Download 1. This cartoon shows Complex I in the ETC, in its two alternative

 1. This cartoon shows Complex I in the ETC, in its two alternative conformations. When Complex I switches
between the two conformations, proton pumping results. Note that the complex has a binding “slot” for Q and four
proton channels.
(13 pts)
Proton
gradient/motive force
Intermembrane space
H
H
Q
NADH
NAD
+
+
H
+
H
+
H
+
H
+
H
+
H
+
E to K
EC to Kin
+
Q
Oxidized/O
NADH
R/
X
Reduced/R
K to C
Kin to Chem
Matrix
NAD
+
a. In the blanks provided, label each side of the
membrane with one of the terms listed to the right:
cytosol
matrix
intermembrane space
cell exterior
1pt each (total 2)
b. The two conformations of Complex I depend on whether Q is reduced or oxidized. Indicate which is which, by
writing “Reduced” or “Oxidized” next to each Q.
3pts
c. Rotenone is a poison that works by blocking electron transfer from proteins with Fe-S centers. These proteins
pass electrons from NADH to Q. On either version of the complex, add a large R to indicate where rotenone acts.
2 pts (can be on either Complex I cartoon)
d. Add a large arrow to the diagram labeled “Proton Gradient.” Orient the arrowhead so that it points in the
+
direction that H will move spontaneously.
2pts
e. Add a sketch of ATP synthase to the right (where the phospholipid icons stop). Include an “E to K” label to
indicate the place in ATP synthase where the energy in an electrochemical gradient is transformed to kinetic
energy. Include a “K to C” label on ATP synthase to indicate where kinetic energy is transformed to chemical
energy in the form of ATP.
2pts sketch
1pt each label
2. At age 45, Mr. Drunk started noticing that he was having the 'happy' effects of alcohol without ingesting any
alcohol. Doctors found high levels of ethanol in his blood tests. They also found that Mr. Drunk had an unusual
population of bacteria in his gut. How can you explain Mr. Drunk's symptoms? (Maximum two sentences)
Answer: The bacteria in Mr. Drunk's gut are probably producing high levels of ethanol due to fermentation in this
anaerobic environment. The ethanol is being absorbed into Mr. Drunk's blood stream and this is the cause of his
symptoms (3 points).
1pt bacteria involved
1pt anaerobic environment in gut
1pt alcohol fermentation
Partial credit? (1 point) given if stating that Mr. Drunk had the genes for the enzyme required to produce ethanol via
fermentation. This is unlikely because the symptoms had a later onset in life and there is no mention about Mr.
Drunk having the symptoms only when doing intense exercising/running out of breath.
(3 pts)
3. When D-isocitrate is converted to α-ketoglutarate in the Krebs cycle,
+
carbon dioxide and a proton are released and NAD is reduced to NADH.
(11 pts)
a. Circle the carbon and oxygen atoms that are released from D-isocitrate.
2pts
b. Draw an arrow to the C-C or C-H bond that provides the high energy
electrons “carried” by the NADH produced in this step.
2pts
NOTE: many people drew an arrow to the bond that breaks when decarboxylation occurs. It is true that
this bond breaks, but by comparing C’s in D-isocitrate v α-ketoglutarate, you should see that that C-C
bond is “replaced” with a C-H bond, which will have equivalent potential energy. The next carbon down,
though, experiences a change with a C-H bond in D-isocitrate changing to a C=O in α-ketoglutarate. The
+
high-energy electrons from that C-H bond and a proton are what reduces NAD to NADH.
c. In terms of fitness, why is it logical that this step in glucose oxidation is regulated, when many other steps are
not? (one sentence or less)
Because there is an exceptionally large drop in potential energy from D-isocitrate to a-ketoglutarate,
stopping this step saves a relatively large amount of energy for later use.
2pts: identify large drop/change in energy (1pt if just refer to decarboxylation or just NADH formation)
2pts: connect the dots to saving energy
Notes:
1pt - identify reaction as occurring early in the Krebs Cycle
1pt - regulation prevents buildup of toxic CO2 (while plausible, this is not actually what happens)
d. Why is it logical to observe that the enzyme responsible for catalyzing this reaction is regulated by ATP, as
opposed to other ions or molecules? (maximum of two sentences)
ATP is a product of glucose oxidation, so if ATP levels are high it means that this reaction should slow or
stop.
3pts: concept of feedback inhibition—note, only 1pt for using the term alone—for full credit, must explain
the concept clearly with or without use of the term
4. Phosphofructokinase (PFK) catalyzes a reaction in glycolysis—the transfer of a phosphate group from ATP to
fructose 6-phosphate, forming fructose 1,6-bisphosphate. PFK also has a regulatory site where ATP binds, but
with lower affinity than the active site.
(10 pts)
a) A mutation changes the conformation of PFK without changing the affinity of the enzyme's regulatory site for
ATP. Predict the effects of this mutation on the function of PFK, and explain your answer, in two sentences or
less.
Two scenarios are possible:
The conformational change affects binding of substrates in the PFK active site, affecting its enzymatic
function (either slowing it down or stopping it completely);
OR
The conformational change did not affect the active site and it does not affect PFK function (this is
unlikely because the conformational change still could compromise the structural scaffold of the protein,
ultimately leading to decreased enzymatic performance).
(full 2 points for either answer).
b) If PFK is deleted (knocked out) from a cell, would the cell still be able to survive? Why or why not? Consider
two conditions when you answer: You supply the mutant cells only with either of the molecules below.
• Glucose:
No, the cell would NOT be able to produce any ATP whatsoever (even through glycolysis, since the
pathway will not even reach the ‘energy payoff phase’).
2pts (1pt No, 1pt logic)
•
Fructose 1,6-biphosphate:
Yes, this is a substrate that enters the pathway downstream of PFK, so glycolysis and the rest of the
glucose oxidation pathway would produce their normal products.
3pts (1pt Yes, 1pt concept of “getting past stopped reaction,” 1pt glycolysis (et al.) would go to
completion
c) You introduced the PFK gene from humans into a knock-out fruit fly egg that is missing the PFK gene, and
observe that as the embryo develops, the fly cells are functioning perfectly when supplied with glucose. Explain
how this could be possible. (maximum 2 sentences)
The PFK produced by human and fly genes are similar enough to perform its function in the pathway independently
of the “species-of-origin”.
3pts: concept of identical/similarity in structure and function … gold stars for comments about evolutionary
conservation/strong selection
5. After a sperm nucleus enters the human egg, cortical granules release proteases, peroxidases, and
polysaccharides to the space just outside the egg membrane. In two sentences or less, explain the role of the
polysaccharides in blocking polyspermy. Your answer should also include a labeled drawing.
(7 pts)
The sugars create a steep osmotic gradient that brings water into the space between the egg membrane
and the zona pellucida. The increased volume forces the zona pellucida up and away from the egg
membrane.
H2O in
Z.p. lifts away
Zona
pellucida
Sugars from cortical
granules
Egg
membrane
Notes:
2pts. for the concept of osmotic change:
•
•
(1pt) Needed to relate concentration to osmosis ("the sugars create osmosis" was no accepted),
or explicitly stating that there was an osmotic change or gradient.
o Saying there was an ion gradient was no accepted, but if they stated it created a chemical
or electrochemical gradient was accepted.
(1pt) Water was drawn to this change.
o The concentration is the reason the water moves; saying water was attracted to the sugars
was not accepted since the same outcome can be done with ions, salts, anything of
concentration.
2pts. water enters/moves in between the 2 structures (needed to be explicit in writing or drawing, but only
if it was very well labeled):
•
•
If it was stated that the sugars go there, but not saying that the water goes there, then that netted
1pt.
No points were awarded if it was stated it moved into the cell without specification; if the
response was ambiguous and mentioned the water correctly oriented in respect to one structure
("the water enters past the zona"), then 1pt.
1pt the zona (or other names) lifts/comes off:
•
Saying it expanded was not accepted unless the drawing is very clearly interpreted to the zona
lifting off (found very rarely).
2pts for one of the three drawings (all or none since I never found one that warranted partial credit):
Drawing 1: only accepted if they had a correct explanation of what the sugars eventually did in the long
run (bring in water, help the zona lift off, etc).
Drawing 2: always awarded points for showing water move in unless there was no information written
down about it (all that it took was a quick mention of water movement).
Drawing 3: only accepted if mention that water is the reason it lifts off.
•
Drawings that you had that you believe fit the above, but are poorly labeled were not accepted
since I couldn't concisely figure out what was happening without assuming too much that wasn't
in your picture or writing.
6. Some bacteria are obligate fermenters. They also have an ATP synthase in their plasma membranes that
spins opposite the direction that ATP synthase spins in species that do cellular respiration—it hydrolyzes ATP to
ADP + Pi. This reaction occurs inside the cell. Their membranes also have many co-transporters. These proteins
carry a proton and an important nutrient together, from outside the cell to the inside.
(8 pts)
a. Do these cells have an ETC? (circle one)
YES
NO
2pts
Fermentation
b. Where does the ATP come from that ATP synthase hydrolyzes? ____________________________________
2pts
c. Explain the relationship between ATP synthase in these species and the function of the co-transporters, in two
sentences or less. NOTE: nutrients are usually being brought into the cell against a concentration gradient.
ATP synthase creates a proton gradient that H+ co-transporters can use to bring nutrients into the cell
against their concentration gradient.
2pts recognizing proton gradient
2pts co-transport of nutrients against a gradient
7. Consider the signaling pathway diagrammed to the right.
(10 pts)
Signal
a. Label the following four components:
Response
Cell-cell signal
2
nd
messenger
Receptor
Receptor
1pt each (4 total)
b. Is there an amplication step in this pathway? If so, circle it. If not, explain why
amplication is not occurring in one complete sentence.
2pts
c. UDP-1 is a transmembrane protein that functions as a proton channel whenResponse:
activated. What effect does this have on the proton gradient? Explain your
answer in one sentence or less.
It decreases the proton gradient across the inner membrane, because
protons will flow through it instead of ATP synthase.
1pt decrease gradient
2pts protons flow through it
1pt some recognition of impact on ATP synthase
nd
2
messenger
8. Consider oogenesis, diagrammed to the right.
(10 pts)
Oogonium
a. Label the following structures:
Polar body
Primary oocyte
Oogonium
Secondary oocyte
1pt each (4 total)
NOTE: No
points awarded for a label that ambiguously points to >1 cell.
Primary
oocyte
Secondary
b. Five cell processes are general in development: proliferation, movement/elongation, oocyte
programmed death, cell-cell interactions, and differentiation. Choose any two, and
describe how each is relevant to oogenesis.
Process
Cell proliferation
Polar
body
Role in oogenesis
Oogonia divide by mitosis to produce many primary
oocytes.
In mammals, ova are released from the ovary and move
Movement/elongation down the fallopian tube to the uterus.
(OK to mention movements of primary and secondary
oocytes within ovary during maturation.)
Programmed cell
death
Polar bodies die
Cell-cell interactions
… not too much here from Bio200 stuff (it’s 220!), but
progression through oogenesis is regulated by
hormones (LH, FSH, estrogen)
Differentation
The oocyte is a specialized cell, characterized by
haploid genome, extremely large size, egg cell
receptors for sperm, etc.
1pt for process
2pts for reasonable and accurate association with events, each
6pts total for part (b)
Other notes from the grader:
For the role of cell proliferation in oogenesis, up to 2 points were deducted for mentioning meiosis, which
due to polar body formation does not ultimately increase the total number of cells.
For the role of differentiation in oogenesis, up to 2 points were deducted for mentioning polar bodies vs.
oocytes/ova. This is not really an example of differentiation since the polar bodies soon undergo
apoptosis.
In general, points were deducted for failing to demonstrate a thorough understanding of the role of the
process in oogenesis. No credit for just reciting generic definitions of cellular processes.
9. Female fruit flies with two knock-out alleles at the bicoid gene have offspring that lack
anterior (head) structures—they have posterior structures where anterior structures should
be—and die. The in situ hybridization on the right shows the distribution of bicoid mRNA
High/H
(stained orange) in a normal egg. Use b to symbolize knock-out alleles; B to symbolize
normal alleles.
(13 pts)
Anterior
a. What is the genotype of a mother who produces only
mutant (“double-posterior”) offspring?
Low/L
Posterior
bb
- BB
2pts
NOTE: Either full credit or no credit. If two answers were input, no credit. Homozygous or heterozygous
were accepted for full credit.
Bb
+ b. What was the genotype of this mother’s mother?
BB
2pts; same notes as part a)
c. Add an arrow to the in situ at top right, indicating the concentration gradient in bicoid mRNA. Add “low” and
“high” labels.
1pt
NOTE: High and low labels must be in correct place and arrow oriented from high to low for any credit.
d. In the “empty egg” below, use dots or shading to predict the distribution of bicoid protein.
2pts
NOTE: Concentration (dots) should be focused at anterior edge and lessen on a gradient to the posterior
edge. Answers were largely correct or incorrect, but partial credit was possible if the dots were
concentrated at the anterior but not properly draw for the rest of the egg.
e. Bicoid protein is a transcription factor. Based on the data provided here, state a hypothesis for which genes it
regulates and how. (maximum one sentence)
Bicoid either enhances the expression of “anterior genes” or silences expression of “posterior genes,” or
both.
2pts … full credit for stating just one of three possibilities
f. Explain why high versus medium versus low concentrations of bicoid protein could lead to differential gene
expression.
The binding of a transcription factor to a regulatory sequence in DNA will occur more often at high
concentration, leading to increased expression of target genes.
1pt concept of TF binding to regulatory site (enhancer, silencer, promoter, promoter-proximal element)
2pts binding frequency correlates with concentration
1pt connect dots to differential gene expression/determination
Notes:
Full credit for and of the following: 1) Bicoid enhances the expression of “anterior genes,” 2) Bicoid
silences the expression of “posterior genes,” or 3) Both statements. If “what genes it regulates” was
answered or “how,” one point was given. Statements like “it regulates bicoid genes” or “it
regulates/promotes certain genes” received no credit.
10. BMP4 and noggin are examples of cell-cell signals that interact to influence development.
a. What is the fate of a dorsal ectodermal cell that receives only BMP4?
(7 pts)
Dorsal skin
2pts
b. What is the fate of a dorsal ectodermal cell that receives BMP4 + noggin?
Incorporated into neural tube
2pts (neuron, spinal cord, neurulation all ok for full credit)
c. If a researcher treated ventral ectodermal cells with noggin and observed no change in the cells’ fate, what
would be the simplest explanation (at the molecular level)? Answer in one complete sentence.
Ventral ectoderm cells do not have the noggin receptor
Other full credit possibilities are:
•
3pts
11. Identify one outcome of each of the following four developmental events.
a. Cleavage in frogs or chickens
Sequester cytoplasmic determinants,
produce blastula, divide up egg volume
b. Gastrulation in frogs or chickens
Establish three embryonic tissue layers;
establish three body axes
c.
Formation of neural tube, formation of
somites/lateral plate mesoderm/notochord
Organogenesis—specifically, neurulation in chickens
d. Body folding in chickens
(8 pts)
Formation of ventral/belly skin/ectoderm,
formation of gut (folding of endoderm into
tube), form “tube-within-a-tube”
Notes:
2pts (each): Correct outcome
1pt: Correct outcome, but also added another outcome that was incorrect/contradictory OR the outcome
was too vague.
0pts: Incorrect outcome. No points were given for "blood vessel formation" or "blood vessels connect
yolk with embryo"