CHEM 7784 – Biochemistry Lab
LAB #2 – Biochemistry Laboratory Skills Workshop
A. Graphing of Data
Absorbance at 595 nm
During this course, you will be asked to summarize the data from your experiments in the form of a graph.
Any graph submitted as part of a laboratory report is required to be constructed using a computer program
such as Microsoft Excel. No hand-drawn graphs will be accepted as a part of a formal report. Therefore,
you will need to be familiar with graphing software. Using the data below, a sample graph was made using
Microsoft Excel.
Concentration Absorbance
Calibration Curve for Protein Assay
µg/cuvette
595 nm
1.5
0
0.01
1.25
2
0.05
1
5
0.12
0.75
10
0.25
0.5
25
0.63
0.25
50
1.24
0
Slope =
2.48E-02
Y-intercept =
3.33E-03
Correlation coeffieient (r) = 0.999927933
0
10
20
30
40
50
60
Protein Concentration (µg/cuvette)
Notice that the graph contains a title and labels for both of the axes. Since this is a linear graph, each of the
axes is divided into equal intervals. Also, the range of the axis “fits” the range of data – there is not a large
amount of left over white space! The data has been plotted on the graph, and the line best fitting the data
has been drawn in. This graph represents a calibration curve for a protein assay similar to one you will be
doing in future labs. At times, you will be asked to provide statistical data to accompany the graph that may
include the mean, standard deviation, %RSD, %EDA, slope, y-intercept, and correlation coefficient values. If
you are unfamiliar with these terms, please refer to any statistics textbook or the help area of Microsoft
Excel or the graphing program you are using.
B. Solutions
1. Molar Solutions (Unit = M = moles/liter)
A 1.0 Molar (1.0 M) solution is equivalent to 1 formula weight (g/mole) of chemical dissolved in enough
water to make 1 liter. The formula weight (FW) is given on the label of a chemical bottle, or use molecular
weight (MW) if formula weight is not given.
Here are some examples:
1. Prepare 2 liters of a 0.2 M (0.2 mole/liter) solution of a chemical having a FW of 154.2 g/mole.
How many grams of the chemical do I need to use?
?grams = 2 liters x 0.2 mole/liter x 154.2 g/mole = 61.68 g of chemical.
Thus, add 61.8 grams to about 1.5 liters of water to dissolve, then make up the final volume to 2.0
liters with distilled water.
2. You have a 5 mg/ml solution of a protein with a molecular weight of 66,000. How many moles/liter
protein are there in your solution?
5 mg/ml x (1 gram/1000mg) x (1000ml/ 1 liter) = 5 gram/liter solution of protein
5 g = 1 liter x ?mole/liter x 66,000 g/mole. Rearranging the equation gives you:
? mole/liter = 5 g
= 7.6 x 10-5 mole/liter(M) = 7.6 x 10-2 mM = 76µM
1 liter x 66,000 g/mol
2. Percent Solutions
Many chemical solutions are mixed as percent concentrations. When working with a dry chemical, the
percent is calculated as grams dry chemical in 100 ml. This is known as a %weight/volume or %w/v formula.
For example, a 0.9% NaCl solution is 0.9 g NaCl in 100 ml water.
When the formula includes a liquid chemical, then the percent solution is volume added chemical in ml
made up to 100 ml with solvent. A 10% acetic acid solution in water is 10 ml glacial acetic acid added to
90 ml water. This is known as a volume/volume or %volume/volume or %v/v formula.
C. Dilutions
A dilution is the process of combining measured volumes of a concentrated solution of analyte with
a buffer (solvent) to make less concentrated solutions. It is a process repeated constantly in the laboratory
and any laboratory worker should know how to make dilutions. There are several ways to make dilutions.
1. Dilution Factors
Most people are using this dilution method when they make up orange juice from the frozen product. One
can of orange juice is diluted with four cans of water. One can of concentrated juice is diluted to a final
volume of 5 cans by adding 4 cans of water. This is a 1:5 (1 to 5; or 1/5) dilution. In this class, 1:5 always
means 1 volume of concentrated material diluted to a final volume of 5.
The table below gives some examples:
Dilution
Vol. Concentrate
Vol. Water
1:2
1 ml
1 ml
1:3
1 ml
2 ml
1:3
3 ml
6 ml
1:4
1 ml
3 ml
1:100
1 ml
99 ml
1:1.5
1 ml
0.5 ml
Final Vol.
2 ml
3 ml
9 ml - This is a 3:9 dilution, but note that = 1:3
4 ml
100 ml
1.5 ml
In the example below, you have a protein solution (a stock solution) at 2 mg/ml. You need to make the
following levels: 400 μg /ml, 100 μg /ml, 20 μg /ml, 5 μg /ml and 1 μg /ml. You need 50 ml of each
concentration. The first thing to do is to get everything into the same units: 2 mg/ml x 1000 μg/mg = 2000
μg/ml. Then it helps to put everything into a table.
Final Conc.
400 μg /ml
100 μg /ml
20 μg /ml
5 μg /ml
1 μg /ml
Calculate Dilution
400/2000=1/5
100/2000=1/20
20/2000=1/100
5/2000=1/400
1/2000=1/2000
Dilution
1:5
1:20
1:100
1:400
1:2000
Vol stock
10 ml
2.5 ml
0.50 ml
0.125 ml
0.025 ml
Vol buffer
40 ml
47.5 ml
49.50 ml
49.875 ml
49.975 ml
Final Vol
50 ml
50 ml
50 ml
50 ml
50 ml
How do we accurately measure 0.125 mL or 0.025 mL?
2. Serial Dilutions
In the section above we used one or two concentrated solutions to make up all the dilutions. Each dilution
is made independently (except for the levels made from the 20 μg /ml level) from another dilution. If an
error is made it only affects one dilution. When serial dilutions are done, each dilution is made from a more
concentrated dilution. A common use of serial dilutions is to create a set of dilutions each one half of the
concentration of the previous level.
Initial Concentration
Vol solution
Vol buffer
Final Concentration
100 mg/ml
5 ml of 100 mg/ml
5 ml
50 mg/ml
50 mg/ml
5 ml of 50 mg/ml
5 ml
25 mg/ml
25 mg/ml
5 ml of 25 mg/ml
5 ml
12.5 mg/ml
12.5 mg/ml
5 ml of 12.5 mg/ml
5 ml
6.25 mg/ml
6.25 mg/ml
5 ml of 6.25 mg/ml
5 ml
3.125 mg/ml
The advantage of this method is ease of pipeting. The disadvantages are that an error in one dilution causes
errors for all the remaining dilutions and you cannot always simply cut the concentration in half.
3. Dilutions Using V1C1=V2C2 Method
In some cases you will have a limited quantity of an expensive material. You will need to prepare a diluted
sample of this material and use the minimum amount. You have an expensive protein that you have
purchased. You have 100 μl at a concentration of 7.22 mg/ml. From this material, you need to make 4 ml of
a 50 μg/ml solution. To determine the amount of the expensive protein you need, use the formula:
V1C1=V2C2 where V = volume and C = concentration
The first thing to do is to get everything in units of mg/ml for concentration and ml for volume. Then we
have:
V1 x 7.22 mg/ml = 4 ml x 0.05 mg/ml, where V1 = the unknown volume of expensive protein.
We rearrange terms so that V1 =
4 ml x 0.05 mg/ml
7.22 mg/ml
Then V1 = 0.028 ml of expensive protein. To make the dilution, you would take 0.028 ml (28 μl) of the
expensive protein solution and add it to 3.972 ml buffer to get a final volume of 4 ml.
 IN-LAB EXERCISE: You are to make a series of dilutions of “standards” following two different methods
(dilution factors and serial dilutions) of the provided stock solution at the following dilutions (1:5, 1:10,
1:20, 1:40, 1:80, 1:100). The total volume of each standard is to be 100mL. Of course for serial dilutions
you will have to use some of the 100mL of most of the solutions in order to make further dilutions. The
stock solution is a mixture of yellow and red food coloring in water and thus can be dumped down the
drain when you are finished. Assume that the stock solution is at a concentration of 100% and that the
unit of concentration on subsequent dilutions is percent (%). To check the accuracy of your dilutions as
well as the two methods, you are to analyze the series of standards on a Spec-20 by measuring the %T
values at 550nm. Obtain the appropriate data, construct calibration curves (Beers Law Plots), and
perform statistical analysis of each standard curve.
D. Preparation of Buffers
Preparation of buffers and other chemical formulations is an essential part of laboratory work. Buffer
formulations can be given as molar solutions or percent solutions.
There are some general guidelines for the preparation of buffers. If water is the solvent, prepare buffers
with fresh double-distilled or deionized water. Remove powders from their parent containers by shaking
the container or with a disposable spatula. Using the same spatula with different chemicals can result in
cross contamination. Wear gloves when handling chemicals and observe safety precautions for handling
hazardous chemicals.
To make a buffer, add ~3/4 volume of liquid components to a beaker or flask and then add the weighed
ingredients. Mix to dissolve using a clean stir bar. When all the ingredients have dissolved, check and adjust
the pH. Remove the stir bar with a magnet from outside the container. Carefully add water to the final
volume needed after pouring the buffer into a graduated cylinder or volumetric flask. Determine the
volume by looking at the bottom of the meniscus. Place buffer in a clean container, seal and label with
name of buffer, date prepared, and name of preparer(s). Depending on the buffer constituents, buffers
can be filtered or sterilized for long term storage.
 IN-LAB EXERCISE: Your group is responsible for making two buffer solutions per group which will be
used in laboratory exercises later during the semester. The total volume of each buffer solution should
be 100mL and each buffer solution should be prepared as directed (0.1M). Professor Bensley will assign
the actual buffer solutions that you will be making during the lab session.
Note that there are several web sites that contain buffer calculators. You will be provided information
here but in the future (such as on the first exam?!), you will be expected to use the HendersonHasselbalch equation to calculate buffer recipes.
o
0.1M Acetate Buffer: (Dissolved in 100.0 mL DI water)
 pH 5.0 = 0.216 gram Acetic Acid and 0.871 gram Sodium Acetate Trihydrate
 pH 6.0 = 0.032 gram Acetic Acid and 1.288 grams Sodium Acetate Trihydrate
 Adjust the final pH by adding either acid or basic salt as needed
o
0.1M Phosphate Buffer: (Dissolved in 100.0 mL DI water)
 pH 7.0 = 0.584 gram Monosodium phosphate, monohydrate and 0.820 gram Sodium
phosphate dibasic, anhydrous
 pH 7.5 = 0.260 gram Monosodium phosphate, monohydrate and 1.153 grams Sodium
phosphate dibasic, anhydrous
 Adjust the final pH using either a 0.1M HCl solution or a 1.0M NaOH solution (other groups
will be making these solutions).
o
Borate Buffer: Composed of 0.1M HCl and 0.025 M Borax (Na2B4O7•10 H2O MW=361.2 g/mole)
 pH 8.0 =50.0 mL of Borax Solution and 20.5 mL of 0.1M HCl
 pH 8.5 =50.0 mL of Borax Solution and 15.2 mL of 0.1M HCl
 For each, bring to 100.0 mL total volume with DI water
 Adjust the final pH by adding either HCl or Borax as needed
o
0.1M Glycine Buffer: (Dissolved in 100.0 mL DI water)
 pH 9.0 = 0.750 gram of Glycine and 1.36 mL of 1.0M NaOH (MW=40 g/mole) solution
 pH 10.0 = 0.750 gram of Glycine and 6.13 mL of 1.0M NaOH (MW = 40 g/mole) solution
 Adjust the final pH by adding Glycine or NaOH as needed.
 Note that these are approximate volumes and the pH of all buffer solutions should be verified using
the supplied digital pH meters. If the pH of the buffer needs to be adjusted, add either the acid or
the base solution until the desired pH is obtained.
E. Assignment (15 points total)
1. Your laboratory notebook will be turned in and should contain: (5 points)
a. The two calibration curves obtained for Part C of the lab.
b. The formula of the best fit line, the correlation coefficient values for each curve
c. A short discussion on which of the two methods is the more accurate one and why you believe this
to be the case.
2. Answer the questions on the following sheets and show work for all calculations for credit!! You will
hand in the worksheet that follows. (10 points)
Both the notebook and the following sheet are due in class on Monday 9/8.
CHEM 7784 – Biochemistry
Name
LAB #2 – Biochemistry Lab Skills Workshop
Date
Answer the following questions. Show all work to receive credit. Due in class next Monday.
1. How many mL of 0.1M sodium acetate should be added to 35.0 mL of 0.50M acetic acid to make a buffer of
pH 4.7? (1 pt) [pKa of acetic acid is 4.75]
2. Use the Henderson-Hasselbalch equation to calculate how many mL of 0.10M HCl should be added to 100.
mL of 0.10 M Tris base (Trishydroxylmethyl aminomethane) to make a buffer of pH 8.8? [pKa of Tris Base is
8.08] (1 pt) HINTS-Think about ratios and what the definition of molarity is.
3. The pKa for phenolphthalein is 9.3 at room temp.
a) Calculate ratio of its anionic form to acid form at pH 7.80 and at pH 10.2. (1 pt)
b) Using the answer from question #3a, explain the color change within this pH range. That is, why does it
change from clear to pink? (1 pt)
4. Ibuprofen (Advil) has a pKa of 4.9. What is the ratio of A¯ to HA in:
(a) the blood (pH = 7.4) (1/2 point)
(b) the stomach (pH = 1.4) (1/2 point)
5. Calculate the pH of the solution that results from the addition of 0.050 moles of HNO3 to a buffer that is
made by combining 0.500 L of 0.275M HC3H5O2 (Ka = 1.30 x 10¯5) and 0.500 L of 0.275M NaC3H5O2 Assume
addition of the nitric acid has no effect on volume. (1.5 pts)
6. You need to produce a buffer solution that has a pH of 4.68. You already have a solution that contains 15.0
mmol of acetic acid. How many millimoles of sodium acetate will you need to add to this solution? The pKa
of acetic acid is 4.75. (1.5 pts)
7. What is the pH when 25.0 mL of 0.200 M of CH3COOH has been titrated with 35.0 mL of 0.100 M NaOH? (1
pt) HINT-Write out the balanced chemical equation for this neutralization reaction and use this to think
about what happens to the moles of acid and base respectively during the titration.
8. Fifty percent of a weak acid is in an ionized form in a solution with pH of 4.50, what is the pKa value for the
weak acid? (1 pt)
Challenge Problem: (Extra Credit +3 points for correct answer)
A beaker with 100.0 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of
acid and conjugate base in this buffer is 0.1000 M. A student adds 5.600 ml of a 0.4800 M HCl solution to the
beaker. What will be the final pH of the solution? The pKa of acetic acid is 4.75.