Problem 14 - Clinical Chemistry

Question

A 24-hour urine has a total volume of 1,200 mL. A 1:200 dilution of the urine specimen gives a creatinine result of \(0.8 \mathrm{mg} / \mathrm{dL}\). The serum value is \(1.2 \mathrm{mg} / \mathrm{dL}\). What is the final value of creatinine in \(\mathrm{mg} / \mathrm{dL}\) in the undiluted urine sample? a. 160 b. \(0.8\) c. 960 d. 860

Step-by-Step Solution

Answer

160 mg/dL

Step 1: Identify Given Information

Identify the key information given in the problem: - Volume of urine: 1,200 mL- Dilution factor: 1:200- Creatinine result of diluted sample: 0.8 mg/dL

Step 2: Understand What is Asked

Determine what needs to be calculated from the given problem: The final concentration of creatinine in the undiluted urine sample in mg/dL.

Step 3: Recall the Dilution Formula

The dilution formula is: \[ \text{Concentration of undiluted sample} = \text{Concentration of diluted sample} \times \text{Dilution factor} \]

Step 4: Plug in the Values

Replace the given values into the formula: \[ \text{Concentration of undiluted sample} = 0.8 \times 200 \]

Step 5: Perform the Calculation

Calculate the concentration of undiluted sample: \[ 0.8 \times 200 = 160 \text{ mg/dL} \]

Key Concepts

creatinine concentration

Understanding creatinine concentration is crucial in clinical chemistry. Creatinine, a waste product from muscle metabolism, filters through the kidneys into the urine. By measuring its concentration in urine and blood, clinicians evaluate kidney function. The concentration in the urine can change based on factors like hydration or renal health. Creatinine concentration in urine is usually measured in mg/dL, and it's crucial to consider how various influencing factors can affect the concentration. Knowing this helps in interpreting results accurately and understanding the health status of the patient.

urine analysis

Urine analysis, or urinalysis, is a test conducted on a urine sample to check for various substances. This test can indicate hydration levels, kidney function, and the presence of diseases like diabetes. A 24-hour urine collection is often used for more precise measurements of substances like creatinine. During this process, all urine excreted in 24 hours is collected in a single container. This cumulative sample provides a more accurate measure of certain components compared to a single, random urine sample. For the creatinine analysis cited in the exercise, a 24-hour urine sample gives a thorough insight into the daily excretion rate of creatinine, helping in the accurate assessment of kidney function.

dilution factor

In clinical chemistry, the dilution factor is an important calculation that indicates how much a sample has been diluted. For instance, in this exercise, the urine is diluted at a ratio of 1:200, which means one part urine is mixed with 199 parts diluent (often water or another solution). This dilution makes it manageable to measure high concentrations of analytes like creatinine without overwhelming the measuring equipment. The formula to calculate the concentration of the undiluted sample is: \[\text{Concentration of undiluted sample} = \text{Concentration of diluted sample} \times \text{Dilution factor}\] By applying this formula, you convert the measured concentration back to its original state before dilution, helping clinicians determine the accurate levels of substances in the urine.

clinical chemistry

Clinical chemistry is a field focused on analyzing bodily fluids for diagnostic purposes. It encompasses the study and measurement of various chemicals, enzymes, and molecular substances in the blood, urine, and other fluids. One of the key components assessed is creatinine, whose levels in blood and urine provide insights into kidney function. The exercise demonstrates a typical procedure in clinical chemistry involving dilution, measurement, and calculation, reflecting how laboratory results are interpreted. With the right calculations, clinicians can make informed decisions about diagnosis and treatment based on precise and accurate data from these tests.