Exam 2 Flashcards
(66 cards)
1
Q
What are the precursors for purine synthesis?
A
a. Aspartate: supplies one nitrogen
b. Glutamine: supplies two nitrogens
c. Glycine: supplies the 2 carbons and a nitrogen
d. CO2: supplies a carbon
e. Formate (N10 formyl THF): supplies two carbons
2
Q
What are the three general steps of purine synthesis?
A
a. Activate ribose, synthesize 5 membered ring, synthesize 6 membered ring
3
Q
When is ribose added?
A
a. The first step in de novo purine synthesis is the activation of ribose by the
addition of pyrophosphate to ribose to form PRPP.
4
Q
How is ribose activated and added?
A
a. Ribose is activated by pyrophosphate addition to form PRPP. The rest of the
purines are built off of ribose by the addition of C and N atoms. ADP and GDP
inhibit this reaction.
5
Q
What is special about PRPP in this reaction?
A
a. PRPP activates the second reaction in a process called feed forward.
6
Q
What is the first purine made?
A
IMP
7
Q
How are the various forms of THF made and interconverted?
A
a. NADPH oxidations and reductions along with changes to a certain carbon group.
Most oxidations occur during the conversion of serine to glycine.
8
Q
Describe the regulation of purine synthesis.
a. ) where are the committed steps?
b. ) how is the balance of AMP and GMP achieved?
c. ) how do folic acid antagonists impair purine synthesis?
A
a.) The first committed step is the addition of glutamine to PRPP to form
phosphoribosyl amine. All guanine and adenine nucleotides inhibit this
reaction.
b.) GTP is a substrate in AMP synthesis and ATP is a substrate in GMP
synthesis.
c.) Methotrexate inhibits purine synthesis and therefore cell division.
Sulfonamides inhibit the conversion of PABA to folic acid. Bacteria are
more susceptible to sulfonamides because they make folic acid.
9
Q
Describe formation of AMP and GMP from IMP.
A
AMP: The addition of aspartate and GTP to IMP creates AMP. AMP has nitrogen
where IMP has oxygen.
b. GMP: The addition of NAD, glutamine, and ATP creates GMP. GMP is the same
structure as IMP with an extra nitrogen.
10
Q
Describe the formation of nucleoside di- and triphosphates, adenylate (AMP) kinase and
ATP generation in muscle.
A
a. Nucleoside diphosphates: Adenylate kinase and guanylate kinase convert NMP
to NDP. ATP is the phosphate donor in both of these reactions. These reactions
are reversible.
b. Nucleotide triphosphates: Nucleoside diphosphate kinase works on any NDP to
convert it to NTP. Any nucleoside triphosphate can act as the phosphate donor.
These reactions are reversible.
c. ATP generation in muscles uses the adenylate kinase reaction: AMP + ATP ← →
2 ADP to synthesize ATP. This reaction can run in the direction of ATP synthesis.
11
Q
What is the role of PRPP?
A
PRPP donates a phosphoribosyl group to create AMP, GMP, and IMP. Reaction
is base + PRPP→ NMP
12
Q
What are the two major phosphoribosyl transferases
A
APRT and HGPRT
13
Q
Explain Lesch-Nyhan disease, gout and uric acid.
A
A complete deficiency in HGPRT causes Lesch-Nyhan disease. This disease
causes an accumulation of uric acid and leads to gouty arthritis. The gene is on
the X chromosome and only appears in males. Uric acid is accumulated due to
the inability to salvage purines, so purines accumulate. Side effects of this
disease are mental retardation, aggressive behavior, and self mutilation.
14
Q
Describe general nucleic acid degradation in the intestines
A
a. Nucleic acids are ingested and degraded to nucleotides by nucleases and
phosphodiesterases
b. Nucleotides are degraded to nucleosides by specific nucleotidases
c. The nucleosides are degraded by nucleosidases (to produce the base + ribose)
or by nucleoside phosphorylases (to produce the base + ribose-1-P).
d. The ribose can be used for energy generation, but the base cannot.
15
Q
Describe general intracellular purine catabolism
A
a. Nucleotidases convert nucleotides to nucleosides
b. Adenosine is converted to inosine by adenosine deaminase
c. The three purine nucleosides are converted by purine nucleoside phosphorylase
to the bases hypoxanthine (from inosine), xanthine, and guanine. The enzyme
adds a phosphate to ribose, which is released as ribose-1-P
16
Q
What is the first common intermediate in the degradation of all purines?
A
Xanthine
17
Q
What are the two reactions of xanthine oxidase?
A
a. It converts hypoxanthine to xanthine
b. It converts xanthine to uric acid (this produces hydrogen peroxide)
18
Q
What is the use of allopurinol?
A
Allopurinol inhibits xanthine oxidase, and therefore prevents uric acid
accumulation. This is used to treat gout.
19
Q
Describe SCID, adenosine deaminase deficiency, and resulting metabolic problems.
A
a. SCID is an immune disorder where B and T lymphocytes are not proliferated.
30% of SCID patients have an ADA deficiency. This deficiency leads to an
accumulation of adenosine and deoxyadenosine. Deoxyadenosine is then
phosphorylated by nucleoside kinase and converted into dATP. dATP is an
inhibitor to deoxynucleotide synthesis and therefore stops DNA synthesis and cell
division.
20
Q
What is the function of the purine nucleoside cycle?
A
a. Skeletal muscle uses this reaction to generate a citric acid cycle intermediate.
AMP is deaminated into IMP. IMP is then converted back to AMP by the last two
enzymes in AMP synthesis. The net result of this cycle is a GTP-dependent conversion of aspartate to fumarate and NH4+. Skeletal muscle uses this
mechanism to generate a citric acid cycle intermediate, since it lacks the normal anaplerotic enzymes that convert three-carbon intermediates to four-carbon
intermediates.
21
Q
Briefly describe the process of pyrimidine synthesis
A
a. The pyrimidine ring is synthesized before ribose is added. UMP is made first.
UMP is phosphorylated to UDP by nucleoside monophosphate kinase.UDP is
phosphorylated to UTP by nucleoside diphosphate kinase. CTP is made from
UTP by CTP synthetase, with glutamine as the nitrogen donor. Note the
conversion at the NTP level.
22
Q
What are the precursors?
A
Aspartate and Carbamoyl- P
23
Q
What is the first pyrimidine made?
A
UMP
24
Q
What is the first cytosine derivative made?
A
CTP
25
Describe regulation of pyrimidine synthesis
a. In bacteria, ATCase is feedback-inhibited by CTP, although in other bacteria UTP
feedback inhibits. ATP ( a signal of energy and purines) stimulates ATCase
activity. Therefore, pyrimidines are not made until there are sufficient purines.
Purines first.
b. The regulation in animals is slightly different. The first committed step is CPS-II,
and it is inhibited by UDP and UTP. ATP and PRPP are allosteric activators.
26
Describe metabolic channeling
a. The first three enzymes in mammals are part of one polypeptide, with three
active sites. The last two reactions are also catalyzed by a protein with two active
sites. The multifunctional proteins are more efficient since they prevent the
diffusion of the metabolites away from the complex, which speeds up the
reaction.
27
Describe ribonucleotide reductase
This enzyme replaces the 2’ hydroxyl with a hydride. The nucleotides are at the
diphosphate level. The reduction involves a free radical mechanism. The overall
process requires three proteins: ribonucleotide reductase, thioredoxin, and
thioredoxin reductase.
28
What is the source of reducing power in deoxyribonucleotide synthesis?
NADPH provides the reducing power, which reduces components of thioredoxin
reductase, which reduces thioredoxin, and finally the -SH groups of
ribonucleotide reductase
29
Describe thioredoxin
a. Thioredoxin is a small protein with reactive cysteine sulfhydryl groups
b. Thioredoxin is involved in many sulfide: sulfhydryl transitions
30
What are the substrates of thioredoxin?
ADP, GDP, CDP, and UDP
31
Describe binding of the substrates to the catalytic site
a. Ribonucleotide reductase is an a2b2 enzyme. The alpha subunit carries two types
of regulatory sites and the catalytic site.
b. The catalytic site binds the substrates
32
What are the two regulatory sites?
a. One regulatory site is the substrate specificity site. It binds ATP, dATP, dGTP, and
dTTP, and determines which substrate binds to the catalytic site.
b. The second regulatory site binds either ATP, which activates activity or dATP
which inhibits total activity
33
What is the precursor?
a. dUMP is the precursor for dTMP.
b. dUDP, the product of ribonucleotide reductase, is first converted to dUTP, which
is then cleaved to dUMP by dUTPase. This strange route utilizes an enzyme that
prevents dUTP incorporation into DNA. dUTP concentration is never very high.
c. dCDP is dephosphorylated to dCMP, and then deaminated by dCMP deaminase
which results in dUMP synthesis. Note that dTTP inhibits this reaction
d. The next step is the conversion of dUMP to dTMP by thymidylate synthase
34
What are the THF derivatives (substrate and products)?
The reaction involves a reductive methylation, with N5
,N10-methylene-THF. The
THF cofactor is itself oxidized, to yield dihydrofolate (DHF). A separate enzyme
reduces DHF to THF, which can then accept another C1 unit in the conversion of
serine to glycine.
35
What is the unusual reaction of a THF derivative?
N5N10 methylene THF oxidized to DHF
36
Why is this enzyme a target of inhibitors of DNA synthesis and how do folic acid analogs
block synthesis?
a. This enzyme is a target of inhibitors of DNA synthesis because without it, thymine
deoxyribonucleotides cannot be made. Folic acid analogs block synthesis by
blocking the reaction where THF is oxidized to DHF.
37
How do tautomerizations affect base pairing?
The shift in structure determines whether the nitrogen or oxygen is a
hydrogen-bond acceptor or donor. The oxygens and nitrogens determine
base-pairing properties. The rare form base-pairs incorrectly, forming A:C and
G:T base pairs. Common forms are keto for G and T, and amino for A and C.
38
What bonds in nucleotides and nucleic acids are unstable and under what conditions?
The purine glycosidic linkage in acidic conditions is very unstable.
39
Describe tautomerization mutations
a. This property can be used to induce mutations. Bromouracil is a base analog that
can be incorporated into DNA. It tautomerizes easily, and BrdU, which should
normally base pair with A, frequently base pairs with G.
b. Inosine also readily interconverts between tautomeric forms. This is why it is
used in tRNA anticodons in the third position, and accounts for why one tRNA
can read more than one codon.
40
Know the difference between bases, nucleosides, nucleotides, and their names.
a. Bases: adenine, guanine, cytosine, thymine, uracil, and hypoxanthine
b. Nucleosides (base + sugar): To form the nucleoside names “osine” is added to
the purines and “idine” is added to the pyrimidines. Ex: adenosine, guanosine,
cytidine, uridine, thymidine, and inosine.
41
The glycosidic linkage and its stability
a. The glycosidic linkage is the link between the sugar and the base. Linked through
the nitrogen of the base and the C1 carbon. Purines: glycosidic linkage is to 5
membered ring. The carbon atom involved in the glycosidic linkage is called the
anomeric carbon. The linkage is an N-glycosidic linkage because the anomeric
carbon is attached to the N-1 of the pyrimidine or the N-9 of the purine. The
linkage is in the b-configuration (above the ring) for all nucleosides and
nucleotides. There is rotation around glycosidic linkage.
b. Stability: stable and alkaline (not true in acid). In acid, purine glycosidic linkage is
unstable.
42
Anti and syn configurations: what structure is not allowed and why?
a. The syn configuration has the base over the sugar, while in the anti conformation
the base is not over the sugar and the nucleoside is fully extended. Pyrimidines
favor the anti configuration because the oxygen of the pyrimidine gets in the way.
Purines can adopt either the syn or anti configuration.
43
The biological function of adenosine, effects of caffeine.
a. Adenosine acts as a local hormone. Adenosine circulates in the bloodstream and
influences blood vessel dilation, smooth muscle contraction, neuronal discharge,
neurotransmitter release, and fat metabolism. Hard-working muscle cells release
adenosine, which dilates blood vessels and increases blood flow, O2 delivery, and
nutrient delivery. Adenosine also regulates heartbeats, by blocking the electrical
current that controls the heart, which slows heartbeats. Adenosine is also
involved in sleep regulation. After long periods of wakefulness, brain adenosine
rises, and promotes sleepiness. Elevated levels of adenosine induces
sleepiness. Caffeine counteracts this effect by blocking the interaction of
adenosine with the neuronal receptors.
44
Conventions for writing polynucleotides sequences and structures.
a. Nucleic acids are polynucleotides. They are linear polymers formed by
successive addition of 5’-nucleoside monophosphates to the 3’-OH of the
preceding nucleotide.
45
Classes of nucleic acids and their properties. What are unique structures or components
of each type of RNA?
a. DNA
i. Right handed antiparallel double helix
ii. Eukaryotic chromosomes are condensed with histones that contain
positively charged arginine and lysine
b. rRNA
i. Most abundant RNA
ii. Highly folded by hydrogen bonds
iii. Subunits of ribosomes are held together by Mg+
c. tRNA
i. Highly folded due to hydrogen bonding
ii. Carriers of amino acids for protein synthesis
iii. The amino acid is attached to the 3’ end of the tRNA. The sequence of
the end is CCA, and the amino acid is added to the 3-OH of the terminal
adenine residue
d. mRNA
i. Made from one strand of DNA
ii. Poly A tail in eukaryotes adds stability
iii. One mRNA in prokaryotes may contain several proteins
e. snRNA
i. Abundant in the nucleus of eukaryotes
ii. Plays a role in splicing
f. siRNA
i. RNA interference, gene silencing, RNA degradation, and chromatin
structure
g. miRNA
i. Control developmental timing
46
Nucleic acid stability in acid and alkali.
DNA is unstable in acid, but stable in alkaline conditions. RNA is stable in acid,
but unstable in alkaline conditions.
47
Describe nucleic acid hydrolysis
a. Hydrolysis typically breaks the phosphodiester backbone.
b. Acid hydrolysis: 1 mM HCl hydrolyzes the purine glycosidic linkage in DNA, but
not RNA. The phosphodiester bond is intact, but a gap exists in the DNA.
c. Alkaline hydrolysis: DNA is stable. RNA is not. The phosphate may end up on the
2’ or the 3’ carbon of ribose.
48
Describe nuclease activity
a. Nuclease specificity
i. Some nucleases can be specific to DNA or RNA or both
ii. Some are specific to single or double stranded
iii. Endo/ Exo nucleases
iv. Some nucleases bind to a specific sequence, such as restriction
enzymes.
1. Restriction enzymes are specific endonucleases.
2. The restriction enzymes are classified into three groups, based on
how they bind and where they cut. Type II are the most useful,
they bind where they cut. The type II enzymes do not require ATP.
They cut at symmetric sites. They can leave staggered ends which
can be resealed. Some leave 5’ overhangs, others 3’ overhangs.
Others leave blunt ends. Restriction sites can be used as physical
markers on the DNA.
49
Describe the logic of nucleic acid sequencing by the Sanger method
a. All four deoxynucleotides are added in four reaction tubes. One of these
nucleotides is tagged. In each tube, a different 2’,3’-dideoxynucleotide is added.
Because the dideoxynucleotides lack a 3’-OH, synthesis stops when one is
incorporated. There is very little of the dideoxy, but when incorporated, a chain
terminates. This generates a series of oligonucleotides with different lengths,
which are separated by size
50
What are the structural differences between A-, B-, and Z-DNA. Do these occur inside
cells?
a. B-DNA: eukaryotic DNA
i. Anti-parallel due to H bonding
ii. Helical twist brings bases closer together
iii. The stability of the structure is based on a conspiracy of hydrogen bonds
in base pairs, hydrogen bonds between the hydrophilic exterior and water,
maximal separation of the phosphates (which allows their interaction with
Mg
++ or other cations), and hydrophobic interactions between base pairs.
iv. Major and minor grooves
v. The helical repeat (one twist) is 10 to 10.6 bp and 3.4 nm.
b. A- DNA: never found in cells
i. The pitch (one helical turn) is 2.46 nm and is 11 bp
ii. However, Dehydrated DNA, and DNA:RNA hydrids may assume this
conformation. Double-stranded RNAs assume the A conformation, since
the 2’-OH prevents formation of the B-form
c. Z-DNA
i. The G residues assume the syn rather than anti conformation (Fig 11.12).
To accommodate the change, C residues must flip, and it takes the sugar
with it (Fig 11.13). In the next base pair, the purine is in the anti
conformation.
ii. The structure requires alternating purines and pyrimidines (on one
strand). The sugar phosphate backbone zigzags with a left-handed
orientation. Methylation of C allows Z-DNA formation even if purines and
pyrimidines do not alternate
51
How can cruciforms form?
a. An inverted repeat (must be at least 6 bases) on a single strand can base pair
and form a cruciform. These structures may form in cells, and may be sites for
the binding of proteins. Cruciforms are stabilized by negative supercoiling
52
Describe Hoogsteen base pairs and triplexes
a. There are different ways of hydrogen-bonding. Normal hydrogen-bonding
involves interactions between two six-membered rings. However,
hydrogen-bonding can also involve the five-membered ring. Some alternate
bonding creates the Hoogsteen base pairs. One form is detected in crystals of A
and T. In slightly acidic conditions, C can be protonated, and H-bond differently,
and C can only have two H-bonds with G. These interactions leave atoms that
normally H-bond, free to form additional H-bonding, and allow triplex formation
b. H-DNA (H for protonation): Stretches of CT can bond with a run of GA on another
strand, forming a triple-stranded structure. This type of structure can regulate
some eukaryotic genes by repressing expression. Sharp DNA bends.
53
What are the factors that favor quadruplexes and where are they found?
```
Such structures are G-rich, cyclic, and have Hoogsteen base interactions.
Certains cations (K+, Na+, Ca++) favor these structures via the O6 carbonyl
group. A variety of these structures can form. They are involved in telomeres, Ig
gene rearrangements, in gene regulatory regions, and some human diseases.
The more G’s you have, the more likely you are to have a disease.
```
54
Factors that affect denaturation and renaturation.
a. Denaturation
i. A variety of conditions can destabilize hydrogen bonds and result in
strand separation: changes in pH, temperature, and ionic strength.
Anything that destabilizes H bonds can result in denaturation. Ethanol is a
denaturant. Urea is also a denaturant.
ii. The higher the GC content, the higher the melting point.
iii. As the strands denature, the absorbance at 260 nm increases by up to
40%. This is called the hyperchromic shift.
iv. Increasing ionic strength, increases stability
v. DNA readily denatures in low ionic strength solutions.
vi. At pH > 10, the bases are deprotonated, and H-bonding is disrupted.
Similarly, at pH < 2.3, the bases are protonated, which disrupts
H-bonding.
vii. Small solutes that readily form H bonds also denature DNA. Examples
include urea and formamide
b. Renaturation
i. Denatured DNA will renature if the disrupting agent is removed. The
strand must reassociate in a process called reannealing. Once the proper
register is found (this is slow)(called nucleation), the process can be very
rapid. The process is facilitated by a moderate temperature (just below
melting)(called zippering), which permits rapid testing of different
associations, until the right register is found.
55
Describe supercoils in DNA tertiary structure
a. Negative supercoils adds twists in the direction of unwinding, positive supercoils
add twists in the direction of winding
b. Negative supercoiling favors strand separation
c. Negatively supercoiled DNA is found in cells. It can arrange into a toroid, that is it
can wrap around proteins. This is important for chromosome structure and gene
expression
d. Supercoiled DNA is denser, and sediments faster during ultracentrifugation than
relaxed DNA which has lost supercoiling.
56
Describe topoisomerases
a. Topoisomerases change the linking number, and are important enzymes in
replication.
b. Linking number (L) is the number of times two strands are intertwined. This is a
constant for covalently closed circles. Assuming 10 bp per turn, a 400 bp duplex
has an L of 40. Linking number can be altered only by breaking and rejoining
strands
c. Topoisomerase I cuts one strand, lets the other strand pass through the break,
and reseals the DNA.
d. Topoisomeras
57
Describe gyrases
DNA gyrase adds negative supercoils, and is a type II topoisomerase
58
Nucleosomes and higher order structures of DNA
a. DNA must be condensed to fit inside the cell.
b. The first layer of condensation is nucleosome formation. Proteins of chromatin
(nucleoprotein complex) consists of histones and nonhistone proteins.
c. Histones are abundant, whereas the nonhistone proteins are not. The latter are
often regulators of gene expression.
d. Histones have a net positive charge and contain arginine and lysine
e. The protein core of the nucleosome is called the octomeric core and contains
H2A, H2B, H3, and H4
f. 146 bases are wrapped around the core in 1.65 turns. There are 40 to 60 bp
between nucleosomes. Histone H1 helps in organizing DNA into the
nucleosomes and between nucleosomes.
59
Structures of tRNA and rRNA: major features.
a. tRNA
i. The structure is called a clover leaf with four double-stranded regions.
ii. The acceptor stem is where the amino acid is linked. The amino acid is
added to the 3-OH of a terminal A. The 3’ end of all tRNAs is CCA.
iii. The D loop contains dihydrouridine.
iv. The anticodon loop contains seven unpaired bases, and three of them
form the anticodon. At the 3’ end of the anticodon, a purine is always
present, and it is often alkylated (methylated, etc.). The 5’ end of the
anticodon is always a U.
v. A variable loop is after the anticodon loop, and the last loop is a TψC
loop. The ψ stands for pseudouridine. Ribosomes recognize the TψC
loop.
b. rRNA
i. The rRNAs have extensive base pairing (secondary structure). The 16S
rRNA from an archaebacterium, an eubacterium, and a eukaryotic have a
similar structure (Fig 11.39), but little sequence similarity. It appears that
the structure, and not the sequence, is important.
60
Features of DNA replication: semiconservative, bidirectional, role of helicases and
topoisomerases, semidiscontinuous synthesis, Okazaki fragments
a. Replication occurs bidirectionally meaning in two directions. This means there
are two replication forks. Replication begins at oriC in E. coli.
b. Positive supercoils are introduced in bacterial chromosomes to unwind the DNA.
DNA gyrase introduces negative supercoils. DnaB (a helicase), separates the
strands and SSB keeps the strands separated.
c. Topoisomerases break phosphodiester bonds and change the linking number.
d. DNA replication is semi- discontinuous. This leads to a leading and lagging
strand. It is semi- discontinuous because DNA is synthesized in the 5’ to 3’
direction. The leading strand is able to be synthesized during replication fork
movement. The lagging strand must wait until a sufficient amount of DNA has
been separated and then it can be replicated in the opposite of the replication
fork movement. The fragments made due to the lagging strand are called
Okazaki fragments. An RNA primer is needed for each fragment and DNA ligase
binds these strands together.
61
Properties of DNA polymerases: the various activities, core and holoenzymes, the
γ-complex and its function, the clamp loader and the sliding clamp, the differences
between DNP I and III and their functions.
a. DNA polymerases require a free 3’OH to build on. DNP’s cannot make the first
phosphodiester bond. DNP’s elongate in the 5’ to 3’ direction.
b. DNP I, II, and V function in DNA repair. DNA III functions in DNA replication.
c. DNP I has 3’-5’ exonuclease activity and 5’-3’ exonuclease activity. The first
removes nucloetides from the 3’ end of an elongating chain. This activity is weak
relative to the 5’-3’ polymerization. It removes improper nucleotides, since
polymerization cannot occur if the terminal base is not base-paired. The 5’-3’
exonuclease acts on double-stranded DNA to remove mispaired segments in
front of the replication fork. This enzyme probably also removes the primer that
initiates DNA synthesis.
d. The core enzyme is the simplest form that can polymerize DNA. It is made of 3
subunits (alpha, epsilon, and theta). The other 7 subunits increase DNP activity.
e. The holoenzyme is formed by two core subunits binding to a y- complex. This ycomplex is bound to DnaB. Then, each core subunit binds a beta dimer to create
the holoenzyme.
f. The y- complex assembles the holoenzyme complex onto DNA. The y- complex
is called the clamp loader.
g. The beta subunits are called the sliding clamp.They form a closed ring around
the DNA and hold the core polymerase to the DNA.The sliding clamp accounts
for the great processivity of the enzyme.
62
Summary of DNA Replication
a. DNA gyrase and DnaB unwind the DNA and SSB keeps the strands apart
b. Primase (DnaG) makes a primer off of the lagging strand
c. Each DNP moves in the 5’ to 3’ synthetic direction
d. The sliding clamp is removed from the lagging strand and reattaches when a new
okazaki fragment with primer is found
e. DNP I excises the primer, replaces it with DNA, and DNA ligase seals the nicks.
This sealing requires adjacent 3’ OH and 5’ phosphates
f. The end of DNA replication occurs at a terminus where the two replication forks
meet. The terminus contains 3-4 repeats of a short DNA sequence. One set of
repeats stops clockwise replication, another stops counterclockwise replication. A
protein bound to this region blocks the action of DnaB, which in turn blocks
forward progress of DNP III.
g. DNP III is IMMOBILE. The polymerase is attached to membranes in bacteria and
to the nuclear matrix in eukaryotes.
63
Telomere replication
a. Telomeres form protective caps at the ends of chromosomes. They consist of
short G rich sequences that are tandemly repeated
b. Vertebrate telomeres are TTAGGG
c. Telomerase is an RNA-dependent DNA polymerase that restores this missing
sequence. The 3’ end of the DNA is used as a primer to add the repeats
64
What are the causes of errors during DNA replication?
a. Point mutations: a single base change or the insertion or deletion of one or more
bases
b. Transition mutation: a point mutation where a purine is changed to the other
purine or a pyrimidine is changed to the other pyrimidine. This results in a change
of an A:T base pair to a G:C base pair etc.
c. Transversion: when a purine is substituted for a pyrimidine or vice versa. For
example, A to C would change a A:T base pair to a C:G base pair
```
d. Point mutations often arise due to errors in DNA replication. If bases are in the
incorrect tautomer form, they will pair with the wrong base. The proof reading
function of DNP III usually catches these mismatches
```
65
How can you induce mutations?
a. A base analog can be incorporated into DNA and can be designed to cause
mispairing
b. 5-bromouracil is a thymine analog. It tautomerizes frequently and can base pair
with G instead of A which induces a transition.
c. 2- aminopurine can also tautomerize. It can pair with C instead of T and induces
a transition.
d. Hypoxanthine can arise in situ by oxidative deamination of adenine.
Hypoxanthine base pairs with C, causing a transition.
e. Nitrous acid (HNO2
) deaminates A and C. C deamination creates uracil, which
changes a C:G pair to a U:A pair and eventually to a T:A pair
f. Hydroxylamine reacts with C, converting it to a derivative that base pairs with A
g. Alkylating agents add methyl or ethyl groups to bases, which alter their
H-bonding properties
h. Intercalating agents can cause frameshifts. These agents fit between base pairs,
and doubles the distance between two base pairs. A nucleotide is inserted during
replication. This can cause inactivation of the whole gene because of
inappropriate translation.
66
DNA repair mechanisms.
a. Photoreactivation: Only in bacteria. In the presence of UV, two thymines will form
a cyclobutyl ring. Photolyase fixes this dimer and requires light for activation.
b. NER: NER(nucleotide released first): major form of repair for thymine dimers.
UvrA (locates damage) and UvrB (a helicase) unwinds strands. UvrB makes a
cut on the 3’ end. UvrC makes a cut on the 5’ end. UvrD is a helicase that
removes the cut fragment. UvrB recruits DNA pol1 and DNA ligase reunites DNA.
c. Transcription coupled repair: Mutations do not accumulate in areas of DNA that
are transcribed. Thymine dimer forms and transcription stalls. Mfd protein recruits
repair enzymes and releases RNA pol.
d. BER(base released first): DNA glycosylase (enzyme) breaks glycosidic linkage
between base and ribose. AP site is left. Endonuclease recognizes the missing
base due to distortion in DNA. Exonuclease removes a few bases. DNA Pol 1
and DNA Ligase fix DNA. Many types of DNA glycosylases. BER fixes most of
oxidative lesions. Uracil is the most common messed up base. More uracils
leads to double stranded breaks by this repair mechanism.
e. Mismatch repair: Loss of this repair mechanism can lead to cancer. This is for
NEWLY synthesized DNA. Newly synthesized DNA is not methylated yet. If there
is a mismatch, the cell decides that the unmethylated strand has the error. Chunk
of DNA removed and DNA Pol 3 is used to fix DNA. Nick is the actual sign that
initiates repair. UvrD helicase is used to unwind.
i. 15% of colon cancers are caused by silencing of MLH1 promoter of one
of the genes in mismatch repair. Gene was silenced by methylation.
Leads to cancer because mismatch is not working.
ii. Cancer: treat with something that causes DNA damage and inhibits DNA
repair so cancer dies…. Localized radiation so it doesn’t damage all cells.
Also there are topoisomerase inhibitors
