Wednesday, March 19, 2014

ENZYME REVIEW QUESTIONS IN CLINICAL CHEMISTRY 2



MATCHING TYPE: MATCH COLUMN B WITH COLUMN A  (WRITE CAPITAL LETTERS ON A WHOLE SHEET OF YELLOW PAPER.
ANSWERS

 COLUMN A                                            COLUMN B
SUBSTANCE TESTED
NORMAL VALUES (CHOICES)
3.  AST - E
 A. < 2 mg/dL
4.  ACP - A
 B.  < 1 /umol/L
5.  ALT - E
 C.  15-165 U/L
6.  LDH - I-..Less..than..200.U/L
 D.  90-190 U/L
7.  Glucose   - H
 E.  35- 55 U/L
8. CPK - C
 F.  0.3 ug/L
9. Myoglobin – none – (0 to 85 nanograms per milliliter (ng/mL)
10. Troponin I -F
 G.  <0 .5="" p="" umol="">
H.  80 -120 mg/dL

I. None of the above


SUBSTANCE TESTED
Clinical Significance (CHOICES)
11.  ALP - A
 A. bone disease
12.  ACP – B, C, D, E
 B.  Paget’s disease
13.  I.P. - A
 C.  Breast cancer
14  LDH – E,G,H
 D.  Prostate cancer
15.  Glucose    - F
 E.  Megaloblastic anemia
16. AST – G, H
 F. DM
17. ALT - G
18. LDH –E,G,H
19. CK - H
 G.  hepatitis
H.  Myocardial infarction

I. None of the above



SUBSTANCE TESTED
Substrate (CHOICES)
20.  ALP – B, C, F, G
 A. creatine
21.  ACP – B, C, F, G
 B.  PNPP
22.  I.P. - I - not an enzyme
 C.  Betaglycero phosphate
23  LDH - E
 D.  alpha ketoglutarate
24.  Glucose    - I -
not an enzyme
 E.  lactate
25. AST – D, G,
 F.  PNP
26. ALT – D
27  TB - I -
not an enzyme
28. CK - A
29. myglobin -I -
not an enzyme
30. Troponin I and T -I -
not an enzyme
 G.  aspartate
H. umbilleferone phosphate

I. None of the above


  1. Discuss 3 Precautions for ACP test.(Specific) READ,,NOTES
  2. Discuss 3 precautions for ALP test. .(Specific)
  3. Discuss 3 precautions for LDH test. .(Specific)





Tuesday, March 18, 2014

Clinical Chemistry Review Questions - Blood Gas Analysis



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:

  1.  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

  1. Based on the laboratory results given in question no.2.  Why is the HCO3 level so low?
    1. Because of rapid respiration
    2. Because of persistent diarrhea
    3. Because it compensates the respiratory aspect
    4. A & C
    5. C & D
    6. None of the above
    7. All of the above

  1. Why does the patient have rapid respiration?
    1. To restore normal pH
    2. To decrease pCO2
    3. To restore 20:1 HCO3 to H2CO3 ratio
    4. A & C only
    5. All of the above
    6. 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

  1. pH= 7.36, HCO3= 23 mmol/L, pCO2= 45 mmHg=M
  2. pH= 7.55, HCO3 = 28 mmol/L, pCO2 = 59 mmHg=H
  3. pH= 7.15, HCO3 = 9 mmol/L, pCO2 = 25 mmHg=G
  4. pH = 7.8, HCO3 = 18 mmol/L, pCO2 = 30 mmHg=F
  5. pH= 7.48, TCO2= 39 mmol/L, pCO2 = 28 mmHg=H
  6. 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:

  1. FEVER –RESULTS TO METABOLIC ACIDOSIS – LUNGS WOULD COM[PENSATE- PCO2 INC.EXCRETION, H+ INCREASED EXRETION.
  2. SWEATING –SAME AS NO. 11
  3. ANXIETY –RESULTS TO HYPERVENTILATION CAUSING RESP.ALKALOSIS- KIDNEYS WOULD COMPENSATE BY DECREASING HCO3, H+ RETENTION AND INCREASING EXCRETION,
  4. PHYSICAL EXERTION –SAME AS NO. 13
  5. COPD – MOST CAUSE HYPOVENTILATION CAUSING RESPIRATORY ACIDOSIS, KIDNEYS WOULD COMPENSATE BY INCREASING RETENTION OF HCO3 AND DECREASING ITS EXCRETION, H+ INCREASED EXCRETION.

  1. 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.

  1. 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?

  1. 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.

  1. 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

  1. 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




SEARCH ENGINE

Google
 

COPYRIGHT

Except for the outside links, printing and copying of the articles contained herein should cite this site as the source. For commercial purposes, the permission of the author should be sought first. You can leave a comment in any of the posts. Thank you.