Gene Expression I: General Mechanisms and the Properties of Transcription Factors Flashcards

1
Q

Explain how the expression of genes determine the properties and functions of cells and tissues

A

Each differentiated cell type has a characteristic gene expression pattern. Differentiated cells produce progeny cells with the same expression pattern of paternal cells. The “different” cells have the same genome, but their proteomes and transcriptomes are different. The knowledge of what genes to express are passed down

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2
Q

The human genome contains approximately ________ genes

A

25,000 genes

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3
Q

A single cell expresses between ________ and _______ transcripts simultaneously

A

10,000 and 20,000 transcripts

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4
Q

Transcriptome

A

All mRNAs expressed by a cell

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5
Q

Proteome

A

All proteins expressed by a cell

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6
Q

In what 3 ways are genes regulated?

A

1) Regulated in terms of SPACE. Cells in the region of a developing embryo in the head need to express genes to form the brain so they express different mRNAs for this 2) Regulated in terms of TIME: Some genes need to be on or off during development 3) Regulated in terms of QUANTITY: Skin is made of stratified squamous epithelial cells with keratin where cells in the mouth are the same cells but without keratin.

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7
Q

Diagram the structure of a gene showing start of transcription, coding region of exons and introns, the promoter region containing the proximal regulatory elements, the positioning of distal regulatory elements (sequences) and insulator elements.

A

The regions that regulate the gene are larger than the coding region itself. The regulatory regions tend to be 6-20 nucleotides in length and they are very specific meaning they bind to a specific regulatory element.

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8
Q

Promoter

A

The promoter consists of the core promoter and the proximal promoter regions.

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9
Q

Core Promoter Region

A

The core promoter comprises DNA sequences within -40 to +40 if the transcriptional start site. It includes elements like the TATA box, BRE, INR and DPE

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10
Q

Proximal Promoter Region

A

Comprises the DNA sequence within -200 to +50 of the transcriptional start site. It contains elements such as the CCAAT box. These elements act to promote transcription in and orientation and distance DEPENDENT manner. The regulatory regions that are further away tend to be distance and orientation INDEPENDENT. In other words if they are flipped or moved upstream they should still work just fine.

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11
Q

Distal Regulatory Elements

A

These are enhancers and silencers which are located further away from the transcriptional start site. They bind to transcription factor proteins that interact through DNA looping with basal transcription machinery. These elements appear to be distance and orientation INDEPENDENT relative to the transcriptional start site.

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12
Q

Insulator Elements

A

These are boundary elements at the ends of gene regions that function to insulate or block the influence of either positive or negative DNA elements from affecting adjacent genes. They prevent the migration of heterochromatin into an open gene until being expressed.

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13
Q

Explain the role of the Mediator complex

A

Since some of the regulatory elements are far away, a mediator complex is involved in helping to transmit the signals to polymerase. We know mediator is essential because it will transfer all of these different signals whether positive or negative, to the RNA polymerase and it will then function based on the net affect. If primaritly inhibitors, it will stop, if promoter, we get more. Mediator is made up of ~25 different proteins and they change based on cell type and cell context.

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14
Q

What are the different regulatory elements and their characteristics

A

1) enhancers: these promote transcription
2) Silencers (repressors): These inhibit transcription

They are short DNA sequences that bind transcription factors/regulatory proteins

They can be located up to 50kb from the transcriptional start site

Can be upstream or downstream of the transcriptional start site or within introns or exons

In the proximal promoter region, they are orientation and distance DEPENDENT and outside this region they are INDEPENDENT

Each element may bind multiple transcription factors.

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15
Q

Recognize that ____1_____ of histones generates regions of heterochromatin (condensed, inert chromatin) while ____2____ generates euchromatin (open, active chromatin). Histone ____3_____ can either signal gene repression or expression, depending on the residue methylated and the number of methyl groups added.

A

1) Deactylation
2) Acetylation
3) Methylation

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16
Q

Transcription Initiation

A

Generally speaking, there is going to be at least one element that will be exposed and able to be engaged by a transcription factor. When this occurs, the transcription factor begins to loosen up the heterochromatin.

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17
Q

Alterations in Chromatin Structure

A

Nucleosome remodeling, nucleosome removal, histone replacement, and certain types of histone modifications favor transcription initiation. These alterations increase the accessibility of DNA and facilitate the binding of RNA polymerase and the general transcription factors.

Binding of a transcription regulator can recruit:

1) Recruitment of chromatin remodeling complexes: They are ATP dependent complexes that slide the nucleosome and expose DNA
2) Histone chaperones: Some remove histones, some replace histones. Some histones that prefer to be in heterochromatin may just be replaced.
3) Histone modifying enzymes: Enzymes that are going to acetylate or methylate lysine in the N-terminal tails of histone.

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18
Q

Histone Acetylation

A

Lysine is usually acetylated. When it is acetylated, an acetyl group (CH3-C=O) is added to the nitrogen and removes the positive charge that previously interacted strongly with the negatively charged DNA as well as with adjacent histones. Thus, acetylation generates euchromatin (open chromatin structure)

19
Q

Histone Deacetylation

A

The removal of the acetyl group from the lysine which will generate heterochromatin by restoring the positive charge and thus the interactions with adjacent histone tails and DNA.

20
Q

Histone Methylation

A

Methylation occurs on multiple different levels to lysine as well as arginine. They can be mono-,di-, and trimethylated, each of which give a different “message”.

21
Q

HAT Enzyme

A

Histone acetyl Transferase. It catalyzes the acetylation reaction which leads to euchromatin (open structure)

22
Q

HDAC Enzymes

A

Histone De-Acetylases. They catalyze the deacetylation reactions and lead to the more condensed heterochromatin.

23
Q

Serine Kinases

A

Add a phosphate group to serine on the histone tail to also give a certain message. Each modification type attracts proteins

24
Q

Proteins containing Bromodomains

A

These proteins bind to acetylated lysines in histones

25
Q

Proteins containing chromodomains

A

These proteins bind to methylated lysines (or arginines) in histones.

26
Q

Explain how the regulation of chromatin structure by histone acetylation/deacetylation, methylation/demethylation, and ATP dependent chromatin remodeling complexes or enzymes (SWI/SNF proteins) act cooperatively with transcription factor proteins to regulate gene expression

A

When a transcription activator binds, for example, certain processes will take part with all of these. Thus, acetylation will occur on lysines to loosen the interactions, methylation will occur to recruit other transcription factors, chromatin remodeling complexes will be recruited to loosen up the chromatin, all working together for the same common goal. The opposite holds true if a repressor binds.

27
Q

What are 4 Properties of Transcription factors?

A

1) Transcription factors have modular design with an activation domain, dimerization domain, and DNA binding domain
2) The DNA binding binding domain contains a structural motif with amino acids that interact with a unique DNA element exposed by the major and/or minor groove of the DNA helix
3) Commonly, but not always, transcription factors are homo- or heterodimers.
4) Dimers bind to a palindrome in the DNA, which contains asymmetry appropriate for dimer binding to two successive turns in the DNA helix

28
Q

Explain the concept of modularity (module structure) of the transcription factor proteins

A

Transcription factors tend to have 3 regions:

1) DNA binding domain (DBD): It interacts with the DNA
2) Dimerization Domain (Dimer): It is involved in dimerization (either homo- or heterodimer formation)
3) Activation Domain (AD): It indicates that this is the domain that is functional or going to produce the downstream results. It may be that it is activating OR repressing transcription. Just because it is called the activating domain doesnt mean it is an activator. It simply means it is the functional region.

In terms of modularity, transcription factors are multi-module proteins, containing these three modules that carry out a particular function.

29
Q

Explain the role of activation domains, DNA binding domains, and dimerization domains in transcription factors

A

Transcription factors tend to have 3 regions:

1) DNA binding domain (DBD): It interacts with the DNA
2) Dimerization Domain (Dimer): It is involved in dimerization (either homo- or heterodimer formation)
3) Activation Domain (AD): It indicates that this is the domain that is functional or going to produce the downstream results. It may be that it is activating OR repressing transcription. Just because it is called the activating domain doesnt mean it is an activator. It simply means it is the functional region.

30
Q

Where do transcription factors tend to bind in regard to the region of DNA?

A

Transcription factors tend to bind to the major groove because there is more information contained there. This is how they get their high specificity that is required.

31
Q

Explain how proteins containing a helix-turn-helix motif bind to DNA

A

Helix-turn-helix motifs are dimers that contain a recognition helix and a support helix. The recognition helix is where the DNA binding domain is located. It is the helix that will fit into the major groove and bind to the DNA. The support helix is there for structural support. The dimerization domain can dimerize through Beta-strands or alpha-helices. However, the two helix-turn-helix motifs will be dimerized and separated by exactly 3.4nm or 34A so they can both fit into the major groove of DNA. The transcription factors bind as dimers in order to enhance the binding interactions. The interactions are synergistic, meaning that they are not additive, they multiply, making it much more specific. Furthermore, they bind and recognize palindromes and pseudopalindromes (a palindrome separtated by a region that is not a palindrome)

32
Q

Describe the role of transcription factor dimerization and the role of a palindrome sequence in promoting dimer binding

A

By being a dimer, the transcription factor’s binding affinity is synergistic, meaning that the affinity is now multiplied, not added. This drastically increases the affinity the transcription factor has for DNA and thus making it very specific with high affinity. The role of the palindrome is that it allows for the two dimers to bind, one binding to each strand. The major groove of the DNA thus usually contains palindromes to bind these monomers.

33
Q

Pseudopalindrome

A

a palindrome separtated by a region that is not a palindrome. It spaces the palindrome out.

34
Q

Imperfect Palindrome

A

It is two “different” palindromic sequences on each strand that allow for the binding of heterdimers.

35
Q

Explain the difference in structure between a helix-turn-helix and a helix-loop-helix motif transcription factor

A

The helix-turn-helix is composed of a short linker sequence that creates a turn. In the helix-loop-helix, there is a loop instead of a linker. The flexibility of the loop allows for one helix to fold back and park against the other thereby forming the dimerization surface.

36
Q

Characteristics of Helix-turn-Helix transcription factors

A

1) each monomer contains a recognition alpha helix joined by a short turn from a linker to a second alpha helix. The second helix supports the recognition helix by hydrophobic interactions
2) The HTH motif is embedded in a domain that may be an all alpha helix, all beta sheet, or an alpha/beta type of domain. The type of dimer interface interaction will depend on the transcription factor protein.
3) In dimer HTH proteins, the recognition helices from both monomers bind to adjacent turns of the major groove. The DNA site is palindromic.

37
Q

Helix-Loop-Helix Proteins

A

Composed of 2 alpha helices as well except instead of being composed of a short linker that creates a turn, there is a loop.

They work as both heterodimers and homodimers and the dimerization region is generated by a leucine zipper motif.

When we form a heterodimer, they bind to imperfect palindromes. They only form heterodimers though within the transcription factor family. They do not just bind to any protein via a leucine zipper.

It is a dimer structure

Each monomer contains a recognition helix joined by a loop to a second helix which contains a leucine zipper motif

THe recognition helices from each monomer bind to adjacent turns of the major groove (usually palindrome)

Example of HLH are Myc, Mad, and Max

38
Q

Recognize and describe the function of a leucine zipper motif

A

A leucine zipper is where leucine or another hydrophobic amino acid is every 6-7 amino acids so when we form the alpha helix structure we will end up with a series of leucines on the edge and form a hydrophobic edge in which the two can now hydrophobically interact via hydrophobic interactions and vander waals

39
Q

Describe the properties of a zinc finger domain and how zinc finger containing transcription factors bind to multiple sites in DNA

A

They are recognized by zinc being able to form ligand bonds with some of the amino acids. ONLY TWO AMINO ACIDS CAN FORM A LIGAND BOND WITH A METAL, CYSTEINE AND HISTIDINE. End up with 2 histidines separated by 2-3 amino acids, and 2 cysteines separated by 2-3 amino acids. These pairs are then separated by 12-15 amino acids. We can have any combination of cysteine or histidine so long as there are 4 total. In the example looked at in class, a zinc finger was holding an alpha helix and a beta sheet together. The zinc fingers often associate in groups and because there are multiple zinc fingers working together the recognition element is going to be slightly larger.

**It is a motif commonly characterized by a Zn atom coordinated by two Cys and two His or any combination of the two

***There is an intervening polypeptide chain of 12-15 amino acids present an alpha helix segment and a beta sheet segment.

***The alpha helix is the recognition helix that binds to the DNA

***A typical transcription factor has multiple Zn finger motifs that bind in tandem to adjacent turns of the major groove of DNA (like diagram with three together)

40
Q

Recognize a bZIP transcription factor and describe how it binds to DNA

A

the b indicates that basic amino acids will be apart of the DNA Binding Domain. They have lysines and arginines in the DNA binding domain that are important (this is the basic part)

ZIP represents the leucine zipper that is part of the domain.

They can function as either homo or heterodimers and we can see how the sequences in the DNA might differ. If there is a heterodimer for example you might have an imperfect palindrome that is being recognized because each monomer recognizes a slightly different sequence.

The dimer forms an alpha helix scissor structure, each monomer is a long alpha helix binding to half of the palindromic element in the DNA

DNA binding module contains multiple basic amino acids

Dimerization is through a leucine zipper (coiled-coil hydrophobic interaction)

41
Q

DNase I footprinting assays

A

DNA Footprinting assays answer the question of whether or not a transcription factor bind to a specific regulatory element or not. First, you isolate out that DNA region with the regulatory element and then label it with radioactivity on the 5’ end. Then, you incubate the double stranded DNA with a protein lysate. Then, you cleave the DNA with a nuclease. Only want enough nuclease in order to cut the DNA once per DNA strand so only put in a limiting amount of nuclease. By cutting it relatively infrequently, you will get successively larger pieces of DNA. When these are ran out on a gel, they look like a ladder. Then, if a protein is bound to that region, it would have protected that region of DNA and thus it wouldnt have been cut by the nuclease. This is very limiting information though because we dont know what protein it was.

42
Q

Electrophoretic Mobility Shift Assay (EMSA) or Gel Retardation Assay

A

This gives us a bit more information. You take a small fragement of radioactive DNA and ask if there is a transciption factor bound to it. The DNA is incubated with proteins and then ran out on a gel. The fragment with no proteins will run off to the bottom of the gel but if there is a protein bound to it it will be retarded within the gel. This gives some information about the size of the protein. You can then do a supershift assay which is adding an antibody to tell exactly what transcription factor is bound because it would cause the band to increase in size and thus be higher on the gel or more retarded.

43
Q

Chromatin Immunoprecipitation (CHIP)

A

Here we are asking whether or not a particular transcription factor binds to a regulatory region within a gene. Advantage to this technique is that we are working with a living cell and thus can see that the binding of this transcription factor actually occurs within the cellular context. You expose the cell to formaldehyde which cross-links the DNA and proteins (binds them together). Then you lyse the cell and sonicate the DNA to break it up into small fragments of DNA. Then use our immunoprecipitation techniques by taking an antibody that interacts with the protein which will then pull the protein down. Can then use an acid to reverse the cross-linking and then use PCR to amplify this DNA region and identify what part of the genome it is.

44
Q

Reporter Assay

A

This tells us functional activity where all of the other tests have only told us about proteins binding to DNA, they havent told us that it actually does anything. A reporter plasmid contains the gene for luciferase which will fluorese. We can use our recombinant DNA techniques and remove the enhancers from the genome and insert it upstream from a basal promoter and luciferase gene. If it is an enhancer, we would see increased luciferase and thus more fluoresence. If not, it would be the same. If it is a repressor we would see decreased!