Chem+153B+-+Homework+3
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University of California, Los Angeles *
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Course
153B
Subject
Chemistry
Date
Feb 20, 2024
Type
Pages
6
Uploaded by LieutenantCrownDragonfly3
Name _______________ ID _____________
Chem 153A –
Homework 3
due 8/28 @ 11:59 pm 1.
Envision a 508 bp
double-stranded piece of closed, circular DNA with an average helical twist of 34°
. The path of the backbone is shown in the figure on the right. a.
What is your best estimate of the linking number of this loop? b.
A 23 bp
segment of alternating purine/pyrimidine residues in our loop spontaneously changes to the Z conformation (helical twist angle of -30°
). The DNA backbone remains intact throughout this transition. Estimate the new writhe of the loop. 2.
Envision a 1059 bp
-long piece of closed, circular DNA with an average helical twist angle of 35°
. The path of the backbone is shown in the figure on the right. Contained within this sequence is a 21 bp
long AT-rich region that has a tendency to melt, forming a denatured bubble. a.
Determine the twist, writhe, and linking number for this DNA loop prior to
bubble formation. b.
Determine the twist, writhe, and linking number for this DNA loop after
bubble formation (assume the helical twist angle in the bubble is 0°
and that the DNA backbone remains intact). c.
Is the formation of this 21 bp bubble favored, disfavored, or neither? Mia
Kim
005896550.
->
r
T
=
58(34)
=
n48
-I
W
=
-
3
-
L
=
48
-
(
3)
=
45
&
&
445
T
=
(23)(34)+138
=
44
36
45.81
-1.9
=
T
W
Not
44
45
=
change
W
=
1
A
-1
A
-
=1035)
=
163
W
=
L
=
103-3
=
10
&
T
=
1059
21335
+
0)
=
10
360
L
=
100
=
10
1
+
W
W
=
-
1
formation
of
bubble
was
favored
because
II
was
decreased
from
1-31-1-1)
⊥
Name _______________ ID _____________
3.
The following strand of circular dsDNA (twist = -1, writhe = 0) is acted upon by a type I topoisomerase. What are the two possible outcomes? Draw both and note the resulting twists and writhes (hint: Look closely at the diagram!). 4.
The following strand of circular dsDNA (twist = 0) is acted upon by a type II
topoisomerase. What are the two possible outcomes? Draw both and note the resulting twists and writhes.
5.
In prokaryotes (like E. Coli) what determines where replication starts? How is replication initiated? -lest
⑨
*
right
&
Es
T
=
-2
T
=
0
W
=
0
W
=
0
W
=
-
2
-
->
&
S
T
08
W
0
=
8
4
-
origin
of
replication
has
several
sequences
with
affinity
to
the
replication
initiation
factor
DnaA.
It
binds
to
these
sequences,
creating
supercoiling
that
open
up
a
replication
bubble.
Draft
also
recruits
draB
helicases
which
binds
to
SSDNA
in
the
replication
bubble
and
begins
unwinding
the
double-stranded
DNA
at
each
replication
fork.
Name _______________ ID _____________
6.
PDB DNA/Protein Scavenger Hunt
–
Search for 6FBI on rcsb.org and visualize with Pymol or rcsb’s 3D View. Use this structure of klentaq DNA polymerase bound to template DNA to answer the following: a.
Which strand is being synthesized? Use a description or a screenshot for your answer. b.
Which nucleotide is about to be added to the newly synthesized strand? c.
Find the two metal ions vital to the polymerization mechanism. Take a screenshot showing the groups they’re interacting with
d.
Isolate the duplex base pair most recently synthesized by klentaq. Two amino acid side chains make sequence-independent H-bond contacts with two atoms in the base pair. Draw the base pair shown in this scene along with the two amino acid side chains, representing proper geometry, accurate distances, and any intermolecular interactions. e.
Explain how the H-bonds you added in part d allow the DNA polymerase to distinguish between WC and non-WC base pairs. Consider how the positions of the two H-bond acceptors in the bases makes these H-bonds sequence-
independent. Why is access to the minor groove useful here? The
orange
strand
A
guanosine
phosphate
is
in
position
to
be
added
to
the
new
strand.
This
is
the
guanine-cytosine
right
below
the
about
to
be
added
guanosine
triose
phosphate.
Arg573
and
Gin
(34
are
interacting
with
the
minor
groove
of
this
base
pair.
There's
H-bond
between
quanidium
group
of
Arg373
side
chain
and
the
or
atom
of
cytosine
and
an
H-bond
between
amide
group
of
the
GIn734
side
chain
and
the
N3
atom
of
guanine.
All
watson-Crick
base
pairs
have
2
H-bond
acceptor
in
the
same
part
of
the
minor
groove
edge.
Due
to
their
symmetry
around
the
dyad
axis,
the
exchange
positions
whenever
a
purine:
pyrimidine
pair
is
swapped
for.
a
pyrimidine:
purine
pair.
Most
non-Watson-Crick
base
pairs
lack
H-bond
acceptors
in
one
or
both
of
these
positions.
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Name _______________ ID _____________
7.
How does the baseline intracellular pyrophosphate concentration promote DNA synthesis? What would happen if this concentration was increased –
and why? 8.
List (and explain) the 5 paradigms of DNA replication. 9.
Explain the two roles of topoisomerases in DNA replication. 10.
How do the metal ions in the active sites of DNA polymerases reduce the activation barrier to phosphoester bond formation? DNA
Synthesis
produces
pyrophosphate
as
a
product.
Give
this,
the
relative
concentration
of
pyrophosphate
can
impact
the
favorability
of
DNA
synthesis.
Conveniently
the
baseline
concentration
of
pyrophosphate
is
low
enough
(around
0.5MM)
so
the
DNA
synthesis
proceeds
forwards
by
Le
Chatelier's
principle.
If
the
pyrophosphate
concentration
were
to
be
increased
(to
200
times
the
dNTP
concentration)
the
rate
of
DNA
degradation
would
match
the
rate
of
synthesis,
as
we
have
reached
the
equilibrium
concentration.
as
helicase
unwinds
the
DNA
duplex,
it
decreases
the
natural
positive
twist
of
DNA
(reduce
it
to
of
Due
to
topological
constraints,
this
increase
writhes
eventually
increasing
the
free
energy
needed
to
continue
unwinding
the
strand.
&one
role
of
topoisomerases
to
move
a
heat
of
the
replication
fork,
uncoiling
the
DNAs
as
to
keep
the
free
eners
be
them,
we
are
stranded
break
is
required.
The
metal
ions
stablize
the
substrates
(dNTP
and
DNA
strand)
and
hold
them
(as
well
as
the
nascent
covalent
bond
between
them
?
In
optimal
geometry,
which
stablizes
the
reaction,
as
well
as
the
eventual
transition
state.
They
also
stablize
the
oxyanion
form
of
the
nucleophile
(3
hydroxyl)
that
initiate
the
reaction.
Name _______________ ID _____________
11.
Suppose that the double-stranded DNA shown below is part of the circular E. Coli chromosome. One DNA polymerase III is currently traversing this sequence left to right (and has been for more than 2 kilobases). a.
Indicate which strand is the template being used for DNA synthesis. b.
Which strand is the lagging strand for this replication fork? c.
Indicate the direction of movement for the replication fork on the diagram below. d.
The asterisk below indicates the 5’ end of a primer adjacent to an Okazaki fragment. Write the first 6 base pairs of its sequence. e.
A fraction of a second prior to the arrival of Pol III, the DnaB helicase moves through this area. At one moment it is bonded to the area boxed in below (I’ve boxed both strands but it may not be associated with both!). The “top” subunit of the helicase hydrolyzes ATP and moves to the “bottom” of the hexamer. Circle the base pairs it associates with at this point. 12.
Describe the proofreading mechanism of Pol I. Why is movement by the nascent (newly created) DNA critical for the editing process as carried out by this enzyme? *
top
Strand
bottom
strand.
5
CCA
AAA
3
c.
replication
fork
template
(and
leading
strand)
i
&
ACC
lagging
strand.
wherever
a
mismatched
base
pair
is
formed
the
DNA
duplex
is
destablized.
The
movement
of
the
DNA
allows
the
duplex
to
slide
into
a
region
of
the
channel
where
the
mismatched
strand
can
dissociate
and
flip
into
the
3
to
5'
exonuclease
active
site.
This
active
site
degrades
the
mismatched
base
pair,
allowing
for
synthesis
to
replace
it.
Name _______________ ID _____________
13.
What are Okazaki fragments and why are they necessary? Draw a diagram incorporating the lagging strand to assist your explanation. 14.
Why is the β
2
sliding clamp a necessary feature of DNA Pol III? What role does it serve? one
strand
(the
3'
tos'
leading
strand)
can
have
DNA
added
in
the
5'
to
'direction
(which
is
necessary
for
polymerizations
but
the
lagging
strand
(5
to
3')
is
in
the
wrong
orientation
to
add
DNA
towards
the
replication
fork.
Okazaki
fragments
any
short
strands
of
replicated
DNA
formed
on
the
lagging
strand
by
discontinuous
replication
events,
each
primed
by
distinct
RNA
primer.
They
solve
the
issue
of
the
lagging
strand
as
each
is
polymerized
in
a
5'ts
3'
direction
(away
from
the
replication
fork)
but
they're
created
sequentially
moving
in
a
3
tos'
direction.
DNAtaki
fragment
≘
·
mer
5/
5
I
DNA
helicase
3
DNA
is
made
continuosly
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0 H
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▼
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A.
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D.
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pH = 1
pH = 5
CH3
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H
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.
¹0-2-0
Explanation
O=P-O-CH₂,0
O=P-O-CH₂, 0.
1-
-0-2-0
O
OH
O=P-O-CH₂_0.
0
Check
N.
N
N.
N
O=P-O-CH₂ 0.
O
NH₂
NH₂
N
OH OH
0=
N.
NH
N
O
O=P-O-CH₂,0.
N.
(1
N
NH₂
'N
N
N
NH
NH₂
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Name of base at 3' end:
0
Base sequence abbreviation:
0
Type of molecule:
(Choose one)
Name of base at 3' end:
0
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C
с
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1
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CH
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H―
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O H N
U
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1
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1
i
Find the pH range to within 1 pH unit over which the polypeptide below will have an overall
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(In class on Friday, I worked out the changes in charge that the N-terminal -NH3+ and the C-terminal
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-
Lys
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Glu
Ala
-
A C
*H3N
CH3
.SH
H
K
H
HO.
E
-
Asn
His
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Tyr
N
H
Y
O=0
■NH3*
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ΤΗΝ
NH
H
Тон
JOH
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10.5
4.1
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C
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I
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HC
=
||
HN
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✔
|||
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HC
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|
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Chapter 11 HW - General, Organic, and Biological Chemistry for Health
Resources
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Hint
Guanine is a component of several important molecules, including guanosine triphosphate (GTP), DNA, and RNA.
NOV
Draw guanine.
G
Select
Draw
c
Templates
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C
H
N
0
21
M
MacBook Pro
114
2Q
ZA
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I am given an unknown sample of an amino acid. On testing it, I find that its R group is polar, and contains a benzene ring.
is most likely to be:
O a. Threonine
O b. Phenylalanine
O c. Tryptophan
O d. Histidine
e. Tyrosine
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Combine these amino acids into a tripeptide. Add or remove atoms and bonds as needed.
Select
Draw
Rings
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/ | C NOHS
C-O - H
0 - H
O - H
N - C - H
H
CH,
H,C
CH,
он
SH
HC
CH
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a.
Но
Но-с.
CH2-CH-NH-CH3 HCI
ČH3
b.
3. Draw the tripeptide Alanine-Cysteine-Serine (R1 = CH3, R2
conditions and list the net charge on the protein (5 pts. e.
%3D
a. pH = 1
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264&OpenVellumHMAC-ab644d9786730f4650f72638a7bd0c5b#10001
Part A
L
Submit
Part B
amylose and cellulose
Match the words in the left column to the appropriate blanks in the sentences on the right.
a(16)
B(14)
a(12)
glucose
a(14)
8(16)
a(12)
fructose
Request Answer
a(12)
B(16)
a(16)
cellulose and chitin
Match the words in the left column to the appropriate blanks in the sentences on the right.
glucose
-tose also called - Search
a(14)
fructose
8(14)
B(12)
x | +
Both are linear polysaccharides composed entirely of
linkages are
but in cellulose, they are
Both are linear polysaccharides with
2
DELL
and chitin contains N-acetylglucosamine.
A Q
Reset Help
In amylose, the glycosidic
glycosidic bonds. Cellulose contains
he
Reset Help
ENG
Constants
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Are the following compounds correct?
These may OR may not be correct. Please look them over and correct them if needed.
Please list what was done incorrectly if applicable.
b. Is the following nucleotide found in DNA?
If not, please explain why.
a. Is the following tRNA correct?
NH2
VALINE
<5'
N.
3'-
0=P-0-CH2
ОН ОН
GUA
a.
b.
DNAA=T
G=C
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7. Consider the DNA segment
5'-ATGACGGTTTACTAAGCC-3'
a. Write the mRNA sequence that would be obtained from the segment's
transcription.
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- Combine these amino acids into a tripeptide. Add or remove atoms and bonds as needed. Select Draw Rings More Erase / // III 0 H Atom Decorati || H H H H H C H 1 1 XXXF О-Н N H N C H H H CH₂ H₂C H CH₂ H₂C CH₂ 4 → 0arrow_forwardWhat is the sequence of bases in the strand of DNA that is complementary to the one shown below? 5'-T-T-C-A-C-T-G-A-C-3' a.) 3'-U-U-C-A-C-U-G-A-C-5' b.) 3'-T-T-C-A-C-T-G-A-C-5' c.) 5'-A-A-G-T-G-A-C-T-G-3' d.) 5'-T-T-C-A-C-T-G-A-C-3' e.) 5'-G-T-C-A-G-T-G-A-A-3'arrow_forwardThe image below is a positive ionization mode Electrospray ionization-mass spectrum (ESI-MS) spectrum for a peptide, obtained using high resolution mass spectrometry. If ionization occurs through de-protonation (assuming that the mass of a proton is 1.0 Da), what is the neutral mass of the peptide, in Da? 627.271 627.604 100 - 627.938 628.271 628.604 628.938 627.0 628.0 629.0 mass/charge (m/z) 628.271 626.271 1878.813 O 1884.813 % Abundancearrow_forward
- ▼ ▼ Part C -A-G-G-C-A-A-T-C-G- Express your answer as a sequence of nucleotides separated by dashes and start and end your answer with a dash (e.g., -A-T-C-...-G-). Submit Request Answer Part D -A-C-T-G-G-C-A-G-T-C-A-G- Express your answer as a sequence of nucleotides separated by dashes and start and end your answer with a dash (e.g., -A-T-C-...-G-). Submit Request Answerarrow_forwardNow consider alanine as a whole molecule. Use the Ką values from the chart above, and your understanding of the relationships among Ka, pH, and charge to select the predominant form of alanine that would exist at the pH indicated below. Note some choices may not be used and some may be used more than once. A. NH3+ D. NH3+ pH = 1 pH = 5 CH3 I C H pH = 7 pH = 11 CH3 + C I H COOH COO- B. NH₂ E. CH₂™ *-+000 COO- H NH₂ A D CH3 + C | H с COOH C. NH₂ F. NH₂ CH3 + - C H CH₂™ H COO- COOHarrow_forward- - Biological Macromolecules Understanding the structure of nucleic acid strands For each molecule in the table below: Classify the molecule using the drop-down list. • Name the base at the 3' end of the molecule. • Write the standard base sequence abbreviation for the molecule. . ¹0-2-0 Explanation O=P-O-CH₂,0 O=P-O-CH₂, 0. 1- -0-2-0 O OH O=P-O-CH₂_0. 0 Check N. N N. N O=P-O-CH₂ 0. O NH₂ NH₂ N OH OH 0= N. NH N O O=P-O-CH₂,0. N. (1 N NH₂ 'N N N NH NH₂ Type of molecule: (Choose one) Name of base at 3' end: 0 Base sequence abbreviation: 0 Type of molecule: (Choose one) Name of base at 3' end: 0 Base sequence abbreviation: MacBook Pro C X © 2023 McGraw Hill LLarrow_forward
- Given an amino acid sequence, you perform a FASTA search and find one match with an E value of le What does the low E value tell you about the the homology between the matched (target) sequence and the query sequence? The two sequences are not homologs. Both sequences are from the same organism. An E value of this size does not prove or disprove homology. The two sequences are homologs.arrow_forwardModify isoleucine to show its structure at pH 1 and pH 13. Modify each amino acid by adding or removing atoms or bonds and by adding charges where appropriate. pH 1 pH 13 Select Draw Rings More Erase Rings More Erase C с CH₂ CH₂ H₂C H₂N 1 CH₂ | CH | 310 H― H O H N U с || 0 H Q2 Q Select Draw / ||| ||| H₂C H₂N 1 CH₂ | CH - n O H | H 0 H N H Q2Qarrow_forwardIdentify the following antioxidant molecule. A. alpha tocopherol hydroquinone B.dehydroascorbic acid CH3 CH3 C. alpha tocopherol semi-quinone \ / D. ascorbic acid C = C E. alpha tocopherol quinone / \ : O – C C – O CH3 \\ // \ / C – C C – CH2–CH2–CH2–CH–CH2–CH2–CH2–CH–CH2–CH2–CH2–CH–CH3 / \ / l l l CH3 CH2– CH2…arrow_forward
- Deoxyadenosine monophosphate (dAMP) and guanosine monophosphate (GMP) are nucleotides. The similarities between dAMP and GMP are that they both have? -an alpha (central) carbon.-the same R group.-a phosphate group.-a pentose (5 sided) sugar-an amino group-a nitrogenous base.arrow_forwardThe following diagram represents a(n): Top view 8 Side view Ο Ομισο 1000 CO O CO 100 OP CO POR an a-helix. an antiparallel a-helix. a parallel a-helix. an antiparallel b-pleated sheet. a parallel b-pleated sheet. 1080 O 22 Q O=Oarrow_forwarda) Find the pH of 0.1 M of the different form of histidine species. (See image attached) b) What is the principal species at pH , pH 5, pH 8, and pH 11?arrow_forward
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