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Camille Rose F. Yling BSMT -4 MTAP October 22, 2021 4. Given the following data, calculate the coefficient of variation for glucose. Analyte: Glucose Mean: 76 mg/dL Standard deviation: 2.3 a. 3.0% Rationale: The formula for coefficient of variation of standard deviation divided by mean and multiplied by 100%. Since we have the given number, our standard deviation will be 2.3 divided by our mean which is 76 mg/dl multiplied by 100. The answer is 3.02% or 3.0%, the correct answer is A [ CITATION Bis181 \l 13321 ]. b. 4.6% Rationale: B is not the answer. Based on the formula, if we have the standard deviation of 2.3 which will be divided by our mean of 76mg/dl times 100 equal 3.02% or 3.0% instead of 4.6% [ CITATION Bis181 \l 13321 ]. c. 7.6% Rationale: Like letter B, 7.6% is also not the answer since we have the given number for standard deviation of 2.3 and mean of mean of 76mg/dl. If we compute using our formula for coefficient of variation, we will have the 3.0% [ CITATION Bis181 \l 13321 ]. d. 33.% Rationale: The same with b and c, 33% is a higher number which is also not the answer since we have the

given number for standard deviation of 2.3 and mean of mean of 76mg/dl. If we compute using our formula for coefficient of variation, we will have the 3.0% [ CITATION Bis181 \l 13321 ]. 5. A cholesterol QC chart has the following data for the normal control: x = mean of data x = 137 mg/dL x = 1,918 mg/dL 2 SD = 6 mg/dL N = 14 The coefficient of variation for this control is: a. 1.14% Rationale: A is not the answer. Based on the formula, if we have the standard deviation of 3 which will be divided by our mean of 137mg/dl times 100 equal 2.19% instead of 1.14% [ CITATION Bis181 \l 13321 ]. b. 2.19% Rationale: First, to compute for the coefficient of variation, we will divide the SD by 2 which will be 3. Now we will use the formula of standard deviation divided by mean and multiplied by 100%. Using our given, 3 divided by 137 is 0.0218 multiplied by 100, the answer will be 2.19%. c. 4.38% Rationale: Like letter A, 4.38 % is also not the answer since we have the given number for standard deviation of 3 and mean of mean of 137mg/dl. If we compute using our formula for coefficient of variation, we will have the 2.19% [ CITATION Bis181 \l 13321 ]. d. 9.49%

Rationale: The same reason with a and c. The 9.49% is not the answer if we use the formula for the coefficient of variation [ CITATION Bis181 \l 13321 ]. 7. In addition to the number of true negative (TN), which of the following measurements is needed to calculate specificity? a. True positives Rationale: True positive is termed use for a person who tested positive and has the disease. This is not use for the specificity since specificity is used to identify the number of people without the disease. The TP however is used for the computation for the sensitivity [ CITATION Bis181 \l 13321 ]. b. Prevalence Rationale: It does not involve in the specificity since it refers to the proportion of the person with the disease in a given population during a specific period of time [ CITATION Bis181 \l 13321 ]. c. False negatives Rationale: False negative on the other hand defines as a person who tested negative and has the disease. This is not also use for the specificity but in sensitivity which used to identify the number of people with the disease. Therefore, letter c is not the correct answer [ CITATION Bis181 \l 13321 ]. d. False positives Rationale: Specificity is the measurement used to identify the number of people without the disease wherein formula involved is true negatives divided by the sum of

true negatives and false positives. The false positives refer to the person which tested positive but do not have the disease. Since we are looking for specificity, false positive is included in the computation [ CITATION Bis181 \l 13321 ]. 8. The component of a spectrophotometer that is responsible for detecting transmitted light and converting light energy to electrical energy is the: a. Detector Rationale: The answer is letter A. Detector is also known as the photodetector, a part of spectrophotometer that converts the light energy after it detects the electromagnetic radiation that is transmitted by the solution into electrical signal [ CITATION Bis181 \l 13321 ]. b. Analytical cell Rationale: Analytical cell is not part of spectrophotometer. This is a device that is used for the analysis of the cell. Spectrophotometer only composed of light source, entrance slit, monochromator, exit slit, cuvette, detector and reading device [ CITATION Bis181 \l 13321 ]. c. Monochromator Rationale: Unlike letter A, the monochromators transmit the bandpass or spectral bandwidth by dispersing the light into wavelengths[ CITATION Bis181 \l 13321 ]. d. Readout device Rationale: This device is the final part of the spectrophotometer where the results produced by electrical energy from the detector will be

displayed on the screen for either printout or recording [ CITATION Bis181 \l 13321 ]. 10. Which of the following instruments is used in the clinical laboratory or in reference laboratories to detect beta and gamma emissions? a. Fluorometer Rationale: This is under the principle of fluorescence wherein excitation is required for absorption of radiant energy. In fluorometer, a UV light is used. The excitation light will pass through the primary filter and then the analyte is measured by specific wavelength. Then, the excitation light is absorbed by the atoms of the analyte producing a light that passes to the secondary filter and detector [ CITATION Bis181 \l 13321 ]. b. Nephelometer Rationale: Unlike letter D, the nephelometer is not used for the detection of beta and gamma emissions. It applies the principle of the measurement of light scattered in a solution wherein the amount of scatter is directly proportional to the number and size of particles present in the solution [ CITATION Bis181 \l 13321 ]. c. Spectrophotometer Rationale: Contrary to Scintillation counter, the spectrophotometer is used to detect the concentration of a certain analytes in a solution. Primarily, it measures light transmitted by the analyte in a solution. The light transmitted is directly proportional to the concentration of analytes in a

solution [ CITATION Bis181 \l 13321 ]. d. Scintillation counter Rationale: Scintillation counter detect beta and gamma emissions using a fluor. This substance is responsible for the conversion of radiation energy to light energy. For gamma detection, it utilizes the large crystal of sodium iodide containing a small amount of thallium as an activator whereas for beta counting, they used the fluor dissolved in organic solvents [ CITATION Bis181 \l 13321 ]. 11. In flow cytometry, the side scatter is related to the: a. DNA content of the cell b. Granularity of the cell Rationale: Flow cytometry uses the principle of scattering light produces by the cell when it is transported under fluidic pressure. The detection occurs when the cell passed one by one through a laser beam which the measurement involves the forward light scattering, side light scattering and light emitted from fluorescent labels. For the granularity of the cell, we use the side light scattering [ CITATION Bis181 \l 13321 ]. c. Size of the cell Rationale: For the size of the cell, it corresponds to the forward light scattering. The light produces by the cell is detected by the photomultiplier tubes. This is also important to measure to determine if blood cells are abnormal in shape or not [ CITATION Bis181 \l 13321 ]. d. Number of cells Rationale: The number of cells is based on the changes of electrical

resistance where the flow of the current is interrupted as the cell pass through the electrical field producing a voltage pulse where it is detected and measured. The number of pulses is proportional to the number of cells counted [ CITATION Bis181 \l 13321 ]. 27. A decreased LDH/HBD ratio is significant for: a. Myocardial infarction Rationale: High level of HBDH or the a-hyrdoxybutyrate dehydrogenase can be found in the heart muscle tissue which is why it can be used for the diagnosis of myocardial infarction. LDH isoenzyme primarily the LD1 and LD2 ratio is also use for diagnosing myocardia infarction. Therefore, decreased in LDH/HBD ration is significatant for myocardial infarction. b. Liver disease Rationale: Unlike in myocardial infarction, for the diagnosis of liver disease we usually measured the AST and ALT level together with other liver enzymes. c. Both of these Rationale: The answer is only letter A and not letter B. d. None of these Rationale: Among the choices, letter A is the best answer. 39. The creatinine clearance test is routinely used to assess the glomerular filtration rate. Given the following information for an averagesize adult, calculate the creatinine clearance. Urine creatinine - 120 mg/dL

Plasma creatinine - 1.2 mg/dL Urine volume for 24 hours - 1520 mL a. 11 mL/min Rationale: This is not the answer since the formula for creatinine clearance is Urine Concentration (Us) multiplied by Urine Volume (Vs) divided by the product of Plasma Concentration (Ps) and the time in minutes. If we will compute, the answer will be letter d. b. 63 mL/min Rationale: Like letter a, b is also not the answer since the formula for creatinine clearance is Urine Concentration (Us) multiplied by Urine Volume (Vs) divided by the product of Plasma Concentration (Ps) and the time in minutes. If we will compute, the answer will be letter d. c. 95 mL/min Rationale: This is not the answer since the formula for creatinine clearance is Urine Concentration (Us) multiplied by Urine Volume (Vs) divided by the product of Plasma Concentration (Ps) and the time in minutes. If we will compute, the answer will be letter d. d. 106 mL/min Rationale: The formula for creatinine clearance is Urine Concentration (Us) multiplied by Urine Volume (Vs) divided by the product of Plasma Concentration (Ps) and the time in minutes. Therefore, we will have: 120 md /dl x 1520 mL 1.2 mg/dl x 1440 mins ¿ 105.5 mL/min . Or 106 ml/min.

50. Which phenotype in the Frederickson classification of hyperlipoproteinemia is characterized by an increase in cholesterol, an increase in beta lipoproteins, normal triglycerides, and the absence of chylomicrons? a. I Rationale: In contrast to the results given in the question, type 1 or the familial LPP lipase deficiency is not the answer. In type 1, the triglyceride and chylomicrons are increased while the cholesterol is either within the normal range or increase also [ CITATION Bis181 \l 13321 ]. b. II Rationale: There are five classifications if Frederickson classifications of hyperlipoproteinemia. The increase of cholesterol but normal triglyceride describes the type 2 hyperlipoproteinemia. However, this type 2 is divided in type 2a which is the familial hypercholesterolemia and type 2b for combines hyperlipidemia. Among the two, type 2a is most likely associated to the results given in the question since in type 2b, there is an increase in both triglycerides and cholesterol [ CITATION Bis181 \l 13321 ]. c. III Rationale: Type 3 or the familial dysbetalipoproteinemia is also not the answer since the triglyceride in this condition is increased contrary to the given result in the question which is normal [ CITATION Bis181 \l 13321 ]. d. IV

Rationale: The same with A and C, the type 4, or also known as hypertriglycedemia has an increased in triglyceride instead of normal. Its cholesterol is either within the normal range or also increase [ CITATION Bis181 \l 13321 ]. 53. “Floating beta lipoprotein” refers to: a. B-VLDL Rationale: Beta-VLDL is refers to floating beta lipoprotein” and known as the “abnormally migrating BetaVLDL”. Although it has the density of the VLDL, during electrophoresis, it migrates the same with the LDL which is why it is termed as the “floating beta lipoprotein”. It is commonly found in type 3 hyperlipoproteinemia or the dysbetalipoproteinemia [ CITATION Bis181 \l 13321 ]. b. HDL Rationale: They commonly known as good cholesterol instead of “floating beta lipoprotein”. They contain primarily of protein, small amount of cholesterol, phospholipids, and small amount of triglyceride. This lipoprotein is synthesized in the liver and small intestine [ CITATION Bis181 \l 13321 ]

c. VLDL Rationale: Contrary to the question, this lipoprotein is the major carrier of endogenous triglycerides in the body and not referred as “floating beta lipoprotein” although synthesized in the liver, containing apo B-100, it also contains apo E, and apo Cs. Like chylomicrons, they also caused turbidity in fasting hyperlipidemic

plasma specimens [ CITATION Bis181 \l 13321 ]. d. B-LDL Rationale: There is no beta-LDL only LDL which refers as bad cholesterol since it transports cholesterol to the cell. LDL contains more cholesterol than other apo B–containing lipoproteins. It is synthesized from the metabolism of VDLD in the circulation [ CITATION Ric173 \l 13321 ][ CITATION Bis181 \l 13321 ] 78. A patient’s T4 is 10 mg/dL. The THBR is 24% (normal=30%). What is the FT4I? a. 8 Rationale: The formula for the FT4I is THBR/Normal average multiply by the T4. Using the given numbers, we will have the computation: 24 % 10 mg x =8 30 % dl Therefore, the correct answer is letter A. b. 6 Rationale: This is not the answer since the formula for for the FT4I is THBR/Normal average multiply by the T4. Using the given numbers. If we will compute, the answer will be letter A. c. 4 Rationale: Like letter b, c is also not the answer since the formula for for the FT4I is THBR/Normal average multiply by the T4. Using the given numbers. If we will compute, the answer will be letter A. d. 2 Rationale: Like letter b and c, d is also not the answer since the formula for for the FT4I is

THBR/Normal average multiply by the T4. Using the given numbers. If we will compute, the answer will be letter A. 79. The principal estrogen produced during pregnancy is: a. Estrone Rationale: Unlike estriol, estrone involves after menopausal stage of a women. It is present in the female body after the menopause. This is the weaker form of estrogen and can be used for form other type of estrogen if necessary [ CITATION Bis181 \l 13321 ]. b. Estradiol Rationale: Estradiol is the major estrogen produces in the ovary. Its main function is in the ovulation and thickening of the uterus lining. This hormone is responsible for the maturity of the egg cell in the ovary which is then released after ovulation. Then, it will now cause the uterus lining to thicken for the preparation for the implantation of fertilized egg cell [ CITATION Bis181 \l 13321 ]. c. Estriol Rationale: Estriol on the other hand is the one that increases during the pregnancy. This hormone is synthesized in the placenta. Its primary function is to regulate the blood flow in the uteroplacental and its vascularization for the preparation of the body as well as the uterus for the pregnancy [ CITATION Bis181 \l 13321 ]. d. Corticosterone

Rationale: This is not involved in pregnancy. However, it is the primary corticosteroid of the adrenal gland. It is secreted in the adrenal cortex in response to stress as well as it contributes to the metabolism of the body [ CITATION Bis181 \l 13321 ]. 82. Which of the following polypeptide hormones may be described as having alpha chains that are biochemically identical but beta chains that as biochemically unique? a. FSH, TSH, ACTH, TRH Rationale: Although FSH and TSH is included, the remaining two are not. The TRH or the thyroid regulating hormone has one 3 amino acids or known as tripeptides. These are the glutamic acid, histidine and proline whereas ACTH or the adrenocorticotropic hormone has only 39-amino acid peptide [ CITATION Bis181 \l 13321 ]. b. LH, ACTH, HCG, TRH Rationale: The TRH and ACTH is also not included in the polypeptide hormones having alpha chains that are biochemically identical but beta chains that as biochemically unique. The TRH or the thyroid regulating hormone has one 3 amino acids or known as tripeptides. These are the glutamic acid, histidine and proline whereas ACTH or the adrenocorticotropic hormone has only 39-amino acid peptide [ CITATION Bis181 \l 13321 ]. c. TSH, LH, TRH, HCG Rationale: The TRH or the thyroid regulating hormone has one 3 amino acids or known as tripeptides. These

are the glutamic acid, histidine and proline. The C-terminal amino acid is the proline while the N- terminal is the glutamic acid. Therefore, it is not included in polypeptide hormones having alpha chains that are biochemically identical but beta chains that as biochemically unique [ CITATION Bis181 \l 13321 ]. d. HCG, FSH, TSH, LH Rationale: Among the choices, letter D is the correct answer. HCG or also known as the human chronic gonadotropin which has a 92 amino acid alpha-subunit identical to the LH, FSH and TSH. However, the beta-subunit of this hormone is different from the three since it consists of 145 amino acid betasubunit. The luteinizing hormone on the other hand has 120 amino acid beta-subunit whereas folliclestimulating hormone has 111 amino acid beta-subunit. Thyroid stimulating hormone on the other hand has 118 amino acid betasubunit [ CITATION Bis181 \l 13321 ]. 93. Caffeine is an metabolite of this drug:

important

a. Acetaminophen Rationale: Also known as Tylenol which is used as analgesic. Caffeine is not the metabolite of this drug. When it is ingested, it will be metabolized by the liver. Its metabolite is N-acetyl-pbenzoquinone imine which is highly toxic to liver if overdose by this drug [ CITATION Bis181 \l 13321 ]. b. Digoxin

Rationale: The metabolite of the digoxin is the dihydrodigoxin. This drug is under the cardioactive drugs which is used to treat congestive heart failure [ CITATION Bis181 \l 13321 ]. c. Theophylline Rationale: This drug is under the bronchodilator drug. Its metabolite is the caffeine. It is used for the treatment of asthma and other COPD. It is administered orally and metabolized by the liver then filtered by the kidney [ CITATION Bis181 \l 13321 ]. d. Phenobarbital Rationale: Contrary to the caffeine, the active metabolite of phenobarbital is the aripiprazole. This type of drug is antiepileptic drug which is used to control seizures. It is also metabolized by the liver then filtered by the kidney [ CITATION Bis181 \l 13321 ]. 95. The formation of this crystal in urine, although not a constant finding is an important diagnostic clue of ethylene glycol poisoning: a. Uric acid Rationale: Uric acid crystal can be seen as yellow brown crystals with rosettes or wedges under the microscope. Unlike calcium oxalates, this is crystals is associated to increased levels of purines and nucleic acids which are commonly seen in patient with Lesch-Nyhan syndrome and gout [ CITATION Str141 \l 13321 ]. b. Ammonium biurate Rationale: This crystal is described as “thorny apples” because of its

appearance when viewed in the microscope. This crystal can be in old specimens and can be associated with high level of ammonia in the urine produced by urea-splitting bacteria [ CITATION Str141 \l 13321 ]. c. Triple phosphate Rationale: Also known as the ‘coffin lid” crystals. Commonly seen in alkaline urine. They are normal crystals which has no clinical significance, however unlike in calcium oxalate, increase in its number is seen when there is a presence of urea-splitting bacteria [ CITATION Str141 \l 13321 ]. d. Calcium oxalate Rationale: It has three appearances. It can be envelopes in dihydrate form and oval or dumbbell in monohydrate form. This is mostly seen in acidic urine. The monohydrate form is significant in ethylene glycol poisoning also known as antifreeze poisoning [ CITATION Str141 \l 13321 ].

References

Bishop, M., Fody, E., & Schoeff, L. (2018). Clinical Chemistry, Principle, Techniques and Correlations, 8th Edition. WoltersKluwer . McPherson, R. M., & Pincus, M. M. (2017). HENRY’S CLINICAL DIAGNOSIS AND MANAGEMENT BY LABORATORY METHODS, 23e. St. Louis, Missouri: Elsevier Inc. Strasinger, S. (2014). Urinalysis and body fluids, Sixth Edition....