Sample Diluent Volume (mL) Sample Volume (mL) Absorbance 280 nm A .990 .010 0.57

Question

Sample

Diluent Volume (mL)

Sample Volume (mL)

Absorbance 280 nm

A

.990

.010

0.571

B

.990

.010

0.520

C

.667

.333

0.913

D

.667

.333

0.424

1. Estimate the total protein concentration in each sample using Beer’s law (?280 = 0.80 (mg/ml)-1 cm-1). Remember to take into account the dilution of each sample in your calculation when determining the concentration of protein in each sample.

2. How does the total protein concentration compare between the untreated and ammonium sulfate-treated NS and IS samples? Explain your results.

Sample

Diluent Volume (mL)

Sample Volume (mL)

Absorbance 280 nm

A

.990

.010

0.571

B

.990

.010

0.520

C

.667

.333

0.913

D

.667

.333

0.424

Explanation / Answer

absorbance = elc, e: molar absorptivity constant

l: path lenth (icm)

c: concentration

A 0.571= 0.80*1*c

c= 0.714 mg/ml

sample had 0.01ml of protien, so according to the concentration,if 1 ml has 0.714mg then 0.01ml has 0.0714mg

B. 0.520= 0.8*1*c

c= 0.65mg/ml

so, 0.065mg

C. 0.913=0.8*!*c

c= 1.14mg/ml

actual concentration is 1.14*0.333= 0.47mg

D. 0.424 = 0.8*1*c

c= 0.53mg/ml

actual concentration is 0.53*0.333= 0.176mg

since the solution was 1ml concentration for A, B, C and D is 0.0714mg/ml, 0.065mg/ml, 0.47mg/ml and 0.176mg/ml respectively

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