Question

INTRODUCTION Albumin is synthesised in the liver at a rate that is dependent on dietary protein intake. Little is lost from the body by excretion. It is catabolized in various tissues, where it is taken up by cells by pinocytosis. Albumin has a vast capacity for ligand binding due to the large number of charges on each molecule as well as the very large number of molecules available. This binding property of albumin is related to one of its functions, namely transport of non – polar molecules such as bilirubin and free fatty acids; hormones such as thyroxine and cortisol and drugs such as salicylate and warfarin. Changes in serum levels of albumin can be a result of a large number of pathological conditions. Note that the normal reference range for albumin in human blood is 34-54 g/L. Albumin also has the ability to bind a wide variety of organic anions, including complex dye molecules. The shift in the absorption maximum of the dye allows the resulting colour to be measured. In the presence of excess dye, all of the albumin molecules take part in the reaction. A wide variety of dyes have been employed for the measurement of albumin including bromocresol green (BCG) which we are using in this practical. Refer Lecture notes 2 (week 1) for further details.

OBJECTIVES/AIMS In this practical, students will learn how to operate the absorbance spectrophotometer, determine the optimum wavelength for albumin, and construct a standard curve. During the online practical class, by watching the links, participating in the simulation lab activity and reading the information provided within the notes, you should learn and understand: • How to use the spectrophotometer to determine the optimum wavelength • How to construct a calibration curve (standard curve) and to determine the concentration • How to calculate the dilution factor

PART B Experimental plan: You are provided with a standard solution of albumin, which is 60 g/L. This is known as the primary standard. Your task is, complete the table below to prepare 3 other standards with 15, 30 and 45 g/L concentration, respectively with the final volume of 1 mL. These are known as secondary standards. Note: 15 g/L tube has been completed and calculation steps are provided below. C1V1 = C2V2 (Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration and V2 is the final volume). In this scenario, we know C1 is 60 g/L, C2 is 15 g/L (for making 15 g/L secondary standard) and V2 is 1 mL (1 mL=1000 µL). The unknown factor is V1. 1ml - V1 = volume of water needs to make up 1 mL secondary standard solution (i.e. to make up the final volume as 1 mL). Now apply this formula to the 15 g/L secondary standard tube. C1V1=C2V2 60 g/L x V1 = 15 g/L x 1000 µL V1 = (15 g/L x 1000 µL )/60 g/L V1 = 250 µL. 1000 µL – 250 µL = 750 µL deionised water.

You must calculate for 30 g/L and 45 g/L secondary standards and complete Table 1 below. Table 1: Secondary standards to be used in the Albumin assay

Microfuge Tube	60 g/L Albumin Standard (µL)	Deionised water (µL)
15 g/L	250	750
30 g/L
45 g/L

The following sample tubes, as shown in Table 2 below, have been prepared, and absorbance values were recorded as described below. 10 µL of deionized water (tube 1) or samples (2-7) as detailed in Table 2 were mixed well with 2 mL of BCG reagents and incubated at room temperature for 5 minutes. Sample from tube 1 (blank) was transferred into a cuvette and placed in the spectrophotometer, which was set to the optimal wavelength, 630 nm. This blank sample’s absorbance was set to 0.000. The absorbance of tubes 2-7 was read from the spectrophotometer and results were recorded (see Table 2, last column

Table 2: Albumin Assay

Tub	Sample	Sample volume (µL)	BCG reagent (mL)	Absorbance
1	Blank (deionized water)	10	2.0	0.0000
2	15 g/L albumin std	10	2.0	0.1008
3	30 g/L albumin std	10	2.0	0.2091
4	45 g/L albumin std	10	2.0	0.3561
5	60 g/L albumin std	10	2.0	0.4299

1) Aim and hypothesis (1 mark).

. 2) Insert Table 1 and show the calculation for both 30 g/L and 45 g/L (1 mark)

3) Construct a calibration curve (standard curve). Place the concentration of the standards on the X axis and the absorbance on the Y axis, with a line of best fit going through zero (2 marks).

4) From your standard curve, can you determine the concentration of the following unknown samples?

If yes you can, then write the answer, and if you can’t determine, then explain why (1 mark).

Sample 1: Absorbance value is 0.3108

Sample 2: Absorbance value is 0.6219

Answer 1

Ans1:

Aim: Estimation of protein in unknown sample

Hypothesis: Calculate the amount of protein in unknown sample by generating the standard curve

Ans 2:

For 30g L: (30*1000)/60= 500 ul standard + 500 ul Deionized water

For 45g/L: (45*1000)/60= 750 ul standard + 250 ul Deionized water

Microfuge Tube	60 g/L Albumin Standard (µL)	Deionised water (µL)
15 g/L	250	750
30 g/L	500	500
45 g/L	750	250

Ans 3:

Standard Calibration curve 0.5 0.45 y = 0.0076 -0.0096 R2 = 0.9858 0.4 0.35 0.3 OD 630nm 0.25 02 0.15 0.1 0.05 0 10 O 20 30 40 50 60 70 Conc. g/L

Ans 4:

Yes I can determine the value for unknown sample. By putting the value in equation displayed on chart

Sample 1: Absorbance is 0.3108

So here Y=3108 and we have to find x means concentration of unknown sample1

y=0.0076x-0.0096

Therefore: (0.3108+0.0096)/0.0076

= 42 g/L for sample 1

Same way calclation for sample 2

Sample 2: Absorbance is 0.6219

So here Y=0.6219 and we have to find x means concentration of unknown sample2

y=0.0076x-0.0096

Therefore: (0.6219+0.0096)/0.0076

= 83g/L for sample 2