Biochemistry 3723 Exp. 5 Protein Determination Sept 4, 2002
1.Learn to use the conventional and diode array spectrophotometers
2.Learn to calculate concentration from spectrophotometric data
3.Appreciate the advantages/disadvantages of common methods used to measure protein concentration.
I.Theory--a little physics.
A.Spectrum--light absorbed at s that represent energies of excitation of electrons in molecule.
1.Theoretically should be sharp bands of absorption, but many vibrational levels gives broad band spectrum that we’re used to seeing.
2.Examples: NADH, Vitamin B12.
1.A = E C l
a.A = absorbance (= – log T/To) no units; call them absorbance units
b.C = Concentration (g/liter, moles/liter etc.)
c.l = path length; is the distance light travels through the sample. (1.000 cm unless otherwise stated).
d.E = extinction coefficient; a proportionality constant
i.Units must cancel units of conc. and path length (liters/mol-cm, or liters/gram-cm. etc); also called absorptivity.
ii.Characteristic of compound and wavelength at which measured.
e.Resulting graph: linearity and usefulness of graph
ii.loss of linearity due to interactions of substrate or lack of reagent
C.Instrumentation-- read in your textbook
vs. Diode Array Spec.
2.Important difference: diode array measures all wavelengths at once.
3.Consider single beam vs. double beam. (Both conventional and diode array instruments we have are single beam instruments)
a.Double beam spec splits light through two samples
b.One of the samples is always the blank; its absorbance is subtracted out
c.Single beam spec. You must measure absorbance of a blank. Subtract it out.
4.Cuvettes: specially made cells/tubes.--path length well defined. Different materials have different characteristics. See book.
D.Determination of Protein Concentration- many assays
1.Absorption spectrum--Sensitive, non-destructive. Several methods
a.Estimate: C (mg/ml) ≈ A280; but depends on AA composition.
i.If you have pure protein, can establish exactly.
ii.BSA: 1.00 mg/ml sln has A280 = 0.620
iii.Lysozyme: 1.00 mg/ml sln has A280 = 2.46
b.A280/A260ratio tells ~ nucleic acid contamination
i.Protein has A280/A260 ~ 1.75
ii.Nucleic acid has A280/A260 of ~ 0.5
c.A280 due to tyr and trp mainly; very sensitive to AA composition.
A260 due to purines, pyrimidines (NAs)--an interference
A230 due to peptide bond--independent of AA composition (Abs max of peptide bond is ~214 nm). (See graph above)
d.Groves et al: C (mg/ml) =1.55 A280 – 0.76 A260
e.Kalb and Bernlohr: C (g/ml) = 183 A230 – 75.8 A260
f.These formulas, as you will see, don't work very well for pure proteins. Lysozyme gives high values and BSA low values due to Tyr and Trp composition of protein. Work better for mixture of proteins.
3.Biuret Assay. Show chemistry
a.Complex of Ns (-amino groups) from peptide with alkaline Cu+2
b.Name--based on similar reaction that biuret undergoes
c.Color intensity relatively constant with varying AA composition
d.Requires relatively much protein (mg). Is destructive.
4.Bradford Assay-- Coomassie Blue dye
a.Absorbance shift of dye when binds protein. (450 to 595 nm).
b.Varies with AA composition
c.Acid reagent; destructive to protein
d.Sensitive and rapid (µg)
5.There are many other methods!
A.Two proteins used
1.Lysozyme--14,000 Da, (= 14 kDa) high in tyr & trp. 10.00 mg/ml std.
2.Bovine serum albumin (BSA) 66 kDa, low in tyr & trp. 10.00 mg/ml
3.Your unknown--is it lysozyme or BSA and what is its concentration? (NOT 10 mg/ml)
B.Take spectrum of samples.
1.A230 must be below 1.50 for Kalb & Bernlohr equation to be valid
2.Dilute standards and unknown in water before taking spectrum as instructed in protocol.
3.Note difference between lysozyme and BSA in spectrum--due to difference in tyr & trp content and a contaminant in the lysozyme.
4.From A280/A230 ratio decide whether unknown is BSA or lysozyme
5.Using Groves et al and Kalb and Bernlohr equations calculate concentration of unknown. Remember to correct for dilution.
Note on Dilution: A 1/80 dilution means 1 part of the material you are diluting to 80 parts total, so you would add 1 part material plus 79 parts of diluent. (Either 1 ml + 79 ml, or 0.1 ml + 7.9 ml, or 0.05 ml +3.95 ml, etc.) Also the convention of writing 1:80 is confusing! That really means 1 part plus 80 parts, which is a 1/81 dilution. Always write dilution as a fraction and there will never be any confusion about what the dilution really is.
1.2 Standard curves; one using lysozyme and one with BSA. Do you expect them to be the same or different?
2.Measure unknown and calculate concentration from each standard curve. See below for calculations.
3.On conventional spectrophotometers. You will have to do all the calculations. Demo of Spec 70s in lab.
D.Calculations--for Biuret Assay
1.Calculate amount of standard in each tube.
a.Beer’s law: A = E C l; should calculate concentration of each standard.
(10.0 mg/ml) (0.200 ml) = (x mg) (5.00 ml)
x= 0.400 mg/ml (conc.)
2..Graph Absorbance (y axis) vs. amount (or conc.) of standard (x axis).
3.For unknown, measure absorbance and then read amount off x axis
4.Use this value to calculate unknown concentration.---do sample
say got 1.28 mg/ml final concentration of unknown from 0.500 ml unknown sample:
(0.500 ml) (x mg/ml)= (5.00 ml) (1.28 mg/ml) and
x= 12.8 mg/ml
5.Average all usable values--- what is a usable value?
E.Bradford Assay. Will do next lab
1.Do 2 standard curves; one using lysozyme and one with BSA. Do you expect them to be the same or different?
2.Must calculate concentration or amount of protein in each standard to enter in computer program.
a.Do calculations before you come to class on Thursday and enter in Table 2, p 51.
(1.40 µg/µl) 20.0 µl) = (x µg/µl) 5100 µl);
x= 5.49 µg/µl.
3.Measure unknown and calculate concentration from each standard curve. Calculations similar to those for Biuret.
4.Do on diode array spectrophotomers. Standard curves and concentrations calculated by the computer. See Appendix III of lab manual.
5.A peculiarity of the computer program: Standard concentrations must be expressed as values > 1
III. Practical for today
1.Do on diode-array spectrophotometer
2.Must use quartz cuvette. Expensive
3.Dilute all proteins as instructed
4.Blank instruments against water
5.A230 MUST be below 1.50 (which it will be if you dilute correctly).
1.Do on conventional spectrophotometer
2.Since must sit for > 30 minutes after mixing, I recommend you start this one before starting absorption spectrum.
3.Don’t dispose of samples until I have checked your data! Then dispose in bottle provided (Cu shouldn’t go down sink)
1.How would you make 100. µl of a 1/75 dilution of a sample?
Solution: (x µl) = 100. (1/75); x = 1.3 µl.
So take 1.3 µl of sample and add 100.-1.3= 99 µl of diluent.
Note. It is important to take 1.3 µl (not 1 µl)
2.What is the dilution factor if you mix 0.35 ml of sample with 5.00 ml of H2O?
Solution: the dilution factor is (5 .00 + 0.35)/0.35 = 15.3 fold.(really 15 fold to the correct sfs)
B.Calculating concentration from UV absorbance.
1.A protein sample gives the following absorbance readings:
A230 =1.25; A260 = 0.040; A280 = 0.267
Determine the concentration of this sample by three different methods.
a.By estimation : concentration is ~ .27 mg/ml
b.By Groves: 1.55 (.267) - .76 (.040) = 0.383 mg/ml
c.By K and B: 183(1.25) – 75.8 (.040) =226µg/ml =.226 mg/ml
2.Assuming the above protein is pure lysozyme, calculate the concentration using the known A280 for lysozyme.
Solution: A 1.00 mg/ml solution of pure lysozyme has an A280 = 2.64.
Thus: (x mg/ml)/(0.267) = (1.00 mg/ml)/(2.64)
x = 0.101 mg/ml, which is much lower than the other methods give because of lysozyme's high content of tyrosine and tryptophan
C.Standard Curve calculations—a favorite exam question! This is off of the first exam from last semester:
You have a urine sample from a patient you suspect of being diabetic. You add 20.00 mL of 50 mM tricine buffer pH 7.8 to 5.00 mL of the urine sample. You then do a glucose oxidase assay (a colorimetric test we will learn about later in the semester) of the urine sample along with a series of standards made from a 1.00 mg/mL glucose solution with the following results:
3723 F02glucose std
ml / diluted urine
ml / 50 mM
ml / glucose
ml / final
(units?) / A450
0.00 / ----- / 1.000 / 5.00 / 0.000
0.200 / ---- / 0.150
0.500 / ---- / 0.390
0.800 / ---- / 0.595
1.000 / ---- / 0.780
----- / 0.300 / 0.275
----- / 0.750 / 0.710
----- / 1.000 / 0.825
a.Fill in the table with amounts of reagents used.
b.Calculate final [glucose] and enter in table.
c.Determine the glucose concentration in the urine sample.
d.Calculate the molar extinction coefficientfor glucose (181 g/mol)in this assay.
Solution: a. All tubes must have the same volume (6.00 ml in this case) and the same amount of color reagent (5.00 ml). So add tricine buffer to each sample to make up the difference in volume. (0.800, 0.500, 0.200, 0.000, 0.700, 0.250, and 0.000 ml respectively). And add 5.00 ml glucose reagent to each tube.
b.The final [glucose] can be calculated as amount or concentration.
Amount: 0.200 ml (1.00 mg/ml) = 0.200 mg etc.
(0.200, 0.500, 0.800, 1.00 mg, respectively)
Concentration: (0.200 ml)(1.00 mg/ml) = (6.00 ml)(x mg/ml)
x = (0.200ml)(1.00mg/ml)/(6.00 ml)
x = 0.0333 mg/ml, etc.
(0.0333, 0.0833, 0.133, 0.167 mg/ml, respectively)
c. To determine glucose in samples you need to generate a standard curve, using amount or concentration of glucose standards on x axis and Absorbance on y axis:
Solution: On next page is the graph of Absorbance vs. mg/ml of each standard (called a standard curve). Using the standard curve and the known absorbance of each sample, determined the mg/ml in each final sample: 0.275 = .0625 mg/ml
0.710 = .152 mg/ml
0.825 = off the standard curve: don't use.
Then you have to calculate the original concentration from the final concentration of each sample:
(0.0625 mglml)(6.00 ml) = (0.300 ml)(x)
x = 1.25 mg/ml
(0.152 mg/ml)(6.00 ml) = (0.750 ml)(x)
x = 1.216 mg/ml
And then average the two values: (1.25 + 1.216)/2 =1.233 =1.23 mg/ml
The glucose concentration in DILUTED urine is 1.23 mg/ml. The dilution factor is 25/5. So the original concentration is 1.23 x 5 = 6.15 mg/ml.