Module I Flashcards

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1
Q

What do the letters dNTP stand for?

A

Deoxyribonucleoside triphosphate (these are the precursor building blocks to nucleic acids/nucleotides

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2
Q

Which form of DNA is predominant in vivo?

A

B-DNA

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3
Q

What is the significance of the major and minor grooves of DNA? How are they formed?

A

The sugar-phosphate backbone of DNA isn’t spaced equally, resulting in a major and minor groove. The Major groove exposes the base letters, allowing binding proteins to “read” the exposed nucleotides. Specifically, the solvent-exposed N and O atoms in the nucleotides form H-bonds with the binding protein’s amino acid side chains, and the difference in the donor-acceptor patterns tells the binding protein which pair of nucleotides it’s attached to.

The width of the minor groove is dependent on which nucleotides are present in the surrounding DNA. For example, a short section of adenine (an A-tract) forms a narrow minor groove. The Minor groove is thought to be harder to read because the H-bond patterns are the same regardless of which way the base pair is flipped, meaning that the only difference in pattern would be between AT and GC pairs. However, there are some DNA-binding proteins that can recognize the minor groove, such as the TATA-binding protein (TBP), which binds at the TATA box and plays a role in initiating gene transcription in eukaryotes.

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4
Q

Why does G+C count affect the melting temperature of DNA?

A

Because G+C base pairs are held together with 3 H-bonds instead of the 2 found in A+T base pairs, it takes more energy to cleave the base pairs apart, and therefore a greater temperature during melting.

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5
Q

What are some other methods to denature DNA that aren’t heat?

A

Lowering salt concentration (this removes the cations that shield the negative charges on two strands from each other)

High pH (disrupts H-bonds)

Organic solvents (disrupts H-bonds)

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6
Q

Why does cooling time matter for renaturation?

A

If denatured single strand DNa is cooled rapidly, it will remain single stranded. However, if it is cooled slowly, the complementary sequences will find each other and form a new double helix.

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7
Q

What happens when we change the values of one of the variables in the Cot curve equation?

A

Cot Curve Equation: C/C。= 1/(1+tKC。)

This whole equation is really just a ratio. Increasing one variable makes all other variables increase.

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8
Q

Why could Cot ½ of one species be greater than that of another?

A

The larger the genome size, the longer it will take for any sequence to encounter its complementary sequence. So unless the two species in question have identical genome sizes, the Cot ½ of each species will be different.

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9
Q

Why do some species have Cot curves that follow the ideal model, while others vary?

A

In the graph of a Cot curve, the fraction of DNA that reanneals faster represents repeated DNA sequences. Because they are found more often in the genome, they have a greater probability of finding a matching sequence and reannealing faster. The remaining fraction of the genome reanneals more slowly because it is made of unique sequences and has a lower probability of finding its one matching sequence.

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10
Q

What is the significance of supercoiling in vivo?

A

Negative Supercoiling
Puts Potential Energy into DNA, making it easier to pull the helix apart during replication
Plays a role in replication, transcription, and recombination
Transitions between B-DNA and Z-DNA may be triggered by negative supercoiling

Positive Supercoiling
Found in thermophilic archaea, as it is more energetically stable and resistant to degradation at higher temperatures

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11
Q

Under what conditions does B-DNA form? What is its applications?

A

High humidity (95%)

Low salt concentrations

Predominant helical form in vivo

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12
Q

Under what conditions does A-DNA form? What is its applications?

A

Low humidity (75%)

High salt concentrations

Found in RNA and in vitro

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13
Q

Under what conditions does Z-DNA form? What is its applications?

A

High MgCL2, NaCl, or ethanol

Found only in the presence of methylated cytosine

Some evidence for in vivo, but is predominantly the product of bored lab technicians.

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14
Q

How does DNA denaturation depend on G+C content and salt concentration?

A

High G+C content → Increased melting temperature

Low salt concentration → Increased denaturation speed

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15
Q

How does melting affect relative absorbance at 260 nm?

A

Once the strands have been separated/melted, the UV absorbance at 260nm increased by 40%, a phenomenon called hyperchromicity.

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16
Q

What are the different types of unusual DNA secondary structures?

A

Slipped structure, cruciform structure, triple DNA helix, G-Quadruplex

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17
Q

How are slipped structures formed, where are they expressed, and how are they used?

A

Forms at tandem repeats in which misalignment leads to single stranded loops

Found upstream of regulatory sequences

Possible recognition sites for binding proteins

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18
Q

How are cruciform structures formed, where are they expressed, and how are they used?

A

Forms when AT rich regions unwind and form a stem-loop cruciform.

For a sequence to form a cruciform, it must be a palindrome, or inverted along its 5’ →3’ directionality

Found in vitro in plasmids and bacteriophages

May potentially be a regulatory element of replication and gene expression, but could also potentially lead to genetic instability and cancer

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19
Q

How are triple helix structures formed, where are they expressed, and how are they used?

A

Forms when a third strand intertwines

Found in mirror repeat symmetrical sequences

The third strand is either intramolecular (from the same helix) or extramolecular (from a separated DNA molecule)

Uses Hoogsteen base pairs between AT and GC to form

Requires a lower pH (between 4-5) to form

Found in Friedreich’s ataxia

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20
Q

How are G-Quadruplex structures formed, where are they expressed, and how are they used?

A

Four stranded structure formed in stretches of tandem guanines

Uses Hoogsteen bonds to form the planar G-Tetrad

Stabilized with a central monovalent cation.

Multiple G-tetrads then stack and interconnect with loops, forming the full G-Quadruplex

Found in telomeres and promoter regions

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21
Q

What’s the difference between a classic Watson-Crick G+C nucleotide pair and a GU Wobble?

A

A GU wobble uses 2 H-bonds, and the classic G+C base pair requires 3 H-bonds.

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22
Q

How does RNA differ from DNA in primary structure? What about secondary structure?

A

Primary → RNA uses uracil in place of DNA’s thymine

Secondary → RNA forms a single stranded helix, DNA forms a double stranded helix

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23
Q

One tertiary structure found in RNA is a tetraloop, in which four nucleotides form a loop. What kind of interactions stabilize this structure?

A

Hydrophobic and Van der Waals interactions, and base stacking

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24
Q

Why was the term ribozyme created? How is it different from an enzyme?

A

The classic definition of an enzyme doesn’t always apply to RNA-enzyme molecules. Enzymes never permanently change their shape and are used over and over ad nauseum. But RNA-enzymes often self-cleave, meaning that they are used only once. So a new term was created to describe these misfit molecules: ribozymes.

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25
Q

The Hammerhead Ribozyme is a ribozyme that self-cleaves at a phosphodiester linkage with the 2’ hydroxyl group acting as the attacking nucleophile. Why does this happen only in the Hammerhead Ribozyme? Phrased another way, why don’t all RNA molecules spontaneously self-cleave?

A

RNA molecules don’t spontaneously self-cleave because of the way the 2’ OHs are folded into the RNA’s 3D tertiary structure. The molecules simply aren’t close enough together to trigger this self-cleaving reaction. It’s only when the hammerhead ribozyme brings these two sites closer together that they can react and cleave apart.

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26
Q

What are noncanonical pairs, and why are they useful for the RNA molecule?

A

Noncanonical pairs are any nucleotide pairs outside the classic Watson-Crick AT GC binary. They are useful for the RNA molecule because they widen the major groove and make it more accessible to ligands.

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27
Q

How does the RNaseP RNA differ in E. coli and in humans?

A

E. coli → True enzyme, no chemical transformation

Humans → Protein-controlled ribozyme/catalytic RNP enzyme

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28
Q

The “RNA World Hypothesis” states that life first existed with RNA, and that only later did DNA and proteins evolve from RNA. What is the evidence for this hypothesis?

A

Proteins can’t replicate themselves

RNA has all the structural prerequisites for replication

RNA genomes are widespread among viruses

RNA can do everything proteins can do and more

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29
Q

What is coaxial stacking? List two examples where coaxial stacking can be seen in RNA.

A

Coaxial stacking occurs when bases of two separate stems stack, align, and form what ~appears~ to be a continuous helix, but they aren’t actually bonded to each other. They are pseudo-continuous. This can be found in tRNA. The stacked arms form its L shape.

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30
Q

Why does RNA need hyper specific binding proteins to create its secondary and tertiary structures?

A

Because RNA is negatively charged, formation of tertiary structure requires charge neutralization via either basic proteins, or monovalent or bivalent metal ions.

Also, only a few variations of secondary and tertiary structure in RNA leads to any sort of function. Therefore, the more binding proteins and chaperones the RNA has at its disposal to ensure it gets to a functional structure, the better.

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31
Q

Why do RNA genomes typically have a very high mutation rate?

A

There is a fitness trade off between replication rate and replication fidelity. For organisms with RNA genomes, their high mutation rate indicates that it was more evolutionarily advantageous to multiply their genomes quickly than it was to multiply their genomes accurately.

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32
Q

What structural features allow proteins to fulfill their functions?

A

The functional groups found on amino acids

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33
Q

What are some advantages of intrinsic disorder within proteins? What are some disadvantages?

A

Advantages → conformational flexibility lets the protein fulfill more than one function

Disadvantages → these proteins are more prone to misfolding and aggregation

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34
Q

Why is phosphorylation the most important part of post-transcriptional regulation? What are some of the things it can do?

A

Phosphorylation can:
Cause change in a protein’s shape

Mask or unmask a catalytic/active site or functional domain

Provide a recognition point or binding motif

Promote dissociation in multiprotein complexes

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35
Q

How do histone structures compare between archaea and eukaryotes?

A

Archaean genomes are more compact with fewer intergenic regions. Some have histones, but they are only found in a single histone fold domain, and they lack N and C terminal tail extensions. Additionally, DNA wraps only once around archaeal histones.

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36
Q

Why is mitochondrial inheritance almost always predominantly maternal?

A

The mitochondria of sperm cells is not located in the head of the cell. Because only the head of the cell enters the egg in fertilization, the paternal mitochondrial genome is not inherited in the offspring.

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37
Q

What are some atypical mitochondrial inheritance patterns and where have these patterns been documented?

A

mtDNA leakage from sperm

Some fungi, slime molds, and plants have biparental inheritance

Some land plants and algae have strictly parental inheritance

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38
Q

How does heteroplasmy impact mitochondrial diseases?

A

When not all of the copies of mtDNA in a mitochondria are identical, any harmful mutations within them are spread unevenly across the different copies of the mitochondrial genomes. If a greater percentage of those genome copies have the harmful mutations, the severity of the symptoms for the disease it causes will increase.

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39
Q

Compare and contrast the eukaryotic genome with the bacterial genome.

A

Bacteria have circular DNA that is condensed into a nucleoid. They also carry some plasmids, but those are replicated and transcribed separately from the genome.

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40
Q

How does DNA wrap around histones? (What kind of supercoils does the DNA form?)

A

DNA wraps around histones twice and restrains it into negative (lefthand) supercoils.

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41
Q

Major interactions between DNA and core histones are probably electrostatic. What evidence supports this hypothesis?

A

Histones can be removed from DNA by high salt concentrations.

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42
Q

Why do chromatin regions naturally repel each other, despite the presence of core histones, and how is this problem solved by cellular machinery?

A

The negative charges of DNA’s phosphate backbone are only partially shielded by the positive charges of core histones, making chromatin naturally repellent of itself. This issue is solved by using cations, linker histones, and miscellaneous positively charged proteins to shield the backbone’s negative charges

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43
Q

How does the DNA condense further after being wrapped into nucleosomes?

A

The condensin complex and ATP hydrolyzing enzymes further condense the chromatin.

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44
Q

What decides the location of the centromere on each chromosome?

A

The formation of a specialized chromatin structure, a nucleosome with a centromere-specific histone H3 variant called CENP-A or CenH3, decides centromere location.

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45
Q

DNA binding proteins gain access to the chromatin at three different levels of regulation. Name each of these levels, then provide at least one example of their regulation.

A
  1. DNA Sequence
    - Wrapping of the DNA around the histone core octamer inhibits access to this part of the DNA sequence
    - Wrapping of the DNA around linker histones allows access to the DNA sequence
    - Segments within a chromosome can switch between different transcriptional states
  2. Chromatin
    - Changes in how the DNA is wrapped around the histones results in uncoiling and decondensation
    - Nucleosome positions are determined partially by differing affinities for DNA sequences
    - Can be displaced and recruited by competing or cooperating with other protein factors
  3. Positioning of the chromosomes in the nucleus
    - Nucleosome can be actively moved and displaced by remodeling complexes
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46
Q

In what type of DNA does gene silencing occur?

A

Heterochromatin

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47
Q

What is HP1⍺ and what characteristic of its structure allows it to fulfill its function?

A

HP1⍺ (heterochromatin protein 1⍺) spreads along long regions of the genome. Its intrinsically disordered regions promote phase separation which allow it to silence genes.

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48
Q

What is the C-Value paradox and what are some of its current accepted resolutions?

A

The C-Value paradox states that the amount of DNA in a haploid genomes doesn’t seem to correspond with the complexity of the organism. Some accepted resolutions are that most genomic DNA consists of various classes of repeats that originate from the replication of “jumping genes”, and that the majority of the human genomes is just intergenic regions (regions that aren’t under selective pressure, where mutations are maintained and transmitted, and that consist of unique and medium-to-highly repetitive sequences).

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49
Q

What does mtDNA typically code for?

A

Proteins responsible for oxidative phosphorylation and the mitochondrial electron transport chain.

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50
Q

What types of symptoms do mitochondrial diseases have and why? In carriers of these diseases, at what point are the symptoms expressed?

A

Symptoms include degeneration of the brain and muscles, muscle paralysis, and blindness, really any tissue that heavily relies on the mitochondria’s oxidative phosphorylation. In carriers of these diseases, symptoms are usually expressed once over 60% of the mtDNA is mutated.

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51
Q

Some examples of mitochondrial diseases include Leber’s Hereditary Optic Neuropathy and Kearn-Sayre Syndrome. How are these diseases inherited?

A

Maternally

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52
Q

What is heteroplasmy as it relates to mitochondrial and chloroplastic DNA, and how does it affect the functioning of these organelles?

A

Heterplasmy is when not all of the mitochondria or chloroplasts within a cell have the same genome. Some of the mitochondrial or chloroplastic genomes have mutations. The functioning of these organelles depends on the severity and magnitude of the mutations. Cell functioning is dependent on the total number of mutated mitochondria or chloroplasts and their genome’s severity of mutations.

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53
Q

Compare and contrast archeal genome organization with eukaryotic genome organization.

A

Archaeal genomes are more compact than eukaryotic genomes, with fewer intergenic regions. Some have histones, but these histones have only a single histone fold domain, lack N and C terminals tail extensions, and the DNA wraps only once around them. Additionally, where eukaryotic histones can only heterodimerize, archaeal histones can heterodimerize AND homodimerize.

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54
Q

Why can’t DNA polymerase initiate DNA synthesis de novo?

A

Because dNTPs are present usually only in low concentrations, DNA polymerases wouldn’t have evolved to bind to 2 dNTPs at the same time. Therefore, DNA polymerase can’t form the initial phosphodiester bond, and needs other cellular machinery to make that first bond in DNA synthesis.

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55
Q

What characteristics must primers have?

A

Nucleotide primers must be able to bind to the 3’ OH- group, so primers must have a free 3’ OH-. Primers must also be made of small RNA segments.

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56
Q

How does DNA polymerase replicate both the leading and lagging strands of DNA despite being able to add primers only in the 5’→3’ direction?

A

DNA replication is semi-discontinuous. The leading strand is synthesized start-to-end, and the lagging strand is synthesized in short, discontinuous fragments, and later glued together to form one long strand.

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57
Q

What are some differences in genome replication between bacteria and eukaryotes?

A

Bacteria
- One single, well defined replication origin
- Recruits helicase to the origin immediately after initiation proteins accumulate there

Eukaryotes
- Multiple replication origins
- Separate origin selection from initiation with a pre-replication complex, in order to prevent overreplicaiton of the genome

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58
Q

Origin sequences tend to be rich in which sequences? Why?

A

Origin sequences tend to be rich in AT sequences because they are bound with 2 hydrogen bonds, unlike GC sequences which are bound with 3 H-bonds. The smaller number of bonds requires less energy to break, making AT rich sequences energetically ideal to start replication.

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59
Q

Where are mammalian replication origins located?

A

In active or open chromatin, however much of the genome is capable of supporting low levels of replication initiation events. Mammals lack an easily identifiable consensus origin sequence.

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60
Q

What is the rate of genome replication dependent on?

A

The number of origins used and the rate at which they fire.

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61
Q

Topoisomerase I doesn’t create free ends when cleaving the phosphodiester bond between two adjacent nucleotides. How does it do this?

A

It binds to a circular DNA molecule with one negative supercoil and unwinds the double helix. Instead of creating fee ends, it becomes covalently attached to one of the two broken ends of the DNA. Which end depends on the specific enzyme. While winding the broken ends, the enzyme then passes the unbroken strand through the break and ligates the cut ends, thereby increasing the linking number of the DNA by one. The oxygen of the tyrosine hydroxyl group in the active site of the topoisomerase attacks a DNA phosphorus, forming a covalent phosphotyrosine link between the DNA and the enzyme, and breaking a DNA phosphodiester linkage at the same time. Rejoining of the DNA strand occurs by the reverse process. The oxygen of the free DNA 3’-OH group attacks the phosphorus of the phosphotyrosine link, breaking the covalent bond between the protein and the DNA, and reforming the phosphodiester linkage between adjacent nucleotides in the DNA chain.

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62
Q

What happens when gyrase is inhibited?

A

There is a decrease in replication initiation at the origin. This is because negative supercoiling makes it easier to separate the two strands of the double helix at the origin.

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63
Q

How do Type II topoisomerases work?

A

Once cut, the ends of DNA are separated and a second DNA double helix is passed through the temporary gap, also called a DNA linked “protein gate.” The DNA is then relegated. This reaction allows the Type II topoisomerases to increase or decrease the linking number of a DNA by two units and promotes chromosome disentanglement.

Example: Wrapping of the DNa around gyrase places one part of the DNA double helix, the T segment, over another part of the helix, the G segment. Upon ATP binding, one subunit of gyrase (GyrB) forms a dimer that captures the T segment, while at the same time the G segment is transiently cleaved. Hydrolysis of one ATP then allows GyrB to rotate, and the other subunit (GyrA) opens. The T segment drops into this opening and moves through the cleaved G segment. Religating of that G segment puts negative supercoils into the DNA. Finally, the T segment is released from the gyrase and hydrolysis of a second ATP resets gyrase back into its original shape.

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64
Q

What is the significance of complementary base pairing?

A

Complementary base pairing, or Watson-Crick base pairing, allows the 1’ carbons on the two strands to be exactly the same distance apart. This precise spacing results in the regular, symmetric framework of the DNA double helix. Watson-Crick base pairing also explains Chargaff’s rules, or the relationships among the molar concentrations of the different bases.

65
Q

Chemically, what does it mean when we say that the barrel strands of DNA dissociate and release an antibody? See DNA origami nanomachines from Focus Box 2.1

A

The barrel-shaped nanorobot was created with hinges to allow the structure to open and close. The barrel is held in the closed position by the base pairing of two complementary DNA strands, and acts as a lock inside which a specific antibody is enclosed. The target molecule then acts as the key that binds to one strand of DNA in the double stranded lock, making the barrel strands dissociate, spring open, and release the antibody.

66
Q

Explain the polarity of a DNA strand and its significance.

A

Polarity of DNA refers to the fact that the ends of a DNA or RNA chain are distinct and have different chemical properties. The 5’ end has a free phosphate group, and the 3’ end has a free hydroxyl group.

Because of the asymmetric structure of a nucleotide, there is built-in polarity when the nucleotides are linked to form DNA and RNA strands. The 5’ → 3’ polarity of a nucleic acid strand is an extremely important property, and defines the process of transcription and translation, as strands can be “read” only in the 5’ to 3’ direction.

Additionally, the hydrogen bonding between complementary base pairs can occur only if the DNA double helix is antiparallel (one strand runs 5’ → 3’ and the other 3’ → 5’)

67
Q

Explain the chemical polarity of nucleotides and their significance.

A

The nitrogenous bases by themselves are nonpolar and thus hydrophobic meaning they are insoluble in the aqueous cell environment. However, once covalently attached to a sugar and a phosphate to form a nucleotide, they become soluble in water, but there are still strong constraints on the overall conformation of DNA in a solution. The paired bases tend to stack on top of each other in base stacking, eliminating gaps between the bases and excluding the maximum amount of water from the hydrophobic interior of the double helix.

68
Q

Explain the significance of the phosphate group.

A

The phosphate functional group gives DNA and RNA the property of an acid. They form linking ester bonds that are stable yet easily broken by enzymatic hydrolysis. Further, after the phosphodiester linkage is formed, one oxygen atom of the phosphate group is still negatively ionized, making it extremely insoluble in lipids. This ensures that the nucleic acids are retained within the cell.

69
Q

What is the difference between denaturation, renaturation, and hybridization of double stranded DNA molecules?

A

Denaturation is when a double stranded DNA molecule is “melted” into its two component strands. Renaturation is when those two original strands reanneal back into the double helix. Hybridization is when a new single strand of DNA anneals with one of the original complementary strands, forming a new, hybrid, molecule.

70
Q

How has DNA origami been used for drug delivery?

A

Origami DNA has been used to create vehicles for drug delivery. In one example, a barrel-shaped nanorobot uses double stranded DNA to form the lock of an antibody-containing barrel. When the nanorobot is in the presence of the target molecule, or key, the bonds between the base pairs are broken, opening the lock and releasing the antibody.

71
Q

What are supercoils and why do they form?

A

Supercoils are twisted, three-dimensional tertiary DNA structures.

Supercoils form when DNA molecules lose or gain a few turns/twists. This usually happens when the DNA locally unwinds during replication, or when certain proteins bind. If the two ends aren’t free to rotate and release the tension, a small change in the number of turns can cause the DNA to coil through space instead of following a straight path.

Generally, changes in the average number of base pairs per turn of the double helix will be counteracted by the formation of an appropriate number of supercoils in the opposite direction. Undertwisting produces negative supercoils, overtwisting produces positive supercoils. The introduction of supercoils restores the total number of original turns of the helix.

72
Q

Supercoiled DNA is inherently less stable than relaxed DNA, but yet virtually all DNA in prokaryotic and eukaryotic cells exist in a supercoiled state. Why/How?

A

In eukaryotes, DNA is “restrained” when it is supercoiled around DNA-binding proteins (histones), because they have been stabilized by the energy of interaction between the proteins and the DNA. Also, supercoiling is involved in many genetic processes like replication, transcription, and recombination.

Circular DNA molecules from both bacteria and eukaryotes are usually negatively supercoiled. It has been shown that the relaxed circular DNA correlated with reduced activity in replication and transcription in bacteriophages, suggesting that negative supercoiling plays a role in the regulation of these processes.

Negative supercoiling around architectural proteins facilitates chromosome condensation in bacteria.

73
Q

What type of helix does RNA form and why? How does this helix convey information to binding and transcriptional/translational proteins?

A

Right-handed A-type double helix with 11 base pairs per turn. The 2’-hydroxyl group of the ribose sugar in RNA hinders the formation of a B-type helix. A-type RNA helices with Watson-Crick pairs have very deep and narrow major grooves that aren’t well suited for ligand interactions. However, the minor groove is shallow and broad, making it accessible to ligands, and contains the 2’-hydroxyl groups, which are good H-bond acceptors.

74
Q

What are base triples and what is their function in RNA?

A

Base triples occur when a third base interacts with either a Watson-Crick pair or a reverse Hoogsteen pair. Both base triples and noncanonical pairs are important mediators of RNA self-assemble and of RNA-protein + RNA-ligand interactions.

75
Q

What are retroviruses and what is their significance?

A

Retroviruses, after infecting an animal cell, quickly converts their RNA genome into a DNA one via reverse transcription. They then integrate their DNA into the host cell’s genome, where it remains permanently. Retroviruses are impossible to completely remove after infection and integration. Some examples of retroviruses include chicken pox/shingles and HIV-1.

76
Q

What is the significance of DNA compaction to transcription?

A

The wrapping of DNA around the histone core octamer is a transcription regulation mechanism and inhibits DNA-binding proteins from accessing this part of the DNA sequence, whereas the linker DNA region is accessible to DNA-binding proteins.

Segments within a chromosome can switch between different transcriptional states by changing which sequences are wrapped around core octamers and which are linker regions.

Heterochromatin → condensed chromatin, suppressed transcription
Euchromatin → open/decondensed chromatin, allows for transcription

77
Q

Researchers added mouse sperm nuclei to extracts from eggs of the African clawed frog that did not contain histones. They found that chromatid-like structures with central axes of condensin could be assembled, although the “nucleosome-depleted” chromatids were more fragile. Why did this happen?

A

The chromatid structures formed because chromosomes can form almost normally without histones because it is the condensin rings that promote DNA interactions and define the overall shape of the chromosomes, and not the histones. The histones add compaction and protection, which is why the chromatids were so fragile, but their absence does not prevent chromosome formation.

78
Q

What happened to the genome of original free-living organisms that were engulfed to create mitochondria and chloroplasts?

A

Intercompartmental DNA transfer. DNA may be released during disruption of organelle membranes during cell division and cell stress, followed by insertion into the nuclear genomes, which occurs in a process known as nonhomologous end-joining. Intercompartmental DNA transfer most often occurs from chloroplasts/mitochondria to the nucleus.

79
Q

Why is the deletion of CENP-A lethal?

A

CENP-A (also called CenH3) is a centromere-specific histone variant and is essential for centromere function. Without CENP-A, cells cant localize kinetochore proteins and segregate properly. CENP-A is involved in kinetochore formation and comprise the base. This rigid base resists the microtubule pulling forces at the centromere during mitosis. Without CENP-A, the cell cannot allocate its replicated DNA into its daughter cells.

80
Q

What determines nucleosome position?

A

Differing affinities for DNA sequences are the main determinants of nucleosome position, but nucleosomes can also be actively moved and displaced by remodeling complexes.

81
Q

What does cpDNA typically code for?

A

Enzymes involved in photosynthesis

82
Q

How is replication initiation different in bacteriophages, linear plasmids, and some viruses?

A

In bacteriophages, linear plasmids, and some viruses, instead of using an RNA primer a priming nucleotide is provided by a protein that binds to DNA. A deoxynucleoside monophosphate (dNMP) covalently attaches to a specific serine, threonine, or tyrosine residue in the protein.

83
Q

What are some similarities in genome replication between bacteria and eukaryotes?

A

All cellular genomic DNA is replicated by the replisome, a multiprotein machine with the same basic enzymatic functions: helicase, primase, DNA polymerase, and 3’ → 5’ exonuclease.

84
Q

What is the significance of the major groove of DNA? How is it formed?

A

The sugar-phosphate backbone of DNA isn’t spaced equally, resulting in a major and minor groove.

The Major groove exposes the base letters, allowing binding proteins to “read” the exposed nucleotides. Specifically, the exposed solvent-accessible N and O atoms in the nucleotides form H-bonds with the binding protein’s amino acid side chains, and the difference in the donor-acceptor patterns tells the binding protein which pair of nucleotides it’s attached to.

85
Q

What is the significance of the minor groove of DNA? How is it formed?

A

The sugar-phosphate backbone of DNA isn’t spaced equally, resulting in a major and minor groove.

The width of the minor groove is dependent on which nucleotides are present in the surrounding DNA. For example, a short section of adenine (an A-tract) forms a narrow minor groove. The Minor groove is thought to be harder to read because the H-bond patterns are the same regardless of which way the base pair is flipped, meaning that the only difference in pattern would be between AT and GC pairs. However, there are some DNA-binding proteins that can recognize the minor groove, such as the TATA-binding protein (TBP), which binds at the TATA box and plays a role in initiating gene transcription in eukaryotes. Additionally, the RFC (sliding clamp) spiral matches the minor grooves of the double helix, allowing it to lock onto the primed DNA in a screwcap arrangement.

86
Q

What is the difference between denaturation, renaturation, and hybridization of double stranded DNA molecules?

A

Denaturation is when a double stranded DNA molecule is “melted” into its two component strands. Renaturation is when those two original strands reanneal back into the double helix. Hybridization is when a new single strand of DNA anneals with one of the original complementary strands, forming a new, hybrid, molecule.

87
Q

What advantages does the cell get from negative/left hand supercoiling?

A

Negative Supercoiling:
Puts potential energy into the DNA

Makes it easier to pull the two strands of the double helix apart

Negative supercoiling around architectural proteins facilitates chromosome condensation in bacteria

88
Q

What advantages does the cell get from positive/righthand supercoiling?

A

Positive Supercoiling:
Occurs ahead of replication forks and transcription complexes

Makes it harder to open the double helix
Blocks DNA processes, may play a role in gene regulation/silencing

Adaptation to high temperatures in thermophilic archaea; at high temperatures the positive supercoils are stable, energetically favorable, and prevent the DNA from denaturing

89
Q

What aspects of histones give it such a high affinity for DNA?

A

Histones are positively charged

90
Q

Is leading and lagging strand synthesis coordinated?

A

No!

91
Q

Describe the process of replication initiation by the bacterial replisome.

A
  1. Helicase encircles the lagging strand and uses ATP to unwind the dsDNA
  2. Two sliding clamps are loaded onto DNA by a clamp loader machine
  3. The clamp loader has ATPase activity and uses the energy from ATP hydrolysis to open and close the sliding clamps around the DNA
  4. Three DNA Polymerase III molecules are tethered to the clamp loader via its flexible tau domains (two of which are also tethered to the DNA by the sliding clamps)
  5. One DNA Polymerase III-sliding clamp complex extends the leading strand while the other DNA Poly. IIIs copy the lagging strand
  6. Release of DNA Poly III is triggered by either the completion of an Okazaki fragment or by collision with the previous fragment.
  7. New sliding clamps are assembled onto RNA primed sites, upon which DNA Poly III hops aboard to make the next Okazaki fragment
92
Q

What role does DNA Polymerase I play in bacterial replication? What about DNA Polymerase III? What aspects of their structures allow them to fulfill these roles?

A

While it is the most abundant polymerase, DNA Polymerase I is not the main replicative polymerase in E. coli. DNA Poly. I plays a role in primer removal and gap filling between Okazaki fragments. It’s DNA Polymerase III that catalyzes genome replication. This gives it a unique ability to start replication at a broken phosphodiester linkage

DNA Polymerase I is made of two fragments: the Klenow fragment, which has 5’ → 3’ polymerase activity, and the other fragment, which as 3’ → 5’ exonuclease/proofreading activity (can remove a dNMP from the end of a DNA chain by breaking the terminal phosphodiester bond)

DNA Polymerase III is made of 10 different polypeptide subunits, one of which has 5’ → 3’ activity and another which has 3’ → 5’ exonuclease activity.

93
Q

What is the DNA looping mechanism and why is it referred to as the “trombone model” of replication?

A

As the replication fork advances, the lagging strand polymerase remains associated with the replisome, resulting in a DNA loop during Okazaki fragment synthesis. The loop grows as an Okazaki fragment is extended, and upon completion the polymerase is released from the DNA to begin the next fragment. Because the DNA loop grows and disappears repeatedly, it’s called the “trombone model”.

94
Q

How are Okazaki fragments connected into a continuous new strand?

A

DNA ligase catalyzes the formation of a phosphodiester linkage between the adjacent fragments

95
Q

How and why does primase join the replication machinery?

A

Primase joins the replication machinery only by interacting with helicase, which ensures that primase makes primers only at the replication fork.

96
Q

How are bacterial and eukaryotic Topoisomerase I’s different?

A

Bacterial Topoisomerase I can relax only negatively supercoiled DNA. Eukaryotic Topoisomerase I can relax both negatively and positively supercoiled DNA.

97
Q

How is replication initiation different in bacteria and eukaryotes?

A

In bacterial cells, as soon as the initiator proteins accumulate at the origin, DNA helicases are recruited to the origin and initiation begins. Eukaryotic cells, however, separate origin selection from initiation through the formation of a prereplication complex.

98
Q

How do eukaryotes prevent the genome from being overreplicated?

A

Eukaryotic cells use a separates origin selection and initiation into two separate events using a prereplication complex.

99
Q

Why might leading strand synthesis not be as continuous as previously thought?

A

The term “continuous” doesn’t mean that the polymerase speeds along without ever slowing down.

Polymerases may pause, dissociate, and reassociate with the replisome

Movement of E. coli DNA poly III can be blocked by a damaged site on the DNA template

Sometimes DNA poly III collides with RNA polymerase and is stalled
DNA polymerases 𝝳 and ε may pause and backtrack to fix errors (proofreading)

100
Q

How does the replication machinery get past the nucleosomes to interact with the replication origins?

A

…yet to be determined.

Current theories:
Post-translational modifications of histones and chromatin remodeling factors may loosen the chromatin and allow the nucleosomes to disassemble

Presence of G-Quadruplex motifs may favor the formation of nucleosome-free regions or they might recruit origin-binding proteins

101
Q

How is the origin recognition complex (ORC) assembled?

A
  1. ORC forms a complex with protein Cdc6
  2. ORC-Cdc6 recruits a single Mcm2-7 helicase into a complex with Cdt1
  3. A second, identical Mcm27 + Cdt1 complex forms an inactive double hexamer
  4. Other accessory features, like Cdc45 and GINS, are loaded to form a preinitiation complex
  5. The inactive Mcm2-7 double hexamer is now transformed into a single active CMG helicase complex and encircles the leading strand at each of the two divergent replication forks
102
Q

How do eukaryotes ensure that DNA replicates only once per cycle?

A

They use a replication licensing system, of which Mcm2-7 is the main protein. ORC’s ATP hydrolysis stimulates assembly of the prereplication complex. When ORC can’t hydrolyze ATP, that complex can’t be assembled, preventing the synthesis of new DNA.

103
Q

How is licensing by Mcm2-7 regulated?

A

CDK (cyclin-dependent kinase) activity. When CDK activity is low (in M and G1 phases), Mcm2-7 can be loaded onto the origins to prepare for replication. When CDK activity is high (in S and G2 phases), the licensed origins fire. High CDK activity blocks the relicensing of origins that have already fired, aka no more Mcm2-7 can be loaded onto origins.

*The mode of CDK’s regulation differs for each protein in the licensing complex, and between yeast and vertebrates.

104
Q

Which of the 17 distinct eukaryotic DNA polymerases are referred to as the Replicative Polymerases and what type of DNA are they used in?

A

DNA Polymerases ⍺, 𝝳, and ε are used in chromosomal replication
DNA polymerase 𝛄 is used strictly for mtDNA replication

This distinction distinguishes them from the remaining polymerases, which are involved in the repair processes.

105
Q

How are mismatched pairs detected and removed?

A

The selection of the correct nucleotides is largely dependent on the geometries of the Watson-Crick pairs. While AT and GC have similar geometries, those of mismatched pairs are quite different and result in esoteric hindrance at the active site that inhibits efficient catalysis. This pauses the polymerase, and the 3’ end of the new strand is transferred to the exonuclease domain, where the mismatched nucleotides are excised and released as a dNMP.

106
Q

Give a brief overview of maturation, the final steps in DNA replication.

A

RNA Primer Removal:
- The nascent flap re-equilibrates to a single nucleotide that binds to FEN-1
- FEN-1 (flap exonuclease 1) removes the RNA-containing-5’-flap
A stretch of unreplicated DNA is left behind

Gap Fill-In:
- The remaining gap left by primer removal is filled in by DNA Poly 𝝳

Okazaki Fragments:
- DNA ligase I catalyzes the formation of new phosphodiester bonds and joins the fragments together

107
Q

How is the RNA-containing-5’-flap kept to a minimal size?

A

The rate of DNA synthesis slows as each nucleotide is displaced, acting like a progressive molecular brake

The 3’ → 5’ exonuclease activity of DNA polymerase 𝝳 can trim the flap

108
Q

What is the mechanism of poison-type topoisomerase-targeted anti-cancer drugs?

A

In nature, breaks created by topoisomerase are brief, but in the presence of the poison drug, the broken DNA is trapped as a stable intermediate bound to topoisomerase, with a high chance of being converted to a permanent break in the genome. These drugs preferentially target cancer cells because they are rapidly growing and usually contain higher levels of topoisomerases. The higher the cellular level of topoisomerases, the deadlier the drugs become. Cancer cells often have impaired DNA repair pathways, so they are more susceptible to these DNA damaging agents.

109
Q

Although not completely understood, mtDNA replication is widely accepted to follow the Strand Displacement model. What are the mechanisms for this model? What evidence suggests that it may not be?

A

Replication is unidirectional around the circle, with one replication fork for each of the two strands, called the light strand (L) and the heavy strand (H). There’s one priming event and one origin per strand. Replication begins at the D (displacement) loop. The H strand is used as a template to make a new L strand. About ⅔ through, it passes by the L strand’s origin, opening it and initiating the synthesis of a new H strand from the L strand. RNase MRP removes the RNA primers.

The Strand Coupled Model for mammalian mtDNA replication, in which the coupled leading H strand and lagging L strand synthesis represent a semidiscontinuous, bidirectional mode of replication. Evidence for this is that when mtDNA replication intermediates were analyzed with two-dimensional agarose gel electrophoresis, patterns arose that suggested that the lagging L strand initiates at multiple sites, indicating discontinuous synthesis of short Okazaki fragments.

110
Q

What is the end replication problem?

A

DNA polymerase requires a short RNA primer and proceeds only 5’ to 3’. When the final primer is removed from the lagging strand at the end of a chromosome, a small region, the telomere, is left unreplicated, and there is no upstream strand onto which DNA polymerase can fill the gap. This should make the chromosomes shorten with each round of replication, however this clearly doesn’t happen.

111
Q

What kind of enzyme is telomerase?

A

An RNP made of RNA and telomerase reverse transcriptase.

112
Q

How does telomerase maintain telomeres?

A

Telomerase causes the 3’ template of the lagging strand to elongate, resulting in the addition of one telomeric repeat. The telomerase repositions and repeats this process.

113
Q

Where is telomerase located?

A

In Cajal bodies until it is recruited to the telomeres during S phase.

114
Q

How is telomerase activity regulated?

A

Shelterin forms a folded chromatin structure and prevents telomerase from accessing the telomeres

Telomeric DNA forms a t-loop structure in which the 3’ tail invades and binds to a C strand sequence in the dsDNA to form the loop. Without the unpaired 3’ end, telomerase cannot bind

G-Quadruplexes tend to form in telomeres, which can also block telomerase

115
Q

What is shelterin?

A

A 6-protein structure that protects the ends of the chromosomes and prevents telomerase from accessing the telomeres.

116
Q

What is the difference between a spontaneous mutation and an induced mutation?

A

Spontaneous → result of natural processes in cells, ex: DNA replication error, reactive oxygen species produced during aerobic metabolism

Induced → result of interactions between DNA and outside agents of mutagens that causes damage

117
Q

Are all DNA mutations detrimental? Does all unrepaired DNA damage lead to mutations that promote disease or lead to cell death?

A

No! While high-fidelity DNA replication is beneficial for maintaining genetic information over many generations, low-fidelity DNA replication is the basis of evolution and generates genetic diversity.

118
Q

Why is the study of genetic mutations important?

A

Mutations drive evolutionary change

Mutations have deleterious (and occasionally beneficial) consequences to an organism or its descendants, such as heritable genetic disorders or acquired diseases

Mutant organisms are important tools in characterizing the genes involved in different cellular processes

119
Q

What’s the difference between a transition mutation and a transversion mutation, and which type is more common?

A

Transition: pyrimidine → pyrimidine; purine → purine
Transversion: pyrimidine → purine; purine → pyrimidine

Spontaneous nucleotide substitutions are most likely to be transition mutations.

120
Q

What are the consequences of silent mutations?

A

None. Multiple codons can code for the same amino acid, so it remains the same.

121
Q

What are the consequences of missense mutations?

A

Caused by nucleotide substitutions in protein-coding regions, the resulting changed amino acids can alter the biological properties of the protein. Missense mutations often cause disease by loss of protein function or disrupting key protein interactions. However they could also cause a gain of function and allow interactions with new partners.

Ex: sickle cell disease. A single amino acid changes as a result of an A → T transversion, altering the hemoglobin’s shape, causing the dysfunctional sickle cell shape.

122
Q

What are the consequences of nonsense mutations?

A

Caused by a nucleotide substitution that creates a new stop codon, nonsense mutations cause premature chain termination during protein synthesis. The protein fragment is almost always nonfunctional.

Ex: resistance to thyroid hormone in humans. Nonsense mutations cause premature truncation of the thyroid hormone receptor, disrupting the ligand-binding domains, making the receptor unable to bind with thyroid hormone.

123
Q

What are the consequences of frameshift mutations?

A

Caused by insertions or deletions of DNA. If the length of the insertion or deletion isn’t in exact multiples of three, the mutation shifts the reading frame, and all amino acids downstream of the mutation are altered.

Ex: cystic fibrosis. Deletion of three base pairs in the cystic fibrosis transmembrane conductance regulator gene results in the loss of the codon for phenylalanine, resulting in a defect in chloride ion transport, causing the disease.

124
Q

What are the consequences of trinucleotide repeat expansions? Which genes are especially sensitive to this?

A

Trinucleotide repeat expansions can adopt triple helix conformations and assume other unusual secondary structures, which can interfere with replication and transcription. Genes that code for the nervous system and neurological functions seem to be especially sensitive to trinucleotide repeat expansions, as many disorders associated with this type of mutation have at least some aspect of neurological dysfunction associated with them (ex: Fragile X syndrome, Huntington’s disease, Kennedy’s disease, Friedreich’s ataxia, spinocerebellar ataxia type I, myotonic dystrophy, oculopharyngeal muscular dystrophy)

125
Q

What is the most basic definition of a mutagen?

A

Any chemical or physical agent that causes an increase the the rate of mutation above spontaneous background level.

126
Q

What are the consequences of deamination?

A

Deamination is the chemical conversion process of a changing single base pair to another. Ex: the replacement of the amino group on cytosine with oxygen, converting it to uracil.

Deamination is the most frequent type of hydrolytic damage, and can occur spontaneously from the action of water or can be induced by a chemical mutagen. Usually causes only minor structural distortions in the DNA double helix, and is not likely to completely block replication or transcription processes.

127
Q

What are the consequences of UV-induced lesions?

A

Most common type of UV-induced lesions are thymine dimers, aka cyclobutane-pyrimidine dimers, in which a cyclobutane ring is generated by links between carbon atoms 5 and 6 of adjacent thymines. Thymine dimers distort the structure of the DNA double helix because the covalent bonds between the thymines disrupt the complementary base pairs on between the adjacent strands. This structural distortion can block polymerases and impede transcription and replication.

128
Q

What is the mechanism of translesion synthesis, also known as the DNA damage tolerance pathway?

A

Polymerase switching model: More favored model. When a replicative polymerase reaches the lesion, it falls off or simply jumps and continues replication downstream.

Gap-filling model: DNA replication machinery leaves a single-stranded in a gap opposite the lesion, and then translesion polymerases fill the gap after the replication fork moves on.

129
Q

How do error-prone polymerases differ from high-fidelity polymerases?

A

The error-prone polymerases have a more spacious active site, while the high-fidelity polymerases fingers close tightly on the newly formed base pairs to check for mismatches.

130
Q

How do error-prone polymerases deal with lesions that have altered the pairing properties of pre-existing bases?

A

The polymerases can cause nucleotide substitution mutants by inserting incorrect nucleotides opposite the lesions. They can also generate frameshift mutations by inserting a correct nucleotide after skipping past the lesion.

131
Q

How does DNA Polymerase η protect the cell from UV damage?

A

DNA Polymerase η has an even wider active site than other error-prone polymerases, and can accommodate two nucleotides at the same time. Van der Waals forces and H-bond interactions hold the thymine-thymine dimer in place so that they can be paired with adenine, fixing the lesion.

132
Q

How does photoreactivation reverse UV radiation damage?

A

DNA photolyase directly repairs pyrimidine dimers by using energy (photons in the blue light to near UV wavelength range) to break the covalent bonds holding two adjacent pyrimidines together.

It has two cofactors buried in its globular tertiary structure: one is a pigment (folate) that absorbs blue to near UV light; the other is FADH-. The folate acts as a “solar panel”, absorbing light and transferring it to FADH-, which transfers an electron to the thymine-thymine dimer and splits the dimer before returning the electron back to FADH-.

133
Q

How do methyltransferases reverse DNA damage?

A

Methyltransferasses repair O6-methylguanine by catalyzing the removal of the methyl group from the damaged guanine. They bind to the minor groove of the double helix, widening it and allowing the damaged nucleotide to flip into the enzyme’s active site.

Once the methyltransferase accepts the damaged guanine’s methyl group, it cannot be used again.

134
Q

How is SPRTN regulated and what would happen if its regulation mechanism was removed?

A

SPRTN is a DNA-dependent metalloprotease and removes DNA-protein cross-links during DNA synthesis. It is regulated by a ubiquitin switch, a type of post-translational modification. When SPRTN is monoubiquitinated, it is excluded from the chromatin. But when DNA-protein cross-links are formed, they trigger deubiquitination and SPRTN’s subsequent self-cleavage.

135
Q

What are the consequences of defective SPRTN gene?

A

If the SPRTN gene is defective, then the cell/organism will have no way to remove DNA-protein cross-links. Ex: Ruijs-Aalfs syndrome (Spartan syndrome), a disease characterized by chromatin instability, premature aging, and early-onset hepatocellular carcinoma.

136
Q

What key feature do all repair systems that remove damaged DNA share?

A

The key these pathways share is their multiple dynamic protein interactions, or their ordered handoff of DNA from one protein or complex to another.

They also share the same basic path:
The repair machinery senses and accesses the damaged DNA.
The repair machinery then removes the damaged DNA.
The resulting gap is filled by DNA polymerase.
PCNA recruits chromatin assembly factors which deposit histones onto the new DNA.

137
Q

How does the repair machinery gain access to the DNA?

A

Upon sensing the damage in the DNA, nucleosomes are loosened by the combined action of histone acetylation and chromatin remodeling.

138
Q

How does hOGG1 differentiate between a normal GC base pair and an oxoG-C pair?

A

hOGG1 binds to DNA nonspecifically, but it has a series of two pockets to discern between damaged and undamaged GC pairs. All Gs are transiently extruded into hOGG1’s G-specific pocket, but only oxoGs are allowed to access the secondary oxoG pocket.

139
Q

What are abasic nucleotides and how are they removed?

A

Abasic nucleotides are missing their nitrogenous base. They are fixed with either short or long patch repairs, both of which are characterized by a handoff of the DNA from the endonucleases that excise the abasic nucleotide to DNA polymerase and ligase for repair synthesis.

140
Q

What is the difference between base excision repair and mismatch repair?

A

Base excision repair fixes damaged bases. Mismatch repair corrects the mistakes that occur during DNA replication but aren’t caught by polymerase proofreading.

141
Q

What are the consequences of mismatch repair deficiency for humans?

A

Hereditary deficiencies in mismatch repair increases the rate of gene mutations and our susceptibility to certain types of cancer (ex: lynch syndrome, aka hereditary nonpolyposis colon cancer)

142
Q

How do the proteins involved in mismatch repair determine which strand is the daughter strand with the mistake?

A

Still under investigation. Proposed models include:

The 5’ end of an Okazaki fragment and the polymerase machinery associated with the 3’ end of the nascent strand may provide markers for strand discrimination when mismatch repair occurs in the final stages of replication

It has been shown that a single ribonucleotide in the vicinity of the mismatch can act as an initiator of mismatch repair

In E. coli the mistake strand is identified by the absence of methyl groups on GATC sequences

143
Q

How does hereditary nonpolyposis colorectal cancer (HNPCC) progress/occur/develop?

A
  1. An individual inherits one allele for the inactivation of the mismatch repair genes.
  2. The second, wild-type allele is spontaneously lost or inactivated, usually due to the probability of mutation being increased in dividing cells
  3. The two inactivated alleles in dividing colon cells lead to a completely defective mismatch repair pathway
  4. Mistakes accumulate during DNA replication, increasing the spontaneous mutation rate. Mutations in tumor suppressor genes and oncogenes subsequently accumulate, leading to deregulated cell growth
  5. The defects in the mismatch repair genes also result in an accumulation of mutations in the DNA’s microsatellite regions, leaving the cell unable to bind to growth inhibitors and inhibiting apoptosis pathways.
144
Q

How do humans and other placental mammals repair DNA damage caused by UV radiation?

A

We CANNOT use photolyases. Every other organism can, but that ability was lost for us. Instead, we use nucleotide excision repair, which is carried out by six repair factors.

145
Q

How do E. coli’s small number of nucleotide excision repair proteins scan millions of base pairs to find the rare sites of DNA damage?

A

Proteins that mediate this damage recognition bind transiently to DNA and slide along it by either diffusion or directed motion, scanning the genome (they can also jump from one DNA molecule to another. Fun!)

146
Q

What are the consequences of defects in nucleotide excision repair?

A

An inability to properly cope with UV-induced DNA lesions, resulting from an unusually high sensitivity to UV light. Any lesions caused by UV-light lead to an increased mutation rate, as they cannot be fixed.

147
Q

What type of DNA damage is the most harmful?

A

Double-strand breaks

148
Q

How are double-strand breaks fixed?

A

With either homologous recombination or nonhomologous end-joining.

Homologous recombination plays a major role in repair in prokaryotes and single-celled eukaryotes, but in mammalian cells double strand breaks are primarily repaired through nonhomologous end joining.

149
Q

How does nonhomologous end joining work?

A

NHEJ rejoins double strand breaks via direct ligation of the DNA ends without any sequence homology requirements.

Three steps, of which there’s some flexibility to their order: nucleolytic action, polymerization, and ligation.

Broken DNA ends are recognized. A scaffold forms that holds the broken ends in close proximity. The nuclease, the polymerases, and the ligase complex is recruited. The excess or damaged DNA is trimmed, any gaps are filled in, and ligase rejoins the ends.

150
Q

What are some negative consequences of NHEJ?

A

Two broken ends can be ligated together by repair machinery regardless of whether they came from the same chromosome

NHEJ frequently results in insertions or deletions at the break site

But evolutionarily, anything is better than potentially lethal breaks in the genome, so these mutations are selected for

151
Q

What is the difference between an RNA enzyme and an RNP enzyme?

A

An RNA enzyme can function catalytically by itself, without anything else. An RNP enzyme needs a protein to function. The RNA molecule is still the catalyst, but it needs the protein to hold it in its proper shape.

152
Q

Explain the polarity of a DNA strand and its significance.

A

Polarity of DNA refers to the fact that the ends of a DNA or RNA chain are distinct and have different chemical properties. The 5’ end has a free phosphate group, and the 3’ end has a free hydroxyl group.

Because of the asymmetric structure of a nucleotide, there is built-in polarity when the nucleotides are linked to form DNA and RNA strands. The 5’ → 3’ polarity of a nucleic acid strand is an extremely important property, and defines the process of transcription and translation, as strands can be “read” only in the 5’ to 3’ direction.

Additionally, the hydrogen bonding between complementary base pairs can occur only if the DNA double helix is antiparallel (one strand runs 5’ → 3’ and the other 3’ → 5’)

153
Q

What features of a sequence of dsDNA would cause it to have a high T(m)?

A

Sequences that are GC rich will have higher T(m)s because GC is bound with three hydrogen bonds, and therefore needs more energy to fully denature.

154
Q

What aspects of a dsDNA sequence would make it unlikely to reanneal into its original structure after denaturation?

A

Sequences that are composed of tandem repeats, especially ATAT tandem repeats, are least likely to form their original structure. ATAT tandem repeats have the potential to form slipped structures with compensating single-stranded loops in the alternate strands, which is certainly not part of the original B-DNA double helix.

155
Q

Assume that you measure the relative absorbance at A(260) of a sample of purified genomic DNA and a sample of purified mRNA, both from human cells. Which sample would you expect to have the greater relative absorbance of UV light?

A

I would expect the single stranded mRNA to have a greater relative absorbance at 260 nm. Base stacking in dsDNA quenches the bases’ capacity to absorb light. Since RNA is single stranded, it would have a greater relative absorbance.

156
Q

What structural features of RNA make it a more versatile molecule than DNA?

A

Because RNA is single stranded, it can fold into many different unique secondary and tertiary structures and base pair with other RNA molecules or DNA.

The ribose 2’-OH is a good hydrogen bond acceptor and acts as a nucleophile in self-splicing ribozymes.

RNA also contains many modified bases that provide binding sites for proteins.

157
Q

A portion of the catalytic center of a large ribozyme is susceptible to misfolding and the formation of an alternative helix. What are some possible consequences of this alternate helix for the ribozyme’s function?

A

The alternative helix could obstruct the catalytic center or not allow the catalytic center to form in a functional shape, potentially reducing or blocking the substrate from binding to the ribozyme. The shape change could also allow a different substrate to bind to the catalytic center.

158
Q

What is the significance of DNA compaction to cell divison?

A

If you tried to separate decondensed chromosomes for cell division, the chromosomes would be a tangled mess and wouldn’t be able to form the centromeres and kinetochore microtubules necessary for cell division. Chromosomes must be condensed to facilitate cell division.