Chemistry Sample Questions PDF

132 CHEMISTRY SAMPLE QUESTIONS 1. A patient is administered an oral glucose 6. A 65 year old male visits his physician tolerance test with the following results: complaining of fatigue, shortness of breath Fasting serum glucose- 128 mg/d...

132 CHEMISTRY SAMPLE QUESTIONS 1. A patient is administered an oral glucose 6. A 65 year old male visits his physician tolerance test with the following results: complaining of fatigue, shortness of breath Fasting serum glucose- 128 mg/dl and difficulty breathing. He is 6 feet tall 2 hour post-load serum glucose - 238 and weighed 200 pounds for many years, mg/d but has recently gained 30 pounds. His These results correlate with ankles and hands are swollen and he A. Normal complains of feeling bloated. His blood B. Diabetes mellitus pressure is markedly elevated. C. Hypoglycemia What test should the physician order? D. Malabsorption A. Troponin 2. A physician calls to ask assistance in B. Arterial blood gases choosing a test to monitor a diabetic C. BNP or NT pro-BNP patient's long term control. Which of the D. Myoglobin following would you suggest? 7. Which results will typically not be affected A. C-peptide of insulin in a hemolyzed sample? B. Glycosylated hemoglobin(s) A. Acid phosphatase C. Fasting plasma glucose B. Sodium D. Postprandial plasma glucose C. Iron D. Potassium 3. Which of the following sets of test results indicates the greatest risk for coronary 8. Which enzyme is decreased in insecticide artery disease? poisoning due to organophosphates? Total Cholesterol HDL cholesterol LDL cholesterol A. Alkaline phosphatase mg/dL mg/dL mg/dL B. Amylase A. 145 55 90 C. Cholinesterase B. 165 60 105 D. Creatine kinase C. 245 60 95 9. The biuret method for determining serum D. 345 30 205 total protein is dependent upon 4. Elevated conjugated billirubin in both A. Amino acid content serum and urine, decreased urine B. Number of peptide bonds urobilinogen, and decreased fecal urobilin C. Nitrogen content D. Protein precipitation is characteristic of A. Cirrhosis 10. An arterial blood sample is received in the B. Hemolytic jaundice laboratory 45 minutes after collection with C. Hepatitis a bubble in the syringe. The technologist D. Obstructive jaundice should 5. Calculate the LDL cholesterol from the A. Perform the test immediately following data: B. Reject the sample because the p02 Total cholesterol = 250 mg/dL will be falsely 1' HDL cholesterol = 40 mg/dL C. Reject the sample because the pH Triglyceride = 210 mg/dL will be falsely 't D. Reject th e sample because the pCO2 A. 140 mg/dL will be falsely 1' B. 168 mg/dL C. 210 mg/dL 11. A patient in diabetic ketoacidosis would D. 237 mg/dL exhibit a/an A. ,t. pCO2 , ,t. HCO 3, ,t. pH B. tpC02, tHCOj , pH + C. Normal pCO2 , ,t HCO3 , t pH D. Normal pCO2, t HCO3, t pH  133 12. The best laboratory test for detecting D. To enhance the effect of the drug cystic fibrosis is A. Lipase 17. Electrolyte values on 4 patients ore as B. Sodium follows: C. Sweat chloride 1. Na+ 149 mmol/L; c1- 102 mmol/L; D. Trypsin HC03 26 mmol/L 13. A 25 year old female visits her physician 2. Na+ 153 mmol/L; c1- 105 mmol/L; with the following symptoms: feeling tired HC03 28 mmol/L all of the time, recently gaining IS pounds, 3. Na+ 150 mmol/L; c1- 103 mmol/L; swollen neck, dry skin, hoarseness, and HC03 25 mmol/L delayed reflexes. The physician ordered 4. Na+ 151 mmol/L; c1- 104 mmol/L; thyroid tests with the following results: HC03 27 mmol/L TSH = 3.0 mIU/L (ref. 0.2-4.0 mIU/L) Based on these results, what do you Free T4 = 0.4 ng/dL (ref. 0.8-1.8 ng/dL) conclude about your electrolyte analyzer? A. There is a problem with the chloride The physician performed a TRH stimulation analysis. test with the following results: B. There is a problem with the 30 minute TSH = 6.0 mIU/L bicarbonate analysis. 60 minute TSH = 15.0 mlU/L C. All analyses seem to be accurate. The results indicate that the patient is D. There is a problem with the sodium analysis. suffering from: A. Primary hyperthyroidism 18. Why should potassium levels be B. Primary hypothyroidism monitored prior to and during C. Secondary or "pituitary" administration of IV insulin? hypothyroidism A. Insulin causes potassium to move into D. Tertiary or "hypothalamic" cells which may cause a drop in hypothyroidism potassium levels. 14. A blood ammonia level was ordered on a B. Insulin. concentrates potassium in patient with Reye·s syndrome. The results the seru:m causing hyperkalemia. C. IV fluids dilute electrolyte values on the clotted sample were normal. The causing hyponatremia and most likely explanation for these results is hypokalemia. A. Inappropriate sample collection D. Exogenous insulin causes hemolysis B. Incorrect diagnosis which falsely elevate potassium C. Patient is under treatment levels. D. QC was out of acceptable limits 15. A patient with ,t- serum levels of creatine kinase (CK), aldolase (ALO) and aspartate aminotransferase (AST) but normal levels of alanine aminotransferase (ALT) most likely has A. Hepatitis B. Muscular dystrophy C. Myocardialinfarction D. Pulmonary infarction 16. In drug testing using thin layer chromatography, why is a volatile organic solvent mixed with urine specimens? A. To preserve the drug B. To remove interferences C. To extract the drugs 134 ANSWERS AND RATIONALE 1. B 4. D The American Diabetes Association (ADA) Obstruction in the bile duct results in criteria for classification and diagnosis of conjugated hilirubin hacking up into the diabetes is as follows: circulation. Because it is water soluble, conjugated biliruhin is excreted into the One of the four criteria must be met to urine. Since bile flow to the intestines is diagnose Diabetes mellitus obstructed , urobilinogen and urobilin are 1. Diabetic Symptoms and random found in the feces in less than normal glucose2 200 mg/dl amounts. Because of the liver damage, 2. Fasting serum glucose z 126 mg/dJ options A and C would result in increased 3. 2-hour postload serum glucose z 200 levels of unconjugated and conjugated mg/dl bilirubin in the serum. The incr ea sed 4 Ale greater than or equal to 6.5% conjugated fraction would be excreted in the urine. However fecal excretion of In this case, two of the criteria are met. urobilinogen and urobilin would also be The oral glucosfi tolerance test is not increased as the intestinal bacteria broke necessary when the fasting glucose is z down the increased hiliruhin. Option B 126 mg/dl. would result in increased unconjugated 2. B bilirubin. Since the liver is undamaged , it could conjugate at the normal rate and there Glycosylated hemoglobin is the specific would be no excess to b e excreted in the hemoglobin fraction to which glucose urine. There would be excess fecal molecules become irreversibly attached. urobilinogen due to the increased total Results of glycosylated hemoglobin are biliruhin which would cause an increase in proportional to the average glucose level urine urohilinogen. during the previous 1 to 3 month period. Option A is used to evaluate causes of fasting 5. B hypoglycemia. Options C and D give The Friedewald formula is: information about glucose levels for only a short period of time. LDL chol = total chol - HDL chol - triglycerides 3. D 5 = 250 - 40 - (210/5) = 168 The treatment guidelines established by the Adult Treatment Panel of the National 6. C Cholesterol Education Program include recommendations that goals for cholesterol The patient is exhibiting symptoms of (assuming no other risk factors are present) congestive heart failure ( CHF). BNP ( or NT should be less than 200 mg/dL for total pro-BNP) levels correlate linearly to the New cholesterol,