Solutions to Chapter 1 Selected Problems
Students should cross out units with a pencil to make sure the answer units are correct.
1. a. The typical cell =50 μm at 10,000 X magnification:
50 μm x 10,000 = 500,000 μm
convert this to meters: 500,000 μm x
1m
= 0.5 m
1,000 ,000 m
which is about 19.5 inches.
b. This is hypothetical, muscle cells are not spherical; neither is actin.
Volume of a sphere = (4/3) π r3
So the cell’s volume = (4/3) 3.14 (25 x 10-6m)3
= (4.18)(15,625 x 10-18 m3)
= 6.5 x 10-14 m3
Volume of a spherical actin = (4/3) 3.14 (1.8 x 10 -9 m)3
= (4.18)(5.8 x 10-27m3)
= 2.4 x 10-26 m3
6.5 x 10 14 m 3
So, how many actin volumes in a cell volume =
= 2.7 x 1012 actin volumes / cell.
26
3
2 .4 x 10 m
Does this make sense at all? Note that the sentence above does not say actin molecules, but
actin volumes. You could not cram 2.7 x 1012 spheres into a cell: this does not take
into account the unused packing space between spheres. In doing so, it would be less
and a more complex geometry problem to solve. That is why “volumes” was used.
c. This part just replicates the same logic in b and is solved the same way. Just to check, you
should come out with 3.6 x 104 mitochondrial volumes/cell.
d. So here we are stuffing the cell with glucose. A 1 mM solution of glucose would have a
thousand less molecules / L than a 1 mole/L concentration. At 1 molar, there are Avrogado’s
number of molecules / L = 6.02 x 1023 molecules/L.
The cell volume above is in units of m3. There are 106 mL / 1 m3; or 103 L / m3
Lets change cell volume to Liters: 6.5 x
In 1 Liter of 1 mM glucose there are:
10-14
m3
10 3 L
x
= 6.5 x 10-11 L per cell
3
m
(6.02 x 1023 molecules/M )x(10-3 M) = 6.02 x 1020 molecules/L.
So, in one cell: (6.02 x 1020 molecules/L) x (6.5 x 10-11 L) = 3.9 x 1010 molecules.
e. Concentration of hexokinase is 20 μM, the concentation of glucose is 1 mM. To get the ratio
these numbers need to be in the same units: 1 mM is the same as 1000 μM.
So, 1000 μM glucose / 20 μM hexokinase = 50 glucoses per 1 hexokinase.
3. Escherichia coli is about 2 μm long. We have to get the length of the E. coli chromosome (it is
going to be enormous compared to the cell).
Here is what we have: chromosome molecular weight = 3.1 x 10 9 g / mole.
the average N-tide pair molecular weight = 660 g / mole.
the length of an N-tide pair in DNA is 0.34 nm (= 3.4 Å)
First lets find the number of N-tide pairs in the chromosome, then simply multiply that times
the length of a N-tide pair.
3.1 x 10 9 g / mole
Number of N-tides in the chromosome =
= 4.69 x 106 N-tide pairs/chromosome
660 g / mole
Now just multiply number of N-tide pairs times each N-tide pair length:
4.69 x 106 N-tide pairs x 0.34 nm / N-tide pair = 1,594,600 nm = 1,595 µm
So the length of DNA in an E. coli is 797 times longer than the cell (only 2 µm long) ! This
means the DNA (a string of N-tides) has be pretty well wound up inside the cell.
5. Axonal transport involves two things: the nerve electrical impulse and the vesicles carrying the
neurotransmitters that are released into the neuromuscular synapses. Here we are looking only at
vesicle transport. Given a 2 m long nerve, a vesicle moving at 1 µm/sec will take how long to travel
from the nerve soma (in the spinal column) to the toe (where the synapse resides). This problem is
not entirely anatomically correct: 1. muscles that move toes are primarily in the calf, 2. the soma are
lower in spinal column, and 3. hardly anyone has 2 m long legs! But, lets do the problem anyway:
10 6 m
The length of travel, 2 m needs to be converted to µm: 2 m x
= 2 x 106 µm
m
2 x 10 6 m
So,
= 2 x 106 sec
1m / sec
How many seconds in a day? = 8.64 x 10 4 sec/day
2 x 10 6 sec
Thus, it would take =
= 23.1 days
8 .64 x 10 4 sec/ day
Go ahead move your toes, the molecules that turned on those muscles took some part of a
month to get to the synapse. The nerve impulse got there FAST !
8. Chiral centers: in isoproterenol there is only one, the only carbon bound to 4 different groups:
the one just outside but bound to the benzene ring, an alcohol, a hydrogen and the rest of the
molecule. The importance of this question is that a small difference in molecular structure can have
amazingly different properties. This is based on how this molecule fits into a protein binding site.
11. Chiral centers: again the same logic is used as in problem 8. This one reinforces the concept
that chiral isomers can have identical physical properties (C, H, N analysis, melting point, solubility,
and others) but be different in shape, this is so important in binding to a protein.
12. This looks at molecules, GTP, Methionine enkephalin (a 5 amino acid peptide), and a
membrane phospholipid, phosphatidylcholine.
a. GTP is made up of three phosphates: two acid anhydride bonds, one phosphate ester bond
to ribose and the glycoside bond between ribose and guanine. NOTE that these structures
ALWAYS show the molecule at pH 7. So if the molecule has weak acid groups they should be
represented at the form at pH 7: either associated or dissociated. The structure in the text is
correct. Organic chem structures do not show the molecules as they exist in water.
b. Methionine enkephalin is made up of 5 amino acids. The text structure has a mistake: the
amino group on the left should be -NH3+ . It should be -NH2 only in Organic Chem texts, not in
biochemistry, the beautiful chemistry of life in water.
c. Phosphatidylcholine has a choline attached to the phosphate (ester) and the phosphate is
also attached to glycerol (ester). The glycerol is attached to two fatty acids (both as esters).
The 16 carbon fatty acid is saturated, the 18 carbon fatty acid is unsaturated.