Exam 2 Flashcards

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1
Q

Describe basic gene structure

A

Promoter: Upstream regulatory region that binds transcription initiation complex.
5’ Untranslated Region (5’ UTR): Precedes the coding sequence, may contain regulatory elements.
Coding Sequence (Exons): Sequences that are transcribed into mRNA and translated into a protein.
Introns: Non-coding sequences removed from the primary transcript during RNA splicing.
3’ Untranslated Region (3’ UTR): Follows the coding sequence, may contain regulatory elements.
Terminator: Downstream sequence that signals transcription termination and release of mRNA.

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2
Q

coding sequence (continuous vs. non-continuous)

A

coding sequence contains DNA sequence used to produce the specific
protein encoded by the gene (will contain codons)

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3
Q

regulatory elements (promoter, terminator, enhancer, silencer)

A

regulatory elements are sequences of DNA that control transcription of a gene.
Usually proteins bind to these sequences to modulate transcription

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4
Q

Regulatory Elements:

A

Promoter: Upstream region that binds transcription initiation complex, initiates transcription.
Terminator: Downstream sequence that signals the end of transcription and releases RNA.
Enhancer: Can be upstream, downstream, or within introns. Binds transcription factors to increase transcription rate.
Silencer: Can be upstream, downstream, or within introns. Binds repressor proteins to inhibit gene expression

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5
Q

Describe transcription factor (general, activator, repressor)

A

Transcription factors are proteins that regulate transcription of a gene by
binding regulatory elements in a sequence specific manner

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6
Q

Transcription Factors:

A

General:

Proteins that bind to regulatory DNA sequences
Control transcription initiation, elongation, or termination

Activator:

Binds enhancers or promoters
Recruits transcription machinery
Increases transcription rate

Repressor:

Binds silencers or promoters
Blocks transcription machinery binding
Decreases or prevents transcription

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7
Q

Describe transcription start site (+1)

A

The nucleotide position where RNA polymerase begins transcribing a gene into mRNA.
Denoted as +1 in the DNA sequence.
Located within the promoter region, typically around 25-30 base pairs downstream of the promoter.
Marks the start of the transcribed region.
Accurate identification is crucial for determining promoter elements and gene structure

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8
Q

Describe proximal, distal, upstream, downstream

A

Proximal: Close to the gene or regulatory element being described. A proximal promoter is located near the transcription start site.
Distal: Far away from the gene or regulatory element. A distal enhancer can be thousands of base pairs away from the gene it regulates.
Upstream: The DNA sequence located before (5’ to) the gene or regulatory element. Upstream regulatory regions like promoters are found preceding the gene.
Downstream: The DNA sequence located after (3’ to) the gene or regulatory element. Downstream sequences like terminators follow the gene coding region.
In summary:

Proximal = close to
Distal = far from
Upstream = before
Downstream = after

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9
Q

coding strand

A

Coding strand (sense/non-template strand)
Same sequence as mRNA (except T→U)
Template for mRNA synthesis during transcription
Contains protein-coding sequence
Codons read 5’ to 3’ direction
Specifies amino acid sequence for translation
Other strand is template (antisense/non-coding)

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10
Q

template strand

A
  • Template strand (antisense/non-coding strand)
  • Complementary to coding/sense strand
  • Used as template by RNA polymerase during transcription
  • mRNA sequence is complementary to template strand sequence
  • Does not directly encode protein sequence
  • Contains anti-codons complementary to mRNA codons
  • Provides template for mRNA synthesis
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11
Q

5’

A

Refers to the 5’ carbon in the sugar of a nucleic acid
Denotes the start/upstream end of a polynucleotide chain
Has a free phosphate group at the 5’ end
Read/synthesized in 5’ to 3’ direction
mRNA has 5’ cap for stability and translation

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12
Q

3’

A

Refers to the 3’ carbon in the sugar of a nucleic acid
Denotes the end/downstream end of a polynucleotide chain
Has a free hydroxyl group at the 3’ end
DNA synthesis occurs in 5’ to 3’ direction
mRNA has 3’ poly-A tail for stability and export

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13
Q

sense and antisense

A

Here are concise bullet points about sense and antisense for your cheat sheet:

Sense (Coding Strand):
- Same sequence as mRNA (except T→U)
- Template for mRNA synthesis
- Contains protein-coding sequence
- Codons read 5’→3’ direction
- Specifies amino acid sequence

Antisense (Template Strand):
- Complementary to sense/coding strand
- Used as template by RNA polymerase
- mRNA sequence complementary to antisense
- Contains anti-codons complementary to mRNA
- Does not directly encode protein

In summary:
- Sense = Coding, template for mRNA
- Antisense = Non-coding, template strand

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14
Q

coding region (coding
sequence)

A

Coding Region/Sequence:
- Portion of a gene that contains codons
- Codons specify amino acid sequence of protein
- Transcribed into mRNA
- Begins with a START codon (AUG)
- Ends with a STOP codon (UAA, UAG, UGA)
- Introns removed by splicing in eukaryotes
- Exons joined to form mature mRNA
- Does not include regulatory regions (promoters, etc.)
- Read in 5’ to 3’ direction during translation

In summary:
- Codes for protein product
- Transcribed region between START and STOP codons
- Exons make up coding sequence after splicing

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15
Q

Describe what happens and how in each stage of bacteria transcription
o Initiation, Elongation, and Termination

A

Initiation:

RNA polymerase binds to the promoter region of a gene
Promoter provides a recognition site for RNA polymerase
Transcription factors may assist RNA polymerase binding
DNA strand separation occurs, forming an “open complex”
RNA polymerase begins synthesizing RNA using ribonucleoside triphosphates
The first few nucleotides bind and form the transcription initiation complex

Elongation:

RNA polymerase moves along the template DNA strand
Ribonucleotides are added one by one to the 3’ end of the growing RNA transcript
The RNA transcript is extended in the 5’ to 3’ direction
RNA polymerase unwinds the DNA helix to expose the template strand
Transcription factors may regulate this process
RNA transcript is released from the exit channel of RNA polymerase

Termination:

RNA polymerase recognizes a terminator sequence in the DNA
This triggers a conformational change that destabilizes the transcription complex
The newly synthesized RNA transcript is released from RNA polymerase
RNA polymerase dissociates from the DNA template
Rho protein may aid in termination at certain terminator sequences

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16
Q

Generic overview transcription

A

Initiation: RNAP binds to promoter
Elongation: RNAP polymerizes mRNA complementary to DNA template sequence
Termination: A hairpin in the mRNA causes RNAP to dissociate from DNA

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17
Q

Transcription elongation:

A

RNAP adds RNA complementary to DNA
template and catalyzes a phosphodiester bond in 5’ to 3’ direction

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18
Q

RNA polymerase (RNAP)

A

enzyme that
polymerizes (bonds together) RNA nucleotides
that are complementary to the template sequence

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19
Q

Transcription initiation:

A

1) Sigma factor part of holoenzyme will
bind to the promotor in a sequence specific manner
Consensus sequence is the
most commonly found
sequence of nucleotides or
amino acids (across organisms
or within a genome)
Bacteria consensus promoter
sequence at -10 and -35 shown.
gene A
5’
3’
promoter RE
2) RNAP unwinds DNA double-helix and
3) sigma factor dissociates

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20
Q

Transcription termination:

A

terminator sequence causes hairpin to
form, which causes RNAP to dissociate

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21
Q

transcription elongation:

A

RNAP adds RNA complementary to DNA
template and catalyzes a phosphodiester bond in 5’ to 3’ direction

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22
Q

Know the parts of bacteria RNAP (RNA polymerase): core enzyme and sigma factor and
their functions

A

Bacterial RNA polymerase (RNAP) consists of two main components: the core enzyme and the sigma factor. Here are the parts and their functions:

Core Enzyme:
- Consists of 5 subunits: 2 α subunits, 1 β subunit, 1 β’ subunit, and 1 ω subunit
- Catalyzes the synthesis of RNA using the DNA template
- Contains the active site for nucleotide polymerization
- Responsible for elongation of the RNA transcript

Sigma (σ) Factor:
- A dissociable subunit that associates with the core enzyme
- Primary sigma factor (e.g., σ70 in E. coli) is required for initiation of transcription
- Recognizes and binds to specific promoter sequences on DNA
- Enables core enzyme to locate and bind to promoters
- Helps melt/unwind DNA for transcription initiation
- Dissociates from core enzyme after initiation

In summary:
- Core enzyme catalyzes RNA synthesis and elongation
- Sigma factor enables promoter recognition and initiation
- Together, they carry out the three stages of transcription

The sigma factor provides promoter specificity, while the core enzyme has the catalytic activity for RNA polymerization.

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23
Q

RNA polymerase

A

Consists of 5 subunits (2α, 1β, 1β’, 1ω)
Catalyzes RNA synthesis (nucleotide polymerization)
Responsible for elongation of RNA transcript

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24
Q

sigma factor

A

Dissociable subunit that binds core enzyme
Recognizes and binds specific promoter sequences
Enables core enzyme to bind promoters
Required for transcription initiation
Dissociates after initiation

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25
Q

holoenzyme

A

The complete transcription initiation complex
Consists of RNA polymerase core + sigma factor
Sigma factor confers promoter specificity
Holoenzyme binds promoter and initiates transcription
After initiation, sigma released and core does elongation

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26
Q

transcription initiation

A

Promoter Recognition:

Sigma factor of RNA polymerase holoenzyme recognizes and binds to promoter DNA sequence
Promoters have -10 and -35 regions that sigma factor binds to

Closed Complex Formation:

RNA polymerase holoenzyme binds to promoter, forming a stable closed complex

Open Complex Formation:

DNA strands are unwound/melted open around the transcription start site (+1)
Helped by sigma factor and transcription factors binding upstream

Initiation:

RNA polymerase begins synthesizing RNA transcript complementary to the coding strand
Incorporates first few ribonucleotides using DNA template strand
Forms an initiation complex or open complex

After initiation:

Sigma factor is released from the core RNA polymerase enzyme
Core enzyme proceeds into the elongation phase

Key players:

Promoter DNA sequence
Sigma factor for recognition
RNA polymerase holoenzyme
Transcription factors

The initiation phase sets up the transcription bubble and allows transition into elongation.

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27
Q

transcription
elongation

A

Here are the key points about the transcription elongation phase in bacteria:

Elongation Complex:
- After initiation, sigma factor dissociates
- Core RNA polymerase enters elongation phase as the elongation complex

RNA Synthesis:
- RNA polymerase moves along DNA template strand
- Adds ribonucleotides one by one to the 3’ end of growing RNA transcript
- Synthesizes in the 5’ to 3’ direction
- Unwinds DNA helix to expose template strand

Transcription Bubble:
- Region of unwound DNA around active site
- Maintains ~12-14 bp of DNA unwound

Proofreading:
- RNA polymerase has intrinsic 3’->5’ exonuclease activity
- Removes misincorporated ribonucleotides for high fidelity

Regulation:
- Elongation factors can modulate RNA polymerase’s activity
- Transcription-coupled repair mechanisms act during elongation

Termination Signals:
- Specific DNA sequences/structures signal termination
- Rho protein aids termination at some terminators

The elongation complex faithfully copies the DNA template into an RNA transcript until it reaches a termination signal.

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28
Q

transcription termination

A

Here are the key points about transcription termination in bacteria:

Intrinsic Termination:
- Occurs at specific DNA sequences that form a hairpin loop followed by a U-rich region
- Hairpin causes RNA polymerase to pause and dissociate from the DNA template
- Common in prokaryotes and some eukaryotes

Rho-dependent Termination:
- Involves the rho protein binding to the C-rich rho utilization (rut) site on the nascent RNA
- Rho translocates along the RNA, removing RNA polymerase from the DNA
- Common in prokaryotes

Steps:
1) Termination signal sequence is transcribed
2) RNA polymerase pauses/stalls
3) Conformational change destabilizes the transcription complex
4) Completed mRNA transcript is released
5) RNA polymerase dissociates from the DNA template

Terminators ensure efficient termination and recycling of RNA polymerase after genes are transcribed.

Termination prevents:
- Interference with downstream genes
- Depletion of RNA polymerase and ribonucleotide pools

Proper termination is critical for gene regulation and transcriptional control.

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29
Q

terminator sequence

A

DNA sequence that signals the end of transcription
Contains a hairpin loop followed by a U-rich region
Hairpin causes RNA polymerase to pause/stall

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30
Q

mRNA hairpin

A

Stem-loop structure formed by the nascent mRNA at the terminator
G-C base pairs in the stem, loop at the end
Hairpin causes physical disruption of transcription complex
Facilitates release of the mRNA transcript

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31
Q

Gene expression

A

is the process by which a gene gets turned on/off in a
cell to make RNA and proteins. Gene expression may be measured by
looking at the mRNA transcribed (mRNA expression), or the protein
translated from the mRNA (protein expression)

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32
Q

What binds to the promotor?

A

sigma factor

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33
Q

Where does RNAP begin transcribing?

A

TSS

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34
Q

The mRNA that would be produced from gene P above will only contain the coding sequence. T/F

A

F

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35
Q

Some parts of the mRNA will not be translated by the ribosome. T/F

A

T

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36
Q

If a repressor were bound to the silencer, would transcription likely occur? Y/N

A

N

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37
Q
  1. The terminator, silencer, and promoter are transcription factors. T/F
A

F

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38
Q
  1. General TF like sigma & activators are need to transcribe a gene at high levels. T/F
A

T

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39
Q

Bacterial Genomes

A
  • 1 circular chromosome (& circular
    plasmids)
  • Not-membrane bound in nucleus
    (nucleoid region)
  • Double-stranded DNA
  • Circular chromosome compacted nearly
    1000x to fit in the nucleoid region
  • Bacterial histone-like proteins (bend
    instead of wrap DNA)
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40
Q

Bacteria Genome Organization

A
  • Smaller in size (#base pairs) than euk
  • High proportion of protein-coding
    sequence (only ~10% non-coding)
  • Coding sequences are continuous
    (contiguous) *no introns
  • Both DNA strands can contain the
    coding sequence
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41
Q

Operons

A

are clusters of genes transcribed as one unit. They have one
promoter and one mRNA is produced.
Each gene’s coding sequence is translated from the single mRNA. thrA, thrB,
and thrC proteins will be made from the single mRNA.
These proteins often function together in the cell.
Operons are common in bacteria and much less common in eukaryotes.

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42
Q

Inducible Operons

A

The lac operon genes will produce enzymes that break down lactose when
lactose is present.
Inducible operon is off until the inducer molecule ”induces” the operon by inhibiting
the repressor (through binding it) *Inhibit an inhibitor = ON

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43
Q

Repressible Operons

A

Tryptophan operon contains genes that work together to synthesize the amino acid tryptophan when
it is absent

Repressible operons are on until the co-repressor binds the repressor.

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44
Q

ligand

A

molecule that binds
a protein (e.g. Iron is the ligand
for Fur repressor)

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45
Q

Regulatory elements can be anywhere

A
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46
Q

The coding strand of a chromosome is not always the same strand of DNA
depending on which gene you are examining. T/F

A

T

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47
Q
  1. Bacteria DNA exists as chromatin. T/F
A

T

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48
Q
  1. To regulate the transcription of a gene, an enhancer, silencer, and operator cannot
    all be used for the same gene. T/F
A

F

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49
Q
  1. Bacteria genes must exist in an operon. T/F
A

F

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50
Q
  1. Enhancer and silencer sequences can overlap with promoter sequences. T/F
A

T

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51
Q
  1. Silencers and enhancers can be upstream or downstream. T/F
A

T

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52
Q

Describe gene expression and understand why expression is a highly regulated process

A
  • Gene expression is the process by which the instructions in a gene are used to synthesize functional gene products (proteins or non-coding RNAs)
  • It involves two main steps:
    • Transcription: DNA is transcribed into RNA
    • Translation: RNA is translated into proteins
  • Gene expression is highly regulated for several reasons:
    • Different cells express different genes for specialized functions
    • Genes are expressed at the right time and level based on the cell’s needs
    • Regulating expression prevents wasteful overproduction of proteins
    • It allows the cell to respond to environmental signals and changes
  • Regulation occurs at multiple levels:
    • Transcriptional regulation (activating/inhibiting transcription)
    • RNA processing and stability
    • Translational regulation
    • Post-translational modifications of proteins
  • Precise gene regulation is crucial for normal development, cellular differentiation, and maintaining proper cellular function
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53
Q

Describe the characteristics of bacteria genomes (contrasting with euk)

A

Bacterial Genomes:
- Single circular chromosome (some bacteria have additional small circular plasmids)
- Relatively small genome size (few million base pairs)
- Lack membrane-bound nucleus
- DNA is naked, not associated with histones
- Genes are densely packed with little non-coding DNA
- Operons allow coordinated regulation of related genes
- Absence of introns (genes are continuous)
- Rapid replication and cell division

Contrasting with Eukaryotic Genomes:
- Multiple linear chromosomes
- Much larger genome sizes (billions of base pairs)
- DNA packaged into nuclei with histones
- Significant amount of non-coding DNA (introns, regulatory regions)
- Genes are discontinuous (exons separated by introns)
- Complex transcriptional regulation at individual gene level
- Presence of membrane-bound organelles like nucleus, mitochondria
- Slower replication and cell cycle

Overall, bacterial genomes are simpler, more compact, and lack the complexity of eukaryotic regulation and nuclear organization.

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54
Q

Describe operons: inducible and repressible types

A

Operons:
- Operons are clusters of genes that are co-transcribed from a single promoter
- Allow coordinated regulation of related genes involved in the same pathway
- Characteristic of bacterial genomes and some archaea

Inducible Operons:
- Genes are normally turned off (not transcribed)
- Presence of an inducer molecule activates transcription
- Inducer binds to a repressor protein, inactivating it
- Allows RNA polymerase to bind and transcribe the operon

Example: lac operon in E. coli
- Induced by lactose to produce enzymes for lactose metabolism
- Lack of glucose also induces (glucose prevents induction)

Repressible Operons:
- Genes are constitutively transcribed by default
- Presence of a corepressor molecule represses/turns off transcription
- Corepressor binds to a repressor protein, activating its binding to operator

Example: trp operon in E. coli
- Repressed when tryptophan levels are high (corepressor present)
- Allows regulation of tryptophan biosynthesis

This coordinate regulation allows bacteria to efficiently express genes when needed and conserve resources otherwise.

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55
Q

Understand how both types of operons work and their regulatory elements &
transcription factors
co-repressors, repressors, activators, co-activators

A

Inducible Operons:
- Regulated by repressor protein that binds to operator, blocking transcription
- Inducer molecule binds to and inactivates repressor
- This allows RNA polymerase to bind promoter and transcribe operon

Repressible Operons:
- Regulated by repressor protein + corepressor molecule
- Corepressor binds repressor, increasing its affinity for operator
- Activates repressor binding, blocking RNA polymerase

Regulatory Elements:
- Promoter - RNA polymerase binding site to initiate transcription
- Operator - Repressor protein binding site that blocks transcription
- Inducer/Corepressor - Small molecules that modulate repressor activity

Transcription Factors:
- Repressors - Bind operator to repress transcription (e.g. LacI, TrpR)
- Activators - Bind DNA and recruit RNA polymerase (e.g. CRP for lac)
- Coactivators - Help activators bind/function (e.g. CAP for lac)
- Inducers - Inactivate repressors, allowing transcription
- Corepressors - Activate repressors to block transcription

This interplay between activators, repressors, inducers, and corepressors allows bacteria to precisely control operon expression in response to environmental/metabolic conditions.

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56
Q

Be able to recognize what type of operon regulation is occurring given
scenarios/information

A

Scenario 1:
Certain genes are normally not expressed. However, when a specific small molecule is added to the bacterial culture, those genes become expressed.

This describes an inducible operon. The small molecule acts as an inducer, inactivating the repressor protein and allowing transcription of the operon.

Scenario 2:
A set of enzymes involved in an amino acid biosynthesis pathway are produced continuously. But when the amino acid is added to the growth medium, production of those enzymes stops.

This is an example of a repressible operon. The amino acid acts as a corepressor, binding to and activating the repressor protein to block transcription of the operon.

Scenario 3:
Genes encoding enzymes for lactose metabolism are expressed only when lactose is present and glucose is absent in the growth medium.

This matches the regulation of the famous lac operon, which is an inducible operon. Lactose acts as the inducer, while glucose prevents induction (glucose repression).

Scenario 4:
A repressor protein is bound to the operator region, blocking transcription. Adding a specific metabolite causes the repressor to change shape and fall off the operator.

This scenario depicts an inducible operon where the metabolite is the inducer that inactivates the repressor protein bound at the operator.

The key is to identify if a small molecule is allowing transcription (inducer) or blocking it (corepressor) to distinguish inducible vs. repressible regulation.

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57
Q

Be able to read a transcriptional profile (gene expression data with the colored blocks)
for what is turned on/off

A

Here are some tips for reading a transcriptional profile or gene expression data represented by colored blocks to determine which genes are turned on or off:

  1. Look at the scale/legend to understand what the colors represent:
    • Red typically indicates high expression/upregulation
    • Green usually means low expression/downregulation
    • Black or gray often represents baseline/no change
  2. Compare across conditions/time points:
    • Bright red blocks indicate that gene is highly expressed in that condition
    • Bright green blocks mean the gene is repressed/low expression
  3. Look for patterns within rows (same gene):
    • A row with mostly red means that gene is generally highly expressed
    • A row with mostly green suggests that gene has low expression levels
  4. Focus on differences between conditions within a row:
    • A switch from red to green indicates the gene was downregulated
    • Change from green to red means the gene became upregulated
  5. Subtle shades indicate degrees of change:
    • Darker reds = very high expression
    • Lighter greens = modest downregulation
  6. Black/gray blocks typically mean no significant change in expression.

The key is correlating the color intensity to expression levels based on the scale and looking for contrasting patterns between conditions/time points to identify up/downregulated genes. With practice, these visual profiles provide a powerful way to analyze gene expression data.

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58
Q

Understand that transcriptional regulation can occur in many combinations of
regulatory elements and transcription factors

A
  • Transcriptional regulation is a complex process involving the interplay of multiple components:
    • Regulatory DNA sequences (promoters, enhancers, silencers, insulators)
    • Transcription factors (activators, repressors, co-regulators)
    • Chromatin remodeling complexes
    • DNA methylation/epigenetic modifications
  • Promoters alone can have multiple binding sites for:
    • Activators that recruit RNA polymerase
    • Repressors that block polymerase binding
    • General transcription factors required for initiation
  • Enhancers are distal regulatory regions that can bind:
    • Activators to increase transcription
    • Repressors to decrease transcription
    • Interactions with promoters via DNA looping
  • Transcription factors can act cooperatively:
    • Multiple activators synergize for stronger activation
    • Activators and repressors compete for binding
    • Co-activators/co-repressors modulate activity
  • Chromatin structure and epigenetics add another layer:
    • Open chromatin allows transcription factor access
    • Closed/methylated chromatin blocks binding
    • Chromatin remodelers alter accessibility
  • Cell type and environmental signals dictate which factors are present
  • The precise combination of sequences, binding factors, chromatin state determines the transcriptional output

This modularity and cooperativity of regulatory components allows incredibly precise spatiotemporal control of gene expression programs.

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59
Q

Another take away is that enhancers and silencers can overlap with promoters and
there is nothing that says enhancers have to be upstream and silencers downstream.
They can be either

A
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60
Q

histone-like proteins (I would say bacteria DNA is chromatin),

A
  • Bacteria lack histones but have histone-like proteins called nucleoid-associated proteins (NAPs)
  • NAPs help compact and organize bacterial DNA into a nucleoid structure
  • The nucleoid functions like a simpler form of eukaryotic chromatin
  • Examples of bacterial NAPs: HU, H-NS, IHF, Fis, Dps
  • NAPs bind DNA, induce bends/loops, and condense the DNA molecule
  • The bacterial nucleoid, organized by NAPs, is analogous to eukaryotic chromatin organized by histones
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61
Q

nucleoid-region

A

The nucleoid region is a condensed area within a prokaryotic cell where the genetic material, typically in the form of a circular chromosome, is located. It lacks a membrane-bound nucleus found in eukaryotic cells.

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62
Q

operon

A

An operon is a functional unit of genetic material found in prokaryotic cells consisting of a cluster of genes under the control of a single promoter. It includes structural genes that encode proteins, along with regulatory elements such as an operator and a promoter. The operon allows for coordinated gene expression, typically in response to environmental signals.

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63
Q

inducible operon

A

An inducible operon is a genetic regulatory system in prokaryotic cells where gene expression is normally off but can be turned on in response to specific environmental signals or molecules.

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64
Q

inducer molecule

A

An inducer molecule is a substance that initiates or enhances the expression of specific genes in an organism by interacting with regulatory proteins such as transcription factors.

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65
Q

repressible operon

A

A repressible operon is a genetic regulatory system in bacteria where the transcription of a group of genes is typically active, but can be inhibited (repressed) when a specific molecule, usually a product of the metabolic pathway controlled by those genes, is present in abundance.

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66
Q

co-repressor

A

A co-repressor is a molecule that binds to a repressor protein, enhancing its ability to inhibit the expression of specific genes. This binding typically occurs in regulatory systems where the presence of the co-repressor is associated with certain environmental or cellular conditions.

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67
Q

operator

A

An operator is a DNA sequence found within the promoter region of a gene that serves as a binding site for regulatory proteins, such as repressors or activators. The binding of these proteins to the operator can either inhibit or facilitate the initiation of transcription of the adjacent gene(s).

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68
Q

untranslated region (UTR)

A

The untranslated region (UTR) is a segment of mRNA (messenger RNA) that lies outside the coding sequence of a gene. It occurs both upstream (5’ UTR) and downstream (3’ UTR) of the coding region. While not translated into protein, UTRs play important roles in regulating gene expression, mRNA stability, and localization within the cell.

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69
Q

co-activator

A

A co-activator is a protein that interacts with transcription factors to enhance the transcriptional activity of specific genes. Co-activators typically do not directly bind to DNA but instead facilitate the assembly of transcriptional machinery or modify chromatin structure to promote gene expression. They play crucial roles in regulating various cellular processes by modulating gene transcription.

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70
Q

After you learn eukaryotic genomes and transcription be able to compare and contrast
bacteria and eukaryotic process

A

Similarities:
* Both involve proteins that bind and compact DNA
* Serve to organize and package the genetic material
* Help condense long DNA molecules into more compact structures

Differences:
* Eukaryotes use histone proteins, bacteria use histone-like NAPs
* Eukaryotic chromatin is more complex, with multiple levels of organization
* Bacterial nucleoid is a simpler, more loosely packaged structure
* Histones form nucleosomes, the basic unit of eukaryotic chromatin
* NAPs don’t form nucleosome-like structures in bacteria
* Chromatin undergoes more dynamic changes (condensation/decondensation) during cell cycle
* Bacterial nucleoid is more static, less regulated compaction

In essence:
* Both use proteins to compact and organize DNA
* But eukaryotic chromatin is a more complex, dynamic, and regulated process
* Bacterial nucleoid formation by NAPs is a simpler, more static process

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71
Q

Be able to do deduce from scenarios or diagrams what type of operon
(inducible/repressible) is being used or would be used. Like the iron-stealing operon example.

A

Here are some concise points to help deduce the type of operon (inducible or repressible) from scenarios or diagrams:

Inducible Operon:
- Genes are normally OFF or expressed at low levels
- Presence of an inducer molecule/signal turns ON expression
- Often used for catabolic operons (breaking down substrates)
- Example: lac operon induced by lactose, allows bacteria to metabolize lactose

Repressible Operon:
- Genes are normally ON or constitutively expressed
- Presence of a corepressor molecule/signal turns OFF expression
- Often used for anabolic operons (biosynthesis pathways)
- Example: trp operon repressed by tryptophan, stops production when enough tryptophan

Deducing from Scenarios/Diagrams:
- If operon is OFF normally, needs a molecule/signal to turn ON → Inducible
- If operon is ON normally, needs a molecule/signal to turn OFF → Repressible
- Look for inducers (substrates, metabolites) that activate expression → Inducible
- Look for corepressors (end-products) that repress expression → Repressible
- Iron-stealing operon likely inducible, turned ON by iron starvation signal

So analyze the normal/default state, regulatory molecules involved, and whether they activate or repress expression to deduce inducible vs repressible.

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72
Q

Be able to look at a transcriptional profile and understand how it shows genes being
turned on/off

A
  • High signal/intensity = gene is expressed (turned on)
  • Low/no signal = gene is not expressed (turned off)
  • Compare conditions to see induction (turned on) or repression (turned off)
  • Clustered patterns indicate co-regulated genes
  • Sharp increases/decreases = genes turned on/off in response
  • Consistent expression = constitutive (always on)
  • Changes over time = dynamic regulation
  • Correlate with known regulators/pathways for insights
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73
Q

Operon Summary

A
  • Inducible operons are OFF (not transcribed) because repressor IS bound to the
    operator in the absence of inducer molecule. Should make sense, operon off
    without the inducer.
  • When inducer molecule present it binds to repressor and causes
    repressor to dissociate from operator which induces the operon
    (transcribed)
  • Repressible operons are ON (transcribed) because repressor is NOT bound to
    the operator in the absence of co-repressor molecule. Should make sense,
    operon on without the co-repressor.
  • When co-repressor molecule present it binds to repressor and causes
    repressor to bind the operator which represses the operon (not
    transcribed)
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74
Q
  1. Termination of transcription will happen:
A

b) somewhere soon after AAUAAA

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75
Q

Eukaryotic genomes

A
  • Membrane-bound (nucleus)
    chromosomes
  • Linear chromosomes
  • Compacted and wrapped around
    proteins called histones (8 histones
    make up a nucleosome)
  • MOST compacted during
    mitosis/meiosis
  • Lots of non-coding sequence (intergenic – between genes)
  • AND non-coding sequence within genes (introns)
  • Eukaryotic genes are NOT continguous; coding sequence (exons) is broken up with introns
  • Genes go in both ‘directions’ – meaning the coding strand can be on the top or bottom DNA strand of the
    double helix (same in bacteria)
  • Lots of repetitive sequences! (don’t worry about what theses are)
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76
Q

Post-transcriptional processing : mRNA modification

A

Eukaryotic mRNA made is in a pre-mRNA state – not actual message used to make
protein yet.
Pre-mRNA modified w/ Cap, polyA Tail, and spliced in the nucleus to become
mature mRNA!
Bacteria modify the 5’ end of
mRNAs too – but let’s just say it’s more
extensive in eukaryotes.
5’ cap - Modified guanine nucleotide added to 5’ of mRNA (technically happens
during txn). Function: bind ribosomes & protect mRNA from degradation

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77
Q

polyA tail

A

50-300 A’s; added after
termination. Function: protects from
degradation; Aids in nuclear export
of transcript

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78
Q

Post-transcriptional process: RNA splicing in eukaryotes

A

Notice there is an Un-Translated Region (UTR) both on the 5’ and 3’ end of the mRNA
Bacteria have UTRs as well.
Served regulatory functions as well.
Splicing technically happens co-/post-transcriptionally. Do
not worry about the specific order of processing like 5’ cap,
polyA, splicing. We will lump them as post-transcription.

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79
Q

Post-transcriptional regulation: Alternative splicing

A

MANY ways to splice mRNA
Each of these would
be translated into
a different protein!
from one MRNA
multiple variations
of the coding seq
that give unique
proteins.

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80
Q

Affinity & concentration

A

Different TFs (proteins) may also have different affinity for the same regulatory sequence.
Affinity is the strength/weakness at which an interaction occurs.
There may also be different amounts of the proteins available in the cell!

regulating seg. differences can affect TF
affinity.

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81
Q

Post-transcriptional regulation: RNA interference

A

Regulatory small non-coding RNAs (ncRNA) can regulate gene expression.
*usually bind in the 5’ or 3’ UTRs of mRNA transcripts; 20-25 nt long

short interfering RNA (siRNA) = perfect match – degrade mRNA (post-transcription reg)

micro RNA (miRNA) = imperfect match – block translation (post-transcription reg)

Bacteria do not have the proteins for siRNA but have ways to regulate their mRNAs post-
transcriptionally using the UTRs as well

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82
Q

Amino acid functional groups interact with major or minor
groove of DNA in sequence specific way

A

a) A-T, T-A, G-C, & C-G base pairs
present different shapes and chemical
groups in the grooves of DNA
Regulatory sequences are bound in a double-stranded
manner via a “hydrogen bonding” code in the major
and minor grooves

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83
Q

The same regulatory sequence can vary within genomes and be a
form of transcriptional regulation!

A

Bigger the letter the more often it is
found in the TATA box of the particular
type of genes.

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84
Q

Describe eukaryotic transcription steps initiation, elongation, and termination

A

Initiation:

RNA polymerase II binds to promoter region of gene
Transcription factors assist binding and initiation
General transcription factors recruit RNA pol II to promoter
TFIIH unwinds DNA to allow transcription bubble formation

Elongation:

RNA pol II moves along template strand, synthesizing RNA
Transcription elongation factors assist and regulate process
RNA pol II unwinds DNA ahead, re-anneals behind as it moves
5’ cap added to nascent mRNA
RNA splicing occurs - introns removed, exons joined

Termination:

Termination signals in DNA sequence trigger process
RNA pol II dissociates from DNA template
Polyadenylation factors add poly-A tail to 3’ end of mRNA
Fully processed, mature mRNA is released for translation

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85
Q

Transcription Initiation is always about recruiting RNAP to promoter
* Need TBP + general TFs to bind promoter = pre-initiation complex – how
is this different than bac?
* RNAP binds to pre-IC and all becomes the initiation complex
* Once initiation complex, RNAP unwinds DNA = initiation stage done

A

Recruiting RNA Polymerase II (RNAP II) to the promoter is key for initiation
TATA-binding protein (TBP) binds to TATA box in promoter
TBP assists binding of other general transcription factors (GTFs)
Together they form pre-initiation complex (pre-IC)
Difference from bacteria:
In bacteria, RNAP alone can bind promoter and initiate
In eukaryotes, RNAP II requires pre-IC with TBP and GTFs first
RNAP II joining pre-IC:
Pre-IC recruits RNAP II to promoter
RNAP II joins, forming full initiation complex (IC)
Initiation completion:
Within IC, RNAP II unwinds DNA around transcription start site
This formation of transcription bubble completes initiation stage

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86
Q

Understand DNA looping is one way distal elements can regulate transcription
initiation in both euk and bac

A

DNA Looping in Eukaryotes:
- Distal enhancers can be thousands of base pairs away from promoter
- DNA loops bring enhancers into close proximity with promoter
- This enhancer-promoter looping is mediated by transcription factors
- Allows regulatory proteins bound at enhancers to interact with transcription machinery

DNA Looping in Bacteria:
- Distal cis-regulatory sequences located upstream or downstream of promoter
- DNA is looped to allow binding of transcription factors near promoter
- Loop formation assisted by nucleoid-associated proteins
- Allows transcription factors bound at distant sites to influence initiation complex

General Points:
- DNA looping overcomes linear distance between regulatory elements and promoters
- Enables communication between distal binding sites and transcription machinery
- Allows integration of multiple regulatory signals for controlled initiation

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87
Q

Eukaryotic transcription elongation is just like in bacteria except bacteria
translate mRNA as it is being made. Euk cannot do this b/c transcription in
nucleus and translation in cytosol.

A
  • The elongation process itself is similar in eukaryotes and bacteria
    • RNA polymerase moves along DNA template, synthesizing mRNA
    • Transcription elongation factors assist and regulate the process
  • Key difference is the coupling to translation:
    • In bacteria, translation can begin as mRNA is being transcribed (coupled)
    • In eukaryotes, transcription occurs in nucleus but translation in cytoplasm
    • So eukaryotic mRNA cannot be translated as it is being transcribed
  • This separation is due to:
    • Nuclear envelope barrier in eukaryotic cells
    • mRNA must be fully transcribed, processed, exported to cytoplasm
    • Only then can translation on mature mRNA occur in the cytoplasm
  • In bacteria (no nuclear envelope):
    • Ribosomes can directly access and translate the nascent mRNA
    • Allowing co-transcriptional translation to occur efficiently
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88
Q

Termination uses a terminator sequence. In euk causes mRNA to be cut off and
released from RNAP. In euk right before terminator is a polyA signal which is
used post-transcriptionally to add polyA tail. How is this different than bac?

A

Eukaryotes:
- Termination guided by terminator sequence in DNA
- This causes RNAP to dissociate, releasing the nascent mRNA
- Just upstream of terminator is a poly(A) signal sequence
- This poly(A) signal is not used during transcription termination itself

  • After termination, the poly(A) signal sequences guide:
    • Endonucleolytic cleavage of the 3’ end of the mRNA
    • Addition of a poly(A) tail to the 3’ end by poly(A) polymerase

Bacteria:
- Termination can occur through rho-independent or rho-dependent mechanisms
- Rho-independent uses RNA hairpin followed by U-rich sequence
- No poly(A) tail is added to bacterial mRNAs
- The 3’ end is simply the last transcribed portion

Key Differences:
- Eukaryotes use separate terminator and poly(A) signals
- Poly(A) tail added as a post-transcriptional processing step
- Bacteria lack poly(A) tails and processing of the 3’ end
- Termination signals are encoded differently

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89
Q

coding vs. non-coding sequences in genome (different than coding and template
strands!!)

A

Coding Sequences:
- Regions of DNA that get transcribed into mRNA
- The mRNA is then translated into proteins
- Include exons of genes that code for proteins
- Make up a small fraction of the genome (~1-2% in humans)

Non-Coding Sequences:
- Regions of DNA that do not code for proteins
- Not transcribed into mRNA that gets translated
- Include introns within genes
- Also regulatory regions like promoters, enhancers, silencers
- Repeat sequences, telomeres, centromeres etc.
- Comprise the vast majority of the genome (~98% in humans)

Key Points:
- Coding sequences are the protein-coding exons
- Non-coding includes introns, regulatory regions, other genomic sequences
- Non-coding sequences can still be transcribed into non-coding RNAs
- But they don’t get translated into proteins
- The non-coding portion is much larger than the coding portion

So in summary - coding = exons that get translated into proteins
- non-coding = introns, regulatory, other non-translated sequences

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90
Q

contiguous vs. non-contiguous genes

A

Here are the key points about contiguous versus non-contiguous genes:

Contiguous Genes:
- Genes that are located right next to each other on the same chromosome
- The coding sequences are contiguous/uninterrupted
- No non-coding regions or other genes in between

Non-Contiguous Genes:
- Genes that are separated by stretches of non-coding DNA
- The coding sequences are non-contiguous/interrupted
- Interspersed with introns, regulatory regions, other genes etc.

In both cases:
- The coding sequences (exons) are transcribed into mRNA
- Introns and other non-coding regions are removed during splicing

Key Differences:
- Contiguous genes don’t require splicing of the transcript
- Non-contiguous genes require splicing to remove interrupting sequences
- Contiguous genes are relatively rare in eukaryotes
- Most eukaryotic genes are non-contiguous due to presence of introns

So in summary:
- Contiguous = Coding sequences are together without interruptions
- Non-contiguous = Coding sequences are separated by non-coding regions
- Most eukaryotic genes are non-contiguous and require splicing

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91
Q

polyA signal

A

A polyadenylation signal, often abbreviated as polyA signal, is a specific nucleotide sequence in mRNA precursor molecules (pre-mRNA) that signals the addition of a polyadenylate (poly-A) tail during mRNA processing. This signal is recognized by proteins involved in mRNA maturation, leading to the cleavage of the pre-mRNA and the addition of a string of adenine nucleotides (poly-A tail) to the mRNA’s 3’ end. The poly-A tail plays roles in mRNA stability, transport, and translation.

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92
Q

RNA pol II vs I/III

A

RNA Polymerase II (RNA Pol II):
- Transcribes protein-coding genes
- Produces precursors of mRNA that will be translated into proteins
- Includes promoter recognition, initiation, elongation, termination
- Transcripts are capped, spliced, polyadenylated

RNA Polymerase I (RNA Pol I):
- Transcribes ribosomal RNA (rRNA) genes
- rRNA is a major constituent of ribosomes
- Found in nucleolus region of nucleus

RNA Polymerase III (RNA Pol III):
- Transcribes transfer RNA (tRNA) genes
- Also transcribes 5S rRNA, some small nuclear RNAs
- Transcripts are relatively short

Key Differences:
- RNA Pol II - mRNA transcription for protein synthesis
- RNA Pol I/III - transcription of non-coding RNAs
- Different promoter recognition and regulation mechanisms
- RNA Pol II transcripts are extensively processed

So in summary:
- RNA Pol II for protein-coding mRNAs
- RNA Pol I for large rRNA transcripts
- RNA Pol III for small stable RNAs like tRNAs
- Different roles but all essential for gene expression

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93
Q

TATA box

A

The TATA box, also known as the Goldberg-Hogness box, is a DNA sequence found upstream (usually around 25-30 base pairs upstream) of the transcription start site of many eukaryotic genes. It consists of the sequence “TATAAA” or a similar variant. The TATA box serves as a binding site for the TATA-binding protein (TBP), a component of the transcription factor IID (TFIID) complex. Binding of TBP to the TATA box helps to initiate the assembly of the RNA polymerase II transcription initiation complex, facilitating the start of transcription. The TATA box is crucial for the accurate positioning of the transcriptional machinery and the regulation of gene expression.

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94
Q

TATA
binding protein (TBP)

A

The TATA-binding protein (TBP) is a key component of the transcription factor IID (TFIID) complex in eukaryotes. It binds specifically to the TATA box, a DNA sequence found upstream of many eukaryotic genes, and plays a central role in the initiation of transcription by RNA polymerase II. TBP helps recruit other transcription factors and RNA polymerase II to the promoter region of genes, facilitating the assembly of the pre-initiation complex and the initiation of transcription. It is essential for the accurate regulation of gene expression.

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95
Q

general transcription factor

A

A general transcription factor is a protein involved in the initiation of transcription of protein-coding genes in eukaryotes. These factors are required for the assembly of the transcriptional machinery at the promoter region of genes. They include proteins like the TATA-binding protein (TBP) and other factors associated with RNA polymerase II. General transcription factors help to position RNA polymerase II correctly at the transcription start site and facilitate the initiation of transcription. They are essential for the transcription of most protein-coding genes and are distinct from regulatory transcription factors that bind to specific DNA sequences to modulate gene expression.

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96
Q

pre-initiation complex

A

The pre-initiation complex (PIC) is a multi-protein complex that forms at the promoter region of a gene during the initiation stage of transcription in eukaryotes. It includes general transcription factors, such as TATA-binding protein (TBP), and other proteins that help to position RNA polymerase II and initiate transcription. The assembly of the pre-initiation complex is a crucial step in gene expression, marking the beginning of transcription of a specific gene.

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97
Q

initiation complex

A

The initiation complex refers to the ensemble of proteins assembled at the promoter region of a gene to initiate transcription. In prokaryotes, the initiation complex typically involves RNA polymerase along with sigma factor, which helps recognize the promoter sequence. In eukaryotes, the initiation complex includes RNA polymerase II, general transcription factors (such as TATA-binding protein and TFIIA, TFIIB, etc.), and other regulatory proteins. The formation of the initiation complex marks the beginning of transcription and is a key step in gene expression.

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98
Q

Describe eukaryotic RNA post-transcriptional processing: 5’ cap, 3’ polyA tail, splicing
* know functions of 5’ cap and 3’ polyA tail

A

5’ Capping:
- 7-methylguanosine cap added to 5’ end of nascent mRNA
- Functions:
- Protects mRNA from degradation
- Assists nuclear export of mRNA
- Recruits ribosomes for translation initiation

3’ Polyadenylation:
- Poly(A) tail of ~200 adenines added to 3’ end
- Functions:
- Protects mRNA from degradation
- Facilitates transport from nucleus
- Enhances translation by stabilizing mRNA

Splicing:
- Removal of non-coding intron sequences from pre-mRNA
- Joining of coding exon sequences into mature mRNA
- Occurs in nucleus before mRNA export
- Requires spliceosome complex of snRNPs and proteins

Additional points:
- All three processes occur in nucleus before export
- Allow proper maturation and regulation of eukaryotic mRNAs
- Increase stability, translation efficiency of mature transcripts
- Splicing enables combination of exons, protein diversity

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99
Q

Describe alternative splicing in eukaryotes
* alt. splicing as a form of post-transcriptional regulation

A
  • Most eukaryotic genes contain multiple exons and introns
  • During splicing, different combinations of exons can be included in the mature mRNA
  • This produces multiple different mRNA transcripts from the same gene
  • Mechanisms of alternative splicing:
    • Exon skipping - certain exons are omitted
    • Mutually exclusive exons - only one of two exons is included
    • Alternative 5’ or 3’ splice sites are used
    • Intron retention - an intron may remain in the mRNA
  • Allows a single gene to code for multiple different protein isoforms
  • Expands the protein diversity encoded by the genome
  • Alternative splicing is tightly regulated
    • By splicing regulatory proteins
    • Binding of these proteins influences exon inclusion
    • Allows splicing to respond to developmental/environmental cues
  • Functions as a form of post-transcriptional gene regulation
    • Controls which protein isoforms are produced
    • Modulates protein function, localization, interactions
  • Very prevalent in humans, contributes to protein diversity
    • Up to 95% of human genes may undergo alternative splicing

So in summary, alternative splicing is a crucial mechanism that allows production of multiple protein variants from one gene through differential exon inclusion.

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100
Q

Describe RNA interference and two ways it regulates gene expression
* siRNA and mRNA stability – post transcriptional regulation
* miRNA and translation inhibition – post-transcriptional regulation

A
  1. siRNA and mRNA Stability (Post-Transcriptional Regulation):
    - siRNAs (small interfering RNAs) are ~20 nucleotides long
    - Derived from long double-stranded RNA precursors
    - The siRNA associates with the RISC complex
    - RISC uses siRNA as a guide to bind and cleave complementary mRNA targets
    - This degradation of the mRNA prevents its translation into protein
    - Acts as a post-transcriptional gene silencing mechanism
  2. miRNA and Translation Inhibition (Post-Transcriptional Regulation):
    - miRNAs (microRNAs) are also ~22 nucleotides long
    - Derived from stem-loop precursors encoded in the genome
    - The miRNA associates with the RISC complex
    - RISC uses the miRNA to bind to partially complementary sites in 3’UTRs of mRNAs
    - This does not degrade the mRNA, but represses its translation into protein
    - Allows miRNAs to fine-tune protein output post-transcriptionally

In both cases:
- RNAi pathways involve small non-coding RNAs
- These guide Argonaute-containing RISC complex to targets
- Results in post-transcriptional regulation of gene expression
- siRNAs degrade mRNAs, miRNAs inhibit translation without degradation

So in summary, RNAi exploits small RNAs to regulate genes post-transcriptionally by either mRNA degradation (siRNA) or translational repression (miRNA).

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101
Q

Describe where and how DNA binding proteins recognize specific DNA sequences (both
bacteria & eukarya)

A

Bacteria:
- Many bacterial transcription factors contain helix-turn-helix motifs
- The α-helices in this motif fit into the major groove of DNA
- Specific amino acids make contacts with exposed DNA bases
- This allows sequence-specific recognition of promoter regions

Eukaryotes:
- Eukaryotic transcription factors use several DNA binding motifs:
- Zinc finger proteins - zinc ions help position α-helices in DNA major groove
- Leucine zipper - dimerization allows recognition of longer sequences
- Helix-loop-helix - HLH motifs bind DNA and allow dimerization

  • Other motifs like homeodomain, HMG box also facilitate DNA binding

General Mechanisms:
- α-Helices in the protein motifs are key for DNA major groove binding
- Amino acid side chains make sequence-specific contacts with exposed bases
- Dimerization of DNA binding domains increases binding specificity
- Minor groove interactions and DNA bending also contribute

Additional Points:
- Cooperative binding with other proteins enhances specificity
- Post-translational modifications can modulate DNA binding affinity
- Chromatin structure impacts accessibility of binding sites in eukaryotes

So in essence, DNA binding proteins use a variety of structural motifs that allow their α-helices and amino acids to specifically read out and bind target DNA sequences.

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102
Q

Understand how regulatory sequence variation and TF affinity and concentration can be
used to regulate transcription initiation. (both bacteria & eukarya)

A

Regulatory Sequence Variation:
- Mutations in promoter or enhancer/operator sequences
- Can increase or decrease affinity for TF binding
- Alters recruitment of transcription machinery and initiation frequency

TF Affinity:
- Determined by TF structure and DNA sequence motif
- Higher affinity allows tighter TF binding to regulatory regions
- Increases probability of initiation complex assembly

TF Concentration:
- Abundance of specific TFs is a major determinant
- Higher TF levels increase occupancy at regulatory sequences
- More efficient recruitment of polymerase for initiation

In Bacteria:
- Promoter mutations affect binding of σ factors and RNAP
- Operator mutations modulate repressor/activator binding
- TF levels controlled by regulation of their expression

In Eukaryotes:
- Promoter and enhancer mutations impact TF and coactivator binding
- Variation in activation domains affects recruitment ability
- TF levels controlled by complex regulatory mechanisms

So in both systems, changes to regulatory sequences, TF-DNA binding affinities, and cellular TF concentrations can all be leveraged to fine-tune transcription initiation rates for different genes under different conditions.

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103
Q

splicing

A

Splicing is the process in eukaryotic cells by which introns, non-coding regions of a gene, are removed from the precursor messenger RNA (pre-mRNA) transcript and the exons, coding regions, are joined together to form mature mRNA. This process occurs in the cell nucleus and is catalyzed by the spliceosome, a complex of RNA and protein. Splicing is essential for generating functional mRNA transcripts that can be translated into proteins.

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104
Q

alternative splicing

A

Alternative splicing is a process in eukaryotic gene expression where different combinations of exons within a pre-mRNA transcript can be joined together during splicing, resulting in multiple mRNA isoforms derived from a single gene. This process allows for the production of diverse protein products from a limited number of genes and contributes to cellular diversity and complexity. Alternative splicing can generate mRNA transcripts with different coding sequences, leading to proteins with distinct functions, localization, or regulatory properties.

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105
Q

intron

A

An intron is a non-coding segment of a gene’s DNA that is transcribed into pre-mRNA but is removed during RNA processing and does not appear in the final mRNA molecule. In eukaryotic cells, introns interrupt the coding sequences (exons) of genes and are spliced out during mRNA maturation. Although they do not encode protein sequences, introns can have regulatory functions and contribute to genetic diversity through alternative splicing.

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106
Q

exon

A

An exon is a coding region of a gene’s DNA that is transcribed into mRNA and appears in the final, mature mRNA molecule after introns have been removed during RNA processing. Exons contain sequences that encode specific amino acids, and they are joined together during splicing to form the coding sequence of the mRNA. Exons determine the protein-coding information of a gene and are crucial for synthesizing functional proteins.

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107
Q

UTR

A

The untranslated region (UTR) is a segment of mRNA (messenger RNA) that lies outside the coding sequence of a gene. It occurs both upstream (5’ UTR) and downstream (3’ UTR) of the coding region. While not translated into protein, UTRs play important roles in regulating gene expression, mRNA stability, and localization within the cell.

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108
Q

pre-mRNA

A

Pre-mRNA, or precursor messenger RNA, is an initial transcript produced during transcription in eukaryotic cells. It contains both exons (coding regions) and introns (non-coding regions). Pre-mRNA undergoes processing, including capping, splicing, and polyadenylation, to become mature mRNA. This processing involves the removal of introns and the addition of a 5’ cap and a poly(A) tail. The mature mRNA is then transported to the cytoplasm for translation into protein.

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109
Q

mature mRNA

A

Mature mRNA is the final processed form of mRNA that is ready for translation into protein in eukaryotic cells. It is derived from pre-mRNA through processing steps including capping, splicing, and polyadenylation. Mature mRNA contains only the exons (coding regions) of the gene and has a 5’ cap structure and a poly(A) tail added to it. It serves as the template for protein synthesis during translation.

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110
Q

5’ G-cap

A

The 5’ G-cap, also known as the 5’ cap, is a modified nucleotide structure added to the 5’ end of eukaryotic mRNA molecules during RNA processing. It consists of a guanine nucleotide linked to the mRNA via a 5’-5’ triphosphate bridge, followed by additional methylations. The 5’ cap plays crucial roles in mRNA stability, translation initiation, and mRNA export from the nucleus to the cytoplasm.

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111
Q

polyA tail

A

The poly(A) tail is a stretch of adenine nucleotides added to the 3’ end of eukaryotic mRNA molecules during RNA processing. It typically consists of multiple adenine residues, forming a “tail” structure. The poly(A) tail plays important roles in mRNA stability, facilitating mRNA export from the nucleus, and enhancing translation efficiency. It also protects the mRNA from degradation by exonucleases.

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112
Q

RNA interference

A

RNA interference (RNAi) is a biological process in which small RNA molecules, such as small interfering RNA (siRNA) or microRNA (miRNA), regulate the expression of genes by inhibiting the translation of mRNA or by promoting the degradation of specific mRNA molecules. RNAi acts as a natural defense mechanism against viruses and transposons and also plays essential roles in regulating gene expression during development and in response to environmental stimuli.

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113
Q

siRNA

A

Short interfering RNA (siRNA) is a class of small RNA molecules typically around 20-25 nucleotides in length. siRNAs are involved in the RNA interference (RNAi) pathway, where they mediate sequence-specific gene silencing by targeting complementary mRNA molecules for degradation or by inhibiting their translation. In experimental settings, synthetic siRNAs can be introduced into cells to specifically silence target genes, making them powerful tools for studying gene function and potential therapeutic agents for treating diseases caused by aberrant gene expression.

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114
Q

miRNA

A

MicroRNA (miRNA) is a class of small non-coding RNA molecules typically around 21-25 nucleotides in length. miRNAs are involved in post-transcriptional regulation of gene expression by binding to complementary sequences in the 3’ untranslated regions (UTRs) of target mRNA molecules. This binding can lead to mRNA degradation or inhibition of translation, thereby modulating the expression of target genes. miRNAs play critical roles in various biological processes, including development, differentiation, proliferation, and apoptosis, and dysregulation of miRNA expression has been implicated in many diseases, including cancer and neurodegenerative disorders.

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115
Q

major groove

A

The major groove is one of the two larger grooves found in the double helical structure of DNA, the other being the minor groove. It is characterized by a wider space between the two strands of the DNA helix and provides access to the nitrogenous bases of the nucleotides. The major groove is important for interactions with proteins, such as transcription factors and other DNA-binding proteins, which recognize specific sequences of DNA bases by fitting into and making contacts within this groove. It plays a crucial role in various cellular processes, including gene regulation, replication, and repair.

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116
Q

minor groove

A

The minor groove is one of the two smaller grooves found in the double helical structure of DNA, the other being the major groove. It is characterized by a narrower space between the two strands of the DNA helix and provides access to the edges of the nitrogenous bases of the nucleotides. While not as spacious as the major groove, the minor groove still plays a significant role in DNA-protein interactions, as certain DNA-binding proteins can recognize specific DNA sequences by fitting into and making contacts within this groove. The minor groove is also involved in various cellular processes, including gene regulation, replication, and repair.

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117
Q

affinity

A

Affinity refers to the strength of interaction between two molecules, such as a ligand and a receptor, or an enzyme and its substrate. It quantifies how readily molecules bind to each other and is typically measured by the equilibrium binding constant (Kd) in biochemistry. High affinity indicates a strong binding interaction, with molecules staying bound for longer periods or requiring lower concentrations for binding, whereas low affinity indicates weaker binding, with molecules dissociating more readily or requiring higher concentrations for binding. Affinity plays a crucial role in various biological processes, including signal transduction, enzyme catalysis, and molecular recognition.

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118
Q

Bacteria vs eukaryotic

Promoter
RNA Pol(s)
General TFs
Enhancers/silencers
Activators/repressors
Transcription Elongation
Transcription Termination
Post-transcriptional
5’/3’ UTR
Genome feature differences
Chromosome shape
Gene structure

A

Bacteria:
- Promoters for RNAP binding/initiation
- Single RNAP enzyme
- Sigma factors act as general TFs
- No enhancers/silencers
- Activators and repressors control transcription
- Transcription elongation by RNAP alone
- Rho-dependent or -independent termination
- No post-transcriptional modifications like capping/polyadenylation
- 5’/3’ UTRs present but no splicing

Eukaryotes:
- Promoters with TF binding sites
- 3 RNAP enzymes (I, II, III)
- Multiple general TFs like TFIID
- Enhancers and silencers regulate genes
- Activators, repressors, chromatin factors
- Transcription elongation involves many factors
- Termination with poly(A) site, processing factors
- 5’ capping, 3’ polyadenylation, splicing of introns
- Long 5’/3’ UTRs with regulatory roles

Genome Features:
- Bacterial genome is circular chromosome
- Eukaryotic genome has linear chromosomes
- Bacterial genes are simple, contiguous
- Eukaryotic genes have split exon/intron structure

In general, eukaryotic transcription has more protein factors involved and additional regulatory complexity from chromatin, RNA processing, and genome organization compared to the streamlined bacterial system.

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119
Q

Understand transcription and gene regulation in the context of chromatin
* Describe how heterochromatin vs. euchromatin controls gene expression
* Describe which parts of DNA would be more accessible – linker DNA vs.
nucleosomal DNA

A

Heterochromatin vs Euchromatin:
- Heterochromatin is densely packed, transcriptionally inactive
- Euchromatin is more open and accessible for transcription
- Heterochromatin limits binding of transcription factors/machinery
- Euchromatin allows transcriptional machinery access to DNA

  • Highly condensed heterochromatin marks genes for silencing
  • More open euchromatin structure permits active transcription

Nucleosomal DNA vs Linker DNA:
- Nucleosomes wrap DNA around histone octamers
- The linker DNA between nucleosomes is more exposed
- Transcription factors/complexes can more easily access linker DNA
- Nucleosomal DNA is less accessible when wrapped on histones

  • Nucleosome positioning/remodeling exposes or hides sequences
  • Activating sequences preferentially located in linker regions

Additional Regulation:
- Histone modifications (acetylation, methylation) affect chromatin
- ATP-dependent remodeling complexes alter nucleosome positioning
- Higher order folding into chromatin fibers also impacts accessibility

In summary:
- Heterochromatin is restrictive while euchromatin permits transcription
- Linker DNA is more accessible than nucleosomal DNA for binding
- Chromatin remodeling and histone modifications regulate accessibility

So the chromatin environment acts as a major regulator of transcription through modulating the physical accessibility of DNA sequences.

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120
Q

Describe epigenetics
* Understand how histone modifications regulate gene expression via chromatin
remodeling
* Describe epigenetic inheritance

A

Histone Modifications and Chromatin Remodeling:
- Histones can be covalently modified (e.g. acetylation, methylation)
- These modifications alter the charge and structure of nucleosomes
- Affect how tightly DNA is wrapped around histones
- Recruit or prevent binding of chromatin remodelers/transcription factors

  • Histone acetylation generally leads to more open, transcriptionally active chromatin
  • Histone methylation can lead to active or inactive states depending on residue
  • ATP-dependent chromatin remodeling complexes can:
    • Slide or remove nucleosomes to expose DNA sequences
    • Regulate access of transcription machinery to genes
  • Histone modifications and chromatin remodeling control gene expression epigenetically

Epigenetic Inheritance:
- Epigenetic marks like DNA methylation and histone modifications are mitotically inherited
- Patterns are propagated through cell divisions by enzymes
- Allows stable propagation of gene expression states

  • In some cases, epigenetic states can be inherited transgenerationally
  • Allows transmission of environmentally-induced epi-mutations

So in summary:
- Epigenetics refers to heritable modifications that regulate gene expression
- Without altering the underlying DNA sequence itself
- Histone modifications and chromatin remodeling are key epigenetic mechanisms
- Epigenetic information can be mitotically and sometimes transgenerationally inherited

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121
Q

euchromatin

A

Euchromatin is a loosely packed form of chromatin found in the nucleus of eukaryotic cells. It is characterized by a more open and accessible structure, allowing for active gene transcription and gene expression. Euchromatin is typically associated with genes that are actively transcribed to produce RNA, and it contains a higher density of genes and regulatory elements compared to heterochromatin. The less condensed nature of euchromatin facilitates the binding of transcription factors and other regulatory proteins to DNA, enabling efficient gene expression.

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122
Q

heterochromatin

A

Heterochromatin is a densely packed form of chromatin found in the nucleus of eukaryotic cells. It is characterized by a tightly condensed structure, which makes it less accessible for gene transcription and gene expression. Heterochromatin is typically associated with regions of the genome that are transcriptionally inactive or contain repetitive DNA sequences, such as centromeres and telomeres. The condensed nature of heterochromatin prevents the binding of transcription factors and other regulatory proteins to DNA, thereby silencing gene expression in these regions.

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123
Q

nucleosome

A

A nucleosome is the fundamental unit of chromatin, the complex of DNA and proteins that make up chromosomes in eukaryotic cells. It consists of a segment of DNA wrapped around a core of histone proteins. The core histones, comprising two copies each of histones H2A, H2B, H3, and H4, form an octamer around which approximately 147 base pairs of DNA are wound in about 1.65 turns. Nucleosomes serve to compact and organize the long strands of DNA within the cell nucleus and play crucial roles in gene regulation, DNA replication, and repair.

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124
Q

histone

A

Histones are a family of highly conserved proteins found in eukaryotic cell nuclei. They are responsible for packaging and organizing DNA into structural units called nucleosomes, which form the building blocks of chromatin. Histones play a crucial role in regulating gene expression by controlling access to DNA and influencing chromatin structure. They have a high density of positively charged amino acids, allowing them to interact with the negatively charged phosphate backbone of DNA. In addition to their structural role, histones can undergo various post-translational modifications, such as acetylation, methylation, and phosphorylation, which further regulate chromatin structure and gene expression.

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125
Q

nucleosomal DNA

A

Nucleosomal DNA refers to the DNA segment that is wrapped around the histone octamer core to form a nucleosome. It consists of approximately 147 base pairs of DNA that are wound around the histone proteins in about 1.65 turns. Nucleosomal DNA plays a crucial role in the organization and compaction of chromatin structure within the cell nucleus. Additionally, it is subject to various epigenetic modifications and influences gene expression by regulating the accessibility of DNA to transcription factors and other regulatory proteins.

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126
Q

linker
DNA

A

Linker DNA, also known as linker histone, refers to the DNA segment that connects adjacent nucleosomes in chromatin. It is the stretch of DNA that extends between two nucleosomes and is not tightly wrapped around histone proteins. Linker DNA contributes to the higher-order organization of chromatin by providing flexibility and allowing nucleosomes to be spaced apart. It also plays a role in regulating chromatin structure and gene expression by influencing the accessibility of DNA to regulatory proteins.

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127
Q

condensed chromatin

A

Condensed chromatin refers to regions of chromatin that are tightly packed and highly compacted within the cell nucleus. It is characterized by a dense arrangement of nucleosomes and other associated proteins, resulting in limited accessibility of the DNA to transcription factors and other regulatory proteins. Condensed chromatin is typically transcriptionally inactive and is often associated with heterochromatic regions of the genome, such as centromeres and telomeres. The compaction of chromatin into condensed structures helps to organize and protect the genome and plays a role in regulating gene expression and other nuclear processes.

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128
Q

decondensed chromatin

A

Decondensed chromatin refers to regions of chromatin that are less compacted and more open within the cell nucleus compared to condensed chromatin. It is characterized by a looser arrangement of nucleosomes and other associated proteins, allowing for increased accessibility of the DNA to transcription factors and other regulatory proteins. Decondensed chromatin is often associated with transcriptionally active regions of the genome, where genes are actively transcribed to produce RNA. The relaxation of chromatin structure into a decondensed state facilitates gene expression and other nuclear processes such as DNA replication and repair.

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129
Q

methylation

A

Methylation is a biochemical process where a methyl group (CH3) is added to a molecule, typically occurring on DNA, RNA, proteins, or small molecules. In the context of DNA, DNA methylation involves the addition of a methyl group to the cytosine base of DNA, typically at cytosine-phosphate-guanine (CpG) dinucleotide sites. DNA methylation plays a crucial role in regulating gene expression, genome stability, and chromatin structure. In gene regulation, methylation of CpG islands in gene promoter regions can lead to transcriptional repression by inhibiting the binding of transcription factors or recruiting proteins involved in gene silencing. In addition to DNA, methylation can also occur on histone proteins and RNA molecules, influencing various cellular processes and gene expression patterns.

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130
Q

acetylation

A

Acetylation is a biochemical process in which an acetyl group (-COCH3) is added to a molecule, typically occurring on proteins or histone proteins. In the context of histone proteins, acetylation involves the addition of an acetyl group to the lysine residues of histone tails. This modification is catalyzed by histone acetyltransferase (HAT) enzymes. Histone acetylation is associated with relaxed chromatin structure and increased gene expression, as it neutralizes the positive charge of lysine residues, weakening the interaction between histones and DNA, and facilitating access of transcription factors to DNA. Acetylation of non-histone proteins, such as transcription factors, can also regulate their activity and stability, influencing gene expression and various cellular processes.

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131
Q

chromatin
remodeling

A

Chromatin remodeling refers to the dynamic process by which the structure of chromatin, the complex of DNA and proteins in the nucleus of eukaryotic cells, is altered to regulate gene expression and other nuclear processes. It involves the repositioning, eviction, or restructuring of nucleosomes along the DNA, thereby modulating the accessibility of DNA to transcription factors and other regulatory proteins. Chromatin remodeling complexes, such as ATP-dependent chromatin remodelers, use energy from ATP hydrolysis to drive the movement or restructuring of nucleosomes, allowing for changes in chromatin organization and gene regulation. Chromatin remodeling plays crucial roles in diverse biological processes, including transcriptional activation and repression, DNA repair, and genome stability.

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132
Q

HAT

A

HAT stands for Histone Acetyltransferase. It is an enzyme that catalyzes the addition of an acetyl group (-COCH3) to lysine residues on histone proteins. This process, known as histone acetylation, typically occurs on the N-terminal tails of histones and is associated with relaxed chromatin structure and increased gene expression. HAT enzymes play a crucial role in regulating gene expression by modifying chromatin structure, making DNA more accessible to transcription factors and other regulatory proteins. They are involved in various cellular processes, including development, differentiation, and response to environmental cues.

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133
Q

HDAC

A

HDAC stands for Histone Deacetylase. It is an enzyme that catalyzes the removal of acetyl groups from lysine residues on histone proteins. This process, known as histone deacetylation, typically results in a more condensed chromatin structure and decreased gene expression. HDAC enzymes play a crucial role in gene regulation by reversing the effects of histone acetylation, leading to transcriptional repression. They are involved in various cellular processes, including development, differentiation, and response to environmental cues. HDAC inhibitors have been studied as potential therapeutics for various diseases, including cancer and neurodegenerative disorders, due to their ability to modulate gene expression patterns.

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134
Q

HMT

A

HMT stands for Histone Methyltransferase. It is an enzyme responsible for catalyzing the addition of methyl groups (-CH3) to lysine or arginine residues on histone proteins. This process, known as histone methylation, can result in different outcomes depending on the specific amino acid residue modified and the number of methyl groups added. Histone methylation can either activate or repress gene expression, depending on the site and degree of methylation, and is involved in regulating various cellular processes, including development, differentiation, and disease. HMTs play crucial roles in establishing and maintaining histone methylation patterns, which contribute to the epigenetic regulation of gene expression.

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135
Q

what processes of transcription
would be impossible if the gene above were in the heterochromatin state?

A

a) transcription factor binding to regulatory elements for initiation
b) elongation

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136
Q

What state of chromatin will transcribed genes most likely be in then?

A

b) euchromatin

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137
Q

DM

A

demethylases (removes)

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138
Q

Bacteria have compacted chromosomes as well and also histone-like proteins
but they do not use this. process to regulate it (as far as I know)

A
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139
Q

DNA can also directly be methylated and is another
form of epigenetics

A
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140
Q

Summary of regulatory points

A

Transcription (all at initiation)
- chromatin accessibility of the gene
*euk only (though bac do have
”chromatin”)
- DNA methylation *both
- activators, repressors, silencers,
enhancers, operators *both
Post-transcriptional
- alternative splicing of mRNA *euk only
- mRNA modification *both
- but polyA and 5’cap are *euk only
-5’ and 3’ UTR regulation *both
- small nc-RNAs binding to UTRs *both
but siRNA only euk.

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141
Q

Know which histone modifications and how they affect the accessibility of DNA
sequences.
***Really good to have big picture synthesis and connection of all parts of gene expression and
its regulation so far and how each step happens (process) and points of control (regulation) at
each step.

A

Transcription Initiation:
- Regulated by promoter sequences and bound transcription factors
- Chromatin accessibility controlled by histone modifications/remodeling
- e.g. H3K9ac increases accessibility, H3K9me3 decreases accessibility
- DNA looping brings distal enhancers/silencers into proximity of promoters
- Concentrations and activities of transcription factors modulate initiation

Elongation:
- Regulated by elongation factors and chromatin structure
- Histone modifications affect elongation efficiency
- e.g. H3K36me3 prevents initiation in gene bodies
- Chromatin remodelers clear nucleosomes for polymerase progression

Termination:
- Regulated by terminator sequences, poly(A) signals
- Proper chromatin context required for termination factors to bind

RNA Processing:
- 5’ capping, 3’ polyadenylation, splicing are regulated
- Splicing regulated by splicing factors binding signals in RNA
- Alternative splicing expands protein diversity

Epigenetic Regulation:
- Histone modifications and DNA methylation control chromatin state
- Inherited patterns maintain active/repressed expression states
- Chromatin remodelers expose or conceal regulatory regions

So in summary, every stage of eukaryotic gene expression is subject to multiple layers of regulation, with chromatin structure and epigenetic marks playing a central role in modulating accessibility and recruitment of the required machinery.

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142
Q

Know the features of the genetic code and what they mean – triplet nucleotide codon,
redundancy, NOT ambiguous, universality, wobble position, reading frame, and types of point
mutations (missense, nonsense, silent, and frameshift)

A

Triplet Nucleotide Codons:
- The genetic code is read in triplets of 3 nucleotides (codons)
- Each codon specifies a particular amino acid or stop signal

Redundancy:
- Multiple codons can code for the same amino acid
- This redundancy is described as the code being degenerate

Non-ambiguity:
- Each codon specifies only one amino acid (or stop)
- The code has no ambiguity in its meanings

Universality:
- The genetic code is nearly universal across all living organisms
- A few exceptions in certain viruses/prokaryotes

Wobble Position:
- The third position of a codon is more flexible in base-pairing
- Allowssome non-canonical base pairings (wobble)

Reading Frame:
- The codons are read in a continuous sequence without overlaps
- Frameshifts result in completely different proteins

Point Mutations:
- Missense - Single nucleotide change results in different amino acid
- Nonsense - Single nucleotide change introduces premature stop codon
- Silent - Single nucleotide change has no effect on amino acid
- Frameshift - Insertion/deletion shifts reading frame of all codons after

So in summary: Triplet codons, redundancy, non-ambiguity, near universality, wobble in third position, maintenance of continuous reading frame, and potential for different point mutation types are key features of the genetic code.

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143
Q

Even if activators and repressors are present, they may not be able to bind to their
regulatory elements depending on chromatin state. T/F

A

T

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144
Q

Even in the euchromatin state, nucleosomes have to be temporarily removed for
transcription elongation to occur. T/F

A

T

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145
Q

Heterochromatin is the compacted state of chromatin and histone acetylation causes
heterochromatin to form. T/F

A

F

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146
Q

What proteins are required to convert heterochromatin to euchromatin?

A

demethylase and acetyltransferase

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147
Q

We would expect an actively transcribed gene to be in a heterochromatin state. T/F

A

F

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148
Q

Epigenetic changes are a normal cellular process but can be caused by environmental
exposures. T/F

A

T

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149
Q

Epigenetic changes can occur at the histone and DNA level. T/F

A

T

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150
Q

Different cell types like skin vs. brain cells express different genes. One way to produce
this is to remodel chromatin such that skin specific genes are in euchromatin and other cell-specific genes in heterochromatin. T/F

A

T

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151
Q

Different cell types like skin vs. brain cells express different genes. One way to produce
this is to remodel chromatin such that skin specific genes are in euchromatin and other cell-
specific genes in heterochromatin. T/F

A

T

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152
Q

Different cell types like skin vs. brain cells express different genes. One way to produce a
skin cell is to express skin-specific transcription factors in skin cells only. T/F

A

T

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153
Q

The difference between histone methylation and DNA methylation is that histone
methylation promotes transcription while DNA methylation represses transcription. T/F

A

F

154
Q

Epigenetic inheritance might explain the concept of ”environmental effects” from our
genetics module. T/F

A

T

155
Q

DNA to RNA (Transcription), RNA to Protein (Translation), Protein to Active Protein (Post-translational modification)

A

Each step from DNA to functional product can be regulated to allow
cells to
- specialize
- conserve energy by not making unneeded products
- change in response to conditions

156
Q

Transfer RNAs (tRNA)

A

-very much like a translator:
one end anticodon which complementary to
the codon and the other end has the
corresponding amino acid
Ex: Arg-tRNA has Arginine anticodon of
5’UCG and Arg amino acid attached to 3’ end

157
Q

Wobble Position

A

This serine tRNA can
recognize both 5’ UCC
and 5’ UCU codons.
So two different codons
give same amino acid =
redundancy
Base pairing in the 3’ mRNA position can be non-standard with
the 5’ tRNA anticodon. This allows fewer tRNAs to
accommodate all 61 codons and gives redundancy in the code

158
Q

tRNA synthetases

A
  • tRNA transcribed in nucleus, exported
    to cytoplasm
  • Need to get correct amino acid
    attached to tRNA with correct
    anticodon
  • Aminoacyl-tRNA synthetase (enzyme!)
    matches tRNA to correct amino acid
    * 20 synthetases – one for each 20 aa
    * Recognize anticodon and aa
  • Aminoacyl-tRNA or a charged tRNA – is
    tRNA with its amino acid attached
159
Q

Translation Initiation

A

Initiation in Bacteria:
- Ribosome binding site on mRNA recruits small ribosomal subunit
- Start codon (AUG) on mRNA is recognized
- Initiation factors assist binding of initiator tRNA to start codon
- Small subunit joins with large subunit around mRNA
- Forms complete 70S initiation complex ready for elongation

Initiation in Eukaryotes:
- 5’ cap on mRNA recruits small 40S ribosomal subunit
- Scanning for start codon (AUG) facilitated by initiation factors
- Initiator Met-tRNA binds to start codon in 40S subunit
- Large 60S subunit joins, forming complete 80S initiation complex
- Additional initiation factors required compared to bacteria

Key Components:
- mRNA with ribosome binding site/start codon
- Small and large ribosomal subunits
- Initiator aminoacyl-tRNA (Met-tRNA or fMet-tRNA)
- Initiation factors assist recruitment and joining

So in essence, the initiation phase involves the recruitment of the ribosomal subunits and initiator tRNA to the start codon of the mRNA, facilitated by initiation factors, to form the completed translation initiation complex.

160
Q

Translation: Initiation Bacteria vs. Eukaryotes

A

Shine-Dalgarno Seq
binds ribosome through
complementary bp
5’-methyl-Gcap
f-Met first AUG
Rest of AUGs are Met Normal Methionine

161
Q

Translation: Elongation *translocation reqs. GTP

A

A – aminoacyl-tRNA site 4. next aminoacyl-tRNA enters & binds codon
P – peptidyl-tRNA site 5. Peptidyl transferase activity catalyzes peptide bond between the
existing polypeptide chain in P site to the amino acid in A site. The polypeptide will be
attached to the tRNA in the A site
E – exit 6. Ribosome moves down mRNA, now the ”empty” tRNA P (it’s not aminoacyl tRNA
anymore) is going to be moved to the E site displacing the previous tRNA
All same bacteria & eukaryotes.

162
Q

Translation: Termination

A

When a stop codon is encountered a release factor enters the
ribosome and causes it to release the peptide and disassemble
Release factor is a PROTEIN!

163
Q

Describe, label, and recognize steps of translation stages – initiation, elongation, termination

A

Initiation:
1) Ribosome Binding - Small ribosomal subunit binds to Ribosome Binding Site (RBS) on mRNA
2) Start Codon Recognition - Start codon (AUG) on mRNA is recognized
3) Initiator tRNA Binding - Initiator tRNA (Met-tRNAi or fMet-tRNA) binds to start codon
4) Large Subunit Joining - Large ribosomal subunit joins, forming complete initiation complex

Elongation:
1) Aminoacyl-tRNA Entry - Aminoacyl-tRNA corresponds to next codon enters A-site
2) Peptide Bond Formation - Peptide bond formed, adding new amino acid to polypeptide
3) Translocation - Ribosome moves one codon along, deacylated tRNA exits E-site
4) Repeat of 1-3 until stop codon reached

Termination:
1) Stop Codon in A-site - A stop codon (UAA, UGA, or UAG) enters the A-site
2) Release Factor Binding - Release factor binds and hydrolyzes peptidyl-tRNA bond
3) Polypeptide Release - Completed polypeptide is released from the ribosome
4) Ribosome Recycling - Subunits dissociate, preparing for next translation cycle

So in summary - initiation involves start codon recognition and ribosome assembly, elongation cycles of decoding, peptide bond formation and translocation, and termination triggered by stop codons to release the completed polypeptide.

164
Q

Know the differences in bacteria vs. eukaryotic initiation (elongation & termination are same)
Don’t worry about f-Met vs. Met  focus on the difference on how mRNA interacts with
the small ribosomal subunit

A

Bacteria:
- The small 30S ribosomal subunit directly binds to the Ribosome Binding Site (RBS) on the mRNA
- The RBS is a purine-rich sequence around 8 nucleotides upstream of the start codon
- This direct RBS binding helps recruit the small subunit to the correct start site

Eukaryotes:
- The small 40S subunit first binds to the 5’ cap of the mRNA
- It then scans along the mRNA linearly in a 5’ to 3’ direction
- Aided by initiation factors like eIF4, the 40S subunit searches for the start codon
- No RBS sequence, the start codon itself is recognized during scanning

Additional Points:
- In bacteria, RBS interaction helps simple positioning of 30S at start
- In eukaryotes, more complex scanning and start codon recognition mechanism
- More initiation factors required in eukaryotes to facilitate this process
- But the overall initiation complex assembly is conceptually similar

So the key difference is that bacterial mRNAs have a RBS that the small subunit directly recognizes and binds, while eukaryotic initiation relies on 5’ cap binding followed by linear scanning along the mRNA to locate the start codon.

165
Q

Describe and label a tRNA and its parts – 5’ vs. 3’, anticodon, amino acid attachment site.

A

Here is a labeled diagram of a tRNA (transfer RNA) molecule with its key parts:

5' End
 \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
|                                                |
|           \_\_\_\_\_\_                               |
|          /      \                              |
|         /        \                             |
|        |          |                            |
|        |          |            Anticodon      |
|        |        \_\_\\_\_\_            |||         |
|        |       /      \           XXX         | 
|        |      /        \                      |
|        |      \        /                      |
|        |       \\_\_\_\_\_\_/                       |
|        |                                      |
|        |                                      |
|\_\_\_\_\_\_\_\_|\_\_\_\_\_\_\_\_\_\_\_\_           \_\_\_\_\_\_\_\_\_\_\_\_\_\_|
                      \         /
                       \       /
                        \     /
                         \   /
                          | |
                          | |
                   Amino Acid Attachment
                         Site
                          |
                         3' End

Key Parts:
- 5’ End and 3’ End - Ends of the tRNA molecule
- Anticodon - A triplet sequence of 3 nucleotides that base pairs with the complementary codon on mRNA
- Amino Acid Attachment Site - The site where the specific amino acid is covalently attached to the tRNA
- Cloverleaf structure - The tRNA folds into this cloverleaf secondary and tertiary structure

During translation, the anticodon of the charged aminoacyl-tRNA base pairs with the codon on the mRNA in the ribosome’s A-site. This allows the correct amino acid attached to the 3’ end of the tRNA to be added to the growing polypeptide chain.

166
Q

Understand the tRNA charging process with tRNA synthetases

A

Here are the key points about the tRNA charging process involving tRNA synthetases:

Role of tRNA Synthetases:
- Enzymes that attach the correct amino acid to its corresponding tRNA molecule
- One specific tRNA synthetase exists for each amino acid

Charging Process:
1) tRNA synthetase binds to its cognate tRNA
2) It recognizes and binds the appropriate amino acid
3) A bond is formed between the amino acid and the 3’ end of the tRNA
4) This joins the amino acid to the tRNA, producing an aminoacyl-tRNA or charged tRNA

Specificity Mechanisms:
- Each tRNA synthetase has a specific amino acid binding pocket
- It selects the correct amino acid based on size, shape, chemical properties
- It also has an anticodon-binding domain to recognize the appropriate tRNA

Double-Sieve Mechanism:
- The active site acts like a double sieve or filter
- The smaller sieve excludes amino acids too large
- The larger sieve stops amino acids that are too small

Error Checking:
- Proofreading mechanisms check the amino acid is correctly attached
- Incorrectly charged tRNAs are hydrolyzed and released
- Ensures fidelity of the charging process

So in summary, extremely specific tRNA synthetases catalyze the attachment of the correct amino acids to their corresponding tRNA molecules through a high-fidelity, error-checked charging reaction.

167
Q

codon

A

A codon is a sequence of three nucleotides (triplet) in mRNA that corresponds to a specific amino acid or serves as a start or stop signal during protein synthesis. Each codon encodes a specific amino acid, except for three codons (UAA, UAG, and UGA) that serve as stop signals, terminating translation. The genetic code, which is the set of rules by which nucleotide sequences are translated into amino acid sequences, dictates the correspondence between codons and amino acids. Codons are read sequentially during translation by the ribosome, with each codon specifying the addition of a specific amino acid to the growing polypeptide chain.

168
Q

redundancy

A

Redundancy in genetics means that multiple codons can code for the same amino acid, providing flexibility and robustness in protein synthesis.

169
Q

universality

A

Universality in genetics refers to the fact that the genetic code, the set of rules that translate nucleotide sequences into amino acid sequences, is nearly identical across all living organisms. This means that the same codons typically code for the same amino acids in diverse species, from bacteria to humans. This universality of the genetic code underscores the fundamental unity of life and allows for the exchange of genetic information between different organisms through processes such as horizontal gene transfer and genetic engineering.

170
Q

wobble position

A

The wobble position refers to the third nucleotide position in a codon. It is called “wobble” because there is flexibility in base pairing between the third nucleotide of the codon and the corresponding nucleotide in the anticodon of the tRNA molecule during translation. This flexibility allows a single tRNA to recognize multiple codons that code for the same amino acid, enhancing the efficiency of protein synthesis.

171
Q

reading frame

A

The reading frame is the specific way nucleotides in mRNA are grouped into codons during translation, crucial for accurate protein synthesis.

172
Q

small ribosomal subunit

A

The small ribosomal subunit is a component of the ribosome responsible for binding to mRNA during translation.

173
Q

large ribosomal subunit

A

The large ribosomal subunit is a component of the ribosome responsible for catalyzing the formation of peptide bonds during translation.

174
Q

Shine-
Delgarno sequence

A

The Shine-Dalgarno sequence is a short nucleotide sequence found in prokaryotic mRNA molecules, usually located upstream of the start codon (AUG). It functions as a ribosome binding site, allowing the ribosome to recognize and initiate translation at the correct position on the mRNA.

175
Q

tRNA

A

tRNA, or transfer RNA, is a type of RNA molecule responsible for transporting specific amino acids to the ribosome during protein synthesis (translation). Each tRNA molecule carries a specific amino acid at one end and contains a three-nucleotide sequence called an anticodon at the other end, which base-pairs with the corresponding codon on mRNA. tRNA plays a crucial role in deciphering the genetic code and ensuring that the correct amino acids are incorporated into the growing polypeptide chain during translation.

176
Q

anticodon

A

The anticodon is a sequence of three nucleotides found on transfer RNA (tRNA) molecules. It pairs with a complementary codon on messenger RNA (mRNA) during translation, ensuring that the correct amino acid is added to the growing polypeptide chain.

177
Q

tRNA synthetase

A

tRNA synthetase is an enzyme responsible for attaching specific amino acids to their corresponding transfer RNA (tRNA) molecules during protein synthesis. This process, known as aminoacylation or tRNA charging, ensures that the correct amino acid is loaded onto the appropriate tRNA molecule, facilitating accurate translation of the genetic code.

178
Q

charged-tRNA (aminoacyl tRNA)

A

Charged tRNA, also known as aminoacyl tRNA, is a transfer RNA (tRNA) molecule that has been covalently attached to its corresponding amino acid by an enzyme called aminoacyl-tRNA synthetase. The amino acid is linked to the 3’ end of the tRNA molecule, forming an aminoacyl-tRNA complex. Charged tRNA molecules are crucial for protein synthesis, as they deliver the correct amino acid to the ribosome during translation, ensuring the accurate incorporation of amino acids into the growing polypeptide chain.

179
Q

release factor

A

Release factor is a protein that binds to the ribosome during translation termination. It promotes the hydrolysis of the bond between the final tRNA molecule and the completed polypeptide chain, releasing the newly synthesized protein from the ribosome.

180
Q

E site

A

The E site, or exit site, is a ribosomal site where a transfer RNA (tRNA) molecule exits the ribosome after donating its amino acid during translation.

181
Q

P site

A

The P site, or peptidyl site, is a ribosomal site where the tRNA molecule carrying the growing polypeptide chain is located during translation.

182
Q

A site

A

The A site, or aminoacyl site, is a ribosomal site where a charged tRNA molecule binds during translation, bringing in the next amino acid to be added to the growing polypeptide chain.

183
Q

polypeptide

A

A polypeptide is a linear chain of amino acids linked together by peptide bonds. It is a precursor to a protein and can fold into a specific three-dimensional structure to perform a particular biological function. Polypeptides are synthesized during translation, where amino acids are joined together in a specific sequence dictated by the mRNA template. Multiple polypeptide chains can associate to form a functional protein molecule.

184
Q

peptide bond

A

A peptide bond is a covalent bond that forms between the amino group (-NH2) of one amino acid and the carboxyl group (-COOH) of another amino acid during protein synthesis. This bond forms through a condensation reaction, resulting in the loss of a water molecule. Peptide bonds link amino acids together to form polypeptide chains, which are the building blocks of proteins.

185
Q

amino acid

A

An amino acid is a small organic molecule that serves as the building block of proteins. It contains an amino group (-NH2), a carboxyl group (-COOH), and a side chain (R group) attached to a central carbon atom. There are 20 standard amino acids commonly found in proteins, each with a unique side chain that determines its chemical properties. Amino acids are linked together by peptide bonds to form polypeptide chains during protein synthesis. They play essential roles in various biological processes, including enzyme catalysis, cell signaling, and structural support.

186
Q

Old 141 objective – be able to translate a sequence using codon table and identify
mutation type if given a mutant sequence

A

Translation Steps:
1) Identify the start codon (AUG) - this indicates the reading frame
2) Split the sequence into triplet codons starting from the start codon
3) Use codon table to look up each codon and write down the corresponding amino acid
4) Continue until you reach a stop codon (UAA, UGA, UAG)
5) The amino acid sequence is the translation of that coding sequence

Mutation Identification:
If provided a mutant sequence:
1) Translate the original and mutant sequences
2) Compare amino acid sequences to identify the changed amino acid(s)
3) Determine the codon change that led to the amino acid change
4) Categorize the mutation type:
- Missense - Single nucleotide change resulting in different amino acid
- Nonsense - Single nucleotide change introducing premature stop codon
- Silent - Single nucleotide change with no amino acid change
- Insertion/Deletion - Frameshift mutation altering all codons after

Example:
Original: 5’- AUGGCCUUGGAC -3’
Met Ala Leu Asp
Mutant: 5’- AUGGCCUUAGAC -3’
Met Ala STOP

This is a nonsense mutation, where a single C->A change in the 3rd codon (UGG->UGA) introduces a premature stop codon.

So in summary, translate using the codon table, identify differences versus a mutant sequence, and categorize as missense/nonsense/silent/frameshift.

187
Q

Understand how tRNA abundance and codon usage and bias can regulate translation
rate

A

tRNA Abundance:
- Different tRNA species are present in varying abundances in the cell
- Highly abundant tRNAs can more rapidly deliver their amino acids during translation
- Limiting tRNA abundances for particular codons can slow down that codon’s translation

Codon Usage:
- Codons for the same amino acid are not used equally in gene sequences
- Some synonymous codons are used more frequently than others
- Codon usage is biased and varies between organisms and gene sequences

Codon Bias:
- In any gene, some synonymous codons are used more frequently
- This bias matches the most abundant tRNA species in the cell
- Allows more efficient and rapid translation of these abundant codons

Effects on Translation Rate:
- Regions with abundant tRNAs/preferred codons are translated faster
- Regions with rare tRNAs/codon bias are translated more slowly
- This can create “ramps” or “peaks” of slower translation rates
- Believed to help with protein folding, localization, regulation

So in essence, the abundance of tRNA isoacceptors, the codon usage patterns of genes, and the resulting codon biases together influence the speed at which codons are translated, allowing regulation of translation elongation rates for different regions.

188
Q

Describe post-translational modifications and the ways they can regulate protein: shape,
activity, location, and stability
* Understand that multiple PTMs may be used together to regulate a protein
* Connect PTMs to chromatin remodeling! – Histones are proteins that get
methylated and acetylated and phosphorylated!

A

Shape/Conformation:
- Phosphorylation can induce conformational changes by adding charged groups
- Disulfide bonds formed by oxidation stabilize protein folding
- Prolyl isomerization changes the shape around proline residues

Activity:
- Phosphorylation often activates or inactivates enzymes
- Acetylation can increase or decrease protein binding affinities
- Lipidation (e.g. prenylation) can activate membrane-bound proteins
- Glycosylation modulates activity of extracellular proteins

Localization:
- Lipidation targets proteins to cell membranes
- Glycosylation aids protein targeting to organelles like ER/Golgi
- Nuclear localization signals direct nuclear import after acetylation

Stability:
- Ubiquitination marks proteins for proteasomal degradation
- Methylation can protect proteins from degradation
- Deamidation and oxidation often destabilize proteins

Multiple PTMs can work together:
- A protein may require several coordinated modifications for full regulation
- e.g. Phosphorylation primes substrate for subsequent ubiquitination

Histones and Chromatin:
- Histone tails are heavily modified by acetylation, methylation, phosphorylation
- These PTMs regulate chromatin compaction and gene expression

So in summary, a wide array of covalent PTMs act as molecular switches, altering protein properties like shape, binding, localization and half-life in a dynamic and combinatorial fashion, including on histone proteins.

189
Q

codon usage bias

A

Codon usage bias refers to the tendency for certain codons to be used more frequently than others in the genetic code of an organism.

190
Q

rare codon

A

A rare codon is a codon that is used less frequently than other synonymous codons for encoding the same amino acid within the genetic code of an organism.

191
Q

common codon

A

A common codon is a codon that is used more frequently than other synonymous codons for encoding the same amino acid within the genetic code of an organism.

192
Q

post-translational modification

A

Post-translational modification refers to the chemical modification of proteins after they have been synthesized by ribosomes during translation. These modifications can include the addition of functional groups such as phosphate, methyl, acetyl, or lipid groups, as well as cleavage of specific peptide bonds or the attachment of sugar moieties (glycosylation). Post-translational modifications play crucial roles in regulating protein function, stability, localization, and activity, and they are involved in various cellular processes, including cell signaling, metabolism, and gene expression.

193
Q

phosphorylation

A

Phosphorylation is a post-translational modification process in which a phosphate group (-PO4) is added to a protein molecule, typically to a serine, threonine, or tyrosine residue. This modification is catalyzed by enzymes called protein kinases and plays a key role in regulating protein function and cellular signaling pathways. Phosphorylation can affect protein activity, stability, subcellular localization, and interactions with other molecules, thereby influencing various cellular processes such as cell growth, proliferation, differentiation, and apoptosis.

194
Q

acetylation

A

Acetylation is the addition of an acetyl group (-COCH3) to a molecule, often proteins, through a post-translational modification process.

195
Q

methylation

A

Methylation is the addition of a methyl group (-CH3) to a molecule, such as DNA, RNA, proteins, or small molecules, through a post-translational modification process.

196
Q

ubiquitination – don’t worry about the specific
name for adding a fatty acid tail,

A

Ubiquitination is a post-translational modification process in which a small protein called ubiquitin is covalently attached to a target protein, typically marking it for degradation by the proteasome or regulating its activity, localization, or interactions with other proteins. Ubiquitination is catalyzed by a cascade of enzymes, including ubiquitin-activating enzymes (E1), ubiquitin-conjugating enzymes (E2), and ubiquitin ligases (E3). It plays crucial roles in regulating various cellular processes, including protein degradation, DNA repair, cell cycle progression, and signal transduction.

197
Q

kinase

A

A kinase is an enzyme that adds a phosphate group to a substrate, typically a protein, nucleotide, or lipid.

198
Q

phosphatase

A

A phosphatase is an enzyme that removes a phosphate group from its substrate, typically a protein, nucleotide, or lipid.

199
Q

ubiquitin ligase

A

A ubiquitin ligase is an enzyme that catalyzes the attachment of ubiquitin molecules to target proteins, marking them for degradation or altering their function, localization, or interactions.

200
Q

deubiquitinating enzyme

A

A deubiquitinating enzyme (DUB) is an enzyme that removes ubiquitin molecules from proteins, reversing the ubiquitination process.

201
Q

Know DNA replication steps and process

A
  1. Initiation
    • Origins of replication are recognized and bound by initiator proteins
    • Helicase unwinds the DNA double helix, forming a replication bubble
    • Single-stranded binding proteins (SSBs) bind and stabilize the unwound strands
  2. Primer Synthesis
    • RNA primase synthesizes an RNA primer complementary to each parental strand
  3. Elongation
    • DNA polymerase III (bacteria) or polymerases α, δ, ε (eukaryotes) bind to primers
    • Polymerases synthesize new complementary daughter strands in 5’ to 3’ direction
    • On leading strand, synthesis is continuous
    • On lagging strand, synthesis occurs in short Okazaki fragments
  4. Ligation
    • On lagging strand, DNA ligase seals nicks between Okazaki fragments
  5. Termination
    • Two replication forks meet and replication is complete
    • Specific termination sequences allow displacement of final RNA primer
  6. Primer Removal and Filling of Gaps
    • RNA primers are removed by RNase H and DNA polymerase
    • Gaps are filled and nicks sealed by DNA polymerase and DNA ligase

So in summary: Initiation at origins, primer synthesis, polymerase elongation on leading/lagging strands, Okazaki fragment ligation, termination when forks meet, and processing of primers/gaps to complete replication.

202
Q

Understand why an RNA primer is needed for DNA synthesis

A

Specifically:

1) DNA polymerases can only synthesize DNA in the 5’ to 3’ direction by adding deoxynucleotides to the 3’-OH end of an existing strand.

2) At the start of replication, there is no pre-existing 3’-OH onto which the polymerase can add nucleotides.

3) To provide this initial 3’-OH group, a short RNA primer is synthesized by the primase enzyme.

4) The RNA primer has a free 3’-OH end that the DNA polymerase can then extend from by adding DNA nucleotides complementary to the template strand.

5) RNA is used instead of DNA because only RNA polymerases can initiate de novo synthesis. DNA polymerases cannot start without a pre-existing strand.

6) Once elongation begins, the DNA polymerase can continue adding DNA nucleotides in the 5’->3’ direction using the parental template.

So in essence, an RNA primer provides the free 3’-OH end that DNA polymerases require to initiate DNA synthesis, since they cannot start de novo. The RNA primer is extended by DNA nucleotides and is later removed and replaced.

203
Q

Understand why RNA primer requirement creates a lagging strand that is synthesized
differently

A
  • DNA replication is semi-conservative - each new double helix has one old and one new strand
  • DNA polymerase can only synthesize new DNA in the 5’ to 3’ direction
  • Leading strand is synthesized continuously by DNA polymerase moving along with replication fork
  • Lagging strand is synthesized discontinuously in short Okazaki fragments due to antiparallel orientation
  • DNA polymerase requires an RNA primer to initiate synthesis of each Okazaki fragment
  • RNA primers are laid down by primase enzyme and extended into Okazaki fragments by DNA polymerase
  • Later, RNA primers are removed and gaps filled by DNA polymerase, then sealed by DNA ligase
  • This process creates slightly more work and potential for errors on the lagging strand
204
Q

Know the few differences between bacteria and eukaryotic DNA replication – there’s 1
difference each in initiation, elongation, and termination

A

Initiation:
* Bacteria - Single origin of replication (ori) on circular chromosome
* Eukaryotes - Multiple origins of replication (ori) on linear chromosomes

Elongation:
* Bacteria - One DNA polymerase III holoenzyme handles leading and lagging strands
* Eukaryotes - Different polymerases for leading (pol ε) and lagging (pol δ) strands

Termination:
* Bacteria - Termination occurs when two replication forks meet and terminate
* Eukaryotes - Termination zones lacking specific termination sequences, forks dissociate randomly

205
Q

Be able to draw and label all steps and all proteins required and what they are doing (Replication)

A

Initiation:
- Origin of Replication (ori)
- Helicase (unwinds DNA double helix)
- Single-Stranded DNA Binding Proteins (ssb) (bind to unwound DNA, preventing rewinding)
- Primase (lays down RNA primers on lagging strand template)

Elongation:
- Leading Strand - DNA Polymerase III (synthesizes continuously)
- Lagging Strand:
- DNA Polymerase III (extends from RNA primer, forming Okazaki fragments)
- RNA Primer (allows Pol III to initiate synthesis)

  • DNA Ligase (seals nicks between Okazaki fragments after RNA primers removed)

Termination:
- Two Replication Forks meet and terminate

Other Proteins:
- Topoisomerase (relieves torsional strain ahead of replication forks)
- RNase H (removes RNA primers)
- DNA Polymerase I (replaces RNA primers with DNA)

206
Q

Describe the Hayflick Limit, the end replication problem of linear chromosomes, and
why it occurs

A

Hayflick Limit:
- Limits the number of times a cell population can divide
- Cultured human cells can only divide 40-60 times
- Cellular senescence occurs after this limit

End Replication Problem:
- DNA polymerase cannot completely replicate the ends of linear chromosomes
- It leaves a short overhanging sequence at the 5’ end unreplicated
- This causes chromosomes to shorten with each division cycle

Why it Occurs:
- DNA polymerase requires an RNA primer to initiate replication
- RNA primers are removed, leaving a gap at the 5’ end
- On linear DNA, there is no way to replace the sequence lost from the 5’ end
- Repeated rounds lead to progressive shortening of telomeres until critically short
- This triggers cellular senescence at the Hayflick Limit

207
Q

Explain how end replication problem limits cell division and its consequence on
chromosome length

A
  • During DNA replication, DNA polymerase cannot completely replicate the ends of linear chromosomes
  • A small portion at the 5’ end remains unreplicated after removal of the RNA primer
  • With each cell division cycle, this unreplicated portion gets progressively shorter
  • This causes the telomeres (chromosome ends) to shorten with every round of replication
  • Telomeres protect the coding regions of the chromosome from being truncated
  • As telomeres shorten, coding sequences start getting deleted, making cell inviable
  • Once telomeres become critically short, cells trigger replicative senescence (Hayflick Limit)
  • This permanently arrests cell division to avoid genomic instability
  • The progressive shortening of telomeres due to the end replication problem is a cellular molecular clock
  • It limits how many times a cell can divide before losing viability
  • This has implications in aging, stem cell replicative potential, and cancer development

So in summary, the end replication problem causes telomere attrition with each division cycle, ultimately limiting the number of times a cell can replicate before losing chromosome integrity and viability.

208
Q

Describe telomeres and their function and how they are the counter in Hayflick’s limit

A

Telomeres:
- Repetitive, non-coding sequences at the ends of linear chromosomes
- Consist of same short DNA sequence repeats bound by shelterin protein complex
- In humans, telomeres are TTAGGG repeats ranging from 8-15 kilobases long

Functions of Telomeres:
- Protect coding DNA at chromosome ends from degradation and fusion events
- Distinguish natural chromosome ends from DNA double-strand breaks
- Allow complete replication of coding DNA by providing disposable overhangs

Telomeres and the Hayflick Limit:
- With each cell division, telomeres shorten due to the end replication problem
- Cells divide until telomeres erosion reaches a critical short length
- Extremely short telomeres trigger DNA damage response pathways
- This permanently arrests the cell cycle, causing replicative senescence
- The progressive shortening of telomeres acts as a cellular “replicative clock”
- Telomere length dictates the maximum number of divisions (Hayflick Limit)

So telomeres prevent genome instability, but their shortening with each replication cycle limits how many times a cell can divide before losing viability and hitting the Hayflick Limit of cellular senescence.

209
Q

Explain how telomerase lengthens telomeres

A

Here’s an explanation of how the enzyme telomerase lengthens telomeres:

  • Telomerase is a specialized reverse transcriptase enzyme that adds telomeric repeats to the ends of chromosomes.
  • It contains an RNA template complementary to the telomere repeat sequence (e.g. TTAGGG in humans).
  • Telomerase carries its own RNA that provides a template to extend the 3’ overhang of the telomere.
  • It binds to the 3’ overhang and uses its RNA template to synthesize complementary telomeric DNA repeats onto the end.
  • This compensates for the shortening caused by the end replication problem during cell division.
  • Telomerase acts in a repeating cycle, rebinding and re-extending the telomere after each round of replication.
  • This allows indefinite extension of the telomere, preventing it from eroding critically short.
  • Most somatic cells do not express telomerase, so telomeres shorten with each division eventually leading to senescence.
  • Telomerase is expressed in stem cells and some cancer cells, allowing them to maintain telomere length and divide indefinitely.

So in summary, telomerase solves the end replication problem by using its RNA template to add back the telomeric repeats lost during replication, preventing telomere shortening.

210
Q

Understand how telomere maintain impacts regular cells and abnormal cells like cancer
cells

A

Regular Cells:
- Most somatic cells do not express telomerase
- Telomeres shorten with each cell division due to end replication problem
- Once telomeres become critically short, cells enter replicative senescence (Hayflick Limit)
- This limits the replicative potential and lifespan of cells
- Prevents abnormal cells from dividing indefinitely, acting as a tumor suppressor mechanism

Cancer Cells:
- Many cancer cells reactivate or upregulate telomerase expression
- Telomerase maintains telomere length by adding back repeats after replication
- This allows cancer cells to divide indefinitely without hitting replicative senescence
- Circumvents the Hayflick Limit and enables unlimited replicative potential
- Facilitates immortalization, a hallmark of cancer cells
- Suggests telomerase inhibition could be a therapeutic target in cancer treatment

So in summary:
- Lack of telomerase in normal cells limits replicative potential as a protective mechanism
- Telomerase reactivation in cancer cells confers unlimited replicative ability by maintaining telomeres
- This key difference in telomere maintenance separates mortal normal cells from immortal cancer cells

211
Q

Be sure you know that end replication problem does NOT apply to CIRCULAR
CHROMOSOME

A
212
Q

Be sure you know that every newly synthesized DNA strand will be made by leading and
lagging synthesis. None are made of completely leading or completely lagging synthesis.

A
  • Leading strand is synthesized continuously in the 5’ to 3’ direction by DNA polymerase
  • Lagging strand is synthesized discontinuously in 5’ to 3’ direction as Okazaki fragments
  • For any given replication fork, there is one leading strand and one lagging strand template
  • The new leading strand is synthesized continuously from start to end
  • The new lagging strand starts with RNA primer and Okazaki fragment, synthesized discontinuously
  • When the replication fork meets the next origin, roles reverse for leading/lagging
  • The new strands get synthesized partially as previous leading, partially as previous lagging

So in essence, all newly replicated double-stranded DNA molecules contain regions synthesized via both leading and lagging strand mechanisms at alternating sections along their length. No single new strand is purely leading or purely lagging in its synthesis.

213
Q

Understand why eukaryotic chromosomes often have many more origins of replication than
bacteria chromosome

A
  1. Chromosome Size:
    - Eukaryotic chromosomes are much larger (millions to billions of base pairs) compared to a single bacterial circular chromosome (millions of base pairs)
    - More origins are required to replicate the larger eukaryotic chromosomes in a reasonable time frame
  2. Linear vs Circular:
    - Eukaryotic chromosomes are linear
    - Replication can only proceed bi-directionally from each origin
    - More origins are needed along the length of linear chromosomes
  3. Nuclear Membrane:
    - In eukaryotes, replication occurs in the nucleus
    - The nuclear membrane acts as a barrier limiting the minimum number of replication forks
  4. Chromosome Compaction:
    - Eukaryotic chromatin is highly compacted, making it harder for replication machinery to travel long distances
    - Multiple closely-spaced origins help replication traverse compacted chromatin regions
  5. Cell Cycle Regulation:
    - Eukaryotic cell cycles are more complex and strictly regulated
    - Multiple origins ensure coordinated and complete replication within the time constraints

In contrast, the smaller circular bacterial chromosome can be replicated from a single origin more easily within their simpler cell cycle. So the larger size and linear nature of eukaryotic chromosomes necessitates having many more replication origins.

214
Q

origin of replication

A

The origin of replication is a specific sequence of DNA where the replication of the DNA molecule begins. It serves as the starting point for the assembly of the replication machinery and the initiation of DNA synthesis during DNA replication. Each DNA molecule typically has multiple origins of replication to ensure efficient and timely replication of the entire genome.

215
Q

DNA replication bubble

A

A DNA replication bubble is a region of DNA where the double helix unwinds and opens up to form two single strands during DNA replication. It is characterized by the formation of a replication fork, where new DNA strands are synthesized in opposite directions. The replication bubble expands bidirectionally as DNA replication proceeds, with each replication fork moving outward from the origin of replication.

216
Q

single-stranded DNA binding protein (SSBP)

A

A single-stranded DNA binding protein (SSBP) is a protein that binds to single-stranded DNA (ssDNA) to stabilize and protect it from degradation or reannealing during DNA replication, repair, or recombination processes. SSBPs prevent the reformation of double-stranded DNA (dsDNA) and help to keep the DNA in a suitable conformation for enzymatic processing. They play crucial roles in various cellular processes, including DNA replication, DNA repair, and homologous recombination.

217
Q

replication fork

A

A replication fork is a structure formed during DNA replication where the double helix unwinds and separates into two single strands. It is characterized by the presence of two replication forks moving in opposite directions from the origin of replication. At each replication fork, the parental DNA strands serve as templates for the synthesis of new DNA strands, leading to the formation of two daughter DNA molecules. The replication fork is the site of active DNA synthesis and is associated with various proteins and enzymes involved in DNA replication, such as DNA polymerases, helicases, and single-stranded DNA binding proteins.

218
Q

DNA polymerase

A

DNA polymerase is an enzyme responsible for synthesizing new DNA strands by catalyzing the addition of nucleotides to a growing DNA chain during DNA replication. DNA polymerases require a template strand and a primer strand to initiate DNA synthesis and extend the primer in the 5’ to 3’ direction. These enzymes are essential for accurate and efficient replication of the genome and play key roles in DNA repair and other DNA metabolic processes. Multiple DNA polymerases exist in cells, each with specific functions and properties.

219
Q

primase

A

Primase is an enzyme responsible for synthesizing short RNA primers on single-stranded DNA templates during DNA replication. These RNA primers provide a starting point for DNA polymerases to initiate DNA synthesis. Primase synthesizes RNA primers by catalyzing the addition of ribonucleotides to the DNA template, creating a short stretch of RNA that serves as a primer for DNA polymerase. Once the RNA primer is synthesized, DNA polymerase extends it by adding deoxyribonucleotides, ultimately forming a new DNA strand complementary to the template. Primase plays a crucial role in the initiation of DNA replication and is essential for the synthesis of both the leading and lagging strands of the replication fork.

220
Q

helicase

A

Helicase is an enzyme that catalyzes the unwinding of the double-stranded DNA helix during DNA replication, transcription, and repair. It uses the energy from ATP hydrolysis to break the hydrogen bonds between the complementary base pairs, separating the two DNA strands and creating a replication fork. Helicase plays a crucial role in the initiation of DNA replication by creating the single-stranded DNA templates required for the synthesis of new DNA strands by DNA polymerases. It also participates in various other DNA metabolic processes, such as DNA repair and recombination, by facilitating the unwinding of DNA duplexes to allow access for other proteins and enzymes.

221
Q

RNA primer

A

An RNA primer is a short segment of RNA nucleotides that serves as a starting point for DNA synthesis during DNA replication. Primers are synthesized by the enzyme primase, which adds complementary RNA nucleotides to the template DNA strand. These RNA primers provide a free 3’ hydroxyl group for DNA polymerase to initiate DNA synthesis, as DNA polymerase can only elongate an existing strand and cannot initiate synthesis de novo. RNA primers are later removed and replaced with DNA nucleotides by DNA polymerase, resulting in the synthesis of a continuous DNA strand.

222
Q

sliding clamp

A

A sliding clamp, also known as a clamp loader, is a protein complex that helps to hold DNA polymerase onto the template DNA strand during DNA replication. It encircles the DNA double helix and forms a ring-shaped structure that can slide along the DNA strand. The sliding clamp acts as a processivity factor, allowing DNA polymerase to remain attached to the DNA template for extended periods and catalyze the addition of nucleotides to the growing DNA strand without dissociating. This enhances the efficiency and speed of DNA replication by preventing the polymerase from falling off the template DNA strand.

223
Q

leading strand

A

The leading strand is the DNA strand that is synthesized continuously in the 5’ to 3’ direction during DNA replication. It is synthesized in the same direction as the movement of the replication fork and requires only one RNA primer for initiation. DNA polymerase synthesizes the leading strand continuously by adding nucleotides to the 3’ end of the RNA primer as the replication fork progresses. Since the leading strand is synthesized continuously, there is no need to repeatedly start and stop DNA synthesis, resulting in faster and more efficient replication compared to the lagging strand.

224
Q

lagging strand

A

The lagging strand is the DNA strand that is synthesized discontinuously in the 5’ to 3’ direction during DNA replication. It is synthesized in the opposite direction to the movement of the replication fork, which necessitates the use of multiple RNA primers for initiation. DNA polymerase synthesizes short fragments of DNA, called Okazaki fragments, on the lagging strand, starting from each RNA primer and extending in the direction away from the replication fork. After synthesis, the RNA primers are removed, and the gaps between the Okazaki fragments are filled in by DNA polymerase and sealed by DNA ligase, resulting in the synthesis of a continuous DNA strand.

225
Q

Okazaki fragment

A

An Okazaki fragment is a short segment of DNA nucleotides synthesized discontinuously on the lagging strand during DNA replication. These fragments are typically around 100-200 nucleotides in length in prokaryotes and shorter in eukaryotes. Each Okazaki fragment is initiated by the synthesis of an RNA primer by primase, which provides a starting point for DNA polymerase to elongate the fragment. DNA polymerase then synthesizes DNA nucleotides in the 5’ to 3’ direction, away from the replication fork, until it encounters the RNA primer of the adjacent Okazaki fragment or reaches the end of the DNA template. After synthesis, the RNA primers are removed, and the gaps between the Okazaki fragments are filled in by DNA polymerase and sealed by DNA ligase, resulting in the synthesis of a continuous lagging strand.

226
Q

DNA polymerase I

A

DNA polymerase I is a type of DNA polymerase enzyme found in bacteria, including the well-studied Escherichia coli. It plays multiple roles in DNA replication, repair, and recombination.

In DNA replication, DNA polymerase I is involved in removing RNA primers from Okazaki fragments on the lagging strand and replacing them with DNA nucleotides during synthesis. This process, known as primer removal and gap filling, ensures the continuity of the lagging strand.

Additionally, DNA polymerase I is involved in DNA repair processes, such as base excision repair and nucleotide excision repair, where it fills in gaps left after damaged nucleotides are removed.

Furthermore, DNA polymerase I also possesses 5’ to 3’ exonuclease activity, allowing it to proofread and correct errors in newly synthesized DNA strands.

Overall, DNA polymerase I plays essential roles in maintaining the integrity and fidelity of the bacterial genome.

227
Q

DNA ligase

A

DNA ligase is an enzyme that seals nicks in DNA strands by catalyzing the formation of phosphodiester bonds between adjacent DNA fragments.

228
Q

Hayflick Limit

A

The Hayflick limit is the maximum number of times a human cell population can divide before it stops dividing and enters a state of senescence or cell death.

229
Q

end replication problem

A

The end replication problem refers to the challenge of replicating the ends of linear DNA molecules, where the last RNA primer cannot be replaced with DNA nucleotides, leading to gradual shortening of the chromosome ends with each round of replication.

230
Q

telomere

A

Telomeres are repetitive DNA sequences at the ends of chromosomes that protect them from degradation and fusion, and their shortening with each cell division contributes to aging.

231
Q

telomerase

A

Telomerase is an enzyme that adds repetitive DNA sequences to the ends of linear chromosomes, known as telomeres, to counteract their shortening during DNA replication and cellular division.

232
Q

Describe the cell cycle phases

A

G1 (Gap 1) Phase:
- Cells increase in mass and size
- Carry out metabolic functions
- Check for any DNA damage before proceeding

S (Synthesis) Phase:
- DNA replication occurs
- Chromosomes are duplicated

G2 (Gap 2) Phase:
- Cell continues growing
- Produces proteins for mitosis
- Checks for any DNA damage before mitosis

M (Mitosis) Phase:
- Nuclear envelope breaks down
- Chromosomes condense and are separated
- Cytoplasm divides forming two daughter cells

Cytokinesis:
- Final physical separation of the cytoplasm
- Occurs after mitosis is complete

G0 (Quiescent) Phase:
- Cell exits the cycle, no growth/division
- Some cells enter this resting state
- Can re-enter cycle under right conditions

The key control checkpoints occur at the G1/S and G2/M transitions, ensuring proper replication and segregation before advancing the cycle. Proper regulation of this cycle is critical for growth and reproduction.

233
Q

Describe the 3 purposes of the cell cycle

A

The cell cycle serves three main purposes:

  1. Cell Growth and Replication:
    - Allows cells to replicate their DNA (S phase) and divide to produce genetically identical daughter cells (M phase)
    - Enables growth, repair, and replacement of cells in multicellular organisms
    - Facilitates asexual reproduction in unicellular organisms
  2. Genetic Regulation and Integrity:
    - Has checkpoints (G1/S and G2/M) to ensure DNA is replicated and repaired properly before division
    - Prevents propagation of genetic errors or damaged DNA to daughter cells
    - Maintains proper ploidy levels and chromosomal organization
  3. Cellular Differentiation and Development:
    - Cell division allows transformation of a single-celled zygote into a complete organism (embryogenesis)
    - Regulated cell cycles allow stem cells to differentiate into various specialized cell types
    - Precisely timed cycles coordinate developmental processes like morphogenesis

So in summary, the three key purposes are:
1) Replicating genetic material for growth/reproduction
2) Regulating genetic fidelity and integrity
3) Enabling cellular differentiation for development

Proper regulation of the cell cycle is crucial for the survival, propagation and development of all living organisms.

234
Q

Describe cyclin-CDKs and how they are regulated

A

Here are the key points about cyclin-CDK complexes and how they are regulated during the cell cycle:

Cyclin-CDK Complexes:
- Cyclins are regulatory proteins that undergo cyclical synthesis and degradation
- CDKs (Cyclin-Dependent Kinases) are serine/threonine kinase enzymes
- Cyclins activate their respective CDK partners by binding to them

Functions:
- Cyclin-CDK complexes phosphorylate various substrate proteins
- This drives progression through different cell cycle phases
- Different cyclin-CDK complexes are active at specific phases

Regulation:
- Cyclin levels fluctuate - synthesized and degraded at specific times
- This periodically activates and inactivates different CDKs
- CDK activity is also regulated by activating/inhibitory phosphorylation
- CDK inhibitor proteins (CKIs) can bind and inhibit CDK activity

Key Complexes:
- G1/S: Cyclin D-CDK4/6, Cyclin E-CDK2 (initiate DNA replication)
- S: Cyclin A-CDK2 (continues DNA replication)
- G2/M: Cyclin B-CDK1 (promotes mitosis entry)
- M: Cyclin B-CDK1 (drives through metaphase/anaphase)

So in summary, cyclical synthesis/degradation of cyclins activates specific CDK partners at the right times, driving orderly progression through cell cycle phases. Multiple regulatory mechanisms tightly control this process.

235
Q

Understand cyclin-CDK activity determines cell cycle phase

A

Here are the key points to understand how cyclin-CDK activity determines which cell cycle phase a cell is in:

  • Different cyclin-CDK complexes are active and regulate different phases of the cell cycle.
  • The specific cyclins that are present determine which CDKs are activated, as cyclins are the regulatory subunits.
  • In G1 phase, Cyclin D-CDK4/6 and Cyclin E-CDK2 are active, preparing the cell for DNA synthesis.
  • In S phase, Cyclin A-CDK2 is the active complex driving DNA replication.
  • In G2 phase, Cyclin A-CDK1 is active, allowing the cell to grow and prepare for mitosis.
  • At the G2/M transition, Cyclin B-CDK1 becomes active, triggering breakdown of the nuclear envelope.
  • During M phase, Cyclin B-CDK1 continues to be active, facilitating chromosome condensation and segregation.
  • The periodic cyclical synthesis and degradation of cyclins leads to the sequential activation of different CDKs.
  • This regulated cyclin oscillation creates an orderly progression through the cell cycle phases.

So in essence, the specific cyclin-CDK complexes that are active act as molecular markers, determining which phase of the cycle a cell is currently in based on their substrate activities. Their precise regulation is critical for proper cell cycle progression.

236
Q

Understand how cyclin-CDK activity ensures the cell cycle proceeds in only one direction

A
  1. Cyclical synthesis and degradation of cyclins:
    - Cyclins are synthesized and degraded in a specific temporal order
    - This allows activation of different CDK partners sequentially, not simultaneously
    - Prevents skipping phases or reversal of the cycle
  2. Substrate specificity of cyclin-CDK complexes:
    - Each cyclin-CDK complex phosphorylates a distinct set of substrates
    - This triggers phase-specific events like DNA replication, chromosome condensation etc.
    - Creates a unidirectional cascade of events
  3. Irreversibility of some events:
    - Some cyclin-CDK initiated events are irreversible once started
    - E.g. Breakdown of nuclear envelope in M phase
    - This commits the cell to completing that phase and the entire cycle
  4. Checkpoints and regulation:
    - Checkpoints restrict transitions until prerequisites are met
    - CDK activity is regulated by inhibitors and activating/inactivating phosphorylation
    - This precise control prevents reversal or skipping of phases
  5. Oscillation of cyclin levels:
    - Cyclin levels rise and fall in a wave-like pattern
    - Successive cyclin-CDK waves push the cycle in a unidirectional manner

So in summary, the ordered appearance, substrate specificity and regulated activity of cyclin-CDKs, coupled with the irreversibility of some events, ensures the cell cycle follows a strict unidirectional sequence of phases without reversal or phase skipping.

237
Q

cell cycle

A

The cell cycle is the series of events that take place in a cell leading to its division and duplication. It consists of interphase (G1, S, and G2 phases), during which the cell grows and replicates its DNA, and mitotic (M) phase, where the cell divides into two daughter cells through mitosis and cytokinesis.

238
Q

G0

A

G0 phase is a resting phase in the cell cycle where cells temporarily exit the cell cycle and enter a quiescent state. Cells in G0 are not actively dividing but may re-enter the cell cycle in response to specific signals or stimuli.

239
Q

G1

A

G1 phase is the first gap phase of the cell cycle, occurring after mitosis and before DNA synthesis (S phase). During G1 phase, the cell grows in size, synthesizes proteins, and prepares for DNA replication.

240
Q

S

A

S phase is the synthesis phase of the cell cycle, occurring after G1 phase and before G2 phase. During S phase, DNA replication takes place, resulting in the synthesis of an identical copy of the cell’s genetic material.

241
Q

G2

A

G2 phase is the second gap phase of the cell cycle, occurring after DNA synthesis (S phase) and before mitosis (M phase). During G2 phase, the cell continues to grow and prepares for cell division by synthesizing proteins and organelles needed for mitosis.

242
Q

M phase

A

M phase, or mitotic phase, is the phase of the cell cycle during which cell division occurs. It consists of two main stages: mitosis and cytokinesis. Mitosis is the process by which the nucleus of the cell divides into two daughter nuclei with identical genetic material. Cytokinesis is the division of the cytoplasm, resulting in the formation of two separate daughter cells.

243
Q

cyclin

A

Cyclin is a family of proteins that regulate the progression of the cell cycle by activating cyclin-dependent kinases (CDKs). Cyclins bind to CDKs, forming complexes that phosphorylate target proteins and drive the transitions between different phases of the cell cycle. Cyclin levels fluctuate during the cell cycle, peaking at specific stages to coordinate cell cycle events.

244
Q

cyclin-dependent kinase (CDK)

A

Cyclin-dependent kinase (CDK) is an enzyme that regulates the cell cycle by phosphorylating target proteins in complex with cyclin proteins.

245
Q

Understand how to think about gene regulation from a genome-wide view

A
  1. Global Regulatory Mechanisms:
    - Epigenetic modifications (DNA methylation, histone modifications) can influence gene expression across the entire genome.
    - Chromatin remodeling complexes can alter chromatin accessibility and structure, impacting multiple genes.
    - Global transcription factors and signaling pathways can coordinate the expression of gene networks.
  2. Regulatory Landscapes:
    - The genome contains regulatory regions (promoters, enhancers, insulators) that influence the expression of nearby genes.
    - These regulatory elements can interact with each other and with distant genes through chromatin looping.
    - The organization and interaction of these regulatory elements create complex regulatory landscapes.
  3. Cellular Context:
    - Gene regulation is highly context-dependent, varying based on cell type, developmental stage, and environmental conditions.
    - Each cell type has a specific gene expression profile determined by its unique combination of regulatory factors and chromatin state.
  4. Systems-Level Analysis:
    - High-throughput techniques (ChIP-seq, RNA-seq, ATAC-seq) can map regulatory elements, chromatin states, and gene expression levels across the entire genome.
    - Integration and analysis of these large-scale datasets can reveal global regulatory patterns and networks.
  5. Computational Approaches:
    - Bioinformatics tools and machine learning algorithms can identify regulatory motifs, predict regulatory interactions, and model gene regulatory networks.
    - Systems biology approaches can integrate diverse data types to understand the complex interplay of regulatory mechanisms.

To truly understand gene regulation from a genome-wide perspective, we need to consider the interplay of global regulatory mechanisms, the organization of regulatory landscapes, the cellular context, and leverage high-throughput data and computational approaches to uncover the complex systems-level regulatory networks that govern gene expression patterns across the genome.

246
Q

Apply gene expression and regulation to understand how multiple types of regulation
work together to ensure genome duplicated only once using two regulatory features of
cell cycle control:
* cell cycle progression in one direction; no reversing

A
  1. Cyclin-CDK regulation:
    • Cyclical synthesis and degradation of cyclins activate specific CDK partners at precise times
    • This creates an irreversible, unidirectional wave of cyclin-CDK activity driving the cycle forward
    • Substrate specificity of cyclin-CDKs triggers phase-specific, irreversible events like DNA replication
  2. Cell cycle checkpoints:
    • Checkpoints act as regulatory control points between phases
    • They monitor completion of prerequisites (e.g. DNA replication) before allowing progression
    • Checkpoints are enforced by inhibiting cyclin-CDK activity until requirements are met

Together, these regulatory mechanisms work in the following way:

  • In G1 phase, cyclin D/E-CDK complexes prepare the cell for S phase, but cannot initiate DNA replication directly.
  • The G1/S checkpoint ensures cyclin E-CDK2 is only activated once prerequisites (cell size, nutrients) are fulfilled.
  • Once activated, cyclin E-CDK2 and cyclin A-CDK2 initiate DNA replication by phosphorylating replication initiation factors.
  • However, their precisely timed cyclin oscillations ensure DNA replication occurs only once per cycle.
  • The S phase checkpoint monitors completion of replication before activating cyclin B-CDK1 for mitosis.
  • Cyclin B-CDK1 then triggers a cascade of irreversible events like nuclear envelope breakdown, committing to mitosis.
  • This unidirectional cyclin-CDK oscillation coupled with checkpoints coordinating phase transitions prevents re-replication or reversal.

So in essence, the orderly activation of cyclin-CDKs drives unidirectional progression, while checkpoints ensure genome duplication occurs exactly once before division, maintaining genomic integrity.

247
Q

Understand how to think about gene regulation from a genome-wide view

A
  1. Global Regulatory Mechanisms:
    - Epigenetic modifications (DNA methylation, histone modifications) can influence gene expression across the entire genome.
    - Chromatin remodeling complexes can alter chromatin accessibility and structure, impacting multiple genes.
    - Global transcription factors and signaling pathways can coordinate the expression of gene networks.
  2. Regulatory Landscapes:
    - The genome contains regulatory regions (promoters, enhancers, insulators) that influence the expression of nearby genes.
    - These regulatory elements can interact with each other and with distant genes through chromatin looping.
    - The organization and interaction of these regulatory elements create complex regulatory landscapes.
  3. Cellular Context:
    - Gene regulation is highly context-dependent, varying based on cell type, developmental stage, and environmental conditions.
    - Each cell type has a specific gene expression profile determined by its unique combination of regulatory factors and chromatin state.
  4. Systems-Level Analysis:
    - High-throughput techniques (ChIP-seq, RNA-seq, ATAC-seq) can map regulatory elements, chromatin states, and gene expression levels across the entire genome.
    - Integration and analysis of these large-scale datasets can reveal global regulatory patterns and networks.
  5. Computational Approaches:
    - Bioinformatics tools and machine learning algorithms can identify regulatory motifs, predict regulatory interactions, and model gene regulatory networks.
    - Systems biology approaches can integrate diverse data types to understand the complex interplay of regulatory mechanisms.

To truly understand gene regulation from a genome-wide perspective, we need to consider the interplay of global regulatory mechanisms, the organization of regulatory landscapes, the cellular context, and leverage high-throughput data and computational approaches to uncover the complex systems-level regulatory networks that govern gene expression patterns across the genome.

248
Q

Understand how growth factors control cell cycle entry
* Understand the Restriction Point and growth factor sensitivity/insensitivity

A

Growth Factor Control of Cell Cycle Entry:
- Cells require extracellular mitogenic growth factors to drive quiescent cells into the cell cycle
- Growth factors bind to cell surface receptors, activating intracellular signaling pathways
- These pathways lead to increased levels of G1 cyclins (Cyclin D) and CDK activation

Restriction Point:
- The restriction point or “start” is a key control point late in G1 phase
- Before this point, the cell is sensitive to growth factor levels and can exit the cycle
- After the restriction point, the cell becomes insensitive to growth factors
- It is now committed to completing the current cell cycle

Growth Factor Sensitivity vs Insensitivity:
- In early G1, the cell is growth factor sensitive - it requires mitogenic signals to progress
- If growth factors are removed before the restriction point, the cell exits the cycle back to G0
- After passing the restriction point in late G1, activation of Cyclin E-CDK2 commits the cell cycle
- Now the cell cycle progresses autonomously, insensitive to growth factor levels

So in summary:
- Growth factors drive quiescent cells past the restriction point in G1
- This commits the cell to replicating its DNA and dividing
- The restriction point separates the growth factor sensitive and insensitive periods
- It acts as a crucial irreversible switchpoint controlling cell proliferation

Understanding this restriction point control allows regulation of abnormal proliferative disorders.

249
Q

Be able to describe in how the regulation and activity of cyclin-CDKs ensures the genome is
replicated once and only once (proceeds in one direction and never reverses) and no DNA
damage is passed on

A
  1. Cyclin-CDK oscillation drives unidirectional progression:
    • Different cyclin-CDK complexes are activated sequentially in an oscillating pattern
    • This creates an irreversible, forward-driving cascade of events
    • E.g. Cyclin E/A-CDK2 for S phase entry, Cyclin B-CDK1 for mitosis
  2. Substrate specificity and irreversible events:
    • Each cyclin-CDK complex phosphorylates specific substrates for that phase
    • Some events initiated are irreversible once started (e.g. nuclear envelope breakdown)
    • This commits the cell to completing that phase and the entire cycle without reversal
  3. Regulated cyclin synthesis/degradation:
    • Cyclins have burst of synthesis followed by scheduled degradation
    • This ensures a cyclin-CDK is only active during its designated phase
    • Prevents re-replication by not allowing Cyclin E/A re-accumulation in S phase
  4. DNA replication licensing and checkpoints:
    • Origins of replication are “licensed” only once per cycle by controlling factors
    • The G1/S and intra-S checkpoints verify replication has not been initiated/completed
    • This ensures replication is not re-initiated if blocked/stalled
  5. DNA damage checkpoints:
    • The G2/M checkpoint prevents entry into mitosis if DNA is damaged
    • Gives time for DNA repair before segregating chromosomes
    • Halts the cycle to avoid passing on mutations to daughter cells

So the precise, transient regulation of cyclin-CDKs, coupled with licensing mechanisms and checkpoints, allows one round of replication per cycle while ensuring genomic integrity by repairing damage before division. This unidirectional, irreversible system maintains ploidy.

250
Q

What are all the details from the breakouts for then?  Be able to connect and explain how
various types of transcriptional, post-transcriptional, translational, and post-translational
regulation work together to produce cell cycle regulation – grow and divide when conditions
favorable, proceed one direction only, Restriction point, DNA damage checkpoints

A

Transcriptional Regulation:
- Transcription factors like E2F control expression of cyclin genes in a cell cycle-dependent manner
- E.g. E2F activates cyclin E/A for G1/S transition when conditions are favorable

Post-Transcriptional Regulation:
- Regulation of mRNA stability and localization impacts cyclin/CDK levels
- E.g. Cell cycle-regulated miRNAs can destabilize cyclin mRNAs

Translational Regulation:
- Regulation of translation initiation/elongation affects synthesis of cyclins/CDKs
- E.g. mTOR pathway mediates translation in response to growth factors

Post-Translational Regulation:
- Cyclin degradation, CDK inhibitor binding, CDK activating phosphorylation
- Critical for coordinating cyclin-CDK oscillations and activities

These multi-layered regulations work together:
1) To produce cyclins/CDKs only when extracellular conditions (growth factors/nutrients) are favorable for growth/division.

2) To generate precisely timed waves of cyclin-CDK activities that drive unidirectional, irreversible transition between phases.

3) To enforce the Restriction point in G1 - a point of no return after which the cycle proceeds autonomously.

4) To implement critical DNA damage checkpoints that halt the cycle until DNA lesions are repaired before segregating chromosomes.

So transcriptional/post-transcriptional control regulates cyclin synthesis, while translation and post-translational mechanisms facilitate the periodic cyclin-CDK activities - all converging onto the core cell cycle engine in response to internal/external cues. This integrated regulatory network ensures controlled proliferation and genomic integrity.

251
Q

quiescent state

A

The quiescent state, also known as G0 phase, is a period in the cell cycle where cells temporarily exit the cycle and remain in a non-dividing, resting state.

252
Q

target gene

A

A target gene is a specific gene that is regulated by a transcription factor or other regulatory protein, resulting in changes in its expression levels or activity.

253
Q

growth factor

A

A growth factor is a signaling molecule that stimulates cell growth, proliferation, and differentiation by binding to specific receptors on the surface of target cells and activating intracellular signaling pathways.

254
Q

restriction point

A

The restriction point, also known as the R point, is a key regulatory checkpoint in the G1 phase of the cell cycle. It represents the point beyond which progression through the cell cycle becomes independent of external growth signals. Once the cell passes the restriction point, it commits to completing the cell cycle and entering S phase, regardless of subsequent changes in external conditions.

255
Q

growth factor
sensitive/insensitive

A

A growth factor-sensitive cell is one that responds to the presence of a growth factor by initiating cell proliferation, growth, or differentiation. In contrast, a growth factor-insensitive cell does not respond to the same growth factor stimulus and remains quiescent or exhibits limited growth and proliferation.

256
Q

Describe stages of cancer and the Hallmarks of Cancer

A

Stages of Cancer:

1) Initiation - First mutagenic event leads to transformation of a normal cell to a tumor cell.

2) Promotion - Tumor cells with activated oncogenes/inactivated tumor suppressors start proliferating.

3) Progression - Additional mutations accumulate, tumor becomes invasive and metastatic.

Hallmarks of Cancer (Acquired Capabilities):

1) Sustained Proliferative Signaling - Cancer cells produce their own growth signals.

2) Evading Growth Suppressors - Defective pathways that normally inhibit cell proliferation.

3) Resisting Cell Death - Evading apoptosis pathways allows survival and accumulation of mutations.

4) Enabling Replicative Immortality - Upregulating telomerase to escape Hayflick limit.

5) Inducing Angiogenesis - Promoting growth of new blood vessels to supply nutrients.

6) Activating Invasion and Metastasis - Ability to invade local tissue and spread to distant sites.

7) Reprogramming Energy Metabolism - Preferentially using glycolysis (Warburg effect).

8) Evading Immune Destruction - Avoiding immune surveillance and destruction.

9) Genome Instability - Enabling mutational capability via defective DNA repair mechanisms.

10) Tumor-Promoting Inflammation - Inflamed tumor microenvironment supports other hallmarks.

These acquired capabilities, driven by genetic/epigenetic changes, allow cancer cells to survive, proliferate, and disseminate - the essence of the neoplastic disease process.

257
Q

Define proto-oncogene, oncogene, tumor suppressor

A

Proto-oncogene:
- A normal cellular gene that codes for proteins involved in regulation of cell growth and differentiation
- When mutated or overexpressed, it has the potential to become an oncogene that can cause cancer

Oncogene:
- A gene that has undergone mutation or dysregulation, causing a gain-of-function
- Oncogene products (proteins) allow uncontrolled cell growth and division
- Oncogenes are derived from proto-oncogenes by mutation or increased expression
- Examples: RAS, MYC, HER2/neu

Tumor Suppressor Gene:
- A gene that codes for proteins that inhibit cell division, repair DNA mistakes, or induce apoptosis
- Mutation or inactivation (loss-of-function) of these genes allows uncontrolled proliferation
- Both copies must be inactivated to fully disable tumor suppressor function (“two-hit” hypothesis)
- Examples: TP53, RB1, BRCA1/2, APC

In summary:
- Proto-oncogenes are normal genes that can be converted to oncogenes by mutation
- Oncogenes promote cell growth and division in an unregulated manner
- Tumor suppressor genes normally act as brakes on cell division and survival
- Inactivation of tumor suppressors removes these brakes, enabling cancer development

The interplay between activated oncogenes and inactivated tumor suppressors disrupts the normal regulation of the cell cycle and drives the uncontrolled proliferation characteristic of cancer.

258
Q

Describe types of DNA damage – UV  thymidine dimerreplication stress; double
strand breakschromosomal rearrangements: translocation, inversion, deletion

A

1) UV-Induced Thymidine Dimers:
- UV radiation from sunlight causes covalent linkages between adjacent thymines
- Forms cyclobutane pyrimidine dimers or 6-4 photoproducts
- Distorts the DNA helix and blocks replication/transcription

2) Replication Stress:
- Stalled or collapsed replication forks during DNA replication
- Can be caused by DNA lesions, depleted nucleotide pools, or tightly bound proteins
- Leads to exposure of single-stranded DNA vulnerable to breakage

3) Double-Strand Breaks (DSBs):
- Both strands of DNA helix are severed
- Highly cytotoxic if unrepaired
- Caused by ionizing radiation, reactive oxidants, replication errors

4) Chromosomal Rearrangements:
- Translocations: Swapping of genetic material between non-homologous chromosomes
- Inversions: Reversing the orientation of a chromosomal segment
- Deletions: Loss of a chromosomal segment

These chromosome abnormalities result from misrepair of DSBs or replication errors and can lead to:
- Loss of genomic material
- Gene fusions
- Dysregulation of gene expression
- Genomic instability promoting cancer

Proper DNA repair mechanisms like nucleotide excision repair, mismatch repair, homologous recombination, and non-homologous end joining are critical for maintaining genomic integrity by fixing these diverse DNA lesions.

259
Q

Describe the DNA damage response and damage checkpoints – basic steps sensing,
activate DDR kinase, DNA damage checkpoint: arrest cell cycle + repair + cell death

A

1) Sensing DNA Damage:
- Sensor proteins detect specific types of DNA lesions (e.g. ATR for ssDNA, ATM for DSBs)
- This initiates a signal transduction cascade

2) Activating DDR Kinases:
- Sensor proteins activate downstream transducer kinases like ATR, ATM, CHK1, CHK2
- These phosphorylate numerous substrate proteins to amplify the damage signal

3) DNA Damage Checkpoints:
- Key substrates are checkpoint proteins that induce cell cycle arrest
- Allow time for DNA repair before proceeding with replication/division

4) Cell Cycle Arrest:
- Checkpoints activate cell cycle inhibitors that inactivate cyclin-CDK complexes
- G1/S checkpoint prevents initiation of DNA replication
- Intra-S checkpoint stalls ongoing replication
- G2/M checkpoint prevents entry into mitosis

5) DNA Repair Pathways:
- Various repair mechanisms are activated in parallel (e.g. NER, MMR, HR, NHEJ)
- Repair proteins are recruited to damage sites by DDR proteins

6) Cell Fate Decision:
- If damage is repaired, checkpoints are deactivated and cell cycle restarts
- If damage is excessive/irreparable, apoptosis or senescence pathways are induced

So in summary, the DDR senses the DNA lesion, activates checkpoints to arrest the cell cycle, allows repair pathways to fix the damage, and then either restarts the cycle or triggers cell death - preventing propagation of mutations to daughter cells.

260
Q

Describe importance of DNA damage response in preventing cancer and mutator
phenotype

A

Preventing Mutations and Genomic Instability:
- The DDR detects and repairs various types of DNA lesions before they become mutations
- This prevents accumulation of oncogenic mutations in genes like proto-oncogenes and tumor suppressors
- It also fixes DNA breaks and errors that can lead to chromosomal abnormalities

Averting Mutator Phenotype:
- The mutator phenotype refers to a high spontaneous mutation rate in cancer cells
- It is often caused by defects in DNA repair pathways like mismatch repair (MMR)
- An intact DDR helps maintain low baseline mutation rates by repairing mistakes

Enforcing Cell Cycle Checkpoints:
- The DDR triggers cell cycle checkpoints to halt progression in presence of DNA damage
- This provides time for repair and prevents replication or segregation of damaged DNA
- Failure of these checkpoints leads to propagation of mutations

Inducing Apoptosis or Senescence:
- If the damage burden is too high, the DDR can induce apoptosis or senescence
- This eliminates severely mutated cells that could potentially become cancerous
- It prevents proliferation of genomically unstable cells

By coordinately detecting, signaling and promoting repair of DNA lesions, while eliminating heavily damaged cells, the DDR acts as a critical anti-cancer barrier. Defects in DDR pathways are observed in many cancer types and promote tumorigenesis through genomic instability.

261
Q

tissue border

A

A tissue border refers to the boundary between adjacent tissues or regions within an organism. It delineates the spatial separation between different types of tissues, such as epithelial and connective tissues, or between different functional regions within an organ. Tissue borders play important roles in maintaining tissue organization, function, and integrity, as well as regulating cellular interactions and signaling between adjacent tissues.

262
Q

benign tumor

A

A benign tumor is a mass of cells that grows slowly and remains localized without invading surrounding tissues or spreading to distant sites in the body. Unlike malignant tumors (cancer), benign tumors are typically well-defined, non-invasive, and do not metastasize to other parts of the body. While benign tumors are not cancerous, they can still cause health problems depending on their size, location, and effects on nearby tissues or organs.

263
Q

malignant tumor

A

A malignant tumor, or cancer, is a mass of abnormal cells that grow uncontrollably, invade surrounding tissues, and can spread to other parts of the body through metastasis.

264
Q

cancer

A

Cancer is a disease characterized by the uncontrolled growth and spread of abnormal cells in the body.

265
Q

invasive

A

Invasive refers to the ability of cells, tissues, or tumors to penetrate and spread into surrounding tissues or organs.

266
Q

metastasis/metastatic cancer

A

Metastasis, or metastatic cancer, refers to the spread of cancer cells from the primary tumor to distant sites in the body, where they form secondary tumors.

267
Q

oncogene

A

An oncogene is a gene that has the potential to cause cancer when mutated or overexpressed. These genes typically encode proteins involved in cell growth, proliferation, or survival, and their abnormal activation can drive uncontrolled cell growth and contribute to the development of cancer.

268
Q

tumor suppressor

A

A tumor suppressor is a gene that encodes a protein involved in inhibiting cell growth, proliferation, or survival. These proteins help regulate the cell cycle, repair damaged DNA, and promote programmed cell death (apoptosis), thereby preventing the development of cancer. Loss or inactivation of tumor suppressor genes can lead to uncontrolled cell growth and contribute to the formation of tumors.

269
Q

nucleotide crosslink (T-T)

A

A nucleotide crosslink, specifically a T-T crosslink, refers to a covalent bond formed between two adjacent thymine (T) bases on the same DNA strand due to exposure to certain environmental agents, such as ultraviolet (UV) radiation. These crosslinks can distort the DNA structure and interfere with DNA replication and transcription, potentially leading to mutations or DNA damage if not repaired properly.

270
Q

apoptosis = programmed cell death

A

Apoptosis is a programmed cell death process that occurs in multicellular organisms as a mechanism to remove unwanted or damaged cells.

271
Q

DNA damage checkpoint (G1/S, S phase, G2/M)

A

DNA damage checkpoints, including the G1/S, S phase, and G2/M checkpoints, are regulatory mechanisms in the cell cycle that monitor the integrity of the DNA and prevent the progression of cell cycle stages if DNA damage is detected.

272
Q

mutator phenotype

A

A mutator phenotype refers to a condition in which cells have an increased rate of mutation compared to normal cells, resulting from defects in DNA repair mechanisms or other factors that promote genomic instability.

273
Q

Practice gene regulation and apply to understand one example of how chromosomal
rearrangements due to DNA DSBs can lead to cancer development
1. Be able to understand if given a DIFFERENT rearrangement with different gene(s)
and what the consequence would be. A new example is on PS6.
2. Be able to understand that some mutants/rearrangements only matter in certain
cell types if the genetic change happens to a cell type specific regulatory
element.

A
  1. For a given chromosomal rearrangement involving different genes:
    - I will consider the normal functions of those genes and how their dysregulation could contribute to hallmarks of cancer
    - For example, if the rearrangement creates a fusion oncogene by joining two proto-oncogenes, it could lead to uncontrolled proliferative signaling
    - Or if it deletes/truncates a tumor suppressor gene, it removes that critical brake on cell division/survival
  2. I understand that the consequences of some mutations/rearrangements may be cell type-specific due to differential gene regulation:
    - Some genes are only expressed/required in certain cell types based on the cell’s transcriptional program
    - Their regulatory regions like enhancers/promoters may also be cell type-specific
    - So a rearrangement disrupting regulatory elements would impact gene expression differently across cell types
    - This could explain why some cancers originate from and are limited to a particular cell lineage

Additionally, I’ll consider how the genetic lesion may alter chromatin topology/organization and impact regulatory landscapes genome-wide in that cell type. The complex interplay between DNA sequence changes and transcriptional regulatory programs determines the ultimate oncogenic outcome.

274
Q

Practice gene regulation and apply to understand how the DNA Damage Response is
activated in response to DSBs in G1/S and G2/M – it’s similar for both
- Understand why the DNA damage response inhibits (e.g. cyclin-CDK) or activates
particular proteins (repair proteins, cell cycle arrest proteins, etc.) to properly
regulate cell cycle in presence of damage

A

G1/S Checkpoint Activation:
- In G1, sensor proteins like the MRN complex detect DSBs and activate the kinase ATR
- ATR phosphorylates and activates the effector kinase CHK1
- Active CHK1 inhibits CDC25A phosphatase that normally removes inhibitory phosphorylations on cyclin-CDK complexes
- This prevents activation of cyclin E/A-CDK2, blocking initiation of DNA replication
- CHK1 also activates the transcription factor p53, which induces p21 (CDK inhibitor)
- p21 inhibits cyclin E/A-CDK2, reinforcing the G1/S cell cycle arrest

In parallel:
- ATR and CHK1 activate repair proteins like BRCA1 to initiate DNA repair pathways
- They also induce expression of genes involved in nucleotide metabolism, dNTP production

G2/M Checkpoint Activation:
- In G2, DSBs are sensed by ATM and the MRN complex
- Activated ATM/ATR phosphorylate CHK1/CHK2 effector kinases
- CHK1/2 inhibit CDC25C, preventing its removal of inhibitory phosphates on cyclin B-CDK1
- This maintains cyclin B-CDK1 in an inactive state, preventing entry into mitosis
- p53 is also activated, inducing proteins like 14-3-3 sigma that sequesters cyclin B-CDK1

Simultaneously:
- ATM/ATR activate DNA repair pathways like homologous recombination and non-homologous end joining
- Cell cycle arrest allows time for repair before segregating damaged chromosomes

So in summary, upon DSB detection, the DDR activates specific transcriptional programs and post-translational mechanisms to inhibit cyclin-CDK activities, induce cell cycle checkpoints, and simultaneously activate DNA repair pathways - an integrative response ensuring genomic integrity is maintained.

275
Q

Given a scenario/mutation or change in cell cycle regulation or DNA damage response
explain the consequence. Also be able to classify genes/proteins as oncogenes or tumor
suppressors given their function in the scenario.
1. Remember we got rid of proto-oncogene and just using normal/mutant
oncogene and normal/mutant tumor suppressor

A

For any given scenario, I will:

1) Identify the specific gene(s)/protein(s) impacted and their normal roles in cell cycle control or DNA repair pathways.

2) Classify whether the gene is typically an oncogene or tumor suppressor based on its function:
- Oncogenes: Promote cell proliferation/survival when mutated/overactive
- Tumor suppressors: Inhibit cell division/induce apoptosis when mutated/inactivated

3) Determine if the scenario represents:
- An activating mutation in an oncogene
- An inactivating mutation/loss in a tumor suppressor

4) Explain the downstream consequences by reasoning how that genetic change would disrupt normal regulation:
- For oncogene mutation: Uncontrolled proliferative signaling, evading growth suppression, etc.
- For tumor suppressor mutation: Inability to induce cell cycle arrest, apoptosis resistance, etc.

5) Discuss how the specific defect enables hallmark capabilities of cancer like sustained proliferation, evading cell death, genomic instability, etc.

6) For cell cycle/DDR mutations, analyze how the regulatory Circuit is rewired, like:
- Inability to activate checkpoints
- Failure to inhibit cyclin-CDK activities
- Impaired DNA repair pathways

276
Q

positive and negative feedback loops – be able to recognize them

A

Positive Feedback Loop:
- In a positive feedback loop, the initial change/stimulus leads to a response that further amplifies or reinforces the original stimulus.
- This causes an increasing, self-perpetuating cycle that continues until some limiting factor intervenes.
- Example: Blood clotting process - clotting factors activate enzymes that produce more clotting factors, amplifying the clotting cascade.

How to recognize:
- The response or output of the loop acts to increase/intensify the original input signal
- Creates a cyclical, amplifying effect that accelerates a particular direction/outcome
- Often requires an external brake or limitation to terminate the self-amplifying cycle

Negative Feedback Loop:
- In a negative feedback loop, the initial change/stimulus elicits a response that counteracts or negates the original stimulus.
- This creates a self-correcting cycle that opposes and reverses any deviation from the desired set-point.
- Example: Regulation of blood glucose levels - high glucose triggers insulin release which acts to lower glucose levels.

How to recognize:
- The output or response reduces/opposes the original input stimulus
- Creates a cyclical, dampening effect that stabilizes a particular parameter around a set value
- Promotes equilibrium and homeostasis by reversing deviations in either direction

So in summary:
- Positive loops amplify and reinforce changes in one direction until limited
- Negative loops counteract changes to maintain constant values/states
- Analyzing whether the cycle output enhances or diminishes the initial stimulus reveals the loop type

277
Q

DO NOT MEMORIZE ATM-p53 pathway. Be able to read the diagram and connect and apply all
the previous content in shown in that diagram from gene expression to PTMs to microRNAs to
cell cycle control to cancer.

A
278
Q

tRNA abundance often correlates with codon usage
tRNAs for common codons are more abundant
tRNAs for rare codons are less abundant

A
279
Q

Summary of translational regulation

A

1) There are many proteins
that work with ribosome
that can be regulated and
promote/inhibit translation
initiation and elongation
IMPORTANT! Codon usage is not something cells actively
change! This comes from genome evolution!
tRNA abundance b/c they are transcribed can be
regulated like gene expression.

280
Q

Post-translational regulation: Post-translational
Modifications (PTMs)

A
  1. Phosphorylation - add phosphate to serine, threonine, or tyrosine aa
  2. Acetylation – add acetyl to lysine aa
  3. Methylation – add methyl to lysine aa
  4. Glycosylation – add sugar to aa
  5. Ubiquitination – add of ubiquitin (small protein) to lysine aa
  6. Palmitoylation - Add fatty acid cysteine aa

Enzymes
add | remove

Kinase | phosphatase
AcTransferase | deacetylase
MeTransferase | demethylase
don’t care
Ub Ligase | Deubiquitinating Enz
don’t care

281
Q
  1. Are PTMs added to proteins covalently or through weak electrostatic interactions?
  2. Are the PTMs permanent modifications of proteins?
  3. Do PTMs need enzymes to be added to proteins?
A

Covalently
No
Yes

282
Q

Summary of regulatory points

A

Transcription (all at initiation)
- chromatin accessibility of the gene
*euk only (though bac do have
”chromatin”)
- DNA methylation *both
- activators, repressors, silencers,
enhancers, operators *both
Post-transcriptional
- alternative splicing of mRNA *euk only
- mRNA modification *both
- but polyA and 5’cap are *euk only
-5’ and 3’ UTR regulation *both
- small nc-RNAs binding to UTRs *both
but siRNA only euk

283
Q
  1. Codon usage bias can occur because the genetic code is redundant. T/F
A

T

284
Q

Codon usage bias is when a particular codon for an amino acid is used more than other codons for the
same amino acid. T/F

A

T

285
Q

The presence of common codons in a protein coding sequence will make gene expression more
efficient. T/F

A

T

286
Q
  1. Codon usage is something cells can actively change to regulate gene expression. T/F
A

F

287
Q
  1. Protein post-translational modifications are used to remodel chromatin. T/F
A

T

288
Q
  1. Protein PTMs are covalently added but they are not necessarily permanent. T/F
A

T

288
Q
  1. Can we think of PTMs as being an ON/OFF switch for protein activity? Y/N
A

Y

289
Q

RNA vs. DNA polymerases

A

RNA polymerases do not need primers

RNA polymerase is like an enzyme
w/TWO hands – each can grab a free
nucleotide and bring them together.
Can start a strand from scratch

DNA polymerases need an RNA primer

Primer is a short strand of complementary
RNA or DNA. It provides an existing
3’OH for DNA polymerase to add to

DNA polymerase is like an enzyme
w/ONE hand – can only grab one free
nucleotide and add to existing strand

289
Q

DNA Replication Initiation

A

Purpose – melt the dsDNA at origin to produce a replication bubble that contains specific
proteins needed for replication elongation

290
Q

The Cell Cycle Controls Euk. Cell Division

A

G1 phase – cell grows
S phase – Synthesis: DNA replication
G2 phase – cell grows more &
prepares for mitosis
Mitosis – division of duplicated
chromosomes
Cytokinesis – division of cytoplasm
G0 – cell has exited cell cycle and not
dividing

290
Q

DNA Replication Elongation Leading Strand

A

Purpose – replicate DNA by synthesizing complementary bases to the template (parent) strand

291
Q

DNA Damage: UV Damage

A

UV damage causes covalent crosslinks
between two adjacent thymidines
Thymidine dimers cause a kink in helix
and block DNA polymerases but not helicases

291
Q

What does the Cell Cycle control?

A

1) Ensures replicate genome once
and only once.
1a Cell Cycle occurs in one direction
only – no reversing!
1b Once the decision to divide is made
cells cannot exit the cell cycle until
complete
2) Separate replicated genome into
two nuclei (mitosis)
3) No DNA damage or unwanted
changes to the genome passed
on

291
Q

Cyclin-CDKs control cell cycle progression

A
  • Cyclin-dependent kinases (CDK) phosphorylate cell cycle proteins to
    allow cells to progress through the cell cycle
  • Cyclin – protein regulatory subunit of cyclin-dependent kinases (CDK)
    required for CDK activity
292
Q

6 Hallmarks of cancer

A

proto-oncogene – a gene whose normal
function promotes cell division or survival
oncogene – mutant form of a proto-
oncogene that causes too much cell division
or inappropriate survival
tumor suppressor – a genes whose normal
function is to inhibit cell division, induce
cell death when necessary, or repair DNA
damage

293
Q

There is a unique
Cyclin-CDK for each
phase

A

Cyclin-CDK of previous phase controls activity of next phase’s cyclin
and ensures proceed in one direction only

– each cyclin-CDK regulates the activity of the next cyclin.
– cannot control the cyclin for the previous phase
– Ensures cell cycle proceeds in one direction

294
Q

DNA Damage: Replication Stress from UV TT

A

Replication Stress – any type
of damage or alteration that
causes the excessive ssDNA
formation due to DNA
polymerase blockage

294
Q

DNA Damage: DSBs can cause chromosomal rearrangements & deletions

A
  1. Chromosomal Translocations:
    - DSBs on two different chromosomes can be incorrectly joined by error-prone non-homologous end joining (NHEJ) repair
    - This creates an abnormal fusion chromosome, known as a translocation
    - Translocations can lead to formation of chimeric fusion genes with dysregulated function
    - Example: BCR-ABL fusion in chronic myeloid leukemia is an oncogenic translocation
  2. Inversions and Deletions:
    - A single DSB that is mis-rejoined can cause inversion of the DNA sequence orientation
    - Or if the broken ends fail to rejoin, it results in a chromosomal deletion
    - Both events can remove or disrupt coding regions, regulatory elements like enhancers/promoters
    - Loss of critical genes like tumor suppressors facilitates cancer development
  3. Gene Dysregulation:
    - Chromosomal rearrangements can also reposition genes near different regulatory regions
    - This can abnormally activate proto-oncogenes or silence tumor suppressor genes
    - Example: Deletion of CDKN2A locus deregulates cell cycle inhibitors in many cancers
  4. Genomic Instability:
    - Improper DSB repair also generates vulnerabilities for further breakage events
    - This genomic instability fuels acquisition of compounding mutations
    - The “mutator phenotype” is a hallmark enabling other cancer hallmarks

So in summary, the major consequence of mis-repaired DSBs is the generation of chromosomal abnormalities that deregulate gene expression, disrupt critical genes, produce fusion oncogenes, and enable genomic instability - all of which are driver events in cancer initiation and progression.

295
Q

DNA Damage: Double-Strand Breaks (DSB)

A

Gamma irradiation, X-rays – causes DNA double
strand breaks; the phosphate sugar backbone is
broken on both strands.

One of most harmful types of DNA damage bc
large pieces of genome can be lost or
rearranged

295
Q

Mutations in DNA damage response enables accumulation of mutation

A

mutator phenotype: increased rate of mutations and survival of cells with mutations

296
Q

DNA Damage Response
protects our genome

A
  1. Proteins sense DNA damage: Specialized proteins detect DNA damage within the cell.
  2. ATM, ATR, and DNA-PK kinases: These are enzymes that play key roles in the DNA damage response. They activate various cellular responses when DNA damage is detected.
  3. DNA damage checkpoints: Upon sensing DNA damage, the cell cycle can be arrested at specific checkpoints to allow time for repair before cell division proceeds.
  4. Repair the damage: Various mechanisms exist within the cell to repair DNA damage, such as base excision repair, nucleotide excision repair, and homologous recombination.
  5. Apoptosis: If the damage is too severe and cannot be repaired, the cell may undergo programmed cell death, or apoptosis, to prevent the propagation of damaged DNA.
296
Q

All tumors are considered cancer. T/F

A

F

297
Q

DNA Damage Response inhibit cyclin-CDKs to keep
cells from progressing through cell cycle = DNA Damage Checkpoints

A

In addition to “pausing” the cell cycle, DNA Damage Response activate repair mechanisms
to fix damage and/or kill the cell if too much damage occurs

298
Q

Some transcription factors are cell type specific and only expressed in certain cell types. T/F

A

T

298
Q

Chromosomal rearrangements are the result of DNA damage. T/F

A

T

299
Q

DNA Damage Checkpoints Ensures No DNA damage is passed down

A

Damage checkpoints will arrest the cell cycle by inhibiting cyclin-CDK so DNA
can be repaired. Once repaired, the cell cycle will resume. If damage is too
extensive cell will initiate programmed cell death – apoptosis. This is the last cell
cycle purpose

300
Q

Chromosomal rearrangements can delete or swap the regulatory elements controlling gene
expression. T/F

A

T

301
Q

The DNA Damage Response (DDR) arrests the cell cycle by inhibiting cyclin-CDK activity. T/F

A

T

302
Q

To think about the consequence of #1-2, we need to know whether the regulatory element
uses a cell type-specific TF and what cell type we are considering. T/F

A

T

303
Q

We should memorize how ATM and p53 signaling works. T/F

A

F

304
Q

When the last RNA primer is removed on the lagging strands there is a problem because
DNA pol I cannot fill in the gap. T/F

A

T

304
Q

When replication reaches the end of the linear template, BOTH newly synthesized DNA
molecules have a problem. T/F

A

T

305
Q

The result of the problem is that ends of linear chromosomes have unreplicated DNA. T/F

A

T

305
Q

It is always the 5’ end of the new DNA that was synthesized as the lagging strand that has
a problem. T/F

A

T

306
Q

What function might telomere sequences provide to combat the end replication
problem?

A

protective cap/ buffer that can be lost before genes are lost.

307
Q

TELOMERE REPLICATION

A
  1. Telomerase has a
    repetitive RNA
    template
    complementary to
    telomeres
  2. Telomerase possesses
    DNA pol activity

The RNA template bound to telomerase serves TWO functions:
a) brings telomerase only to telomeres bc the RNA template can only bp
with telomere sequence
b) serves as the template for telomerase to synthesize telomere repeat

308
Q

One purpose of the cell cycle is to ensure no DNA damage is present or passed on. T/F

A

T

309
Q

Synthesize across lectures. The telomerase gene has a promoter with many CG’s. Think back
to Lec10-1. In a somatic cell that should divide a limited number of times would the promoter
be methylated or unmethylated? What about a cancer cell with high telomerase expression?

A
  1. Telomerase is a reverse transcriptase that adds telomeric repeats to chromosome ends. Its expression allows unlimited replicative potential.
  2. In normal somatic cells that should divide only a limited number of times (Hayflick limit), telomerase expression is tightly repressed.
  3. Promoters with high CG density can be regulated by DNA methylation - a repressive epigenetic mark.

Putting this together:
In a normal somatic cell undergoing senescence, the telomerase promoter would likely be heavily methylated to keep its expression turned off. This methylation prevents the cell from activating telomerase and becoming immortal.

Conversely, in a cancer cell with high telomerase expression (a hallmark of most cancers), the telomerase promoter would be unmethylated or hypomethylated.

This lack of repressive methylation, combined with activating mutations in the promoter and/or transcriptional machinery, allows for aberrant upregulation of telomerase in cancer cells.

The unmethylated, active state of the telomerase promoter in cancer enables telomere maintenance and unlimited replicative potential - one of the key acquisitions that permits unconstrained proliferation of the malignant cells.

So in summary, proper methylation-mediated silencing of the telomerase promoter enforces the mortality of normal cells, while its hypomethylation in cancers contributes to the immortality and uncontrolled growth of these cells by permitting telomerase expression.

310
Q

Cyclin-CDK refers to just one protein. T/F

A

F

310
Q

CDK is a kinase, which means it is an enzyme that can phosphorylate proteins. T/F

A

T

311
Q

The cell cycle ensures the genome is copied exactly once each round of cell division. T/F

A

T

312
Q

The identity of the cell cycle phase is essentially determined by which cyclin-CDK has the
highest activity. T/F

A

T

313
Q

Restriction Point (R) is a point of no return. Must complete S
phase and the cell cycle. Cell is committed!

A

Restriction Point is the threshold of S cyclin concentration that commits the cell to
enter and complete the cell cycle. EVEN IF conditions change and nutrients are not
available the cell will continue to progress after the restriction point.
Before the restriction point cells are growth factor sensitive but after the restriction point
cells are growth factor insensitive

313
Q

Each cyclin-CDK will activate specific genes
needed in each phase. It could do this by
controlling the transcription of these genes. T/F

A

T

314
Q

Since cyclin-CDK is a kinase, which of these
will it likely post-translationally modify to
activate genes?

A

activators and repressors

315
Q

Let’s consider the transition from G1 phase to S phase (blue box). For the change from low
to peak S-CDK activity (yellow line) which might be true?
a. Turn on gene expression of S cyclin
b. Turn on gene expression of CDK
c. Post-translationally modify S-CDK to active form
d. Turn off a negative regulator of S cyclin expression
e. Turn on a positive regulator of S cyclin expression
f. all of these would increase S-CDK activity

A

f

315
Q

Summary: Each cyclin is post-translationally regulated by
protein stability & localization

A

PTMs used to enact post-translational regulation varies among cyclin-CDK.
Localization varies as well: cytoànucleus vs nucleus à cyto vs. cyto<–> nucleus
Do not memorize. Understand

316
Q

Cells grow & divide only when conditions favorable

A
317
Q

Cells grow & divide in response to external growth factor signals

A

Growth factors activate a
kinase that phosphorylates
the transcription factor
c-myc.
Phosphorylated c-myc can
bind to its regulatory
element

When growth factor is
absent, a phosphatase is
active that
dephosphorylates c-myc.
c-myc can NOT bind to its
regulatory element

317
Q

DNA Damage Checkpoints Ensures No DNA damage is passed down

A

Damage checkpoints will arrest the cell cycle by inhibiting cyclin-CDK so DNA
can be repaired. Once repaired, the cell cycle will resume. If damage is too
extensive cell will initiate programmed cell death – apoptosis. This is the last cell
cycle purpose

318
Q

Post-translational modifications of transcription factors can control transcription of a
gene. T/F

A

T

318
Q

Cell cycle progression, going from one phase to the next, is controlled transcriptionally
by cyclin-CDKs. Where one cyclin-CDK activates transcription of the next phase’s cyclin.
T/F

A

T

319
Q

Spindle proteins would be a likely phosphorylation target of which cyclin-CDK?

A

M-CDK

319
Q

Cyclin proteins are synthesized in the cytoplasm and would need to be imported into the
nucleus to phosphorylate transcription factors. T/F

A

T

320
Q
  1. Imagine Protein A activates Protein B which inhibits Protein C. To activate protein C:
A

A must be inhibited

320
Q

If cells were always responsive to external conditions like growth factor, the cell cycle
could reverse directions and result in replicating parts of the genome more than once. T/F

A

T

320
Q

If S-CDK proteins were not destroyed at the same time that M-CDK activity increases,
DNA replication might occur again during mitosis. T/F

A

T

320
Q

Genes that work together in cell cycle phases often have very similar regulatory
elements controlling their transcription. T/F

A

T

320
Q
  1. Phosphorylation is always activating. T/F
A

F

321
Q

Chromatin remodeling enzymes (e.g. HDAC, HAT, HM, DM) can be localized to specific
genes by interacting with transcription factors that can bind to DNA sequences regulating
those genes. T/F

A

T

322
Q

Imagine in a parallel universe that DNA polymerases evolved to be able to function just like RNA polymerases except they still incorporate DNA nucleotides. Answer the following questions:

  1. Linear chromosomes would no longer get shorter each round of cell division True .
  2. Primase would still be needed for DNA replication False .
A
322
Q

The leading strand only requires a single priming event.

A

True

322
Q
  1. There is a restriction point at the boundary of every cell cycle phase. T/F
A

F

323
Q

The main difference between bacteria and eukaryotic DNA replication initiation is ___.

A

only one origin in bacteria while 100s-1000s of origins in eukaryotes

323
Q

A scientist is trying to express a synthetic gene sequence they created. They have the following data points:

The gene is transcribed.
The mature mRNA does not exist very long once it is produced.
What could explain their data?

A

An siRNA is binding to the mRNA’s UTR.

323
Q

One scientist estimates there are 20,000 proteins produced by the human genome. A second scientist contests this and estimates there are 50,000 proteins produced by the human genome. What is a plausible explanation for the different estimates?

A

one scientist includes alternative splicing

324
Q

Which of the following, if missing, would usually prevent translation initiation in eukaryotes?

A

5’G-cap

324
Q
  1. A bacteria genome consists of one linear chromosome. False
  2. Eukaryotic genomes tend to have more non-coding sequence than bacteria genomes. True
  3. Bacteria genomes are usually smaller than eukaryotic genomes. True
  4. Bacteria genes are continuous rather than non-continuous. True
  5. Bacteria genes are always in operons. False .
  6. Eukaryotes can transcribe and translate a gene simultaneously. False
A
324
Q

Which of the following would NOT result in negatively regulating transcription?

A

histone acetylation

324
Q

Four bacteria proteins are controlled by an operon.
All four proteins are needed to synthesize compound Z.
Compound Z is required when oxygen is absent.
Compound Z is not needed when oxygen is present.
Answer the following questions.
1. How many promoter(s) control the expression of the four proteins? 1
2. What will assemble (bind) at the promoter(s) if transcription is active? sigma factor and RNAP
3. For this particular scenario, what would be the optimal way to control the operon? repressor with oxygen as a co-repressor
4. You do an experiment and delete the operator from the operon. Now, you expect that in the presence of oxygen, transcription of the four genes would be on .
5. When RNA pol transcribes the operon, how many transcripts will be produced for the four genes? 1

A
325
Q

Only one of the two replication forks from an origin of replication will produce Okazaki fragments.

A

False

326
Q

We can take a unique gene Z from species 1 and insert it into species 2, which never had the sequence for gene Z. Assume we put all the necessary transcription regulatory elements required, and species 2 transcribes gene Z. Species 2 will be able to translate the mRNA and produce the correct polypeptide.

A

True

327
Q

Gene 1 encodes for a protein that contains 50 histidines in the protein sequence. Scientists did an experiment where they created three versions of Gene 1 and inserted it into species X.

Version 1: 25 histidines use the codon CAU and the other 25 histidines use CAC.

Version 2: 5 histidines use the codon CAU and the other 45 histidines use CAC.

Version 3: 45 histidines use the codon CAU and the other 5 histidines use CAC.

The scientists find version 2 produced more protein than version 3. Answer the following questions:

  1. Which codon is the rare codon? CAU
  2. If the scientists inserted the three versions of gene 1 into species Z, will the rare codon necessarily be the same No could be different. ?
  3. Two different codons exist for histidine because of redundancy in the genetic code .
  4. When the scientists made the three versions of the gene, they made what type of mutations? silent
A
328
Q

When a linear chromosome is replicated, only one of two duplicated chromosomes has unreplicated DNA at the ends.

A

False

329
Q

Imagine the S cyclin-CDK could control transcription of the M cyclin. Cells would now be able to enter the cell cycle and divide uncontrollably.

A

False

330
Q

DNA virus genomes can only be replicated when the host genome is replicating.

In a cell that is NOT actively cycling through the cell cycle, how could a virus induce cell cycle entry even in the absence of growth factor? Inhibit a repressor of G1 cyclin transcription

A

Inhibit a repressor of G1 cyclin transcription

331
Q

Telomere protein 1 binds to telomere sequences of chromosomes and binds single-stranded binding protein (SSBP). This causes the telomere to fold over. The second drawing shows what happens when telomeres get too short - it cannot be folded over..

  1. In which case would the DNA damage response be activated? short telomere length .
  2. Which kind of DNA damage would a short telomere be similar to?
    [ Select ]
  3. Which correctly describe function(s) of the DNA damage response?
    [ Select ]
  4. Telomere protein 2 inhibits DNA repair proteins at telomeres. Because DNA repair is inhibited at telomeres, the DNA damage response does not turn off when telomeres get short. This situation explains which of the following Hayflick’s Limit

Answer 1:
Correct!
short telomere length
Answer 2:
Correct!
replication stress
Answer 3:
Correct!
A and C
Answer 4:
Correct!
Hayflick’s Limit

A
332
Q

The most detrimental type of DNA damage is

A

double-stranded DNA breaks

333
Q

A scientist growing normal non-cancerous human cells in the lab forgets to add growth factor to the culture media. The cells are initially all in various phases. Answer the following questions.

Cells in G1 will go to G0 (or stay in G1) True .
Cells in S phase will arrest until growth factor is added back False .
Cells in G2 will complete the cell cycle and then go to G0 True .
Cells in M will arrest until growth factor is added back False .
Eventually all the cells will end up in G0/G1 True .
Answer 1:
Correct!
True
Answer 2:
Correct!
False
Answer 3:
Correct!
True
Answer 4:
Correct!
False
Answer 5:
Correct!
True

A
334
Q

Transcribing a DNA sequence to produce the complementary mRNA sequence is a pre-requisite
objective from BIOL141. Here is a problem to make sure you can do it. Be sure you label the 5’
and 3’ end of your mRNA transcript and that you transcribe the correct strand.

A

3’ tactagtggtttttggcgttggacgtgagcggg 5’ template strand
5’ atgatcaccaaaaaccgcaacctgcactcgccc 3’ coding strand
5’ augaucaccaaaaaccgcaaccugcacucgccc 3’ mRNA transcript

335
Q

T/F Bacteria RNAP does not need a general transcription factor to bind to promoters
directly while Eukaryotic RNAP requires general transcription factors.

A

False. Sigma factor
needed, and it is a general TF

336
Q

T/F Eukaryotic transcription and translation are coupled.

A

F, need to be explored from
nucleus, spliced, and modified

337
Q

T/F Bacterial genes are always organized in operons.

A

F, can be one gene controlled by one
promoter

338
Q

T/F Eukaryotic genomes are much larger and contain more coding sequence than non-
coding sequence.

A

F, first part true, but more non-coding than coding in euk

339
Q

T/F RNAP synthesizes RNA 3’ to 5’ and uses the template strand 5’ to 3’.

A

F, RNAP synthesizes
5’ to 3’. Since it must make a strand complementary to its template, the template must be 3’ to
5’ b/c strands of nucleic acids are always anti-parallel

340
Q

T/F Transcription factors unwind DNA in order to bind specific sequences.

A

F, bind double-
stranded DNA in major or minor grooves.

341
Q

T/F Bacterial chromosomes are naked and not compacted.

A

F, they have histone-like proteins
and are compacted

342
Q

You delete an important histone acetylase gene in a mouse model and discover severe
developmental defects. Explain why deletion of the gene likely led to these defects. Be thorough
and detailed in your explanation. Pretend you are trying to show off your knowledge!

A

Acetylated histones causes chromatin to loosen (or to be converted to euchromatin) and be more
accessible by TFs. This allows genes to be transcribed. Without the histone acetylase, chromatin
could not be remodeled to euchromatin and might remain methylated or heterochromatinized.
This would prevent TFs from being able to bind and keep genes off during development that
need to be on.

343
Q

In XX individuals, one of the X chromosomes is inactivated in every cell – none of the genes
are expressed. The histones of the inactivated X chromosome is most likely

A

heavily methylated

344
Q

The primary difference between conservative and semiconservative DNA replication models
is that ________.

A

parental strands are reunited in the conservative model but remain with newly synthesized
daughter strands in the semiconservative model

345
Q

The leading and the lagging strands differ in that ________.

A

the leading strand requires a single RNA primer, and the lagging strand requires multiple
priming events

346
Q

Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a
consequence of which of the following?

A

Gaps left at the 5’ end of the lagging strand

347
Q

Telomere shortening puts a limit on the number of times a cell can divide. T/F

A

T

348
Q

. DNA polymerase cannot create a polynucleotide without a 3’OH created by a primer. T/F

A

T

349
Q

. DNA and RNA polymerases add new nucleotides to the 3’OH of the previous nucleotide. T/F

A

T

350
Q

. Bacteria DNA replication does not have a lagging strand. T/F

A

F

350
Q

. The enzyme that joins Okazaki fragments is primase. T/F

A

F

351
Q

Each replication of origin produces two replication forks, causing DNA to be replicated in
both directions from the forks. T?F

A

T

352
Q

. The leading strand is always the top strand being synthesized. T/F

A

F