EXAM 3 Study Guide CHAT

Describe the regulatory steps of the GAL1 Promoter activation and repression system

Repressed state (no galactose): 1.) Gal4 is bound to the UAS_GAL. 2.) Gal80 binds Gal4 and blocks its activation function. 3.) Even though Gal4 is on the DNA, transcription stays off or very low because the transcription machinery is not efficiently recruited.
Activated state (galactose present, glucose absent): 1.) Gal3 senses galactose. 2.) Activated Gal3 interacts with Gal80, preventing Gal80 from inhibiting Gal4. 3.) Gal4 is now free to activate transcription. 4.) Gal4 recruits coactivators, Mediator/chromatin-remodeling activities, general transcription factors, and RNA polymerase II to the core promoter/TATA box. 5.) GAL1 transcription is turned on strongly.
Glucose repression (glucose present): 1.) Mig1 becomes active in the presence of glucose. 2.) Mig1 binds glucose-responsive repression regions in the GAL1 promoter. 3.) Mig1 recruits the Ssn6–Tup1 corepressor complex. 4.) This prevents efficient transcription, so GAL1 is repressed, even if galactose is around.

Describe the cis elements involved in the regulatory steps of the GAL1 Promoter activation and repression system

There are four important cis elements involved in the regulatory steps of the GAL1 Promoter activation system: UAS_GAL (upstream activating sequence), Core promoter / TATA box, Transcription start site (TSS), and Glucose-repression elements. The UAS_GAL (upstream activating sequence) contains Gal4-binding sites, is the main positive regulatory element, and activates transcription. The Core promoter / TATA box is where the general transcription machinery assembles. The Transcription start site (TSS) is where RNA synthesis/transcription begins. Glucose-repression elements are sites that help mediate repression when glucose is present, mainly through repressors such as Mig1.

Describe the trans-acting factors involved in the regulatory steps of the GAL1 Promoter activation system

There are six important trans-acting factors involved in the regulatory steps of the GAL1 Promoter activation system: Gal4, Gal80, Gal3, General transcription factors + RNA polymerase II:, Mig1, Ssn6–Tup1. Gal4 is the sequence-specific activator that binds the UAS_GAL. Gal80 is the repressor that binds Gal4 and blocks its activation domain. Gal3 is the galactose sensor/transducer that relieves Gal80 inhibition in the presence of galactose. General transcription factors + RNA polymerase II assemble at the core promoter to start transcription. Mig1 is a glucose-dependent repressor. Ssn6–Tup1 is the corepressor complex recruited by Mig1.

Explain the role of activators, how they influence transcription as well as the chromatin structure.

Activators are regulatory proteins that bind specific DNA control elements such as enhancers or upstream activating sequences. Their main role is to increase transcription of a target gene. Activators influence transcription by help recruit or stabilize: mediator, general transcription factors (ex. TFIID), RNA polymerase II. This makes assembly of the preinitiation complex (PIC) easier and increases transcription initiation. Activators can influence chromatin structure by often recruiting co-activators that modify chromatin. These can include histone acetyltransferases (HATs), which acetylate histones, and chromatin-remodeling complexes, which reposition or loosen nucleosomes. This makes chromatin more open and accessible, so the transcription machinery can reach the DNA more easily.

Explain the role of co-activators, how they influence transcription as well as the chromatin structure.

Co-activators are proteins that help activators increase transcription, but they usually do not bind DNA directly. They are recruited to genes by DNA-bound activators. Co-activators influence transcription by helping connect activators to mediator, general transcription factors, and RNA polymerase II. This helps assemble or stabilize the preinitiation complex (PIC), increasing transcription. Co-activators influence chromatin structure by often having/recruiting histone acetyltransferase (HAT) activity and chromatin-remodeling complexes. These activities loosen chromatin by acetylating histones or repositioning nucleosomes. This makes DNA more accessible to the transcription machinery.

Explain the role of repressors, how they influence transcription as well as the chromatin structure.

Repressors are regulatory proteins that decrease or shut down transcription of a gene. They usually bind specific DNA regulatory elements such as silencers or repressive control regions. Repressors influence transcription by blocking activators, preventing the recruitment of Mediator, general transcription factors, or RNA polymerase II, and interfere with formation of the preinitiation complex. This lowers or prevents transcription initiation. Repressors influence chromatin structure by often recruiting co-repressors that bring histone deacetylases (HDACs) chromatin-remodeling/compacting complexes. These make chromatin more condensed and less accessible. Tighter chromatin makes it harder for transcription machinery to reach the DNA.

Explain the role of co-repressors, how they influence transcription as well as the chromatin structure.

Co-repressors are proteins that help repressors shut down transcription, but they usually do not bind DNA directly. They are recruited to DNA by DNA-bound repressors. Co-repressors reduce transcription by blocking recruitment or activity of Mediator, general transcription factors, and RNA polymerase II. They also reduce transcription by helping prevent formation of the preinitiation complex. Co-repressors influence chromatin structure by often recruiting or containing histone deacetylases (HDACs) and chromatin-remodeling/compacting complexes. These remove activating histone acetylation and promote tighter nucleosome packing. This makes chromatin more closed and less accessible to transcription machinery.

Explain the role of the Plus 1 site (+1) in transcription

The +1 site is the transcription start site (TSS) and it is the first DNA base that is copied into RNA by RNA polymerase. It marks the point where transcription begins and it determines the 5′ end of the RNA transcript. It matters because it defines exactly where RNA polymerase starts RNA synthesis and its position helps determine the length/sequence of the 5′ untranslated region (5′ UTR) and the rest of the transcript.

Match the letters in the diagram to parts of gene for transcription below (hint: some of the terms below may or may not be repeated) :
Regulatory Sequences
Transcription activators and co-activators
Gene
Promoter
TATA
POl II
General transcription factors
Inr (the initiator element)
General Transcription Factors
TFIID

A: General transcription factors
B: Promoter
C: TFIID
D: TATA
E: PolI II
F: Inr (the initiator element)
G: Regulatory Sequences
H: Transcription activators and co-activators
I: Regulatory Sequences

What is the part of the gene for transcription of A and B?

A: Regulatory Region
B: Coding Sequence

Define the regulatory region of a gene and explain how to be able to point out where it is on the gene (diagram)

.

The regulatory region of a gene is the region of a gene where RNA Polymerase and other accessory transcription modulator proteins bind and interact to control RNA synthesis. It usually includes the promoter. Think of a gene like this: upstream DNA -> regulatory region -> +1 start site -> transcribed region -> coding region -> downstream DNA. To identify it on a diagram, if the +1 site, the regulatory region is usually to the left/upstream. The following are clues that you are looking at the regulatory region: TATA box, CAAT box, GC box, UAS, enhancer, silencer.

Define the promoter of a gene and explain how to be able to point out where it is on the gene (diagram)

The promoter of a gene is the DNA region where the transcription machinery assembles to begin transcription. It provides the binding site for RNA polymerase and general transcription factors. It controls whether the gene is transcribed weakly, strongly, or not at all. The main idea for finding the promoter on a gene is that if you can find +1 site, the promoter is usually immediately before it.

Match the letter with the following regulatory regions below (hint: some terms may be repeated):
Proximal --> Core
Terminator
Enhancer/Silencer
Promoter
Regulatory Sequence

A: Enhancer/Silencer
B: Regulatory Sequence
C: Promoter
D: Proximal / Core
E: Regulatory Sequence
F: Enhancer/Silencer
G: Terminator

Match the letters with the following regulatory regions/elements bellow (hint: some terms may be repeated):
TATA BOX
Enhancer/Silencer
Proximal Promoter
Core Promoter
Gene body exons / introns

A: Enhancer/Silencer
B: Proximal Promoter
C: TATA BOX
D: Core Promoter
E: Gene body exons / introns2

Define the following regulatory region on the gene and its role: core promoter

The core promoter is the DNA region immediately around the transcription start site (+1). This is where RNA polymerase and general transcription factors assemble. It often contains elements like the TATA box, Inr, or other core promoter motifs. Its role is starting transcription.

Define the following regulatory region on the gene and its role: proximal promoter

The proximal promoter is a short region upstream of the core promoter and it contains binding sites for specific transcription factors. Its role is to increase transcription by serving as binding sites for transcription factors (e.g., CAT box, GC box) that load and stabilize RNA polymerase II. Conversely, their role is also to decrease transcription when repressor proteins bind to them, obstructing RNA polymerase or causing promoter-proximal pausing and premature termination.

Define the following regulatory region on the gene and its role: enhancers

Enhancers are regulatory DNA sequences that bind activators.They can be located upstream, downstream, or inside introns. They can work even when they are far from the promoter. Enhancers’ role is to increase transcription by acting as binding sites for activator proteins, which fold DNA to bring these activators into contact with the promoter, facilitating the recruitment and stabilization of RNA polymerase II. These distant DNA sequences act in three-dimensional space, utilizing mediator complexes to create chromatin loops, bringing the enhancer and gene closer to boost transcription initiation frequency.

Define the following regulatory region on the gene and its role: silencers

Silencers are regulatory DNA sequences that bind repressors. They can be located upstream, downstream, or inside introns. They can work even when they are far from the promoter. Silencers’ role is to Silencers decrease transcription by binding repressor proteins, which physically block RNA polymerase from binding to the promoter, compete with activator proteins, or promote condensed chromatin (heterochromatin) structure, making DNA inaccessible.

Define the transcription initiation site of a gene and explain how to point it out on a gene

The transcription initiation site of a gene, also called the transcription start site (TSS) or +1 site, is the exact DNA nucleotide where RNA polymerase begins making the RNA transcript. The main idea behind being able to point it out on a gene, is that it is just downstream of the promoter, just after the core promoter elements, but before the rest of the transcribed region.

Define the transcription termination site of a gene and explain how to point it out on a gene

The transcription terminator site is the DNA sequence or region that signals RNA polymerase to stop transcription and end the RNA transcript. The main idea behind pointing it out on a gene, is to find the +1 site where transcription starts, follow the gene in the direction of transcription, and the terminator is at the far downstream end, where transcription stops.

How to identify a consensus sequence given a sequence alignment and provide the percent conservation.

For each column: 1.) Count how many A, T, G, C there are. 2.) Pick the most common one -> that is the consensus residue. 3.) Percent conservation =(number of sequences with the most common residue at that position / total number of sequences compared at that position) * 100

What doe the following notions in writing consensus sequences mean?
Y, R, N, ~/~

Y: Pyrimidines = equal T and C
R: Purines = equal A and G
N: no particular base is more common
C/G or C/A or T/A or T/G: one pyrimidine and one purine are equally common

Define the role and characteristics of the -10 region of the prokaryotic promoter during transcription

The -10 region of a prokaryotic promoter, also called the Pribnow box, is a cis-acting DNA element located about 10 base pairs upstream of the transcription start site (+1). It’s role is being a key site recognized by the sigma factor of RNA polymerase. Its main function is to help the DNA unwind (melt) so transcription can begin. This happens because the -10 region is usually A/T-rich, and A-T base pairs separate more easily than G-C base pairs. Its most important characteristic is that it has the following consensus sequence: TATAAT.

Define the role and characteristics of the -35 element of the prokaryotic promoter during transcription

The -35 element is a cis-acting DNA sequence in a prokaryotic promoter located about 35 base pairs upstream of the transcription start site (+1). Its role is serving as an important recognition and binding site for the sigma factor of RNA polymerase. Its main job is to help RNA polymerase initially recognize the promoter and bind in the correct orientation. Its most important characteristic is that it has the following consensus sequence: TTGACA

What are the two forms that prokaryotic RNA polymerase exists as?

The core enzyme and the holoenzyme

Define the holoenzyme form of prokaryotic RNA polymerase and its function

The holoenzyme form of prokaryotic RNA polymerase is the complete promoter-recognizing form of the enzyme. It consists of the core enzyme and a sigma factor. The holoenzyme’s main function is to recognize specific promoter sequences and initiate transcription at the correct site. The sigma factor allows the enzyme to bind promoter elements such as the -35 and -10 regions. Once transcription begins, the sigma factor often dissociates or becomes less tightly associated.

Define the core enzyme form of prokaryotic RNA polymerase and its function

The prokaryotic RNA polymerase core enzyme is a multi-subunit complex responsible for the catalysis of RNA synthesis during transcription elongation. Lacking the sigma subunit, the core enzyme cannot initiate transcription at promoters but efficiently elongates RNA chains by moving along DNA and synthesizing RNA from a DNA template. The function of the The core enzyme is to carry out the catalytic synthesis of RNA from a DNA template. It can unwind and track along DNA during transcription, as well as join ribonucleotides together to make the RNA strand.

What are the components that make up the core enzyme form of prokaryotic RNA polymerase and their respective functions?

The core enzyme is made up of 2 alpha subunits, 1 beta subunit, 1 beta prime subunit, 1 omega subunit. The alpha subunits are essential for the assembly of the enzyme, and the two alpha subunits act as a scaffold. The C-terminal domains of the alpha subunits interact with various regulatory transcription factors to enhance or inhibit gene expression. The beta subunit contains the catalytic site for RNA synthesis and is responsible for binding the incoming ribonucleotide triphosphates (NTPs) and forming the phosphodiester bonds. The beta prime subunit is responsible for DNA binding, and acts like a "clamp" that holds the DNA template securely in the channel so the beta subunit can transcribe it. The role of the omega subunit is not strictly necessary for catalysis, but it is vital for chaperoning. It helps the beta subunit fold correctly and keeps the entire complex stable.

Define sigma factors as they relate to prokaryotic RNA polymerase and its role

Sigma factors are essential bacterial proteins that bind to RNA polymerase (RNAP) to form the holoenzyme, enabling it to recognize specific promoter regions (-10 and -35 sequences) and initiate transcription. They regulate gene expression by guiding RNAP to specific genes, allowing rapid adaptation to environmental changes, such as stress or nutrient limitation.

What are the main cis elements of prokaryotic promoters and their respective functions that makes it different from eukaryotic promoters?

The main cis elements of prokaryotic promoters are the -10 element (Pribnow Box) and the -35 element. The -10 element (Pribnow box), with the consensus sequence TATAAT is essential for binding RNA polymerase (via the sigma factor) and facilitating DNA unwinding. Its AT-rich nature allows for easy separation of the DNA strands, crucial for initiation. The -35 element with the consensus sequence TTGACA has the primary function of serving as the initial binding site for the RNA polymerase holoenzyme, specifically recognized by the sigma factor subunit. It acts as an anchor to recruit RNA polymerase to the DNA, determining the strength of the promoter.

What are the main cis elements of eukaryotic promoters and their function that makes it different from prokaryotic promoters?

The main cis elements of eukaryotic promoters are the TATA Box (Hogness Box), Inr, the CAAT box, the GC Box. enhancers/silencers. The TATA box is a conserved DNA sequence (TATAAA) found in the promoter region of eukaryotic and archaeal genes, typically located 25–35 base pairs upstream of the transcription start site. The TATA box serves as the initiation site for transcription. It binds the TATA-binding protein (TBP), part of the transcription factor complex (TFIID), which subsequently recruits RNA polymerase to begin synthesizing RNA from DNA. The Initiator (Inr) element is a crucial core promoter DNA sequence, typically spanning 2–17 bp around the transcription start site (+1), that directs RNA polymerase II to begin transcription. TFIID, a basal transcription factor, recognizes and binds to the Inr via subunits TAF1 and TAF2. This binding forms the preinitiation complex (PIC). The CAAT box is a conserved eukaryotic promoter element located roughly -75 of the transcription start site. The CAAT box is a promoter element that binds transcription factors (such as CTF or C/EBP) to facilitate the initiation of transcription. The GC box is a distinct, GC-rich nucleotide pattern (consensus sequence often 5'-GGGCGG-3') found usually upstream of the TATA box. It binds the transcription factor Sp1, which helps initiate transcription. Enhancers and silencers are cis-regulatory DNA sequences that bind transcription factors to increase (enhancers) or decrease (silencers) gene expression. Enhancers bind specific transcription factors (activators) that recruit co-activators and RNA polymerase II to the promoter, forming an "enhanceosome". Silencers bind repressor proteins that prevent RNA polymerase from binding to the promoter or induce chromatin to close, making it inaccessible.

What is the core idea behind a luciferase reporter assay?

A luciferase reporter assay tells you how much a DNA sequence or protein condition changes transcription of a reporter gene. More luciferase signal means that there is more transcriptional activation of the reporter, and less luciferase signal means that there is less transcription, or repression.

What is the first general step in interpreting luciferase reporter data properly?

Step 1: Always ask what was changed
Before interpreting any graph, identify what the experiment manipulated:

Was the variable:

a DNA sequence?

-> then the assay is testing cis elements

a protein or a protein mutant/domain deletion?

-> then the assay is testing trans-acting protein function

both?

-> then it is testing whether a protein acts through a specific DNA element

What is the second general step in interpreting luciferase reporter data properly?

Step 2: Read the axes

Usually:

x-axis = constructs or conditions

y-axis = luciferase activity

often shown as:

relative luciferase units

fold activation

normalized luciferase activity

Higher bar = stronger reporter transcription.

What is the third general step in interpreting luciferase reporter data properly?

Step 3: Know the baseline

Interpretation depends on what the construct is compared to.

Common baselines:

empty vector

minimal promoter alone

wild-type promoter

untreated control

reporter without enhancer

wild-type protein

You always interpret relative to the control.

What is the fourth general step in interpreting luciferase reporter data properly?

Step 4: Use the basic interpretation rule

If reporter activity goes up

That usually means the tested sequence/protein has a positive effect on transcription.

If reporter activity goes down

That usually means the tested sequence/protein has a negative effect, or that an important activating function was lost.

What are 4 important patterns to memorize when it comes to interpreting luciferase reporter assay data?

Pattern 1

Mutation lowers activity

-> that sequence/domain is likely needed for activation

Pattern 2

Mutation raises activity

-> that sequence/domain is likely involved in repression

Pattern 3

Deleting a region lowers activity

-> deleted region likely contains a positive element/domain

Pattern 4

Deleting a region raises activity

-> deleted region likely contains a negative element/domain

What are some cis elements that can analyzed using luciferase reporter assays?

Cis elements are DNA sequences in the reporter construct, such as:

core promoter

TATA box

Initiator (Inr)

proximal promoter elements

enhancers

silencers

specific transcription factor binding sites

What are some typical protein manipulations done when testing protein domains using luciferase reporter data?

full-length protein

deletion mutants

point mutants

domain fusions

This tells you which protein domains are needed for activation or repression.

What are three reliable writing templates for interpreting luciferase reporter assay data of cis elements?

“Mutation/deletion of this region reduced reporter activity, indicating that this sequence contains a positive cis-regulatory element required for transcription.”

“Mutation of this motif increased reporter activity, suggesting that it functions as a negative cis-regulatory element.”

“The Inr mutation reduced basal luciferase expression, indicating that the Initiator contributes to core promoter function and transcription initiation.”

What are three reliable writing templates for interpreting luciferase reporter assay data of protein domains?

“Deletion of this domain abolished reporter activation, indicating that the domain is required for transcriptional activation.”

“Loss of repression after deletion suggests that the removed region contains a functional repression domain.”

“The mutant retained the DNA-binding region but lost activation, suggesting that the deleted region is needed for co-activator recruitment rather than DNA binding.”

Describe the stepwise process of 5’ capping in eukaryotes

In eukaryotes, 5′ capping is a co-transcriptional RNA-processing event that occurs very early, usually when the nascent pre-mRNA is about 20–30 nucleotides long.
1. Nascent RNA emerges from RNA polymerase II

The new transcript initially has a 5′ triphosphate end

The C-terminal domain (CTD) of RNA polymerase II helps recruit the capping machinery.
2. RNA 5′-triphosphatase acts
removes the terminal γ-phosphate from the 5′ end of the RNA.

Now the RNA has a 5′ diphosphate end.

3. Guanylyltransferase adds GMP to the 5′ diphosphate end of the RNA.

This forms an unusual 5′–5′ triphosphate linkage

This added guanine is the cap guanosine.
4. Guanine-N7 methyltransferase methylates the cap
Methyl donor: SAM (S-adenosylmethionine)
methylates the N7 position of the added guanine
This produces the 7-methylguanosine cap
This is the basic cap 0 structure.

What are the main players involved in 5’ capping in eukaryotes?

RNA polymerase II CTD: recruits capping enzymes

RNA 5′-triphosphatase: removes γ-phosphate

Guanylyltransferase: adds GMP

Guanine-N7 methyltransferase: makes m⁷G

2′-O-methyltransferases: make cap 1 and sometimes cap 2

SAM: methyl donor for methylation reactions

What are the main modifications in 5’ capping in eukaryotes?

1.) Removal of γ-phosphate

2.) Addition of GMP

3.) Formation of a 5′–5′ triphosphate linkage

4.) N7 methylation of guanine -> m⁷G cap

5.) Possible 2′-O-methylation of the first and second nucleotides

What is constitutive splicing?

Constitutive splicing is the default splicing pattern in which all introns are removed and all exons are joined together in the same way every time.

The same mature mRNA is produced from that pre-mRNA.

No exon choices are changed.

What is alternative splicing?

Alternative splicing is when the same pre-mRNA can be spliced in different ways, so different combinations of exons are included or excluded.

This allows one gene to produce multiple different mRNA isoforms

Those different mRNAs can produce different protein isoforms

What is the difference between constitutive splicing and alternative splicing?

Constitutive splicing always processes a pre-mRNA the same way, whereas alternative splicing allows different mature mRNAs to be made from the same gene.

What is RNA editing and what can it do?

The alteration of the sequence of nucleotides in the RNA after it has been transcribed from DNA but before it is translated into protein.
What it can do:

change one nucleotide to another

insert nucleotides

delete nucleotides

What is substitution editing and a simple example of it?

It is the chemical alteration of individual nucleotides (functionally the equivalent of point mutations but not considered a mutation because the gene is not changed).
Example: Can convert coding triplet to a stop codon to truncate protein.

What kind of editing is mammalian apolipoprotein B Glu codon conversion into a stop codon?

This is substitution editing, specifically:

C-to-U RNA editing

A cytidine in the RNA is chemically deaminated to uridine.

This is a post-transcriptional modification:

DNA is unchanged

RNA sequence is changed

protein output changes

What is the actual editing event that takes place during mammalian apolipoprotein B Glu codon conversion into a stop codon?

At a specific position in apoB mRNA:

original codon: CAA = glutamine

edited codon: UAA = stop codon

This happens because one cytidine (C) in the RNA is converted to uridine (U).
So: CAA -> UAA
That single base change creates a translation stop signal.

What are the functional consequences of the two proteins involved in mammalian apolipoprotein B Glu codon conversion into a stop codon (ApoB-100 and ApoB-48)?

ApoB-100

made in the liver

full-length protein

important for VLDL/LDL metabolism

contains the LDL receptor-binding region in the C-terminal part

ApoB-48

made in the intestine

shorter protein because translation stops early

used in chylomicron assembly

lacks the C-terminal portion of apoB-100

What are the players involved in the substitution editing of mammalian apolipoprotein B Glu codon conversion into a stop codon?

1.) APOBEC-1

2.) ACF / A1CF (APOBEC-1 complementation factor)
3.) The apoB RNA editing site
4.) Cis-acting RNA sequences
5.) The editosome

What is the role of APOBEC-1 in the substitution editing of mammalian apolipoprotein B Glu codon conversion into a stop codon?

This is the key catalytic enzyme.

Role

It is a cytidine deaminase

It converts C -> U in the RNA

What it does chemically

It removes an amino group from cytidine, converting it into uridine.

Importance

Without APOBEC-1, the editing reaction does not occur.

Why is the mammalian apolipoprotein B Glu codon conversion into a stop codon a classic example of substitution editing?

The apoB case is one of the most famous examples because it clearly shows that:

RNA can be changed after transcription

a single nucleotide substitution can drastically alter protein output

editing can be regulated by tissue

the edited RNA can produce a functionally different protein

What kind of RNA editing is glutamate receptor channel protein Gln codon converted to an Arg codon?

A to I substitution editing:
Adenosine Deaminase Acting on RNA (ADAR) enzymes, which convert adenosine residues to inosine in an mRNA molecule by hydrolytic deamination.

Why does the A to I substitution editing of glutamate receptor channel protein Gln codon converted to an Arg codon matters?

The Q/R site lies in the region of the receptor that helps form the ion channel pore.

That means the amino acid at this position strongly affects what ions can pass through the channel.

If the site is unedited: Q

the channel is more Ca²⁺ permeable

the channel tends to have higher conductance

the receptor is more susceptible to Ca²⁺ entry

If the site is edited: R

the positively charged arginine alters pore properties

Ca²⁺ permeability is strongly reduced

channel behavior changes in a protective way for neurons

So the key biological effect is:

Q → R editing makes the AMPA receptor less permeable to Ca²⁺

What is the common chemical mechanism behind substitution editing of A or C Residues?

Most common chemical mechanism: deamination

Deamination means removal of an amino group (-NH2) from a nitrogenous base.

This chemical change alters the identity/base-pairing properties of that nucleotide.

What are two common examples of deanimation in RNA editing by substitution?

1. A-to-I editing

enzyme removes the amino group from adenosine

adenosine becomes inosine

inosine is interpreted like guanosine (G)

So functionally: A -> I = G

2. C-to-U editing

enzyme removes the amino group from cytidine

cytidine becomes uridine

So: C -> U

What modification changes results from editing by substitution involving deanimation?

Because the base is chemically altered, the edited RNA may now have different:

codons

base-pairing behavior

splicing patterns

stability

protein-coding potential

For example:

a codon may specify a different amino acid

or it may become a stop codon

What is the general concept behind RNA editing via substitution for both C->U and A->I editing?

Both usually happen through a deamination-type modification, meaning the base is chemically altered so it behaves like a different base.
For both types, the logic is:

A specific RNA is made from DNA

An editing enzyme recognizes a particular site on that RNA

The enzyme changes one base into another base-like form

That edited base is then interpreted differently during:

translation

splicing

RNA folding

RNA function

What is the main concept behind C->U editing and a classic example of it?

In C-to-U editing, a cytidine in RNA is converted into uridine.

This can change:

a codon

the amino acid sequence

or even create a stop codon

A classic example is apoB mRNA editing, where CAA becomes UAA.

What are the main players involved in C->U editing?

1. Target RNA

The RNA that contains the cytidine to be edited

It must contain the correct nearby cis-acting recognition sequences

2. APOBEC family editing enzyme

The classic catalytic enzyme is APOBEC-1

APOBEC proteins are cytidine deaminases

Their role is to convert C → U

3. Accessory RNA-binding proteins

These help APOBEC edit the correct site

A classic example is A1CF/ACF in the apoB system

Other accessory factors can also help depending on context

4. Cis-acting RNA elements

These are nearby RNA sequences that help the enzyme recognize the proper editing site

In apoB, an important example is the mooring sequence

5. Editosome

The functional RNA-protein complex that performs editing

Includes the catalytic enzyme plus accessory factors

What do the modification in C->U editing means functionally?

Once the cytidine becomes uridine:

the codon may now code for a different amino acid

or may become a stop codon

so the final protein product can change dramatically

What is the main concept behind A -> I editing and what is a classic example of it?

In A-to-I editing, an adenosine in RNA is converted into inosine.

Inosine is important because the cell usually reads inosine like guanosine (G).

So functionally: A -> I = G

That means A-to-I editing can change:

codons

splice signals

RNA structure

miRNA targeting

RNA stability/function

A classic coding example is the GluA2 Q/R site.

What are the main players involved in A->I Editing

1. Target RNA

Usually a pre-mRNA or other RNA containing an editable adenosine

2. ADAR enzymes

The major enzymes are ADARs (adenosine deaminases acting on RNA)

Common members include ADAR1 and ADAR2

They catalyze the conversion A → I

3. Double-stranded RNA structure

ADARs usually do not edit single-stranded RNA sites efficiently on their own

They usually require the target adenosine to be in a double-stranded RNA region

4. Editing complementary sequence (ECS) or paired region

A nearby complementary RNA sequence often base-pairs with the target region

This forms the dsRNA structure recognized by ADAR

5. RNA-processing machinery

Because much A-to-I editing happens in pre-mRNA, splicing and nuclear RNA processing context matter

6. Ribosome / spliceosome / other readers

These cellular machines “interpret” inosine as if it were guanosine

That is how the edit affects protein sequence or RNA processing

What do the modifications in A->I editing means functionally?

Because inosine behaves like G:

a codon can change to encode a different amino acid

splice-site choice can change

RNA pairing properties can change

regulatory interactions can change

What is a snRNP?

A snRNP (small nuclear ribonucleoprotein) is a complex made of:

a small nuclear RNA (snRNA)

associated proteins

snRNPs are major components of the spliceosome, the machinery that removes introns from pre-mRNA.

What are the components of snRNP?

A snRNP contains:

1. snRNA

a small nuclear RNA such as U1, U2, U4, U5, or U6

helps recognize splice sites and participates in the splicing reaction

2. Proteins

common Sm or Sm-like proteins that help stabilize the snRNP

snRNP-specific proteins that give each snRNP its particular function

What are the functions of snRNPs?

snRNPs mainly function in pre-mRNA splicing.

They help:

recognize the 5′ splice site

recognize the branch point

bring splice sites together

form the spliceosome

catalyze intron removal and exon ligation

What does splicing do?

Splicing removes an intron from a pre-mRNA and joins the two surrounding exons together.

It is carried out by the spliceosome, a large RNA-protein complex built from snRNPs and other splicing factors.

What are the main cis elements in the pre-mRNA involved in splicing? (Textbook + ChatGPT)

5′ splice site (donor site/Splice donor)

Located at the 5′ end of the intron

end of the first exon

Usually begins with GU

Marks where the intron starts

Branch point sequence/site

Located (within) inside the intron, upstream of the 3′ splice site

Contains a critical branch-point adenosine (A)

This A performs the first nucleophilic attack

Polypyrimidine tract

Usually between the branch point and the 3′ splice site
Py rich upstream of the splice acceptor

Rich in U and C

Helps recruit key factors

3′ splice site (acceptor site/splice acceptor)

Located at the 3′ end of the intron

beginning of second exon

Usually ends with AG

Marks where the intron ends

Give an overview of the pre-mRNA splicing reaction with the general steps

Pre-mRNA splicing occurs through two site-specific transesterification reactions that result in phosphodiester bond cleavage and ligation. The 5′ and 3′ splice sites consist of the conserved sequence elements shown; the Py tract in the intron is a string of pyrimidine residues.

General Steps
1.) The branch-point 2’-OH attacks the 5’ splice site.
2.) The 5’ splice site is now activated to attack the 3’ splice site.
3.) The intron is released from the spliced mRNA as a lariat

Give an overview of the steps involved in spliceosome assembly on pre-mRNAs involving base pairing to snRNAs

1.) U1 binds to the 5’ splice site; U2 binds to the branch point
2.) The U4-U6-U5 trimeric snRNP displaces U1 at the 5’ splice site, then U4 dissociates
3.) U6 and U2 catalyze attack of the branch point on the 5’ splice site
4.) The 5’ splice site attacks the 3’ splice site, completing the reaction
Notice that in step 2, U6 snRNA base-pairs near the 5′ exon binding site where U1 snRNA was formerly bound. As U4 dissociates, the U2-U6-U5 complex remains assembled on the pre-mRNA. U5 base-pairs to both sides of the splice junction to align the RNA for the splicing reaction, and U2 and U6 base-pair to each other.

Fill in the blank

Splicing occurs through formation of a ___ structure and is a 2-step ___ process.

lariat, transterification

What are the 2 steps in the 2 step Tranesterification process that splicing occurs through the formation of a lariat structure? (lecture)

Step 1.) Transesterification produces a new ester bond at 2 ́ hydroxyl position of the branch point A ribose -> 2 ́, 5 ́ phosphodiester bond formed, phosphodiester bond 3 ́ position is broken
Step 2.) The free 3 ́ hydroxyl can attack P at the 5 ́ end of exon splice acceptor site. Excised lariat intron is processed by debranching enzyme and degraded

What are the main trans-acting factors in the stepwise splicing process and their roles?

U1 snRNP

Recognizes and base-pairs with the 5′ splice site

SF1 / BBP (branch-point binding protein)

Initially recognizes the branch point

U2AF

A heterodimer:

U2AF65 binds the polypyrimidine tract

U2AF35 helps recognize the 3′ splice site AG

U2 snRNP

Replaces SF1 at the branch point

Binds near the branch point so that the A is bulged out

Prepares the branch-point A for reaction

U4/U6.U5 tri-snRNP

A preassembled complex containing:

U4 snRNP

U6 snRNP

U5 snRNP

Roles:

U4: keeps U6 inactive at first

U6: becomes part of the catalytic center

U5: helps align the exons for ligation

SR proteins and other regulatory proteins

Help define splice sites

Promote proper spliceosome assembly

Important especially in exon recognition and alternative splicing

What are the steps in the stepwise spliceosome assembly?

Step 1: Early (E) complex formation

This is the first recognition stage.

What binds:

U1 snRNP binds the 5′ splice site

SF1/BBP binds the branch point

U2AF65 binds the polypyrimidine tract

U2AF35 binds the 3′ splice site

Purpose:

Marks the intron boundaries

Commits the pre-mRNA to splicing

Brings the key cis elements into an organized framework

Step 2: A complex (pre-spliceosome) formation

What happens:

U2 snRNP is recruited to the branch point

U2 replaces SF1/BBP

Important feature:

U2 binds in a way that leaves the branch-point A unpaired/bulged out

Purpose:

Positions the branch-point A for the first reaction

Step 3: B complex formation

What happens:

The U4/U6.U5 tri-snRNP joins the complex

Now the spliceosome contains: U1, U2, U4, U5, U6

Purpose:

Brings in the snRNPs needed to build the active catalytic spliceosome

Step 4: Spliceosome activation

This involves major rearrangements.

What happens:

U1 leaves the 5′ splice site

U6 replaces U1 at the 5′ splice site

U4 leaves, allowing U6 to become active

U2 and U6 interact extensively and form the catalytic core

U5 contacts the exon ends and helps align them

Important concept:

The spliceosome’s catalytic center is largely RNA-based, especially involving U2 and U6 snRNAs.

Purpose:

Convert the spliceosome into its active catalytic form

What is an important note for the chemical steps of the two transesterification reactions involved in splicing?

No net ATP is required for the chemistry itself

ATP is used for assembly and rearrangements, not for the transesterification reactions directly

Summarize the role of each major U sRNP

U1
binds the 5′ splice site
starts splice-site recognition

U2
binds the branch point
positions the branch-point A for catalysis

U4

keeps U6 inactive before activation

U6

replaces U1 at the 5′ splice site
forms the catalytic center with U2

U5

aligns the exon ends for exon ligation

What happens during the first transesterification reaction of splicing? (chatgpt)

Reaction:

The 2′-OH of the branch-point A attacks the phosphate at the 5′ splice site

Result:

The bond between exon 1 and the intron is broken

The 5′ end of the intron is joined to the branch-point A

This creates a lariat intron

Exon 1 is left with a free 3′-OH

Product after reaction 1:

free 5′ exon with a 3′-OH

lariat intron–3′ exon intermediate

Role of cis elements here:

5′ splice site = site of cleavage

branch-point A = nucleophile

U2 positions the branch A

U6/U2 help form the active center

What happens during the second transesterification reaction of splicing? (chatgpt)

Reaction:

The free 3′-OH of exon 1 attacks the phosphate at the 3′ splice site

Result:

Exon 1 and exon 2 are ligated together

The intron lariat is released

Product after reaction 2:

spliced mRNA exon-exon product

released lariat intron

Role of factors here:

U5 snRNP helps align the two exons so ligation can occur accurately

3′ splice site AG is the cleavage/ligation point

What is the Exon Definition concept?

Exon definition is the idea that, in many eukaryotic pre-mRNAs—especially in vertebrates with long introns and short exons—the splicing machinery first recognizes an exon as a unit, rather than recognizing the whole intron first.

What is the core idea behind the Exon Definition concept?

The spliceosome initially identifies an exon by pairing:

the 3′ splice site at the upstream end of the exon with the 5′ splice site at the downstream end of the same exon

So the exon gets “marked” or defined before the final splicing reactions occur.

How does the Exon Definition work?

1. Factors bind around the exon

U2AF and related factors recognize the upstream 3′ splice site, polypyrimidine tract, and branch region

U1 snRNP binds the downstream 5′ splice site

These interactions occur across the exon.

2. The exon is recognized as a legitimate exon

If both splice sites are recognized well, the exon is “defined” as something to keep.

3. Rearrangement to catalytic spliceosome

After exon definition, the complex rearranges so the correct 5′ splice site of one exon is paired with the correct 3′ splice site of the next exon across the intron, allowing actual intron removal.

Why is the Exon Definition concept particularly important to eukaryotes?

In higher eukaryotes:

introns are often very long

exons are relatively short

So it is often easier for the cell to recognize the small exon unit first, instead of trying to identify an entire long intron all at once.

How does the exon definition concept guide splicing in a given tissue?

Exon definition is strongly influenced by tissue-specific RNA-binding proteins.

Key idea

Different tissues express different:

SR proteins

hnRNP proteins

other splicing regulators

These proteins bind cis-regulatory elements on the pre-mRNA, such as:

ESEs = exonic splicing enhancers

ESSs = exonic splicing silencers

ISEs = intronic splicing enhancers

ISSs = intronic splicing silencers

What do proteins like SR proteins and hnRNP proteins do in the concept of the Exon Definition?

If activators bind

They help recruit or stabilize:

U1

U2AF

other spliceosome components

This strengthens exon definition, so the exon is more likely to be included.

If repressors bind

They block splice-site recognition or prevent factor assembly.

This weakens exon definition, so the exon is more likely to be skipped.

How does the exon definition relate to tissue-specific outcome?

Because different tissues express different regulatory proteins, the same exon may be:

well defined and included in one tissue

poorly defined and skipped in another tissue

That is one major basis of alternative splicing.

Example logic

In tissue A, an SR protein binds an ESE → exon recognition is strong → exon included

In tissue B, a repressor binds an ESS → exon recognition is weak → exon skipped

So exon definition is the framework through which tissue-specific splicing regulators decide whether an exon is used.

How many times are micro RNA’s (miRs) cleaved before they become mature miR?

2 Times
Drosha cleavage in the nucleus

Dicer cleavage in the cytoplasm

What is the first step of RNA cleaving in the process of microRNA synthesis?

Step 1: Transcription of the miRNA gene

RNA involved

pri-miRNA (primary miRNA transcript)

Proteins involved

usually RNA polymerase II

What is produced

a long pri-miRNA

contains a stem-loop / hairpin structure

may also have a 5′ cap and poly(A) tail if transcribed by Pol II

Purpose

This is the original precursor RNA that will be processed into a mature miRNA.

What is the second step of RNA cleaving in the process of microRNA synthesis?

Step 2: Nuclear cleavage by the Microprocessor complex

This is the first RNA-cleavage step.

RNA involved

pri-miRNA

Proteins involved

Drosha

an RNase III enzyme

performs the actual cleavage

DGCR8 (called Pasha in some organisms)

an RNA-binding partner of Drosha

recognizes the pri-miRNA hairpin and helps position Drosha correctly

Complex involved

Microprocessor complex = Drosha + DGCR8

What happens

DGCR8 recognizes the hairpin region in the pri-miRNA

Drosha cleaves near the base of the hairpin

this releases a shorter hairpin called the pre-miRNA

Product

pre-miRNA

typically a ~70 nt hairpin with a 2-nt 3′ overhang

Purpose

This cleavage excises the hairpin precursor from the larger pri-miRNA transcript.

What is the third step of RNA cleaving in the process of microRNA synthesis?

Step 3: Export from nucleus to cytoplasm

RNA involved

pre-miRNA

Proteins involved

Exportin-5

nuclear export receptor

recognizes the pre-miRNA hairpin and its 3′ overhang

Ran-GTP

provides energy/directionality for export

What happens

Exportin-5 binds the pre-miRNA

exports it through the nuclear pore into the cytoplasm

Purpose

Moves the pre-miRNA to the cytoplasm for the next cleavage step

What is the fourth step of RNA cleaving in the process of microRNA synthesis?

Step 4: Cytoplasmic cleavage by Dicer

This is the second RNA-cleavage step.

RNA involved

pre-miRNA

Proteins involved

Dicer

another RNase III enzyme

cleaves the loop region of the pre-miRNA hairpin

Accessory proteins

Depending on the organism/cell type, Dicer often works with proteins such as:

TRBP (TAR RNA-binding protein)

PACT

These help Dicer processing and loading into Argonaute.

What happens

Dicer measures from the pre-miRNA end

cleaves the hairpin near the loop

generates a short miRNA duplex

Product

miRNA duplex

about ~21–23 nucleotides

contains:

guide strand

passenger strand

usually has 2-nt 3′ overhangs

Purpose

Produces the short RNA duplex from which the mature miRNA will be selected.

What is the fifth step of RNA cleaving in the process of microRNA synthesis?

Step 5: Loading into Argonaute / RISC assembly

RNA involved

miRNA duplex

Proteins involved

Argonaute (AGO)

core protein of the RISC complex

binds the miRNA and uses it to recognize target mRNAs

RISC-loading machinery

includes Argonaute and associated factors

What happens

the miRNA duplex is loaded into Argonaute

one strand is selected as the guide strand

the other strand is the passenger strand (miRNA*)

Fate of the passenger strand

usually removed and degraded

in some cases may be cleaved or unwound, depending on complementarity and AGO type

Product

mature miRNA-loaded RISC

Purpose

Creates the active silencing complex that can regulate target mRNAs.

What is the sixth step of RNA cleaving in the process of microRNA synthesis?

Step 6: Mature miRNA function

RNA involved

mature guide miRNA

target mRNA

Protein involved

Argonaute

What happens

the miRNA guides Argonaute to target mRNAs by base pairing

this leads to:

translational repression

mRNA destabilization/deadenylation

sometimes mRNA cleavage if complementarity is very high

What are the main RNAs involved in the steps of RNA cleaving in the process of microRNA synthesis?

pri-miRNA = primary transcript

pre-miRNA = Drosha-cleaved hairpin precursor

miRNA duplex = Dicer product

mature miRNA = guide strand in Argonaute

target mRNA = RNA regulated by the mature miRNA

What are the main proteins involved in the steps of RNA cleaving in the process of microRNA synthesis?

RNA polymerase II = transcribes pri-miRNA

Drosha = nuclear RNase III cleavage

DGCR8 = pri-miRNA recognition partner

Exportin-5 = exports pre-miRNA

Ran-GTP = nuclear export factor

Dicer = cytoplasmic RNase III cleavage

TRBP / PACT = Dicer-associated factors

Argonaute (AGO) = effector protein of RISC

miRNAs can reduce gene expression in what two related ways?

1.) translation blocking (translational repression)

2.) mRNA degradation / decay

Both are guided by the same basic principle: a miRNA in Argonaute/RISC base-pairs with a target mRNA, usually in the 3′ UTR.

What is similar between miRNA (miR) blocking and mRNA degradation as controlled by miRs?

Both mechanisms:

are directed by a miRNA

use the RISC/Argonaute complex

require base pairing between the miRNA and target mRNA

reduce the amount of protein produced

usually occur after transcription, so they are post-transcriptional regulation