MCB 252 – Exam 2 - Exam 2 summary and lecture notes PDF

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Exam 2 summary and lecture notes...

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MCB 252 – Lecture 2/15 (red = memorize for exam)  Histone tails can be post-transcriptionally modified o Histone modifications: phosphorylation, ubiquitinylation, methylation or acetylation o These modifications alter gene expression  More Information about histones o Major histones  H1 (linker histone) and core histones (2 H2A, 2 H2B, 2 H3, 2 H4) o Cell has major histones and minor histones (histone variants)  Major histones would be core histones and they are present in most nucleosomes in our body  Minor histones are present only in some nucleosomes  Ex) CENPA is a histone variation of H3. This protein preferentially associates with nucleosomes present on the centromere o Centromeres are highly condensed DNA o CENPA replaces H3 in the nucleosomic centromere. This is because CENPA helps further chromatin condensation  Ex 2) H2AX is a variant present in nucleosomes and they are distributed randomly throughout the genome. o H2AX gets phosphorylated and this gets recognized by DNArepairing machinery (helps when there is DNA damage) o Histone tails stabilize intra- and inter-nucleosomal DNA interaction o The tails of H4 are known to be acetylated during gene activation (H4K16Ac = Histone 4 Lysine 16 Acetylation)  Acetylation neutralized the K basic charge, which loosens contact between tails and other histones or DNA  Positively charged lysine interacts with negatively charged DNA o Interaction helps chromatin condensation  H4 interacts with H2A/H2B of adjacent nucleosomes and this causes nucleosomes to get closer and condense more  ACETYLATION OF LYSINES HELPS OPENING UP OF CHROMATIN, WHICH HELPS TRANSCRIPTION ACTIVATION  A major way these modifications help chromatin compaction is by recruiting specific proteins to specific sites o H3K9Me  acts as a binging site to specific types of proteins  HP1 (heterochromatin protein 1) is a protein that gets recruited to these sites  Heterochromatin interacts with HP1, which interacts with nuclear lamins and nuclear membrane proteins  H3K9Me is present on heterochromatin, and this allows the H3K9Me bind to the heterochromatin (helping it bind to the nuclear periphery)  This binding is facilitated by the interaction with HP1  HP1 has 2 important domains o Chromodomain – required to interact with H3K9Me o Chromoshadow domain – facilitates interaction with other domains that have chromoshadow domains (so it would be able to interact with other HP1 proteins near it)  This allows HP1’s to come together like glue and prevents other proteins from coming to this region

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This helps the inactive chromatin domain (heterochromatin domain) be established and remain condensed because no proteins can get past the HP1 barrier modification examples that facilitate gene activation. o H3K9Me  gene inactivation o H3K9Ac  gene activation Histone Code Hypothesis o By having differential modifications on different regions on the chromatin, the cell can decide whether that particular region needs to be transcriptionally active/inactive or if the compaction of the chromatin needs to be increased/decreased o Proteins can bind to this modified site or read the site (readers of the code) o Enzymes that add the modification are the writers of the code Now back to our example o There are specific enzymes that can add methyl or acetyl groups to the histone tails o Histone H3K9 methyl transferase  (writer of the code) Adds one or multiple methyl groups to the Lysine 9 of the histone 3  The HP1 (reader) recognizes and binds to the site  Once HP1 binds, it uses its chromoshadow domain to recruit more HP1’s. But, the histone methyl transferase also has a chromoshadow domain so it can recruit more proteins like itself. So, an HP1 is recruits and a transferase come as well, causing this reaction to spread.

NEW LECTURE MATERIAL  In eukaryotes, DNA is assembled into chromatin, which regulated access of the transcriptional machinery to genes o Transcription factors can be activators or repressors  Differential gene expression is controlled by changes in the chromatin structure through modification we mentioned earlier o These modifications are present in all types of histones and they encourage different types of cellular processes (not just transcription)  Modifications impacting the structure of chromatin o Histone Acetyl Transferases (HATs)  Also known as Lysine Acetyl transferases (KAT)  Enzymes that add the acetyl group to the lysine  Acetylation helps open up and de-condense  HAT is a positive regulator of transcription  HATS found in: Co-activators (SAGA), GTFs (TFIID) o Histone Deacetylases (HDACs)  Remove the acetyl group from the lysine  Deacetylation causes chromatin to condense  Inactive genes get de-acetylated  HDAC is a negative regulator of transcription (makes genes inactive)  The reason these modifications can influence gene expression is because they can recruit different domains. How can they recruit proteins? o Chromodomains: help bind to H3K9Me regions (lecture before)

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o Bromodomains: these proteins recognize acetylated lysine’s on the nucleosomes and recruit them (bromodomains can positively regulate transcription  open chromatin) o There are many types of chromo and bromo domains, depending on the proteins that recognize the structure  Chromodomains always recognizes methylated sites  Bromodomains always recognize acetylated sites  Bromodomains are found in HATs (SAGA), CRC (Swi/Snf), GTFs (TFIID) Why does the cell require these modifications? o Naked DNA, with no histones, can bind to any and all transcription factors. Therefore, gene expression can be activated. (WTF Ignore this) o Chromatin requires transcription factors to change the chromatin structure or recruit proteins to influence transcription. TF’s can activate or repress.  Early evidence suggest gene activation and chromatin condensation is related o Evidence also suggests that nucleosome depletion results in transcriptional activation  So, nucleosomes have a negative impact of transcription (inhibit)  Removal of nucleosomes results in gene activation  One of the ways that the cells can remove nucleosomes (or histones) is by a protein called SWI2 (or SNF2) A mutation in Swi2 causes a reduction in transcription of 2 unrelated genes o These are the two genes that were down regulated  The HO gene: DNA endonuclease that is required for mating type switching  The SUC2 gene: invertase (for using sucrose as a source of carbohydrates) o The mutations in SWI2/SNF2 could be suppressed by mutations in Histone H4  So, HO gene can be expressed by a mutation in H4  So, SWI2 is a negative regulator of the HO gene o The main function of SWI2 is to remove H4 from the chromatin, so the HO gene can be activated.  If there is no functional SWI2 gene, the H4 will remain bound to the nucleosomes and the HO gene cannot be activated o It was later found that the SWI2/SNF2 protein is a component of a large multi-protein complex that could disrupt or reconfigure nucleosomes and stabilize the binding of TFs to nucleosomal DNA. This complex was named the SWI/SNF complex  SWI2/SNF2 subunit is an ATPase that’s stimulated by DNA and nucleosomes  Needs ATP to push the nucleosomes off the DNA

MCB 252 Lecture 2/18  How do chromatin remodeling complexes function? o They bind to chromatin with the help of its association with modified nucleosomes.  This binding to chromatin is facilitated by transcription factors o After binding, it can push/pull nucleosomes on a chromatin. This results in naked DNA in specific regions. If this naked DNA is near a promoter sequence, the binding of other TFs or GTFs can be facilitated and transcription can be activated.  CRC function for transcription, replication and even DNA damage o So, the main idea is to pull nucleosomes away from the chromatin

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Chromatin remodeling factors associate to certain regions and push the DNA away from the core histone complex, so the DNA is then exposed and TFs can come bind  The two ways that chromatin can do this: octamer sliding and octamer transfer Octamer Sliding o Chromatin remodeling factors will push the nucleosomes away from the core promoter sequence so GTFs can bind, recruit RNA poly II and activate transcription. o Once chromatin remodeling factors (CRC) bind, the ATPase and helicase activity (some subunits of the CRC have helicase activity) will help push the nucleosomes away from the region of interest. This opens up the promoter so RNA Poly II binds Octamer Transfer o This is where the nucleosomes are being removed from the core promoter sequence. The CRC grabs the nucleosomes and pulls them off the site of interest. Studies have mentioned that moving nucleosomes away from a site causes transcription to be activated. But we know that those nucleosomes have been pushed somewhere else and now that region is nucleosome dense and transcription has been inactivated there. Histone Acetyl Transferases (HATs) and CRC play important roles in opening chromatin So, the overall steps for transcription to occur o An Activator must find access to its binding site on chromatin  So, the TF will use its DNA-binding domain to bind to the enhancer sequence of chromatin (even if there are nucleosomes attached to the chromatin) o This activator recruits remodeling complexes (and other factors)  The TF uses its activation domains to recruit CRC to that region. This CRC will change the structure of the chromatin by pushing nucleosomes away and encouraging recruitment of other proteins. This activation domain can also recruit HATs, which also encourage transcription.  HATs and CRC are known as co-activators because they don’t directly bind to DNA (they are recruited to chromatin by TFs), but they influence the transcriptional activity o Next, holoenzyme is recruited to form the pre-initiation complex Binding site on naked DNA is easily accessible. Since the DNA is unwound, the regions for binding are open and available. o Biding site is no longer accessible after nucleosome deposition and chromatin folding o Sometimes, the binding site is located in the linker DNA and its partially accessible o Sometimes, the binding site is displayed properly, even when wrapped around the histone octamer, and it is partially accessible An example of the sequence of events associated with the activation of a promoter is provided by the HO gene o 1. Swi5 finds access to the binding site on DNA o 2. Swi5 recruits the CRC (Swi/Snf complex. This complex starts pushing nucleosomes away from the promoter sequence. o 3. Chromatin remodeling by Swi/Snf leads to the recruitment of the SAGA complex (a specific HAT)  This causes the chromatin to open up and acetyl groups to be added to the lysine’s on the histone tail o 4. Several proteins in the SAGA complex and CRC have bromodomains

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Bromodomains: these proteins bind to acetylated lysine’s on the nucleosomes and recruit them (they positively regulate transcription  open chromatin)  Because of these domains, these proteins can now bind to the acetylated chromatin without the help of transcription factors  So, the TF that initiated this entire process is no longer needed and can be lost or removed from the process o 5. Remodeling and hyper-acetylation of histone tails causes the chromatin to open and causes the recruitment of the TF SBF  SBF recruits holoenzymes and GTFs  This results in transcriptional activation This same sequence of events in yeast o The enhancer sequence in yeast is UAS – upstream activating sequence o GCN4 is like a transcriptional activator (like a TF) and it has a DNA binding domain + an activator domain o GCN5-containing complex = SAGA = co-activator o 1. DNA binding domain of GCN4 recognizes the UAS and the activation domain recruits GCN5 o 2. The GCN5 acetylates the adjacent nucleosomes (because it is a HAT) and this opens up the chromatin  They can also recruit CRC to enhance transcription activation Three types of co-activators o Chromatin remodeling complexes o Histone acetyl Transferases (HATs)  Similar to CRC in that each are complex proteins that are conserved from yeast to humans and neither bind to DNA directly. Both of them have bromodomains. Both can bind to modified histones.  TF are different because they can bind to DNA directly by the DBD  Example of HAT is CREB binding protein  Cyclic AMP response element binding protein  CREB binds to promoters and activates downstream pathways  It is a TF that is a transcription activator o Mediator TF’s always come upstream of activators (like Hats, CRC, mediators and GTFs) Yeast have 2 major HDAC complexes Mammals have multiple HDAC complexes o HDAC are co-repressors  They repress transcription but they don’t directly interact with DNA o If HDAC is mutated or cannot be recruited by Rb (the tumor suppressor gene), the Rb cannot repress its specific target genes and this causes a deregulation of the cell cycle Apart from histone modifications, nucleotide modifications can also influence transcriptional activity o The cytosine residues on DNA can be methylated by DNA methylases (DMT)  DNA methyl transferases add the methyl group – writes of the code o Methylated C are recognized by the protein MeCP2 – readers of the code o MeCP2 recruits HDACs by interacting with a subunit of the complex

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This causes repression of transcription So, these methyl groups on cytosine’s cause transcriptional repression

NEW LECTURE MATERIAL  Epigenetics: an epigenetic trait is a stably heritable phenotype resulting from changes in a chromosome without alterations in the DNA sequence (examples would be methylation of DNA)  First example of epigenetics: Dosage compensation o Dosage compensation: a mechanism used by cells to compensate the levels gene expression between 2 different sexes o X-chromosome inactivation in mammals (mice)  So, females should be XX and males should be XY  Dosage compensation… by inhibiting the expression of most of the genes in one X in females, you compensate the expression of X chromosome gene  So… males would have one active X while females have one active X and the other X is 90% inactive (or transcriptionally inert)  This is called X chromosome inactivation in females  The inactive X chromosome transcribes/produces a non-coding RNA gene called Xist. In the active chromosome, Xist is inactive. o Dosage compensation in flies  X-chromosome hyper activation  In flies, females have 2 active XX but males have a hyperactive X chromosome  So, to compensate, instead of inactivating one X in females, they would hyperactive the X in males MCB 252 Lecture 2/20  X-chromosome inactivation in mammals (mice) o So, females should be XX and males should be XY o Males would have one active X while females have one active X and the other X is 90% inactive (or transcriptionally inert) o The inactive X chromosome transcribes/produces a non-coding RNA gene called Xist. In the active chromosome, Xist is inactive. o This inactive X chromosome is also hypo-acetylated but hyper-methylated  It is H3K27 methylated  It the inactive X also contains specific histone variants like macroH2A  X-inactivation = epigenetic mechanism of dosage compensation o In the initial state of development, both XX are transcriptionally active o Around the 20-cell stage, one X in EVERY cell becomes transcriptionally inactive  This is a random event  So, either the maternal or paternal X chromosome become transcriptionally inactive and heterochromatinized. Once it is randomly decided (let’s assume the paternal X is inactive and maternal X is inert), all the progeny will follow suit. The following generations, all cells with paternal X will be inactive and the ones with maternal X will be inert.  This example can be seen in calico cats. Women have the XX which represent two different colors so they are a mixed color. Males are either orange (orange X) or black (black X)

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Mitotic inheritance is passing down the active or inactive X Cellular memory: the cell needs to remember which X is inactive so it can be passed down  This is also called a position effect mechanism: the position of the genes on a chromatin determine if the gene will be active or inactive in a cell Molecular mechanism for how differential gene expression on X is achieved o Example of inactive X chromosome in mice is Xist, explained next Xist mediated X chromosome inactivation o Xic: X inactivation center o Xist: long non-coding RNA gene o Tsix: anti-sense transcript of Xist o In the early stages of embryogenesis, when both XX are active, the XX chromosomes produce the Tsix non-coding RNA  So, Tsix is transcriptionally active in active XX  At the 20-cell stage, inactivating one X causes Tsix to become inactive  This triggers the Xist gene to become active (tsix and xist have an inverse relationship)  Active xist starts coding the complete X chromosome. The transcription starts at Xist and spreads from there. While spreading, it recruits proteins to the inactive X chromosome.  The Xist non-coding RNA literally acts like a repressor. Because it can recruit the co-repressor proteins like HDAC to be transcription repressors and encourage condensation of chromatin into heterochromatin  Even though it’s not a protein, xist acts like a transcriptional repressor o The other active X is still transcriptionally active (because it has tsix instead of xist) Saccharomyces Cerevisiae life cycle (yeast that makes beer and bakes) o Single celled organism, can grow as haploid or diploid o Diploid yeast cell has an A and alpha factor o Two important things to know… o 1. When these cells are starved, by depriving them of nitrogen, these cells undergo meiosis and produce haploid ascospores  Meiosis creates 4 cells  2 cells have an active alpha gene and the other 2 have an active a gene  Haploid cells have either a or alpha active genes but the diploid genes have both genes active in the same cell  But the a-type haploid can mate/fuse with alpha-type haploid cells and create a diploid with both genes o A-type can NEVER mate with another alpha type o 2. An a-type mother will produce another a-type haploid cell. But once the haploid is created, the mother switches genotype to become an alpha-type mother cell  This is called mating type switches  This also happens from alpha mothers, they switch to a-type More information of this mating type switch o It takes place on chromosome 3  it has the gene locus MAT

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If the haploid cell is a-type, the MAT locus will have the a-gene which is transcriptionally active. If the haploid cell is alpha-type, the alpha-type is present on MAT and is transcriptionally active  MAT locus can never have both a and alpha type gene on it o On either side of the MAT locus, closer to the telomeres, there is an intact copy of the a-gene (HMLa) and the other side has an intact copy of the alpha-gene (HML)  Both genes on the HML and HMLa are transcriptionally inactive  But the gene is active on the MAT locus  They are transcriptionally inert because HML areas have a silencer region near it Silencing genes/proteins  play important role in keeps genes on HML locus inactive o SIR1: acts in complex at the silencer o SIR2: histone deactylase (writer, an HDAC) o SIR3/SIR4: a reader of the histone code o Rap1: repressor activator protein (binds silencer) o ORC: origin recognition complex (binds silencer) How this silencing takes place… o The silencer sequences on HML act as binding sites to Rap1 and ORC  Rap1 can act as transcription activator or repressor, depending on where it binds  In this context, it acts as a silencer  Once bound, Rap1 and ORC recruit proteins SIR1, 3 and 4 to the silencer region  This creates a multi-protein complex. Once built, it causes SIR 2 to bind as well  SIR 2 will remove the acetyl groups on the adjacent nucleosomes. These deacetylated nucleosomes now become binding sites for SIR 3/4 o Bef...