECUTIVE LAB TESTS. (ALL)
ANSWER: NORMAL PATIENT
18. LABORATORY RESULTS:
pH = 7.27
pCO2= 55 mm Hg
pO2= 50 mm Hg
Hgb = 13.5 g/L
O2 Saturation: 79%
HCO3= 28 mmol/L
QUESTIONS:
1. IDENTIFY THE CONDITION. –PARTIALLY COMPENSATED RESPIRATORY ACIDOSIS W/T HYPOXIA
2. DEFEND YOUR ANSWER.
19. LABORATORY RESULTS:
pH = 7.27
pCO2= 58 mm Hg
pO2= 100 mm Hg
Hgb = 13 g/L
O2 Saturation: 98%
HCO3= 23 mmol/L
QUESTIONS:
1. IDENTIFY THE CONDITION. –UNCOMPENSATED RESPIRATORY ACIDOSIS
2. DEFEND YOUR ANSWER-
20. LABORATORY RESULTS:
pH = 7.10
pCO2= 40 mm Hg
pO2= 91 mm Hg
Hgb = 14 g/L
O2 Saturation: 95%
HCO3= 13 mmol/L
QUESTIONS:
1. IDENTIFY THE CONDITION. - UNCOMPENSATED METABOLIC ACIDOSIS
2. DEFEND YOUR ANSWER-REFER TO NOTES
FOR MORE REVIEW QUESTIONS, YOU CAN VISIT CLIN CHEM REVIEWER TO BROWSE MORE QUESTIONS.
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