Wednesday, June 28, 2017
ASPARTATE AMINOTRANSFERASE DETERMINATION Review Questions and Answers
1. State the principle of the test.
AST in the serum catalyzes the transfer of an amino acid group to a keto acid group to form oxaloacetate and L-glutamate. This is measured spectrophotometrically at 505-535 nanometers.
2. Name sources of error in this method.
Hemolyzed serum will increase values 10-15 X
Turbid, lipemic and icteric serum needs serum blanking for accuracy
Altered temperatures could either increase nor decrease values
Prolonged or shortened incubation time at specified conditions could increase or decrease values respectively
3. What is the purpose of allowing the reagents to come to room temperature.
The purpose is to allow the reagents to inactivate the reagents so that they could react properly.
4. Differentiate AST from ALT. Cite specific differences.
CRITERIA AST ALT
Substrate L-aspartate & alpha-ketoglutarate L-alanine & alpha-ketoglutarate
Old name SGOT SGPT
One of the end products oxaloacetate Pyruvate
Amount in inside serum 10-15 times 5-8 times
Major clinical significance heart Liver
5. What is the component of the SGOT substrate?
L- aspartate and alpha-ketoglutarate
Sunday, June 25, 2017
Answers to Normality and Molarity Review Questions
Solving the Normality and Molarity of solutions is quite easy by remembering the relationship of Normality to Molarity.
1. Normality may be equal but is always greater than the Molarity in the same solution.
2. If the valence is 1, Normality is equal to Molarity.
Try solving these problems. Atomic weights: Na = 23; Cl = 35.5, H = 1, Ca = 40, Valences: Ca = 2, NaCl = 1, HCl = 1.
1. If you have dissolved 20 grams of sodium chloride in 1.5 Liter of distilled water, what is the:
1.1. Normality
You have to remember that Normality is the Gram Equivalent Weight of substances. Hence, valence is involved. The formula would be:
N = W/GEW/Liter of solution
GEW = W/MW/valence
So, you have to solve first the MW (sum of atomic weights);
MW = 23 (Na) + 35.5 (Cl) = 58.5
Thus, substituting the values: GEW = 20/58.5/1
GEW = 0.3418803
N = GEW/Liter of solution
N = 0.3418803/1.5
N = 0.2279202
Where:
N - Normality
W = weight of substance
GEW = Gram Equivalent Weight
MW = Molecular Weight
M = Molarity
1.2. Molarity
Since you already know the Normality of the solution, you can make use of the short-cut formula, which is:
M = N/valence
Thus, substituting the values:
M = 0.2279202 x 1
M = 0.2279202
The answer is the same. This follows the rule: If the valence is 1, Normality is equivalent to Molarity.
1.3. Percent solution
You don’t usually need the molecular weights when solving percent solutions. You could use the general formula:
% = weight/total volume x 100
NOTE:
In solving for percent solutions, you have to convert the volume to milliliters. So,
1.5 Liters x 1000 mL/1 Liter = 1, 500 mL
Hence:
% = 20/1,500 x 100
% = 1.33333%
2. What is the Molarity of 1 N Hydrochloric acid?
Again in this problem, since you know the Normality, make use of the short-cut formula:
Thus:
M = N/valence
M = 1/1
M = 1
Again: If the valence is 1, Normality is equivalent to Molarity.
3. What is the Normality of 0.8 M calcium chloride?
Use the short-cut formula:
N = M x valence
N = 0.8 x 2
N = 1.6
1. Normality may be equal but is always greater than the Molarity in the same solution.
2. If the valence is 1, Normality is equal to Molarity.
Try solving these problems. Atomic weights: Na = 23; Cl = 35.5, H = 1, Ca = 40, Valences: Ca = 2, NaCl = 1, HCl = 1.
1. If you have dissolved 20 grams of sodium chloride in 1.5 Liter of distilled water, what is the:
1.1. Normality
You have to remember that Normality is the Gram Equivalent Weight of substances. Hence, valence is involved. The formula would be:
N = W/GEW/Liter of solution
GEW = W/MW/valence
So, you have to solve first the MW (sum of atomic weights);
MW = 23 (Na) + 35.5 (Cl) = 58.5
Thus, substituting the values: GEW = 20/58.5/1
GEW = 0.3418803
N = GEW/Liter of solution
N = 0.3418803/1.5
N = 0.2279202
Where:
N - Normality
W = weight of substance
GEW = Gram Equivalent Weight
MW = Molecular Weight
M = Molarity
1.2. Molarity
Since you already know the Normality of the solution, you can make use of the short-cut formula, which is:
M = N/valence
Thus, substituting the values:
M = 0.2279202 x 1
M = 0.2279202
The answer is the same. This follows the rule: If the valence is 1, Normality is equivalent to Molarity.
1.3. Percent solution
You don’t usually need the molecular weights when solving percent solutions. You could use the general formula:
% = weight/total volume x 100
NOTE:
In solving for percent solutions, you have to convert the volume to milliliters. So,
1.5 Liters x 1000 mL/1 Liter = 1, 500 mL
Hence:
% = 20/1,500 x 100
% = 1.33333%
2. What is the Molarity of 1 N Hydrochloric acid?
Again in this problem, since you know the Normality, make use of the short-cut formula:
Thus:
M = N/valence
M = 1/1
M = 1
Again: If the valence is 1, Normality is equivalent to Molarity.
3. What is the Normality of 0.8 M calcium chloride?
Use the short-cut formula:
N = M x valence
N = 0.8 x 2
N = 1.6
ALANINE AMINOTRANSFERASE (ALT) DETERMINATION Questions and Answers
1. Discuss the principle of the method.
ALT in the serum catalyzes the transfer of an amino acid group to a keto acid group to form pyruvate and L-glutamate. This is measured spectrophotometrically at 505-535 nanometers.
2. Aside from sodium hydroxide, what reagent could also be used to alkalinize the solution?
Potassium hydroxide
3. Why should the distilled water used in dissolving your NaOH pellets be CO2 free?
Because carbon dioxide can affect the transfer of the amino group to the keto acid group due to its carbon content. The carbon atom may act as an acceptor molecule in the reaction. This would falsely decrease your values.
4. Why do we use a semi-log graphing paper in plotting your calibration curve?
Because the it is simpler to use since the graphing paper will find the logarithms of the values beforehand.
Saturday, June 24, 2017
How to Solve Normality and Molarity of Solutions
Solving the Normality and Molarity of solutions is quite easy by remembering the relationship of Normality to Molarity.
1. Normality may be equal but is always greater than the Molarity in the same solution.
2. If the valence is 1, Normality is equal to Molarity.
Try solving these problems. Atomic weights: Na = 23; Cl = 35.5, H = 1, Ca = 40, Valences: Ca = 2, NaCl = 1, HCl = 1.
1. If you have dissolved 20 grams of sodium chloride in 1.5 Liter of distilled water, what is the:
1.1. Normality
1.2. Molarity
1.3. Percent solution
2. What is the Molarity of 1 N Hydrochloric acid?
3. What is the Normality of 0.8 M calcium chloride?
CLICK HERE FOR THE ANSWERS.
1. Normality may be equal but is always greater than the Molarity in the same solution.
2. If the valence is 1, Normality is equal to Molarity.
Try solving these problems. Atomic weights: Na = 23; Cl = 35.5, H = 1, Ca = 40, Valences: Ca = 2, NaCl = 1, HCl = 1.
1. If you have dissolved 20 grams of sodium chloride in 1.5 Liter of distilled water, what is the:
1.1. Normality
1.2. Molarity
1.3. Percent solution
2. What is the Molarity of 1 N Hydrochloric acid?
3. What is the Normality of 0.8 M calcium chloride?
CLICK HERE FOR THE ANSWERS.
Friday, June 23, 2017
CLINICAL CHEMISTRY 2 DIAGNOSTIC QUIZ
MATCHING TYPE:
MATCH COLUMN B WITH COLUMN A
COLUMN A (SUBSTANCE)
1. GLUCOSE
2. CHOLESTEROL
3. TAG
4. BUA
5. CREATININE
6. BUN
7. HDL
8. TB H.
9. B1
10. B2
COLUMN B (CONVERSION FACTOR TO S.I.)
A. 0.357
B. 0.113
C. 0.01128
D. 0.0258
E. 0.059
F. 0.0555
G. 0.1357
H..88.4
I. 0.555
J. 0.0359
K. 17.1
ESSAY:
1. DISCUSS THE PREPARATION OF ALKALINE PICRATE.
MATCH COLUMN B WITH COLUMN A
COLUMN A (SUBSTANCE)
1. GLUCOSE
2. CHOLESTEROL
3. TAG
4. BUA
5. CREATININE
6. BUN
7. HDL
8. TB H.
9. B1
10. B2
COLUMN B (CONVERSION FACTOR TO S.I.)
A. 0.357
B. 0.113
C. 0.01128
D. 0.0258
E. 0.059
F. 0.0555
G. 0.1357
H..88.4
I. 0.555
J. 0.0359
K. 17.1
ESSAY:
1. DISCUSS THE PREPARATION OF ALKALINE PICRATE.
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