HELP WITH PART D) E) Question: STANDARD CURVE PREPARATION Always indicate the vo
ID: 573749 • Letter: H
Question
HELP WITH PART D) E)
Question: STANDARD CURVE PREPARATION Always indicate the volume of water (or additional solution) to be add... STANDARD CURVE PREPARATION Always indicate the volume of water (or additional solution) to be added to complete any dilution.
a) A stock solution of 6mg/ml BSA is provided. Calculate the dilutions for the preparation of the following BSA standards (2.0ml volumes of each): 60, 170, 500, 1500, 3000ug/ml. Also include the dilution factor for each preparation. On a separate sheet of paper with a heading of Calculations, include one complete sample calculation of this dilution calculation.
b) To 0.5ml of your unknown samples (and dilutions of your samples) and to 0.5ml of the standards add 2.5ml of Biuret reagent. Calculate the final concentration of BSA after addition of the Biuret reagent. iv. neatly show one sample calculations.
c) Hypothetical absorbencies with respect to the concentrations in ‘a’ are: 0.16, 0.20, 0.28, 0.45 and 0.70. v. Use computer software such as excel to plot a standard curve using these absorbencies against the concentration of each BSA standard prior to the addition of Biuret reagent (use concentrations as given in step 11a directly) vi. Label the figure as Figure 1 and give it an appropriate descriptive title. The title should be above the figure. vii. Plot a second standard curve using the same absorbencies (0.16, 0.20, 0.28, 0.45 and 0.70) against the concentration of the BSA standard after the addition of the Biuret reagent (as calculated in step 11b) viii. Label this figure as Figure 2 and give it an appropriate descriptive title.
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.
e) A protein sample was prepared with 5.0 x 10-2 ml protein sample B, 0.45ml water and 2.5ml of Biuret reagent . It had an absorbance of 0.22 in the spectrophotometer. With this data, calculate the protein concentration of the original protein B sample, in g/ml, using either standard curve. x. On the calculation sheet (see iii), neatly show all the calculations required to solve for the concentration of the protein sample. Include a title to separate this calculation from others on the page. Refer to the Figure that was used to solve for the concentration.
The standard curve equations are:
y=0.0002x+0.1706
y=0.0011x+0.1706
Explanation / Answer
Ans. #D.I. Step 1: Calculation of unknown [Glucose] 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 OD 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 OD 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 = 10.182 mg/ mL
#D.II. Step 1: Given- standard curve equation: y = 0.0011x + 0.1706
Absorbance of unknown sample = 0.51
Putting the value of y in above standard curve equation-
0.51 = 0.0011x + 0.1706
Or, x = (0.51 – 0.1706) / 0.0011
Hence, x = 308.545
Therefore, [protein] in final solution (whose OD or Abs is taken) = 308.545 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 = 308.545 ug mL-1 x 3.0 mL
Or, C1 = (1697 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 = 1.85 mg/ mL
#E. I. Step 1: Given- standard curve equation: y = 0.0002x + 0.1706
Absorbance of unknown sample solution = 0.22
Putting the value of y in above standard curve equation-
0.22 = 0.0002x + 0.1706
Or, x = (0.22 – 0.1706) / 0.0002
Hence, x = 247
Therefore, [protein] in final solution (whose OD or Abs is taken) = 247 ug/ mL
# Step 2: 5.0 x 10-2 mL (= 0.05 mL) of original protein B solution is mixed with 0.45 mL water and 2.5 mL Biuret reagent.
Final volume of diluted solution (whose OD is taken)
= 0.05 mL + 0.45 mL + 2.5 mL = 3.0 mL
Using C1V1 (original protein solution B, 0.5 mL) = C2V2 (final solution, 3.0 mL)
Or, C1 x 0.05 mL = 247 ug mL-1 x 3.0 mL
Or, C1 = (247 ug mL-1 x 3.0 mL) / 0.05 mL
Hence, C1 = 14820 ug/ mL
Therefore,
[Protein] in original protein A sample = 14820 ug/ mL = 14.82 mg/ mL
#E. II. Step 1: Given- standard curve equation: y = 0.0011x + 0.1706
Absorbance of unknown sample solution = 0.22
Putting the value of y in above standard curve equation-
0.22 = 0.0011x + 0.1706
Or, x = (0.22 – 0.1706) / 0.0011
Hence, x = 44.91
Therefore, [protein] in final solution (whose OD or Abs is taken) = 44.91 ug/ mL
# Step 2:
Using C1V1 (original protein solution B, 0.5 mL) = C2V2 (final solution, 3.0 mL)
Or, C1 x 0.05 mL = 44.91 ug mL-1 x 3.0 mL
Or, C1 = (44.91 ug mL-1 x 3.0 mL) / 0.05 mL
Hence, C1 = 2649.55 ug/ mL
Therefore,
[Protein] in original protein A sample = 2649.55 ug/ mL = 2.65 mg/ mL
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