Question

A 1 to 5 dilution was performed on a serum sample. The diluted sample was analyzed...

A 1 to 5 dilution was performed on a serum sample. The diluted sample was analyzed for glucose and determined to be 100 mg/dL. What is the true/reportable value of glucose?
0 0
Add a comment Improve this question Transcribed image text
Answer #1

The solution is diluted to 5 times , that means a concentration of solution M1 changes to M1/5 .

M1/ 5 = 100mg /dL

M1 (reported value of glucose ) = 500mg/dL. (Answer)

Or

= 500×10-3g/0.1L = 5 g/L (answer)

Or

= 5g/L ÷ 180g/mol =0.02777mol/L

= 0.028 mol/L (Answer)

Add a comment
Know the answer?
Add Answer to:
A 1 to 5 dilution was performed on a serum sample. The diluted sample was analyzed...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 1) if 5 parts sample is added to 10 parts diluents, what is the dilution that...

    1) if 5 parts sample is added to 10 parts diluents, what is the dilution that was just made? 2. What is the ratio if 5 parts sample volume is added to 10 parts diluents volume? 3. In a 1 to 3 dilution, what is the dilution FACTOR? 4. In a 1 to 10 dilution, what is the dilution FACTOR? 5. What is the dilution FACTOR if 40 mcL of sample added to 260 mcL of diluent? 6. What is...

  • 13. A 1/2, fourfold serial dilution was performed. The original concentration of the analyte was 1200...

    13. A 1/2, fourfold serial dilution was performed. The original concentration of the analyte was 1200 μg/dL. What is the concentration of the analyte in the final tube? 7. Calculate the true analyte value given the following information: Analyte diluted using 10 μL into 90 μL diluent; Diluted analyte value = 85 mg/dL.

  • o Calculate the following sample concentrations based upon given dilutions. Final (Diluted) Concentration 1st Dilution 2nd...

    o Calculate the following sample concentrations based upon given dilutions. Final (Diluted) Concentration 1st Dilution 2nd Dilution Concentration Original 1:5 1:4 1:15 1:3 1:3 3 (none) 1:2 1:5 (none) 1:5 167 mg/dL b. 139 mg/dL 15 mg/dL d. 557 mg/dL 23 mg/dL Page 2 of 3

  • 2. The AMR for a method for measuring glucose has been determined by the manufacturer to...

    2. The AMR for a method for measuring glucose has been determined by the manufacturer to be 20-500 mg/dL A. The data for its verification study is shown below. Is the method's linearity verified for this range? Standard Standard [Glucose] mg/dL Absorbance Blank Standard 1 Standard 2 Standard 3 Standard 4 Standard 5 Standard 6500 0.000 0.050 0.099 0.201 0.298 0.402 0.501 50 100 200 300 400 B. The method allows a maximum dilution by a factor of 10. What...

  • A volume of serum sample is diluted with equal amounts of buffer (1:1). Following this, a...

    A volume of serum sample is diluted with equal amounts of buffer (1:1). Following this, a series of five dilutions is made of this first dilution by diluting it 1/10, rediluting 1/10, and then three times more, each resulting solution then being a 1/10 dilution of the previous one in the series. What is the concentration of serum in each solution made (6 in total)?

  • -What is the molarity of 1.5 N CaCl2? -You take 1 mL of patient serum and...

    -What is the molarity of 1.5 N CaCl2? -You take 1 mL of patient serum and add it to 3 mL of water. Next, you take 1 mL of the dilution you just made and add it to 4 mL of water. You perform a glucose test using this final sample and receive a result of 6 mg/dL. a. What is the dilution factor of the final sample? b. Is the patient’s glucose value in the reference range?

  • 12. Given a series of 4 dilutions, each 1/5, what is the concentration in tube No....

    12. Given a series of 4 dilutions, each 1/5, what is the concentration in tube No. 3 if the original concentration was 100 mg/dL? 13. A 1/2, fourfold serial dilution was performed. The original concentration of the analyte was 1200 μg/dL. What is the concentration of the analyte in the final tube? 7. Calculate the true analyte value given the following information: Analyte diluted using 10 μL into 90 μL diluent; Diluted analyte value = 85 mg/dL.

  • 25 mL of a serum sample is to be analyzed by atomic emission spectroscopy for silver (Ag) metal. ...

    Clearly, the answer is 220ppm. However, I need the steps leading to it. 25 mL of a serum sample is to be analyzed by atomic emission spectroscopy for silver (Ag) metal. After thorough mixing, a student transferred 2.5 mL of this serum sample to a 100 mL volumetric flask followed by the addition of 10 mL 0.5 M sulfuric acid to the sample. He then diluted it to volume in the volumetric flask with deionized water. After setting up his...

  • Five milliliters of 5% alcohol are added to 5 mL of urine. Two milliliters of this solution is diluted to 50 mL with wat...

    Five milliliters of 5% alcohol are added to 5 mL of urine. Two milliliters of this solution is diluted to 50 mL with water, and 5 mL of this are used for analysis. How many milliliters of urine are in the analysis sample? You are doing a procedure that calls for 5 mL of undiluted urine. You use 3 mL of a 1/5 dilution of urine instead. The answer obtained is 50 mg/dL. What result should be reported? A procedure...

  • 1-4 Pipette 0.1 ml serum and 0.3 mL DI water to a small tube and mix....

    1-4 Pipette 0.1 ml serum and 0.3 mL DI water to a small tube and mix. I assay this diluted serum and obtain a glucose value of 860 mg/ Liter. The glucose value I will report is ??? How do you know when an assay must be repeated on dilution? How do you find the value for the limit of linearity for a method? I read the absorbance for the patient's sample on instrument #2. I used a calibration curve...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT