Problem Set 1 – Problems on Biomolecules (2 problems):
1. The sequence of amino acids of the enzyme lysozyme is known. Below is a list of amino acids
and the number of each in the lysozyme molecule.
Type

Molecular Weight

Number in Lysozyme

Alanine

89

12

Arginine

174

11

Asparagine

132

13

Aspartic acid

133

8

Cysteine

121

8

Glutamic acid

147

2

Glutamine

146

3

Glycine

75

12

Histidine

155

1

Isoleucine

131

6

Leucine

131

8

Lysine

146

6

Methionine

149

2

Phenylalanine

165

3

Proline

115

2

Serine

105

10

Threonine

119

7

Tryptophan

204

6

Tyrosine

181

3

Valine

117

6

Questions:
a. What is the average molecular weight of the 20 amino acids?
b. What is the molecular weight of lysozyme?
c. What is the average molecular weight of the amino acids in lysozome?
d. How many S-S bonds are possible in lysozyme?
(hint: look at the structures of each amino acid, and figure out which of these have sulphurs
that are FREE to form bonds).
e. Is the net charge on lysozyme positive or negative?
(hint: again, look at the structures of each amino acid, and figure out which of these are
neutral, which are positively charged, and which are negatively charged; then add up the
individual charges for each amino acid to get the net charge of this protein)

2. Proteins (both enzymatic and structural) play a significant role in maintaining the pH in a cell.
Seven of the 20 amino acids commonly found in proteins have an ionizable group. These
groups (side chains) determine the charge on the protein and buffer the cytoplasm of the cell
that contains it.
Populations of proteins that contain ionizable side chains have a probability of being “charged”
or not. The probability of a side chain being charged is related primarily to the side chain
chemistry and the pH of the cellular compartment; other usually less important factors are
temperature, ionic strength, and bulk microenvironments.
The ratio of the charged to uncharged side chains is usually symbolized by K:
K=c/u
where c is the proportion of charged side chains and u is the proportion of uncharged side chains
(so that c + u = 1).
In the laboratory, the pH of a protein’s environment can be altered so that we have equal
numbers of charged and uncharged molecules. We call the pH at which K is 1 (equal number of
charged and uncharged side chains) the pK of the side chain.

Below is a list of the seven amino acids whose charge is altered by the cellular pH:
Acidic

pK

Basic

pK

Aspartic acid

3.87

Lysine

10.53 Cysteine

8.33

Glutamic acid

4.25

Arginine

12.48 Tyrosine

10.07

Histidine

Polar

pK

6.00

Cell compartments can vary in pH from about 4.0 to 8.5, but the pK values of the various amino
acids remain fixed. When the pH of the environment is not equal to the pK of the side chain,
the proportion of charged and uncharged side chains changes to as to satisfy the equation:
pH = pK + log ( c / u )
As an example, for aspartic acid in a protein with a pH environment of 3.87, pH = pK,
so log (c/u) = 0. Consequently, (c/u) = 10 ^ 0 = 1, and c = u. However, when the pH and the
pK are different, the term log (c /u) will not be zero and the ratio (c/u) will be different from 1.
Questions:
a. The pH of healthy cell cytoplasm varies from 7.2 to 7.4. Using this information, and the
information above, complete the following table:

Amino acid in protein

pK of the side chain

Percent Charged at pH Percent Charged at pH
= 7.2
= 7.4

Aspartic acid

3.87

?

?

Histidine

6.0

?

?

Cysteine

8.33

?

?

Arginine

12.48
_________________________________________________

Problem Set 2 – Problems on Cell Structures and Membranes (3 problems):

3.

Lysosomes are little sacs of acid in a cell. Their pH is about 5, and an electron
micrograph suggests they have a diameter of 0.5 µm. The increased hydrogen ion
concentration inside lysosomes is due to the pumping of hydrogen ions across the
lysosomal membrane from the surrounding cytosol, which has a pH of 7.2.

a. Assuming that a lysosome has the shape of a sphere and that there is no
buffering capacity inside the lysosome, how many hydrogen ions were moved
to the inside of the lysosome to lead to an internal pH of 5?
(hint: first determine the volume of a lysosome in liters, then determine [H+] in
moles/L in a lysosome at each pH (5 and 7.2), then determine the number of
moles of hydrogen ions at each pH, and finally determine and compare the
number of hydrogen ions at each pH).

4.

Liposomes are laboratory-prepared artificial membranes. Liposomes can be made in a
variety of sizes and can be made so that they have transmembrane proteins, which form
membrane. Contents of the liposomes can also be known.

For example, let’s say that one lab makes liposomes that are spheres with the diameter of
4 µm and that each liposome has an average of ten protein pores. Each liposome has an
internal potassium ion concentration of 100 mM. Each protein pore transports 3x 10^6
potassium ions per second. The pores stay open an average of 0.3 second and stay closed
an average of 2 seconds; so, each pore opening and closing cycle takes about 2.3 seconds.

a. Assuming that a liposome has the shape of a sphere, how many potassium
ions are in a liposome initially?
(hint: the method here is similar to what you used to solve problem 3 above,
except find the volume of a liposome in µm^3 and the [K+] in mol/µm^3)

b. How much time is required for the potassium ions in the liposome to reach
equilibrium with their environment? Assume that this environment is
relatively large and potassium-free.
(hint: before calculating the total time it would take to reach this equilibrium,
think about how many potassium ions would need to leak out of the liposome in
order to reach this equilibrium – all of them, half of them, none of them, why?)

5. Glycophorin is a single-pass transmembrane protein in red blood cells (RBCs). The
protein component of glycophorin is 131 amino acids long and binds carbohydrates on
the outside (noncytoplasmic side) of glycophorin. Then, approximately 100 modified
sugar residues are attached near the end of each glycophorin; these account for about
60% of this macromolecule’s mass. The average molecular weight of an amino acid is
130 daltons.

a. What is the average molecular weight (in daltons) of each modified sugar
residue on the glycophorin?

b. An RBC contains an average of 6 x 10 ^5 glycophorin molecules. How many
modified sugar residues are found attached to glycophorins in one RBC?

c. How many grams does the protein component of glycophorin weigh in one
RBC?
______________________

Problem Set 3 – Problems on Enzymes and Cellular Respiration (3 problems):

6. Hydrogen peroxide is usually stored in a brown bottle away from sunlight because it
spontaneously (but slowly) decomposes into O2 and water. In the brown bottle, the free
energy of activation ∆G+ = 18 kcal/mol. In the presence of a catalyst, the decomposition
is much faster. For each decrease of 1.36 kcal/mol in ∆G+, the rate of reaction is ten
times faster.
In the presence of catalase, an enzyme found in the blood, ∆G+ = 7 kcal/mol. In the
presence of the inorganic catalyst platinum, ∆G+ = 13 kcal/mol.

a.

How much faster is peroxide decomposition in the presence of a catalase?

b. How much faster is peroxide decomposition in the presence of platinum?

c. How many times faster is peroxide degradation under the influence of a catalase
than degradation under the influence of platinum?

7. You eat a candy bar that has 180 calories. This energy is converted during respiration to
ATP. The reaction ADP + Pi  ATP requires 7.3 kcal/mol. One “dietary” calorie is
equal to 1000 “chemical” calories. Respiration is maximally about 39% efficient in
converting substrate calories to ATP calories.

a.

Assuming that all of the energy in the candy actually gets to the respiratory site
in the cell, how many ATPs could your body make from this candy bar?

8. The first two stages in respiration are glycolysis and Kreb’s cycle. For each molecule of
glucose input,
10 NAD+ molecules are reduced
2 FADs are reduced
4 ADPs are phosphorylated.
The free energy ∆G0’ of the relevant “respiration reactions” is:
Reaction
NADH + H+ + ½ O2  NAD+ + H2O
FADH2 + ½ O2 FAD + H2O
ATP  ADP + Pi
Glucose + 6 O2  6 CO2 + 6H2O

a.

∆G0’ (kcal/mol)
-51.7
-43.4
-7.3
-686

How much of the energy from one mole of glucose is converted into ATP during
the first two stages?

b. How much of the glucose energy is conserved in the reduced coenzymes NAD+
and FAD?

The third stage of respiration is oxidation phosphorylation.
c.

A total of 32 ATPs per glucose molecule are made during oxidative
phosphorylation from the reduce coenzymes. How much of the glucose energy is
conserved in ATP at the end of ALL three stages?

d. What percentage of the total ATP energy is converted by the oxidative
phosphorylation of the reduced coenzymes?
e. What percentage of the glucose energy was lost was heat?
_____________________
Problem Set 4 – Problems on Photosynthesis (3 problems):
9. The fixing of carbon in photosynthesis varies in regard to the amount of ATP needed for each
carbon fixed.
C3 plants use 3 ATPs per carbon
C4 plants use 5 ATPs per carbon
CAM plants use 5.5 ATPs per carbon

If ATP yields 7.3 kilocalories per mole, how many ATP calories are needed to create 1 mole
of glucose (686 kcal/mol) by:
a.
b.
c.
d.

a C3 plant? (hint: first, think about how many carbons are in 1 mole of glucose)
a C4 plant?
a CAM plant?
Which type plant uses the most ATP energy in the making of glucose?

10. When light is shined on a leaf, it causes hydrogen ions to be pumped into discs called
thylakoid lumens. The ions then diffuse out through a protein, and in the process an ATP
molecule is made for every three hydrogen ions. While illuminated, inside the disc, the pH can
be as low as 4. Outside the disc, the pH is about 7.2.

A thylakoid lumen can be modeled as a short cylindrical rod that is 80 Å long and 5000 Å in
diameter.
a. How many hydrogen ions are found in one thylakoid lumen of this size at pH 4?
b. How many are found at pH 7.2?
c. How many more ATP molecules can be made from the disc described above,
AFTER the light is turned off?

11. Light is important in biology for photosynthesis. There are two different ways that light is
described in physics.
In the first description, light travels in waves at a fixed speed c = 2.998 x 108 meters per second.
The wavelength is the distance from peak to peak of a light wave, and corresponds to the color of
the light. The wavelength is given by λ (the Greek letter lambda). The wavelength varies from
400 nm to 700 nm for light in the visible range, with blue light having λ=450 nm and red light
having λ=680 nm.
The frequency is given by ν (the Greek letter nu). The frequency is the number of peaks that
pass a point in a given time. Frequency is related to wavelength by the formula: ν = c / λ
In the second description, light travels in particles called photons or quanta. Using this
description it makes sense to speak of a mole of light as 6.02 x 1023 photons.
The energy of one photon of light is given by
E = (hc) / λ = hν
where h is a conversion factor called Planck’s constant; h = 1.583 x 10-34 calorie seconds.
In the laboratory, light with a very narrow wavelength range can be used for experiments. One
mole of an actinic light (activating light) that has a wavelength of 680 nm was used to excite
chlorophyll, and caused fluorescence measured at a wavelength of 690 nm. The chlorophyll was
isolated, and therefore could do no photochemistry.
a. What is the amount of energy (in kilocalories) in one mole of actinic red light?
b. What is the amount of energy (in kcal) in the light that was fluoresced (assuming
maximal fluorescence)?
c. What is the amount of energy (in kcal) that was lost as heat?
d. What percentage of the red light energy was lost as heat?

A photon of blue light will energize an electron from chlorophyll to a level comparable to a
photon of red light. Suppose blue light energy also caused fluorescence measured at a
wavelength of 690 nm.
e. What percentage of the blue light energy was lost as heat (again assuming maximal
fluorescence)?
___________________________________________________________________________________________________

Problem Set 5 – Problems on Plant Transport/Resource Acquisition, and Plant Response to
Stimuli (3 problems):

12. Plasmodesmata are cytoplasmic connections across plant cell walls that connect adjacent cell
cytoplasms. Some cells have few plasmodesmata connections while others have more; this is
due to genetics, age, and location within the plant.
The density of plasmodesmata within the cell membrane ranges from a high of 25 per square
micrometer to a low of 0.2 per square micrometer. The average plasmodesmata tube is 40 nm in
diameter.

Diagram of plasmodesmata between two plant cells
a. What percentage of the cell membrane surface area is composed of plasmodesmata
at the high density? (Assume the area of one end of a plasmodesmal tube is a circle
– see image above)

b. What percentage of the cell membrane surface area is composed of plasmodesmata
at the low density?

13. Liquid water moves into and out of the cell by diffusion. Water vapor also moves from the
inside of the plant leaf to the ambient air by diffusion, in a process known as transpiration. This
transpiration causes dissolved minerals to be moved long distances in the plant.
The time t in seconds for water to move such a long distance d in meters is given by
t = (d2)/D
where D is the diffusion coefficient. A reasonable value for D is 2.4 x 10-5 m2/sec. (Note: water
vapor diffuses in air much more rapidly than in liquid water).
The path of the diffusion of water vapor from a leaf into the air varies considerably, but a
measured distance of 1 mm is reasonable (1 mm = 10-3 m).

a. How long (in seconds) does it take a molecule of water vapor to be lost by this leaf?
Hairiness of leaves is a genetic trait. Leaf hairs may double the distance water must diffuse.
b. How long (in seconds) would it take to lose water from a hairy leaf?

14. (This problem is split into two different problems – 14 and 15 – to give you more credit for
the graphs.). One environmental factor that plants monitor is the duration of light. The length of
an uninterrupted dark period is often extremely important in the kind of growth (vegetative as
compared to flowering). Long-day (LD) plants require at least a certain day length before they
begin to flower, whereas short-day (SD) plants require at most a certain day length before they
begin to flower.
In the corn belt in the United States, sunrise and sunset define the length of the day. For
example, here are the times of sunrise and sunset of the first day of each month in northern Iowa:

Month
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec

a.

Sunrise
7:22
7:09
6:34
5:44
5:00
4:33
4:35
4:58
5:28
5:56
6:29
7:02

Sunset
4:45
5:19
5:52
6:24
6:54
7:23
7:33
7:14
6:32
5:43
4:58
4:35

Plot the sunrise and sunset times on the same graph. (You can do this either on the computer
with Excel or by hand on graph paper. If you draw this by hand, make sure to scan your graph
and then paste the image into the document that you will be submitting for grading. Please
label your graph and axes properly: title below the graph (Figure 1. Give your own title here), x
axis = X title (in ? – hint: measured in what units?); y axis = Y title (in? units). Typically, in
science, graphs show the independent variable on the x-axis, and dependent variable on the yaxis. So, you will need to figure out which of these is the independent variable, and which is the
dependent variable. The independent variable is the variable that is directly manipulated by a
scientist during an experiment. The dependent variable is not directly manipulated by a
scientist, and instead changes in response to a change in the independent variable.)

b. Plot the duration of light per day and the duration of dark per day on the same
second graph (see (a) for graphing instructions).

15. This problem is a continuation of problem 14 above. Now, suppose you have inherited a
greenhouse and you decided to become a farmer. You don’t want to pay for additional lighting,
so you will use only sunlight. Also, you have a local market only for the following plants:

Plant
Dill
Spinach
Soybean
Cocklebur

SD or LD
LD
LD
SD
SD

Critical Day Length
11
13
15.5
14

a. When during the year would you expect each plant to be induced to flower?
Explain how you determined this (hint: look at your graphs for problem 14).

_________________________________________
Problem Set 6 – Problems on Gene Expression/Repair and Protein Synthesis/Degradation (3
problems):

16. The DNA in a human non-gametic cell contains 6 billion base pairs. It is estimated that
about 10,000 DNA changes occur in each cell in one day. These are quickly repaired so that
only a few (1 to 5) mutations accumulate in one cell in a year.

a. What percentage of the base pairs are altered each day?

b. What percentage of the DNA changes that occur in one cell in one year escape
the proofreading and repair process – calculate this in both cases: if 1 mutation
accumulates in one cell per year, and if 5 mutations accumulate in one cell per
year?

17. In one mammalian cell it is estimated that 10,000 to 20,000 different types of mRNA can be
found. Abundant mRNAs exist in many copies per cell (up to 12,000 copies/cell); but scarce
mRNAs (5 to 15 copies/cell) can also be detected. At any instant, a snapshot of mRNA content
would reveal a total of 360,000 mRNAs. The cell usually has about 10 times as many ribosomes
as mRNAs. Assume 75% of the ribosomes are involved in protein synthesis at any instant in
time.
a. What is the maximum number of proteins that could be in the process of
synthesis at any instant?
b. How many scarce proteins (use 5 copies/cell) could be in the process of being
made? (hint: use that 7.5 ribosomes are reading each of the 5 mRNAs; but explain or
show calculations for why you should use this).
c. How many abundant proteins could be in the process of being made? (hint: use
that 7.5 ribosomes are reading each of the 12,000 mRNAs; but explain or show
calculations for why you should use this).
d. What is the ratio of developing abundant proteins to developing scarce proteins?

18. Protein molecules in the cytosol of a cell have different half-lives t ½. The half-life is the
time needed for 50% of the molecules to be lost or altered. Half-lives are determined by several
factors, one of which is the “marking” of the protein by ubiquitin, which signals that the protein
is intended for digestion by proteases. Ubiquitin binds differently to proteins due to differences
in their amino ends. The following table gives the half-lives t ½ for some proteins with different
amino ends:

Amino Acid End of Protein
MET
SER
THR
ALA
VAL
LEU
PHE
ASP
LYS
ARG

t½
> 20 hrs
> 20 hrs
> 20 hrs
> 20 hrs
> 20 hrs
< 3 min
< 3 min
< 3 min
< 3 min
< 2 min

Half-life is related to the rate K of protein loss by the following equation:
t ½ = 0.693 / K

where 0.693 is the natural logarithm of 2, and K is measured in reciprocal time units.
a.

What is the rate of protein loss per minute when MET is the amino terminal?
(Assume t ½ = 24 hr).

b. Compare this with the rate when ARG is the amino terminal end. (Assume t ½ =
2 min).

c.

What percentage of the protein with a MET amino end exists in the cell after 5
half-lives?

d. What percentage of the protein with an ARG amino end exists in the cell after 5
half-lives?

e. How much faster (as a percentage) is the rate of ARG-amino-end protein
degradation as compared to MET-amino-end degradation?

f. Plot the loss of a population of protein molecules that have a MET amino end.
Use time t on the horizontal axis and size of the population on the vertical axis,
starting with 100% of the population at the starting time t = 0.

g. On the same graph, plot the loss of a protein with ARG as the amino terminus.

h. Hemoglobin exists in the cytoplasm of a red blood cell (RBC). Red blood cells
last about 120 days in the bloodstream. In your initial experiments you find that the
amino terminus of one protein chain in RBCs is either a valine or a leucine (so a
difference of only one methyl group). Which is more likely to be the correct amino
acid terminus of this protein? (Assume t ½ = 24 hr for valine and 2 min for leucine).
_______________________________________

Problem Set 7 – Problems on Microorganisms/Bacteria (2 problems):
19. Assume that you conducted the following experiment in order to examine how bacterial
evolution occurs in jumps:
An E. coli culture was maintained for 10,000 generations over 4 years. The liquid medium was
changed daily to maintain a constant environment. The average size of the cells at the start of
the experiment was 0.35 x 10 -15 L. After 300 generations the size increased to 0.48 x 10 -15 L,
and after another 300 generations the average increased to 0.49 x 10 -15 L. After 1200
generations it increased to 0.58 x 10 -15 L and remained so until the end of the experiment.

Volume ( x 10 -15 L)

E. coli Evolution
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0

500

1000

1500

2000

Generations

2500

3000

3500

a. How many hours long is a generation of E. coli?

b. What was the average size change over the course of the experiment in liters?

c. What was the average size change in the first 300 generations in liters?

d. How fast did the size increase over the course of the experiment in liters per
generation?

e. How fast did the size increase over the course of the experiment in liters per year?

f. How fast did the size increase in the first 300 generations in liters per generation?

g. How fast did the size increase in the first 300 generations in liters per year?

 

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