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Gene Technology > L4-6: Regulation of Gene Expression > Flashcards

L4-6: Regulation of Gene Expression Flashcards

1
Q

Each chromosome consists of two identical _______.

A

chromatids

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2
Q

True or false:

The chromosomes in a pair are identical to each other.

A

False. They are homologous but not identical.

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3
Q

What type of cells are haploid?

A

Gametes

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4
Q

There are approximately _____ different transcription factors in humans.

A

5,000

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5
Q

Which reporter gene is best for analysis of diurnal regulation of genes?

A

Luciferase (Luc) as it is highly unstable, therefore level of mRNA transcripts = level of protein.
GFP is very stable, hence is not a good indicator of diurnal regulation

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6
Q

Each coding region in eukaryotes starts with which codon?

A

ATG (start codon)

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7
Q

Why is codon optimisation necessary

A

Amino acids can be encoded by different codons. Different organisms exhibit bias towards using certain codons over others for the same amino acid.

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8
Q

Construction of expression vector - elements

A
  1. Promoter specific for organ and cell-specific expression
  2. Compartmentalisation - specific signal peptide needed for specific cell compartments
  3. Strong ATG (Kozak consensus sequence)
  4. . Codon optimisation
  5. Terminator
  6. Affinity tag
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9
Q

Expression Vector for Vaccines Production - elements

A
  1. Promoter specific for organ and cell-specific expression
  2. Immunogenicity agent - highly immunogenic element
  3. Compartmentalisation - specific signal peptide needed for specific cell compartments
  4. Strong ATG (Kozak consensus sequence)
  5. Codon optimisation
  6. Terminator
  7. Affinity tag
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10
Q

A common immunogenicity agent used in vaccine vector production is _______

A

Cholera toxin B

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11
Q

Chromatid

A

One half of two identical copies of a replicated chromosome

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12
Q

What are joined chromatids called?

A

Sister chromatids

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13
Q

Centromere

A

Specialised DNA sequence of chromosome that links a pair of sister chromatids.

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14
Q

During mitosis, spindle fibres attach to the centromere via the ________.

A

kinetichore

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15
Q

Chromosomes are connected to the centrosome via their ______ through microtubules

A

centromeres

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16
Q

Kinetochore

A

Microtubule-binding structure that forms on the chromosomal centromere during late prophase; a kinetochore is a disc-shapedproteinstructure associated with duplicatedchromatidsineukaryoticcells where thespindle fibresattach duringcell divisionto pullsister chromatidsapart.

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17
Q

Kinetochore forms on the centromere during which stage?

A

Late prophase

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18
Q

During interphase, genetic material in the nucleus consists of _______

A

loosely packedchromatin

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19
Q

Interphase

A

Divided into three phases: G1(first gap),S (synthesis), andG2(second gap)

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20
Q

Prophase

A
  • Occurs after G2 phase
  • At the onset of prophase, replicated chromatin fibres condense to form discrete chromosomes, consisting of two sister chromatids connected at the centromere
  • The two centrosomes associated with the nuclear membrane migrate to opposite poles of the cell
  • Gene transcriptionceases during prophase and does not resume until late anaphase to early G1phase.
  • Thenucleolusalso disappears during early prophase
  • Mitotic spindle forms between the two centrosomes
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21
Q

Prometaphase

A
  • In animal cells, nuclear envelopefragments and disintegrates into small membranevesicles and microtubules invade the nuclear space (open mitosis)
  • In late prometaphase, kinetochore microtubulesbegin to search for and attach to chromosomalkinetochores. Each chromatid has its own kinetochore
  • A number ofpolar microtubulesfind and interact with corresponding polar microtubules from the opposite centrosome to form the mitotic spindle
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22
Q

Metaphase

A
  • Chromatin condenses further
  • All chromosomes connect to centrosome via kinetochores
  • Microtubules that are bound to kinetochores of sister chromatids radiate from opposite poles of the cell. After the microtubules have located and attached to the kinetochores in prometaphase, the two centrosomes begin pulling the chromosomes towards opposite ends of the cell. The resulting tension causes the chromosomes to align along themetaphase plate, located between the two centrosomes (at approximately the midline of the cell)
    The metaphase checkpoint ensures equal distribution of chromosomes at the end of mitosis.
23
Q

Anaphase

A
  • During anaphase A, cohesins that bind sister chromatids are cleaved, forming two identical daughter chromosomes.
  • As the kinetochore microtubules shorten, newly formed daughter chromosomes are pulled towards opposite ends of the cell by spindle fibres
  • In anaphase B, polar microtubules push against each other, elongating the cell.
  • In late anaphase, chromosomes reach peak condensation levels to aid in chromosome segregation and reformation of nucleus.
24
Q

Telophase

A
  • Polar microtubules continue to lengthen, further elongating the cell
  • A new nuclear envelope forms around each daughter chromosome
  • Enveloped chromosomes begin to decondense
25
Q

Molecular mechanism of offspring genetic diversity

A
  1. Random assortment of chromosomes (meiosis),
  2. Crossing over (chromosomes switch chunks of DNA)
  3. Random fertilisation (chance alone is responsible for which sperm meets which egg)
26
Q

Telomeres are usually ____ in length

A

10-15kb

27
Q

Telomeres are composed of _____

A

tandem repeat sequence TTAGG

28
Q

Function of telomeres

A

Telomeres maintain chromosome stability by protecting ends from exonuclease digestion or recombination and also ensure complete chromosome replication and proper segregation.

29
Q

Telomerase mechanism

A

Reverse transcriptase that carries its own RNA molecule, which is used as a template when it elongates telomeres, which are shortened after each replication cycle.

When RNA primer is removed from the 5’ end of the lagging strand, a strand of parent DNA remains unreplicated. Telomerase binds to the 3’ overhang of ssDNA and ads dNT to the end of the parent DNA, complementary to its primer. Telomerase continues to move down the parent DNA, adding additional repeats.
Primase, DNA polymerase and ligase synthesise the lagging strand in a 5’-3’ direction

30
Q

Chromatin structure and organisation

A

Chromatin is composed of nucleosomes (DNA wrapped around 8 histone proteins) and linker DNA. The histone protein H1 is located outside the nucleosome

31
Q

The nucleosome core ring is __ nm wide and contains __ bp of DNA wrapped ___ times around a central core histone octamer.

A

11nm; ~147bp; 1.7 times

32
Q

Histone octamer composition and assembly

A

The histone octamer is composed of one H3/H4 tetramer and two H2A/H2B dimers:

  1. H3 histone binds to H4 twice to form an H3/H4 tetramer (do not need DNA for binding)
  2. DNA then interacts with the H3/H4 tetramer, after which the H2A/H2B dimers bind (H2A-H2B) — (H3-H4)x2 — (H2A-H2B)
  3. H1 acts as glue to seal the nucleosome from the outside
33
Q

Histone Acetylation

A 
Histone acetyltransferase (HAT) acetylates conserved lysine residues on histone proteins. Histone acetylation generally increases gene expression.
In general, histone acetylation leads to increased transcription and gene expression through disruption of strong ionic interactions between positively charged amine groups on histone proteins and the negatively charged phosphate backbone of DNA. HATs acetylate specific lysine residues on histones, neutralising their charge and disrupting their ionic interactions with DNA. Acetylated lysines also act as interaction sites for other DNA-associated proteins. This causes loosening of chromatin structure; highly acetylated histones form more accessible chromatin, allowing transcriptional machinery access to the DNA.
Acetylation is not always associated with enhanced transcriptional activity, e.g. acetylation of H4K12 has been associated with condensed and transcriptionally inactive heterochromatin.
Histone deacetylases (HDAC) removes the acetyl group on acetylated histone lysine residues, restoring their positive charge and their ability to bind DNA, restoring the condensed chromatin structure. Deacetylation is associated with transcriptionally inactive heterochromatin.
34
Q

Histone Methylation

A

Histone methyltransferases (HMT) methylate R groups of lysine and arginine residues on histone proteins
Methylation of histones can either increase or decrease transcription of genes, depending on which amino acids are methylated and the number of methyl groups attached. Methylated histone tails serve to recruit other proteins and protein complex which regulate chromatin activation/inactivation
Methylation events that weaken the interaction between histones and DNA are associated with increase in transcription, as they allow DNA to unwrap itself from histone proteins and for transcriptional factors to bind
Methylated histones can either repress or activate transcription depending on the site of methylation e.g. dimethylation of lysine 9 on histone H3 (H3K9me3) in the promoter region of genes prevents excessive expression of these genes. In contrast, trimethylation ofhistone H3atlysine4 (H3K4me3) is associated with transcriptionally active euchromatin
Histone demethylases (HDM) remove methyl groups

35
Q

Histone Phosphorylation

A

Hydrolysis of ATP to produce ADP can phosphorylate histone protein, creating negatively charged residues which won’t bind DNA
Hydrolysis of ATP may also create enough energy to change conformation of protein, disrupting the interaction between histones and DNA

36
Q

DNA Methylation

A

The CpG sites are regions of DNA where a cytosine nucleotide is followed by a guanine nucleotide in the linear sequence of bases along its 5’→3’ direction
Methylation of cytosine to 5-methyl cytosine in mammalian DNA
Catalysis by DNA methyltransferases (DNMTs)
Site of methylation is the cytosine of a CpG doublet. This is a palindrome (i.e. present on both strands)

37
Q

General (basal) Transcription Factors (GTF)

A

Basal transcription factors assist RNA polymerase in the recognition of promoter sequences and unwinding the DNA double helix, among other functions.
Required for all transcription events.

38
Q

TATA box binding protein (TBP)

A
  • A subunit of TFIID, binds to the negatively charged phosphates of the TATA box through its positively charged lysine and arginine residues
  • TBP sits aside the TATA box as a molecular ‘saddle’, inducing a sharp bend in the DNA through insertion of four phenylalanine residues into the minor groove between base pairs, partially unwinding the helix, and doubly kinking it.
  • This initiates separation of the DNA strands, and is assisted through the weak interactions between adenine and thymine in this A-T rich region
  • This separation of the strands allows RNA polymerase II to begin transcription
39
Q

Steps of transcriptional activation

A

TBP, a subunit of TFIID, bind the TATA box in the core promoter region
TBP recruits TFIIA, followed by TFIIB to join initiation complex, stabilising the PIC
TFIIB recruits TFIIF bound to RNA polymerase II
TFIIE joins the complex, recruiting TFIIH

40
Q

Transcription Initiation

A
  • Subunits of TFIIH with ATP-ase and helical activity create negative superhelical tension in the DNA, causing ~14 base pairs to unwind, and form a transcription bubble.
  • TFIIH protein complex hydrolyses ATP into ADP.
  • With the release of a free phosphate, CTD:Carboxyl Tail domain of Pol II is phosphorylated by TFIIH on serine residues 2, 5 and 7
  • RNA polymerase II is moved downstream ~40 nucleotides
41
Q

Transcription Elongation

A

Early after transcription initiation, the ternary elongation complex (consisting of RNA pol II, DNA template and RNA transcript) is arrested at a “checkpoint” to ensure proper pre-mRNA capping. In eukaryotes, upon promoter clearance, Kin28 subunit of TFIIH phosphorylates serine 5 on the CTD of RNA polymerase II, leading to the recruitment of capping enzyme (CE) and the nascent RNA becomes capped during this first stage of elongation. The RNA cap is formed by addition of a methylated guanosine to the 5’ end of the RNA through the action of capping enzyme and an RNA (guanine-7) methyltransferase.
Release from this checkpoint involves the kinase action of P-TEFb, which targets Ser 2 on the CTD. Subsequently during elongation, phosphatases such as PTase dephosphorylate Ser 5 residues within the CTD; Ser 5-specific phosphatase Ssu72 is involved in allowing the correct transcript cleavage necessary for efficient termination. Both Ssu72 and the FCP1 phosphatase, which also targets Ser5, are involved in recycling the RNAP II for re-initiation and subsequent rounds of transcription, although the specifics of recycling are currently unknown.

42
Q

5’ Methyl Guanosine Cap

A

Protects transcript from degradation and helps with initiation of translation
After the initial pre-mRNA transcript has been created the first major modification is the addition of a 5’ methyl guanosine cap.
Addition of the 5’ 7-MG cap is important for two reasons: the 5’ caps are recognized by protein factors that initiate translation, and it also helps protect the transcript from nucleases.

43
Q

Poly(A) Tail

A

Once RNA polymerase reaches the terminator region, it releases the capped pre-mRNA molecule
Cleavage factors bind to specific regions in the pre-mRNA segment (GU-rich region and AAUAAA), the 3’ end is reconfigured and stabilising factors bind
Poly-A polymerase binds and the pre-mRNA molecule is cleaved, disassociating the complex
Poly-A polymerase then synthesises the poly-A tail by adding adenines to the cleavage site

44
Q

Enhancers

A

cis-regulatory elements (CREs) that enhance transcription of genes on the same molecule of DNA.

45
Q

Activators

A

TFs/trans-elements that bind enhancers or promoter-proximal regions to increase gene transcription

46
Q

Repressors

A

TFs that bind to silencer region, occupy region where TFs bind

47
Q

Silencers

A

Silencersare CREs that can bindtranscription regulation factors calledrepressors,thereby preventing transcription of a gene.

48
Q

Isolation of CREs can be achieved using which method

A

Chromatin Immunoprecipitation (ChIP)

49
Q

Chromatin Immunoprecipitation (ChIP)

A

ChIP aims to define the cistrome - the set of cis-acting targets of trans-acting factors - and determine whether specific proteins such as TFs are associated with specific genomic regions, such aspromotersor otherDNA binding sites
ChIP also aims to determine the specific genomic location of histone modifications
DNA and associated proteins onchromatinin living cells or tissues are crosslinked (this step is omitted in Native ChIP).
The DNA-protein complexes (chromatin-protein) are then sheared into ~500 bp DNA fragments bysonicationor nuclease digestion.
Cross-linkedDNA fragments associated with the protein(s) of interest are selectively immunoprecipitated from the cell debris using an appropriate protein-specific antibody.
The associated DNA fragments are purified and their sequence is determined. Enrichment of specific DNA sequences represents regions on the genome that the protein of interest is associated within vivo.

50
Q

Promoter Deletion Analysis/Promoter Bashing

A

Technique used to identify promoters/TF-binding regions
Specific point mutations or deletions are made in specific regions of the promoter and the transcription of the gene is measured; the level of transcription can be attributed to the contribution of the region of the promoter
The ability of upstream promoters can be easily assayed by removing segments from the 5’ end, and the same for the 3’ end of the strand for downstream promoters.
Procedure:
Region of DNA thought to act as promoter is cloned and extracted; this ensures promoter is the only factor affecting expression
Region is sequenced to identify differences in WT and mutated promoters
Region is digested with appropriate nucleases to systematically remove elements of the region
Gene expression levels are measured through a reporter gene ligated to the promoter region

51
Q

Isolation of mRNAs from Total RNAs can be achieved through

A

Oligo-dT purifications of mRNA will result in extraction of mRNA transcripts with polyA tails only; Only mRNAss contain a poly-adenylated tail.

52
Q

Creating a cDNA library

A
  1. Oligo dT primer + reverse transcriptase, dNTPs

2. Treat with NaOH to destroy RNA

53
Q

Protein-Protein Interaction Analysis: Yeast Two-Hybrid System

A

A transcription factor is split into its constituents elements: DNA binding domain (DBD), specific to the TF, and an Activation Domain (AD) that is generic across TF. The DBD is fused to a protein, termed bait. The AD is fused to a different protein, termed prey.
In the yeast two hybrid system, two plasmids are transformed into yeast, one plasmid containing the sequence for the DBD and bait protein, as well as a selectable marker such as a gene coding for Leucine production. The other plasmid encodes for the AD, prey and a different selectable marker such as tryptophan. The two plasmids are inoculated and transformed into a yeast strain deficient in both of the selectable markers and depends upon them for growth. This mixture is then plated and incubated on a medium deficient in both selectable markers, thereby selecting for transformants with both plasmids only. Within the yeast, the DBD-domain will bind to its target sequence, a reporter gene such as Luc or GFP. If the prey and bait proteins bind, their interaction will bring the AD and BDB in close proximity and transcription will occur, and the reporter gene will be transcribed and translated.

54
Q

Integration Pattern: Position Effect

A

If gene of interest is integrated in transcriptionally inactive heterochromatin, it will not be transcribed
Genome editing for targeted transgene integration requires creating a double-stranded break in the target sequence to increase efficiency of homological recombination as well as flanking (to 0.001%) as well as adding flanking regions to your cassette identical to the flanking regions of the target sequence.

Gene Technology (3 decks)

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