Select Solutions to End of Chapter 4 Problems
First Class
1. a. The length of the bond indicates it has properties between a single bond and a double bond: but
it is more close (0.05 Å) to being a double bond than that of a single bond (0.17 Å).
b. Pauling and Corey showed that the peptide bond is planar, and can not rotate like a single bond
and carboxyl and amino groups are in trans configuration.
2. a. Wool is just about all α-helix with a complete turn of the helix being 5.4 Å. Remember that this
structure is held together by H-bonds and dipole interaction, these are weak bonds. Steaming allows
stretching by breaking some of these weak bonds, but upon return to cooler temperatures, these
broken weak bonds reform.
b. Wool is in a α-helical coiled coil. In wool processing, the fibers are combed and stretched to
straighten it to β-structure. Moist heat in washing and drying allows the fibers to go back to the αhelical structure. Silk is different, it is just about all β-structure AND composed of amino acids with
small R groups, the structure of silk is much more stable than wool.
3. Calculating the rate of growth of hair (15-20 cm/year). We know that hair is almost all α-helix
which means one turn of the helix is 5.4 Å long and is composed of 3.6 amino acids. So the length of
an amino acid in α-helix is:
5.4 Å/𝑡𝑢𝑟𝑛
3.6 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖𝑑𝑠/𝑡𝑢𝑟𝑛
= 1.5 Å / amino acid = 1.5 x 10-10 m / amino acid.
At a growth rate of 20 cm/year…lets convert this to m /sec :
20 𝑐𝑚/𝑦𝑒𝑎𝑟
𝑑𝑎𝑦𝑠
ℎ𝑟𝑠
𝑚𝑖𝑛
𝑠𝑒𝑐
(365𝑦𝑒𝑎𝑟 ) (24𝑑𝑎𝑦)(60 ℎ𝑟 )(60min
)
= 6.3 x 10-7 cm / sec = 6.3 x 10-9 m / sec
So, the rate to insert an amino acid into a growing hair α-helix is:
6.3 x 10-9 m/sec / 1.5 x 10-10 m/amino acid = 42 amino acids/sec
So, if it took you say 5 minutes to do this problem, each of your hairs grew 12,600
amino acids longer.
4. This problem uses circular dichroism (CD) to measure secondary structure. In the “lecture” the
figure shows that the angular change when circularly polarized light through a protein is characteristic
and different for α-helix, β-structure, and random conformation. In this problem there are two,
artificially synthesized proteins, each one having only one amino acid: poly-glu and poly-lys. The
data shows the CD angles change based on pH. Poly-glu is in α-helix from very low pH’s but at pH 5-
6 it makes a big change to random coil. Poly-lys is just the opposite: in random coil from low pH’s
until pH around 8-9 and it then goes into α-helix at high pH’s
Why?
It is all about the R group differences: what is the charge of the glutamate R group above pH 5-6?
And, what is the charge of the lysine R group below pH 8-9? You can explain this by electrical
repulsion or either all positively charged R groups or all negatively charged R groups. That repulsion
is strong enough to overcome the H-bonds that form the α-helix. Note that random conformation
does not have one set specific rotation, it’s random !
Second Class
8. Bacterial rhodopsin, a 26,000 MW membrane protein (it can convert light energy into the proton
motive force). It has 7 membrane spanning α-helical segment that are in the membrane (thickness of
about 45 Å). How many amino acids in one transmembrane segment? And, how much of the protein
actually in the membrane?
So, first how long in 1 amino acid in the α-helix:
3.6 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖𝑑𝑠/𝑡𝑢𝑟𝑛
5.4 Å/𝑡𝑢𝑟𝑛
= 0.67 amino acid / Å
then, (45 Å) x (0.67 amino acids /Å) = 30 amino acids
How much of the protein in the membrane. The number of amino acids in the protein can be
estimated to be:
26,000 𝑑𝑎𝑙𝑡𝑜𝑛𝑠
110 𝑑𝑎𝑙𝑡𝑜𝑛𝑠/𝑎𝑎
= 240 amino acids total.
7 membrane spanning segments = 7 x 30 aa/segment = 210 amino acids.
Thus, there is 210 amino acids in membrane / 240 total amino acids = 0.875 or 87.5% of the
protein is in the membrane.
9. Myoglobin is a whole protein as a single polypeptide, so it is a complete 3-dimensional structure
possessing a domain and motifs. No quaternary structure.
10. Interpreting the Ramachandran plots. This is super easy: look at the proteins, each one is
dominated by one type of secondary structure: “a” is mostly β-structure, and “b” is mostly α-helix.
Now check out page 124, Fig 4-9a to see where on the Ramachandran plot each secondary structure
would be.
11. One of the enzymes secreted by Clostridium perfringens is a protease that cuts collagens which
have a high concentration of this sequence. Collagens are connective tissue that hold cells and
tissues together. These collagenases break down collagens and allow the bacterium to penetrate
into tissue. These collagenases are not harmful to bacteria for two reasons: 1) bacteria do not have
collagen, and 2) excreted enzymes are not allowed to fold to the active conformation in the
cytoplasm. That is, they are bound with chaperone proteins which hold them as a long string of
amino acids (unfolded) to transport proteins in the membrane, the transport the inactive collagenase
to the outside of the bacterium where is can fold into it’s active conformation.
12. Number of polypeptides in a large protein can be determined by measuring the amount of Nterminal amino acid(s). In this case, using FDNB to label the N-terminal amino acids, it produced only
one DNP-amino acid: DNP-val (protein sample => 5.5 mg DNP-val).
a. All we need to do is find the moles of DNP-val formed from the number of moles of the
protein (132,000 daltons) in a 660 mg sample.
b. Lets do it ! First moles of protein = (0.66 g protein) / (132,000 g/mole) = 5 x10 -6 moles of
protein.
Now, moles of DNP-valine (MW DNP-val = 283 daltons).
Thus the moles of DNP-val = 0.0055 g / (283 g/mole) = 1.9 x 10-5 moles.
So then, 1.9 x 10-5 moles DNP-val / 5 x 10-6 moles protein = 3.9 or 4 N-terminals per protein.
Thus, there are 4 polypeptides that make up this protein.
c. The simplest would be do run an SDS- PAGE gel with molecular weight standards. It is
likely that they all come out to be 33,000 daltons ( 33 kD) if they were all identical forming only
one band. But, it could be that they came from different cistrons, and in that case there would
likely be different MW bands even though they all had valine as the N-terminal amino acid.