Genetics, Lecture 10 (Lecture Notes)

8/8/2019 Genetics, Lecture 10 (Lecture Notes)

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13RNA Structureand TranscriptioNabeelAmjad Ham1/11/20


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RNA Structure and Transcription II

The doctor started the lecture talking about the first examination and he was holding

good news about it; each student has been given 3 credits (bonus marks), and

accordingly there is a student, Ghassan Shboul, who has got 30 out of 30. Congrats

Ghassan and go on.

The average for both sections is 17.6!

You can find the slides for this lecture in the file no.4 (Transcription). This lecture

covers the slides (18-44)

- Slide (18) shows the types of RNA polymerases in eukaryotics (btw its u-karyotics

not e-u-karyotics!!):

* We have RNA polymerase I, II, and III, and mitochondrial RNA polymerase,

whereas we have one type of RNA polymerase in prokaryotics, and we saw, in the

previous lecture, how (sigma) factor and the core polymerase together constitute

the holoenzyme.

* Each type of these polymerases is responsible for the synthesis of specific types of

RNA molecules:

RNA polymerase I: (located in the nucleolus) is responsible for the synthesis of 18S,

28S, and 5.8S rRNAs

RNA polymerase II: (located in the nucleus nucleoplasm) is responsible for the

synthesis of mRNA, hnRNA, and snRNA (U1, U2, U4, U5).

RNA polymerase III: (located in the nucleus nucleoplasm) is responsible for the

synthesis of tRNA, 5S rRNA, U6 snRNA, and 7SL RNA.

Mitochondrial RNA polymerase: (located in the mitochondria) is responsible for the

synthesis of all RNA molecules which are expressed within the mitochondria.

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* It was found that these RNA polymerases could be differentiated from one another

by their sensitivity or resistance to a natural molecule called -amanitin which is

extracted from a type of mushroom called Amanita phalloides.

* It was found that: 1) RNA pol I is resistant to -amanitin (its not inhibited by -

amanitin), 2) RNA pol II is highly sensitive to -amanitin, and 3) RNA pol III has low

sensitivity to -amanitin.

* So by using -amanitin we can differentiate between these polymerases.

* -amanitin will inhibit RNA pol II rapidly and as a consequence transcription of genes

will be stopped, so be careful not to eat any type of mushroom. (Id like to lol but I see

that its not a joke!!)

- Slide (19) shows the structure of eukaryotic mRNA. But before that, which RNA pol

is responsible for the synthesis of mRNA?! Its RNA pol II.

* These are the general characteristics of mature eukaryotic mRNA:

1- The circledpart is the 5 end which is also

called the Cap end.

2- The down-arrowed part is the 5 untransl-

ated region.

3- The up-arrowed part is the initiation codon

(AUG) for translation.

4- The light-colored part is the translated region.

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5- The left-arrowed part is the termination codon or signal (UGA) for translation.

(i.e., the translation stops upon this codon).

6- Then we have the 3 untranslated region.

7- And finally the mRNA ends up with the (AAUAAA) polyadenylation signal whose

role is to add a poly (A) tail to the mRNA at the 3 end.

- Recall that not all parts of mRNA will be translated.

- Questions about this slide from the students:

Q1: not heard, but it was about exons and introns.

A: there are no introns in mRNA; it contains exons, and untranslated regions.

Q2: what are the untranslated regions (what do they contain)?!

A: they are promoters, regulatory sequences, and other sequences.

Q3: So the untranslated regions are transcribed but not translated?!

A: yes, they are transcribed from DNA into mRNA but not translated from mRNA into

protein.

Q4: not heard, but I think it was about the source of the whole mRNA.

A: This all has come from the template, except this. (not sure what does this refer

to, but I had a rapid look at Wikipedia and I think he meant the Cap and the Poly (A)

Tale ).

Q5: not heard, but I think its about the polyadenylation signal.

A: Its a signal for a specific enzyme, as you well see, to come and cleave in this region

and to add a poly (A) about 200 AMPs to form the poly (A) tale.

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Q6: what is the significance of the poly (A) tale for the mRNA molecule?!

A: stabilization, and preventing degradation of mRNA molecule. And the same thing for

the cap (5 end).

Efff, questions never end!! There was another question whose answer was the same as

Q6. Really students questions are impossible to hear, so sorry for the complicated

thing. (Dear colleagues, when you ask, please, raise your voice to the max. taking in

consideration that there is a poor guy here trying to hear you!)

- Slide (20) shows the complexity of mRNA molecules in eukaryotic cells.

* the complexity concept here is the same as studied in the DNA lectures, (depends

on size, and repeating).

* We have high, intermediate, and low abundant classes. The abundance means how

many copies of that mRNA are present per cell.

* The high abundant molecules have 12,000 copies per cell, and there are 9 different

species of these mRNA molecules (low complexity).

* The intermediate abundant molecules have 300 copies per cell, and there are 700

different species of these mRNA molecules (intermediate complexity).

* The low abundant molecules have 15 copies per cell, and there are 11,500 different

species of these mRNA molecules (high complexity).

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Laugh with Dr Nabeel:

Dr Nabeel asked Mr Ali to close the doors claiming that there are students coming

in after half an hour of the lectures beginning. Damn it, Im still on minute 16:09!!

What a lie!


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BUT to be honest, the doctor said that the high abundant molecules are high-

complexity molecules, too. And after looking at slides (the notes of the slides) I see

that this is wrong.

SO high abundant low-complexity.

* The abundance of mRNA molecules depends on the transcription rate, (directly

proportional), e.g: the high-abundance mRNAs have high rates of transcription so they

are always available to be expressed.

- Slide (21) represents the promoter region of a eukaryotic gene. (this structure is a

DNA not an mRNA)

* The region labeled with (+1) is the beginning of transcription. We are concerned in

this slide with the promoter region and we also have specific sequences known as

transcription elements

* The promoter is a specific sequence of DNA that is consensual for all eukaryotic

genes, located just upstream (and only upstream) the beginning of the transcription

unit (the transcription unit starts from +1 region and continues downstream till the

end of the gene). (upstream means toward the 5 end, or to the left).

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P

TE


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* The promoters function is to give (allow) the basal expression of the gene which is

the amount of the genes expression without regulation (positive, and negative

effectors.) just the basal level.

* The transcription elements are also specific sequences of DNA located upstream,

downstream, or within the gene. They will regulate the expression of the gene.

so the promoter is for the basal expression, and TEs will regulate (increase, or

decrease) the rate of transcription of a specific gene.

* These TEs are located in all genes and they will bind to specific proteins in each

tissue to affect the expression of specific genes in theses tissues.

* TEs will perform their function wherever they found (upstream, downstream, or

within the gene); they could be as far from the promoter region or the transcription

unit as 50 kb (50,000 base pairs) upstream or downstream the promoter region.

* The TEs include different types of DNA sequences, one of them is the so-called

enhancers. These enhancers, as the name states, are DNA sequences that can

increase (enhance) the rate of transcription. Another type is what we call silencers

which are specific DNA sequences that can decrease (inhibit) the rate of

transcription. Also we have what is called the response elements that will bind

specifically to some proteins to increase or decrease the rate of expression.

- Slide (22) shows the possible places of TEs (enhancers, silencers, or response

elements); they could be located in a close proximity of the promoter, they could be

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tens of kilo bases away from the promoter (upstream, or downstream), or they could

be within the gene.

Many unheard questions were asked, I think, about the promoter region and the

answers were the same; the promoter is located within the 5 untranslated region and

that means it will be transcribed but not translated.

- Slide (23): Sometimes in some genes youll see another sequence (TE) called LCR (or

Locus Control Region).

* When there are genes expressed in coordination with each other, these LCRs will

control the expression of these coordinated genes. Supposedly we have a locus on a

chromosome containing 3 genes; 1,2, and 3, and these genes must be expressed in a

sequence; 1, then 2, and finally 3, or 1,3, and 2, etc. What controls the expression of

these genes on this locus is a specific sequence of DNA called LCR.

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* locus refers to the SITE of the gene where it is located on the chromosome. And

these loci are some positions that will bind to LCRs which in turn will control the

expression of the coordinated genes.

- Slide (24) shows the regulatory elements for a typical eukaryotic gene, we have the

promoter, and the TE.

* We have numbered regions; +1, -25,

-50, -80, -95, and -130.

* These numbered regions within the promoter have consensual sequences allover the

eukaryotic genes. Each one of them has invariant nucleotides in its sequence. Also

each one is bound to a specific protein, literally, a transcription factor TF.

* The significance of these sequences is to form the Pre-initiation Complex of

transcription, PIC.

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* PIC is composed of RNA pol and other initiation transcription factors in order to

initiate transcription. So the whole mechanism helps RNA pol bind to specific region

and start transcription at +1 region.

* Without these transcription factors RNA pol will not be able to bind to the promoter

to start transcription at +1 region.

* Now we turn to the structures of these sequences and the specific proteins they

bind:

1- TATA box: at position -25, and it is rich in T, and A nucleotides. (recall from the

last lecture that in prokaryotes it was at position -10 and we called it the Pribnow

box). It has the sequence of TATAAAA, the underlined nucleotides are invariant ones

allover the TATA boxes in all eukaryotic promoters. It is located approximately 25-30

base pairs upstream the +1 region. It determines the exact starting site. As well as it

binds to a protein called TATA Binding Protein TBP which is part of one of the

important transcription factors that is called TFIID (described later).

2- GC box: there are, at least, two GC boxes, and they are important; they bind to a

specific protein called Specificity Factor1 Sp1.

3- CAAT box: at -80, it binds to CAAT box Transcription Factor CTF.

4- Octamer: a transcription element located at -130. It binds a protein called Octamer

Transcription Factor OTF.

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* These proteins that will bind to these sequences (in the promoter) are very

important to help RNA pol bind and initiate transcription, and if any of these

sequences has been mutated a lot of diseases will result, and well see later on some

examples on these diseases.

- Slide (25): We have general transcription factors, and specific transcription factors

for specific tissues. And well see in a minute that the transcription factor TFIID is

the most important to initiate transcription.

* II in TFIID is related to RNA pol II; TFIID will bind to RNA pol II

* TFIID is a multisubunit protein which will bind to the TATA box by the TATA

binding protein (TBP) to initiate the formation of transcription complex (apparatus),

and once TBP binds to the TATA box many general transcription factors will also bind

to the complex in order to form pre-initiation complex (PIC) of transcription.

-Slides (26-30) demonstrate what happens during the whole process (formation of

PIC).

* Here we dont have RNA pol II yet.

* So TBP subunit will bind to the TATA box,

then the other general transcription factors

(TBP Associated Factors TAFs); A, B, E, F, H,

and J will bind to the TFIID.

* This is the starting point of formation of PIC.

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* Once that is done it will be easy for RNA

pol II to come and bind to this complex because

there is a specific binding between these

proteins and RNA pol II, and at the same time

RNA pol II will bind to that region of DNA to

which this complex is being bound.

* So RNA pol II without these proteins will be unable to recognize the promoter

region to start transcription at +1 region.

* This is the PIC and these new parts are

accessory proteins (specific TFs), among them

are those that bind to TEs that could be

anywhere on the gene, and this binding is for

regulation (either positive, or negative), and as

we said these TEs could be away from the

promoter region about 100 kb upstream or downstream. These proteins will increase

the efficiency of RNA pol II to start transcription.

* Now, the proteins (factors) that bind to the TEs are specific in each tissue, and this

is the cause of differentiation of gene expression (the basis of tissue

differentiation), this explains why our liver cells are different from our brain cells.

* The mechanism of differentiation in brief: each type of cells has specific type of

proteins (factors), not found in other types of cells, that will bind to the TEs, and this

will lead to the expression of specific genes in that type of cells.

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* Once these proteins bind they must interact

with the PIC, in order to increase the initiation

tendency of RNA pol II, so there must be what

is called DNA looping, in order to make these

proteins in a close proximity to the promoters

binding proteins, and once this looping occurs

there will be an interaction between the elements and the complex, and there will be

an effect on RNA pol II (conformational change, phosphorylation-dephosphorylation,

or acetylation-deacetylation) to start the process of transcription.

- Reminder: We are talking about transcription so we are concerned with, and only

with, RNA pol II because its the only type of polymerases that can synthesize mRNA.

* TFIIH has many functions:

1) 5 3 and 3 5 helicase activities ; to open (unwind) the two DNA strands for

RNA pol II to start transcription.

3) Kinase activity; to phosphorylate RNA pol II.

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Recall when we say TFIIH we mean the following:

TF: Transcription Factor.

II: related to RNA pol II.

H: the factor that we are talking about.


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In the future this sample might double, and we are not supposed to memorize each and

every single one of them, except those I mention. We are concerned with AP1, and

Sp1. (in the upcoming pages you will discover that he meant ER instead of Sp1. Good,

keep going!)

* But before we start talking about these transcription factors, there is a piece of

information you should know; once RNA pol II leaves the PIC and starts transcription

some of these transcription factors stay at the promoter region, and at the same time

some elongation transcription factors will bind to RNA pol II to help it in elongation

till it reaches the termination signal where RNA pol II will be dissociated, mRNA will

be released, and transcription will be ended (terminated).

* Within the chromosome, the genes are there, and TEs are scattered everywhere. If

these TEs work on any gene there will be transcription of that gene all the time, in

addition to the basal expression which is found all the time without regulation and we

call it Constitutive Expression. So these TEs will bind to the promoter region so that

the gene will be expressed always.

* It was found that for each gene there are boundaries (sequences that TEs will not

go beyond them), and that means the TEs which are specific for a gene will not go and

activate transcription of other genes because of these restrictive boundaries.

- Slide (32): We have transcription factors composed of 1) Basic region, and 2) Leucine

zipper. We call these TFs Basic region-leucine zipper (bZIP) transcription factors.

* Leucine zipper: functions in dimerization; it is composed of Leucine amino acid

residues that will help the TF work (act) in a dimer (they will not be active in the

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monomer form, they will be active only in the dimer form two-subunit-form). So this

leucine zipper will form a structure that will help combine two subunits together (by

hydrophobic forces) in a dimer form to be functional (stabilized structure).

* The basic region: important for DNA binding; since it is composed of basic amino

acids (Lysine, and Arginine). (Arginine has Guanidino group rather than the simple Amino group; it

is more complicated). How does this basic region work?! See next.

* The DNAs surface is negatively charged, hence these basic regions help TFs bind

the DNA (basic region is positive and DNAs surface is negative electrostatic

interactions).

* This is the whole story explained: we have TEs which are DNA sequences that will

bind to proteins (TFs), and these proteins have specific characteristics including;

Basic region, and Leucine zipper.

* The forces generated by Leucines are hydrophobic forces since Leucine is a non-

polar amino acid

*An example on this family of TFs that has these two features is AP1 (Activator

Protein- 1): it will activate transcription, it is composed of two subunits; one subunit is

called the Fos subunit, and the second subunit is called the Jun subunit, or both

subunits are Jun subunits (Jun-Jun dimer). These Fos, and Jun proteins are proto-

oncogenes protein products.

* Fos and Jun proto-oncogenes are important in regulating growth and development of

cells.

* AP1 is involved in the regulation of gene expression, and controlled by various growth

factors, hormones, tumor promoters, neuronal stimulation, and cellular stress.

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- Slides (33-38) show the mechanism by which the AP1 activates the expression of a

gene depending on the structural features that AP1 has (Leucine zipper, and the basic

region).

* This (the figure) is AP1 which is composed of a

Fos subunit, and a Jun subunit.

* As you can see, there are basic regions import-

ant for the binding to DNA, and there are also

Leucine-rich regions which are composed of

-helices, and, as you can see, one leucine residue

represented by this ball (circle) is present at

every 7th position in these -helices. So you see leucines on both monomers (subunits).

* Because of the hydrophobic interactions

between Leucine zippers, (look at the figure)

there will be dimerization.

* The basic regions will be for DNA binding.

* If you dont have these structural features

for AP1 protein, transcription will not take place;

efficiency of transcription for that gene will

remain at the basal level (no increase, nor activa-

tion of transcription for that specific gene). All

these features constitute the active AP1 dimer.

* What weve talked about is the structural features of AP1, but how does this

structure work?! See next.

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* We have a specific DNA sequence; it is a

palindrome (i.e., you can read one strand in the

53 direction in the same way you read the

other strand in the 53 direction). So we

need a dimer rather than a monomer.

* The AP1 protein will bind to the DNA in the

minor or major groove using the basic regions.

* Once it binds to this specific sequence (if it

is far, it will loop to become closer) there is an

enzyme called Jun N-terminal kinase JNK

that will phosphorylate Jun, then other TFs will

be phosphorylated, resulting in the activation of RNA pol II. (the doctor said that Jun

itself has the kinase activity but the information above are as mentioned in the slides).

- Slide (39) indicates another TF which is the Estrogen Receptor factor ER.

- Slide (40): Some transcription factors are called Zinc fingers which are another

example on TFs that regulate eukaryotic transcription.

* This zinc component of the TF, as a metal, is important; it acts as a cofactor to

activate the TF, in order to activate the PIC (remember: PIC is the pre-initiation complex).

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Dear Dr Nabeel,

Please, Im begging you not to standnear the speakers, you caused a

migraine to me!!


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* Zinc finger (see the figure) means:

We have a zinc atom coordinated with four

amino acid residues; two Cysteines, and two

Histidines. We call this zinc finger

C2H2 zinc finger. Sometimes it is C4 zinc

finger (4 cysteines), and sometimes it is H4

zinc finger (4 histidines).

* This Zn atom will stabilize the structure.

* Usually, Zn finger is found in this form (at the right in the figure); which is

composed of an -helix and an anti parallel -pleated sheet.

- Slide (41) shows the role of Zn fingers in ER.

* This protein (in the figure) represents an

ER with its main features:

We have the N-terminal at the left-most,

the C-terminal at the right-most, the trans-

activation domain, the DNA binding domain

(region of Zn fingers) C4 + C5, and a domain

for hormone binding, dimerization, and trans-

activation.

* In the DNA binding domain there is a C4 + C5 zinc finger pair; one finger contains

four cysteines and the other contains five cysteines.

- Slide (44): (it is not clear, so refer to the original slide). Once the hormone binds to

the ER (at the hormone binding domain) the whole complex will enter the nucleus, and

bind to a specific sequence called Response Element, to activate a gene which is

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important for an enzyme used in Estrogen synthesis.

* These features are significance for Estrogen synthesis; if there is a defect in this

gene Estrogen will not be synthesized.

* The doctor didnt mention anything about slides (42, and 43), so here are the slides,

read them (they are easy!)

The End

Sorry for the delay, but honestly its not my fault, since I received the record

yesterday.

To all those who look for the light in the darkness, when

you feel like giving up remember that what doesnt kill you

makes you stronger.

Again, I dedicate this work to you.

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Amjad Hamad

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Genetics, Lecture 10 (Lecture Notes)