The biruet reagent is prepared by dissolving 1.5g of copper sulfate and 6g sodiu

Question

The biruet reagent is prepared by dissolving 1.5g of copper sulfate and 6g sodium potassium tartrate in 500 mL oh H2O. Add 300mL of 10% NaOH, and bring the final volume up to 1L. It will keep indefinitely if 1g potassium iodide is added to inhibit the reduction of copper.

d) A sample prepared with 0.50ml of protein sample A and 2.5ml of Biuret reagent gave an absorbance of 0.51. Calculate the protein concentration of the original protein A sample, in g/ml, twice: once with each standard curve. ix. On the calculation sheet (see iii), neatly show all the calculations required to solve for the concentration of the protein sample. Remember to perform this calculation twice: once with each standard curve. Include a title to separate this calculation from others on the page.

The equations from the standard curve graphs are:

y=0.0002x+0.1706

y=0.0011x+0.1706

Explanation / Answer

Ans. #1. Using standard curve equation y = 0.0002x + 0.1706

Step 1: Calculation of unknown [protein] using standard graph.

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression or standard curve equation) equation y = 0.0002x + 0.1706 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 0.0002 units on X-axis (concentration) plus 0.1706.

Given, absorbance of unknown, y = 0.51

Putting y = 0.51 in first standard curve equation

            0.51 = 0.0002x + 0.1706

            Or, 0.51 – 0.1706 = 0.0002x

            Or, x = 0.3394 / 0.0002

            Hence, x = 1697

Unit of x in the standard graph = ug/ mL

Hence, [protein] in the final solution (whose abs is taken) = x = 1697 ug/ mL

Step 2: Given, 0.50 mL of protein sample A is mixed with 2.5 mL Biuret reagent.

So, final volume of protein solution (whose abs is taken) = 0.50 mL + 2.5 mL = 3.0 mL

It is the final solution (of volume 3.0 mL) in which the calculated [protein] = 1647 ug/mL.

Now, Using   

            C1V1 (original protein solution A, 0.5 mL) = C2V2 (final solution, 3.0 mL)

            Or, C1 x 0.5 mL = 1647 ug mL-1 x 3.0 mL

            Or, C1 = (1697 ug mL-1 x 3.0 mL) / 0.5 mL

            Hence, C1 = 10182 ug/ mL

Therefore,

[Protein] in original protein A sample = 10182 ug/ mL = 0.010182 g/ mL

#2. Using standard curve equation y = 0.0011x + 0.1706

Step 1: Calculation of unknown [protein] using standard graph.

Given, absorbance of unknown, y = 0.51

Putting y = 0.51 in first standard curve equation

            0.51 = 0.0011x + 0.1706

            Or, 0.51 – 0.1706 = 0.0011x

            Or, x = 0.3394 / 0.0011

            Hence, x = 308.55

Hence, [protein] in the final solution (whose abs is taken) = x = 308.55 ug/ mL

Step 2: Using           

            C1V1 (original protein solution A, 0.5 mL) = C2V2 (final solution, 3.0 mL)

            Or, C1 x 0.5 mL = 308.55 ug mL-1 x 3.0 mL

            Or, C1 = (308.55 ug mL-1 x 3.0 mL) / 0.5 mL

            Hence, C1 = 1851.27 ug/ mL

Therefore,

[Protein] in original protein A sample = 1851.27 ug/ mL = 0.001827 g/ mL

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