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Wednesday, August 22, 2012

Clinical Chemistry 3 Answers to Case Analysis, Problem Solving in Endocrinology and Toxicology

IV. CASE ANALYSIS: (10 pts.)

A 32 YR. OLD FEMALE PATIENT COMPLAINED OF DIZZINESS, DYSPNEA AND NAUSEA. SHE HAD RED RASHES ALL OVER HER BODY, AND HER BREATH WAS FRUITY.

THE LAB RESULTS WERE:
:
•    FT3 - 30 ng/dL
•    FT4 - 2 ng/dL
•    TSH - normal
•    WCC – 5,100/ mm3
•    RCC – 5, 113, 100/ mm3
•    Total cholesterol – 1.87 mmol/L
•    Triglycerides – 1.3 mmol/L
•    Glucose = 3.8 mmol/L
•    Hemoglobin = 150 mg/dL
•    Hematocrit 52 vol%

QUESTIONS FOR CASE ANALYSIS: (10 pts.)

1. WHAT FURTHER TESTS WOULD YOU RECOMMEND? DEFEND YOUR ANSWER.

ANSWER:


All of the values given are within normal concentrations, so thyroid dysfuntction could be discounted, if other diagnostic tests and physical symptoms are not evident.  

Alcohol testing should be done, like Gas Chromatography or Enzymatic. A breath analyzer test for alcohol, or QED test for alcohol in saliva could also be performed. GGT could also be tested for alcohol long term abuse.


2. WHAT WOULD BE THE MOST LIKELY DIAGNOSIS? DEFEND YOUR ANSWER.

Red rashes without any occurrence of fever and a normal WBC most likely indicates allergy. Since it is coupled with dyspnea and dizziness, allergy is further emphasized.

The fruity odor could not be because of hyperglycemia beause of diabetes mellitus, so there’s one reason, which is alcohol consumption.

These all point out to alcohol allergy and intoxication.


V. PROBLEM (5 PTS. EACH)

1. WHAT IS THE HALF OF A DRUG WITH AN ORIGINAL CONCENTRATION OF 500 mg/dL, AND IS REDUCED 50 mg/dL FOR EVERY 30 MINUTES.

ANSWER =150 minutes or 2.5 hours

500 mg/dL - 30 minutes = 450 mg/dL
450 mg/dL – 60 minutes = 400 mg/dL
400 mg/dL -  90 minutes = 350 mg/dL
350 mg/dL – 120 minutes = 300 mg/dL
300 mg/dL – 150 minutes = 250 mg/Dl

Or 450 mg/dL minus 400 = 50 mg/Dl

50 / 30 = 250/X;
   
    50X = 250 x 30

    250 x 30
X = _____________
           50
   
X = 150 minutes or 2.5 hours


2. SOLVE FOR THE HALF-LIFE OF PARACETAMOL WITH AN ORIGINAL CONCENTRATION OF 600 mg/dL, AND WHICH HAS BEEN REDUCED TO 400 mmol/L AFTER 120 MINUTES.

Before you start solving the problem, you can convert either the mg/dL to mmol/L or vice versa. (If you have tried converting the units, you get 3 points. Since the atomic weight of paracetamol is not given. You should have shown conversion formula, anyhow, to get full points. That’s why it helps to keep reading. Board Topnotchers are those who research and read.)

Use the following conversion units:

1 g = 1,000 mg
100 ml= 1 dL
10 dL = 1 L
1 mole = 1,000 mmol

You can convert from mmol/L to mg/dL by:

400 mmol/L  x  1 mole     x   MW paracetamol (in grams)  x  1,000 mg x 1 L_= P in mg/dL
                      1,000 mmol          1 mole                                      1 g             10 dL

You can also convert mg/dL to mmol/L by:

600 mg/dL  x  1 g      x   1 mole                                x   1,000 mmol x 10 dL_= P in mg/dL
                   1,000 mg  MW paracetamol (in grams)        1mole            1L

MW OF PARACETAMOL = 151.16

(C8H9NO2)

C = 12, H=2, N= 14, O = 16

SINCE THE VALUE IS EXTREMELY HIGH, IT IS ENOUGH THAT YOU HAVE SHOWN THE CORRECT STEPS.

VI. TRANSCRIPTION: (1 pt.)

DASDRUG ABUSE SCREENING
ETEMERGENCY TOXICOLOGY
TDMTHERAPEUTIC DRUG MONITORING
D-9THCDELTA-9-TETRAHYDROCANNABINOL
NSAIDNON-STEROIDAL ANTI-INFLAMMATORY DRUGS
HLHALF LIFE
NADHNICOTINAMIDE ADENINE DINUCLEOTIDE REDUCED FORM
EMIT ENZYME MULTIPLIED IMUNOASSAY TECHNIQUE
ELISAENZYME LINKED-IMMUNOSORBENT ASSAY
ANS 8 –ANILINO-1 NAPHTHALENE  SULFONIC ACID

Clinical Chemistry 3 - Endocrinology -Toxicoloy Exam Questions


ANSWERS TO CLINICAL CHEMISTRY 03 MIDTERM EXAMS 2012-2013
IDENTIFICATION
1.__half-life__________________________________
2.__pituitary gland_____________________________
3.__cortisol binding protein______________________
4.__T3 or triiodothyronine_______________________
5.__RT3U, THBR_ ______________________________
6._Thyroid Binding Globulin_or Thyroxine Binding Globulin
7.__anterior pituitary___________________________
8.__ultrasound________________________________
9.__8-anilino-1 naphthalene sulfonic acid____________
10.__acetaldehyde______________________________

MULTIPLE CHOICE
1.       C             26. E       51. B      76. G
2.       A             27. A      52. D      77. F
3.       B             28. A      53. D      78. F
4.       C             29. D      54. C      79. BONUS
5.       D             30. A      55. B      80. BONUS
6.       C             31. D      56. A      81. F
7.       B             32. A      57. B      82. F
8.       D             33. B      58. B      83. F
9.       C             34. D      59. B      84. G
10.   D             35. D      60. C      85. F
11.   A             36. A      61. B      86. E
12.   A             37. A      62. D      87. F
13.   E              38. A      63. BONUS 88. A
14.   A             39. D      64. D      89. A
15.   D             40. A      65. B      90. B
16.   B             41. C      66. D      91. C
17.   D             42. B      67. C      92. A
18.   B             43. A      68. B      93. E
19.   D             44. A      69. C      94. F
20.   D             45. D      70. D      95. A
21.   D             46. D      71. D      96. A
22.   A             47. C      72. B      97. B
23.   D             48. A      73. D      98. H
24.   E              49. A      74. A      99. F
25.   B             50. D      75. A      100. C
MATCHING TYPE

1.      FT3         =          A                                             A. 10-40 ng/dL
2.      FT4        =          G                                             B. 60-220 ng/dL
3.      TT4        =          D                                             C. 0.02-0.03%
4.      TT3        =          B                                             D. 4.5 – 13 ug/dL
5.      TSH       =          E                                              E. 0.5 – 5.0 u U/mL                                       
6.      ETHANOL DETECTION        = C                  F. 0.2-0.3%
7.      TBG     =           I                                               G. 1.2 –2.5 ng/dL
8.      THBR    =          K                                             H. 15-24 ug/mL
9.      FT3I       =          H                                             I. LESS THAN 10 ng/mL
10.  FT4I       =          L                                             J. 0.8-1.35
K. 25 % -30 %
L. 4.5 – 12 ug/Dl