Biol 211g a20 – cellular and organismal biology
Problem Set 1 - Problems on Biomolecules (2 completions):
1. The continuity of amino shrewds of the enzyme lysozyme is notorious. Underneath is a register of amino shrewds
and the estimate of each in the lysozyme monad.
Number in Lysozyme
a. What is the mean glaciscular impressiveness of the 20 amino shrewds?
b. What is the glaciscular impressiveness of lysozyme?
c. What is the mean glaciscular impressiveness of the amino shrewds in lysozome?
d. How comprehensive S-S bonds are potential in lysozyme?
(hint: appear at the structures of each amino shrewd, and outline out which of these accept sulphurs
that are FREE to outline bonds).
e. Is the net carry on lysozyme settled or disclaiming?
(hint: frequently, appear at the structures of each amino shrewd, and outline out which of these are
neutral, which are settledly carryd, and which are disclaimingly carryd; then add up the
individual carrys for each amino shrewd to get the net carry of this protein)
2. Proteins (twain enzymatic and structural) resemble a telling role in detaining the pH in a cell.
Seven of the 20 amino shrewds regularly endow in proteins accept an ionizable knot. These
groups (party securitys) individualize the carry on the protein and buffer the cytoplasm of the cell
that comprehends it.
Populations of proteins that comprehend ionizable party securitys accept a likelihood of nature “charged”
or not. The likelihood of a party security nature carryd is skinred primarily to the party security
chemistry and the pH of the cellular compartment; other usually hither influential contents are
temperature, ionic ability, and bigness microenvironments.
The agreement of the carryd to unfull party securitys is usually symbolized by K:
wless c is the adaptation of carryd party securitys and u is the adaptation of unfull party securitys
(so that c + u = 1).
In the laboratory, the pH of a protein's environment can be altered so that we accept correspondent
numbers of carryd and unfull monads. We flatter the pH at which K is 1 (correspondent estimate of
full and unfull party securitys) the pK of the party security.
Below is a register of the seven amino shrewds whose carry is altered by the cellular pH:
Cell compartments can dissimilate in pH from environing 4.0 to 8.5, but the pK computes of the several amino
acids rest urban. When the pH of the environment is not correspondent to the pK of the party security,
the adaptation of carryd and unfull party securitys shifts to as to content the equation:
pH = pK + log ( c / u )
As an copy, for aspartic shrewd in a protein delay 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 divergent, the ordain log (c /u) gain not be naught and the agreement (c/u) gain be divergent from 1.
a. The pH of hale cell cytoplasm varies from 7.2 to 7.4. Using this notification, and the
notification aloft, exhaustive the forthcoming consultation:
Amino shrewd in protein
pK of the party security
Percent Full at pH Percent Full at pH
Problem Set 2 - Problems on Cell Structures and Membranes (3 completions):
Lysosomes are brief sacs of shrewd in a cell. Their pH is environing 5, and an electron
micrograph suggests they accept a bisection of 0.5 µm. The acceptiond hydrogen ion
distillation inparty lysosomes is due to the pumping of hydrogen ions athwart the
lysosomal membrane from the enstagnation cytosol, which has a pH of 7.2.
a. Inflated that a lysosome has the outline of a ordain and that tless is no
buffering talents inparty the lysosome, how comprehensive hydrogen ions were advanced
to the inparty of the lysosome to transfer to an inverge pH of 5?
(hint: foremost individualize the work of a lysosome in liters, then individualize [H+] in
moles/L in a lysosome at each pH (5 and 7.2), then individualize the estimate of
moles of hydrogen ions at each pH, and finally individualize and parallel the
estimate of hydrogen ions at each pH).
Liposomes are laboratory-prepared concocted membranes. Liposomes can be made in a
variety of bulks and can be made so that they accept transmembrane proteins, which outline
membrane. Contents of the liposomes can to-boot be notorious.
For copy, let’s say that one lab effects liposomes that are ordains delay the bisection of
4 µm and that each liposome has an mean of ten protein pores. Each liposome has an
inverge potassium ion distillation of 100 mM. Each protein pore transports 3x 10^6
potassium ions per cooperate. The pores cling unconcealed an mean of 0.3 coopeadmonish and cling closed
an mean of 2 cooperates; so, each pore unconcealeding and stagnation cycle follows environing 2.3 cooperates.
a. Inflated that a liposome has the outline of a ordain, how comprehensive potassium
ions are in a liposome moderately?
(hint: the way less is concordant to what you used to reresolve completion 3 aloft,
except meet the work of a liposome in µm^3 and the [K+] in mol/µm^3)
b. How comprehensive span is insist-upond for the potassium ions in the liposome to thrust
makeefficacy delay their environment? Assume that this environment is
relatively capacious and potassium-free.
(hint: precedently wary the sum span it would follow to thrust this effectweight,
apprehend environing how comprehensive potassium ions would failure to tend out of the liposome in
ordain to thrust this effectefficacy – all of them, half of them, none of them, why?)
5. Glycophorin is a single-pass transmembrane protein in red class cells (RBCs). The
protein rudiment of glycophorin is 131 amino shrewds hanker and binds carbohydrates on
the outparty (noncytoplasmic party) of glycophorin. Then, approximately 100 qualified
sugar residues are sturdy cdestroy the end of each glycophorin; these aggregateity for environing
60% of this macromolecule’s bulk. The mean glaciscular impressiveness of an amino shrewd is
a. What is the mean glaciscular impressiveness (in daltons) of each qualified sugar
residue on the glycophorin?
b. An RBC comprehends an mean of 6 x 10 ^5 glycophorin monads. How comprehensive
qualified sugar residues are endow sturdy to glycophorins in one RBC?
c. How comprehensive grams does the protein rudiment of glycophorin search in one
Problem Set 3 - Problems on Enzymes and Cellular Respiration (3 completions):
6. Hydrogen peroxide is usually stored in a brown bottle abroad from suncapricious owing it
spontaneously (but reluctantly) decomposes into O2 and breathe-into. In the brown bottle, the unimpeded
distillation of activation ∆G+ = 18 kcal/mol. In the bias of a catalyst, the resolution
is comprehensive steadfaster. For each subdue of 1.36 kcal/mol in ∆G+, the admonish of reaction is ten
In the bias of catalase, an enzyme endow in the class, ∆G+ = 7 kcal/mol. In the
bias of the inanimate catalyst platinum, ∆G+ = 13 kcal/mol.
How comprehensive steadfaster is peroxide resolution in the bias of a catalase?
b. How comprehensive steadfaster is peroxide resolution in the bias of platinum?
c. How comprehensive spans steadfaster is peroxide suspension inferior the bias of a catalase
than suspension inferior the bias of platinum?
7. You eat a candy bar that has 180 calories. This distillation is converted during respiration to
ATP. The reaction ADP + Pi ATP insist-upons 7.3 kcal/mol. One “dietary” calorie is
correspondent to 1000 “chemical” calories. Respiration is maximally environing 39% prolific in
converting substadmonish calories to ATP calories.
Assuming that all of the distillation in the candy in-effect gets to the respiratory site
in the cell, how comprehensive ATPs could your collectiveness effect from this candy bar?
8. The foremost two grades in respiration are glycolysis and Kreb’s cycle. For each monad of
10 NAD+ monads are unworthy
2 FADs are unworthy
4 ADPs are phosphorylated.
The unimpeded distillation ∆G0’ of the applicable “respiration reactions” is:
NADH + H+ + ½ O2 NAD+ + H2O
FADH2 + ½ O2 FAD + H2O
ATP ADP + Pi
Glucose + 6 O2 6 CO2 + 6H2O
How comprehensive of the distillation from one glacis of glucose is converted into ATP during
the foremost two grades?
b. How comprehensive of the glucose distillation is conserved in the unworthy coenzymes NAD+
The third grade of respiration is oxidation phosphorylation.
A sum of 32 ATPs per glucose monad are made during oxidative
phosphorylation from the subdue coenzymes. How comprehensive of the glucose distillation is
conserved in ATP at the end of ALL three grades?
d. What percentage of the sum ATP distillation is converted by the oxidative
phosphorylation of the unworthy coenzymes?
e. What percentage of the glucose distillation was obsolete was ardor?
Problem Set 4 - Problems on Photostructure (3 completions):
9. The fixing of carbon in photostructure varies in behold to the quantity of ATP failureed for each
C3 sets use 3 ATPs per carbon
C4 sets use 5 ATPs per carbon
CAM sets use 5.5 ATPs per carbon
If ATP yields 7.3 kilocalories per glacis, how comprehensive ATP calories are failureed to engender 1 glacis
of glucose (686 kcal/mol) by:
a C3 set? (hint: foremost, apprehend environing how comprehensive carbons are in 1 glacis of glucose)
a C4 set?
a CAM set?
Which stamp set uses the most ATP distillation in the making of glucose?
10. When capricious is shined on a leaf, it causes hydrogen ions to be pumped into discs flattered
thylakoid lumens. The ions then discursive out through a protein, and in the mode an ATP
monad is made for integral three hydrogen ions. Opportunity illuminated, inparty the disc, the pH can
be as low as 4. Outparty the disc, the pH is environing 7.2.
A thylakoid lumen can be modeled as a incomplete cylindrical rod that is 80 Å hanker and 5000 Å in
a. How comprehensive hydrogen ions are endow in one thylakoid lumen of this bulk at pH 4?
b. How comprehensive are endow at pH 7.2?
c. How comprehensive aggravate ATP monads can be made from the disc descriptive aloft,
AFTER the capricious is peevish off?
11. Capricious is influential in biology for photosynthesis. Tless are two divergent ways that capricious is
descriptive in physics.
In the foremost denomination, capricious travels in triumphs at a urban despatch c = 2.998 x 108 meters per cooperate.
The triumphtediousness is the tediousness from peak to peak of a capricious triumph, and corresponds to the complexion of
the capricious. The triumphtediousness is consecrated by λ (the Greek missive lambda). The triumphtediousness varies from
400 nm to 700 nm for capricious in the manifest collocate, delay bluish capricious having λ=450 nm and red capricious
having λ=680 nm.
The abundance is consecrated by ν (the Greek missive nu). The abundance is the estimate of peaks that
pass a top in a consecrated span. Abundance is skinred to triumphtediousness by the outlineula: ν = c / λ
In the coopeadmonish denomination, capricious travels in particles flattered photons or quanta. Using this
denomination it effects meaning to pronounce of a glacis of capricious as 6.02 x 1023 photons.
The distillation of one photon of capricious is consecrated by
E = (hc) / λ = hν
wless h is a shift content flattered Planck’s firm; h = 1.583 x 10-34 calorie cooperates.
In the laboratory, capricious delay a very reconnoitret triumphtediousness collocate can be used for illustrations. One
glacis of an actinic capricious (activating capricious) that has a triumphtediousness of 680 nm was used to excite
chlorophyll, and caused fluorescence measured at a triumphtediousness of 690 nm. The chlorophyll was
isolated, and hence could do no photochemistry.
a. What is the quantity of distillation (in kilocalories) in one glacis of actinic red capricious?
b. What is the quantity of distillation (in kcal) in the capricious that was fluoresced (assuming
c. What is the quantity of distillation (in kcal) that was obsolete as ardor?
d. What percentage of the red capricious distillation was obsolete as ardor?
A photon of bluish capricious gain energize an electron from chlorophyll to a roll approximate to a
photon of red capricious. Judge bluish capricious distillation to-boot caused fluorescence measured at a
wavetediousness of 690 nm.
e. What percentage of the bluish capricious distillation was obsolete as ardor (frequently inflated maximal
Problem Set 5 - Problems on Set Transport/Resource Acquisition, and Set Vindication to
Stimuli (3 completions):
12. Plasmodesmata are cytoplasmic cohereions athwart set cell walls that coless neighboring cell
cytoplasms. Some cells accept few plasmodesmata cohereions opportunity others accept aggravate; this is
due to genetics, age, and location delayin the set.
The inobservance of plasmodesmata delayin the cell membrane collocates from a noble of 25 per clear
micrometer to a low of 0.2 per clear micrometer. The mean plasmodesmata tube is 40 nm in
Diagram of plasmodesmata betwixt two set cells
a. What percentage of the cell membrane exterior area is secure of plasmodesmata
at the noble inobservance? (Assume the area of one end of a plasmodesmal tube is a circle
– see shadow aloft)
b. What percentage of the cell membrane exterior area is secure of plasmodesmata
at the low inobservance?
13. Liquescent breathe-into advances into and out of the cell by evacuation. Breathe-into mist to-boot advances from the
inparty of the set leaf to the ambient air by evacuation, in a mode notorious as transpiration. This
transpiration causes dissolved minerals to be advanced hanker tediousnesss in the set.
The span t in cooperates for breathe-into to advance such a hanker tediousness d in meters is consecrated by
t = (d2)/D
wless D is the evacuation coefficient. A sound compute for D is 2.4 x 10-5 m2/sec. (Note: breathe-into
mist discursives in air comprehensive aggravate astride than in liquescent breathe-into).
The footfootpath of the evacuation of breathe-into mist from a leaf into the air varies considerably, but a
measured tediousness of 1 mm is sound (1 mm = 10-3 m).
a. How hanker (in cooperates) does it follow a monad of breathe-into mist to be obsolete by this leaf?
Hairiness of leaves is a genetic feature. Leaf hairs may enfold the tediousness breathe-into must discursive.
b. How hanker (in cooperates) would it follow to destroy breathe-into from a hairy leaf?
14. (This completion is rive into two divergent completions – 14 and 15 - to concede you aggravate honor for
the graphs.). One environmental content that sets warner is the prolongation of capricious. The tediousness of
an incessant ebon end is repeatedly very-abundantly influential in the skin of augmentation (vegetative as
compared to fair). Long-day (LD) sets insist-upon at meanest a unfailing day tediousness precedently they
begin to perfection, when-in-fact incomplete-day (SD) sets insist-upon at most a unfailing day tediousness precedently they
begin to perfection.
In the corn hem in the United States, sunrise and sunset bound the tediousness of the day. For
example, less are the spans of sunrise and sunset of the foremost day of each month in northern Iowa:
Plot the sunrise and sunset spans on the selfselfcorresponding graph. (You can do this either on the computer
delay Excel or by index on graph brochure. If you induce this by index, effect sconsultation to reconnoitre your graph
and then paste the shadow into the muniment that you gain be submitting for grading. Please
label your graph and axes properly: distinction underneath the graph (Figure 1. Concede your own distinction less), x
axis = X distinction (in ? – hint: measured in what units?); y axis = Y distinction (in? units). Typically, in
science, graphs demonstration the defiant fickle on the x-axis, and relying fickle on the yaxis. So, you gain failure to outline out which of these is the defiant fickle, and which is the
relying fickle. The defiant fickle is the fickle that is immediately manipulated by a
scientist during an illustration. The relying fickle is not immediately manipulated by a
scientist, and instead shifts in vindication to a shift in the defiant fickle.)
b. Frame the prolongation of capricious per day and the prolongation of ebon per day on the selfsame
coopeadmonish graph (see (a) for graphing instructions).
15. This completion is a sequence of completion 14 aloft. Now, judge you accept inherited a
greenhouse and you unwavering to beseem a citizen. You don’t failure to pay for subjoined capriciousing,
so you gain use barely sunlight. Also, you accept a persomal negotiate barely for the forthcoming sets:
SD or LD
Critical Day Length
a. When during the year would you rely-on each set to be redundant to perfection?
Explain how you individualized this (hint: appear at your graphs for completion 14).
Problem Set 6 - Problems on Gene Expression/Repair and Protein Synthesis/Degradation (3
16. The DNA in a cosmical non-gametic cell comprehends 6 billion low pairs. It is estimated that
environing 10,000 DNA shifts arise in each cell in one day. These are promptly relited so that
barely a few (1 to 5) mutations learn in one cell in a year.
a. What percentage of the low pairs are altered each day?
b. What percentage of the DNA shifts that arise in one cell in one year escape
the prooflection and relit mode – proportion this in twain cases: if 1 mutation
accumulates in one cell per year, and if 5 mutations learn in one cell per
17. In one mammalian cell it is estimated that 10,000 to 20,000 divergent stamps of mRNA can be
found. Comprehensive mRNAs insist in comprehensive copies per cell (up to 12,000 copies/cell); but sparse
mRNAs (5 to 15 copies/cell) can to-boot be detected. At any flash, a snapshot of mRNA content
would unearth a sum of 360,000 mRNAs. The cell usually has environing 10 spans as comprehensive ribosomes
as mRNAs. Assume 75% of the ribosomes are complicated in protein structure at any flash in
a. What is the ultimatum estimate of proteins that could be in the mode of
structure at any flash?
b. How comprehensive sparse proteins (use 5 copies/cell) could be in the mode of nature
made? (hint: use that 7.5 ribosomes are lection each of the 5 mRNAs; but illustreprimand or
demonstration calculations for why you should use this).
c. How comprehensive comprehensive proteins could be in the mode of nature made? (hint: use
that 7.5 ribosomes are lection each of the 12,000 mRNAs; but illustreprimand or demonstration
calculations for why you should use this).
d. What is the agreement of developing comprehensive proteins to developing sparse proteins?
18. Protein monads in the cytosol of a cell accept divergent half-lives t ½. The half-life is the
span failureed for 50% of the monads to be obsolete or altered. Half-lives are individualized by several
factors, one of which is the “marking” of the protein by ubiquitin, which signals that the protein
is contrived for digestion by proteases. Ubiquitin binds divergently to proteins due to differences
in their amino ends. The forthcoming consultation concedes the half-lives t ½ for some proteins delay divergent
Amino Shrewd End of Protein
> 20 hrs
> 20 hrs
> 20 hrs
> 20 hrs
> 20 hrs
< 3 min
< 3 min
< 3 min
< 3 min
< 2 min
Half-life is skinred to the admonish K of protein dropping by the forthcoming equation:
t ½ = 0.693 / K
wless 0.693 is the intrinsic logarithm of 2, and K is measured in reflexive span units.
What is the admonish of protein dropping per diminutive when MET is the amino ordaininal?
(Assume t ½ = 24 hr).
b. Parallel this delay the admonish when ARG is the amino ordaininal end. (Assume t ½ =
What percentage of the protein delay a MET amino end insists in the cell succeeding 5
d. What percentage of the protein delay an ARG amino end insists in the cell succeeding 5
e. How comprehensive steadfaster (as a percentage) is the admonish of ARG-amino-end protein
suspension as paralleld to MET-amino-end suspension?
f. Frame the dropping of a population of protein monads that accept a MET amino end.
Use span t on the absolute axis and bulk of the population on the upright axis,
starting delay 100% of the population at the rouseing span t = 0.
g. On the selfselfcorresponding graph, frame the dropping of a protein delay ARG as the amino ordaininus.
h. Hemoglobin insists in the cytoplasm of a red class cell (RBC). Red class cells
last environing 120 days in the classstream. In your moderate illustrations you meet that the
amino ordaininus of one protein security in RBCs is either a valine or a leucine (so a
difference of barely one methyl knot). Which is aggravate slight to be the chasten amino
shrewd ordaininus of this protein? (Assume t ½ = 24 hr for valine and 2 min for leucine).
Problem Set 7 - Problems on Microorganisms/Bacteria (2 completions):
19. Assume that you conducted the forthcoming illustration in ordain to discuss how bacterial
evolution arises in jumps:
An E. coli amelioration was detained for 10,000 races aggravate 4 years. The liquescent balance was
changed daily to detain a firm environment. The mean bulk of the cells at the rouse of
the illustration was 0.35 x 10 -15 L. Succeeding 300 races the bulk acceptiond to 0.48 x 10 -15 L,
and succeeding another 300 races the mean acceptiond to 0.49 x 10 -15 L. Succeeding 1200
generations it acceptiond to 0.58 x 10 -15 L and rested so until the end of the illustration.
Volume ( x 10 -15 L)
E. coli Evolution
a. How comprehensive hours hanker is a race of E. coli?
b. What was the mean bulk shift aggravate the road of the illustration in liters?
c. What was the mean bulk shift in the foremost 300 races in liters?
d. How steadsteadsecure did the bulk acception aggravate the road of the illustration in liters per
e. How steadsteadsecure did the bulk acception aggravate the road of the illustration in liters per year?
f. How steadsteadsecure did the bulk acception in the foremost 300 races in liters per race?
g. How steadsteadsecure did the bulk acception in the foremost 300 races in liters per year?