Solutions to Selected End of Chapter 3 Part 2 Problems
8. Protein size: a protein with 682 amino acids, what is the approximate molecular weight? If you
take the average molecular weight of all 20 protein amino acids (Table 3.1 in text) you get 128
daltons. Now…in making the protein, each amino acid is added to form the peptide bond through a
dehydration reaction (removing a water, thus each amino acid is 18 daltons lighter) so the average
molecular weight of the amino acid in the protein is 110 daltons (128 – 18 = 110).
So, the size of the protein is 682 amino acids x 110 daltons/amino acid = 75,020 daltons.
Because this is an approximation, then the best answer is 75,000 daltons.
9. Quantitative amino acid analysis to calculate protein molecular weight. The example is BSA
(bovine serum albumin, a protein used a lot in biochemistry and clinical assays) that contains 0.58%
tryptophan by weight. Tryptophan has a molecular weight = 204 daltons, it is the largest amino acid.
So, trp in a protein molecular weight is 204 – 18 = 186 daltons.
a. The relationship of weight by weight is the same a molecular weight by molecular weight.
This can be expressed where n is the number of trp/protein is assumed to be 1 (this calculates
the minimum MW):
𝑤𝑒𝑖𝑔ℎ𝑡 𝑡𝑟𝑝
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑝𝑟𝑜𝑡𝑖𝑒𝑛
=
𝑛 (𝑀𝑊 𝑡𝑟𝑝)
0.58 𝑔
𝑀𝑊 𝑃𝑟𝑜𝑡𝑒𝑖𝑛
100 𝑔
=
1 (186 𝑑𝑎𝑙𝑡𝑜𝑛𝑠)
𝑀𝑊 𝐵𝑆𝐴
Calculate the minimum MW BSA = 32,069 daltons.
b. Gel filtration chromatography indicated the MW of BSA is about 70,000 daltons. Gel
filtration like polyacrylamide gel electrophoresis can estimate MW, but the accuracy is usually
plus or minus thousand or so. It is obvious that 70,000 is bigger than roughly twice as big as
32,000, but not 3 times as large. So divide the minimum MW into the gel filtration MW gives
2.2. This suggests that BSA has 2 trp per BSA molecule (there can’t be a part of trp, it either
has the whole amino acid or not).
10. Proteins with quaternary structure. Data: Gel filtration MW about 400,000 daltons (= 400 kD).
Now, gel electrophoresis in presence of SDS reveals three proteins: 180 kD, 160 kD and 60kD. This
says right away that there are at least 3 subunit proteins that make up the whole protein, and that
SDS releases the subunits from each other suggesting that they were held together by hydrophobic
interactions and H-bonds. To check that out, add the subunit MWs together, does 180 kD + 160 kD +
60 kD = 400 kD?
A second electrophoresis was done: SDS and dithiothreitol. Dithiothreitol reduces disulfide bonds to
sulfhydryls. The result was 160 kD, 90 kD and 60 kD subunits. Note that in this electrophoresis, the
180 kD subunit did not show, but a new size, 90 kD was present. This means that the 180 kD subunit
in the SDS alone electrophoresis was composed of two 90 kD proteins held together with the help of
at least one (or more) disulfide bond(s). So the complete subunit composition of the 400 kD protein
is: one 160 kD subunit, two 90 kD subunits bound by disulfide bond(s), one 60 kD subunit.
11. Electric charge of a peptide varies with pH. The peptide is:
D–H–W–S–G–L–R–P–G
The ionizable groups are the N-terminal amino acid, C-terminal amino acid, and ONLY the R
groups of the middle amino acids.
a. What are the charges at pH 3, 8, and 11? Diagram shows only ionizable charges: the
amino terminal, R groups of Asp, His and Arg, and carboxy terminal.
+ D–H–W–S–G–L–R–P–G-
At pH 3
0
+
-
At pH 11
+
+D–H–W–S–G–L–R–P–G–
At pH 8
0
overall charge = 0
+
0 D–H–W–S–G–L–R–P–G-
overall charge = +2
0
overall charge = -1
+
b. You can see that the actual pI is going to be near 8. So, what ionizable groups are closest
to 8? The amino terminal (9.67) and histidine R group (6.0). So pI is easy:
pI =
9.67+6.0
2
= 7.8
13. What is the isoelectric point of Histones that make up the nucleosomes, the structure that DNA is
wound around in the nucleus. DNA is made up of nucleotides held together by phosphate di-esters
leaving each phosphate with a negative charge. So to get DNA wound around the nucleosomes, the
histones have to have lots of positive charges. The important concept here is that it has to be R
groups of amino acids that are not the N- or C-terminal. Which ones would have a positive charge at
physiological pH? Check out Table 3.1 for the amino acids R, K and H.
15. Classical protein purification (we will do one of these in class). These always begin with a crude
extract: the grinding up of cells to make a cell free homogenate of a tissue containing all the proteins
of the sample.
a. Then after each procedure you need to calculate three things:
1. percent yield for each step (did you loose protein by any of these steps?). This is the
protein activity after each procedure divided by the protein activity in the crude extract,
expressed as %,
2. specific activity which is the activity of your protein divided by the total amount of
protein left after that procedure in units/mg protein, and
3) purification factor (how many more times is the sample purified after each step
compared to the crude extract) which is the specific activity after each procedure
divided by the original crude extract specific activity.
Now you can check to see if the calculated values in the table are correct, and then calculate
the rest, all those blanks.
Procedure
Crude Extract
Salt ppt
Acid ppt
Ion Exchange
Affinity Chrom
Gel Fil Chrom
Protein, mg
20,000
5,000
4,000
200
50
45
Activity, units
4 x 106
3 x 106
1 x 106
8 x 105
7.5 x 105
6.75 x 105
Yield, %
100
75
25
Sp. Act. Units/mg
200
600
250
Fold Purification
1.0
3.0
1.25
b. Which step is the best (largest increase in fold purification)? It should be the Ion exchange
step if your calculations are correct.
c. Which are the least effective? It should be the acid ppt, it destroyed some of your protein
and decreased purification, just the opposite of what you are trying to accomplish.
d. Which step is your protein going to be as pure as possible? It should be affinity
chromatography. The Gel Filtration Chromatography did not improve on its specific activity
and in this case is not needed. It is likely pretty pure, but you can check that by doing an
analytical native PAGE and an SDS PAGE. The PAGE results should show mainly one
protein band on the native PAGE, and possible subunits on the SDS PAGE.
16. Dialysis. Data: your purified protein was in a 0.02M HEPES buffer that had 0.5 M NaCl. You
need to dialyze this to remove the NaCl for the next steps in your research. So, you put 1 mL of your
sample into a dialysis membrane bag, and put it into 1 L of 0.02M HEPES with no NaCl.
a. After dialysis, what is the concentration of NaCl in the sample? Your calculation should
show that this has reduced the NaCl to 0.5 mM. Note that the concentration of HEPES on
both sides of the membrane are the same, so that its concentration does not change, and the
protein does not change concentration because the molecule are much bigger than the
effective pore size in the dialysis membrane.
b. A different dialysis strategy was done with the same but fresh dialysis buffer, but first using
only 100 mL of dialysis buffer, then after completion, the dialysis was done again with another
fresh 100 mL of dialysis buffer. What is the NaCl concentration after this two successive stage
dialysis? Your calculation should show it to be 0.05 mM.