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2nd Exam Answer Key
Biology 0200 – 2011
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)
A
The amount of product present in an enzyme catalyzed reaction can be
determined by measuring the amount of light absorbed by a sample in which the reaction
is taking place. The graph above represents change in light absorbance over time for two
samples. One sample contains excess substrate and enzyme in solution (dotted line). A
second sample (solid line) contains enzyme and substrate in the same amounts as the first
sample. An unidentified compound, molecule “X”, has also been added to the second
tube. Molecule “X” behaves most like:
A) an activator
B) a competitive inhibitor
C) a non-competitive inhibitor
D) additional substrate
E) a denaturing compound
2)
D
The reactions of glucose metabolism and the reactions of photosynthesis
have all of the following in common except:
A) movement of electrons along an electron transport chain
B) formation of a proton (H+) gradient
C) ATP production
D) generation of CO2
E) ATP consumption
3)
D
Which of these 4 descriptions of the composition of dATP is accurate?
A) A 6-carbon sugar, phosphate groups, and adenine.
B) Ribose, phosphate groups, and adenosine.
C) Alanine, deoxyribose, and phosphate groups.
D) A 5-carbon sugar, phosphate groups, and adenine.
E) None of the above.
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2nd Exam Answer Key
Biology 0200 – 2011
4)
D
Photo-51, Rosalind Franklin’s X-ray diffraction of
DNA, was one of the keys to solving the structure of DNA. However, by
itself it did not reveal everything about the molecule. Which if the following
aspects of DNA structure is not directly implied by the pattern in photo-51?
A) Its 20Å diameter.
B) Its helical three-dimensional shape.
C) Two anti-parallel strands.
D) Specific base pairing (A with T and G with C).
E) None. Actually, the pattern implies each of the features in answers A-D.
5)
A
Which statement about the tertiary structure of proteins is not true?
A) Tertiary structure is not affected by primary structure
B) Tertiary structure can be stabilized by hydrogen bonds
C) Covalent bonds sometimes are very important in stabilizing tertiary structure
D) A tertiary structure may contain both α-helix and β-sheet.
E) The tertiary structure of myoglobin includes an oxygen-binding region
6)
E
When a pea plant with round yellow seeds was self fertilized, it produced
offspring in the following proportions:
Phenotype
Round,
Yellow
Round,
Green
Shrunken, Yellow
Shrunken, Green
Number of Offspring
99
33
33
11
Given these results, which of the following statements must be true?
A) Recombination has occurred
B) The genes controlling the two traits (seed color & seed shape) are not linked
C) The plant produces all gamete types with roughly equal frequency
D) Green is recessive to yellow
E) All of the above.
7)
C
During translation, a covalent bond exists between each amino acid
added to the growing polypeptide chain and
A) mRNA
B) rRNA
C) tRNA
D) the small subunit of the ribosome
E) the large subunit of the ribosome
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2nd Exam Answer Key
Biology 0200 – 2011
8) The Peptide Bond
(14 points)
Show below are the structures of two important amino acids (aspartate & alanine).
A) First, circle the alpha-carbon in each of these amino acids: [4 points]
+2 points for correct α -carbon on ASP
+2 points for correct α-carbon on ALA
B) Draw the structure of the dipeptide N-asp-ala-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 Carboxyl on Asp 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 ala).
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.
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2nd Exam Answer Key
Biology 0200 – 2011
9) Metabolism [22 points]
Malonate (-OOC-CH2COO-) is a specific inhibitor of the enzyme succinate dehydrogenase in
the Krebs cycle (This enzyme catalyzes the reaction where succinate is converted to fumarate.
There’s an arrow labeled “Q #9” on the Krebs cycle diagram to make it easy to find). The
following data on the conversion of pyruvate to CO2 were obtained when an extract of pigeon
breast muscle mitochondria was incubated with the substances listed, under aerobic conditions
(i.e., in air). (NOTE: this preparation contains intact mitochondria and all enzymes and
cofactors to carry out all reactions of glycolysis and the Krebs cycle)
flask
1.
2.
3.
substances added (micromoles)
pyruvate malonate
fumarate
0.05
----0.05
1.0
--0.05
1.0
0.025
CO2 produced
in reaction (micromoles)
0.15
<0.001
0.075
A) Flask 1 produced 0.15 micromoles of CO2 from just 0.05 micromoles of pyruvate. Why did it
produce exactly 3 times as much CO2 as the pyruvate starting material? [4 points]
It produces 3 times as much because each molecule of pyruvate produces three (3)
molecules of CO2 as it moves into and through the Krebs cycle. +4 points
B) Flask 2 contained the malonate inhibitor. Why did Flask 2 fail to produce significant
amounts of CO2? Be sure to address the fact that the reaction blocked by the inhibitor actually
occurs after the Krebs cycle reactions that release CO2! [6 points]
The inhibitor (malonate) blocks the formation of fumarate.
(+2 points for recognizing that this particular blockage is important, but that’s all if they
don’t address the second part of this question, as below)
If fumarate is not formed, then oxaloacetate cannot be produced, and new molecules
cannot enter the Krebs cycle (so CO2 production stops)
(+4 points for recognizing that the reason that CO2 production stops is because new
molecules cannot enter the cycle).
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2nd Exam Answer Key
Biology 0200 – 2011
C) Like Flask 2, Flask 3 also contained the malonate inhibitor. Yet, it produced significant
amounts of CO2 while Flask 2 did not. Why? [6 points]
Because of the presence of fumarate.
(+3 points for pointing out that fumarate is the key)
By supplying fumarate in the mixture, oxaloacetate can be produced, allowing new
molecules to enter the Krebs cycle, even though one of the reactions is blocked by
the inhibitor.
(+3 points for pointing out that the added fummarate allows new Krebs intermediates to
be produced. Not necessary to name the intermediates)
D) Why did Flask 3 produce exactly 0.075 micromoles of CO2, and not 0.15 micromoles like
Flask 1? Remember that both flasks had exactly the same amount of pyruvate substrate (0.05
micromoles). [6 points]
Each fumarate molecule allows one pyruvate to move into the Krebs cycle. Since each
pyruvate yields 3 CO2 molecules, .025 micromoles of fumarate allows 3 times that
amount (exactly .075 micromoles) of CO2 to be produced.
(+6 points for connecting the amount of fumarate to the 3:1 stoichiometry of the
reactions)
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2nd Exam Answer Key
Biology 0200 – 2011
10) DNA Replication [15 points]
Answer the two questions below based on these "facts:"
• You have been assigned to a NASA lab studying bacteria-like organisms recovered from a deep
space probe. These organisms contain DNA, and you have decided to perform a version of the
Meselson-Stahl experiment with them.
• Cells were grown for weeks on "heavy" 15N medium, density-labeling their DNA
• At t=0 of the experiment, cells were transferred to light 14N medium
• Density-gradient centrifugation was used to determine the density of DNA isolated from the
growing organisms at 30 and 60 min. after the transfer to 14N (light) medium. The results:
Summary of Results:
Density of light 14N DNA: 1.60 g/cc
Density of heavy 15N DNA: 1.70 g/cc
0 min: 100% of DNA at density of 1.70 g/cc
30 min: 100% of DNA at density of 1.65.
60 min: 100% of DNA at density of 1.625.
a) From these data, what pattern of replication does this double-stranded DNA follow?
(possibilities are semi-conservative, conservative, and dispersive) [5 points]
Dispersive. (+5 points)
b) Explain how these experimental results fit the pattern you have chosen by drawing two DNA
molecules after a single round of replication beginning with one double-stranded DNA
molecule. Use solid lines to show N-15 DNA, and dashed lines for N-14 DNA. The starting
condition at T=0 is already drawn for you: [10 points]
Grading:
+3 points for proper number of chromatids (4.)
+4 points for showing old and new DNA as part of same strand.
+3 points for showing dispersive replication correctly.
• Award 5 points for a diagram that correctly shows Semiconservative or Conservative.
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2nd Exam Answer Key
Biology 0200 – 2011
11) Gene Mapping [21 points]
Clear wing, Black eye, and Hairless (c, b, and h) are linked,
recessive traits carried on chromosome 3 of the Narragansett
quahaug fly. A clear-winged, black-eyed fly is crossed with a
hairless fly.
A) Their F1 offspring were 97 wild type quahaug flies. What is
the genotype of these F1 flies?? [4 points]
Cc Bb Hh
Genotype:
or, [cbH / CBh], etc…
(heterozygous for all 3 traits)
Grading: +4 points
10 of these flies were then mated with flies that exhibit all three traits (clear-winged, black-eyed,
and hairless). They produced a total of 1,000 F2 offspring:
clear-wings, black eyes
hairless
clear wings
black eyes, hairless
wild type
clear wings, black eyes, hairless
black eyes
clear wings, hairless
cbH
CBh
cBH
Cbh
CB
cbh
CbH
cBh
415
411
61
58
22
26
4
3
• Use these data to construct a map of these three genes in two steps.
B) First, calculate recombination frequencies for each pair of genes. Circle the recombination
frequencies so that the grader can locate them. [12 points]
Parental Allele Combinations: cbH and CBh
C-B
B-H
C-H
61+58+4+3= 126
61+58+22+26= 167
22+26+4+3= 55
12.6%
16.7%
5.5%
Grading: +4 points for each correct recombination frequency. (NOTE: the allele
combinations and the identification of parental alleles written on the key are not part of
the required answer)
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2nd Exam Answer Key
Biology 0200 – 2011
C) Next, use the frequencies you have calculated to construct a genetic map. Show the order and
relative distances between the 3 genes on the map, and circle the map so the grader can locate it.
[5 points]
H
C
5.5%
B
12.6%
Grading: +5 pts. for correct map. Deduct -3 for any mistake in location of an allele.
Award full credit, +5 points, for correct mapping of incorrect frequencies.
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