Urinalysis/Body Fluids PDF

81 URINALYSIS/BODY FLUIDS...

81 URINALYSIS/BODY FLUIDS by Michele Zitzmann* Kidney Nephron Distal Tubule lomeru 1,1 Cortex /' H+ I Medulla \ Proximal ! Tubule Glucose Salts - - - -'I-"! Amino Acids Ureter u::: ollecting J ") Loop of Henle Diagrams adopted with permission from Modem Urine Chemis/JY Manua/;Siemens Healthcare Diagnostics Inc~ 1996 Renal Physiology 2. on-selective filter of plasma NEPHRON substances of less than 70,000 MW 1. Major functional unit of the kidney 3. Water, glucose, electrolytes, amino acids , urea, uric acid, creatinine, and 2. Approximately I million per kidney ammonia comprise the glomerular filtrate 3. Composed of glomerulus and r enal tubules 4. Filters 120 ml/min, or one-fifth of renal GLOMERULUS plasma 1. Coil of capillary vessels TUBULES FUNCTION Reabsorbs water, Sodium Chloride, Bicarbonate, Potassium, Calcium, Amino Acids, Proximal Tubule Phosphates, Protein, and Glucose Glucose - Threshold Substance -Reabsorb at 160-180 mg/di or less Secretes - Sulfates, glucuronides, Hydrogen ions and drugs loop of Henle Reabsorbs water Descending No solutes reabsorbed Loop of Henle Reabsorbs Solutes (Sodium, Chloride, Calcium, and Magnesium Ascendin2 No water reabsorbed Distal & Collecting Reabsorbs Sodium Tubules Secretes Potassium, Ammonia and Hydrogen ions Potassium ions exchanged for sodium ions *od ginal contribution hy LilJhy Spence 82 URINE VOLUME VOLUME DISEASE AND CAUSES Normal Average of 1200-1500 ml/day Polyuria >2500 ml/day Diabetes insipidus, Diabetes mellitus, diuretics Caffeine, alcohol, excessive water intake Oliguria 1 mg/dl measures from Urmary Comtituents adequacy of collection 4. Mathematical formula: Correlate Physical, Chemical Creatinine Clearance = U x V at,(/. Mictbllcopit Re.suits p l ' ' I u = Concentration of urine creatinine mg/dL V = Volume of urine in ml/time in minutes Pregnancy Testing P ;;; Concentration of plasma creatinine mg/dL 1. Human chorionic gonadotropin (hCG) , a glycoprotein composed of alpha and 5. Norma] Creatinine Clearance= 120 beta subunits, is secreted by the ml/min for adults ; with age t placenta. The appearance and rapid rise in concentration of hCG in the 6. Correction for Kidney Mass (Kid11ey mother's serum and urine make it an Mass Proportional to Body Size) excellent marker for confirming pregnancy. Clearance = U x V x 1.73 p A a. hCG can be detected as early as 8- 10 days after ovulation (1 day after A = Body Surface Area implantation) 7. Example: b. hCG peaks at 8-10 weeks of pregnancy Plosmo Creotinine = 1.5 mg/dL c. First morning specimen preferred Urine Creatinine = 120 mg/dl (more concentrated) Urine Volume = 1.2 L Surface Area = 1.30 mm3 Renal Function Tests Collection Time = 24 hours 1. Used to test for glomerular filtration and tubular function Volume = 1200 ml = 0.833 1440 min 2. Tests for glomerular filtration a. Clearance tests 120 mg/dL x (0.833 ml/min) = 66.67 ml/min b. B2-Microglobulin 1.5 mg/ dl c. Cystatin C CREATININE CLEARANCE TEST 66.67 x 1.73 = 88.71 ml/min 1.30 1. Most commonly u sed clearance test to assess GFR (amount of blood filtered of a particular substance in a given time)  93 ESTIMATED GRF Renal Diseases 1. Assists in detecting chronic kidney disease 2. More sensitive than creatinine Findings in Renal Disease clearance 3. Calculation based on serum cr eatinine, DISEASE LAB FINDINGS patient's age, gender , and e thnicity Acute Glomerulonephritis Gross hematuria, Smoky B2-MICROGLOBULIN Turbidity, RBC Casts, Waxy 1. Useful marker of renal tubular Chronic Hematuria; All Types of function. Glomerulonephritis Casts, but Only 0cc to Few 2. +plasma concentrations indicate a RBC Casts reduced GFR (not specific) Acute Pyelonephritis Turbid, Pos Nitrite, Pos Leukocyte Esterase, CYSTATIN C WBC Casts, Bacteria 1. May provide an equal or better Chronic Pyelonephritis Pos Nitrite, Pos Leukocyte detection of adverse changes in GFR Esterase; AlI Types of Casts, 2. Disadvantages: but Only 0cc to Few WBC Casts a. Higher cost than creatinine clearance Nephrotic Syndrome May See "Free" Fat Droplets; b. Possible variable results among Fatty Casts; Oval Fat Bodies individuals Cystitis/Lower Urinary Tract Bacteria; WBC's Infection Body Fluids CEREBROSPINAL FLUID (CSF) 2. Normally clear, colorless 1. Reasons for analysis: a. Meningitis 3. Xanthochromia - pink, orange, or b. Encephalitis yellow CSF supernatant (usually due to c. Syphilis.hemoglobin) d. Brain abcess / Tumor 4. Important to differentiate between e. lntracranial hemorrhage intracranial hemorrhage and traumatic f. Leukemia / Lymphoma with CNS collection involvement DISTRIBUTION OF CSF TUBES LABORATORY SECTION(S) POSSIBLE TESTS Tube #1 Chemistry and/or Serology Protein, Glucose, Lactate, VDRL, Latex agglutination tests Tube #2 Microbiology Gram stain, culture, India ink Tube #3 Hematology Cell count and differential DIFFERENTIATION BETWEEN HEMORRHAGE VS. TRAUMATIC TAP HEMORRHAGE TRAUMATIC TAP Appearance All tubes equally red (bloody) Subsequent clearing of blood in each tube Supernatant Xanthochromic Clear Presence of Clots No Yes, due to fibrinogen 94 Differentiating Causes of Meningitis - CSF Studies PROTEIN GLUCOSE WBC POPULATION LACTATE Bacterial ++ t Neutrophils + Viral ♦ N Lymphocytes N Fungal + N-t Lymphocytes and Monocytes ♦ 2. Referen ce ranges a. Volume - 2-5 ml b. Count - 20-250 million/ml c. Motility - > 50% d. Morphology - < 30% Abnormal Differentiate Traumatic Tap Forms from lntracramal. e. Viability > 75% live forms Hemmorrbage Azoosperm1a - no sperm ❖ Oligospermia - < 20 milhonlml 5. Normal Analyte Values a. Protein - 15-45 mg/dl b. Glucose - 60-70% plasma glucose c. Cell Count - 0-5 WBCs/microliter( ul) d. Differential - 70% lymphocytes; ( I 30% monocytes Analyze Seminal FJ,p.d t~ ID ~ Norm,.Jity , JI 1 /Ji ~ l, 6. CSF Electropheresis - oligoclonal banding= multiple sclerosis ( t glucose and ,t. protein) SEROUS FLUIDS MANUAL CELL CALCULATION FOR BODY FLUIDS 1. Reasons for analysis: a. Sepsis # Cells Counted X Dilution b. Malignancy # Squares Counted X Depth (0.1) c. Systemic disease SEMINAL FLUID SYNOVIAL FLUID 1. Reasons for semen analysis 1. Reasons for analysis: a. Infertility (most common) a. Sepsis b. Post vasectomy b. Hemorrhage c. Forensic medicine - presence of c. Crystal induced inflammation acid phosphatase confirms presence of semen in alleged rape cases; 2. Normal appearance: clear, pale yellow flavin in semen fluoresces on 3. Monosodium urate crystals= gout clothing under UV light 4. Calcium pyrophosphate crystals pseudogout Sammy Sperm Baby Shannon ( normal) (spcrmatld) 5. Compensated polarization: ~ a. Calcium pyrophosphate appears V blue when parallel to compensator and yellow when perpendicular b. Monosodium urate appears yellow when parallel to compensator and blue when perpendicular Slim (pinhead) 6. Cell count normal ranges: Sylvester a. < 200 WBC/µL amol' hous head b. < 2000 RBC/µL  95 Transudates vs. Exudates b. Phosphatidylglycerol (PG) ❖ Lipid component ofpulmonary TRAN SU DATE EXUDATE surfactants ❖ Not usually detected until 35 Color Colorless Yellow-White, (inflammation); weeks of gestation Red-Brown (hemorrhage); Yellow-Brown (bilirubin); ❖ Absence does not rule out mature Milky-Green (chylous fluid) fetal lungs Turbidity Clear, Watery Cloudy, Viscous c. Lamellar body counts Specific Gravity < 1.015 > 1.015 ❖ Secreted frlto alveolar lumen at 20-24 weeks of gestation Protein < 3 g/dl > 3 g/dl ❖ Anmiotic fluid is analyzed on LO (Lactic < 200 IU > 200 IU automated instrument hy using Dehydrogenase) the platelet count value Cell Count < 1000/µI > 1000/µI ❖ A dvantages ,t. Associated Congestive Infections and Malignancies ~ small sample volume With Heart Failure, ~ short turnaround time Changes in Hidrostatic ~ low cost ressure ~easily interpreted PLEURAL PERICARDIAL AND PERITONEAL FLUID 3. Other amniotic fluid testing a. Alpha-fetoprotein (AFP) 1. Effusion - build-up of fluid in a body ❖ ,I- associated with neural tube cavity due to a pathologic process disorders (such as Spina Bifida) 2. Cell count and differential performed b. Bilirubin in same manner as CSF ❖ Reliable estimate of fetal. red cell destruction due to maternal AMNIOTIC FLUID antibody 1. Reasons for analysis c. Foam stability index a. Fetal wdl-Leing; d. Fluorescence Polarization Assay b. Fetal lung maturity SWEAT 2. Fetal lung maturity tests 1. ,t. sweat electrolytes, sodium and a. Lecithin/Sphingomyelin (LIS) ratio chloride, confirms the diagnosis of ❖ Measures the phospbolipids cystic fibrosis lecithin and sphingomyelin to DETECTION OF MALIGNANCY assess fetal lung maturity 1. Benign tissue cells must be ❖ H the LIS ratio is> 2. 0, fetal lungs differentiated from malignant cells in are usually mature (diabetic body fluids. mothers - PG must be present forFLM) DIFFERENTIATING BENIGN AND MALIGNANT CELLS BENIGN CELLS MALIGNANT CELLS Fluids with cells suspicious for Distinguishable cell borders in clumps Poorlv defined malignancy Flat clusters - not ball-like Ball-like spherical clusters always should be Individual cells Often cells within cells (cannibalism) referred to May be large, but not giant, tend to be uniform Often bizarre, monstrous - non-uniform cytology and for N/C ratio low Frequently high N/C ratio pathologist's Smooth, clear membrane Irregular nuclear membrane review. Even chromatin Uneven chromatin The cells should Can be multinucleated but nuclei uniform Multinucleated forms with varied be reported as nuclear sizes atypical and Nulei round or oval Nuclear molding (mosaic) suspicious for Nucleus even, no clefts Nuclear clefts malignancy on May be vacuolated MaY. be vacuolated with vacuoles over the differential nucleus cell report. 96 Summary of Body Fluids FLUID SOURCE APPEARANCE LAB TESTS CLINICAL SIGNIFICANCE Serous Ultrafiltrate of plasma; Normal; clear and fills organ cavities pale yellow PLEURAL Thoracic Cavij Turbid white eelIs, bacteria Cell Counts: +Red Cells trauma, malignancy (around lungs Bloody trauma, malignancy +Neutrophils - bacteria Milky. chylous fluid + Lymphocytes tuberculosis, malignancy Glucose: + Tuberculosis, rheumatoid inflammation, malignancy pH: t Tuberculosis, malignancy, esophageal rupture Amylase: +Pancreatitis CEA ,t.. Malignancy (care!noembryonic antigen): PERICARDIAL Percardial Cavity Turbid - infection, Cell Counts: +Red Cells tuberculosis, tumor, cardiac (around heart) malignancy puncture Bloody - tuberculosis, tumor, +Neutrophils- bacterial endocarditis cardiac puncture Milky - lymphatic drainage Glucose: t Bacterial infection CEA: +Malignancy PERITONEAL Peritoneal Cavity Turbid - peritonitis, cirrhosis Cell Counts: +Red Cells trauma (ascites/ (around abdomen) Bloody trauma +Neutrophils - peritonitis Milky. chylous fluid Green bile Glucose: tTubercular peritonitis, malignancy Amylase: +Pancreatitis, GI perforation Alkaline Phosphatase: +Intestinal perforation Urea/Creatinine: +Ruptured bladder Other Other Other Fluids Fluids Fluids Sodium Chloride SWEAT Sweat Glands (of skin) Clear and colorless (sweat testt. +in Cystic Fibrosis Cell Counts: SYNOVIAL Synovial Membrane (around Clear, pale yellow, and ,t.. Red Cells hemorrhage joints) viscous +Neutrophils (> 25%) sepsis Bloody - hemorrhagic "- Lvmphocvtes non-septic inflammation arthritis Milky crystals, cells Crystals: Uric Acid (monosodium urate) gout Green tinge bacteria Deep yellow inflammation Ropes Clot Test: t clot arthritis GASTRIC HCI and Pepsin (secreted in Depends on diet Titratable Acidity: t in pernicious anemia (no intrinsic factor) stomach) ,t- in duodenal ulcer, Zollinger-Ellison Syndrome AMNIOTIC Amniotic Sac Clear and colorless normal Bilirubin: +in HON (indicates red cell destruction) Yellow-greeen bilirubin, reel cell destruction US Ratio: >2.0 = fetal lung maturity Phosphatidylglycerol: +indicates fetal maturity (similar to US ratio) Creatinine: Fetal age Alpha Fetoprotein: +indicates neural tube disorders  97 URINALYSIS/BODY FLUIDS SAMPLE QUESTIONS I. Which of the following changes occur when a 5. The reagent strip reaction used to test for the urine specimen is left at room temperature for presence of glucose is based on the principle longer than 1-2 hours? of A. Ketones t A. A buffered reaction of mixed B. Biliruhin + enzyme indicators. C. Bacteria t B. Acid precipitation of glucose salts. D. Glucose+ C. Douhle sequential enzyme reactions. 2. A urinalysis on a 3 year old revealed a positive D. Glucose producing a change in pH copper reduction test and a negative glucose in a buffered system. oxidase test. How would these results be interpreted? 6. An elevated urine urobilinogen and a negative A. A strong oxidizing agent is present. test for urine bilirubin may indicate which of B. Copper reduction tests are more the following conditions? sensitive than glucose oxidase tests. A. Acute hepatic toxicity C. Galactose is present. B. Biliary obstruction D. Glucose and 1actose are both C. Hemolytic disease present. D. Urinary tract infection 3. Chemical and microscopic analysis of a urine 7. Calculate the creatinine clearance of a specimen yields the following results 24-hour urine specimen using the following data: Protein: 4+ RBC 0-2 /hpf Urine creatinine 500 mg/L WBC 0-5 / hpf Plasma creatinine 8 mg/L Few hyaline casts Urine volume l.5 L Moderate fatty, waxy, and granular casts Many oval fat bodies A. 6.5 ml/min These results are consistent with which B. 0.75 ml/24 h ours diagnosis? C. 65 ml/min A. Hepatic insufficiency D. 85 ml/min B. Glomerulonephritis C. Nephrotic syndrome 8. An US ratio of 2.7 indicates D. Pyelonephritis A. Fetal distress. B. Fetal lung maturity. 4. An abnormally high specific gravity may be C. Fetal red cell destruction. seen in which of the following conditions? D. Inadequate fetal pulmonary A. Chronic renal disease surfactants. B. Diabetes insipidus 9. Cerebrospinal fluid results reveal an ,t.. protein, C. Metabolic acidosis normal glucose, normal lactate and.f. D. Radiographic dye injection lymphocytes. These results most likely indicate A. Bacterial meningitis. B. Fungal meningitis. C. Multiple Sclerosis. D. Viral meningitis. 98 ANSWERS AND RATIONALE 1. A 5. C When a specimen is left unpreserved at room The reagent strip r eaction combines two temperature, several changes take place. Option separate reactions catalyzed by two different A is correct because ketones will t due to enzymes. Glucose is combined with oxygen in bacterial metabolism of acetoacetate to acetone. the presence of the enzyme glucose oxidase. Option B is incorrect because bilirubin t as it is Gluconic acid and hydrogen peroxide are formed photooxidized to biliverdin. Option C is from this reaction. Hydrogen peroxide formed incorrect because bacteria multiply causing an ,t. in the first reaction oxidizes a chromogen in the result. Option D is incorrect because glucose t presence of the enzyme peroxidase to produce a as bacteria utilize the glucose for glycolysis. color change. Thus, this method is termed a double sequential enzyme reaction. 2. C 6. C The copper reduction method will detect any reducing sugar, whereas, the glucose oxidase Increased hemoglobin degradation associated method is specific for glucose. Option A is with hemolytic conditions such as transfusion incorrect because oxidizing agents give false reactions and sickle cell disease results in the positive results with the glucose oxidase method production of large amounts of unconjugated and not the copper reduction method. Option B bilirubin which is presented to the liver to be is incorrect because the glucose oxidase method conjugated. This large amount of bilirubin then detects minimum glucose levels of 100 mg/dl, goes to the intestine and is broken down by while the copper reduction method only detects bacteria into urobilinogen. Because larger minimum levels of 200 mg/dl. Option D is amounts of bilirubin are excreted into the incorrect because if these both sugars were intestine and larger amounts of urobilinogen are present, both tests would be positive. made, increased amounts of urobilinogen are reabsorbed into the cir culation. As a r esult, the 3. C urinary bilirubin remains negative but the urinary urobilinogen increases above its normal Hallmarks of the nephrotic syndrome include level. Option A would result in the production of fatty casts, oval fat bodies, and markedly increased levels of bilirubin and urobilinogen increased protein in the urine. Option A does due to liver dysfunction. Option B is incorrect not usually demonstrate cast formation. Option because bilirubin, due to the obstruction, could B would show a positive blood result and many not be converted to urobilinogen. Less than red cell casts. Option D would have many normal amounts of urobilinogen would be WB C's and white cell casts. excreted in the kidney. ( The dipstick will show normal urobilinogen because it cannot detect 4. D lower than normal amounts.) Option D does not Radiographic dyes are water soluble affect bilirubin metabolism. substances that are readily excreted in urine that cause a significantly increased specific gravity. 7. C In Option A, the kidneys are no longer able to To calculate the creatinine clearance: concentrate or dilute urine causing a fixed specific gravity usually around 1.010. Option B UxV is incorrect because persons with diabetes p insipidus are unable to produce concentrated urine due to defective ADH production. In Where: Option C, the body r eacts to maintain acid-base U = the concentration of urine balance by inducing the kidneys to eliminate H+ creatinine in mg/ml ions through the urine which does not affect the V= the volume of urine in ml/min specific gravity. p= the concentration of plasma creatinine in mg/ml  99 The first step is to convert the r esults to the correct units. Because the volume of urine must be in ml/min, convert the volume of urine to ml then divide the ml by 1440 (n umber ofminutes in 24 hours). In this case, 1.5 L = 1500 ml = 1.04 ml/min. If the height and weight of the patient are known a correction for the kidney mass is calculated. Use this formula: U x V x 1.73 p A Where: A= the surface area of the body. The surface area of the body can be determined by a nomogram which can he found in several refer ence books. = 500 mg/L x 1. 04 mL/min 8mg/L = 520 mLJmin 8 = 65.1 mL/min 8. B Lecithin and sphingomyelin are produced by the fetal pulmonary system in a relatively constant rate until the 35th week of gestation, at which time the concentration of sphingomyelin decreases and the concentration of lecithin increases in the amniotic fluid. Before this time, the L/S ratio is usually less than 2.0 which is associated with immaturity of the fetal lungs. When the US ratio r eaches 2.0 or greater it is indicative of fetal lung maturity. 9. D These results are indicative of viral meningitis. Option A would show an ,t. in neutrophils. Option B would show an ,t- lactate and possible monocytes. Option C would show a t glucose and oligoclonal banding with electrophoresis. I@" Beable to identify cells, casts, bacteria and yeast from graphic images. Correlate microscopic findings with chemical tests and disease states.