Chapter 11: DNA Replication & Recombination Flashcards Preview

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Flashcards in Chapter 11: DNA Replication & Recombination Deck (126):
1

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.

2

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

DNA polymerase catalyzes the addition of nucleotides to a growing DNA chain.

3

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.

4

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

5

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.

6

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.

7

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.

8

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.

9

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

10

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

11

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

12

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

13

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

14

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

15

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.

16

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

17

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.

18

All of the following are related to telomeres:

telomerase enzyme

links to the aging process

short tandem repeats located at the ends of telomeres

19

True or False?

Telomeres are found in eukaryotes and prokaryotes.

FALSE

Telomeres are found in eukaryotes but not prokaryotes.

20

Structures located at the ends of eukaryotic chromosomes are called ________.

telomeres

21

True or False?

Telomerase is an RNA-containing enzyme that adds telomeric DNA sequences onto the ends of linear chromosomes.

True

22

Which structures can be involved in recombination?

Chromatids of homologous chromosomes

Chromatids of homologous chromosomes can recombine during meiosis.

23

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.

24

Which process does not occur during recombination?

DNA polymerization

Recombination does not include the synthesis of new DNA.

25

Which processes do occur during recombination?

Ligation

Nicking of the sugar‑phosphate backbone

Strand displacement

26

True or False?

DNA replicates conservatively, which means that one of the two daughter double helices is "old" and the other is "new."

FALSE

27

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.

28

DNA replication occurs in the 5' to 3' direction; that is, new nucleoside monophosphates are added to the 3' end.

TRUE

29

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.

30

DNA polymerase III adds nucleotides ________.

to the 3' end of the RNA primer

31

replication fork

The Y-shaped region of a chromosome associated with the site of DNA replication.

32

Okazaki fragment

The short, discontinuous strands of DNA produced on the lagging strand during DNA synthesis.

33

polymerases

Enzymes that catalyze the formation of DNA and RNA from deoxynucleotides and ribonucleotides, respectively.

34

leading strand

During DNA replication,the strand synthesized continuously in the direction of the replication fork.

35

lagging strand

During DNA replication,the strand synthesized in a discontinuous fashion, in the direction opposite of the replication fork. See also Okazaki fragment.

36

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.

37

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).

38

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.

39

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.

40

primer

In nucleic acids,a short length of RNA or single-stranded DNA required for initiating synthesis directed by polymerases.

41

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.

42

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.

43

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?

44

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.

45

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.

46

Chp. 11: DNA Replication & Recombination:

Overall Questions:
-How does DNA replicate?
-How is replication done accurately?
-How does recombination occur?

47

In the S phase of the cell cycle, ___ occurs

In the S phase of the cell cycle, DNA synthesis occurs

48

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.

49

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”

50

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

51

Conservative replication

New & old strands re-anneal w/one another following replication

52

Testing of Watson & Crick Hypothesis of semi-conservative replication of DNA

if one could differentially label the old & new strands in some way....

Differential labeling of the DNA strands is the key of the Meselson-Stahl experiments

53

Meselson-Stahl Experiment

Relied upon differential labeling of old & new DNA strands provided by the availability of “heavy” Nitrogen (15N) that could be incorporated into DNA

Grow E. coli in presence of
“heavy” 15N, heavy N incorporated in all DNA strands

Centrifuge DNA after 3 generations, 2 DNA bands observed, One “light” 14N band w/ more DNA, and
intermediate band with less DNA

Results supported the semi-conservative nature of DNA replication

54

Eukaryotic Replication

Examined using autoradiography using 3H-thymidine in the bean Vicia faba

Similar experiment to Meselson-Stahl, they used 3H as a way to label DNA

Observe results in this case by visualization using autoradiography

55

Eukaryotic Replication:
Vicia faba Experiment

Grow V. faba in 3H-thymidine,
label all chromatids after Metaphase

Following replication II, only 1 chromatid will still be labeled if no recombination occured.

With recombination, you will observe reciprocal labeling of chromatids

Results supported semi-conservative replication in eukaryotes

Later work on other eukaryotic organisms supports the ubiquity of semi-conservative replication

56

Origin of replication (ori):

Where replication of DNA begins.

57

Replication fork

where DNA strands are unwound.

58

Replicon

a length of DNA replicated at one replication fork.

The unit of DNA replication, beginning with DNA sequences necessary for the initiation of DNA replication.

In bacteria, the entire chromosome is a replicon.

59

Replication Was First Studied in Prokaryotes

E. coli

Used radioisotopes and autoradiography

Observed one origin of replication

245 bp region where replication originated (oriC)

Whole chromosome is replicated in one replicon

Results supported bi-directional replication, moving away from oriC in both directions

60

radioactive isotope

An unstable isotope with an altered number of neutrons that emits ionizing radiation during decay as it is transformed to a stable atomic configuration.

61

autoradiography

Production of a photographic image by radioactive decay.

Used to localize radioactively labeled compounds within cells and tissues or to identify radioactive probes in various blotting techniques.

62

Summary of early findings

DNA replication is semi-conservative in nature.

DNA replication begins at a defined point (oriC) in E. coli.

DNA replication is bi-directional in E. coli

63

Kornberg Experiments

Looking for enzyme that was able to replicate DNA in an in vitro system.

Isolated an enzyme he called DNA polymerase I (DNA Pol I).

Major in vitro requirements: dNTP’s, template DNA (with partial complement)

GET DNA EXTENSION

64

in vivo

Literally, in the living; occurring within the living body of an organism.

65

Join 5’ P to 3’ OH, Release 2P

DNA pol I synthesizes this reaction

66

Kornberg cont...

Kornberg sought to determine the accuracy of the newly synthesized DNA.

Indirect method which examined base composition found equivalent composition.

This result indirectly supported accuracy, but did not prove it.

Kornberg sought to show that DNA pol I could create “biologically active DNA”.

Biologically active DNA is capable of supporting metabolic activities and directs reproduction of the organism from which it was originally duplicated.

67

Problems of DNA pol I in vitro

Much slower rate of synthesis than in vivo.

Worked better on single stranded template than double stranded template.

Also, DNA pol I degraded DNA (exonuclease activity)

68

Kornberg’s reasoning

IF DNA pol I could make DNA that was biologically active, THEN it must be the major catalyzing force of DNA synthesis in the cell

69

øX174 phage (bacterial virus)

Single stranded circular DNA (+) strand infects E. coli

Phage inserts + strand
DNA into E. coli

Replicates (-) strand in E. coli, resulting in ds duplex or replicative form (RF)

the + strand directs - strand
replication

RF serves as template for its own replication

Only (+) strands are produced

(+) strands packaged into viral coats to form mature virus particles

70

Kornberg’s Phage Experiment

Relied on in vitro method to discriminate newly made (-) strand and subsequently made (+) strand

Final (+) strands used to infect E. coli

Hypothesis: IF these final (+) strands were capable of producing mature phage particles following transfection, THEN DNA pol I replicated DNA that was biologically active

Synthetically made + strands added to E. coli protplasts (transfection),
NEW PHAGE ARE PRODUCED

Final (+) strand of øX174 infected E. coli and produced infectious mature phage particles.

71

Kornberg’s Phage Experiment: Conclusion

DNA pol I had demonstrated biological activity of the DNA that it replicated

DNA replication was accurate, any alteration in the 5386 bases of øX174 would have probably rendered it non-viable

72

Kornberg’s Phage Experiment: More questions ....

E. coli polA1 mutants found, lacked DNA polymerase I but still survived, but susceptible to UV and radiation damage to DNA

Therefore, there must be at least one other polymerase active in E. coli

Also, DNA pol I may play some role in DNA repair

73

Other DNA polymerases

DNA polymerase II and III

Similarities with DNA pol I, and differences

All can elongate DNA from a primer, 5’-3’ polymerization

All have 3’-5’ exonuclease activity

DNA pol I has 5’-3’ exonuclease activity, the others do NOT

74

DNA polymerase III holoenzyme

Dimer of 10 different subunits

Molecular weight of 900,000Da (900kDa=molecular weight)

THINK BIG STRUCTURE, most proteins have mass between 10-100kDa

α, ε, θ subunits make up core enzyme
α does 5’--> 3’ polymerization
ε does 3’-->5’ exonuclease

Second group of 5 subunits form γ complex involved in enzyme loading

Energy for this comes from hydrolysis of ATP

ß subunit is “clamp” that prevents core enzyme from falling off during polymerization

t facilitates dimerization between 2 core polymerases

75

Subunit α

5’-->3’ polymerization

76

Subunit ε

3’-->5’ exonuclease

77

Subunit θ

Core assembly

78

exonuclease

An enzyme that breaks down nucleic acid molecules by breaking the phosphodiester bonds at the 3'- or 5'-terminal nucleotides.

79

DNA polymerase III core enzyme

α, ε, θ subunits make up core enzyme

80

DNA polymerase III non-core subunits

Second group of 5 subunits form γ complex involved in enzyme loading

ß subunit is “clamp” that prevents core enzyme from falling off during polymerization

t facilitates dimerization between 2 core polymerases

81

Subunit γ complex (5 subunits)

Clamp enzyme loading

Second group of 5 subunits form γ complex involved in enzyme loading

82

Subunit ß

Sliding clamp

prevents core enzyme from falling off during polymerization

83

Subunit t

Core dimerization

facilitates dimerization between 2 core polymerases

84

Replisome

another cell machine

DNA polymerase III holoenzyme and other proteins at replication fork

Huge complex of proteins

85

How is DNA unwound so that replication can happen?

In E. coli, oriC origin of replication

245 bp

Consists of repeating seqs of 9 and 13 bases

Sequences recognized by multimer of protein DnaA

Binding of DnaA allows subsequent binding of DnaB and DnaC, further opens the helix

DnaB & DnaC then bind to further open helix

NOTE: ALL STEPS REQUIRE ATP HYDROLYSIS

86

Binding by DnaA’s results in _____ formation

Binding by DnaA’s results in helix bubble formation

87

How is DNA unwound so that replication can happen?
cont...

Binding of DnaA allows subsequent binding of DnaB and DnaC, further opens the helix

All require ATP as energy source to break H bonds

Open configuration then stabilized by single-stranded binding proteins

88

How is the tension created by unwinding of DNA relieved?

Unwinding creates coiling tension: supercoiling

Supercoiling is relaxed by DNA gyrase enzyme (a DNA topoisomerase)

DNA gyrase makes single or double stranded breaks, strands resealed

Requires ATP

89

DNA polymerases cannot initiate replication, so what is the primer used to allow replication?

DNA pol I, II, & III cannot initiate polymerization

Short segment of RNA (5-15 nucs) made complementary to DNA

Done by an RNA polymerase (RNA primase)

Allows DNA pol III to proceed

90

DNA polymerases work 5’--3’

DNA strands are antiparallel

One goes 5’-->3’, the other goes 3’-->5’

DNA pol III only synthesizes 5’--3’

Therefore, can only proceed continuously on one of the strands (LEADING STRAND)

91

If DNA polymerases work 5’--3’, then how is the other strand replicated?

Other strand must be synthesized discontinuously (LAGGING STRAND)

Lagging strand synthesis done in segments

RNA primers are added to lagging strand

DNA polymerization begins from these fragments

Proceeds until the next RNA primer encountered

Called Okazaki fragments

Okazaki fragments: 1000-2000bp in E. coli

92

Leading and Lagging strands are synthesized simultaneously

The synthesis going on in boxed area of enzyme is in 5’->3’ in
both leading and lagging strands, achieved by physical looping
of the lagging strand prior to entry into polymerase

93

How are the initial primers removed?

DNA polymerase I removes the RNA primer
(5’-->3’ exonuclease activity, UNIQUE to DNA pol I)

94

After primer removal, how is the gap filled?

DNA polymerase I also fills in the primer region with DNA

95

After gap is filled, how are the ends of the gaps joined?

Done by DNA ligase

Catalyzes phosphodiester bond to seal the gap

96

How is replication proofread to insure accuracy?

Synthesis is very good, but not perfect

All polymerases possess 3’-->5’ exonuclease activity

Detects and excise mismatches

97

Overall Replication Model

Helicase/primase at replication fork

ssbp stabilize open helix

Gyrase removes tension

Beta clamp binds DNA

Simultaneous leading and lagging strand synthesis

RNA primers removed, gaps filled, Okasaki fragments joined

98

Eukaryotic DNA Replication

Same general features as prokaryotic replication, but differences exist

Rate of eukaryotic synthesis is slower

Larger amounts of DNA must be replicated

Studied first in yeast (model system)

Multiple origins of replication, called autonomously replicating sequences (ARS) in yeast

Contains consensus seq of 11bp

ARS bound in G1 of cell cycle by specific proteins

Forms origin recognition complex (ORC)

Replication begins in S phase after involvement of other signals (kinases)

Polymerases bind, replication begins

Signals also insure that replication only occurs once in S phase

Histone proteins must be disassociated prior to synthesis and re-associated after

99

histones

Positively charged proteins complexed with DNA in the nucleus. They are rich in the basic amino acids arginine and lysine,and function in coiling DNA to form nucleosomes.

100

Eukaryotic DNA Replication cont....

Pol α and δ are key eukaryotic DNA polymerases

Synthesis begins with pol α (low processivity)

Then, DNA “polymerase switching” leads to pol δ (high processivity) to continue replication (100X rate increase)

Okazaki fragments smaller (100-150nucs)

Multiple origins of replication result in reasonable length of time to replicate larger genomes

101

Eukaryotic Chromosome Telomere Replication

Antiparallel nature of DNA leads to problem at the ends of chromosomes (telomeres)

Inability to fill the final gap will result in shortening of the ends of the chromosome during each round of replication

Telomere replication solved by an elegant mechanism

First observed in Tetrahymena

Ends of chromosomes are repeated sequences

Eg., 5’-TTGGGG-3’ in Tetrahymena

102

Eukaryotic Chromosome Telomere Replication cont...

Special enzyme (Telomerase) contains protein and RNA (ribonulceoprotein)

RNA serves as guide and template

RNA has seq. 3’-AACCCC-5’

Basepairing overlap followed by reverse transcription of the overlap

Leads to extension of the lagging strand

Telomerase function found in all eukaryotic organisms studied

Humans telomeres are 5’-TTAGGG-3’

Human telomerase has RNA with seq 3’-AAUCCC-5’

Telomerase not active in most somatic cells --->> ends degrade over time

103

telomerase

The enzyme that adds short, tandemly repeated DNA sequences to the ends of eukaryotic chromosomes.

104

Telomerase: Key to Aging?

Telomeres shorten with age

Cells from older humans undergo fewer replications before undergoing senescence

Telomerase not present in somatic cells

Could we reverse aging by adding telomerase to cells?

What about negative effects of added telomerase?

Telomerase is active in malignant cells

Could telomerase be an anticancer drug target?

105

Eukaryotic DNA Recombination

General (homologous recombination)

Like replication, recombination is directed by specific enzymes

Series of enzymatic processes accomplish process

106

Gene Conversion

Consequence of DNA recombination

Recombination leads to mismatch if wild type and mutant type are in the recombined region

Mismatch can be excised on either strand

Subsequent repair can lead to conversion to mutant or wild type

107

gene conversion
definition

The process of nonreciprocal recombination by which one allele in a heterozygote is converted into the corresponding allele.

108

recombination

The process that leads to the formation of new allele combinations on chromosomes.

109

The DNA double helix is composed of two strands of DNA; each strand is a polymer of DNA nucleotides. Each nucleotide consists of a sugar, a phosphate group, and one of four nitrogenous bases. The structure and orientation of the two strands are important to understanding DNA replication.

The DNA double helix is constructed from two strands of DNA, each with a sugar-phosphate backbone and nitrogenous bases that form hydrogen bonds, holding the two strands together. Each DNA strand has two unique ends. The 3' end has a hydroxyl (-OH) group on the deoxyribose sugar, whereas the 5' end has a phosphate group. In the double helix, the two strands are antiparallel, that is, they run in opposite directions such that the 3' end of one strand is adjacent to the 5' end of the other strand.

110

In DNA replication in bacteria, the enzyme DNA polymerase III (abbreviated DNA pol III) adds nucleotides to a template strand of DNA. But DNA pol III cannot start a new strand from scratch. Instead, a primer must pair with the template strand, and DNA pol III then adds nucleotides to the primer, complementary to the template strand. Each of the four images below shows a strand of template DNA (dark blue) with an RNA primer (red) to which DNA pol III will add nucleotides.

In the example above, DNA pol III would add an adenine nucleotide to the 3' end of the primer, where the template strand has thymine as the next available base. You can tell which end is the 3' end by the presence of a hydroxyl (-OH) group.
The structure of DNA polymerase III is such that it can only add new nucleotides to the 3' end of a primer or growing DNA strand (as shown here). This is because the phosphate group at the 5' end of the new strand and the 3' -OH group on the nucleoside triphosphate will not both fit in the active site of the polymerase.

111

DNA replication always begins at an origin of replication. In bacteria, there is a single origin of replication on the circular chromosome. Beginning at the origin of replication, the two parental strands separate, forming a replication bubble. At each end of the replication bubble is a replication fork where the parental strands are unwound and new daughter strands are synthesized. Movement of the replication forks away from the origin expands the replication bubble until two identical chromosomes are ultimately produced.

Antiparallel elongation at the replication forks. Keep in mind that the two strands in a double helix are oriented in opposite directions, that is, they are antiparallel.

DNA polymerase III can only add nucleotides to the 3' end of a new DNA strand. Because the two parental DNA strands of a double helix are antiparallel (go from 3' to 5' in opposite directions), the direction that DNA pol III moves on each strand emerging from a single replication fork must also be opposite.
For example, in the replication fork on the left, the new strand on top is being synthesized from 5' to 3', and therefore DNA pol III moves away from the replication fork. Similarly, the new strand on the bottom of that same replication fork is being synthesized from 5' to 3'. But because the bottom parental strand is running in the opposite direction of the top parental strand, DNA pol III moves toward the replication fork.
In summary, at a single replication fork, one strand is synthesized away from the replication fork, and one strand is synthesized toward the replication fork. When you look at both replication forks, note that a single new strand is built in the same direction on both sides of the replication bubble.

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Topoisomerase breaks a covalent bond in the backbone of one parental strand.

Single-strand binding proteins bind to parental DNA immediately after the helicase, preventing the two single strands from joining andre-forming a double helix

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topoisomerase

A class of enzymes that converts DNA from one topological form to another. During replication, a topoisomerase, DNA gyrase, facilitates DNA replication by reducing molecular tension caused by supercoiling upstream from the replication fork.

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The newly added nucleotide forms a bond with the hydroxyl (-OH) group on the 3' end of the primer.

As DNA replication continues and the replication bubble expands, the parental double helix is unwound and separated into its two component strands. This unwinding and separating of the DNA requires three different types of proteins: helicase, topoisomerase, and single-strand binding proteins.

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At each replication fork, helicase moves along the parental DNA, separating the two strands by breaking the hydrogen bonds between the base pairs. (This makes the two parental DNA strands available to the DNA polymerases for replication.) As soon as the base pairs separate at the replication fork, single-strand binding proteins attach to the separated strands and prevent the parental strands from rejoining.

As helicase separates the two parental strands, the parental DNA ahead of the replication fork becomes more tightly coiled. To relieve strain ahead of the replication fork, topoisomerase breaks a covalent bond in the sugar-phosphate backbone of one of the two parental strands. Breaking this bond allows the DNA to swivel around the corresponding bond in the other strand and relieves the strain caused by the unwinding of the DNA at the helicase.

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Helicase

Breaks H-bonds between bases

Binds AT the Replication Fork

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Topoisomerase

Breaks covalent bonds in DNA backbone

Binds AHEAD of the Replication Fork

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Single-Strand Binding Protien

Prevents H-bonds between bases

Binds AFTER the Replication Fork

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Polymerase 1 elongates its strand toward the replication fork, but polymerase 2 elongates its strand away from the replication fork.

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The role of RNA primers in DNA replication:
DNA polymerase III cannot initiate the synthesis of a new daughter strand; it can only add nucleotides to an existing strand that is paired with aparental (template, dark blue) strand. The enzyme primase creates a short RNA primer (red) that is complementary to the parental strand. Thisprimer serves as a starting point for DNA pol III, which adds nucleotides to the 3' end of the primer.

Synthesis of the first primer on the lagging strand
As a replication bubble forms at an origin of replication, the primers for the leading strands are produced first. In the image below, you can seethat the first parts of the leading strands have already been synthesized by the time the first lagging strand primers are produced.The reason for this lag is that about 1500 nucleotides (the average length of a bacterial lagging strand segment) of parental DNA must beexposed at the replication fork before lagging strand synthesis can begin. This makes room for the first lagging strand segment between thefirst primer and the origin of replication. DNA polymerase III produces the first lagging strand segments (not shown) by adding nucleotides to the3' ends of the lagging strand primers.

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As the two parental (template) DNA strands separate at a replication fork, each of the strands is separately copied by a DNA polymerase III (orange), producing two new daughter strands (light blue), each complementary to its respective parental strand. Because the two parental strands are antiparallel, the two new strands (the leading and lagging strands) cannot be synthesized in the same way.

Because DNA polymerase III can only add nucleotides to the 3' end of a new DNA strand and because the two parental DNA strands are antiparallel, synthesis of the leading strand differs from synthesis of the lagging strand.
The leading strand is made continuously from a single RNA primer located at the origin of replication. DNA pol III adds nucleotides to the 3' end of the leading strand so that it elongates toward the replication fork.
In contrast, the lagging strand is made in segments, each with its own RNA primer. DNA pol III adds nucleotides to the 3' end of the lagging strand so that it elongates away from the replication fork.
One side of the origin of replication, a new strand is synthesized as the leading strand, and on the other side of the origin of replication, that same new strand is synthesized as the lagging strand. The leading and lagging strands built on the same template strand will eventually be joined, forming a continuous daughter strand.

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LEADING STRAND ONLY

Made continuously

Only one Primer needed

Daughter strand elongates toward the replication fork

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LAGGING STRAND ONLY

Made in segments

Multiple Primers needed

Daughter strand elongates away from the replication fork

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BOTH LAGGING STRAND and LEADING STRAND

Synthesized 5' to 3'

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As soon as the replication bubble opens and the replication machinery is assembled at the two replication forks, the two primers for the leading strands are produced. The production of the first primers on the lagging strands (those closest to the origin of replication) is delayed slightly because the replication forks must open up further to expose the template DNA for the lagging strands.

After completion of the first segments of the lagging strands, additional template DNA must be exposed before the second primers can be produced. And after completion of the second segments, additional template DNA must be exposed before the third primers can be produced. In summary, because of the way the replication bubble expands, the lagging strand primers near the origin of replication were produced before the primers near the replication forks.

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In contrast to the leading strand, the lagging strand is synthesized as a series of segments called Okazaki fragments. The diagram below illustrates a lagging strand with the replication fork off-screen to the right. Fragment A is the most recently synthesized Okazaki fragment. Fragment B will be synthesized next in the space between primers A and B.

Synthesis of the lagging strand is accomplished through the repetition of these steps:

Step 1: A new fragment begins with DNA polymerase III binding to the 3' end of the most recently produced RNA primer, primer B in this case, 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.
These steps will be repeated as the replication fork opens up. Try to visualize primer C being produced to the right (closest to the replication fork). Fragment C would be synthesized and joined to fragment B following the steps described here.