Organization of DNA

Lecture 3

• Large DNA molecules must be packaged to fit inside the cell and still be functional !

Supercoiling• mt DNA and the DNA of most prokaryotes are

closed circular structures. These molecules may exist as relaxed circles or as supercoiled structures in which the helix is twisted around itself in three-dimensional space.

• Supercoiling results from strain on the molecule caused by under- or overwinding the double helix:

• Negatively supercoiled DNA: if the DNA is wound more loosely than in Watson-Crick DNA. This form is required for most biologic reactions.

• Positively supercoiled DNA: if the DNA is wound more tightly than in Watson-Crick DNA.

• Topoisomerases – can change the amount of supercoiling.

– make transient breaks in DNA strands by alternately breaking and resealing the sugar-phosphate backbone.

– E.g., in Escherichia coli, DNA gyrase (DNA topoisomerase II) can introduce negative supercoiling into DNA, whereas DNA topoisomerase I can relax the supercoils

Nucleosomes and Chromatin

• Nuclear DNA in eukaryotes is found in chromatin associated with histones and nonhistone proteins.

• The basic packaging unit of chromatin is the nucleosome • Histones are rich in lys and arg a positive charge.• Two copies each of histones H2A, H2B, H3, and H4

aggregate to form the histone octamer.• DNA is wound around the outside of this octamer to

form a nucleosome (a series of nucleosomes is sometimes called "beads on a string").

Nucleosome and Nucleofilament Structure in Eukaryotic DNA

Chromosome structure

(Nucleofilament)

Organization of human DNA, illustrating the structure of nucleosomes.

Structural organization of eukaryotic DNA.

• Histone H1 is associated with the linker DNA found between nucleosomes to help package them into a solenoid-like structure, which is a thick 30-nm fiber.

• Further condensation occurs to eventually form the chromosome. Each eukaryotic chromosome contains one linear molecule of DNA.

Electron micrograph of a protein-depleted human metaphase chromosome, showing the residual chromosomescaffold and loops of DNA. Individual DNA fibers can be best seen at the edge of the DNA loops. Bar = 2μ. (From Paulson JR,Laemmli UK [1977] The structure of histone-depleted metaphase chromosomes. Cell 12:817–828.

Cells in interphase contain two types of chromatin:

• Euchromatin is loosely packaged and transcriptionally active.

• Heterochromatin is tightly packaged and inactive.

• Euchromatin generally corresponds to looped 30-nm fibers.

• Heterochromatin is more highly condensed.

An Interphase Nucleus

• Gene expression requires that chromatin be opened for access by transcription complexes (RNA polymerase and transcription factors).

• Chromatin-modifying activities include:– Histone acetylation– Histone phosphorylation

• During mitosis, all the DNA is highly condensed to allow separation of the sister chromatids. This is the only time in the cell cycle when the chromosome structure is visible.

• Chromosome abnormalities may be assessed on mitotic chromosomes by karyotype analysis (metaphase chromosomes) and by banding techniques (metaphase, prophase or prometaphase), which identify aneuploidy, translocations, deletions, inversions, and duplications.

1. Cytosine arabinoside (araC) is used as an effective chemotherapeutic agent for cancer, although resistance to this drug may eventually develop. In certain cases, resistance is related to an increase in the enzyme cytidine deaminase in the tumor cells. This enzyme would inactivate araC to form

A. cytosine

B. cytidylic acid

C. thymidine arabinoside

D. uracil arabinoside

E. cytidine

2. A double-stranded RNA genome isolated from a virus in the stool of a child with gastroenteritis was found to contain 15% uracil. What is the percentage of guanine in this genome?

A. 15

B. 25

C. 35

D. 75

E. 85

3. Endonuclease activation and chromatin fragmentation are characteristic features of eukaryotic cell death by apoptosis. Which of the following chromosome structures would most likely be degraded first in an apoptotic cell?

A. Barr bodyB. 10nm fiberC. 30 nm fiberD. CentromereE. Heterochromatin

DNA Replication and Repair

OVERVIEW OF DNA REPLICATION

• Genetic information is transmitted from parent to progeny by replication of parental DNA

• During DNA replication, the two complementary strands of parental DNA are pulled apart. Each of these parental strands is then used as a template for the synthesis of a new complementary strand (i.e., semiconservative).

• During cell division, each daughter cell receives one of the two identical DNA molecules.

Replication of Prokaryotic and Eukaryotic Chromosomes

• The bacterial chromosome is a closed, double-stranded circular DNA molecule having a single origin of replication.

• Separation of the two parental strands of DNA creates two replication forks that move away from each other in opposite directions

• Replication is, thus, a bidirectional process.

• The two replication forks eventually meet, resulting in the production of two identical circular molecules of DNA.

• Each eukaryotic chromosome contains one linear molecule of DNA having multiple origins of replication.

• Bidirectional replication occurs at each origin.• Completion of the process results in the

production of two identical linear molecules of DNA.

• DNA replication occurs in the nucleus during the S phase of the eukaryotic cell cycle.

• The two identical sister chromatids are separated during mitosis.

COMPARISON OF DNA AND RNA SYNTHESIS

• The overall process of DNA replication requires the synthesis of both DNA and RNA.

• These two types of nucleic acids are synthesized by DNA polymerases and RNA polymerases

3'→5'-Exonuclease activity enables DNA polymerase III to “proofread” the newly synthesized DNA strand.

Polymerases and Nucleases

• Polymerases are enzymes that synthesize nucleic acids by forming phosphodiester (PDE) bonds. Nucleases are enzymes that hydrolyze PDE bonds

• Exonucleases remove nucleotides from either the 5' or the3' end of a nucleic acid.

• Endonucleases cut within the nucleic acid and release nucleic acid fragments.

Table 1-2-1. Comparison of DNA and RNA Polymerases

*Certain DNA and RNA polymerases require RNA templates. These enzymes are most commonly associated with viruses.

Similarities include:• The newly synthesized strand is made in the 5'3'

direction.• The template strand is scanned in the 3'5' direction.• The newly synthesized strand is complementary and

antiparallel to the template strand.• Each new nucleotide is added when the 3' hydroxyl

group of the growing strand reacts with a nucleoside triphosphate, which is base-paired with the template strand. Pyrophosphate (PPi, the last two phosphates) is released during this reaction.

Differences include:• The substrates for DNA synthesis are the dNTPs, whereas

the substrates for RNA synthesis are the NTPs.• DNA contains thymine, whereas RNA contains uracil.• DNA polymerases require a primer, whereas RNA

polymerases do not. That is, DNA polymerases cannot initiate strand synthesis, whereas RNA polymerases can.

• DNA polymerases can correct mistakes ("proofreading"), whereas RNA polymerases cannot. DNA polymerases have 3'5' exonuclease activity for proofreading.

Steps of DNA Replication in E. coli

1. The base sequence at the origin of replication is recognized and bound by the dna A protein. The two parental strands of DNA are pulled apart to form a "replication bubble".

2. Helicase uses energy from ATP to break the hydrogen bonds holding the base pairs together. This allows the two parental strands of DNA to begin unwinding and forms two replication forks.

3. Single-stranded DNA binding protein (SSB) binds to the single-stranded portion of each DNA strand, preventing the strands from reassociating and protecting them from degradation by nucleases.

Elongation of the leading and lagging strands (E. coli)

4. Primase synthesizes a short (about 10 nucleotides) RNA primer in the 5'3' direction, beginning at the origin on each parental strand. The parental strand is used as a template for this process. RNA primers are required because DNA polymerases are unable to initiate synthesis of DNA, but can only extend a strand from the 3' end of a preformed "primer".

5. DNA polymerase III begins synthesizing DNA in the 5'3' direction, beginning at the 3' end of each RNA primer. The newly synthesized strand is complementary and antiparallel to the parental strand used as a template. This strand can be made continuously in one long piece and is known as the "leading strand".

• The "lagging strand" is synthesized discontinuously as a series of small fragments (about 1,000 nucleotides long) known as Okazaki fragments. Each Okazaki fragment is initiated by the synthesis of an RNA primer by primase, and then completed by the synthesis of DNA using DNA polymerase III. Each fragment is made in the 5'3' direction.

• There is a leading and a lagging strand for each of the two replication forks on the chromosome.

Discontinuous synthesis of DNA.

6. RNA primers are removed by DNA polymerase I. This enzyme removes the ribonucleotides one at a time from the 5' end of the primer (5'3' exonuclease). DNA polymerase I also fills in the resulting gaps by synthesizing DNA, beginning at the 3' end of the neighboring Okazaki fragment.

7. Both DNA polymerase I and III have the ability to "proofread" their work by means of a 3'5' exonuclease activity. If DNA polymerase makes a mistake during DNA synthesis, the resulting unpaired base at the 3' end of the growing strand is removed before synthesis continues.

Removal of RNA primer and filling of the resulting “gaps” by DNA polymerase-I

8. DNA ligase seals the "nicks" between Okazaki fragments, converting them to a continuous strand of DNA.

• DNA gyrase (DNA topoisomerase II) provides a "swivel" in front of each replication fork. As helicase unwinds the DNA at the replication forks, the DNA ahead of it becomes overwound and positive supercoils form. DNA gyrase inserts negative supercoils by nicking both strands of DNA, passing the DNA strands through the nick, and then resealing both strands again.

• DNA topoisomerase I can relieve supercoiling in DNA molecules by the transient breaking and resealing of just one of the strands of DNA. Replication is completed when the two replication forks

meet each other on the side of the circle opposite the origin.

Formation of a phosphodiester bond by DNA ligase.

• Quinolones are a family of drugs that block the action of topoisomerases. Nalidixic acid kills bacteria by inhibiting DNA gyrase.

• Inhibitors of eukaryotic topoisomerase II (etoposide, teniposide) are becoming useful as anticancer agents.

9. The mechanism of replication in eukaryotes is believed to be very similar to this. However, the details have not yet been completely worked out.

Eukaryotic DNA Polymerases

• DNA polymerase δ elongates leading and Okazaki fragments during replication.

• DNA polymerase α initiates synthesis and has primase activity.

• DNA polymerase γ replicates mitochondrial DNA.• DNA polymerases β and ε are thought to participate

primarily in DNA repair. DNA polymerase ε may substitute for DNA polymerase δ in certain cases.

Telomerase

• Telomeres are repetitive sequences at the ends of linear DNA molecules in eukaryotic chromosomes.

• With each round of replication in most normal cells, the telomeres are shortened because DNA polymerase cannot complete synthesis of the 5' end of each strand.

• This contributes to the aging of cells, because eventually the telomeres become so short that the chromosomes cannot function properly and the cells die.

• Telomerase is an enzyme in eukaryotes used to maintain the telomeres. It contains a short RNA template complementary to the DNA telomere sequence, as well as telomerase reverse transcriptase activity (hTRT). Telomerase is thus able to replace telomere sequences that would otherwise be lost during replication. Normally telomerase activity is present only in embryonic cells, germ (reproductive) cells, and stem cells, but not in somatic cells.

• Cancer cells often have relatively high levels of telomerase, preventing the telomeres from becoming shortened and contributing to the immortality of malignant cells.

Mechanism of action of telomerase

Telomere Ends are Loops Not “naked” ends

• Telomeres and TRF1 (telomere repeat-binding factor) protein stabilize ends for two reasons (replication gaps & DNase degradation)

Reverse Transcriptase

• Reverse transcriptase is an RNA-dependent DNA polymerase that requires an RNA template to direct the synthesis of new DNA. Retroviruses, most notably HIV, use this enzyme to replicate their RNA genomes.

• DNA synthesis by reverse transcriptase in retroviruses can be inhibited by AZT, ddC, and ddI.

Eukaryotic cells also contain reverse transcriptase activity.

• Associated with telomerase (hTRT).

• Encoded by retrotransposons (residual viral genomes permanently maintained in human DNA) that play a role in amplifying certain repetitive sequences in DNA

Bridge to Pharmacology• Quinolones and DNA Gyrase: Quinolones and

fluoroquinolones inhibit DNA gyrase (prokaryotic topoisomerase II), preventing DNA replication and transcription. These drugs, which are most active against aerobic gram-negative bacteria include:

• Nalidixic acid• Ciprofloxacin• Norfloxacin• Resistance to the drugs has developed overtime; current

uses include treatment of gonorrhea, and upper and lower urinary tract infections in both sexes.

Steps and Proteins Involved in DNA Replication

DNA REPAIR

• The structure of DNA can be damaged in a number of ways through exposure to chemicals or radiation.

• Incorrect bases can also be incorporated during replication.

• Multiple repair systems have evolved, allowing cells to maintain the sequence stability of their genomes.

• If cells are allowed to replicate their DNA using a damaged template, there is a high risk of introducing stable mutations into the new DNA.

• Thus any defect in DNA repair carries an increased risk of cancer.

• Most DNA repair occurs in the G1 phase of the eukaryotic cell cycle. Mismatch repair occurs in the G2 phase to correct replication errors.

Repair of Thymine Dimers

• Ultraviolet light induces the formation of dimers between adjacent thymines in DNA (also occasionally between other adjacent pyrimidines).

• The formation of thymine dimers interferes with DNA replication and normal gene expression.

• Thymine dimers are eliminated from DNA by a nucleotide excision-repair mechanism.

Steps in nucleotide excision repair:

• An excision endonuclease (excinuclease) makes nicks in the phosphodiester backbone of the damaged strand on both sides of the thymine dimer and removes the defective oligonucleotide.

• DNA polymerase fills in the gap by synthesizing DNA in the 5'3' direction, using the undamaged strand as a template.

• DNA ligase seals the nick in the repaired strand.

Thymine Dimer Formation and Excision-Repair

Nucleotide Excision Repair in Eukaryotes

• People with Xeroderma Pigmentosum (XP) have defective NER systems – at least eight different genes for NER

• They can not repair bulky DNA damage (e.g., T-dimers)• 1000 times more likely to develop skin cancer• Sun (UV) increases cancer in people with XP (T-dimers)

Global Genome NER Model

Table 1-2-3. DNA Repair

Repair of Deaminated and Missing Bases

• Cytosine can become deaminated spontaneously or by reaction with nitrous acid to form uracil. This leaves an improper base pair (G-U), which is eliminated by a base excision repair mechanism. Failure to repair the improper base pair can convert a normal G-C pair to an A-T pair.

Steps in base excision repair:• A uracil glycosylase recognizes and removes the uracil base,

leaving an apyrimidinic (AP) site in the DNA strand.• An AP endonuclease nicks the backbone of the damaged strand at

the missing base.• Additional nuclease action removes a few more bases, and the gap

is filled in by DNA polymerase.• DNA ligase seals the nick in the repaired strand.

Repair of Deaminated and Missing Bases

Base Excision Repair in E. coli

• Removes damaged base (e.g., O6-ethylguanine)

Mismatch Repair Fixes Heteroduplex DNA

• Heteroduplex DNA when two non-pairing bases on complementary strands

• Involves mut genes in E. coli• Involves msh genes (mutS homologes) in eukaryotes

PROBLEM:• Which strand is correct and which strand has the mutated

base?SOLUTION:• Methylation of A bases provides clues as to which is strand is

correct

Mismatch Repair (MMR)

• Mismatch repair deals with correcting mismatches of the normal bases. It can enlist the aid of enzymes involved in both base-excision repair (BER) and nucleotide-excision repair (NER) as well as using enzymes specialized for this function.

• Recognition of a mismatch requires several different proteins including one encoded by MSH2.

• Cutting the mismatch out also requires several proteins, including one encoded by MLH1.

• Mutations in either of these genes predisposes the person to an inherited form of colon cancer. So these genes qualify as tumor suppressor genes.

DNA Is Methylated in E. coli

• Fixes errors during replication!• “Old” DNA is methylated (CH3) at A in 5’-GATC-3’• Takes 10-20 min after DNA replication for “new” strand

to be methylated• Repair systems use “delay” to select old/correct strand

(methylated) from the new/defective strand (not methylated)

DNA Mismatch Involves Excision Repair

Repairing Strand Breaks

• Ionizing radiation and certain chemicals and viral infections can produce both single-strand breaks (SSBs) and double-strand breaks (DSBs) in the DNA backbone.

• Breaks in a single strand of the DNA molecule are repaired using the same enzyme systems that are used in Base-Excision Repair (BER).

Double-Strand Breaks (DSBs)

• There are two mechanisms by which the cell attempts to repair a complete break in a DNA molecule:

• Direct joining of the broken ends. This requires proteins that recognize and bind to the exposed ends and bring them together for ligation. They would prefer to see some complementary nucleotides but can proceed without them so this type of joining is also called Nonhomologous End-Joining (NHEJ).

• Errors in direct joining may be a cause of the various translocations that are associated with cancers.

Homologous Recombination

• Here the broken ends are repaired using the information – on the intact sister chromatid (available in G2 after chromosome

duplication), or on the – homologous chromosome (in G1; that is, before each chromosome

has been duplicated). This requires searching around in the nucleus for the homolog — a task sufficiently uncertain that G1 cells usually prefer to mend their DSBs by NHEJ, or on

– the same chromosome if there are duplicate copies of the gene on the chromosome oriented in opposite directions (head-to-head or back-to-back).

• Two of the proteins used in homologous recombination are encoded by the genes BRCA-1 and BRCA-2.

Resolving Holliday Junctions • 50% chance of resolving as Crossover Recombinant• 50% chance of resolving as Noncrossover Recombinant

Non-cross over recombinant

Cross over recombinant

Meiotic Recombination in Eukaryotes

Key differences:- Exonuclease- DNA polymerase

Diseases Associated with DNA Repair

• Inherited mutations that result in defective DNA repair mechanisms are associated with a predisposition to the development of cancer. Some examples:

• Xeroderma pigmentosum is an autosomal recessive disorder, characterized by extreme sensitivity to sunlight, skin freckling and ulcerations, and skin cancer. The most common deficiency occurs in the excinuclease enzyme.

• Hereditary nonpolyposis colorectal cancer results from a deficiency in the ability to repair mismatched base pairs in DNA that are accidentally introduced during replication.

Tumor Suppressor Genes and DNA Repair

• DNA repair may not occur properly when certain tumor suppressor genes have been inactivated through mutation or deletion:

• The p53 gene encodes a protein that prevents a cell with damaged DNA from entering the S phase. Inactivation or deletion associated with LiFraumeni syndrome and many solid tumors.

• ATM gene encodes a kinase essential for p53 activity. ATM is inactivated in ataxia telangiectasia, characterized by hypersensitivity to x-rays and predisposition to lymphomas.

• BRCA-1 (breast, prostate, and ovarian cancer) and BRCA-2 (breast cancer) required for p53 activity.

1. It is now believed that a substantial proportion of the single nucleotide substitutions causing human genetic disease are due to misincorporation of bases during DNA replication. Which proofreading activity is critical in determining the accuracy of nuclear DNA replication and thus the base substitution mutation rate in human chromosomes?

A. 3' to 5' polymerase activity of DNA polymerase δB. 3' to 5' exonuclease activity of DNA polymerase γC. Primase activity of DNA polymerase αD. 5' to 3' polymerase activity of DNA polymerase IIIE. 3' to 5' exonuclease activity of DNA polymerase δ

2. The proliferation of cytotoxic T-cells is markedly impaired upon infection with a newly discovered human immunodeficiency virus, designated HIV-V. The defect has been traced to the expression of a viral-encoded enzyme that inactivates a host-cell nuclear protein required for DNA replication. Which protein is a potential substrate for the viral enzyme?

A. TATA-box binding protein (TBP)B. Cap binding protein (CBP)C. Catabolite activator protein (CAP)D. Acyl-carrier protein (ACP)E. Single-strand binding protein (SBP)

3. The deficiency of an excision endonuclease may produce an exquisite sensitivity to ultraviolet radiation in Xeroderma pigmentosum. Which of the following functions would be absent in a patient deficient in this endonuclease?

A. Removal of introns

B. Removal of pyrimidine dimers

C. Protection against DNA viruses

D. Repair of mismatched bases during DNA replication

E. Repair of mismatched bases during transcription

4. The anti-Pseudomonas action of norfloxacin is related to its ability to inhibit chromosome duplication in rapidly-dividing cells. Which of the following enzymes participates in bacterial DNA replication and is directly inhibited by this antibiotic?

A. DNA polymerase I

B. DNA polymerase II

C. Topoisomerase I

D. Topoisomerase II

E. DNA ligase