Download Multiple Choice - 28 points total

Multiple Choice -
28 points total
In each of the questions, select the answer which you believe is most nearly correct. If more than
one answer seems to be correct, pick the single answer you believe to be the best.
1)
E
Which of the following statements is not true about the cytoskeleton?
A) It is involved in chromosome separation in mitosis and meiosis
B) It is made up of microfilaments, intermediate filaments and microtubules
C) Cytoskeletal proteins include actin, myosin, and keratin
D) The principal proteins in muscle cells are cytoskeletal proteins
E) None of the above (statements A-D are all true)
2)
C
Which of the following statements about translation is true?
A) The anticodon is part of the rRNA in the small ribosomal subunit
B) Amino Acids are bound to tRNA by hydrogen bonds
C) During the process, the growing polypeptide chain is covalently attached to RNA
D) Stop codons are found near the end of ribosomal RNA (rRNA) molecules
E) None of the above
Note: Answer C is correct because during translation the growing polypeptide strand
is always covalently attached to a tRNA molecule.
3)
C
All of the following are true statements regarding enzymes except:
A) Enzymes catalyze reactions by stabilizing the transition state.
B) Enzymes bind directly to their substrates at an active site.
C) Enzymes change the overall energetics of reactions to make them more favorable.
D) Conformational changes in enzymes can affect reaction rates.
E) Enzymes lower the activation energy of reactions.
4)
D
The C4 pathway represents an important adaptive strategy used by some
photosynthetic plants. Which of the following is the primary benefit of this pathway?
A) It allows ATP to be produced regardless of light availability.
B) It eliminates the need for the light independent reactions of photosynthesis.
C) It prevents the damaging effects of photolysis.
D) It minimizes photorespiration by facilitating CO2 uptake.
E) It increases the range of wavelengths over which chlorophyll can absorb light.
2nd Exam Answer Key
Biology 0200 – 2013
5)
A
Cheneyase is a protein consisting of 3 polypeptide subunits, two of which are
strongly hydrophobic and one of which (subunit stent) is composed of just 20 amino
acids. The amino acid sequence of the stent subunit is:
N-gly-gly-lys-lys-ala-gly-arg-gly-lys-ser-gly-gly-lys-lys-ala-gly-gly-arg-lys-ser-C
(gly = glycine, lys = lysine, ala = alanine, arg = arginine, ser = serine)
Which of the following statements about the Cheneyase stent subunit is true?
A) It is not joined to the other subunits by disulfide bonds
B) It is too short to form an α-helix
C) It does not have a tertiary structure
D) It does not have a secondary structure
E) From the information given here, we can tell nothing about stent
Note: Since this polypeptide does not have any cysteines, it cannot be involved in
disulfide bonds, so answer A is correct.
6)
B
Which of the following statements is true regarding the Calvin Cycle?
A) The Calvin Cycle does not occur in C4 plants
B) The Calvin Cycle consumes CO2 , ATP, and NADPH.
C) The Calvin Cycle produces O2 , ADP, and NADP+
D) The enzymes of the Calvin Cycle are bound to thylakoid membranes
E) None of the above
7)
E
Which of the following statements is true regarding the β-sheet pattern of protein
secondary structure?
A) The structure is stabilized by hydrogen bonds between adjacent bases
B) The bending of the chain makes a complete circle at 5.8Å intervals
C) The structure is stabilized by disulfide (S-S) bonds
D) The structure is stabilized by hydrogen bonds between R-groups
E) None of the above
Note: Many students picked answers A or D, since the beta-sheet is indeed
stabilizied by hydrogen bonds. However, answer A is incorrect since polypeptides do
not have “bases.” Nitrogenous bases are found in DNA and RNA, not protein. And
the hydrogen bonds in a beta-sheet are between C=O and N—H groups, not R groups,
so answer D is also incorrect.
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2nd Exam Answer Key
Biology 0200 – 2013
8) The Peptide Bond
(14 points)
The Martian Rover has just analyzed three novel compounds that
your colleague suggests are amino acids. But you are certain that
your colleague (an engineer, of course) is wrong. He should have
paid more attention in Bio 0200!
A) Of the three compounds shown below, one is not an amino
acid! Which one, A, B, or C? In a single sentence, explain why
the compound you have selected is not an amino acid. (4 points)
Compound A
Compound B
Compound A is not an Amino Acid. (+2 points)
Why? Because it does not have a true alpha-carbon.
Compound C
(+2 points)
B) Draw a peptide bond between the two genuine amino acids. You may draw the dipeptide in
any order you wish. (10 points)
As noted in the question, the 2 amino acids may be drawn in any order.
Grading:
• Deduct 5 points for a drawing that gets the peptide bond correct, but
uses an incorrect amino or carboxyl group to form the bond.
• Deduct 5 points for any error in the bond itself (the shaded region
below). Note that the bond region may be drawn in either of the resonant forms.
Note: If a student has mis-identified the non-Amino Acid in Part ! (and lost 4 points
there), but uses the incorrect Amino Acid to draw an accurate peptide bond, give full
credit (10 points) for this part of the question.
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2nd Exam Answer Key
Biology 0200 – 2013
9) Metabolism (20 points)
A) You are working at a winery in the Napa Valley during spring break, and given the task of
monitoring sugar content during the fermentation process for a prize batch of grapes. Given your
familiarity with fermentation, you figure you’re a natural for the job. Your measurements during
the first 12 hours of the process looks like this:
Your supervisor calls you into her office, and wonders what might have caused the sudden drop
in sugar content that began at about 10
hours after the vessel was sealed.
Unfortunately, a worker unsealed the
vessel at the 12 hour point, making it
open to the air. You do not discover the
unsealed vessel until the next morning
(12 hours later). Predict, by drawing a
line on the graph from 12 hours to 24
hours, what will happen to the sugar
content in the unsealed vessel. (Just
draw the line. No explanation needed
here!) (6 points for correct line)
Note: Since Oxygen has been
admitted into the vessel, the yeast will switch over to oxidative respiration, which
uses up sugars as a much lower rate than fermentation. Since that’s how the yeast
were using sugar during the first 10 hours, the line you draw should match the slope
of the line in the first 10 hours.
B) Impressed by your success in predicting what would happen in the opened fermentation
vessel in part A (above), the wine master notes that she has a supply of dinitrophenol (DNP) in
her lab. As you may remember from lab, this is a compound that makes the inner mitochondrial
membranes (of humans and yeast) permeable (leaky) to protons. She wonders what would have
happened if DNP were added at the 12 hour point to the fermentation vessel. Would the sugar
content
1) continue to drop rapidly
2) rise
3) level off and remain constant
Pick one of these three answers. And then briefly explain your choice. (8 points)
The sugar content would continue to drop rapidly (Choice #1) +2 points
Explanation: The yeast are active in fermentation (also OK: anaerobic respiration)
during the rapid drop in sugar content (+3 points). Since no electron transport takes
place during fermentation, DNP will have no effect on ATP synthesis or sugar
utilization (+3 points) and fermentation will continue.
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2nd Exam Answer Key
Biology 0200 – 2013
C) Impressed still further with your predictive powers, the wine master notes that there is a lag
period (visible on both of the graphs on the previous page) in the first 10 hours of fermentation
during which very little alcohol is produced. She has heard that a competing winery has found a
way to eliminate the lag period by flushing out dissolved gases from the air in the grape juice.
She orders you to try to do the same.
Using the materials you have at hand (compressed cylinders of oxygen, carbon dioxide, and
nitrogen gas which you could bubble through the vat to displace other dissolved gases), can you
suggest of a way to eliminate the lag period in the next batch of wine so that fermentation can
begin immediately after the vessel is sealed? Explain your choice of gas, and why it just might
work. (6 points)
Either nitrogen or carbon dioxide would be an appropriate choice of gas. (+2 points)
Why might it work? Because fermentation begins only when oxygen is eliminated from
the fermentation vat. So, removing oxygen by bubbling either of these gases through
the vat would allow fermentation to begin immediately (+4 points)
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2nd Exam Answer Key
Biology 0200 – 2013
10) DNA
(16 points)
The famous Hershey-Chase experiment helped to convince James Watson and Francis Crick that
DNA was the genetic material. As you know, the experiment involved bacteriophage, viruses
that infect bacteria. Alfred Hershey and Martha Chase first grew these viruses in bacteria that
were themselves grown in two different media, one of which contained radioactive sulfur (35S)
and one of which contained radioactive phosphorous (32P). They then harvested the radiolabeled
virus particles, and allowed them to infect two new cultures of unlabeled bacteria.
A) Why did they choose these two isotopes (as opposed to, say, radioactive nitrogen, oxygen, or
carbon)? (2 points)
They chose these two because DNA has no sulfur in it, and protein has very little
phosphorous. (+2 points)
(Note: It’s not necessary to say that this makes it possible to determine which of
the two macromolecules enters the cell)
B) Describe (just a sentence or two) the results of the experiment with 32P-labeled viruses.
(2 points)
Radioactive phosphorous entered the bacteria when the viruses infected them.
(+2 points)
C) Describe (just a sentence or two) the results of the experiment with 35S-labeled viruses.
(2 points)
Very little radioactive sulfur entered the cells when the viruses infected them.
(+2 points)
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2nd Exam Answer Key
Biology 0200 – 2013
D) Shown below is a diagram of a DNA replication fork. As you can see, the two original
strands are in the process of separating, and the fork is proceeding from right to left. However,
the location of the newly-replicated DNA is not shown. (10 points)
Draw the newly-replicated DNA strands into the diagram. Be sure to include and to label:
• newly-synthesized DNA strands
• DNA polymerase enzymes
• Okazaki fragments
• DNA Ligase
• The 3’ and 5’ ends of the newly-synthesized DNA
Grading:
+2
+2
+2
+2
+2
points
points
points
points
points
for
for
for
for
for
“new” DNA on both strands
showing DNA polymerase at the 3’ ends of growing strands
showing Okazaki fragments on the upper strand only
DNA ligase joining Okazaki fragments
antiparallel pairing of old and new strands
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2nd Exam Answer Key
Biology 0200 – 2013
11) Gene Mapping
(22 points)
You are studying a true-breeding line of Drosophila with three
mutations. They are Blue Eyes, Yellow Wings, and Hairless (hairless
flies lack the bristles that normally cover a fly’s body). All three of these
alleles are recessive to their wild-type counterparts.
Use the symbols b, y, and h for the mutant alleles and B, Y, and H for the wild-type alleles
to answer the following questions:
A) A hairless female with blue eyes and yellow wings is crossed with a wild-type male. They
produce 126 F1 offspring:
• 60 wild-type females
• 66 hairless, blue-eyed, yellow-winged males
Briefly explain what these results suggest about the chromosomal location of the B, Y, and H
alleles: (4 points)
It suggests that all three alleles (B, Y, and H) are located on the X chromosome.
(also OK: it suggests that they are sex-linked). +4 points
Note: +2 points for identifying the genes as linked (but not realizing sex linkage)
B) Intrigued by these results, you select a group of
the F1 females and cross them with more wildtype males (the wild-type males come from the
same group that served as parents to the F1
generation above). This time the results are
dramatically different. 1000 offspring from this
F2 generation have the following characteristics:
500 wild-type females
160 hairless, blue-eyed, yellow-winged males
162 wild-type males
25 yellow-winged males
23 blue-eyed hairless males
55 hairless males
58 yellow-winged, blue-eyed males
9 blue-eyed males
8 yellow-winged hairless males
Immediately you recognize that these data will allow you to map these 3 genes (but that will
come in part C). First, however, briefly explain why all of the females in this F2 generation are
wild-type, while fewer than half of the males in the F2 generation are: (6 points)
The F2 females inherited one of their X chromosomes from their fathers, all of
whom were wild type males. This means that they inherited a dominant allele for each
of the three genes, making them wild type. (+3 points)
The F2 males, on the other hand, inherited their lone X chromosome from their
mothers, who were heterozygous for each of the three genes. (Also OK: Only one of
their mother’s X chromosomes carried the three wild type alleles. Or, it’s OK to say
that only one of mom’s X’s carries the 3 recessive alleles, or that recombination
produced the other male phenotypes) (+3 points)
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2nd Exam Answer Key
Biology 0200 – 2013
C) Taking the data from Part B (shown again at
right), construct a genetic map of these three genes.
First, calculate recombination frequencies between
each pair of genes. Show your work, and circle your
results for each of the three frequencies.
Then, take the three frequencies and construct a map
showing the relative locations of the three genes.
(12 points)
Recombination Frequencies:
500 wild-type females
160 hairless, blue-eyed, yellow-winged males
162 wild-type males
25 yellow-winged males
23 blue-eyed hairless males
55 hairless males
58 yellow-winged, blue-eyed males
9 blue-eyed males
8 yellow-winged hairless males
(2 points for each correct frequency)
B-Y = 25 + 23 + 9 + 8 = 65 / 500 = 0.13 = 13%
Y-H = 25 + 23 + 55 + 58 = 161 / 500 = 0.322 = 32.2%
B-H = 55 + 58+ 9 + 8 = 130 / 500 = 0.26 = 26%
Genetic Map:
(6 points for correct map)
Grading Notes:
• If a student divides the proper recombinant numbers by 1000 instead of
500, deduct just 2 points from this portion of the problem (max value now 10 points).
Note: The reason we use 500 instead of 1000 here is because no crossing-over
occurred in the X chromosomes inherited by the 500 female flies. Why? Because
that X chromosome came from their father, who was XY. Since he had just one X
chromosome, there was no opportunity for crossing over with another X. Only the
X’s that the 500 males inherited from their mothers could have undergone
recombination.
• If a student correctly maps incorrect recombination frequencies (leading to
deductions in the first part of the problem), give full 6-point credit for the mapping
portion of the problem.
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