Download LS1a Problem Set #2

Name: __________________
TF Name: _______________
LS1a Fall 2014
Problem Set #3
Due Monday 10/20 at 6 pm in the drop boxes on the Science Center 2nd Floor
I.
Basic Concept Questions
1. (7 points) Shown below is a segment of DNA that is hybridized to a segment of RNA. On the
structure below, do the following:
a. (1 point) Fill in the hydrogen bonds between base pairs using dashed lines.
b. (1 point) Identify which strand is DNA.
c. (1 point) Label the 5’ and 3’ ends of both strands.
d. (2 points) On one of the ribose sugars, label the 2’, 3’, and 5’ carbons.
e. (2 points) Place a circle around the base which is unique to RNA. Place a box
around the base that is unique to DNA.
3’ end
5’ end
2’
3’
5’
3’ end
DNA
5’ end
2. (15 points) Shown below are diagrams of a DNA replication bubble and a transcription
bubble.
a. (2 points) Identify each diagram below as either “DNA replication” or “transcription.”
b. Label:
i. (2 points) The 5’ and 3’ ends on the strands acting as templates.
ii. (2 points) The DNA coding strand of the transcription bubble.
iii. (2 points) Indicate the most likely position of the promoter on the diagram of
transcription.
iii. (2 points) The leading and lagging strands for the replication bubble.
iv. (2 points) Draw arrows indicating the direction that the transcription bubble and the
replication forks progress.
Transcription
coding strand
3’
5’
promoter
leading
strands
DNA replication
5’
3’
5’
3’
lagging
strands
c. (3 points) In class we have covered three nucleotide polymerases: DNA polymerase,
which catalyzes DNA replication; RNA polymerase, which catalyzes transcription; and
the poly(A) polymerase, which synthesizes the poly-A tail for a maturing eukaryotic
transcript. In the chart below, indicate whether each enzyme uses either DNA or RNA as
its template, synthesizes it as a product, or both. If the enzyme uses the indicated
nucleic acid as a template, write “T” in the box, and if the enzyme polymerizes the
nucleic acid as a product, write “P” in the box.
DNA
RNA
DNA Polymerase
T, P
RNA Polymerase
T
P
2
Poly(A) Polymerase
P
3. (8 points) Scientists can use PCR to amplify genes from double-stranded DNA templates.
One example of a double-stranded DNA template is drawn below, in which one strand is
drawn as a solid strand and its complementary strand is drawn as a dotted strand. Five short
sequences from portions of the solid (upper) strand are shown.
a. (4 points) Provide the sequences of two six-base DNA primers that can be used to
amplify the 1400 base-pair region BC (but not a longer region). Label the 5’ and 3’ ends
of each primer and indicate which strand (dotted or solid) your primers will bind to.
5’-GGTACC-3’ binds to dotted (and solid)
5’-CCAGTT-3’ binds to solid
b. (4 points) Draw all the different types of newly synthesized DNA strands that would be
exponentially amplified after 30 cycles of PCR using the primers from part (a). Indicate
each product as follows:
i.
Show each product as a dotted or solid line, depending on whether its sequence
matches that of the dotted or solid strand, respectively.
ii. Label the 5’ and 3’ ends of each product strand.
iii. For each product strand, include the primer sequence used to generate it at the 5’
end.
iv. Indicate the region(s) that each product contains (i.e., A, B, C, and/or D).
For example, a solid-strand spanning region AB would be drawn as:
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4. (6 points) Shown below on the left is a simplified diagram of a DNA replication fork. The
DNA sequence at both ends of the two template strands is continuous and only a small
portion is shown. Also shown on the right is a short, four-base segment of DNA that is
bound to a template sequence in the replication fork.
5’
3’
3’
5’
5’
3’
a. (4 Points) Label the location in the replication fork where this four-base DNA segment
would be bound, and indicate the 5’ and 3’ ends of the segment. Also label the 5’ and 3’
ends of both template strands.
b. (2 points) Does this four-base segment represent part of the leading strand or the
lagging strand? Draw an arrow to indicate the direction in which the segment would be
extended.
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5. (13 points) A piece of eukaryotic DNA that contains a gene was purified and bound to the
mature mRNA that was transcribed from it. The resulting structure is illustrated in the
diagram below.
a. (2 points) Which DNA strand (coding or template) is shown?
Template
b. (2 points) Which strand (A or B) is the mature mRNA?
B
c. (4 points) On the diagram, label introns, the poly-A tail, and the general location in
which you would expect to find the promoter.
d. (2 points) How many exons are in this transcript?
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e. (3 points) Circle the features below that are transcribed from the DNA template.
5’ Cap
The Promoter
The Poly-A Tail
The Branch Point Adenosine
Introns
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6. (10 points) The sequences shown below are the -35 and -10 sequences found in six E. coli
promoters. These sequences are on the coding strand and read 5’ to 3’.
trp
tRNATyr
lac
recA
rrnDI
araB
-35
TTGACA
TTTACA
TTTACA
TTGATA
TTGTGC
CTGACG
-10
TTAACT
TATGAT
TATGTT
TATAAT
TATAAT
TACTGT
a. (2 points) The -10 and -35 regions of prokaryotic promoters are consensus sequences,
meaning that their exact sequences can vary so long as most of the nucleotides match
that of an idealized sequence. This idealized sequence is comprised of the nucleotides
most commonly found at each position within that sequence. Based on the sequences
above, write the “ideal” consensus sequences for both the -10 and -35 regions shown
above.
-35: 5’ TTGACA 3’
-10: 5’ TATAAT 3’
b. (2 points) Shown below is a nucleotide sequence from the promoter of a prokaryotic
gene. Identify (box and label) both the -35 and -10 sequences of this promoter based on
your answer to part a.
5’GACCTATGGAGATTATAGTAGAGTGCTTCTATCATGTCAATACCACTACGGT 3’
3’CTGGATACCTCTAATATCATCTCACGAAGATAGTACAGTTATGGTGATGCCA 5’
-10
-35
c. (2 points) Which strand, top or bottom, is the template strand?
The top strand is the template.
d. (4 points) In which direction, left or right, would RNA polymerase transcribe upon
binding to the promoter? Briefly explain your answer.
To the left: the -10 sequence is located to the left of the -35 sequence in this
representation so the +1 sequence would be on the bottom strand to the left of the 10 and transcription would extend 5’ to 3’ to the left.
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II. Applied Concept Questions
7. (16 points) Shown below are four nucleoside triphosphate analogs that can inhibit DNA
synthesis or RNA synthesis by interfering with the enzymes involved in catalyzing these
reactions. Two of the analogs contain a non-hydrolyzable (“uncleavable”) P-CH2-P linkage
instead of hydrolyzable (“cleavable”) P-O-P linkages. [Note: for any of these analogs to
inhibit DNA polymerase or RNA polymerase, the analog would have to be able to enter the
enzyme’s active site.]
a. (4 points) Circle the option that best describes how Analog 1 would affect DNA
synthesis. Briefly explain why the choice you circled is correct.
i. It would not affect DNA synthesis.
ii. It would inhibit DNA synthesis, but it would not be incorporated into the growing
strand.
iii. It would terminate DNA synthesis and it would be incorporated into the growing
strand.
iv. DNA synthesis could still occur, but it would be less thermodynamically favorable.
Analog 1 would be able to incorporate into the growing strand because the linkage
between the α and β phosphates is hydrolyzable. The overall reaction would be
less favorable though, since the DNA synthesis reaction using this analog releases
a pyrophosphate-analog that is not hydrolyzable. This reduces the thermodynamic
favorability of this reaction as the cleavage of pyrophosphate drives the DNA
synthesis reaction by Le Chatelier’s Principle.
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b. (4 points) Circle the option that best describes how Analog 2 would affect DNA
synthesis. Briefly explain why the choice you circled is correct.
i. It would not affect DNA synthesis.
ii. It would inhibit DNA synthesis, but it would not be incorporated into the growing
strand.
iii. It would terminate DNA synthesis and it would be incorporated into the growing
strand.
iv. DNA synthesis could still occur, but it would be less thermodynamically favorable.
Analog 2 would inhibit DNA synthesis because it contains no 3’ OH. It can be
incorporated into the strand because the linkage between the α and β phosphates
can be hydrolyzed. Once the analog is incorporated, the polymerase would be
unable to add another nucleotide because it lacks the 3’ OH that is used as a
nucleophile. Consider the breakout from lecture 11 involving 2’,3’ dideoxy CTP.
c. (4 points) Circle the option that best describes how Analog 3 would affect DNA
synthesis. Briefly explain why the choice you circled is correct.
i. It would not affect DNA synthesis.
ii. It would inhibit DNA synthesis, but it would not be incorporated into the growing
strand.
iii. It would terminate DNA synthesis and it would be incorporated into the growing
strand.
iv. DNA synthesis could still occur, but it would be less thermodynamically favorable.
Analog 3 would not be incorporated into a growing DNA strand. During DNA
synthesis, the P-O-P linkage between the α and β phosphates must be broken, but
this analog prevents this step from occurring, preventing it from being added to
the growing strand. This molecule is still capable of fitting into the DNA
polymerase active site though, so it would compete with normal nucleoside
triphosphates for binding to the polymerase, thereby inhibiting DNA synthesis.
d. (4 points) Circle the option that best describes how Analog 4 would affect DNA
synthesis. Briefly explain why the choice you circled is correct.
i. It would not affect DNA synthesis.
ii. It would inhibit DNA synthesis, but it would not be incorporated into the growing
strand.
iii. It would terminate DNA synthesis and it would be incorporated into the growing
strand.
iv. DNA synthesis could still occur, but it would be less thermodynamically favorable.
Analog 4 contains a 2’ OH, which would prevent it from being used in DNA
synthesis (see questions 10a and 10b). Since it cannot fit into the DNA polymerase
active site, it would have no effect on DNA synthesis. (It would, however, inhibit
RNA synthesis for the same reason that Analog 2 inhibits DNA synthesis. Consider
the breakout from lecture 12 involving cordycepin.)
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8. (11 points) You decide to conduct an experiment using purified E. coli RNA polymerase,
ribonucleotides (CTP, GTP, UTP, and ATP), and a single-stranded DNA template that contains
a promoter. The ATP contains a radioactive phosphorus isotope [32P] in the -phosphate
position. You discover that the resulting transcript is radioactive.
a.
(2 points) Where in the transcript is the radioactive phosphorus?
Radioactivity will only be found on the most 5’ nucleotide.
b.
(4 points) The experiment is repeated three more times, using 32P in the -phosphate of
either UTP or GTP or CTP rather than in the -phosphate position of ATP. In all three
cases, no radioactivity is found in the resulting transcript. Briefly explain how these four
experiments can be consistent.
Adenine is the only nucleotide that occupies the 5’ most position on the transcript,
which is the only position that retains its -phosphate group since it does not form
phosphodiester bond to a nucleotide on its 5’ end. Since UTP, GTP, and CTP are
incorporated later in the RNA stand, the synthesis reaction releases the -phosphate
as each of these nucleotides is incorporated, preventing these transcripts from
becoming radioactive.
c.
(3 points) Which phosphate group would have to contain the radioactive phosphorus in
the UTP, GTP, or CTP substrates in order for the transcript to be radioactive? Briefly
explain your answer.
The -phosphate group, because the -phosphate group is retained in the
phosphodiester bond between nucleotides.
d.
(2 points) What base is at the +1 position of the template strand of the DNA template?
Thymine
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9. (9 points) As we covered in lecture, the first proof-reading step in DNA synthesis depends
upon DNA polymerase’s ability to recognize correctly matched base pairs in the active site.
However, DNA polymerase does not recognize specific bases, rather it recognizes the
geometry of a base pair.
One way in which DNA polymerase recognizes appropriate base-pair (A-T, C-G) geometry is
by inserting two amino acid residues into the minor groove of the DNA before it is
covalently added to the 3’-end of the growing chain (shown below):
Template nucleotide
Nucleoside triphosophate about
to be added to the 3’-end of
growing chain
Arginine
Glutamine
a. (2 points) What type of intermolecular interactions are arginine and glutamine making
with the nascent DNA base pair?
Arginine is donating a hydrogen bond to (or making an ion-dipole interaction with) the
cytosine moiety.
Glutamine is donating a hydrogen bond to the guanine moiety.
(“Moiety” means “part of the molecule.”)
Please do not write below this line.
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b. (7 points) Shown below are three DNA base pair “mismatches” and one representative
cognate base pair. Identify the bases involved in the three base pair mismatches (i-iii)
below. Briefly explain why it is important for DNA polymerase to interact with these
particular atoms in the minor groove if the polymerase needs to accurately incorporate
incoming dNTPs as they pair with the template without recognizing specific bases.
thymine-guanine
thymine-thymine
adenine-guanine
Arginine and glutamine are both interacting with the universal hydrogen bond
acceptors in the minor groove: glutamine is donating a hydrogen bond to one of the
two universal hydrogen bond acceptors in the minor groove; arginine is either making
an ion:dipole interaction with the other universal hydrogen bond acceptor OR you can
say it is donating a hydrogen bonding to it. These two hydrogen bond acceptors on the
DNA bases are called “universal” because all four correct base-pair combinations (A-T,
T-A, G-C, C-G) have hydrogen bond acceptors in those spatial two positions.
While these universal hydrogen bond acceptors are present in all possible base-pair
combinations, they are only in the appropriate location to form hydrogen bonds to
the arginine and glutamine when the correct base pairs with the template. Otherwise
incorrect base pairing changes the distance between the universal hydrogen bond
acceptors in the case of mismatches. If an incorrect dXTP enters the active site
opposite the template strand, the enzyme will not be able to bond to it as tightly
because its minor groove hydrogen bond acceptor will be out of position to make an
ion:dipole interaction with the polymerase’s arginine residue.
This allows the enzyme to make sure that a correct base-pair is being formed
regardless of which bases are involved, as the minor groove hydrogen bond acceptors
are present in the same locations for each correctly formed base pair.
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10. (5 points) DNA polymerase has to select against incorporating ribonucleotide triphosphates
while it is synthesizing DNA. Scientists discovered that mutating a single residue, from a
tyrosine to a valine, dramatically reduced the enzyme’s preference for deoxyribonucleotides
over ribonucleotides.
Below are two images of the same three molecules as they would be found interacting in
DNA polymerase. On the left they are in a “stick” representation which shows the
connectivity of the atoms. One right they are shown as “spheres,” which show how much
space each atom occupies. (Green, carbon; red, oxygen; blue, nitrogen; orange, phosphorus;
magenta, hydrogen bonds). The structure shows an incoming deoxythymidine triphosphate
(“dTTP”) forming hydrogen bonds with the template adenosine as the thymidine is about to
be added to the growing strand of complementary DNA.
dTTP
dTTP
Adenosine
Adenosine
OH
OH
Tyrosine
Stick Representation
Tyrosine
Space-filling Representation
a. (1 point) On the stick representation, draw where the 2’ hydroxyl would be located on
thymidine if it were a ribonucleotide.
b. (4 points) Briefly explain why changing this tyrosine to a valine reduces the enzyme’s
preference for deoxyribonucleotides over ribonucleotides (If you have any trouble
visualizing how these three molecules interact, please consult the “10b.pse” PyMOL file
on the Problem Set page of the course website.)
The tyrosine functions as a “steric gate,” preventing ribonucleotides from entering the
active site in the optimal position required for the DNA synthesis reaction (i.e., the
tyrosine prevents the previous nucleotide’s 3’ hydroxyl from aligning with the
incoming nucleotide’s alpha phosphate if the incoming nucleotide is a ribonucleotide.)
The tyrosine prevents rNTPs from fitting in the DNA polymerase active site in the
appropriate orientation required for the synthesis reaction to occur, so that the
ribonucleoside triphosphate would simply diffuse out of the enzyme’s active site
without being able to fully enter the active site, and therefore without any reaction
having occurred.
If tyrosine is mutated to valine, which is a much smaller amino acid than tyrosine, a
ribonucleoside could fit into the active site in a physical orientation more conducive to
the synthesis reaction occurring. The 2’ hydroxyl group of the ribonucleoside
triphosphate has more space to fit into the active site once tyrosine is mutated to
valine.
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