Genetics, Lecture 11 (Lecture Notes)

8/8/2019 Genetics, Lecture 11 (Lecture Notes)

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Genetics Lecture 11Wednesday, 3/11/2010Done By: Shayma' Naghnaghia

Transcription Lectures (3)(Transcription factors, Mutations affecting promoters,RNA processing)

* Transcription Factors *(continued)

Last time, I've started talking about the regulation of eukaryotic gene

expression. I've already talked about the Leucine Zipper transcription

factor and the Zinc Finger transcription factor. I've also talked about AP1

as an example of a transcription factor that has a Leucine Zipper structure;

without this structure, it will not take its quaternary structure nor its

function. The same thing for the Estrogen Receptor, it's an example of a

transcription factor that has a zinc finger as part of its structure; and

because of this zinc finger structure, it takes the correct tertiary and

quaternary structures, and thus it's functional to do its job.

(Slide 43)

In this slide, we can see the

gene that we are concernedwith and the transcription

factor (the estrogen receptor).

The estrogen receptor is a

dimer and it has 2 zinc fingers;

these zinc fingers are used to

read the sequence in the major

groove.

Note that the 2 zinc fingersread the same sequence

(AGGTCANNNTGACCT). Is it

possible that the 1st zinc finger

read the same sequence as the 2nd zinc finger? How can you explain this?

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>> The sequence is palindromic; this means if you read from 5`to 3`on both

strands, it will be the same thing for the recognition sequence, and this is

because the zinc fingers are dimers and read a specific sequence:

5-AGGTCANNNTGACCT-33-TCCAGTNNNACTGGA-5

(Slide 44)

Here we can see the protein that

carries the estrogen. The estrogen

will leave the protein & enter the

target cell and bind to the

estrogen receptor.

The properties of the estrogen

receptor:

>> It binds the DNA (the gene) by

entering the nucleus and

recognizing a specific sequence

called Estrogen Response Element

(ERE) via zinc fingers. Without

these zinc fingers in the estrogen

receptor, it will not bind to the gene that is responsible to synthesize

estrogen.

So, zinc fingers are one of the needs that will help thetranscription factor to activate specific genes forspecific synthesis of molecules or hormones; like

estrogen.

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* Mutations Affecting Promoters *

Previously, I mentioned that promoters are very important .They are

important to initiate transcription and help RNA polymerase to bind andstart transcription at (+1), as well as other transcriptional elements that will

bind to specific proteins; these proteins will interact with the promoter

binding region in order to initiate transcription.

What will happen if there are mutations in the promoter?

If this mutation was very lethal (it changed the sequence of the promoter),

RNA-polymerase will not be able to bind, so transcription will stop.

Now, we'll see an example of a mutation in the promoter and the

consequences of this. (Slides 45 & 46)

Here we see the promoter for a gene called Factor IX. What is theimportance of Factor IX?

>> Blood coagulation; it's important as a blood clotting factor.

The gene of Factor IX is located on the X chromosome. Its transcribed

region has 8 exons.

The promoter for this gene has overlapping binding sites for the Androgen

Receptor (AR) and Hypatocyte Nuclear Factor-4 (HNF4).

- The transcription factor HNF4 will bind to the region from -27 to -15.- The transcription factor AR will bind to the region from -36 to -22.

* Note: HNF4 & AR are both Zinc Fingers *

Once they bind, they will activate the promoter, RNA-polymerase will bind,

and transcription will start.

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If a mutation happens in the promoter region, what will you expect? If this

mutation stops RNA polymerase, what will you expect to happen?

>> Transcription will be blocked.

Let's talk about two mutations that happen in the promoter region of FactorIX gene to see the results:

The first mutation:It occurs at the -20 region. The transcription factor for this gene will not

be transcribed, as a result the patient will have Hemophilia; specifically

Hemophilia B Leyden.

Because of this mutation, HNF-4 will be unable to bind, and thus RNA-

polymerase will be unable to bind (or the binding is very low), so the rate oftranscription will be very low and Factor IX won't be produced, so no

clotting will happen. This causes Hemophilia B Leyden.

It was found that this type of Hemophilia will be improved at puberty. The

mutation happens at the -20 region which is the region where only the HNF-

4 binds, so because of this mutation HNF-4 can't bind. But AR is NOT

affected; this mutation is outside its binding site, so it will continue binding

and activating the gene; and the gene will be expressed. This happens when

the concentration of AR is high; this occurs during puberty when the level of

testosterone increases, this increases the activity of AR, as a result

Hemophilia will improve (little Hemophilia).

The second mutation:It occurs at the -26 region. This mutation results in Hemophilia B

Brandenburg, and it will not be improved at puberty because the site of

mutation is in the overlap region between AR & HNF-4, so both factors won't

be able to bind, as a result Hemophilia will occur and won't be improved at

puberty.

Theoretically, we can help the patient by giving him AR or HNF-4. Practically,

we help the patient by giving him clotting factor IX.

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* RNA Processing *

The doctor mentioned the learning objectives and the topics he'll cover in

RNA processing; you can find these in slides 47, 48, 49.

Let's start with the steps of RNA processing (slide 50).

What we see here is a primary transcript mRNA; this means that it's an

immature and nonfunctional mRNA that has exons and introns, not having a

cap nor a tail (it cannot be translated into protein).

In order to be mature and functional, it must be processed and pass throughspecific stages, so here are the major steps of mRNA processing:

1- Capping: the process in which we cover the 5`end.

2- Polyadenylation: adding a poly(A) tail to the 3`end.

3- Splicing: removing introns and joining exons together.

The end result is an mRNA which is capped, having a polyadenylated tail, and

spliced; this is called the mature mRNA which is functional and ready for

translation.

* Capping (Slide 51)Capping occurs during transcription, so it is co-transcriptionally. It

functions in protection against degradation and against nucleases, and for

mRNA stability.

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What does the 5` end have? It has a nucleotide with 3 phosphate groups (C5

of the sugar of the nucleotide at the 5` end has 3 phosphate groups), while

the second nucleotide has only one phosphate group, because PPi is released

when nucleotides bind together.

We have two enzymes playing a role in the process of capping:

- Guanylyltransferase: it transfers a Guanine residue to the 5`end.

- Methyltransferase: it adds a methyl group to the Guanine, and at the

same time it will add another 2 methyl groups to the next 2 nucleotide.

At the end of this process, we'll have mGpppNmpNm (you realize that there

are 3 methyl groups). So, we make a covering for the 5`end of the mRNA by

this methylated Guanine and the other methyl groups.

Another important thing in this capping process is that we get a 5`-5`

phosphodiester bond! We're used to a 5`-3` phosphodiester bond not 5`-

5`. How does this bond form?

>> The 3` of the first nucleotide is busy; it is making a 3`-5` bond with the

nearby nucleotide, so as a result the 5` of Guanine makes a bond with the

free 5` of the first nucleotide.

So, capping is the process of adding methylguanineto the 5` end and adding another 2 methyl groups to

the following 2 nucleotides.

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Not a convincing explanation I guess!

I think the Dr. made a mistake by saying 5'-5'phosphodiester bond; it should be 5'-5' triphosphatebond!The way this bond forms is like this:- One of the three terminal phosphate groups on the 5' endis removed, leaving two terminal phosphates.- GTP is added to the terminal phosphates (by guanylyltransferase), losing two phosphate groups (from the GTP) inthe process.

* This results in the 5' - 5' triphosphate bond *

.. Check the figure on the next page ..


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*

Polyadenylation (Slide 52)By the end of transcription, RNA polymerase reaches a signal AAUAAA; this

is called the polyadenylation signal. There is an enzyme with its accessory

protein called poly(A) polymerase; it will recognize this sequence and then a

nuclease will cleave after the polyadenylation signal (approximately 10-30

nucleotides downstream of the AAUAAA signal), and the poly(A) polymerase

will add poly(A) after this region (about 200 adenylate residues) forming the

tail. This is also to protect mRNA from degradation.

A student asks if the tail acts as a stop signal, and the answer was NO!

* Splicing (Slide 53, 54)How are introns removed?

- Two cleavageligation reactions

- Transesterification reactions: this means that we have a bond (G-G) and

we have 2-OH-A, what happens is exchange of ester bonds; a cleavage

happens to (G-G) bond, then A binds to G. This is the transesterification

reaction; replacement of an ester bond by another ester bond.

What is involved in this reaction is a branch site in introns. Any intron has a

branch site; this site is always available. It has a recognition sequence and

the active nucleotide in this site is Adenosine (A).

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This Adenosine in the branch site uses the 2-OH and makes a nucleophilic

attack to the 5ènd of the 1

st

intron, and it makes a 2`-5` phosphodiesterbond. Are you familiar with this? It's the 1st time you see it! Why didn't it

make a 3`-5` bond?

Since the branch site adenosine is part of the RNA polynucleotide chain, its

3' hydroxyl is already involved in a covalent bond (the 3'-5' phosphodiester

bond). Therefore, it utilizes its 2' hydroxyl for this reaction. Attack by

this 2' hydroxyl breaks the bond at the 5' end of the intron by forming a

bond between the 5' end of the intron and this branch site adenosine.

This leaves the 3ènd of the 1st exon with a free 3ÒH group; this OH

makes a nucleophilic attack to the 3ènd of the 1st intron to make another

transesterfication.

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This transesterification reaction will ligate both exons together; you will end

up with the 1st exon ligated to the 2nd exon and the intron is out. This is

called the Lariat Model of intron splicing. (Lariat = (

Is this process enzymatic or self autocatalysis?

It is self autocatalysis, so this is an example of the RNA acting as an

enzyme; it will cleave itself like any other enzyme cleaving pieces of RNA or

DNA, but this autocatalysis requires accessory proteins to help with the

process.

As a summary for the splicing process:Removal of an intron is carried out by mRNA splicing.This is accomplished by two cleavage-ligationreactions: one that cleaves at the 5' end of the intron

and the second that cleaves at the 3' end of theintron. Each of these reactions is a so-calledtransesterification reaction; one phosphodiester bondis exchanged for another. Thus, because bond energyis preserved, there is no need for additional inputenergy for these reactions. Furthermore, thesereactions are not catalyzed by protein enzymes; theyare mediated by the reacting RNAs themselves.

Recognition of Splice Sites (slide 55)Do you remember the snRNA (small nuclear RNA) and the associatedproteins snRNP (small nuclear ribonucleoproteins)? Now we'll talk about

their role. Examples of snRNA are: U1, U2, U3, U4,

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Remember: the splicing of introns is an engineered process; not haphazard;

there are recognition sequences for the beginning of any intron and another

recognition sequence for its end.

These are known as the 5`recognition sequence (upstream) and the

3`recognition sequence (downstream). These recognition sequences areconsensus in all introns. They have some invariant ribonucleotides in all

introns in eukaryotic genes.

The 5`recoginition sequence of the intron is called the donor site, and the

3` recognition sequence is called the acceptor site.

G/GUAAGU is the "donor site" consensus in all intron donor sites, with GU as

invariant; if you check ALL the donor sites for ALL introns in ALL eukaryotic

genes, you'll find the same GU!

The same thing in all the "acceptor sites"; they have the consensusYYYYYNYAG/G, with AG as invariant in ALL acceptor sites of ALL introns in

ALL eukaryotic genes!

It was found that U1 will recognize the donor sequence that will help the

branch site to bind and cleave there, and U2 will recognize the (A) of the

branch site. Also, there are other (U)s that will recognize other sites.

How does U1 recognize the donor site? Or how does U2 recognize the (A) of

the branch site?

>> This is done by a complementary sequence; these (snRNA)s have

complimentary sequences that will bind to its corresponding site.

(Slides 56, 57, 58)You can see here (figure on next page) the hnRNP (heterogeneous nuclear

ribonucleoproteins) and the snRNPs which are called snurps (snRNA +

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protein). These snurps bind to the exons and introns, why?! What is the

significance of the binding of these snurps to the exons and introns?

Always remember: the 3D structure in molecular genetics is very important.

So, these snurps will help the donor site and the acceptor site to take their

tertiary structure in order to be recognized by the attackers (the attacker

here is the branch site (U2)).

This is what happens:

1- U1 & U2 are bond to the donor site and the branch site, respectively.

2- U4, U5, and U6 will come and bind to specific sites.

3- U4 and U1 are released, and we have U2, U5, and U6 still binding.U5 is the background where the reaction between U2 and the 5`end

of the intron occurs.

The biochemical nucleophilic

conditions are ready by the

presence of U5 which causes U2 to

attack the 5`end of the intron. U2

makes transesterfication and the

intron exits as a lariat, and ligationbetween the 2 exons will occurs.

It was found that there are auto

antibodies against snurps, and that

will lead to a syndrome called

Systemic Lupus Erythematosus

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(SLE); an autoimmune diseases. There could be some proteins or

compliments that neutralize these antibodies.

Frequency of bases in each position of the splice sites

(Slide 59)Did you figure out the

message of all these

numbers?It is not for

memorization at all! :D

It is just a statistical

analysis of the

probability of what

nucleotides we have in

the donor and acceptor

sites of splicing.

This why GU is

invariant in donor sites

(100%) and AG is

invariant in acceptor

sites (100%). The other

ones vary, but have a

higher probability for one of the nucleotides.

Mutations that disrupt splicing (slides 60, 61)Introduction:Now, I want to give you examples of mutations in splicing sites that will

result in genetic diseases. The genetic disease Im going to talk about is -

thalassemia; it is a genetic disease in -globin gene that is responsible to

express the -globin polypeptide chain which is one of the subunits of the

hemoglobin molecule that is important in O2 transport. So, if there is a

mutation in -globin, we'll have a defective -globin subunit, and thus a

defective hemoglobin structure, and thus a defect in O2 transport. This is

called thalassemia, and because thalassemia is a defect in -globin gene, it iscalled -thalassemia. If the defect is in the -globin gene, then it's called -

thalassemia.

The difference between +-thalassemia and -thalassemia:

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In the case of -thalassemia, there are hundreds of mutations discovered to

cause -thalassemia; a specific example of these mutations is in the splicing

sites (or in the introns). In +-thalassemia, little of the -globin gene is

expressed, and thus few -hemoglobin subunits will be found.

In -thalassemia, there's no expression for the -globin gene, so the -hemoglobin subunits won't be found at all.

What you see here is a -globin gene which is composed of 3 exons and 2

introns; this shape of the introns is because they are going to be removed.

The donor site has a consensus invariant sequence GU, and the acceptor site

has a consensus invariant sequence AG.

Now, we'll talk about the splicing of intron 2. As we said, the normal donor

site has GU (invariant) and the acceptor site has AG (invariant). As we know,

U2 will come with U5 and remove this intron and ligate the 2 exons together

(exon 2 & exon 3).

-thalassemia:

It was found that there is a mutation that happens in the acceptor site of

intron 2, so it becomes GG instead of the normal AG; this mutant site won't

be recognized as an acceptor site anymore; instead the Cryptic Splice Site

in intron 2 (which has an AG sequence) will be used. As a result, part of the

intron is retained (stays) in the gene; unfortunately, this retained part has a

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stop codon that stops the process of translation, so giving a truncated -

globin chain; so instead of giving 150 amino acids, it gives only 25 amino acids

and no complete -globin will be formed. This is -thalassemia.

Keep in mind that this Cryptic Splice Site is always found in intron 2, butit's never recognized as an acceptor site unless there is a mutation in the

normal acceptor site.

+-thalassemia:

A mutation occurs in intron 1 (NOT in its acceptor site). We see that the

normal sequence GG became AG (AG, as we mentioned before, is the

consensus sequence of the acceptor site). The normal AG acceptor site is

still present. So here we have 2 choices:

- The mutated AG may be used 90% of the times; the result is that nocomplete -globin will be produced, because the retained part of the intron 1

has a stop codon that stops translation.

- The normal AG will be used 10% of the times; this results in complete

formation of -globin chains.

So, we'll have complete -globin formed, but in a very small amount (10%).

This is +-thalassemia (Intron 1).

There is another type of +-thalassemia that the Dr. didn't explain. Checkslide 61. It's the same idea as the previous one, but it involves the donorsite in exon 1 instead of the acceptor site in intron 1. The mutated donor

site is used 40% of the times, while the normal donor site is used 60% of the

times.

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The previous explanation isn't the Dr.'s exact words; the Dr.missed up the process and made a couple of mistakes.

Nothing is mentioned in the book aboutThalassemia, and Icouldn't find anything clear on the Internet. I explained the

process depending on the figures. Hope the ideas are clear


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Patterns of alternative exon usage (slide 62)Alternative splicing means that from one gene we can produce different

(mRNA)s that will be translated into different proteins, but they're highly

related to each other (isoforms).

There are different mechanisms for alternative splicing:

These are the cassette model, the mutually exclusive model, the internal

acceptor site, and the alternative promoters.

* The Dr. quickly (shalfa2a :P) explained those

mechanisms; unfortunately, we understood nothing!

(Slide 63)This slide shows the Troponin T gene which produces Troponin T (a protein in

the muscles). This gene undergoes alternative splicing (using the different

previous models) to produce 64 different isoforms of Troponin T.

* The END *

Done by:

Shayma B. Naghnaghia

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Genetics, Lecture 11 (Lecture Notes)