Download Multiple Choice - 28 points total In each of the questions

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
ATP serves as a substrate for the enzyme phosphofructokinase (which catalyzes
the third reaction in Glycolysis), but it also serves as an inhibitor of the same enzyme.
This is because
A) ATP is a competitive inhibitor of the enzyme
B) ATP is an allosteric activator of the enzyme
C) ATP blocks the first reaction in the glycolytic pathway
D) Phosphofructokinase is stimulated by oxygen bound to its active site
E) None of the above
2)
A
Given that prokaryotic cells lack many of the organelles found in eukaryotes,
including mitochondria, which of the following statements is correct?
A) Prokaryotes carry out both glycolysis and the Krebs cycle in the cytoplasm.
B) Eukaryotic cells use more energy to oxidize electron carriers during respiration.
C) Eukaryotic enzymes are more efficient in glycolysis than prokaryotic enzymes.
D) Prokaryotic cells always require oxygen, while some eukaryotes do not.
E) None of the above.
3)
D
One of T. H. Morgan’s experiments in which he first noticed the effects of gene
linkage began with a cross between a wild-type fly and a fly that was homozygous
recessive for the black body and vestigial wing alleles (b b, vg vg). The result was an F1
generation with the wild-type phenotypes for both traits. When Morgan crossed an F1 fly
with another homozygous recessive, he expected to produce 4 different phenotypes
[wild-type : black : vestigial: black with vestigial] in which ratio?
A) 9:3:3:1
B) 16:9:9:1
C) 3:3:1:1
D) 1:1:1:1
E) None of the above
4)
C
Lipids are one of the four major classes of macromolecules. Which of the
following are not generally found in the lipid molecules associated with living cells?
A) Fatty acids
B) Carbon to carbon double bonds
C) Nitrogenous bases
D) Glycerol
E) Phosphate groups
Biology 0200 – 2012
Answer Key
5)
C
At one point, James Watson favored a “like with
like” model of DNA structure (shown in diagram at right).
One of the reasons he favored this model was:
A) It could explain Chargaff’s rule
B) It produced a helix of constant diameter
C) It could explain how DNA might replicate
D) It placed sugar-phosphate chains on the inside
E) None of the above
6)
B
Which of the following is not true about the final
Watson-Crick double helix model?
A) The two strands are held together by hydrogen bonds
B) One strand twists in the right-handed direction, the other
left-handed.
C) A-T and G-C base pairs are nearly identical in diameter
D) Nucleotides in the same strand are covalently bonded to
each other
E) None. All of the above statements are true.
7)
A
DNA polymerase, the enzyme that replicates DNA
A) produces Okazaki fragments that must be linked together by DNA ligase
B) can work in both the 3' to 5' and the 5' to 3' directions
C) produces DNA by adding bases to the 5' end of the growing strand
D) produces Okazaki fragments that must be linked together by RNA polymerase
E) produces Okazaki fragments that must be linked together by helicase
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Biology 0200 – 2012
8) The Peptide Bond
Answer Key
(14 points)
Show below are the structures of two important amino acids (Glycine & Asparagine).
A) First, circle the alpha-carbon in each of these amino acids: [4 points]
+2 points for each correctly identified alpha-carbon
B) Draw the structure of the dipeptide N-gly-asn-C. You may abbreviate some of the
chemical groups, such as -NH2 or -COOH. But be sure to show each atom clearly in the region
of the peptide bond:
[10 points]
Grading:
+5 points for selecting proper Amino on Asparagine to form peptide bond.
+5 points for correct peptide bond (shaded area).
Deduct -1 point for minor errors in the structure not affecting the peptide bond.
Deduct -2 points for showing the dipeptide in the wrong order (N-term by Asn).
Note: It’s perfectly OK to show the resonant structure of the bond
(at right), or to show any of the amino or carboxyl groups in their
ionized condition.
Page 3
Biology 0200 – 2012
Answer Key
9) Metabolism [22 points]
A) A culture of green algae is placed in bright light, and carries out photosynthesis at a high
rate. Careful measurements show that the culture is producing 10 microliters of oxygen per
minute and consuming 10 microliters of carbon dioxide per minute.
Your lab supervisor asks you to add 100 micromoles (a substantial amount) of dichlorophenol
indophenol (DCPIP) to the culture. DCPIP is a powerful electron acceptor, which "intercepts"
electrons from Ferredoxin before they can be passed to NADP+. Once DCPIP is reduced by
accepting electrons, it undergoes no further chemical reactions in the culture, but the presence of
this much DCPIP completely blocks the transfer of electrons to NADP+.
Will the addition of DCPIP affect the rate of oxygen production? CO2 consumption? Or both?
Predict what effect DCPIP will have on both rates. You need not make a numerical prediction.
Just say "increase," "decrease," or "remain the same.” Then, explain your reasoning for both
O2 and CO2.
Effect on Oxygen Production:
(5 points)
It would Remain the Same. (+2 points)
Reasoning: Since DCPIP merely substitutes for NADP+, electron flow from water
through the electron transport chain will continue in the presence of DCPIP, and this
means that O2 production will continue as well. (+3 points for reasoning)
Effect on CO2 Consumption:
(6 points)
It would Decrease. (+2 points)
Reasoning: In the presence of DCPIP, it would be impossible to reduce NADP+ to
produce NADPH. Without a steady flow of NADPH, the Calvin Cycle (carbon
fixation) would stop, and therefore so would the consumption of CO2.
(+4 points for reasoning)
Page 4
Biology 0200 – 2012
Answer Key
B) A culture of yeast cells is being grown under aerobic conditions,
and produces 10 microliters of CO2 per minute. The very same culture,
when switched to anerboic conditions produces 60 microliters of CO2
per minute. (You may recognize this exact set of culture conditions,
since it was used for the Week 6 homework assignment!). Your
research supervisor congratulates you upon your successful explanation
of the change in CO2 production when these cells are switched to
anerobic conditions. Nice work! Now, however, she has a new
challenge for you.
DNP (dinitrophenol) is a compound that renders the inner mitochondrial membrane completely
permeable to H+ ions. She asks you to predict the rate of CO2 production per minute in the
presence of DNP under the same two conditions (aerobic and anerboic). Write down a numerical
answer for each condition, in terms of microliters of CO2 per minute. Be sure to show the work
you did in arriving at that figure:
Aerobic:
(5 points)
90 microliters per minute (+2 points)
Work should show something like the following:
• DNP destroys the chemiosmotic gradient. So, the only ATPs that can be made are
2 per glucose in glycolysis and 2 more in the Krebs cycle. As a result the number of
ATPs per glucose would drop from 36 to 4 in the presence of DNP.
• Since only 1/9th the previous amount of ATP is made per glucose, the rate of
glucose utilization will go up by a factor of 9, as will CO2 production.
• 9 x 10 microliters = 90 microliteres per minute.
Anerobic:
(6 points)
60 microliters per minute (+2 points)
Work should show something like the following:
• In the absence of oxygen, the only ATPs that can be produced come from glycolysis
and alcoholic fermentation. Therefore, the presence of DNP makes no difference in
the rate of CO2 production.
• Since only 2 ATPs are produced per glucose in glycolysis, the rate of glucose
utilization will go up by a factor of 18 (36/2). However, only 1/3 of the carbons in
glucose will be released as CO2. The remaining 2/3 will be found in ethanol.
• 18 / 3 = 6.
6 x 10 microliters = 60 microliters per minute. (+4 points)
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Biology 0200 – 2012
10)
DNA Replication
Answer Key
[15 points]
As you know, the Meselson-Stahl experiment provided clear
evidence that DNA replication is semi-conservative. The
experiment involved the separation of "heavy" DNA (labeled
with 15N) from "light" DNA (containing 14N) by
centrifugation in a tube containing a cesium chloride (CsCl)
density gradient. The manner in which heavy and light DNA is
separated in such a tube is shown at right:
A colleague of yours is preparing experiments to analyze DNA that may be recovered in a future
space probe to Titan, one of the moons of Jupiter. There is some expectation that DNAcontaining organisms may be found on Titan, and that they may replicate their DNA in a way
other than the semi-conservative pattern found on earth. Your colleague asks you to help plan
Meselson-Stahl type experiments by anticipating the results they might find.
Using the positions of heavy and light bands in the tube as reference points, sketch the positions
of DNA molecules expected in a CsCl gradient after 15N-labeled DNA has undergone three
rounds of conservative replication in 14N medium. At the right side of the tube, clearly
indicate the percentage of DNA to be expected in each of the bands you have drawn (the
percentages of all the bands in the tube, of course, should add up to 100%):
Page 6
Biology 0200 – 2012
Answer Key
11) Gene Mapping [21 points]
A mad scientist, after watching too many fantasy movies, decides to breed mutant Drosophila in
the model of dragons. Mutants with super powers have the dominant alleles for Flame Breath
(F), Neon Skin (N), Red Eyes (R), and Dragon Wings (D). A homozygous dominant “Dragon
Fly” (Drosophila drago) which expresses all these powers (phenotype is FNRD) is mated with a
homozygous recessive Drosophila which expresses no such powers (phenotype is fnrd). They
mate and create a batch of 100 F1 Drosophila. These F1 Drosophila can shoot flames from their
mouths, have neon skin, red eyes, and dragon wings.
A) What is the genotype of the F1 generation? (5 points)
FNRD / fnrd
(also OK: FfNnRrDd)
(+5 points)
One of the F1 generation flies is mated with a homozygous recessive drosophila (fnrd). They
produce 100 F2 flies with the following characteristics (where the “Dragon” phenotypes are
indicated by upper case letters, and the normal phenotypes by lower case letters:
FNRD
fnrd
fNRD
fnrD
fNRd
FNRd
FnrD
Fnrd
23
21
3
22
3
23
3
4
B) Use these data to construct a genetic map of these 4 genes. As you determine the distances
between genes, show your work, and note these distances (in terms of recombination
frequencies) on your map. (10 points)
Recombinants:
F-N
F-R
F-D
N-R
N-D
R-D
3+3+3+4 = 13
3+3+3+4 = 13
3+22+23+4 = 52
0
22+3+23+3 = 51
22+3+23+3 = 51
Note, since the numbers actually add up
to 102, full credit is given for selecting
the correct numbers for each gene pair.
Page 7
Biology 0200 – 2012
Answer Key
Grading:
6 points for correct calculation of recombination frequencies
4 points for correct map. (Note: Map may show gene D on another
chromosome, or at a very distant position on the same chromosome, separated by a 50, 51,
or 52% recombination rate)
C) Did each gene pair show recombination? Please write yes or no in the table below, and if one
or more of the gene pairs did not show recombination, not why this may have been the case.
(6 points)
F/N
F/R
F/D
N/R
Recombination?
(Yes or No)
Yes
Yes
Yes
No
N/D
R/D
Yes
Yes
Gene Pair
Grading:
Explanation for “No’s”
Genes N & R are actually the same gene.
Or, N&R are so close together that their
recombination rate is less than 1%.
+1 points for the correct five “yes” answers.
+2 points for N-R = “No Recombination.”
+3 points for explanation of no N-R recombination.
Note: All gene pairs except for N-R demonstrate recombination. The one gene that is
unlinked (D) still shows recombination with the other genes (at a rate close to 50%).
Page 8