DNA damage and repair
Study questions
.1. Which of the following statements about DNA polymerase I are correct?
a) It adds deoxyribonucleotide units to the 3'-hydroxyl of a primer.
b) It uses the template strand to select the deoxyribonucleotide unit to add to the growing DNA chain.
c) It contains a 3' Õ 5' nuclease that cleaves phosphodiester bonds to yield 3'-dNMPs and 3'-phosphate-terminated DNA. yield 3'-OH ends.
d) It contains two nuclease activities in the same polypeptide chain that contains the polymerase active site.
e) It can be cleaved with a protease into two fragments, each of which has a nuclease activity.
ANSWER:
A, B, D and ECOMMENTS: Unlike DNA pol I, DNA pol III does not have a 5'-exonuclease and so, it cannot remove RNA primers from Okazaki fragments. However, pol III is more "processive" than pol I and can interact with other proteins necessary for replication, like the primase-helicase complex.
2. What property of DNA allows the repair of some residues damaged through the action of mutagens?
3. How does the repair machinery of E. coli identify a DNA strand that has recently misincorporated a noncomplementary nucleotide during replication in order to repair it?
Through the methyl-directed mismatch repair pathway.4. Which of the following enzymes or processes can be involved in repairing DNA in E. coli damaged by uv light-induced formation of a thymine dimer?
a) DNA ligase seals the newly synthesized strand to undamaged DNA to form the intact molecule.
b) The UvrABC enzyme (excinuclease) hydrolyzes phosphodiester bonds on both sides of the thymine dimer.
c) DNA polymerase I fills in the gap created by the removal of the oligonucleotide bearing the thymine dimer.
d) The UvrABC enzyme recognizes a distortion in the DNA helix caused by the thymine dimer.
e) A photoreactivating enzyme absorbs light and cleaves the thymine dimer to re-form two adjacent thymine residues.
ANSWER:
All are correct.COMMENTS: I did not discuss photoreactivation in the lectures. This enzyme system, which is highly efficient in bacteria, appears to be absent in humans, except perhaps for some lymphocytes. Presence of efficient DNA repair in lymphocytes may help diminish harmful effects on the immune system by the Sun's UV-A rays (325-400nM), which penetrate deeper into the skin layers than do UV-B and UV-C rays (<325nM).
5. Humans suffering from xeroderma pigmentosum develop skin cancers when exposed to sunlight because they have a deficiency in:
a) RNA primase.
b) DNA recombinase.
c) Uracil-N-glycosylase.
d) an enzyme of the excision repair pathway.
e) an enzyme of the mismatch-repair pathway.
ANSWER:
DCOMMENTS:
Refer to Dr. Hamori's lectures.6. If cytosine in DNA is deaminated, the resulting base is:
a) removed by an endonuclease.
b) removed by an exonuclease.
c) removed by a glycosylase.
d) not removed, but ultimately replaced by cytosine in subsequent DNA replication.
e) not recognized by DNA polymerase, and thus inhibits DNA replication.
ANSWER:
C; specifically the Uracil-N-glycosylase.
7. Eukaryotic DNA is highly methylated at the C-5 position of cytosine. The degree of methylation is inversely correlated with gene expression. Although the exact role of C-5 methylation in gene expression is not known, it is known that these C-5-methylated cytosines are a source of mutations. Explain why
ANSWER: Because when 5-methylcytosine is oxidatively deaminated it becomes "thymine", which is a normal DNA base and is not removed by glycosylases. It ultimately leads to transition (C to T) mutations (Me-C:G®
T:G ® T:A & C:G)8. Explain the principles of error-prone DNA repair.
ANSWER: Error-prone repair happens when "untemplated" DNA synthesis proceeds over a modified or an abasic (lost base) nucleotide position in the DNA. Two-thirds of the time, the mutations are transversions.
DNA rearrangements: recombination and transposition.
Study questions.
1. Which of the following statements about genetic rearrangements are correct?
a) They generate new combinations of genes.
b) They can move a segment of DNA from one chromosome to another.
c) They are mediated by the breakage of DNA and the rejoining of the resulting fragments.
d) They generate genome sequence variability, upon which natural selection can act.
e) They can regulate gene expression.
f) They always require extensive regions of sequence similarity between the interacting DNA molecules in order to juxtapose the recombining sites.
ANSWERS:
A, B, C, D and E2. Outline the steps of homologous recombination.
See page 802 in Textbook. You need to know the type of enzymes and other activities used by this process.
3. Which of the following statements about the RecA protein of E. coli are correct?
a) It is an ATP-dependent nuclease that generates single-stranded DNA.
b) It catalyzes an ATP-dependent strand-assimilation reaction in which a single-stranded DNA molecule associates with duplex DNA.
c) It hydrolyzes ATP to promote branch migration.
d) It binds to single-stranded DNA to form a filament.
e) It facilitates the search of duplex DNA for regions with sequence similarity to the invading single-stranded DNA.
ANSWER:
B, C, D and E4. Homologous recombination is likely to require which of the following?
a) DnaB protein g) Single-strand binding protein
b) Topoisomerase I h) DNA-dependent RNA polymerase
c) RecA protein i) DNA polymerase I
d) RecBCD complex j) DNA ligase
e) ATP k) dATP
f) NAD+
ANSWER:
B, C, D, E,, G, I, J, KHomologous recombination involves some DNA repair synthesis between recombining molecules. It also requires some of the proteins used for DNA replication; DnaB is not required.
5. Which of the following statements about transposons is/are correct?
a) They contain insertion sequences.
b) They contain inverted terminal repeat sequences.
c) They contain one or more genes specifying one or more enzymes that catalyze the transposition event.
d) They require the rec gene products to complete their movements between or within genomes.
e) They can lead to the duplication of short sequences of DNA in the recipient genome.
On either side of the inserted DNA.ANSWER:
A, B, C, and E6. The kind of recombination displayed by transposons, sometimes called RecA-independent recombination, has been regarded as having far greater evolutionary significance than does general, or RecA-dependent, recombination. Why do you think this may be the case?
ANSWER:
Because it allows horizontal transfer of new gene cassettes, e.g., groups of genes that specify drug resistance. Homologous recombination has more of a role in reassorting the alleles of homologous genes in pairs of chromosomes.7. Describe four different ways by which an antibiotic resistance gene might move from the genomic DNA of one bacterium to the genomic DNA of another.
ANSWER:
Through plasmid transfer in bacterial conjugation, through a virus by transduction, through uptake of DNA released from dying cells (i.e., transformation) or through transposition (which may be associated with conjugation, transduction or transformation).True or false?
8. Transposases recognize sufficiently extensive sequences surrounding the integration sites so that the transposon avoids becoming integrated into the middle of a gene, because gene disruption could be lethal to the cell. False. Transposons often do integrate inside and inactivate ("knock out") genes.
9. The capacity of plasmids to replicate indefinitely without being part of a host chromosome distinguishes them from transposable elements.
TrueMutation
Study questions.
1. Which of the following nucleotide substitutions are transition mutations?
a) G for A
b) A for C (transversion)
c) C for T
d) T for G
(transversion)ANSWER:
A and C2. Which of the substitutions in question 1 are transversion mutations?
ANSWER:
B and D
3. Explain why most nucleotides that have been misincorporated during DNA synthesis in E. coli do not lead to mutant progeny. Because the editing 3'-exonuclease of DNA polymerase III and the Methyl-directed Mismatch Repair pathway usually remove the errors allowing for replacements with correctly incorporated nucleotides.
4. Match the type of mutation or physiologic consequence in the right column with the appropriate mutagen in the left column. You may have to look up some of these compounds.
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a) 5-Bromouracil 2 |
1) Transversion. |
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b) 2-Aminopurine 2 |
2) Transition. |
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c) Hydroxylamine 2 |
3) Insertion or deletion. |
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d) Acridines 4,5_____________ |
4) Translational frameshift. |
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e) Nitrous acid ________ 2 _______ |
5) Block in replication. |
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f) Benza[a]pyrine________ 4,5_ ____ |
EXPLANATION:
Base analogues like 5-bromouracil and 2-aminopurine incorporate in place of the corresponding natural bases and usually lead to transition mutations when they exist in ionized or native states that have different base-pairing properties from the natural bases. The compound 5-bromouracil incorporates quite well in place of thymine (5-methyluracil) and codes like thymine (i.e., pairs with adenine in subsequent replications); however, it occurs in the ionized form more frequently than does thymine and as a consequence, creates more A:C and C:A misincorporations during replication. On subsequent replications these misincorporations become transition mutations. The compound 2-aminopurine incorporates, at low levels, in place of either guanine (2-amino, 6-oxypurine) and is a highly potent mutagen that causes G:T and T:G misicorporations, i.e., it is also a transition mutagen.Compounds that lead to oxidative deaminations (by reacting with the NH2 groups of A, G, and C) lead to transitions. They in essence change the H-donating side group (NH2) to an H-accepting (=O) side group. Hydroxylamine and nitrous acid belong to this category of mutagens.
Planar, hydrophobic compounds like acridine dyes and Benza[a]pyrine intercallate in DNA and increase the frequency of slippage during DNA replication, thus leading to frameshift mutations. The presence of an intercalating agent in double-stranded DNA may cause a pause in replication that is usually overcome through the slippage that leads to frameshifts.
5. If DNA polymerase makes a mistake, thus creating a pair of incorrectly hydrogen-bonded bases, the mistake is corrected by a special
6. The enzyme that replicates the HIV virus is
____reverse transcriptase, which lacks 3'-exonuclease (editing)_activity and consequently has poor fidelity.7. When acting on replicated DNA, the methyl-directed mismatch repair system in E. coli can distinguish the parental strand from the progeny strand as long as one or both are methylated, but not if both strands are unmethylated. True or false?
ANSWER:
False. This repair system works only if one strand is methylated at GATC sequences and the other not.8. Which of the following statements about DNA polymerase III holoenzyme from E. coli are correct?
a) It elongates a growing DNA chain approximately 100 times faster than does DNA polymerase I.
b) It associates with the parental template, adds a few nucleotides to the growing chain, and then dissociates before initiating another synthesis cycle.
This is what happens with a nonprocessive enzyme.c) It maintains a high fidelity of replication, in part by acting in conjunction with a subunit containing a 3' Õ 5' exonuclease activity.
d) When replicating DNA, it is a molecular assembly composed of at least 10 different kinds of subunits.
The b subunits (b dimer) of this assembly is the "processivity" factor.ANSWER: A, C and D
Transcription
Study Problems.
1. There are several terms that you should add to your vocabulary. These are listed on pp. 842-843 of Voet & Voet. Make sure you do not miss any.
2. Work "Study Exercises" 1-7 at the end of Chapter 25 in Voet & Voet.
3. Work "Problems" 1,3, and 5 at the end of Chapter 25 before you look up the answers at the end of the book.
4. The sigma subunit of E. coli RNA polymerase
A. is part of the core enzyme.
B. binds the antibiotic rifampicin. It is the b subunit that binds rifampicin.
C. is inhibited by a -amanitin
This toxin from certain mushrooms inhibits eukaryotic RNA polymerase IID. must be present for transcription to occur. s
is only required for initiation.E. allows the enzyme to recognize promoter sites.
ANSWER: E
5. Mark which statements are "True" and which are "False" about the primary transcript in prokaryotes.
A. It contains a triphosphate group at its 5' end.
B. It contains an AU-rich sequence complementary to the -10 region of the DNA template.
C. It sometimes contains information for more than one protein.
For example, polycistronic mRNAD. It sometimes contains a hairpin structure near its 3'-end.
Some mRNAs are terminated distally to such hairpin structures.E. It usually contains a poly A sequence near its 3'-end.
No; however, polyA has been detected in many prokaryotic mRNAs and its possible physiological significance is currently under evaluation.ANSWER: A, C and D are true; B and E are false, except as noted.
6. Use the statements in question 5 to mark those that are false about the primary transcript in eukaryotes.
ANSWERS: A = True; B= False; C= False (few exceptions); D= False (usually); E= True
REGULATION OF GENE EXPRESSION
1. Name several processes by which the level of protein may be regulated in the cell.
a. DNA replication, which can increase the number of genes for the protein. This type of regulation is important for the growth of viruses. It is also seen in cases of "gene amplification".
b. Transcription. Particularly the regulation of mRNA synthesis at the level of initiation, i.e., as emphasized in class
c. Post-transcriptional mRNA processing, including conversion of intron containing mRNA to translatable RNA in eukaryotes. In addition, mRNA degradation, conversion of polysistronic mRNA to individual cistronic mRNAs, and other events like "RNA editing". Some of these mechanisms will be discussed in other courses.
d. Translation. Translation, like transcription, may be regulated at the levels of initiation and elongation. RNA-binding proteins are an important component of gene regulatory circuits in biology. Many examples of translational regulation will be discussed in the Second Semester course, Biochemistry 718.
e. Post-translational protein modification. Processing of "polyproteins" by proteases, cleavage of proteins (such as removal of "signal" peptides), phosphorylation-dephosphorlation, acetylation and addition of carbohydrate and lipid moieties are some of the mechanisms that affect function and stability of the protein products of genes.
2. Give examples of cis-acting and trans-acting genetic factors that control transcription.
cis-acting: promoter, operator, silencer, enhancer
trans-acting: RNA polymerase, repressor, TATA Binding Protein, TFIIB, CAP, tryptophan.
You should be able to predict the effects of such factors.
3. Explain regulation of the E. coli lactose operon.
You should be able to summarize the steps involved in activation and repression of this operon in response to the type of sugar available as carbon source for this organism.
4. Explain how the levels of tryptophan in an E. coli cell regulate the biosynthesis of this amino acid through transcriptional repression-derepression and attenuation of the tryptohan operon.
You should be able to describe regulation of this operon and contrast it to regulation of the lactose operon. Is attenuation a cis- or trans-acting phenomenon?
5. Explain the probable roles of the (a) major and minor grooves, (b) side-groups of the purine-pyrimidine base pairs, and (c) deoxyribose phosphate backbone of the promoter in binding the E. coli RNA polymerase (RNAP) holoenzyme.
Unlike "core" RNAP, which binds DNA independently of the nucleotide sequence, the holoenzyme binds a specific sequence. Sequence-independent binding DNA binding proteins like RNAP core and the histones of eukaryotes generally use electrostatic interactions with the phosphate backbone. Histones are bound to the minor-groove side of DNA thus, permitting access to the major groove by sequence-specific DNA binding proteins. E. coli RNAP holoenzyme is able to recognize promoter sequences through the DNA major groove. Sequence-specificity is imparted to the RNAP by sigma factors. The s 70 factor binds the -35/-10 region of the promoter only if it is associated with RNAP core. In E. coli there are different s factors for different classes of promoters. Some s factors bind promoter DNA sequences independently of their association with RNAP and, under the appropriate conditions, recruit RNAP to promoters, i.e., in an analogous fashion to the recruitment of eukaryotic RNAP to promoters by transcription factors. Geometry of the base-pair side-groups in the major groove determine specificity to the amino-acid sequence of the DNA-binding domain of the protein. Many proteins interact with their specific DNA sequences through hydrogen bonding between the side groups of the DNA base-pairs and side-groups of the amino-acid sequence that binds the DNA. The a -helix of proteins fits just right in the major groove of B-form DNA and is often the bearer of the nucleotide-sequence-specificity of the protein. You should be familiar with the different protein "motifs" that have been implicated in recognition of specific DNA sequences in prokaryotes and eukaryotes.