F20

Question 1

Show how At each replication fork, the lagging DNA strand is synthesized in pieces

Answer 1

Question 2

Show how DNA acts as a template for its own replication

Answer 2

Question 3

Show how DNA polymerase adds a deoxyribonucleotide to the 3’ end of a growing DNA strand

Answer 3

Question 4

Show how DNA polymerase contrains separate sites for DNA synthesis and proofreading

Answer 4

Question 5

Show how DNA synthesis occurs at Y-shaped junctions called replication forks

Answer 5

Question 6

show how During DNA synthesis, DNA polymerase proofreads its own work

Answer 6

Question 7

Show how Multiple enzymes are required to synthesize the lagging DNA strand

Answer 7

Question 8

What happens when proofreading takes place, and DNA polymerization must proceed in the 5’-to-3’ direction

Answer 8

Question 9

At what rate does the cell replicate its DNA?

Answer 9

The cell is replicating its DNA at rates as high as 1000 nucleotides per second.

Question 10

Explain semiconservative DNA replication

Answer 10

DNA replication is “semiconservative”
because each daughter DNA double helix is composed of one
conserved (old) strand and one newly synthesized strand.

Question 11

Explain how the DNA synthesis begins

Answer 11

The process of DNA synthesis is begun by initiator proteins that bind to specific DNA sequences called replication origins. Here, the initiator proteins pry the two DNA strands apart, breaking the hydrogen bonds between the bases. Although the hydrogen bonds collectively make the DNA helix very stable, individually each hydrogen bond is weak. Separating a short length of DNA a few base pairs at a time therefore does not require a large energy input.

Question 12

Which base pair is typically found at replication origins?

Answer 12

A-T base pair is held together by fewer hydrogen onds than G-C base pair. A-T-rich stretches of DNA are typically found at replication origins.

Question 13

What happens after an initiator protein binds to DNA at a replication origin?

Answer 13

Once an initiator protein binds to DNA at a replication origin and locally opens up the double helix, it attracts a group of proteins that carry out DNA replication. These proteins form a replication machine, in which each protein carries out a specific function.

Question 14

Explain bidirectional

Answer 14

The two forks move away from the origin in opposite directions, unzipping the DNA double helix and copying the DNA as they go. DNA replication— in both bacterial and eukaryotic chromosomes—is therefore termed bidirectional.

Question 15

How fast to the forks move?

Answer 15

The forks move very rapidly: 100 nucleotide pairs per second in humans

Question 16

Describe the polymerization reaction

Answer 16

The polymerization reaction involves the formation of a phosphodiester bond between the 3ʹ end of the growing DNA chain and the 5ʹ-phosphate group of the incoming nucleotide, which enters the reaction as a deoxyribonucleoside triphosphate. The energy for polymerization is provided by the incoming deoxyribonucleoside triphosphate itself: hydrolysis
of one of its high-energy phosphate bonds fuels the reaction that links the nucleotide monomer to the chain, releasing pyrophosphate. Pyrophosphate is further hydrolyzed to inorganic phosphate (Pi), which makes the polymerization reaction effectively irreversible.

Question 17

Describe why the 5’-to-3’ direction of the DNA polymerization reaction poses a problem.

Answer 17

The 5ʹ-to-3ʹ direction of the DNA polymerization reaction poses a problem at the replication fork. The two strands in the double helix are antiparallel; that is, they run in opposite directions. As a consequence, at each replication fork, one new DNA strand is being made on a template that runs in one direction (3ʹ to 5ʹ), whereas the other new strand is being made on a
template that runs in the opposite direction (5ʹ to 3ʹ). The replication fork is therefore asymmetrical

Question 18

Describe Ozaki fragments and how the DNA strand grows

Answer 18

The DNA strand that appears to grow in the incorrect 3ʹ-to-5ʹ direction is actually made discontinuously, in successive,
separate, small pieces—with the DNA polymerase moving backward with respect to the direction of replication-fork movement so that each new DNA fragment can be polymerized in the 5ʹ-to-3ʹ direction.

The resulting small DNA pieces—called Okazaki fragments —are later joined together to form a continuous new strand. The DNA strand that is made discontinuously in this way is called the lagging strand, because the cumbersome backstitching mechanism imparts a slight delay to its synthesis; the other
strand, which is synthesized continuously, is called the leading strand

Question 19

In which direction do polymerases work?

Answer 19

DNA polymerases work only in the 5’-to-3’ direction - a restriction that allows DNA polymerase to “check its work”.

Question 20

Describe how DNA polymerase is self-correcting

Answer 20

DNA polymerase is so accurate that it makes only about one error in every 10⁷ nucleotide pairs it copies. Although A-T and C-G are by far the most stable base pairs, other, less stable base pairs—for example, G-T and C-A—can also be formed. If allowed to remain, they would result in an accumulation of mutations.

First, the enzyme carefully monitors the base-pairing between each incoming nucleoside triphosphate and the template strand. Only when the match is correct does DNA polymerase undergo a small structural rearrangement that allows it to catalyze the nucleotide-addition reaction.

Second, when DNA polymerase does make a rare mistake and adds the wrong nucleotide, it can correct the error through an activity called proofreading. Proofreading takes place at the same time as DNA synthesis.

Polymerization and proofreading are tightly coordinated,
and the two reactions are carried out by different catalytic domains in the same polymerase molecule.

This proofreading mechanism is possible only for DNA polymerases that synthesize DNA exclusively in the 5ʹ-to-3ʹ direction. If a DNA polymerase were able to synthesize in the 3ʹ-to-5ʹ direction (circumventing the need for backstitching on the lagging strand), it would be unable to proofread. That’s because if this “backward” polymerase were to remove an incorrectly paired nucleotide from the 5ʹ end, it would create a chemical dead end—a strand that could no longer be elongated.

Question 21

How can the polymerase begin a completely new DNA strand?

Answer 21

To get the process started, a different enzyme is needed—one that can begin a new polynucleotide strand simply by joining two nucleotides together without the need for a base-paired end. This enzyme does not, however, synthesize
DNA. It makes a short length of a closely related type of nucleic acid—RNA (ribonucleic acid)—using the DNA strand as a template. This short length of RNA, about 10 nucleotides long, is base-paired to the template strand and provides a base-paired 3ʹ end as a starting point for DNA polymerase. An RNA fragment thus serves as a primer for DNA synthesis, and the enzyme that synthesizes the RNA primer is known as primase.

For the leading strand, an RNA primer is needed only to start replication at a replication origin; at that point, the DNA polymerase simply takes over, extending this primer with DNA synthesized in the 5ʹ-to-3ʹ direction. But on the lagging strand, where DNA synthesis is discontinuous, new primers are continuously needed to keep polymerization going. The movement of the replication fork continually exposes
unpaired bases on the lagging-strand template, and new RNA primers must be laid down at intervals along the newly exposed, single-stranded stretch. DNA polymerase then adds a deoxyribonucleotide to the 3ʹ end of each new primer to produce another Okazaki fragment, and it will continue to elongate this fragment until it runs into the previously synthesized RNA primer.

Question 22

Describe the process of producing a continuous new DNA strand from many separate pieces of nucelic acid

Answer 22

To produce a continuous new DNA strand from the many separate pieces of nucleic acid made on the lagging strand, three additional enzymes are needed. These act quickly to remove the RNA primer, replace it with DNA, and join the remaining DNA fragments together. A nuclease degrades the RNA primer, a DNA polymerase called a repair polymerase replaces the RNA primers with DNA (using the end of the adjacent Okazaki fragment as its primer), and the enzyme DNA ligase joins the 5ʹ-phosphate end of one DNA fragment to the adjacent 3ʹ-hydroxyl end of the next. Because it was discovered first, the repair polymerase involved in this process is often called DNA polymerase I; the polymerase that carries out the bulk of DNA replication at the forks is known as DNA polymerase III.

Unlike DNA polymerases I and III, primase does not proofread its work. As a result, primers frequently contain mistakes. But because primers are made of RNA instead of DNA, they stand out as “suspect copy” to be automatically removed and replaced by DNA. The repair polymerase that makes this DNA, like the replicative polymerase, proofreads as it synthesizes. In this way, the cell’s replication machinery is able to begin new DNA strands and, at the same time, ensure that all of the DNA is copied faithfully.

Question 23

Describe what is needed for DNA replication to occur

Answer 23

For DNA replication to occur, the double helix must be continuously pried apart so that the incoming nucleoside triphosphates can form base pairs with each template strand. Two types of replication proteins—DNA helicases
and single-strand DNA-binding proteins—cooperate to carry out this task. A helicase sits at the very front of the replication machine, where it uses the energy of ATP hydrolysis to propel itself forward, prying apart the double helix as it speeds along the DNA.
Single-strand DNA-binding proteins then latch onto the single-stranded DNA exposed by the helicase, preventing the strands from re-forming base pairs and keeping them in an elongated form so that they can serve as efficient templates.

Question 24

Describe the function of a sliding clapm and a clamp loader

Answer 24

At the replication fork, an additional protein, called a sliding clamp, keeps DNA polymerase firmly attached to the template while it is synthesizing new strands of DNA. Left on their own, most DNA polymerase molecules will synthesize only a short string of nucleotides before falling off the DNA template strand. The sliding clamp forms a ring around the
newly formed DNA double helix and, by tightly gripping the polymerase, allows the enzyme to move along the template strand without falling off as it synthesizes new DNA.

Assembly of the clamp around DNA requires the activity of another replication protein, the clamp loader, which hydrolyzes ATP each time it locks a sliding clamp around a newly formed DNA double helix. This loading needs to occur only once per replication cycle on the leading strand; on
the lagging strand, however, the clamp is removed and then reattached each time a new Okazaki fragment is made. In bacteria, this happens approximately once per second.

Most of the proteins involved in DNA replication are held together in a large multienzyme complex that moves as a unit along the parental DNA double helix, enabling DNA to be synthesized on both strands in a coordinated manner. This complex can be likened to a miniature sewing machine composed of protein parts and powered by nucleoside triphosphate hydrolysis.

Question 25

Explain what happens when the replication fork approachees the end of a chromosome

Answer 25

As the replication fork approaches the end of a chromosome:
the leading strand can be replicated all the way to the
chromosome tip, the lagging strand cannot. When the final RNA primer on the lagging strand is removed, there is no enzyme that can replace it with DNA.

Bacteria avoid this “end-replication” problem by having circular DNA molecules as chromosomes. Eukaryotes get around it by adding long, repetitive nucleotide sequences to the ends of every chromosome. These sequences, which are incorporated into structures called telomeres, attract an enzyme called telomerase to the chromosome ends.
Telomerase carries its own RNA template, which it uses to add multiple copies of the same repetitive DNA sequence to the lagging-strand template. In many dividing cells, telomeres are continuously replenished, and the resulting extended templates can then be copied by conventional
DNA replication, ensuring that no peripheral chromosomal sequences are lost.

In addition to allowing replication of chromosome ends, telomeres form structures that mark the true ends of a chromosome. These structures allow the cell to distinguish unambiguously between the natural ends of chromosomes and the double-strand DNA breaks that sometimes occur
accidentally in the middle of chromosomes. These breaks are dangerous and must be immediately repaired.

Question 26

What does repetitive DNA sequences in telomeres attract?

Answer 26

In addition to attracting telomerase, the repetitive DNA sequences found within telomeres attract other telomere-binding proteins that not only physically protect chromosome ends, but help maintain telomere length.

Question 27

Describe which cells keep their telomerase fully active

Answer 27

Cells that divide at a rapid rate throughout the life of the organism— those that line the gut or generate blood cells in the bone marrow, for example—keep their telomerase fully active. Many other cell types, however, gradually turn down their telomerase activity. After many rounds of cell division, the telomeres in these descendent cells will shrink, until they essentially disappear. At this point, these cells will cease dividing.

Question 28

Is DNA repair long lasting?

Answer 28

Most DNA damage is only temporary, because it is immediately corrected by processes collectively called DNA repair.

Question 29

Which reaction results in major chemical changes in the DNA?

Answer 29

Just like any other molecule in the cell, DNA is continually undergoing thermal collisions with other molecules, often resulting in major chemical changes in the DNA.

Question 30

Describe depurination and deamination

Answer 30

Depurination does not break the DNA phosphodiester backbone but instead removes a purine base from a nucleotide, giving rise to lesions that resemble missing teeth. Another common reaction is the spontaneous loss of an amino group (deamination) from a cytosine in DNA to produce the base uracil.

Question 31

Which types of DNA damage can stall the DNA replication machinery at the site of the damage?

Answer 31

Some types of DNA damage (thymine dimers, for example)
can stall the DNA replication machinery at the site of the damage.

Question 32

Can DNA be altered by replication itself? - If yes explain why.

Answer 32

DNA can also be altered by replication itself. The replication machinery that copies the DNA can—albeit rarely—incorporate an incorrect nucleotide that it fails to correct via proofreading.

Question 33

Describe the basic pathway for repairing damage to DNA in three basic steps

Answer 33

1. The damaged DNA is recognized and removed by one of a variety of mechanisms. These involve nucleases, which cleave the covalent bonds that join the damaged nucleotides to the rest of the DNA strand, leaving a small gap on one strand of the DNA double helix.
2. A repair DNA polymerase binds to the 3ʹ-hydroxyl end of the cut DNA strand. The enzyme then fills in the gap by making a complementary copy of the information present in the undamaged strand. Although they differ from the DNA polymerase that replicates DNA, repair DNA polymerases synthesize DNA strands in the same way. For example, they elongate chains in the 5ʹ-to-3ʹ direction and have the same type of proofreading activity to ensure that the template strand is copied accurately. In many cells, the repair polymerase is the same enzyme that fills in the gaps left after
   the RNA primers are removed during the normal DNA replication process.
3. When the repair DNA polymerase has filled in the gap, a break remains in the sugar–phosphate backbone of the repaired strand. This nick in the helix is sealed by DNA ligase, the same enzyme that joins the Okazaki fragments during replication of the lagging DNA strand

Question 34

How much can DNA mismatch repair?

Answer 34

The replication machine makes approximately one mistake per 10⁷ nucleotides synthesized; DNA mismatch repair corrects 99% of these replication errors, increasing the overall accuracy to one mistake in 10⁹ nucleotides synthesized.

Question 35

Explain mismatch

Answer 35

Whenever the replication machinery makes a copying mistake, it leaves behind a mispaired nucleotide (commonly called a mismatch).

Question 36

Describe newly synthesized DNA and how new and template strands can be distinguished

Answer 36

In bacteria, newly synthesized DNA lacks a type of chemical modification (a methyl group added to certain adenines) that is present on the preexisting parent DNA. Newly synthesized DNA is unmethylated for a short time, during which the new and template strands can be easily distinguished. Other cells use different strategies for distinguishing their parent DNA from a newly replicated strand.

Question 37

Describe how mismatch plays an important role in preventing cancer

Answer 37

An inherited predisposition to certain cancers is caused by mutatoins in genes that encode mismatch repair proteins. Human cells have two copies of these genes (one from each parent), and individuals who inherit one damaged mismatch repair gene are unaffected until the undamaged copy of the same gene is randomly mutated in a somatic cell. This mutant cell—and all of its progeny—are then deficient in mismatch repair; they therefore accumulate mutations more rapidly than do normal cells. Because cancers arise from cells that have accumulated multiple mutations, a cell deficient in mismatch repair has a greatly enhanced chance of becoming cancerous.
Thus, inheriting a single damaged mismatch repair gene strongly predisposes an individual to cancer.

Question 38

What happens when both strands of the double helix are damaged at the same time?

Answer 38

Mishaps at the replication fork, radiation, and various
chemical assaults can all fracture DNA, creating a double-strand break. Such lesions are particularly dangerous, because they can lead to the fragmentation of chromosomes and the subsequent loss of genes.

To handle this potentially disastrous type of DNA damage, cells have evolved two basic strategies. The first involves hurriedly sticking the broken ends back together, before the DNA fragments drift apart and get lost. This repair mechanism, called nonhomologous end joining, occurs in many cell types and is carried out by a specialized group of enzymes that “clean” the broken ends and rejoin them by DNA ligation. This “quick and dirty” mechanism rapidly seals the break, but it comes with a price: in “cleaning” the break to make it ready for ligation, nucleotides are often lost at the site of repair. If this imperfect repair disrupts the activity of a gene, the cell could suffer serious consequences.

Cells have an alternative, error-free strategy for repairing double-strand breaks, called homolgous recombination.

Question 39

Show how Cells can repair double-strand breaks in one of two ways.

Answer 39

Question 40

Show how mismatch repair eliminates replication errors and restores the original DNA sequence

Answer 40

Question 41

Show how Errors made during DNA replication must be corrected to avoid mutations

Answer 41

Question 42

Show how The basic mechanism of DNA repair involves three steps

Answer 42

Question 43

Show how Chemical modification of nucleotides, if left unrepaired, produce mutations

Answer 43

Question 44

Show how The ultraviolet radiation in sunlight can cause the formation of thymine dimers

Answer 44

Question 45

Show how Depurination and deamination are the most frequent chemical reactions known to create serious DNA damage in cells

Answer 45

Question 46

Show how Telomeres and telomerase prevent linear eukaryotic chromosomes from shortening with each cell division

Answer 46

Question 47

Show how Without a special mechanism to replicate the ends of linear chromosomes, DNA would be lost during each round of cell division

Answer 47

Question 48

Show how Proteins are involved in DNA replication

Answer 48

Question 49

Show how DNA topoisomerases relieve the tension that builds up in front of a replication fork

Answer 49

Question 50

Show how DNA synthesis is carried out by a group of proteins that act together as a replication machine

Answer 50

Question 51

Show how Multiple enzymes are required to synthesize the lagging DNA strand

Answer 51

Question 52

Show how RNA primers are synthesized by an RNA polymerase called primase, which uses a DNA strand as a template

Answer 52

Question 53

Explain what nucleotide changes in somatic cells can lead to (in worst case)

Answer 53

Nucleotide changes that occur in somatic cells can give rise to variant cells, some of which grow and divide in an uncontrolled fashion at the expense of the other cells in the organism. In the extreme case, an unchecked cell proliferation known as cancer results. Cancers are responsible for about 30% of the deaths that occur in Europe and North America, and they are caused primarily by a gradual accumulation of random mutations in a somatic cell and its descendants. Increasing the mutation frequency even two- or threefold could cause a disastrous increase in the incidence of cancer by accelerating the rate at which such somatic cell variants arise.

Question 54

Describe what happens when a mutation of a single nucleotide in the human hemoglobin happens

Answer 54

Mutation of a single nucleotide in the human hemoglobin gene can cause the disease sickle-cell anemia. The hemoglobin protein is used to transport oxygen in the blood. Mutations in the hemoglobin gene can produce a protein that is less soluble than normal hemoglobin and forms fibrous intracellular precipitates, which produce the characteristic sickle shape of affected red blood cells. Because these cells are more fragile and frequently tear as they travel through the bloodstream, patients with this potentially life-threatening disease have fewer red blood cells than usual—that is, they are anemic. Moreover, the abnormal red blood cells that remain can aggregate and block small vessels, causing pain and organ failure. We know about sickle-cell hemoglobin because individuals with the mutation survive; the mutation even provides a benefit—an increased resistance to malaria.

The example of sickle-cell anemia, which is an inherited disease, illustrates the consequences of mutations arising in the reproductive germ-line cells. A mutation in a germ-line cell will be passed on to all the cells in the body of the multicellular organism that develop from it, including the gametes responsible for the production of the next generation.

Question 55

Describe why homologous recombination is crucial

Answer 55

Homologous recombination is versatile, and it also has a crucial role in the exchange of genetic information that occurs during the formation of the gametes—sperm and eggs. This exchange, during the specialized form of cell division called meiosis, enhances the generation of genetic diversity within a species during sexual reproduction.

Question 56

Show how Homologous recombination flawlessly repairs DNA double-strand breaks

Answer 56

Question 57

Explain and describe homologous recombination

Answer 57

Homologous recombination most often occurs shortly after a cell’s DNA has been replicated before cell division, when the duplicated helices are still physically close to each other. To initiate the repair, a recombination-specific nuclease chews back the 5ʹ ends of the two broken strands at the break. Then, with the help of specialized enzymes (called recA in bacteria and Rad52 in eukaryotes), one of the broken 3ʹ ends “invades” the unbroken homologous DNA duplex and searches for a complementary sequence through basepairing. Once an extensive, accurate match is made, the invading strand is elongated by a repair DNA polymerase, using the complementary undamaged strand as a template. After the repair polymerase has passed the point where the break occurred, the newly elongated strand rejoins its original partner, forming base pairs that hold the two strands of the broken double helix together. Repair is then completed by additional DNA synthesis at the 3ʹ ends of both strands of the broken double helix, followed by DNA ligation. The net result is two intact DNA helices, for which the genetic information from one was used as a template to repair the other.