CLINICAL CHEMISTRY 2
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. 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:
- 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
- Based on the laboratory results given in question no.2. Why is the HCO3 level so low?
- Because of rapid respiration
- Because of persistent diarrhea
- Because it compensates the respiratory aspect
- A & C
- C & D
- None of the above
- All of the above
- Why does the patient have rapid respiration?
- To restore normal pH
- To decrease pCO2
- To restore 20:1 HCO3 to H2CO3 ratio
- A & C only
- All of the above
- 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
- pH= 7.36, HCO3= 23 mmol/L, pCO2= 45 mmHg=M
- pH= 7.55, HCO3 = 28 mmol/L, pCO2 = 59 mmHg=H
- pH= 7.15, HCO3 = 9 mmol/L, pCO2 = 25 mmHg=G
- pH = 7.8, HCO3 = 18 mmol/L, pCO2 = 30 mmHg=F
- pH= 7.48, TCO2= 39 mmol/L, pCO2 = 28 mmHg=H
- 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:
- FEVER –RESULTS TO METABOLIC ACIDOSIS – LUNGS WOULD COM[PENSATE- PCO2 INC.EXCRETION, H+ INCREASED EXRETION.
- SWEATING –SAME AS NO. 11
- ANXIETY –RESULTS TO HYPERVENTILATION CAUSING RESP.ALKALOSIS- KIDNEYS WOULD COMPENSATE BY DECREASING HCO3, H+ RETENTION AND INCREASING EXCRETION,
- PHYSICAL EXERTION –SAME AS NO. 13
- COPD – MOST CAUSE HYPOVENTILATION CAUSING RESPIRATORY ACIDOSIS, KIDNEYS WOULD COMPENSATE BY INCREASING RETENTION OF HCO3 AND DECREASING ITS EXCRETION, H+ INCREASED EXCRETION.
- 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.
•
3. WHAT IS THE BODY’S COMPENSATORY MECHANISM?
•
4. WHAT ADDITIONAL TEST COULD YOU PERFORM?
ANSWER: FULLY
COMPENSATED RESPIRATORY ACIDOSIS.
- 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.
•
3. WHAT IS THE BODY’S COMPENSATORY MECHANISM?
•
4. WHAT ADDITIONAL TEST COULD YOU PERFORM?
- 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 COMPESATED RESPIRATORY ACIDOSIS W/HYPOXIA
2. DEFEND YOUR ANSWER.
- 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
- 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
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