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Wednesday, September 12, 2018

Blood Gas Analysis Review Questions in Clinical Chemistry


I. MULTIPLE CHOICE (Select the BEST ANSWER)

1.    The ideal anticoagulant for arterial blood gas measurements is:
a.    EDTA
b.    Sodium citrate
c.    Lithium heparin
d.    Potassium oxalate
e.    All of the above
f.    None of the above

2.    The kidneys can compensate for nonmetabolic alkalosis by (excretion, retention) of bicarbonate and (increased, decreased) excretion of NaH2PO4.
a. excretion, increased            e.  None of the above
b. retention, increased
c. excretion, decreased
d. retention, decreased

3.    The normal ratio of carbonic acid to bicarbonate in arterial blood is:
a.     1:10                    d.  20:1
b.    7.4:6.1                    e.  None of the above
c.    1:20

4.    If the patient’s arterial  blood gas results are : pH = 7.15, PCO2= 85 mmHg
HCO3 = 35 mmol/L.  What would be your most likely concusion?
a.    partially compensated nonmetabolic acidosis
b.    partially compensated metabolic acidosis
c.    uncompensated nonmetabolic alkalosis
d.    uncompensated metabolic acidosis
e.    None of the above’

5.    If a patient’s arterial blood gas results are pH= 7.38, PCO2 = 35 mm Hg, HCO3 = 27 mmol/L.  What would be your most likely concusion?
a.    Partially compensated nonmetabolic alkalosis
b.    Partially compensated metabolic alkalosis
c.    Uncompensated respiratory alkalosis
d.    Uncompensated metabolic alkalosis
e.    None of the above

6.    The following are precautions for ABG analysis, except:
a.    The specimen should be collected anaerobically
b.    The best receptacle for collection is a glass syringe pre-treated with heparin
c.    If sample is not to be tested immediately, it should be refrigerated or kept in chilled ice.
d.    The temperature of the patient at the time of collection should be noted and recorded
e.    All of the above
f.    None of the above

II. IN THE TABLE BELOW, FILL IN THE CORRECT CHOICES BY USING THE LETTERS OF YOUR CHOICES.

pH

a.    acidic or decrease
b.    alkaline or increase

Concentrations of HCO3 and PCO2:


a.    HCO3 Normal
b.    HCO3 Increased
c.    HCO3 Decreased
d.    PCO2 Normal
e.    PCO2 Increased
f.    PCO2 Decreased

Resulting Condition

a.    Respiratory alkalosis
b.    Respiratory acidosis
c.    Metabolic alkalosis
d.    Metabolic acidosis

Compensatory Mechanism

a.    Increase retention and decrease excretion of HCO3 , H+ excretion by kidneys
b.    Increase excretion and decrease retention of HCO3, H+ retention by kidneys
c.    Increase retention and decrease excretion of CO2  by the lungs
d.    Increase excretion and decrease retention of CO2 by the lungs



Substance or
Condition
pH
HCO3
PCO2
Resulting condition
Compensatory Mechanism






Hysteria





Morphine use





Alcohol intoxication





Hyperventilation





Hypoventilation





         


III. PROBLEM SOLVING:

1.    DETERMINE THE pH of the blood sample, if the TCO2 = 22 mmol/L, and the PCO2 is 20 mm Hg. ( 4 pts.)

2.    DETERMINE THE DCO2 IF the H2CO3 is 1.50 mmol/L. ( 2 pts.)

3.    DETERMINE THE pH if PCO2 = 20 mm Hg and HCO3 = 30 mmol/L. ( 4 pts.)

4.    DETERMINE THE pH if the TCO 2 = 28 mmol/L and the PCO2 = 15 mmHg. (5 pts.)

5.    DETERMINE THE ANION GAP If Chloride = 130 mmol/L,  Sodium = 150 mmol/L and HCO3 = 25 mmol/L ( 3 pts)


V. TRANSCRIPTION: ( 1 PT. EACH) Transcribe the acronyms.

1.    TLA______________________________________________
       
2.    NCCLS____________________________________________
           
3.    NIST______________________________________________
               
4.    POC_______________________________________________
           
5.    GGT_______________________________________________
           
6.    BGA_______________________________________________

7.    UDPGT_____________________________________________

8.    OCT________________________________________________

9.    H-H  Equation_________________________________________

10.    PCO2________________________________________________

11.    ALP_________________________________________________

12.      ACP_________________________________________________

13.    LDH_________________________________________________           

14.    CK___________________________________________________

15.    AST__________________________________________________



ANSWER THE QUESTIONS AND RECORD THEM. COME BACK NEXT WEEK TO CHECK YOUR ANSWERS AGAINST THE KEY ANSWERS.

Monday, January 2, 2017

Answers and Explanations to Laboratory Mathematics

Let’s check if your answers are correct to these Review Questions in Lab Math. If not, you can review the correct answers, and then take the test again. Understand the concept, so that when the values change, you still know the correct solution.

1. B- The atomic weights are: Na (sodium) = 23, and Cl (chloride) = 35.5
    23+35.5 = 58.5

2. D – The atomic weights are: H (hydrogen) = 1 (x 2) since you have 2 molecules, S (sulfur)=32, and O (oxygen) = 16 (x 4) since you have 4 molecules of oxygen.

    2+32+    64 = 98  

3. E – The atomic weights are: Ca (calcium)=40, Cl(chloride)= 35.5 x 2, H= 1 x 2, O = 16
    40+71+2+16= 129

4. A – The atomic weights are: H = 1, Cl=35.5
    1+35.5 = 36.5

5. C – The atomic weights are: H=1, N (nitrogen) = 14, O (oxygen) = 16 x 3
    1+14+48 = 63

6. A – 50 milliliters; this is solved using the formula, C1V1=C2V2
    Where: C1- concentration 1 (original concentration)
           V1 – volume 1 (original volume)
            C2 – concentration 1 (final concentration)
             V2 – volume 2 (final volume)

    Analyzing the problem, you can substitute the values now.

    (0.5)V1) = (0.1)(250)
    V1 = (0.1)(250)/0.5
    V1 = 50 milliliters of 0.5 N HCl  

7. D – 1.275 milliliters or 1.3; you can solve it using two methods. The first method is through ratio and proportion. Normal Saline Solution (NSS) is 0.85% NaCl; it is the solute. If you want to round it off, to 0.9% you can.

But for this particular problem, we will use 0.85%. The 0.85% NaCl is 0.85 grams of NaCl diluted to 100 milliliters of distilled water up to the 100 mark of a volumetric flask. Since 0.85 grams is to 100, we’ll make this as the basis:

0.85/100 = x/150

Transposing the equation to get x, we come up with this:

X= 0.85 x 150/100
X = 1.275 or 1.3 milliliters

The second method is to use the percent formula, which is:

%=Weight in grams/Total volume of solution X 100.
Substituting the values:

85= W in g/150 x 100

Transpose to isolate W in g and you’ll come up with this formula:
W in g = 0.85 x 150/100 = 1.275 grams of NaCl

Tips to remember:

* When solving for percent solutions in relation to weight and volume, convert first the volume to milliliters if it is not in milliliters.

* If the volume is 100 milliliters, substitute the percent sign with the grams and that is already the weight of the solute.

Examples:

•    5% CaCl2 is actually prepared by weighing 5 grams of CaCl2 and dilute it with distilled water up to the 100 mark in a volumetric flask. The volumetric flask is the most accurate glassware in the preparation of solutions.

•    10% NaOH = 10 grams NaOH diluted with distilled water up to the 100 mark in a volumetric flask.
   
9. C – 222 grams; to determine the weight of a substance when preparing for a Molar solution, use the formula:

    W= DV x DM x MW

    Where:

    W – Weight in grams
    DV- Desired Volume (in liters)
    DM – Desired Molarity
    MW – Molecular Weight

    Substituting the values, we get:

    W= 1 liter x 2 M x 111
    W = 222 grams of CaCl2

10. A – 1.37 N; You can use the formula for Normality, which is N = (Volume of solute/Volume of solution/MW/f)/Volume of solution in liters, and the formula for percent (V/V), which is % = Volume of solute in milliliters/Total volume of solution in milliliters x 100.

With the first formula, the substitution would be:

    N= ((0.5/100)/(36.5/1)/1
    N = (0.005)/36.5
    N = 1.3698 0r 1.37

The volume of solution is assumed to be 100 milliliters because it is a percent solution.
You can also use the conversion units, making use of the percent formula.

Saturday, December 31, 2016

Review Questions in Laboratory Mathematics

Questions in Laboratory Mathematics

Laboratory Mathematics is a necessary part of the Clinical Chemistry section of a Clinical Laboratory. Although, reagents and substances now are commercially prepared, it is still crucial that a Medical Technologist or Laboratory Scientist knows how to prepare the most common solutions used in CC. The need for serial dilutions is also essential in any section of the laboratory.

DIRECTIONS: For each of the questions, there is only ONE correct answer. Select the BEST answer.

Questions in Laboratory Mathematics

Determine the molecular weights of the following substances: Atomic weights: H=1, Na=23, Cl=35.5, S=32, O=16, Ca=40, N=14

1. NaCl
2. H2SO4
3. CaCl2.H2O
4. HCl
5. HNO3

Choices: A. 36.5 B. 58.5 C. 63 D. 98 E. 129 F. 111

6. How many milliliters of 0.5 N hydrochloric acid is needed to prepare 250 milliliters of 0.1 N hydrochloric acid?

A. 50 milliliters
B. 200 milliliters
C. 20 milliliters
D. 60 milliliters
E. 25 milliliters

7. If you are to prepare 150 milliliters of normal saline solution, how much of the solute will you need?
A. 1.60 milliliters
B. 10 milliliters
C. 15 milliliters
D. 1.3 milliliters
E. 1.5 milliliters

8. How will you prepare a serum-NSS dilution of 1:25, if you have 0.50 milliliters of serum?
A. Add 0.50 milliliter of serum to 20 milliliters of normal saline solution.
B. Add 0.50 milliliters of serum to 25 milliliters of normal saline solution.
C. Add 0.50 milliliters of serum to 12 milliliters of normal saline solution.
D. Add 0.50 milliliters of serum to 24 milliliters of normal saline solution
E. None of the above

9. What is the weight of CaCl2 needed to prepare 1 liter of 2 Molar solution?
A. 111 grams
B. 55.5 grams
C. 222 grams
D. 200 grams
E. 75 grams

10. What is the Normality of a 0.5% HCl solution?
A. 1.37
B. 1.80
C. 18.25
D. 10.85
E. 0.18

Answers to be posted NEXT WEEK.

HERE ARE THE ANSWERS . Click the link to check if your answers are correct!

Sunday, September 4, 2016

Answers to Question on Laboratory Mathematics



1. B- The atomic weights are: Na (sodium) = 23, and Cl (chloride) = 35.5
            23+35.5 = 58.5

2. D The atomic weights are: H (hydrogen) = 1 (x 2) since you have 2 molecules, S (sulfur)=32, and O (oxygen) = 16 (x 4) since you have 4 molecules of oxygen.
            2+32+ 64 = 98    

3. E The atomic weights are: Ca (calcium)=40, Cl(chloride)= 35.5 x 2, H= 1 x 2, O = 16
            40+71+2+16= 129

4. A The atomic weights are: H = 1, Cl=35.5
            1+35.5 = 36.5

5. C The atomic weights are: H=1, N (nitrogen) = 14, O (oxygen) = 16 x 3
            1+14+48 = 63

6. A 50 milliliters; this is solved using the formula, C1V1=C2V2
            Where: C1- concentration 1 (original concentration)
                           V1 volume 1 (original volume)
                            C2 concentration 1 (final concentration)
                             V2 volume 2 (final volume)

            Analyzing the problem, you can substitute the values now.
            (0.5)V1) = (0.1)(250)
            V1 = (0.1)(250)/0.5
            V1 = 50 milliliters of 0.5 N HCl    
 
7. D 1.275 milliliters or 1.3; you can solve it using two methods. The first method is through ratio and proportion. Normal Saline Solution (NSS) is 0.85% NaCl; it is the solute. If you want to round it off, to 0.9% you can. But for this particular problem, we will use 0.85%. The 0.85% NaCl is 0.85 grams of NaCl diluted to 100 milliliters of distilled water up to the 100 mark of a volumetric flask. Since 0.85 grams is to 100, well make this as the basis:

0.85/100 = x/150

Transposing the equation to get x, we come up with this:

X= 0.85 x 150/100
X = 1.275 or 1.3 milliliters

The second method is to use the percent formula, which is:

%=Weight in grams/Total volume of solution X 100.
Substituting the values:

85= W in g/150 x 100

Transpose to isolate W in g and youll come up with this formula:
W in g = 0.85 x 150/100 = 1.275 grams of NaCl

Tips to remember:

* When solving for percent solutions in relation to weight and volume, convert first the volume to milliliters if it is not in milliliters.

* If the volume is 100 milliliters, substitute the percent sign with the grams and that is already the weight of the solute.
Examples:
·         5% CaCl2 is actually prepared by weighing 5 grams of CaCl2 and dilute it with distilled water up to the 100 mark in a volumetric flask. The volumetric flask is the most accurate glassware in the preparation of solutions.

·         10% NaOH = 10 grams NaOH diluted with distilled water up to the 100 mark in a volumetric flask.
           
9. C 222 grams; to determine the weight of a substance when preparing for a Molar solution, use the formula:

            W= DV x DM x MW

            Where:
            W Weight in grams
            DV- Desired Volume (in liters)
            DM Desired Molarity
            MW Molecular Weight

            Substituting the values, we get:

            W= 1 liter x 2 M x 111
            W = 222 grams of CaCl2

10. A 1.37 N; You can use the formula for Normality, which is N = (Volume of solute/Volume of solution/MW/f)/Volume of solution in liters, and the formula for percent (V/V), which is % = Volume of solute in milliliters/Total volume of solution in milliliters x 100.

With the first formula, the substitution would be: 

            N= ((0.5/100)/(36.5/1)/1
            N = (0.005)/36.5
            N = 1.3698 0r 1.37 

The volume of solution is assumed to be 100 milliliters because it is a percent solution.
You can also use the conversion units, making use of the percent formula. For Normality your units would be milliequivalent per liter of solution. (MEq/L)