Genetics - Chapter 11

Suppose that E. coli synthesizes DNA at a rate of 100,000 nucleotides per minute and takes 40 minutes to replicate it chromosome. (a) How many base pairs are present in the entire E. coli chromosome? (b) What is the physical length of the chromosome in its helical configuration - that is, what is the circumference of the chromosome if it were opened into a circle?

(a) 40 min of synthesis at a rate of 100,000 bases per min will yield a chain of about 4,000,000 bp (or 4x10^6 bp). (b) (4x10^6 bp) x 0.34 nm/bp = 1.36x10^6 nm or 1.4 mm

Predict the results of the experiment by Taylor, Woods, and Hughes if replication were (a) conservative and (b) dispersive.

(a) Both strands of the newly labeled DNA will go to one sister chromatid, and the other sister chromatid will consist entirely of the original, unlabeled DNA. (b) All of the newly labeled DNA will be interspersed with unlabeled DNA.

Define and indicate the significance of (a) Okazaki fragments, (b) DNA ligase, and (c) primer RNA during DNA replication.

(a) Okazaki fragments are relatively short DNA fragments that are synthesized in a discontinuous fashion on the lagging strand template during DNA replication. Such fragments appear to be necessary because template DNA is not available for 5' to 3' synthesis until some degree of continuous DNA synthesis occurs on the leading strand template in the direction of the replication fork. The isolation of such fragments provides support for the scheme of replication. (b) DNA ligase is required to form phosphodiester linkages in nicks, which are generated when DNA polymerase I removes RNA primer and meets newly synthesized DNA ahead of it. The discontinuous DNA strands are ligated together into a single continuous strand. (c) Primer RNA is formed by primate to serve as an initiation point for the production of DNA strands on a DNA template. None of the DNA polymerases are capable of initiating synthesis without a free 3' -hydroxyl group. The primer RNA provides that group and thus can be used by DNA polymerase III.

Many of the gene products involved in DNA synthesis were initially defined by studying mutant E. coli strains that could not synthesize DNA. (a) The danE gene encodes the alpha subunit of DNA polymerase III. What effect is expected from a mutation in this gene? How could the mutant strain be maintained? (b) the dnaQ gene encodes the epsilon subunit of DNA polymerase. What effect is expected from a mutation in this gene?

(a) The alpha subunit, part of the core enzyme, is responsible for DNA synthesis along the template strand. Without this subunit, the cell would not be able to replicate its chromosome. Strains that are mutant for this enzyme must contain conditional mutations, and growth can be achieved athletes the permissive condition. (b) The epsilon subunit, also part of the core enzyme, is responsible for the 3' to 5' exonuclease activity involved in proofreading. Proofreading would be hampered in such mutant strains and a higher than expected mutation rate would occur.

Several temperature-sensitive mutant strains of E. coli display the following characteristics. Predict what enzyme or function is being affected by each mutation. (a) Newly synthesized DNA contains many mismatched base pairs. (b) Okazaki fragments accumulate, and DNA synthesis is never completed. (c) No initiation occurs. (d) Synthesis is very slow. (e) Supercoiled strands remain after replication, which is never completed.

(a) no repair from DNA polymerase I and/or DNA polymerase III (b) no DNA polymerase I or DNA ligase activity (c) no primate activity; other possibilities are no Dana protein or faulty helicase (d) only DNA polymerase I activity (DNA Pol III has high processivity, sure to the sliding clamp subunit) (e) no DNA gyrase activity

conservative:

-2 newly created strands -parent strands reassociate "conserved"

All of the following are involved in resolution of the Okazaki fragments in E. coli:

-5' to 3' polymerase activity of DNA polymerase I -DNA ligase -5' to 3' exonuclease activity of DNA polymerase I

DNA chain elongation:

-5'-->3' direction -once 1 nucleotide added to 3' end, 2 terminal phosphates cleaved -new -OH end (newly exposed 3' hydroxyl group --> another nucleotide to continue DNA synthesis) -DNA polymerase I adds nucleotides to the growing chain at the 3' end (5'->3' direction)

DNA synthesis in bacterial involves five polymerases (DNA Pol):

-DNA polymerase I -DNA polymerase II -DNA polymerase III -DNA polymerase IV -DNA polymerase V

Meselson-Stahl Experiment:

-DNA replication is semiconservative in bacterial cells -grew E. coli in presence of 15N -after several divisions, N are 15N Heavy isotope... why? -1 more neuron -15N more dense than 14N To distinguish: -sedimentation equilibrium centrifugation 15N toward bottom of tube -transferred to 14N media and DNA allowed to replicate -DNA extracted and subjected to sedimentation equilibrium centrifugation What is expected for semiconservative replication? -single band of intermediate density

List the proteins that unwind DNA during in vivo DNA synthesis. How do they function?

-DnaA binds to AT-rich regions of Orin to initiate strand unwinding -Helicase uses the energy of ATP hydrolysis to unwind DNA strands, which are then stabilized by single-stranded DNA-binding proteins -DNA gyrase, a DNA topoisomerase, relieves supercoiling generated by helix unwinding. -this process involves breaking both stands of the DNA helix

Origin of replication:

-OriC -replication fork (bidirectional) -replicon -OriC (point of origin) & ter (termination sequence)

All of the following proteins and enzymes function at the origin of replication in E. coli:

-SSBPs -helicase -DnaA, DnaB, and DnaC proteins (DNA ligase catalyzes the last reaction in DNA replication; it seals the nick and joins the Okazaki fragments. This reaction does not occur at the origin of replication.)

synthesis of the lagging strand is accomplished through the repetition of the following steps:

-Step 1: A new fragment begins with DNA Pol III binding to the 3' end of the most recently produced RNA primer (primer B) which is closest to the replication fork. DNA Pol III then adds DNA nucleotides in the 5' to 3' direction until it encounters the previous RNA primer (primer A). -Step 2: DNA Pol III falls off and is replaced by DNA Pol I. Starting at the 5' end of primer A, DNA Pol I removes each RNA nucleotide and replaces it with the corresponding DNA nucleotide. (DNA Pol I adds the nucleotides to the 3' end of fragment B.) When it encounters the 5' end of fragment A, DNA Pol I falls off, leaving a gap in the sugar-phosphate backbone between fragments A and B. -Step 3: DNA ligase closes the gap between fragments A and B.

The results of the Meselson-Stahl experiments relied on all of the following:

-a cesium chloride gradient -that a heavy isotope of nitrogen could be incorporated into replicating DNA molecules -a means of distinguishing among the distribution patterns of newly synthesized and parent molecule DNA possible

All of the following are required for synthesis of DNA in vitro by DNA polymerase:

-a primer -a template -deoxynucleoside triphosphate

Why did Watson and Crick predict double helix could serve as a template for synthesis?

-arrangement of nitrogenous bases -complementary of DNA strands -A would "attract" T; G "attract" C -covalently linked into DNA chains -one old strand and one new strand

accuracy of DNA Pol I:

-base composition very similar to template -DNA polymerase I replicated the templates faithfully (Kornberg's conclusion)

All of the following are stages of general (homologous) recombination:

-branch migration -formation of the Holliday structure -strand displacement (Translesion synthesis is a type of replication and repair mechanism that is present in eukaryotes. It is not a stage in homologous recombination.)

helicase:

-breaks H-bonds between bases -binds at the replication fork

topoisomerase:

-breaks covalent bonds in the DNA backbone -binds ahead of the replication fork

Kornberg isolated enzyme, DNA polymerase I:

-catalyzes DNA synthesis -requires a DNA template and all 4 nucleotides - dNTPs (in vitro) -Kornberg isolated 1st enzyme that was able to direct DNA synthesis in a cell-free (in vitro) system

Chemical reaction catalyzed by DNA polymerase I:

-deoxyribose nucleoside triphosphate (A, T, C, or G) -single dNTP added to new strand during DNA synthesis -with addition of DNA polymerase I and Mg++, (dNMP)n+1 -- drives reaction energetically

Distinguish between the holoenzyme and the core enzyme.

-holoenzyme: the active form of the enzyme Pol -"core" portion / core enzyme: the region responsible for actual polymerization

Which of the following is responsible for stabilizing the dimer structure?

-hydrophobic interactions -hydrogen bonds -ionic bonds

lagging strand:

-multiple primers needed -made in segments -daughter strand elongates away from replication fork -synthesized 5' to 3'

Summarize and compare the properties of DNA polymerase I, II, and III.

-none can initiate DNA synthesis on a template -all can elongate an existing DNA strand assuming there is a template strand -polymerization of nucleotides occurs in the 5' to 3' direction, where each 5' phosphate is added to the 3' end of the growing polynucleotide -all three enzymes are large, complex proteins with a molecular weight in excess of 100,000 daltons -each has 3' to 5' exonuclease activity DNA polymerase I: -5' to 3' exonuclease activity -present in large amounts -relatively stable -removal of RNA primer DNA polymerase II: -possibly involved in repair function DNA polymerase III: -essential for replication -complex molecule

semiconservative:

-one old and one new strand

leading strand:

-only one primer needed -made continuously -daughter strand elongates toward replication fork -synthesized 5' to 3'

dispersive:

-parental strands disperse into both strands -both strands contain old and new DNA -cleavage of parental strands

single-strand binding protein:

-prevents H-bonds between bases -binds after the replication fork

In vivo function of DNA Pol III:

-removing the primer -fills gaps after primer removal -DNA repair (exonuclease activity) -DNA polymerases II, IV, and V --> DNA repair

Bacterial DNA Polymerases:

-scientists predicted additional enzymes are involved -4 additional unique polymerases -I, II and III: elongate an existing DNA strand (primer); cannot initiate DNA synthesis -3' to 5' exonuclease activity (all 3) -proofread newly synthesized DNA, remove/replace incorrect nucleotides I - 5' to 3' exonuclease activity -excise nucleotides same direction as synthesis -polymerase I present in greater amounts compared to polymerase III

Three possible modes of DNA replication:

-semiconservative -conservative -dispersive

DNA double helix composition:

-two strands of DNA, each with a sugar-phosphate backbone and nitrogenous bases that form hydrogen bonds, holding the two strands together 3' end has a hydroxyl (-OH) group on the deoxyribose sugar 5' end has a phosphate group

Distinguish between (a) unidirectional and bidirectional synthesis, and (b) continuous and discontinuous synthesis of DNA.

-unidirectional synthesis: replicate strands in one direction only -bidirectional synthesis: replicate strands in both directions. -synthesis of complementary strands occurs continuous 5' to 3' mode on the leading strand template in the direction of the replication fork, resulting in a single, long product strand -synthesis of complementary strand occurs discontinuous 5' to 3' mode on the lagging strand, resulting in shorter product strands called Okazaki fragments

How many origins of replication are used in a bacterial chromosome?

1

Approximately, how many base pairs exist within the human genome?

3 billion

direction of DNA synthesis

5'-->3'

If DNA polymerase III nonfunctional:

Although the replication fork would form normally, new DNA strands would not be synthesized.

Kornberg showed that nucleotides are added to the 3' end of each growing DNA strand. In what way does an exposed 3' -OH group participate in strand elongation?

An exposed 3' -OH group is necessary for the attachment of the next nucleotide. Components of the 3' -OH group form a covalent bond with the 5' - phosphate of the added nucleotide and inorganic pyrophosphate is released.

Which of the following statements describes the results found by Herbert Taylor, Philip Woods, and Walter Hughes?

Both sister chromatids were labeled after one round of replication. (Since replication occurred in the presence of 3H, both chromatids were labeled.)

Describe the role of 15N in the Meselson-Stahl experiment.

By labeling the pool of nitrogenous bases of the DNA of E. coli with the heavy isotope 15N, it would be possible to "follow" the "old" DNA.

All of the following eukaryotic DNA polymerases are the major forms of the enzyme that function in initiation and elongation during eukaryotic nuclear DNA synthesis:

DNA Pol Eukaryotic DNA Pol is found exclusively in mitochondria and is thus not involved in nuclear DNA synthesis.

While many commonly used antibiotics interfere with protein synthesis or cell wall formation, clorobicin, one of several antibiotics in the aminocoumarin class, inhibits the activity of bacterial DNA gyrase. Similar drugs have been tested as treatments for human cancer. How might such drugs be effective against bacteria as well as cancer?

DNA gyrase is a member of a group of enzymes called DNA topoisomerase that "undo" the twists and knots of supercoiling generated by the unwinding of DNA during replication. By inhibiting DNA gyrase, DNA replication is stopped. In bacteria, this reduces the intensity of an infection; in cancer cells, their proliferation is slowed. Because of structural differences between bacterial and eukaryotic DNA gyrases, different drugs are required for bacterial and cancer treatments.

Which enzyme catalyzes the addition of nucleotides to a growing DNA chain?

DNA polymerase

Which DNA polymerase is responsible for the removal of the RNA primers at the origin and athlete the 5' ends of the Okazaki fragments in E. coli?

DNA polymerase I (The 5' to 4' exonuclease activity of DNA polymerase I is responsible for RNA primer removal.)

Which enzyme can proofread DNA using a 3'-->5' exonuclease activity?

DNA polymerase I, II, and III

In prokaryotes, which enzyme catalyzes the formation of phosphodiester bonds during DNA synthesis?

DNA polymerase III

Which enzyme is responsible for in vivo 5' to 3' polymerization?

DNA polymerase III

If single-stranded binding protein (SSB) nonfunctional:

DNA strands would not remain separated; single strands would reanneal to form double-stranded DNA.

If helicase nonfunctional:

DNA strands would not separate.

If topoisomerase nonfunctional:

DNA would become tangled downstream of the replication fork; replication may start, but would stop when it reaches the tangled DNA.

Which characteristics of DNA polymerase I raised doubts that its in vivo function is the synthesis of DNA leading to complete replication?

DeLucia and Cairns discovered a strain of E. coli (polA1) that still replicates its DNA but is deficient in DNA polymerase I activity. Furthermore, it is capable of degrading as well as synthesizing DNA. Such degradation suggested that it functions as a repair enzyme. In addition, the in vitro rate of DNA synthesis using DNA polymerase I was slow, being more effective at replicating single-stranded DNA than double-stranded DNA.

Why is DNA synthesis expected to be more complex in eukaryotes than in bacteria? How is DNA synthesis similar in the two types of organisms?

Eukaryotic DNA is replicated in a manner that is similar to that of E. coli. Synthesis is bidirectional, continuous on one template strand and discontinuous on the other, and the requirements of synthesis (four deoxyribonucleoside triphosphate, divalent cation, template, and primer) are the same. Okazaki fragments of eukaryotes are about 1/10 the size of those in bacteria. Because there is a much greater amount of DNA to be replicated and DNA replication is slower, there are multiple initiation sites for replication in eukaryotes (and increased DNA polymerase per cell) in contrast to the single replication origin in bacteria. Replication occurs at different sites during different intervals of the S phase.

Chromosome length remains the same throughout the life of eukaryotic organisms. (True or false)

False (Telomers of each chromosome shorten after each cell division.)

Which scientist proposed models to demonstrate the process of homologous replication?

Holliday

Now let's look at the specific amino acid interactions at the dimer interface that link the two monomers together into the larger homodimer structure. Which of the following accurately describes the specific interactions that stabilize the dimer structure?

Hydrophobic interactions between phe106 and ile108 of one monomer and ile272 and leu273 of the other monomer. Ionic bonds between arg96 and arg103 of one monomer and glu300 and glu304 of the other monomer.

Which of the following accurately describes the structure of the DNA polymerase III beta subunit sliding clamp protein?

It's a homodimer with monomers arranged head to head and tail to tail (N--C-->N--C-->).

Who studied and isolated enzymes in prokaryotic replication?

Kornberg

Which scientists used density gradient centrifugation to demonstrate semiconservative replication?

Meselson and Stahl

If DNA ligase nonfunctional:

Nicks would remain in the phosphodiester backbone of the DNA molecule.

Which of the following statements regarding DNA replication is true?

On the leading strand, there is continuous synthesis in the 5'->3' direction.

If DNA polymerase I nonfunctional:

RNA primers would not be removed efficiently and wouldn't t be replaced with DNA nucleotides.

If primase nonfunctional:

RNA primers would not form; DNA Pol III could not synthesize DNA.

List and describe the function of the ten subunits constituting DNA polymerase III.

SUBUNIT: -α (alpha), -ε (epsilon), and -θ (theta) FUNCTION: -5' to 3' polymerization -3' to 5' exonuclease -core assembly GROUPINGS: -Core enzyme: Elongates polynucleotide chain and proofreads SUBUNIT: -γ (gamma), -δ (delta), -δ' (delta prime), -χ (chi), and -ν (nu) FUNCTION: -Loads enzyme on template (serves as clamp loader) GROUPINGS: -γ complex SUBUNIT: -β (beta) FUNCTION: -Sliding clamp structure (processivity factor) SUBUNIT: -τ (tau) FUNCTION: -Dimerizes core complex

Compare conservative, semiconservative, and dispersive modes of DNA replication.

The differences among the 3 models of DNA replication relate to the manner in which the new strands of DNA are oriented as daughter DNA molecules are produced. Conservative: The original double helix remains as a complete unit, and the new DNA double helix is produces as a single unit. The old DNA is completely conserved. Semiconservative: Each daughter strand is composed of one old DNA strand and one new DNA strand. Breakage of hydrogen bonds is required. Dispersive: The original DNA strand is broken into pieces and the new DNA in the daughter strand is interpreted among the old pieces. Breakage of the individual covalent, phosphodiester bonds is required for this mode of replication.

Describe the "end-replication problem" in eukaryotes. How is it resolved?

The end-replication problem refers to the difficulties posed in replicating the ends of linear eukaryotic chromosomes. Once primers are removed from the 5' ends, a gap remains, which cannot be filled. This shortens the chromosome with each round of replication, potentially leading to deletion of gene-coding regions. The action of telomerase lengthens the telomere, which is then made double-stranded (except for the end) through conventional DNA synthesis.

Why is an RNA primer considered essential for DNA synthesis by DNA polymerase III?

The enzyme requires a free 3' -OH group.

What are the requirements for in vitro synthesis of DNA under the direction of DNA polymerase I?

The in vitro replication requires a DNA template, a divalent cation (Mg2+), and all four of the deoxyribonucleoside triphosphate: dATP, dCTP, dTTP, and dGTP.

How did Kornberg assess the fidelity of DNA polymerase I in copying a DNA template?

The indirect approach described in the text was the analysis of base composition, which indicated that the composition of the product met the expected composition within experimental error.

Outline the current model for DNA synthesis.

The model involves opening and stabilizing the DNA helix, priming DNA synthesis with RNA primers, and moving replication forks in both directions, which includes elongation of RNA primers in continuous and discontinuous 5' to 3' modes and their removal by the exonucleolytic activity of DNA polymerase I. Okazaki fragments generated in the replicative process are joined together with DNA ligase. DNA gyrase relieves supercoils generated by DNA unwinding.

What was the significance of the polA1 mutation?

The polA1 mutation was instrumental in demonstrating that DNA polymerase I activity was not necessary for the in vivo replication of the E. coli chromosome. Such an observation opened the door for the discovery of other enzymes involved in DNA replication.

The overall rate of eukaryotic replication is essentially the same as it is in prokaryotes, even though eukaryotic genomes are hundreds to thousands of times bigger than their prokaryotic counterparts. Which of the following statements could account for this observation?

There are more replication origins in eukaryotes.

Which of the following statements best describes Okazaki fragments?

They are formed in the lagging strand.

Which of the following cells are least likely to use the enzyme telomerase?

adult neurons

What enzyme initiates replication?

an RNA polymerase enzyme

Taylor-Woods-Hughes:

autoradiography -pinpoints location of radioisotope in cell -photographic emulsion places over cellular material and stored in the dark -develops much like photographic film -Result: presence of dark grains identifies location of newly synthesized DNA -DNA replication is semiconservative in eukaryotes -vicia faba -root tips excellent source of dividing cells -3H-thymidine-label DNA -Replication I: Grew root tips with 3H-thymidine Replication II: -transferred into unlabeled medium -arrested cells at metaphase -colchicine -1st replication-both sister chromatids labeled Replication II: -2nd replication - only one chromatic is labeled

Which structures can be involved in recombination?

chromatids of homologous chromosomes (can recombine during meiosis)

The discovery of Okazaki fragments suggested that DNA synthesis is:

discontinuous

After observing the results of one round of replication, the scientists obtained results from a second round. The purpose of one additional round of replication was to:

distinguish between semi-conservative and dispersive replication

Which enzyme synthesizes DNA on the leading strand of the nuclear chromosomes in eukaryotes?

epsilon

You are studying an E. coli mutant in which DNA replication is only partially complete: initial strand separation has occurred, but new strands are not fully replicated. The replication bubble is present, but the forks are stalled. Which of the following genes is most likely mutated in this strain of E. coli?

gyrA (gyrA codes for one of the subunits of gyrase, a type of DNA topoisomerase. As unwinding proceeds, the coiling tension ahead of the replication fork must be relieved. Since the forks are stalled in this mutant, it is likely that there isa mutation in this gene.)

Which enzyme relaxes the super coils in the DNA helix?

gyrase

Which enzyme opens up the helix for replication?

helicase

Crossing over in prophase I is an example of:

homologous recombination

Discontinuous synthesis during DNA replication generates Okazaki fragments within the:

lagging strand in prokaryotes and eukaryotes

replicon

length of DNA replicated

The function of the beta subunit is to prevent DNA polymerase III from dissociating from the template DNA strand while still allowing the polymerase to slide along DNA as replication progresses. Given this function, the alpha helices in the beta subunit are:

located on the inner surface and oriented perpendicular to the minor and major grooves of DNA.

If the dispersive model of DNA replication were correct, how many bands would have been observed by Meselson and Stahl after two rounds of replication?

one (If the dispersive model were correct, each new molecule of DNA would contain both 14N and 15N. Therefore, one band of DNA, intermediate between the position of heavy and light nitrogen, would have been observed after two rounds of replication.)

Which of the following would result from a third round of replication using the methods of Meselson and Stahl?

one light band and one intermediate band

In eukaryotes, ARSs serve a function similar to that of __________ in prokaryotes.

oriC (Both sequences re origins of replication.)

What is the name of the location where chromosome replication is initiated?

origin of replication (OriC)

Which enzyme creates short RNA polymers to initiate DNA synthesis?

primase

How many replication initiation sites are used in prokaryotes and eukaryotes?

prokaryotes: 1 eukaryotes: many

What is the location where the two strands of a DNA template are separated and unwound for replication?

replication fork

Which term describes a segment of DNA produced following a single replication initiation?

replicon

What process uses RJA as a template for DNA production?

reverse transcription

Which replication process occurs in eukaryotes?

semi-conservative

Which term best describes a state in which cells begin to function less efficiently, and have stopped undergoing cell division?

senescence

Which of the following enzymes is a reverse transcriptase?

telomerase (Telomerase synthesizes DNA repeats from an RNA template. The RNA that serves as a template for this synthesis is part of the telomerase enzyme.)

conservative model:

the 2 parental DNA strands remain together during replication: -1 progeny contains both parental DNA strands -1 progeny contains only newly synthesized DNA

semi-conservative model:

the 2 parental DNA strands separate: -each strand serves as template for new DNA strands -2 DNA double helices (1 parental & 1 new strand)

Where on chromosomes are telomeres found?

the ends of eukaryotic chromosomes

dispersive model:

the parental double helix broken into double-stranded DNA segments that act as templates for synthesis of new double helix molecules: -segments reassemble into complete double jellies, each with parental and new segments interspersed

bidirectional

two replication forks - migrate at opposite directions

A mutation in the rep gene would affect which process in DNA replication?

unwinding the double helix (Helices unwind the double helix, allowing synthesis to proceed along each strand. rep is a gene that codes for DNA helicase.)