Lab Homework #1 (Serial Dilutions) PDF

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Maria Ruggeri lab director. First homework assignment for lab, about serial dilutions....

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HOMEWORK #1 SERIAL DILUTION and STANDARD CURVE – 10 points George is setting up an experiment in which he needs to serially dilute a bovine albumin stock solution with the initial concentration of 8 mg/ml. He has 10 ml of this stock solution and he needs to do a 5-fold serial dilution (1:5) because his stock solution is too concentrated. 1) Help George and fill out Table 1 with the correct dilution factor and volumes he needs to mix, and then calculate the concentrations of each of the four serially diluted solutions.

Table 1. Albumin Concentrations and Absorbance After a 5-fold Serial Dilution

Dilution Factor (1:5)

Concentration (mg/ml) Absorbance (595nm)

Tube 1 (stock)

Tube 2

Tube 3

Tube 4

1:1

1:5

1:25

V (stock) = 10 ml

V (Tube 1) = 2 mL

V (Tube 2) = 2 mL

V (Tube 3) = 2 mL

V (Tube 4) = 2 mL

V (H2O) = 8 mL

V (H2O) = 8 mL

V (H2O) = 8 mL

V (H2O) = 8 mL

8 mg/ml

1.6 mg/ml

0.32 mg/ml

0.064 mg/ml

0.0128 mg/ml

1.5

0.9

0.65

0.43

0.18

1:125

Tube 5 1:625

2) After you complete this table, plot absorbance vs. concentration graph in Excel. Include the linear regression equation and R 2 value of the curve. Don’t forget to label the X and Y axes correctly.

3) Use the graph you created to make a figure. Your figure has to have a caption that includes a specific title and a brief, but descriptive, explanation of the figure (a figure legend).

Figure 1: Graph of the absorbance vs. the concentration of a 5-fold serial dilution of the bovine albumin solution. A bovine albumin solution stock solution was determined to have a concentration of 8 mg/mL. Four 5-fold serial dilutions were performed, and the absorbances of the stock solution and each serial dilution was recorded at a wavelength of 595 nm. A line of best fit was generated for these data.

4) Once George did all the calculations and made the figure, his instructor gave him an unknown

solution to test. The absorbance of this sample was 0.72. What is the concentration of this unknown sample? Use the standard curve in your figure to answer this question. y = absorbance

x = concentration (mg/mL)

y = 0.1363x + 0.4596

Find: concentration (x)

y−0.4596 0.1363 0.72−0.4596 =1.9 mg/mL Substitute y=0.72 into the equation above: x= 0.1363 Rearrange equation to solve for x:

x=

The concentration of the unknown sample is 1.9 mg/mL...