These are answers for questions used in the clinical chemistry review. Since you already have the questions, I will only do the answers. If you have any questions, don't hesitate to e-mail me at Remember that knowing why an answer is wrong is almost as important as choosing the right answer!

Urinalysis answers are here. Good luck!

1. a is not right since "immediate glucose distress" implies an abnormal glucose value NOW, which is not the case. The glycosylated hemoglobin is, however, elevated (should be <9%), indicating elevated blood glucose levels in the preceding 2-3 months. Since the glucose value today is normal, it would not be diagnostic for NIDDM, and the HbA1c is NEVER used for diagnosis! only for therapy follow-up... the two test aren't meant to correlate since they are asking two completely different questions!

2. Pre-diabetes, as defined in 2003 by the American Diabetes Association, is two instances (using two separate fasting samples) of glucose values between 100 and 125 mg/dL, inclusive. “Normal”, therefore, is now considered to be values LESS THAN 100 mg/dL, and “diabetes mellitus” would be two fasting glucose values greater than or equal to 126 mg/dL.

3. Renal threshold is 160-180 mg/dL. therefore, any tubular filtrate level greater than that will not be fully reabsorbed, leading to glucose in the urine. d is the only value greater than 180 mg/dL.

4. Of all the values listed, and with the patient's history, you'd expect the urine ketone, glucose and clinitest values to be increased. Remember that the sensitivity for dipstix glucose tests is about 50 mg/dL, while for clinitest it is 250 mg/dl. Therefore, you can have a positive stix without having a positive clinitest.

5. Spinal fluid, being an ultrafiltrate of plasma, will have glucose values which are 60-70% of blood glucose values. Given this patient's blood glucose, you could accept a CSF glucose value between 54 and 63 mg/dl. Remember that vitamin C negatively interferes with ANY reaction which uses peroxidase as the final color-producing step.

6. Colloidal osmotic pressure in the blood is maintained by albumin. This provides the concentration gradient needed to keep plasma and tissue water balanced between the two compartments. When albumin is decreased, plasma water will move into the tissue space, causing edema.

7. At pH 8.6, ALL proteins have a net negative charge and will migrate toward the anode (remember the "anode" is named for the type of ion attracted to it!). That means the proteins least attracted (least negative) to the anode will be at the cathode. The protein with the greatest net negative charge is albumin, so it is the "most anodic", "least cathodic", while gamma globulins are closest to the cathode, making them "most cathodic", "least anodic". Gamma globulins are decreased in hypogammaglobulinemia. Nephrotic syndrome would show decreased albumin, viral hepatitis would show increased gamma, and emphysema would show decreased alpha 1.

8. With only alpha 1 and 2 showing an increase, that makes this acute inflammation. With dehydration, ALL would be increased; edema would show decreased albumin, chronic infection would show increased gamma.

9. Cystatin-C is replacing creatinine and creatinine clearance as the preferred assessment for glomerular filtration. High sensitivity C-reactive protein levels will rise at early signs of inflammation from damaged coronary arteries, and therefore is a predictor of those at risk for heart attacks. B-type nartiuretic peptide levels will rise with congestive heart failure. Troponin I is a specific cardiac marker of myocardial infarction.

10. Gamma globulins are made in plasma cells, not the liver.

11. Sodium is the major contributor to serum osmo. If the question asked for URINE, you would choose urea. If the question asked for major ANION, you would choose chloride.

12. pattern 1 = Addison's; pattern 2 = hemolysis; pattern 3 = hypoparathyroidism; dehydration would have ALL increased; Cushing's would be the opposite of Addison's; hyperparathyroidism would be the opposite of hypoparathyroidism; vomiting would show decreased chloride and sodium.

13. Wilson's disease = copper; transferrin = iron; B12 = cobalt; wound healing = zinc; thyroid = iodine

14. High risk is increased total cholesterol with decreased HDL.

15. Elevations of chylomicron will be seen if the sample is non-fasting; this will elevate the triglyceride.

16. If you set up the lipoprotein electrophoresis pattern, so that the cathode is to the left and the anode is to the right, the lipoproteins will be separated (left to right) in the order chylomicron (at the application point), LDL (beta), VLDL (pre-beta), HDL (alpha). There is no significant quantity if IDL normally, but it would be found between LDL and VLDL. Remembering this, you can figure out the lipoprotein increased in each type of hyperlipoproteinemia. I=chylomicron; II=LDL; III=IDL; IV=VLDL; V= chylomicron and VLDL. Therefore, Type IV is characterized by increased pre-beta (VLDL). Elevated chylomicrons would be in types I and V, high dietary triglyceride would also be I and V (from chylomicrons), and the patient's sample after refrigeration would appear milky.

17. Amylase is used for diagnosing pancreatic disease.

18. NADH has an absorbance maxima at 340 nm. This means that if your reagent STARTS with high concentrations of NADH, your absorbance at 340 nm will begin at a high level. As the reaction of NADH to NAD proceeds, the absorbance will decrease. Therefore, as the patient analyte concentration increases, you will have a LOWER final absorbance (inversely proportional relationship). Since this patient's final absorbance is lower than the control's, its concentration must be higher.

19. RBC's contain high concentrations of LD, AST, ALT and ACP.

20. This patient's results are a classical pattern for myocardial infarction. Intravascular hemolysis would have the LD iso1 and 2 both elevated, but 2 would be greater than 1. CK is not increased in liver trauma.

21. UDP-glucuronyl transferase catalyzes the conversion of bilirubin glucuronide. Inhibiting that enzyme would decrease water soluble bili concentrations and elevate lipid soluble bilirubin levels. If bilirubin glucuronide is not put into the intestines, stools will have less color, urine and serum bilirubin will be "normal" due to the low amounts and sensitivity levels of the methods used.

22. Creatinine is related to muscle mass; urea to protein intake, urobilinogen to urine color. Urine creatinine would be INVERSELY related to renal disease.

23. Only renal failure will cause all these to be increased. Creatinine would be normal in the other 3. Urea would be decreased in liver disease.

24. Answers for these were a, d, c, b, a....remember that pH will change only if the ratio between bicarbonate and carbonic acid changes. If the bicarb changes, it's metabolic; if carbonic acid changes, it's respiratory.

25. Normal ratio for pH 7.40 is 20 to 1, but since the normal range for pH is 7.35 to 7.45, the normal range extends from 17.78 to 22.4. The only combination with a value outside that range is the third, giving a value of 24 to 1.

26. If the sample is kept anaerobic the whole time, you are not losing pO2 or pCO2 to the atmosphere. Cellular metabolism continues, however, making CO2 increase and O2 decrease. Since CO2 increases, that will increase carbonic acid levels, making the sample more acidic (decreasing pH). Choice c is the only possibility.

27. Metabolic acidosis means the numerator of HCO3/H2CO3 is decreased. One way to compensate for this is to decrease the denominator by decreasing CO2. Hyperventilation would do nicely! Note that (a) would further decrease the bicarb, while (c) and (d) would both ADD acid to the system.

28. Excitation wavelengths excite fluorescent molecules to a higher outer shell electron energy level. When these electrons come back to ground state, they themselves emit SOME the energy as light. This light will be of a lower energy than that which caused the excitation. Since it's lower energy, the wavelengths must be LONGER, since there's an inverse relationship between wavelength number and light energy. This makes the answer (a)

29. Bandpass is the actual group of wavelengths which will pass through the sample when you have chosen one particular wavelength on the instrument. The wavelength you have chosen will be the middle value in the range, and the bandpass will define the WHOLE range. Therefore, it is actually the set wavelength with 1/2 the bandpass on one side and 1/2 on the other. In this instance, 4 nm bandpass means + 2nm. If the set wavelength is 560 nm and the instrument's bandpass is 4 nm, the actual wavelengths passing through the sample will be 560 + 2nm, or 558-562 nm. Bandpass is a characteristic of each instrument, and the narrower the bandpass, the better.

30. Reflectance as a way to use light does not need a "sample container", since the light is usually reflected from a solid surface. Since you are not using absorbed light, you cannot use "Beer's Law". ISEs use potentiometry as their principle, while reflectance does use light photons striking a photocell, generating a current. (d) is the correct answer, and the "dry slide technology" described is that of the Vitros/Ektachem. It is also the principle used in skin bilirubinometers or skin glucometers.

31. Positive predictive value is the number of true positives divided by all those the test called "positive". For this problem, that is 90 divided by 129, or 70%. Negative predictive value would be the true negatives divided by those your test called "negative". Diagnostic sensitivity is the number of true positives divided by all those who actually have the disease, while specificity is the number of true negatives divided by all those who truly do not have the disease.

32. Diagnostic specificity differentiates true negatives from false positives. Diagostic sensitivity discriminates true positives from false negatives. I remember this in an almost backwards way: the "p" in specificity goes with the "n" in true negative, while the "n" in sensitivity goes with the "p" in true positive. It would be nice if the letters corresponded, but even that helps with remembering they don't!

33. Since the hatched area shows those who are both "healthy" but whom the test called "diseased", these are the "false positives" for this test. Note that such false positives don't just have to be people who have HIGH values!

34. You first need to figure out the average value, which would be the two extremes (152 + 168) divided by 2 = 160 mg/dL. Since the range given is 2sd, you can now subtract 168 minus 160 to get 8 mg/dL. Note that you'd get the same answer if you subtracted 152 from 160! This 8 mg/dL is 2sd, therefore 4 mg/dL is 1sd, and 12 mg/dL would be 3sd. The mean + 3sd range would be 160 + 12 mg/dL, or 148 - 172 mg/dL.

35. Since pH is a logarithmic scale, a difference of 4 is actually a difference of 10 to the 4th power, or 10,000. Since the pH is becoming more alkaline going from 4.6 to 8.6, the change in [H+] is divided by 10,000.

36. Nephelometry uses light scatter as the measured signal. The only "cloudy" type of assay listed is the the lipase emulsion. Note that (d) is describing reflectance!

37. The answers could be either (b) or (c). The change in freezing point is predictable, with every mOsm/kg solute causing a change in freezing point (compared to pure water) of -0.00186 degrees centigrade.

38. As written, the answer is (d). If a hydrogen peroxide electrode is used, the signal will be DIRECTLY proportional. If the RATE of oxygen consumption is the parameter measured, the signal will be DIRECTLY proportional to the glucose concentration.

39. The factors to solve this are (10g/100 mL), (1000 mL/Liter) and (mole/171 g), multiplied together  = 0.6 mol/L

40. 25 mmol HCO3   times   0.5 L   times   mmole Mg(HCO3)2   times     146 mg                       = 912.5 mg Mg(HCO3)2
                 L                                                2 mmoles HCO3                       mmole Mg(HCO3)2

41. This is a V1C1=V2C2 problem since the stock material is already in liquid form.
       (95%)(x mL) = (70%)(2000 mL)
                     x = 1474 mL

42. To correct for a dilution, you multiply the diluted concentration value by the reciprocal of the dilution used. Here, the diluted concentrtion is 108 mg/dL, and the reciprocal of a 1:4 dilution would be 4. Therefore, (108 mg/dL)(4) = 432 mg/dL.

43. The dilution 1:5 sets up the relationship which needs to exist between sample volume (1 part) and total volume (5 parts). Any fraction that would reduce in lowest terms to 1/5 will be acceptable. The 200 uL given in the problem is the desired TOTAL VOLUME, therefore it would go in the denominator.

            1/5 = x microliters/200 microliters..........solving for x = 40 uL sample volume needed

OOPS! A second #43! Steady state is usually achieved in 5 half lives of a drug. If the half-life of the drug is 24 hours, steady state will be achieved in 5 times the 1/2 life, or (5)(24) or 5 days.

44. (b) will cause more free drug to be present in the circulation, therefore creating an INCREASED effective plasma drug level.

45. Only (c) does not experience classical negative feedback from hormones produced by target glands.

46. (c) is correct. (b) and (d) have nothing to do with FSH. (a) would cause an increased FSH because of the removal of negative feedback from the gonads.

47. If the adrenal cortex does not respond to externally supplied ACTH, it itself is the problem and not the anterior pituitary. This makes (a) the answer. (b) and (d) would both be characterized by INCREASED cortisol.

48. T3 uptake is a measure of the level of thyroxine binding globulin, since the test involves adding excess labelled REAGENT T3, and measuring how much has become bound.

49. Answers as follows. There is an excellent summary table on pp. 619-622 in the 4th edition of Teitz Fundamentals of Clinical Chemistry:

Antidiuretic hormone

posterior pituitary

reabsorption of water from distal convulted tubule


adrenal medulla

glycogenolysis, lipolysis



prolongation of corpus leuteum function


anterior pituitary

growth of follicles, secretion of estrogens, ovulation (with LH)



secondary male sex characteristics


thyroid gland

inhibition of calcium resorption; lowers blood [Ca++]

parathyroid hormone

parathyroid glands

regulation of calcium and phsophate levels


thyroid gland

stimulation of oxygen consumption and metabolic rate in tissues


pancreas (alpha cells)



adrenal cortex

salt and water balance


pancreas (beta cells)

regulation of carbohydrate metabolism; lipogenesis


ovary, placenta

preparation of uterus for implantation; maintenance of pregnancy

50. (b) is used for monitoring ovrian cancer; (a) is used for liver CA; (c) is for GI/colorectal, pancreatic, lung, breast cancer; (d) is used for bone, liver, leukemia, sarcoma, as well as ovary, lung, trophoblastic, GI and even Hodgkin's disease.