genetics ch 12

Which of the following enzymes adds nucleotides?

A. DNA polymerase B. primase

DNA replication requires all of the following:

A. a template, that is, a strand or strands to be copied. B. the monomers that are assembled into the new strand, including dCTP. C. the enzymes that carry out the process, including primase. D. the enzymes that carry out the process, including topoisomerase.

Models of DNA replication include:

A. conservative, in which the original DNA strands are together at the conclusion of the process.

The final error rate of DNA replication is exceptionally low due to all of the following EXCEPT:

A. its extremely high accuracy in selecting nucleotides complementary to the template strand. B. its ability to back up, cutting out incorrect nucleotides, in a process known as proofreading. C. the ability of other enzymes to identify mismatches and repair them.

The coordination of a large number of ARSs (autonomously replicating sequences) requires that replication start at much the same time, and is enforced by:

A. licensing. B. activating the helicases at the same time.

Eukaryotic DNA replication differs from prokaryotic DNA replication in all of the following EXCEPT:

A. prokaryotic DNA has more efficient translesion synthesis.

Linear chromosomes have telomeres at their tips, which are described by all of the following:

A. they shorten after replication due to the use of primers. B. they can be extended by the reverse transcriptase telomerase. C. their length is a function of age. D. they have conserved sequences within a species

DNA replication ends when:

A. two replication forks run into each other. B. a protein binds a termination sequence and blocks helicase movement.

Primers are synthesized where on the lagging strand of a replication fork?

At the beginning of every Okazaki fragment Correct! Synthesis is discontinuous on the lagging strand, and synthesis of each fragment begins with the formation of a new primer.

(Problem 23, part 2) A circular molecule of DNA contains 1 million base pairs. If the rate of DNA synthesis at a replication fork is 100,000 nucleotides per minute, how much time will replication of this circular chromosome take by rolling-circle replication? Ignore replication of the displaced strand in rolling-circle replication.

10 minutes

In DNA synthesis, nucleotides are added to the ____ end of the growing nucleotide strand.

3'

Which activity of DNA polymerases in E. coli allows them to remove a nucleotide that has been inserted incorrectly?

3' to 5' exonuclease

Which of the following accurately describes the direction of DNA synthesis?

3' to 5' on the template strand and 5' to 3' on the synthesized strand Correct! DNA polymerase reads the template strand 3' to 5' and synthesizes the new strand 5' to 3'.

(Problem 24) A bacterium synthesizes DNA at each replication fork at a rate of 1000 nucleotides per second. If this bacterium completely replicates its circular chromosome by theta replication in 30 minutes, how many base pairs of DNA will its chromosome contain?

3,600,000

(Problem 23, part 1) A circular molecule of DNA contains 1 million base pairs. If the rate of DNA synthesis at a replication fork is 100,000 nucleotides per minute, how much time will theta replication require to completely replicate the molecule, assuming that theta replication is bidirectional?

5 minutes

Which activity of DNA polymerase I in E. coli allows it to remove the primers laid down by DNA primase?

5' to 3' exonuclease

(Problem 21) Phosphorus is required to synthesize the deoxyribonucleoside triphosphates used in DNA replication. A geneticist grows some E. coli in a medium containing nonradioactive phosphorous for many generations. A sample of the bacteria is then transferred to a medium that contains a radioactive isotope of phosphorus (32P). Samples of the bacteria are removed immediately after the transfer and after one and two rounds of replication. What will be the distribution of radioactivity in the DNA of the bacteria in the sample collected after two rounds of replication?

50% of the DNA will have radioactive phosphorus in both strands, and 50% will have radioactive phosphorus in one strand only.

The diagram below illustrates replication on a small portion of which of the following?

A eukaryotic chromosome Correct! Eukaryotic chromosomes are very large and have multiple origins of replication.

(Problem 13) How does replication licensing ensure that DNA is replicated only once at each origin per eukaryotic cell cycle?

A replication licensing factor binds to the origin during G1 phase and is removed during S phase. Without the licensing factor, replication cannot initiate again in the cell.

(Problem 16) How are telomeres at the ends of eukaryotic chromosomes replicated?

A ribonucleoprotein enzyme uses its internal RNA template to synthesize a single-stranded telomeric sequence.

(Problem 6) Which of the following is a major requirement of replication?

A single-stranded DNA template Deoxynucleoside triphosphate substrates Enzymes and proteins required for synthesis of a polynucleotide polymer

Cleavage in the vertical plane of the Holliday junction shown below would result in which arrangement of alleles?

A with b and a with B Correct! This arrangement is recombinant and is produced by cleavage in the vertical plane.

Scientists studied nucleosome assembly following DNA replication by using amino acids labeled with a heavy isotope. Specifically, they grew cells for several generations in medium that contained heavy amino acids so that all of the histone octamers contained heavy amino acids. Then the cells were transferred to medium that contained light amino acids, and the cells were allowed to complete one round of replication and nucleosome reassembly. Histone octamers were then isolated and subjected to density-gradient centrifugation. Which result would you expect if old octamers remained intact throughout the cell cycle and new octamers were assembled solely from newly synthesized histones?

One band of heavy octamers near the bottom of the tube and one band of light octamers near the top of the tube. Correct! There would be no mixing of old (heavy) and new (light) histones during nucleosome reassembly so that only heavy and light forms would exist.

(Problem 10a) What is an important difference in the enzymatic activity between DNA polymerases I and III?

Only DNA polymerase I has 5′-to-3′ exonuclease activity.

At which location(s) on the leading strand of a replication fork are primers synthesized?

Only at the 5' end of the newly synthesized strand Correct! Primase synthesizes a single primer on the leading strand and synthesis occurs continuously 5' to 3' in the direction of movement of the replication fork.

(Problem 38c) A conditional mutation expresses its mutant phenotype only under certain conditions (the restrictive conditions) and expresses the normal phenotype under other conditions (the permissive conditions). One type of conditional mutation is a temperature-sensitive mutation, which expresses the mutant phenotype only at certain temperatures. Strains of E. coli have been isolated that contain temperature-sensitive mutations in the genes encoding different components of the replication machinery. In each of these strains, the protein produced by the mutated gene is nonfunctional under the restrictive conditions. These strains are grown under permissive conditions and then abruptly switched to the restrictive condition. After one round of replication under the restrictive condition, the DNA from each strain is isolated and analyzed. What characteristics would you expect to see in the DNA isolated from each strain with a temperature-sensitive mutation in its gene that encodes DNA polymerase III?

Replication cannot occur.

Forms of DNA replication include all of the following EXCEPT:

B. multiple replication forks in some plasmids.

(Problem 28) Which of the DNA polymerases shown in Table 12.3 has the ability to proofread?

Both DNA polymerase I and DNA polymerase III

(Problem 20) Suppose a future scientist explores a distant planet and discovers a novel form of double-stranded nucleic acid. When this nucleic acid is exposed to DNA polymerases from E. coli, replication takes place continuously on both strands. What conclusion can you make about the structure of this novel nucleic acid?

Both strands on the nucleic acid must be oriented in the same direction.

Because genomes tend to be quite large, it is essential that the DNA replication process be:

C. extremely accurate, with an error rate of less than one in ten million.

DNA polymerase III has high processivity, which means:

C. it functions at a high rate.

Homologous recombination is described by all of the following EXCEPT:

C. it involves chromosomes that do not normally pair with each other.

(Problem 36b, c) Dyskeratosis congenita (DKC) is a rare genetic disorder characterized by abnormal fingernails and skin pigmentation, the formation of white patches on the tongue and cheek, and progressive failure of the bone marrow. An autosomal dominant form of DKC results from mutations in the gene that encodes the RNA component of telomerase. Tom Vulliamy and his colleagues examined 15 families with autosomal dominant DKC (T. Vulliamy et al., 2004, Nature Genetics 36:447-449). They observed that the median age of onset of DKC in parents was 37 years, whereas the median age of onset in the children of affected parents was 14.5 years. Thus, DKC in these families arose at progressively younger ages in successive generations, a phenomenon known as anticipation. The researchers measured telomere length of members of these families; the measurements are given in the adjoining table. Telomere length normally shortens with age, and so telomere length was adjusted for age. Note that the age-adjusted telomere length of all members of these families is negative, indicating that their telomeres are shorter than normal. For age-adjusted telomere length, the more negative the number, the shorter the telomere.

Children of parents with defective telomerase inherit shorter than average telomeres and subsequently cannot maintain telomeres during their own lifetimes.

(Problem 17) Explain how the type of cleavage of the Holliday intermediate leads to noncrossover recombinants and crossover recombinants.

Cleavage in the horizontal plane produces noncrossover recombinants, while cleavage in the vertical plane produces crossover recombinants.

Which proposed model of replication is illustrated in the diagram below?

Conservative

The Meselsohn-Stahl experiment:

D. demonstrated that replicated double helices consisted of one old and one new strand.

Okazaki fragments are produced because:

D. lagging strands are copied discontinuously.

(Problem 15) What is the end-of-chromosome problem for linear replication?

DNA polymerase cannot fill in the gap left after the removal of RNA primers at the ends of chromosomes.

(Problem 11) Why is primase required for replication?

DNA polymerases cannot begin without a primer containing a free 3′-OH.

Expression of a particular mutation in E. coli causes previously initiated replication to halt on the lagging strand of DNA but not on the leading strand. This mutation most likely affects a gene encoding which protein?

DNA primase Correct! Once initiated, DNA synthesis can proceed normally on the leading strand, but it will quickly halt on the lagging strand because of the need to prime the synthesis of new Okazaki fragments.

Consider a mutation in the gene that encodes Tus protein in E. coli. This mutation causes the Tus protein to bind to other sites in addition to Ter sites. What would be the result of this mutation?

DNA replication would terminate at sites it normally would not terminate. Correct! Tus protein terminates replication at the location in which it binds.

(Problem 14a) In what way is eukaryotic replication similar to bacterial replication?

DNA synthesis is 5′ to 3′.

Consider a mutation that makes DNA polymerase α nonfunctional. What would be a likely result of this mutation?

DNA synthesis would not begin because primers would not be incorporated. Correct! DNA polymerase α provides the primase activity necessary for eukaryotic replication.

(Problem 7) What substrates are used in the DNA-synthesis reaction?

Deoxyribonucleoside triphosphates

Which proposed model of replication is illustrated in the diagram below?

Dispersive

Which proposed model of replication would produce the results shown below from a Meselson-Stahl experiment?

Dispersive Correct! All DNA molecules produced by dispersive replication are intermediate in weight.

In the replication fork, leading and lagging strand synthesis are different in that:

E. None of the above is accurate>>> A. the lagging strand is synthesized 3' to 5'. B. the leading strand does not use a primer. C. the leading strand is synthesized as a number of smaller pieces. D. the lagging strand does not use DNA ligase..

Which of the following proteins identifies specific sequences during the DNA replication process?

E. None of the above.>>>A. helicase B. single-stranded binding proteins (SSBPs) C. primase D. DNA ligase

(Problem 27a) What would be the effect on DNA replication of a mutation that destroyed the 3′ to 5′ exonuclease activity in DNA polymerase I?

Fidelity of the polymerase would decrease without this proofreading function.

Expression of a loss-of-function mutation in yeast causes cells and nuclei to enlarge and to accumulate DNA molecules to a level of 4n or higher. A gene encoding which of the following proteins is most likely affected by the mutation?

Geminin Correct! Geminin is involved in licensing of replication origins in eukaryotes and limiting replication from each origin to once per cell cycle.

The ___________ intermediate shown below is thought to be a feature of homologous recombination.

Holliday

(Problem 10b) What is the primary function of DNA polymerase I in bacterial cells?

It replaces RNA primers with DNA.

(Problem 12) What mechanism or mechanisms ensure the accuracy of replication in bacteria?

Mismatch repair following replication Accurate selection of inserted nucleotides by base pairing with the template Proofreading activity by the DNA polymerase

The short lengths of DNA produced by discontinuous replication on the lagging strand of DNA of a replication fork are called ___________ fragments.

Okazaki

How many origins of replication are typically found on a bacterial chromosome?

One

How many bands of DNA would be expected in Meselson and Stahl's experiment after two rounds of dispersive replication?

One Correct! Each round of replication would cause fragments of the original heavy molecule to be dispersed among the products, resulting in production of a single band that runs somewhere between the heavy and light bands.

(Problem 38d) A conditional mutation expresses its mutant phenotype only under certain conditions (the restrictive conditions) and expresses the normal phenotype under other conditions (the permissive conditions). One type of conditional mutation is a temperature-sensitive mutation, which expresses the mutant phenotype only at certain temperatures. Strains of E. coli have been isolated that contain temperature-sensitive mutations in the genes encoding different components of the replication machinery. In each of these strains, the protein produced by the mutated gene is nonfunctional under the restrictive conditions. These strains are grown under permissive conditions and then abruptly switched to the restrictive condition. After one round of replication under the restrictive condition, the DNA from each strain is isolated and analyzed. What characteristics would you expect to see in the DNA isolated from each strain with a temperature-sensitive mutation in its gene that encodes primase?

Replication cannot occur.

(Problem 38e) A conditional mutation expresses its mutant phenotype only under certain conditions (the restrictive conditions) and expresses the normal phenotype under other conditions (the permissive conditions). One type of conditional mutation is a temperature-sensitive mutation, which expresses the mutant phenotype only at certain temperatures. Strains of E. coli have been isolated that contain temperature-sensitive mutations in the genes encoding different components of the replication machinery. In each of these strains, the protein produced by the mutated gene is nonfunctional under the restrictive conditions. These strains are grown under permissive conditions and then abruptly switched to the restrictive condition. After one round of replication under the restrictive condition, the DNA from each strain is isolated and analyzed. What characteristics would you expect to see in the DNA isolated from each strain with a temperature-sensitive mutation in its gene that encodes initiator protein?

Replication cannot occur.

(Problem 30, part 2) If the gene for primase were mutated so that no functional primase was produced, what would be the effect on rolling-circle replication?

Replication could proceed without a primase.

Which of the following proteins are unique to eukaryotes?

Replication licensing factor

One strand of DNA serves as a template for the synthesis of a new strand through _____________ base pairing.

complementary

In ____________ replication, the entire double-stranded DNA molecule is postulated to serve as a template for the production of a whole new molecule of DNA, with the original molecule remaining intact.

conservative

Bidirectional replication from an origin of replication produces two replication _________, one at each end of the replication bubble.

forks

A scientist examining the products of meiosis in a particular fungus with the genotype Aa notices that occasionally the products occur in the ratios 3 A:1 a or 3 a:1 A. This result most likely results from

gene conversion

During replication in E. coli, DNA _________ reduces torsional strain that builds up ahead of the replication fork as the two template strands unwind.

gyrase

Expression of a particular mutation in E. coli causes previously initiated replication to quickly halt and DNA ahead of the two replication forks to become highly twisted. The mutation most likely affects a gene that encodes DNA

gyrase

Following initiation of DNA replication in E. coli, DNA ____________ breaks hydrogen bonds between the two template strands and moves the replication fork forward.

helicase

During homologous recombination, a single-stranded DNA molecule of one chromosome pairs with a single-stranded DNA molecule of another homologous chromosome, forming ____________ DNA.

heteroduplex

Gene conversion results from _____________ formation that takes place during homologous recombination.

heteroduplex

During replication in E. coli, a(n) ___________ protein binds to the origin of replication and causes a short section of DNA to unwind.

initiator

In the diagram below, the newly synthesized DNA on the top template is called the _________ strand and shows __________ synthesis.

lagging; discontinuous Correct! The new strand is being synthesized in the direction opposite that of the movement of the replication fork. Therefore, synthesis is in fragments and is delayed compared to the leading strand on the bottom template.

In eukaryotes, replication ___________ factor ensures that replication is initiated at thousands of origins of replication only once per cell cycle.

licensing

Expression of a particular mutation in E. coli causes half of newly synthesized DNA to persist as fairly short fragments. The mutation most likely affects a gene that encodes DNA

ligase

The enzyme that joins Okazaki fragments by sealing nicks in the sugar-phosphate backbone of newly synthesized DNA is DNA

ligase

(Problem 38b) A conditional mutation expresses its mutant phenotype only under certain conditions (the restrictive conditions) and expresses the normal phenotype under other conditions (the permissive conditions). One type of conditional mutation is a temperature-sensitive mutation, which expresses the mutant phenotype only at certain temperatures. Strains of E. coli have been isolated that contain temperature-sensitive mutations in the genes encoding different components of the replication machinery. In each of these strains, the protein produced by the mutated gene is nonfunctional under the restrictive conditions. These strains are grown under permissive conditions and then abruptly switched to the restrictive condition. After one round of replication under the restrictive condition, the DNA from each strain is isolated and analyzed. What characteristics would you expect to see in the DNA isolated from each strain with a temperature-sensitive mutation in its gene that encodes DNA polymerase I?

Replication occurs, but RNA primers are not removed from the lagging strand.

(Problem 38a) A conditional mutation expresses its mutant phenotype only under certain conditions (the restrictive conditions) and expresses the normal phenotype under other conditions (the permissive conditions). One type of conditional mutation is a temperature-sensitive mutation, which expresses the mutant phenotype only at certain temperatures. Strains of E. coli have been isolated that contain temperature-sensitive mutations in the genes encoding different components of the replication machinery. In each of these strains, the protein produced by the mutated gene is nonfunctional under the restrictive conditions. These strains are grown under permissive conditions and then abruptly switched to the restrictive condition. After one round of replication under the restrictive condition, the DNA from each strain is isolated and analyzed. What characteristics would you expect to see in the DNA isolated from each strain with a temperature-sensitive mutation in its gene that encodes DNA ligase?

Replication occurs, but the lagging strand has multiple nicks.

Which of the following is a feature of the double-strand break model of recombination but is not a feature of the Holliday model?

Replication of DNA Correct! In the double-strand break model, replication is required late to fill in the single-strand gap on the bottom strand of the broken DNA molecule.

(Problem 1) What is semiconservative replication?

Replication resulting in products that consist of one strand of template DNA and one strand that has been newly synthesized

(Problem 30, part 1) If the gene for primase were mutated so that no functional primase was produced, what would be the effect on theta replication?

Replication would be unable to initiate.

Based on the most recent evidence, what would be the most likely result of expressing telomerase in somatic cells in humans?

Shorter life due to promotion of cancer Correct! Expression of telomerase in somatic cells most likely would remove a limit to the ability of cancer cells to divide, allowing them to proliferate without restraint.

________________ prevent the formation of secondary structures within DNA.

Single-stranded binding proteins

Which activity is NOT associated with DNA polymerases in E. coli?

Synthesize a primer Correct! DNA polymerases cannot prime DNA synthesis. They need primase to synthesize an RNA primer.

A mutant yeast is isolated that stops reproducing after several rounds of cell division. Examination of the chromosomes in the mutant line reveals that the chromosomes shorten after each cell division. Which activity is most likely affected by this mutation?

Telomerase Correct! Lack of telomerase would cause the ends of the chromosomes to be shortened after each round of replication, e

(Problem 34) A number of scientists who study cancer treatment have become interested in telomerase. Why would they be interested in telomerase?

Telomerase activity can confer immortality to a cell lineage and is present in most cancer cells.

(Problem 35) The enzyme telomerase is part protein and part RNA. What would be the most likely effect of a large deletion in the gene that encodes the RNA part of telomerase? How would the function of telomerase be affected?

Telomerase would lose the function of its internal template and would be unable to correctly associate with telomeres and synthesize new telomeric DNA.

(Problem 32) Marina Melixetian and her colleagues suppressed the expression of Geminin protein in human cells by treating the cells with small interfering RNAs (siRNAs) complementary to Geminin messenger RNA (M. Melixetian et al., 2004, Journal of Cell Biology 165:473-482). (Small interfering RNAs form a complex with proteins and pair with complementary sequences on mRNAs; the complex then cleaves the mRNA, so there is no translation of the mRNA; pp. 403-404 in Chapter 14). Forty-eight hours after treatment with siRNA, the Geminin-depleted cells were enlarged and contained a single giant nucleus. Analysis of DNA content showed that many of these Geminin-depleted cells were 4n or greater. Explain these results.

The maintenance of the correct number of chromosomes by replication licensing has been disrupted by the loss of Geminin.

(Problem 14b) In what way is eukaryotic replication different from bacterial replication?

The number of origins per chromosome The assembly of nucleosomes

(Problem 31) DNA polymerases are not able to prime replication, yet primase and other RNA polymerases can. Some geneticists have speculated that the inability of DNA polymerase to prime replication is due to its proofreading function. This hypothesis argues that proofreading is essential for the faithful transmission of genetic information and that, because DNA polymerases have evolved the ability to proofread, they cannot prime DNA synthesis. Explain why proofreading and priming functions in the same enzyme might be incompatible.

The proofreading ability of the DNA polymerase requires a 3′-OH to already be present, which would not be present during priming.

(Problem 27c) What would be the effect on DNA replication of a mutation that destroyed the 5′ to 3′ polymerase activity in DNA polymerase I?

There would be a loss of the ability to synthesize DNA in the gaps created by the removal of RNA primers.

Which mode of replication produces two circular molecules directly?

Theta

Which of the following characteristics is true regarding ALL DNA polymerases in E. coli?

They all require a 3' OH to initiate DNA synthesis. Correct! DNA polymerases cannot initiate DNA synthesis on a bare template.

(Problem 2) What experimental methods did Meselson and Stahl use to determine that replication occurs in a semiconservative manner in E. coli?

They grew E.coli in media with one isotope of nitrogen, then switched to media containing a different isotope of nitrogen for the next rounds of replication.

(Problem 26) In Figure 12.8, which is the leading strand and which is the lagging strand?

Top red stand is lagging and bottom red strand is leading.

(Problem 29) How would DNA replication be affected in a bacterial cell that is lacking DNA gyrase?

Torsional strain ahead of the replication fork would cause replication to stop.

How many bands of DNA would be expected in Meselson and Stahl's experiment after two rounds of conservative replication?

Two Correct! After each round of conservative replication, the original heavy molecules will remain along with an increasing proportion of new light molecules. After two rounds, the ratio should be 3 light:1 heavy.

(Problem 33) What results would be expected in the experiment outlined in Figure 12.17 if, during replication, all the original histone proteins remained on one strand of the DNA and new histones attached to the other strand?

Two bands, with the original histones in the band closer to the bottom of the tube

(Problem 19) What is gene conversion?

When one allele is converted to another through repair of a mismatch that occurs following heteroduplex formation in recombination

(Problem 9) Why is DNA gyrase necessary for replication?

Without DNA gyrase, strain would build up ahead of the replication fork as a result of the unwinding process.

(Problem 39) DNA topoisomerases play important roles in DNA replication and in supercoiling (see Chapter 11). These enzymes are also the targets for certain anticancer drugs. Eric Nelson and his colleagues studied m-AMSA, one of the anticancer compounds that acts on topisomerase enzymes (E. M. Nelson, K. M. Tewey, and L. F. Liu. 1984. Proceedings of the National Academy of Sciences 81:1361-1365). They found that m-AMSA stabilizes an intermediate produced in the course of the topoisomerase's action. The intermediate consisted of the topoisomerase bound to the broken ends of the DNA. Breaks in DNA that are produced by anticancer compounds such as m-AMSA inhibit the replication of the cellular DNA and thus stop cancer cells from proliferating. Choose a mechanism for how m-AMSA and other anticancer agents that target topoisomerase enzymes taking part in replication might lead to DNA breaks and chromosome rearrangements.

m-AMSA inhibits topoisomerase after the topoisomerase has generated double-strand breaks in the DNA, thus preventing the breaks from resealing.

Incorrectly paired nucleotides that remain after replication is complete can be removed and corrected by a process called

mismatch repair.

Autonomously replicating sequences in yeast correspond to ________________ on chromosomes.

origins of replication

During replication in E. coli, the enzyme called __________ synthesizes short stretches of nucleotides to get replication started. content hint: This enzyme has RNA polymerase activity.

primase

Which of the following proteins relies on complementary base pairing during the DNA replication process?

primase

Most of the replication errors that arise when DNA polymerase incorporates an incorrect nucleotide are corrected by a process called

proofreading

Telomerase has a(n) _____________ and a(n) _______________ component.

protein; RNA

The process of homologous ______________ is essential for both crossing over and some types of DNA repair.

recombination

An individual unit of replication is called a __________, each of which contains a replication origin.

replicon

Cleavage of Holliday structures in E. coli is carried out by the enzyme

resolvase.

In _______________ replication, each strand of the parent molecule remains intact and serves as a template for the synthesis of a new strand by complementary base pairing.

semiconservative

Most somatic cells are capable of only a limited number of divisions because they lack _____________ activity.

telomerase

The bacterial chromosome shown below is undergoing ___________ replication.

theta

Specialized enzymes called ___________ polymerases can bypass distortions in the DNA template but tend to introduce replication errors because they have lower fidelity compared to other DNA polymerases.

translesion

Which of the following is NOT involved in the unwinding process of replication in E. coli?

Initiator proteins

(Problem 36a) Dyskeratosis congenita (DKC) is a rare genetic disorder characterized by abnormal fingernails and skin pigmentation, the formation of white patches on the tongue and cheek, and progressive failure of the bone marrow. An autosomal dominant form of DKC results from mutations in the gene that encodes the RNA component of telomerase. Tom Vulliamy and his colleagues examined 15 families with autosomal dominant DKC (T. Vulliamy et al., 2004, Nature Genetics 36:447-449). They observed that the median age of onset of DKC in parents was 37 years, whereas the median age of onset in the children of affected parents was 14.5 years. Thus, DKC in these families arose at progressively younger ages in successive generations, a phenomenon known as anticipation. The researchers measured telomere length of members of these families; the measurements are given in the adjoining table. Telomere length normally shortens with age, and so telomere length was adjusted for age. Note that the age-adjusted telomere length of all members of these families is negative, indicating that their telomeres are shorter than normal. For age-adjusted telomere length, the more negative the number, the shorter the telomere.

Parents, in general, have longer telomeres than their children.

(Problem 40) The regulation of replication is essential to genomic stability, and, normally, the DNA is replicated just once every eukaryotic cell cycle (in the S phase). Normal cells produce protein A, which increases in concentration in the S phase. In cells that have a mutated copy of the gene for protein A, the protein is not functional and replication takes place continuously throughout the cell cycle, with the result that cells may have 50 times the normal amount of DNA. Protein B is normally present in G1 but disappears from the cell nucleus in the S phase. In cells with a mutated copy of the gene for protein A, the levels of protein B fail to disappear in the S phase and, instead, remain high throughout the cell cycle. When the gene for protein B is mutated, no replication takes place. Choose a mechanism for how protein A and protein B might normally regulate replication so that each cell gets the proper amount of DNA.

Protein B is a licensing factor required for initiation of replication, and protein A removes B after replication has initiated.

The enzyme primase acts as a(n) ______ polymerase during DNA replication.

RNA

(Problem 27b) What would be the effect on DNA replication of a mutation that destroyed the 5′ to 3′ exonuclease activity in DNA polymerase I?

RNA primers could not be removed from the DNA during replication.

In eukaryotes, DNA polymerase ______ includes primase activity and initiates nuclear DNA synthesis.

alpha

In their experiment, Meselson and Stahl used a technique called ____________________ in order to distinguish between heavy and light DNA.

equilibrium density gradient centrifugation

In their experiment, Meselson and Stahl used two different isotopes of __________ to distinguish between old and newly synthesized DNA molecules.

nitrogen