In a serum urea estimation using diacetyl monoxime reagent, the procedure is as

Question

In a serum urea estimation using diacetyl monoxime reagent, the procedure is as follows: Protein Pptation Step/Mix the following: 0.1 mL serum 0.1 mL 0.5 M NaOH 3.8 mL Zn SO_4/Na_2 SO_4 pptation reagent Stand and centrifuge. Separate supernatant. Colour Development Step: Stock Standard urea:0.6 g/L urea = 10.0 mM Working Standard:Prepare by diluting 2.0 mL Stock Standard to 100 mL. This corresponds to 8.0 mmol/L serum. Take 1.0 mL of sample supernatant, or Working Standard, mix with 3.0 mL Diacetylmonoxime reagent in a test tube. Stand 20 minutes and read absorbances. (a). Prove that the Working Standard corresponds to 8.0 mmol/L serum. (b). Calculate the concentration of an unknown serum sample which gave an absorbance of 0.180, when the Standard gave an absorbance of 0.240.

Explanation / Answer

Ans. A. Using                                     C1V1 = C2V2                        - equation 1

C1= Concentration, and V1= volume of std. solution 1     ;( stock, 10.0 mM)

C2= Concentration, and V2 = Vol. of final solution 2     ; [working standard]

Putting the values in equation 1-

            10.0 mM x 2.0 mL = C2 x 100.0 mL

            Or, C2 = (10.0 mM x 2.0 mL)/ 100.0 mL

            Or, C2 = 0.2 mM

                        = 0.2 m(mol/ L)                                             ; [1 M = 1 mol/ L]

                        = 0.2 mmol/ L

Note: I. 2.0 mL standard stock is diluted to 100.0 mL, it’s a 1 : 50 dilution (as 2:100 = 1: 50). So, 1: 50 dilution of 10.0 mM stock makes it 0.2 mM (as 10/ 50 = 0.2).

So, the diluted solution does not correspond to 8.0 mmol/L (= 8.0 mM) serum.

II. Please re-check the values mentioned in question and the expected answer- both don’t match up. Please also check if any information or correlating factor is missing, too.

For example, the [Urea] in standard stock of commercially available kits is 40 mM. It would thus give [Urea] of working standard = 8.0 mmol/ L = 8.0 mM.

#B. From A, the calculated [Urea] of working standard = 0.2 mM

However, we assume that working standard [urea] = 8.0 mM – as in commercially available diagnostic kits.

So, we have standard [urea] = 8.0 mM

OD of standard = 0.240

OD of sample = 0.180

Now,

            [Urea] in sample = (OD of sample/ OD of standard) x [urea] in standard

                                                = (0.180 / 0.240) x 8.0 mM

                                                = 6.0 mM

                                                = 6.0 mmol/L

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.