Chapter 11: DNA Replication & Recombination Flashcards
True or False?
The data obtained from the Meselson–Stahl experiment after one generation of replication eliminated the dispersive model of DNA replication.
False
The data obtained from the Meselson–Stahl experiment after one generation was consistent with both the semiconservative and the dispersive model of DNA replication. The conservative model of DNA replication was eliminated because it predicted that there would be two bands representing the original DNA at one density and the newly replicated DNA at a different density.
Which enzyme catalyzes the addition of nucleotides to a growing DNA chain?
DNA polymerase catalyzes the addition of nucleotides to a growing DNA chain.
Single‑strand binding proteins stabilize the open conformation of the unwound DNA.
Once helicase unwinds the double helix, single‑strand binding proteins bind to the open DNA and prevent it from winding together again.
The results of the Meselson-Stahl experiments relied on all of the following:
a cesium chloride gradient
a means of distinguishing among the distribution patterns of newly synthesized and parent molecule DNA possible
that a heavy isotope of nitrogen could be incorporated into replicating DNA molecules
The results of the Meselson-Stahl experiments relied on all of the following except _______.
the fact that DNA is the genetic material
This fact had already been established and was not of any consequence in these experiments.
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.
After one round of replication, the results of these two possibilities are indistinguishable. A second round was required to distinguish between these two possibilities.
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
Of the molecules generated in the third round, 75% are completely light, 25% are intermediate.
In the Meselson-Stahl experiment, which mode of replication was eliminated based on data derived after one generation of replication?
conservative
The conservative replication theory says that parental strands reanneal with parental strands, and daughter strands reanneal with daughter strands after DNA replication. This experiment showed that this was not the case.
Both the dispersive and semi-conservative models predicted that after one round of replication, there would be one half-heavy band. In the dispersive model, the original heavy band has been dispersed evenly into the two daughter molecules, both of which also contain ½ light nitrogen. In the semi-conservative model, each new daughter molecule contains one heavy strand (from the parent molecule) and one light strand (newly synthesized). Since the DNA does not denature on the gradient, one half-heavy band would be predicted by both models, which is what was seen. The conservative model predicted that there would be two bands: one heavy (containing both original heavy parent strands) and one light band (both strands newly synthesized). Thus, the first-round data ruled it out.
After two rounds of replication, the semi-conservative model predicts two bands: one light band (daughter molecules with all new nucleotides) and one half-heavy band (containing one of the parent strands and one new, light strand) while the dispersive model predicts only one ¼ heavy band. Since one-half heavy and one light band was seen, the dispersive model was eliminated and the semi-conservative model was supported.
True or False?
In the Meselson and Stahl (1958) experiment, bean plants (Vicia faba) were radioactively labeled so that autoradiographs could be made of chromosomes.
False
What enzyme or function is being affected in Mutant (a), causing many mismatched base pairs?
Mutant (a): Newly synthesized DNA contains many mismatched base pairs.
no DNA repair
What enzyme or function is being affected in Mutant (b), causing the accumulation of Okazaki fragments and failure to complete DNA synthesis?
Mutant (b): Okazaki fragments accumulate, and DNA synthesis is never completed.
no DNA ligase activity
What enzyme or function is being affected in Mutant (c), causing DNA synthesis to never be initiated?
Mutant (c): No initiation occurs.
no Primase activity
What enzyme or function is being affected in Mutant (d), causing DNA synthesis to be very slow?
Mutant (d): Synthesis is very slow.
only DNA polymerase I activity
What enzyme or function is being affected in Mutant (e), causing supercoiled DNA to remain after replication?
Mutant (e): Supercoiled strands remain after replication, which is never completed.
no DNA gyrase activity
What are the requirements for in vitro synthesis of DNA under the direction of DNA polymerase I?
DNA template, a divalent cation (Mg++ ), and all 4 of the deoxyribonucleoside triphosphates: dATP, dCTP, dTTP, and dGTP.
All of the following are differences between eukaryotic and prokaryotic DNA replication:
the rate of DNA synthesis
the number of replication origins
the type and number of polymerases involved in DNA synthesis
Which of the following is NOT different between eukaryotic and prokaryotic DNA replication?
the ability to form a replication fork
Both prokaryotes and eukaryotes form replication forks during DNA replication.
All of the following are related to telomeres:
telomerase enzyme
links to the aging process
short tandem repeats located at the ends of telomeres
True or False?
Telomeres are found in eukaryotes and prokaryotes.
FALSE
Telomeres are found in eukaryotes but not prokaryotes.
Structures located at the ends of eukaryotic chromosomes are called ________.
telomeres
True or False?
Telomerase is an RNA-containing enzyme that adds telomeric DNA sequences onto the ends of linear chromosomes.
True
Which structures can be involved in recombination?
Chromatids of homologous chromosomes
Chromatids of homologous chromosomes can recombine during meiosis.
The process that determines the length of heteroduplex DNA on the chromatids is called branch migration.
True
The crossbridge DNA structure formed after the initial nick is sealed can migrate along the chromatid. This process is called branch migration, and it increases the length of heteroduplex DNA.
Which process does not occur during recombination?
DNA polymerization
Recombination does not include the synthesis of new DNA.
Which processes do occur during recombination?
Ligation
Nicking of the sugar‑phosphate backbone
Strand displacement
True or False?
DNA replicates conservatively, which means that one of the two daughter double helices is “old” and the other is “new.”
FALSE
Which activity of E. coli DNA polymerase I is responsible for proofreading the newly synthesized DNA?
3’ to 5’; exonuclease
If the wrong nucleotide is inserted, normal base pairing will not be observed and the base in error will be removed from the newly synthesized strand before subsequent nucleotides are added.
DNA replication occurs in the 5’ to 3’ direction; that is, new nucleoside monophosphates are added to the 3’ end.
TRUE
Which DNA polymerase is mainly responsible for genome replication in E. coli?
DNA polymerase III
DNA polymerase III is responsible for the synthesis of the bulk of the DNA in E. coli.
DNA polymerase III adds nucleotides ________.
to the 3’ end of the RNA primer
replication fork
The Y-shaped region of a chromosome associated with the site of DNA replication.
Okazaki fragment
The short, discontinuous strands of DNA produced on the lagging strand during DNA synthesis.
polymerases
Enzymes that catalyze the formation of DNA and RNA from deoxynucleotides and ribonucleotides, respectively.
leading strand
During DNA replication,the strand synthesized continuously in the direction of the replication fork.
lagging strand
During DNA replication,the strand synthesized in a discontinuous fashion, in the direction opposite of the replication fork. See also Okazaki fragment.
DNA replication is a central process in biochemistry and must be done with the highest precision. If a mistake is made in the replication of DNA, the error will propagate into future generations.
An error that occurs during replication is called a mutation.
During DNA replication, each strand in the parental duplex serves as the template for the production of a daughter strand by complementary base pairing. Therefore, one cycle of replication will produce two daughter duplexes, each with one parental strand and one newly synthesized strand.
During a second cycle of replication, all four strands in the two duplexes will serve as templates, resulting in four duplexes (eight strands of DNA).
During replication, DNA synthesis occurs in the 5′ to 3′ direction along both template strands.
On one template strand, synthesis proceeds continuously toward the replication fork, generating the leading strand.
On the other template strand, DNA is synthesized away from the replication fork in segments called Okazaki fragments, generating the lagging strand.
Several proteins are involved in DNA replication, including the following:
Helicase breaks the hydrogen bonds between the parental DNA strands and unwinds the double helix.
Single-stranded binding proteins bind to the single strands of DNA, preventing them from reannealing and allowing synthesis to occur on both strands.
DNA polymerase III synthesizes the new strands, but it requires an existing 3′ hydroxyl (—OH) group to add nucleotides.
Primase creates short RNA primers, initiating DNA synthesis on both template strands.
DNA polymerase I removes the RNA primers and replaces them with DNA nucleotides.
On the lagging strand, DNA ligase joins Okazaki fragments by forming phosphodiester bonds between them, thus completing DNA replication.
primer
In nucleic acids,a short length of RNA or single-stranded DNA required for initiating synthesis directed by polymerases.
the DNA sequence from above pattern interpreted as follows -
Right Arrow consider the band which is lowest from top among all the bands, that will be the 1st nucleotide at 5’ end of DNA sequence . Here its the band in A lane. Thus A is the 1st nucleotide of DNA sequence (5’-3’)
Right Arrow then consider the 2nd most lowest band among all the bands, that will be the 2nd nucleotide of DNA sequence. Here T is the 2nd most lowest band and hence is the 2nd nucleotide.
Right Arrow now continue the pattern of reading the pattern and find the DNA sequence
Thus based on above method, the sequenced strand is as follows -
—> 5’- ATGGACCAGTTG -3’
# So the complementary template strand of this sequenced DNA is as follows - ----> 5' - CAACTGGTCCAT - 3'
The shortest fragment generated by the dideoxynucleotide DNA sequencing reaction migrates to the bottom of the gel. That fragment represents the 5’ nucleotide of the sequenced strand.
Therefore, the 5’ to 3’ sequence of nucleotides in the sequenced strand can be determined by reading the bands in all lanes from the bottom of the gel to the top: 5’-ATGGACCAGTTG-3’, in this example.
The template strand is complementary and antiparallel to the sequenced strand: 5’-CAACTGGTCCAT-3’, in this example.
The densities of 14N/14N, 14N/15N, and 15N/15N double helices differ from each other and thus form bands in different positions. 14N/14N forms a band toward the top, 14N/15N in the middle, and 15N/15N toward the bottom.
When a solution of cesium chloride (CsCl) is subjected to high-speed centrifugation, a stable density gradient is formed. Meselson and Stahl found that when cell contents were subjected to centrifugation with a CsCl solution, a band of DNA formed at the CsCl density that matched the density of the DNA. This technique is called density-gradient centrifugation.
The test tubes below show the results of density-gradient centrifugation of five different DNA samples.
DNA from cells grown in 15N contains only 15N/15N double helices.
DNA from cells grown in 14N contains only 14N/14N double helices.
A 1:1 mixture of DNA from cells grown in 14N and cells grown in 15N contains both 14N/14N and 15N/15N double helices.
A 1:1 mixture of DNA from cells grown in 14N and cells grown in 15N, heated and then cooled, contains 14N/14N, 14N/15N, and 15N/15N double helices.
DNA containing one strand of 15N-DNA and one strand of 14N-DNA contains only 14N/15N double helices.
Meselson and Stahl designed an experiment that would allow them to discern whether DNA replication occurs in a dispersive, semiconservative, or conservative manner.
They started with E. coli that had been growing for many generations in medium containing 15N.
They then transferred the bacteria into medium containing only 14N, and allowed the bacteria to undergo two rounds of DNA replication.
After each round of replication, the scientists performed density-gradient centrifugation of the DNA.
The scientists reasoned that each of the 3 models would predict different DNA banding patterns after the two rounds of replication.
Notice that after one round of replication, the dispersive and semiconservative models predict identical results, whereas the conservative model predicts different results. By continuing the experiment through a second round of replication, would all three models lead to different predictions?
When Meselson and Stahl performed this experiment, their results were consistent with the pattern predicted by semiconservative replication, confirming that as the correct model. When DNA replicates, the two parent strands separate, and each strand serves as a template for the synthesis of a new DNA strand.
Note that Meselson and Stahl were able to rule out the conservative model based on the results after one round of replication. However, they could not rule out the dispersive model at that point because that model predicts the same pattern as the semiconservative model. But, the results after two rounds did enable them to rule out the dispersive model.
To confirm the semiconservative model of replication, it was important for Meselson and Stahl to quantify the amount of DNA in each band produced by density-gradient centrifugation. To accomplish this, they took advantage of the fact that DNA absorbs ultraviolet light, and used UV light to photograph each tube.
By scanning the UV photographs with a microdensitometer, graphs like the ones below were produced. The height of each peak in the graph is directly proportional to the concentration of DNA in the corresponding band. Also, the position of each peak reflects the 14N and 15N content of the band.
Suppose that the scientists analyzed the same amount of DNA (10 units) by density-gradient centrifugation after two, three, and four rounds of replication in 14N medium. What would you predict the microdensitometer graph would look like after each round?
After transfer to 14N medium, new DNA strands are made from 14N only. So, with each round of replication, the amount of 14N/14N-DNA increases relative to the amount of 14N/15N-DNA. Thus, the height of the 14N/14N peak (at position 3, closer to the top of the tube) increases with each round of replication, while the 14N/15N peak (at position 5) decreases.
Chp. 11: DNA Replication & Recombination:
Overall Questions:
- How does DNA replicate?
- How is replication done accurately?
- How does recombination occur?
In the S phase of the cell cycle, ___ occurs
In the S phase of the cell cycle, DNA synthesis occurs
Challenge of Replication
Human DNA:
3,000,000,000 base pairs on 23 chromosomes
Must replicate it and do it accurately
Even a one in one million error rate would lead to 3000 errors per replication! Oops!
Therefore the system must be VERY, VERY accurate.
DNA Replication Hypothesis Proposed by Watson & Crick, 1953
“It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material”
semiconservative replication
A mode of DNA replication in which a double-stranded molecule replicates in such a way that the daughter molecules are each composed of one parental (old) and one newly synthesized strand.
Semi-conservative replication: each DNA strand could serve as a template for its own replication.
Result: 1 old strand and 1 new strand in each double stranded DNA molecule after one round of replication
