Showing posts with label exam questions clinical chemistry. Show all posts
Showing posts with label exam questions clinical chemistry. Show all posts

Monday, May 11, 2015

Review Exam Questions in Clinical Chemistry 1

1. In electrophoresis, the basis of separation are the following, EXCEPT:

a. Charges of ions
b. Electrophoretic media
c. Net charge of particles
d. Concentration of ions
e. NIL

2. The movement of ions in electrophoresis is from:
a. Negative to positive
b. Positive to negative
c. Negative to negative
d. Positive to positive
e. NIL

3. The principle of chromatography is:
a. The movement of charged particles in an electric field
b. The involvement the solute and solvent
c. The measurement of emitted light
d. The excitation of ions in solution
e. NIL

4. Chromatography is affected by the following factors, EXCEPT:
a. Density of particles in solution
b. Size of particle
c. Affinity of particles to chromatographic media
d. pH
e. NIL

5. The following precautions are observed in Gravimetry, EXCEPT:
a. The balance should be adjusted to zero
b. The substance to be measured should be hydrated
c. The substance should be uncontaminated
d. The vessels used should be clean and dry
e. NIL

6. The specific substance that causes lipemia is:
a. Lipoprotein lipase
b. Chylomicrons
c. Triglycerides
d. Cholesterol
e. NIL

7. The most predominant carbohydrate is:
a. Fatty acid
b. Glucose
c. Amino acid
d. Triglyceride
e. NIL

8. Chromatography can be utilized in the separation of the following, EXCEPT:
a. Proteins
b. Sugars
c. Drugs
d. Glucose
e. NIL

9. The following are effects of long standing of serum samples not separated from the cells at room temperature in glucose determination:
1. There will be falsely decreased results.
2. The rate of loss of glucose due to long standing at RT is 7 mg/dL/hr. if the serum is not separated from the cells.
3. The rate of loss of glucose may be higher due to possible bacterial contamination.
4. There will be falsely elevated values.
5. There will be no effect on glucose values.

a. 1
b. 1 and 2
c. 1, 2 and 3
d. 1, 3 and 5
e. NIL

10. Convert 6.5 mmol/L of glucose to traditional units.
a. 122.78
b. 123.56
c. 119.07
d. 120.72
e. NIL

11. Which of the following is the reference method for glucose?
a. Somogyi-Nelson
b. Hexokinase
c. Orthotoluidine
d. Nelson Somogyi
e. NIL

12. Which of the following is a characteristic of Type 1 DM
a. Family history of DM
b. Usually occurs at at age 40 and above
c. Ketosis not prominent
d. Thin persons
e. NIL  

HERE ARE the CORRECT ANSWERS.

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




Sunday, March 13, 2011

Clinical Chemistry Review Questions

Clinical Chemistry Review Questions vary in difficulty and complexity. Here are some basic questions which can help you review common principles in Clinical Chemistry.

Select the BEST ANSWER.

1. The most common sample specimen in clinical chemistry is:

a. Plasma c. whole blood

b. Serum d. buffy coat

2. In enzyme analysis, the following should be monitored closely, EXCEPT:

a. Temperature c. pH

b. Concentration of substrate d. non-competitive inhibitor

3. Electrolytes are called amphoteric substances because of this reason:

a. They can either be negatively or positively charged

b. They can be water or non-water soluble

c. They can transform from one energy form to another

d. They are directly transported in the blood stream.

4. The following statements are true of electrophoresis, EXCEPT:

a. It is a method to separate proteins from one another

b. The media can be paper, agar gel or cellulose

c. The principle depends upon their ability to fluoresce

d. Electrophoretic mobility is based on the charges of the ions.

5. In the maintenance of normal blood pH, these two organs are involved:

a. Lungs and heart c. lungs and kidneys

b. Kidneys and heart d. kidneys and liver

Choices for numbers 6 to 10

a. Uncompensated metabolic alkalosis

b. Uncompensated metabolic acidosis

c. Uncompensated respiratory alkalosis

d. Uncompensated respiratory acidosis

e. Partially compensated metabolic alkalosis

f. Partially compensated metabolic acidosis

g. Partially compensated respiratory alkalosis

h. Partially compensated respiratory acidosis

i. Fully compensated metabolic alkalosis

j. Fully compensated metabolic acidosis

k. Fully compensated respiratory alkalosis

l. Fully compensated respiratory acidosis

m. None of the above

6. Given pH – 7.49, pCO2= 40 mmHg, HCO3 = 32 mmol/L, What do these values indicate?

7. Given pH – 6.8, dCO2= 10 mmol/L, TCO2= 22 mmol/L, What do these values indicate?

8. Given pH- 7.43, HCO3= 29 mmol/L, pCO2 – 50 mmHg. What do these values indicate?

9. Given pH- 7.15, HCO3= 25 mmol/L, pCO2 – 60 mmHg. What do these values indicate?

10. Given pH- 7.50, HCO3= 40 mmol/L, pCO2 – 40 mmHg. What do these values indicate?



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