MCBG Session 2 - Introduction to DNA Flashcards

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1
Q

Outline nucleic acids.

A
  • Nucleic acids are required for the storage and expression of genetic information.
  • There are two chemically distinct types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid.
  • DNA, the repository of genetic information, is present not only in chromosomes in the nucleus of eukaryotic organisms, but also in mitochondria and the chloroplasts of plants.
  • Prokaryotic cells, which lack nuclei, have a single chromosome, but may also contain non-chromosomal DNA in the form of plasmids
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2
Q

Outline the structure of DNA.

A
  • DNA is a polymer of deoxyribonucleoside monophosphates covalently linked by 3’→5’–phosphodiester bonds.
  • Except for a few viruses that contain single-stranded (ss) DNA, DNA exists as a double-stranded (ds) molecule, in which the two strands wind around each other, forming a double helix.
  • In eukaryotic cells, DNA is found associated with various types of proteins (known collectively as nucleoprotein) present in the nucleus, whereas in prokaryotes, the protein–DNA complex is present in a nonmembrane-bound region known as the nucleoid.
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3
Q

Outline the bonds found in the DNA molecule.

A
  • Phosphodiester bonds join the 3’-hydroxyl group of the deoxy pentose of one nucleotide to the 5’-hydroxyl group of the deoxy pentose of an adjacent nucleotide through a phosphate group.
  • The resulting long, unbranched chain has polarity, with both a 5’-end (the end with the free phosphate) and a 3’-end (the end with the free hydroxyl) that are not attached to other nucleotides.
  • The bases located along the resulting deoxy ribose–phosphate backbone are, by convention, always written in sequence from the 5’-end of the chain to the 3’-end.
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4
Q

Compare and contrast the DNA found in eukaryotes with that found in prokaryotes.

A
  • Each chromosome in the nucleus of a eukaryote contains one long, linear molecule of dsDNA, which is bound to a complex mixture of proteins to form chromatin.
  • Eukaryotes have closed, circular DNA molecules in their mitochondria, as do plant chloroplasts.
  • A prokaryotic organism typically contains a single, double-stranded, supercoiled, circular chromosome.
  • Each prokaryotic chromosome is associated with non-histone proteins that can condense the DNA to form a nucleoid.
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5
Q

What are plasmids and what do they do?

A
  • Most species of bacteria also contain small, circular, extrachromosomal DNA molecules called plasmids.
  • Plasmid DNA carries genetic information, and undergoes replication that may or may not be synchronized to chromosomal division.
  • Plasmids may carry genes that convey antibiotic resistance to the host bacterium, and may facilitate the transfer of genetic information from one bacterium to another.
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6
Q

Explain the structure of the DNA molecule in terms of the double helix.

A
  • In the double helix, the two chains are coiled around a common axis called the axis of symmetry.
  • The chains are paired in an anti-parallel manner, that is, the 5’-end of one strand is paired with the 3’-end of the other strand.
  • In the DNA helix, the hydrophilic deoxyribose–phosphate backbone of each chain is on the outside of the molecule, whereas the hydrophobic bases are stacked inside.
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7
Q

Outline base pairing in a DNA molecule.

A
  • The bases of one strand of DNA are paired with the bases of the second strand, so that an adenine is always paired with a thymine and a cytosine is always paired with a guanine.
  • The base pairs are perpendicular to the axis of the helix. Therefore, one polynucleotide chain of the DNA double helix is always the complement of the other.
  • Given the sequence of bases on one chain, the sequence of bases on the complementary chain can be determined.
  • The base pairs are held together by hydrogen bonds: two between A and T and three between G and C.
  • These hydrogen bonds, plus the hydrophobic interactions between the stacked bases, stabilise the structure of the double helix.
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8
Q

Explain how the separation of the DNA strands in a double helix might occur.

A
  • The two strands of the double helix separate when hydrogen bonds between the paired bases are disrupted.
  • Disruption can occur in the laboratory if the pH of the DNA solution is altered so that the nucleotide bases ionize, or if the solution is heated.
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9
Q

Briefly outline the cell cycle.

A
  • The events surrounding eukaryotic DNA replication and cell division (mitosis) are coordinated to produce the cell cycle.
  • The period preceding replication is called the G1 phase (Gap1).
  • DNA replication occurs during the S (synthesis) phase.
  • Following DNA synthesis, there is another period (G2 phase, or Gap2) before mitosis (M).
  • Cells that have stopped dividing, such as mature neurons, are said to have gone out of the cell cycle into the G0 phase. Cells can leave the G0 phase and re-enter the early G1 phase to resume division.
  • The cell cycle is controlled at a series of “checkpoints” that prevent entry into the next phase of the cycle until the preceding phase has been completed.
  • Two key classes of proteins that control the progress of a cell through the cell cycle are the cyclins and cyclin-dependent kinases (Cdk)
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10
Q

What are telomeres?

A
  • Telomeres are complexes of noncoding DNA plus proteins located at the ends of linear chromosomes.
  • They maintain the structural integrity of the chromosome, preventing attack by nucleases, and allow repair systems to distinguish a true end from a break in dsDNA.
  • The single-stranded region is thought to fold back on itself, forming a loop structure that is stabilized by protein.
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11
Q

What is the phenomena of telomere shortening?

A
  • Eukaryotic cells face a special problem in replicating the ends of their linear DNA molecules.
  • Following removal of the RNA primer from the extreme 5’-end of the lagging strand, there is no way to fill in the remaining gap with DNA.
  • Consequently, in most normal human somatic cells, telomeres shorten with each successive cell division.
  • Once telomeres are shortened beyond some critical length, the cell is no longer able to divide and is said to be senescent.
  • In germ cells and other stem cells, as well as in cancer cells, telomeres do not shorten and the cells do not senesce.
  • This is a result of the presence of a ribonucleoprotein, telomerase, which maintains telomeric length in these cells.
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12
Q

What is telomerase?

A
  • This complex contains a protein that acts as a reverse transcriptase, and a short piece of RNA that acts as a template.
  • The CA-rich RNA template base-pairs with the GT-rich, single-stranded 3’-end of telomeric DNA.
  • The reverse transcriptase uses the RNA template to synthesize DNA in the usual 5’→3’ direction, extending the already longer 3’-end.
  • Telomerase then translocates to the newly synthesized end, and the process is repeated.
  • Once the GT-rich strand has been lengthened, primase can use it as a template to synthesize an RNA primer.
  • The RNA primer is extended by DNA polymerase, and the primer is removed.
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13
Q

What are reverse transcriptases?

A
  • Reverse transcriptases, as seen with telomerase, are RNA-directed DNA polymerases.
  • A reverse transcriptase is involved in the replication of retroviruses, such as human immunodeficiency virus (HIV).
  • These viruses carry their genome in the form of ssRNA molecules.
  • Following infection of a host cell, the viral enzyme, reverse transcriptase, uses the viral RNA as a template for the 5’→3’ synthesis of viral DNA, which then becomes integrated into host chromosomes.
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14
Q

Outline the inhibition of DNA synthesis by nucleoside analogs.

A
  • DNA chain growth can be blocked by the incorporation of certain nucleoside analogs that have been modified in the sugar portion of the nucleoside.
  • For example, removal of the hydroxyl group from the 3’-carbon of the deoxyribose ring as in 2’,3’-dideoxyinosine (ddI, also known as didanosine), or conversion of the deoxyribose to another sugar such as arabinose, prevents further chain elongation.
  • By blocking DNA replication, these compounds slow the division of rapidly growing cells and viruses.
  • Cytosine arabinoside (cytarabine, or araC) has been used in anticancer chemotherapy, whereas adenine arabinoside (vidarabine, or araA) is an antiviral agent. Chemically modifying the sugar moiety, as seen in zidovudine (AZT, ZDV), also terminates DNA chain elongation.
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15
Q

Outline DNA Damage.

A
  • Despite the elaborate proofreading system employed during DNA synthesis, errors—including incorrect base-pairing or insertion of one to a few extra nucleotides—can occur.
  • In addition, DNA is constantly being subjected to environmental insults that cause the alteration or removal of nucleotide bases.
  • The damaging agents can be either chemicals, e.g. , nitrous acid, or radiation, for example, ultraviolet light, which can fuse two pyrimidines adjacent to each other in the DNA, and high-energy ionizing radiation, which can cause double-strand breaks.
  • Bases are also altered or lost spontaneously from mammalian DNA at a rate of many thousands per cell per day.
  • If the damage is not repaired, a permanent change (mutation) is introduced that can result in any of a number of deleterious effects, including loss of control over the proliferation of the mutated cell, leading to cancer.
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16
Q

Outline DNA repair.

A
  • Cells are remarkably efficient at repairing damage done to their DNA.
  • Most of the repair systems involve recognition of the damage (lesion) on the DNA, removal or excision of the damage, replacement or filling the gap left by excision using the sister strand as a template for DNA synthesis, and ligation.
  • These repair systems thus perform excision repair, with the removal of one to tens of nucleotides.
17
Q

Outline Methyl-directed mismatch repair.

A
  • Sometimes replication errors escape the proofreading function during DNA synthesis, causing a mismatch of one to several bases.
  • Homologous proteins are present in humans.
  • When a mismatch occurs, the Mut proteins that identify the mispaired nucleotide(s) must be able to discriminate between the correct strand and the strand with the mismatch.
18
Q

Outline the repair of damage caused by UV light.

A
  • Exposure of a cell to UV light can result in the covalent joining of two adjacent pyrimidines (usually thymines), producing a dimer.
  • These thymine dimers prevent DNA polymerase from replicating the DNA strand beyond the site of dimer formation.
  • Thymine dimers are excised in bacteria by UvrABC proteins in a process known as nucleotide excision repair.
  • When the strand containing the mismatch is identified, an endonuclease nicks the strand and the mismatched nucleotide(s) is/are removed by an exonuclease.
  • Additional nucleotides at the 5’- and 3’-ends of the mismatch are also removed.
  • The gap left by removal of the nucleotides is filled, using the sister strand as a template, by a DNA polymerase.
  • The 3’-hydroxyl of the newly synthesized DNA is joined to the 5’-phosphate of the remaining stretch of the original DNA strand by DNA ligase.
19
Q

Outline the correction of base alterations

A
  • The bases of DNA can be altered, either spontaneously, as is the case with cytosine, which slowly undergoes deamination (the loss of its amino group) to form uracil, or by the action of deaminating or alkylating compounds.
  • Bases can also be lost spontaneously.
20
Q

Explain the repair of double-strand breaks.

A
  • High-energy radiation or oxidative free radicals (see p. 148) can cause double- strand breaks in DNA, which are potentially lethal to the cell.
  • Such breaks also occur naturally during gene rearrangements.
  • A system to repair it is the non-homologous end-joining repair, in which the ends of two DNA fragments are brought together by a group of proteins that effect their religation.
  • However, some DNA is lost in the process. Consequently, this mechanism of repair is error prone and mutagenic.
  • Defects in this repair system are associated with a predisposition to cancer and immunodeficiency syndromes.