Final Exam

Be able to describe and/or recognize the three functional bases of epigenetic modification

1) A covalent modification of DNA (methylation of a base) 2) A proteinaceous structure that assembles on DNA (heterochromatin) 3) A protein aggregate that controls the conformation of new subunits as they are synthesized (Acetylated histones, Phosphorylated proteins, Methylated proteins, Prions)

Be able to discuss proofreading mechanisms in tRNA synthetases

9. Charged by aminoacyl-tRNA synthetases (amino acids enter the translation pathway through the action of this enzyme). synthetases has 3 binding sites (tRNA site, AA site, and ATP site) 1) ATP reacts with the AA and forms aminoacyl-AMP 2) the 2'-OH or the 3'OH(located at 3'-A76) group on the tRNA bind to the carbonyl carbon on the AA. a. Confirms identity of correct AA tRNA pair via intermolecular forces i. Does not involve recognition of anticodon loop ii. Changes in anticodon loop can result in incorporation of incorrect amino acids without mutations to the DNA b. tRNA synthetases exist as two distinct classes of enzyme i. Contain mutually exclusive structural motifs and domains c. tRNA synthetases possess at least two proofreading mechanisms i. Kinetic proofreading 1. tRNA lacks necessary structure to facilitate linking to the amino acid within tRNA synthetase molecule ii. Chemical proofreading 1. steric hindrance of aa acid entering tRNA synthetase 2. Pass through from active site to "editing" site where incorrect Aminoacyl tRNA can be removed. d. EF-Tu-aminoacyl tRNA-rRNA interactions are critical for accuracy in translation i. Recall that EF-Tu won't release aminoacyl tRNA until correct base pairing occurs between codon and anticodon loop.

Understand the regulatory roles of anti-sense RNA

Antisense-RNA as much as 70% of human genes produce an antisense. constructed by reversing the orientation of the gene with regard to its promoter. 1. Non-coding RNA used to regulate gene expression a. forms a duplex with target RNA i. blocks translation ii. terminates transcription iii. creates a target for endonuclease b. Other ncRNAs i. long intergenic non-coding RNA ii. promoter RNAs (pRNA) iii. enhancer RNA (eRNA) iv. Cryptic Unstable Transcripts (CUTS) 1. Used to drive formation of heterochromatin over specific genes The gene PHO84 in yeast is regulated but CUTS. In addition to the promoter of the 5' end of the gene, there's another promoter on the opposite strand that is unregulated. This promoter require Set1 histone methyltransferase for activity and produces an antisince RNA. Under normal conditions, this RNA is rapidly degraded. In the absence of degradation or in aging cells, the antisense RNA persists. This antisense RNA, or CUTS, Works in trans to recruit histone deacetylase Enzymes that remove acetate groups from histones, thereby causing the chromatin over the gene region to be remodeled and condensed so that the gene can no longer be transcribed.

Be able to describe the role and function of the CRISPR/Cas system in prokaryotic cells

CRISPR/Cas Systems a. Adaptive Immunity against foreign genetic elements i. Following infection: 1. Foreign DNA sequences are captured and integrated into the host CRISPR locus as new spacers. 2. The CRISPR locus is transcribed and processed to generate mature CRISPR RNAs (about 57nts a. each encoding a unique spacer sequence. 3. Each crRNA associates with Cas (made up of eight genes) effector proteins that use crRNAs as guides to silence foreign genetic elements that match the crRNA sequence. 4. CRISPR-Cas systems are highly diverse and are divided into numerous subtypes. a. Type I-E system E. coli uses the Cascade complex for RNA-guided silencing of foreign DNA. i. Composed of Cse1, Cse2, Cas7, Cas5e, and Cas6e subunits and one crRNA ii. Forms a structure that binds and unwinds dsDNA to form an R-loop. iii. Cascade positions the PAM-proximal end of the DNA duplex at the Cse1 subunit and near the site of Cas3 association. iv. Base pairing of the PAM region is unnecessary for target binding but critical for Cas3-mediated degradation. b. Type II CRISPR-Cas systems use an RNA-guided DNA endonuclease Cas9, to generate double-strand breaks in invasive DNA during an adaptive bacterial immune response. i. Cas9-mediated cleavage is strictly dependent on the presence of a protospacer adjacent motif (PAM) in the target DNA. 5. The ability to program Cas9 for DNA cleavage at specific sites defined by guide RNAs has led to its adoption as a versatile platform for genome engineering and gene regulation.

Understand and be able to describe the function and process of methylation in eukaryotes

CpG methylation in eukaryotes i. Methylation occurs at specific sites and specific sequences 1. Bacteria: a. Methylation is associated with restriction enzyme system i. Restriction enzymes generally don't cut methylated DNA within restriction sequence ii. Usually only host bacterial DNA is methylated in the correct, sequence-specific manner. b. Methylation is also associated with replication 2. Eukaryotes a. Methylation is associated with gene inactivation i. Blind cave fish b. Generally, occurs at CpG islands (cytosine and guanine separated by only one phosphate group). i. Often found in 5' regions of genes ii. Contain increased density of CpG nts 1. often contained in short repeats iii. between 2 and 7% of Cs are methylated in animal cells c. Occurs at the 5th carbon of cytosine i. 5-methylcytosine (5mC) d. Most of the methyl groups are found in CG dinucleotides. d. Typically, both strands are methylated i. Known as fully methylated e. Methylation status of sites can change based on when replication occurs relative to remethylation i. Loss of methylation can occur if hemi-methylated DNA is replicated f. Controlled by DNA methyltransferase enzymes i. Methylases and demethylases 1. Sequence-specific 2. Only act unmethylated strand ii. Maintenance methyl transferases (Dnmt1 1. Methylate hemi-methylated DNA 2. Degree of methylation often decreases as cells age a. MMT activity decreases b. epigenetic drift i. Result in unintended changes in expression patterns (might be responsible for phenotypic variation in aging monozygotic twins)

A covalent modification of DNA

CpG methylation in eukaryotes i. Methylation occurs at specific sites and specific sequences 1. Bacteria: a. Methylation is associated with restriction enzyme system i. Restriction enzymes generally don't cut methylated DNA within restriction sequence ii. Usually only host bacterial DNA is methylated in the correct, sequence-specific manner. b. Methylation is also associated with replication 2. Eukaryotes a. Methylation is associated with gene inactivation i. Blind cave fish b. Generally, occurs at CpG islands (cytosine and guanine separated by only one phosphate group). i. Often found in 5' regions of genes ii. Contain increased density of CpG nts 1. often contained in short repeats iii. between 2 and 7% of Cs are methylated in animal cells c. Occurs at the 5th carbon of cytosine producing 5-methylcytosine (5mC) d. Most of the methyl groups are found in CG dinucleotides. d. Typically, both strands are methylated i. Known as fully methylated e. Methylation status of sites can change based on when replication occurs relative to remethylation i. Loss of methylation can occur if hemi-methylated DNA is replicated f. Controlled by DNA methyltransferase enzymes i. Methylases and demethylases 1. Sequence-specific 2. Only act unmethylated strand ii. Maintenance methyl transferases (Dnmt1 1. Methylate hemi-methylated DNA 2. Degree of methylation often decreases as cells age a. MMT activity decreases b. epigenetic drift i. Result in unintended changes in expression patterns (might be responsible for phenotypic variation in aging monozygotic twins)

A proteinaceous structure that assembles on DNA

Heterochromatin formation a. HP1 (heterochromatin protein 1) is key mammalian protein forming mammalian heterochromatin b. Functions by binding to methylated histone H3 i. Di- or tri-methylated at H3K9 (Lysine 9 of H3)(methylated on the nitrogen in lysine) ii. HP1 has two domains that facilitate chromatin formation 1. Chromodomain a. Bind methylated histones 2. Chromshadow domain a. Chromshadow domains self-aggregate, bringing together the nucleosomes to which their proteins are bound b. Condense the chromatin region they are associated with iii. Functions in concert with other proteins to form heterochromatin 1. Histone deacetylase removes acetyl group from K14 2. Permits methylation of K9 (by SUV39H1 methyltransferase) 3. Forms binding site for HP1 4. HP1 molecules self-aggregate along chromatin structure (extension of heterochromatin) 5. state of histone modification is indicative of state of chromatin a. Methylation of K9 is heterochromatin b. Methylation of K4 is euchromatin (hypomethylation leads to heterochromatin formation) Formation of heterochromatin is also facilitated by silent information regulator (SIR) proteins i. Binding of SIR proteins can silence promoters and/or coding regions ii. Generally, functions in a targeted manner 1. Sequence-specific SIR proteins target specific regions of the genome a. Rap! - sequence specific SIR i. Binds CA repeats b. Yeast i. Telomeres ii. HM loci (mutations in SIR2,3, or 4 cause HML and HMR to become activated And also relieve the inactivation of genes that have been integrated near the telomeric heterochromatin) 2. Various SIR proteins bind to specific histones with specific modifications a. Rap1 identifies the DNA sequences of which heterochromatin forms. It recruits Sir3 & Sir4 which interact with unacylated NTD tails on Histones H3 and H4 i. Sir2 - is an HDAC (deacetylase) required for Sir3 & 4 binding b. Mutations to SIR genes in yeast can activate silent mating types (HML and HMR). c. Binding of Sir3 and Sir4 has an inhibitory effect i. From either: 1. Presence of Sir proteins themselves 2. Repressive effect of histone deacetylation d. Evidence of the repressive nature of chromatin structure, including that of Sir-coated chromatin is evidenced by the effect of mutations to the Polycomb (Pc) genes in D. melanogaster i. Mutations to Pc genes relax chromatin 1. Relieve transcriptional repression 2. Can appear as gain-of-function mutations

Be able to describe the role of histone modification, HP1, and SIR proteins in the formation of heterochromatin

Heterochromatin formation a. HP1 (heterochromatin protein 1) is key mammalian protein forming mammalian heterochromatin b. Functions by binding to methylated histone H3 i. Di- or tri-methylated at H3K9 (Lysine 9 of H3)(methylated on the nitrogen in lysine) ii. HP1 has two domains that facilitate chromatin formation 1. Chromodomain a. Bind methylated histones 2. Chromshadow domain a. Chromshadow domains self-aggregate, bringing together the nucleosomes to which their proteins are bound b. Condense the chromatin region they are associated with iii. Functions in concert with other proteins to form heterochromatin 1. Histone deacetylase removes acetyl group from K14 2. Permits methylation of K9 (by SUV39H1 methyltransferase) 3. Forms binding site for HP1 4. HP1 molecules self-aggregate along chromatin structure (extension of heterochromatin) 5. state of histone modification is indicative of state of chromatin a. Methylation of K9 is heterochromatin b. Methylation of K4 is euchromatin (hypomethylation leads to heterochromatin formation) Formation of heterochromatin is also facilitated by silent information regulator (SIR) proteins i. Binding of SIR proteins can silence promoters and/or coding regions ii. Generally, functions in a targeted manner 1. Sequence-specific SIR proteins target specific regions of the genome a. Rap! - sequence specific SIR i. Binds CA repeats b. Yeast i. Telomeres ii. HM loci (mutations in SIR2,3, or 4 cause HML and HMR to become activated And also relieve the inactivation of genes that have been integrated near the telomeric heterochromatin) 2. Various SIR proteins bind to specific histones with specific modifications a. Rap1 identifies the DNA sequences of which heterochromatin forms. It recruits Sir3 & Sir4 which interact with unacylated NTD tails on Histones H3 and H4 i. Sir2 - is an HDAC (deacetylase) required for Sir3 & 4 binding b. Mutations to SIR genes in yeast can activate silent mating types (HML and HMR). c. Binding of Sir3 and Sir4 has an inhibitory effect i. From either: 1. Presence of Sir proteins themselves 2. Repressive effect of histone deacetylation d. Evidence of the repressive nature of chromatin structure, including that of Sir-coated chromatin is evidenced by the effect of mutations to the Polycomb (Pc) genes in D. melanogaster i. Mutations to Pc genes relax chromatin 1. Relieve transcriptional repression 2. Can appear as gain-of-function mutations

Understand the process of heterochromatin formation, and be able to describe it in a general way.

Heterochromatin formation a. HP1 is key mammalian protein forming mammalian heterochromatin b. Functions by binding to methylated histone H3 i. Di- or tri-methylated at H3K9 (Lysine 9 of H3) ii. HP1 has two domains that facilitate chromatin formation 1. Chromodomain a. Bind methylated histones 2. Chromshadow domain a. Chromshadow domains self-aggregate, bringing together the nucleosomes to which their proteins are bound b. Condense the chromatin region they are associated with iii. Functions in concert with other proteins to form heterochromatin 1. Histone deacetylase removes acetyl group from K14 2. Permits methylation of K9 3. Forms binding site for HP1 4. HP1 molecules self-aggregate along chromatin structure 5. state of histone modification is indicative of state of chromatin a. Methylation of K9 is heterochromatin b. Methylation of K4 is euchromatin c. Formation of heterochromatin is also facilitated by silent information regulator (SIR) proteins i. Binding of SIR proteins can silence promoters and/or coding regions ii. Generally, functions in a targeted manner 1. Sequence-specific SIR proteins target specific regions of the genome a. Rap! - sequence specific SIR i. Binds CA repeats b. Yeast i. Telomeres ii. HM loci 2. Various SIR proteins bind to specific histones with specific modifications a. Sir3 & Sir4 interact with unacylated NTD tails on Histones H3 and H4 i. Sir2 - deacetylase required for Sir3 & 4 binding b. Mutations to SIR genes in yeast can activate silent mating types. c. Binding of Sir3 and Sir has an inhibitory effect i. From either: 1. Presence of Sir proteins themselves 2. Repressive effect of histone deacetylation d. Evidence of the repressive nature of chromatin structure, including that of Sir-coated chromatin is evidenced by the effect of mutations to the Polycomb (Pc) genes in D. melanogaster i. Mutations to Pc genes relax chromatin 1. Relieve transcriptional repression 2. Can appear as gain-of-function mutations

Understand and be able to describe the trends in the genetic code as discussed in class

I. Genetic Code a. Comprised of 4^3 = 64 codons i. Number of options (4 bases) to the power of number of positions (3 per codon) ii. 61 codons represent AAs iii. 3 codons are stop codons b. More codons than there are amino acids i. Most amino acids are represented by more than one codon. 1. Methionine and Tryptophan are exceptions ii. Codons that encode the same amino acids are said to be synonymous 1. May fall out of favor, as codons are not used synonymously, and are tRNAs are not produced in equivalent amounts for all codons c. Codons representing chemically similar amino acids tend to have similar sequences i. Often base in position 3 of the codon is not significant as any base can occupy this position and encode the same AA. 1. Sometimes purine vs pyrimidine distinction exists at this position ii. Third-base degeneracy 1. Eight codon families in which all four codons sharing the same first two bases have the same meaning a. Third base has no role in determining AA identity in these cases. third base degeneracy. 2. Seven pairs wherein identity of pyrimidine at 3rd position doesn't matter 3. Five codon pairs in which either purine may be present without altering the identity of the AA. 4. Only three cases exist wherein the identity of the third position confers a unique meaning a. AUG b. UGG c. UGA i. C and U never have unique role in position 3

Understand and be able to describe why X-inactivation is required and the strategies for equalizing gene dose in organism that utilize sex chromosomes.

If X-linked genes were expressed equally in each sex, females would have twice as much of each product as males. X - Chromosome inactivation a. In species where chromosomes determine sex: i. Variation in the number of X chromosomes presents a problem 1. Dosage ii. Expression of X genes must be equalized 1. Genes should be expressed more-or-less in a sex-independent fashion iii. Equalize the level of exposure to X genes 1. Recall that trisomies disrupt gene dosages 2. Dosage compensation equalizes the expression of X-linked genes iv. Equalization can occur via one of at least a few mechanisms 1. Mammals - one of the two female X chromosomes is inactivated during embryogenesis a. Both X chromosomes are active during early embryogenesis b. "Inactive" X chromosome retains about 5% of its activity 2. Drosophila - Expression of a single male chromosome is doubled relative to the expression of the female chromosome. 3. C. elegans expression of each X chromosome is halved relative to the expression of the single male X chromosome b. Entire chromosome is targeted for inactivation i. Entire chromosome becomes heterochromatic 1. Good example of facultative heterochromatin a. Chromatin that is inactive in one cell type, but active in others b. Contrasted to constitutive heterochromatin i. No coding function ii. Centromeres iii. Telomeres iv. Satellite DNAs (consists of very large arrays of tandemly repeating, non-coding DNA. Satellite DNA is the main component of functional centromeres) Calico cats and other animals provide support for the single X hypothesis i. Variegated phenotypes can be explained by differential inactivation of X chromosomes ii. Cells derived from one precursor lineage with one X chromosome inactivated differ in phenotype from cells derived from a second lineage where the other X chromosome is inactivated. iii. Inactivation of X chromosomes generally adheres to the n - 1 rule: 1. Irrespective of the number of X chromosomes, all but one will be inactivated.

Understand and be able to describe the process of imprinting, and how imprinting can affect gene expression, and in some cases, lead to diseases or other abnormalities.

Imprinting a. Methylation patterns are first removed from primordial germ cells b. Sex-specific methylation patterns are imposed during meiosis i. Some genes are methylated (turned off) on one parental chromosome or the other 1. IGF-II turned on (unmethylated) in paternal murine gametes 2. IGF-II turned off (methylated) in maternal murine gametes a. Opposite is true for IGF-IIR i. receptor for IGF-II c. Pattern established during gametogenesis is known as imprinting d. Differential methylation patterns known as hemizygous i. Imprinting can affect organism survival 1. If defective gene is methylated, organism can be unaffected by defective gene 2. Prader Willi a. Different mechanisms of inheriting disorder i. Paternal allele is usually methylated ii. maternal allele is usually unmethylated iii. Good example of dosage dependent effects. e. Imprinted genes can be controlled by a single center i. ICR - imprinting control regions 1. Differential activation of genes controlled by this region 2. Methylation activates gene by blocking binding of a repressive protein an ICR that sits between IGF2 and H19 controls their activation. if the ICR is methylated, then the IGF2 gene is active. If the ICR is unmethylated and covered by CTCF, then the H19 gene is active. the CTCF blocks the IGF2 promoter. CTCF also regulates chromatin by repressing H3K27 tirmethylation at the IGF2 locus independent of repression by DNA hypermethylation

13. Understand and be able describe the activity of riboswitches

Non-Coding RNAs a. Xist is an excellent example of a non-coding RNA that has a very specific function b. Many other forms of non-coding RNAs exist c. 5' UTR of mRNA i. Sequence can influence transcription ii. Often contain element of secondary structure iii. Riboswitch 1. RNA domain that contains a sequence that can change its secondary structure to control its activity a. Often mediated by small metabolites b. Aptamers - regions of RNA that bind metabolite i. Binding of aptamers triggers a structural change in the remainder of the riboswitch (platform) that carries out a specific function 2. GlmS gene encodes enzyme that produce glucosamine-6-P (GlcN6P) from Fructose-6-P and glutamine a. mRNA contains long 5'-UTR i. Sequence contains a ribozyme 1. In this case endonuclease activity that cleaves its own mRNA 2. Accumulation of GlcN6P activates ribozyme a. Disrupts further translation 3. Found predominately in prokaryotes iv. Eukaryotic Riboswitch 1. Neurospora. crassa a. NMT1 gene involved with vitamin B1 metabolite, thiamine pyrophosphate (TPP) synthesis i. Produces an mRNA precursor with a single intron with two different splice-donor sites. ii. Use of these different sites produces mRNA with different stabilities and translation potentials iii. Responds to the presence of vitamin B1 metabolite (TPP). At a low concentration of TPP the proximal splice donor site is chosen and the distal splice donor site is blocked by the ribose switch, the splice produces a functional mRNA. At high TPP concentration, The aptamer undergoes a conformational rearrangement so that the region that was previously bound to the nearby splice site is now bound to TPP. TPP binds the Riboswitch. page 762

A protein aggregate that controls the conformation of new subunits as they are synthesized

Prion-driven structural changes in proteins can be considered epigenetic activities a. Actively used in yeast to regulate protein conformation b. Thus far only observed as disease causing agents in mammals i. Require the presence and expression of a native PrP^c (protease resistant protein) protein with which PrP^sc prion can interact c. Prions nucleate structural changes in partner proteins. however, if the PrP gene is removed the PrP^sc does not spread. if the proteins cannot link due to a lack of GPI then the PrP^sc can not spread. d. mouse experiments page 758.

Understand the role of regulatory RNAs in prokaryotes and how sRNA functions in prokaryotes

RNA can have changes in its intramolecular and intermolecular structure. intramolecular changes result in a change in the RNAs secondary structure (how and where it base pairs with itself). intermolecular changes result from its interaction with the regulator. the regulator is usually a small RNA molecule with extensive secondary structure, but with a single stranded region that is complementary to a single-stranded region in its targets. The regulator can act to prevent a protein from binding or it can make the RNA a target for binding by a nuclease that would degrade the RNA. The regulator may also prevent the region it is bound to from assuming a secondary structure. in prokaryotes, regulators are called small RNAs (sRNAs; 50-200 nts) and act as antisense RNAs do in eukaryotes. sRNAs a. Prevent transcription of genes b. affect processing of RNA product c. Affect translation d. affect mRNA stability 5. Example oxyS RNA expression in E. coli p.770-771 when E. coli is exposed to reactive oxygen (hydrogen peroxide) it actives a gene called OxyR which Controls transcription of genes related to oxidative stress like the expression of oxyS which is a short Trans-acting anti-sense regulator RNA sequence (109 nts) who's secondary structure allows it to bind to certain mRNAs to prevent their translation. b. sRNA regulatory molecule induced upon exposure to H2O2 c. activates expression in some genes i. Activates rpoS gene ii. rpoS inhibited by 2° structure iii. oxyS binding relieves 2° structure and opens ribosome binding site. d. represses expression of other genes i. Inhibits expression of flhA mRNA by base-pairing upstream from START. 6. Typically require association with a protein Hfq binding protein. a. Hfq, which stabilizes the sRNA-mRNA binding, and increases its effectiveness. i. Often associates with translation apparatus The example of oxyS page 770.

Guide RNA

RNA molecule that serves as a template for an alteration made in mRNA during RNA editing

Be able to describe the process of X-inactivation in humans, and the roles of the Xic, Xist, and Tsix, in this process.

Single locus on the X chromosome is sufficient for inactivation i. X inactivation center (Xic) 1. ~450 kb 2. Pairing of Xic loci between two X chromosomes probably serves as the basis for the random choice during X-inactivation 3. Contains necessary components to count all X chromosomes and inactivate all but one X chromosome. 4. Inactivation begins at Xic locus and spreads over the entire chromosome a. When Xic is transferred to autosome, inactivation spreads into autosomal regions 5. Expresses several ncRNAs a. Xist (X inactive-specific transcript) i. Stably expressed only on inactive chromosome b. Tsix is also found in the Xic locus. when it is transcribed in large quantities, it binds to the Xist promoter and halts Xist production. 1. Opposite behavior relative to the rest of the inactive X chromosome, which is turned off. ii. Deletion of Xist prevents X inactivation, but does not interfere with chromosome counting mechanism. iii. Regulated in an antisense manner by Tsix iv. Mechanism: 1. Xist ncRNA is transcribed and coats the X chromosome from which it is synthesized. 2. Initially synthesized by both chromosomes, eventually, synthesized by only the inactive chromosome. 3. Xist RNA-Chromosome interactions are stabilized by proteins 4. Accumulation of Xist on chromosome a. Represses transcription i. Physically blocks transcription machinery b. Triggers recruitment of polycomb proteins i. Trigger histone modifications that facilitate construction of heterochromatin c. Methylation of promoter DNA 5. Once inactivated, heterochromatin structure of Barr body is stable and doesn't require further transcription of Xist

Understand and/or be able to discuss the phenomenon of transgenerational epigenetics

Some organisms pass on/inherent methylation patterns, while others do not i. Plants transmit methylation patterns through each generation 1. Could be the basis for phenotypic differences observed in plants such as Brassica spp. 2. plants utilize the DEMETER (DME) family of 5mC DNA glycosylases to catalyze a direct removal of 5mC from DNA followed By cleavage of the DNA backbone phosphodiester bonds by apurinic/apyrimidinic endonuclease And insertion of the unmethylated dCMP base through the base excision repair. ii. Mammals 1. Genomic patterns typically erased in germ cells, Then reestablished in new patterns by resetting the state of methylation differently in male and female meioses during gametogenesis f. Proteinaceous epigenetic states can be inherited i. Protein complexes can be passed on ii. Acetylated histones are likely conserved (suppose that the H3 - H4 tetramer Is distributed at random to the two daughter duplexes. Each daughter duplex contain some histone octamer's that are acetylated on the H3 and H4 tails, Where as others are unacetylated. To account for the epigenetic affect, we could suppose of the presence of some acetylated histone octamer's provides a signal that causes the unacetylated octomers to be acetylated) 1. Lactose tolerance a. Loss of heterochromatin structure in adult cells could lead to inheritance of heterochromatic structures g. Epigenetic inheritance can skip generations i. Transgenerational epigenetics (in plants a mutation in maintenance methyltransferase new and Aberrant patterns of epigenetic marks accumulate over several generations, leaving these plants dwarfed and sterile). ii. Seems to be related to methylation states 1. Some loci may transmit epigenetic information 2. Could be related to genomic imprinting a. Cells distinguish maternal and paternal chromosomes from each other via different methylation patterns i. Can result in differential expression of genes from maternal and paternal lineages. 1. One type of Prader-Willi syndrome is believed to be caused by aberrant methylation. this epimutation maybe due to an allele that has passed through the male germline without erasure of the silent epigenetic state established in the grandmother. 2. Gene in paternal chromosome that is normally silenced via methylation is expressed. ii. Lack of methyl donors in diet (folate, choline) can result in defects and inheritance of aberrant methylation patterns

Understand the inheritance patterns associated with epigenetic information

Some organisms pass on/inherent methylation patterns, while others do not i. Plants transmit methylation patterns through each generation 1. Could be the basis for phenotypic differences observed in plants such as Brassica spp. 2. plants utilize the DEMETER (DME) family of 5mC DNA glycosylases to catalyze a direct removal of 5mC from DNA followed By cleavage of the DNA backbone phosphodiester bonds by apurinic/apyrimidinic endonuclease And insertion of the unmethylated dCMP base through the base excision repair. if a plant undergoes a mutation in its maintenance methyltransferase, that makes the MMT ineffective, this triggers genome wide activation of alternative epigenetic mechanisms such as RNA directed DNA methylation, DNA methylation inhibition, and retargeting of histone H3K9 methylation. after several generations the plant can become dwarfed and sterile. ii. Mammals 1. Genomic patterns typically erased in germ cells, Then reestablished in new patterns by resetting the state of methylation differently in male and female meioses during gametogenesis f. Proteinaceous epigenetic states can be inherited i. Protein complexes can be passed on ii. Acetylated histones are likely conserved (suppose that the H3 - H4 tetramer Is distributed at random to the two daughter duplexes. Each daughter duplex contain some histone octamer's that are acetylated on the H3 and H4 tails, Where as others are unacetylated. To account for the epigenetic affect, we could suppose of the presence of some acetylated histone octamer's provides a signal that causes the unacetylated octomers to be acetylated) 1. Lactose tolerance a. Loss of heterochromatin structure in adult cells could lead to inheritance of heterochromatic structures g. Epigenetic inheritance can skip generations i. Transgenerational epigenetics (in plants a mutation in maintenance methyltransferase new and Aberrant patterns of epigenetic marks accumulate over several generations, leaving these plants dwarfed and sterile). ii. Seems to be related to methylation states 1. Some loci may transmit epigenetic information 2. Could be related to genomic imprinting a. Cells distinguish maternal and paternal chromosomes from each other via different methylation patterns i. Can result in differential expression of genes from maternal and paternal lineages. 1. One type of Prader-Willi syndrome is believed to be caused by aberrant methylation. this epimutation maybe due to an allele that has passed through the male germline without erasure of the silent epigenetic state established in the grandmother. 2. Gene in paternal chromosome that is normally silenced via methylation is expressed. ii. Lack of methyl donors in diet (folate, choline) can result in defects and inheritance of aberrant methylation patterns

Understand the process and function of telomeric silencing in yeast

The effective telomeric silencing in yeast is analogous to PEV, genes Translocated to a telomeric location show the same sort of variable loss of activity. This results from a spreading effect that propagates from the telomeres. In this case, The binding of Rap1 protein to telomeric repeats triggers the nucleation event, which results in the recruitment of heterochromatin proteins(Sir3 and 4, Sir2 is a deacetylase).

Understand the role of wobble base pairing rules.

Wobble pairings are permitted at the third position 1. Single tRNA molecule may recognize more than one codon via wobble base pairing a. Increases the probability that single base changes will not affect the resulting polypeptide sequence iii. In general, more common amino acids are represented by a larger number of codons. iv. All codons that a particular tRNA recognizes will be identical in the first two positions v. Bases in the third position will adhere to wobble rules 1. Structure of ribosomal A site permits increased flexibility at this third position. 2. Due to wobble rules only 31 tRNAs are required to represent and pair with all 61 codons that encode amino acids. a. Mitochondria has only 22 i. Implies existence of additional base pairing rules and/or modifications.

heterochromatin

densely packed a transcriptionally inactive DNA

Be able to recognize and/or describe the events associated with b. gRNA-based editing

e. Trypanosomes - much more dramatic changes in the mRNA sequence of several genes i. Cox II 1. Frameshift not present in the gene 2. mRNA contains 4 additional nucleotides not present in DNA ii. CoxIII gene 1. More than half of the U residues contained in the mRNA are not present in the gene 2. No run of longer than 7 bases is preserved from the original DNA sequence iii. Editing driven by guide RNA (gRNA) 1. Provides a template for missing U residues 2. gRNAs encoded as individual transcription units 3. often encoded near genes that they target a. CyB represents the gene b. CyB-1 and CyB-2 represent gRNAs involved with editing the Leishmania CyB gene a. The information for the specific insertions of uridines is provided by guide RNA. Guide RNA contains a sequence that is complementary to the correctly edited mRNA. Typically the complementary is more extensive on the 3' side of the edited region and is rather short on the 5' side. Pairing between the guide RNA and the pre-edited RNA leaves gaps where unpaired A residues in the guide RNA do not find complements in the pre-edited RNA. The guide RNA provides a template that allows the missing U residues to be inserted at these positions. the first guid RNA pairs at the 3'-most site, and the edited sequence then becomes a substrate for further editing by the next guide RNA. Editing of uridines is catalyzed by a 20S enzyme complex called the editosome that is composed of about 20 proteins and contains an endonuclease, a terminal uridyl transferase (TUTase), a 3'-5' U-specific exonuclease (exoUase), and an RNA ligase. the editosome binds the guide RNA and uses it to pair with the pre-edited mRNA. the substrate is cleaved at a site that is presumably identified by the absence of paring with the guide RNA; a uridine is inserted or deleted to base pair with the guide RNA, and then the substrate RNA is ligated. Uridine triphosphate (UTP) provides the source for the uridyl residue. it is added by the TUTase activity. Deletion of U residue is mediated by an exoUase, which functions in concert with a 3' phosphate to allow the newly edited RNA construct to religate. 1) hybridize 2) editosome uses its endonuclease to nick 3) editosome uses its terminal uridyl transferase (TUTase) and/or exoUase to add uridine 3'-5' 4) RNA ligase seals the nick 5) the edited RNA becomes a substrate for the next gRNA 3'-5'

histones

four types: H2A,H2B,H3,H4 these combine to make an octamer. each of the histones has an N-terminal tail and a C-terminal tail. the N tail of H3 and H4 are made up of a lot of lysines and arginines. these amino acids can be methylated, acetylated, or phosphorylated. H3 and H4 are more prone to this histone modification. in H3, methylation is most common. in H4, Acetylation is most common. a histone complex combined with DNA is a nucleosome (146bp of DNA wrapped around a histone octamer). multiple nucleosomes are chromatin.

PEV

gene is transferred, either by chromosome or translocation or via transfection and integration, into a position adjacent to heterochromatin, it may become inactive as a result of its location, implying that it has become heterochromatic. it may differ between individual cells in an organism, in which case it results in the phenomenon of position effect variegation (PEV) , In which genetically identical cells have different phenotypes. Genes affected by PEV have two states-active or silenced-depending on their position relative to the boundary of heterochromatin, which can lead to variegated phenotypes. The effective telomeric silencing in yeast is analogous to PEV, genes Translocated to a telomeric location show the same sort of variable loss of activity. This results from a spreading the fact that propagates from the telomeres. In this case, The binding of Rap1 protein to telomeric repeats triggers the nucleation event, which results in the recruitment of heterochromatin proteins.

methylation

gene silencing. addition of a methyl group to a cytosine base using methyltransferase. this typically occurs in cytosine rich islands called CpG islands. this physically impedes translation and may be bound by methyl CpG binding proteins which promotes the binding of other protein that form inactive heterochromatin.

Understand the role of regulatory RNAs in eukaryotes and how RNA silencing functions in eukaryotes.

ii. Eukaryotes 1. Use ncRNAs to control gene expression a. Control at the level of both DNA and RNA i. DNA heterochromatin formation ii. variety of RNA-based mechanisms 2. microRNAs (miRNAs) a. Gene expression regulators b. probably present in all eukaryotes c. At least 1500 known miRNAs in human genome d. participate in RNAi (interference) e. Can come from introns of coding genes f. Can come from ncRNAs g. Can come from pseudogenes i. Believed to have no function previously h. Can also affect translation i. Generally referred to as short temporal RNA (stRNAs) i. May target multiple (100s) of different genes for regulation j. Multiple classes of RNA i. piRNA (piwi-interacting RNAs) ii. siRNA (small interfering RNAs) 1. Both can be used to control expression of TEs (transposable elements) k. Generally produced as precursor pre-miRNAs i. Usually self-complimentary ii. Processed in a two-step reaction 1. Drosha a Nuclear endonuclease. drosha reduces the pre-miRNA to about a 70-bp 2. Fragments exported from nucleus 3. Dicer (endonuclease) a. Produces short double stranded (~22nt) fragment which has a 2nt single-stranded 3' end which is Usually modified with 2'O-Methyl group for stability. Dicer also has an N-terminal helicase activity which unwinds the DS region. miRNA then assembled into Argonaute (Ago) complex. These short, double-stranded RNA fragments are delivered to, or loaded onto, a complex called RISC a. delivers miRNA to its target sequence. The degree of base pairing and the sequence of the ends Of the duplex dictate which of the multiple Ago Family members picks up the RNA duplex and which strand is selected as the passenger strand to be degraded. selection of the class of target by RISC Lies with specific Ago Protein; the specifics are in a target itself is determined by the miRNA. 6. Regulated by a variety of proteins a. MCPIP & Lin28 i. Negative regulators b. KSRP i. Positive regulator iii. RISC 1. Complex of miRNA bound to argonaute proteins 2. Guided to target RNA by miRNA 3. Functions by a. degrading mRNA (plants mostly) b. inhibiting translation of mRNA (animals mostly) c. choice determined by degree of bp between miRNA and target. RISC uses the micro Renée is a guide to scan mRNA by sliding along the RNA looking for a small 2 to 4 nucleotide region of homology that is then extended to an eight base pair seen region in order to initiate full pairing. These regions are usually found in an AU rich region in the 3' UTR of mRNAs, with a few found in the ORF. 4. Usually binds 2-4 nts initially a. Generally, requires 8 nts minimum for full activity 5. Exerts a variety of different effects a. Translation inhibition b. Degradation of message c. Induction of degradation of polypeptide being produced iv. Can sometimes activate genes 3. Ago in Arabidopsis a. Involved with development b. Ago 1 and Ago 10 compete for binding of the same miRNA c. binding to different proteins induces differential patterns of development. iii. RNA silencing in animal cells 1. dsRNA is perceived as being extremely bad by animal cells. The most common sources of the RNA are a replicating virus or a transposable elements. 2. length of dsRNA determines fate a. long dsRNAs trigger complete degradation of all mRNA b. shorter dsRNAs are degraded in a complimentary fashion 3. can spread throughout organisms; not limited to cell of origin

Understand and be able to describe the role of mosaicism as it pertains the to X chromosome in females.

in mammalian females, both x-chromosomes are active during early embryogenesis. as the cells divide, one of the x-chromosomes is silenced (random). Only about 5% remains active. All the daughter cells of a particular cell will have the same x-chromosome inactivated.

Understand the process whereby ribosomes can read through stop codons and produce a longer protein

in retrovirus translation, programmed frameshifting is used to skip forward to a new reading frame, by passing a nonsense codon it does not terminate the transcription. translational bypassing Involves the movement of the ribosome to change the code on that is paired with a peptidyl-tRNA in the P site.The sequence between the two codons is skipped over and is not represented in the polypeptide product. This allows translation to continue past any termination codons in the intervening region. Into the bypass system is that they're identical codons at either end of the skip sequence. These are referred to as the take off and landing sites. The probability that peptidyl-tRNA will dissociate from its codon in the P site is increased by delays in the aminoacyl-tRNA into the A site

euchromatin

less dense transcriptionally active DNA

Be able to recognize and/or describe the events associated with a. Base-level editing

mammalian cells d. Apo lipoprotein-B gene i. Single gene Apo-B 102 carries cholesterol in the blood ii. 29 exons iii. 4563 codons iv. Full protein 1. 512 kDa 2. Expressed in liver v. Truncated form of the protein Apo-B 48 helps with the absorption of lipids in the intestines 1. 215 kDa 2. comprised of NTD of the gene 3. translated from mRNA with a single base change a. C to U at codon 2153 b. CAA to UAA 4. Carried out by apolipoprotein-B mRNA editing enzyme complex (APOBEC) with the removal of an NH2 group (deamination) vi. Glutamate receptor in rat brain 1. Changes in a glutamate codon to arginine 2. Affects conductivity of the channel that is encoded 3. Also results from deamination 4. A to I 5. Performed by adenosine deaminaes acting on RNA (ADARs) 6. D. melanogaster has at least 16 known targets of ADARs 7. Generally, changes functionally significant amino acids vii. Systems for sequence specificity likely vary between individual mRNA targets 1. In glutamate receptor evidence suggests that deaminase targets an adenine in an imperfectly base-paired region between an exon and an intron in the unspliced mRNA

acetylation

occurs at the N and C terminal tails of the histone proteins. acetyl groups are placed on these tails by histone acetyl transferase (HAT) and are removed by histone deacetylase (HDAC). acetylation leads to the uncoiling of DNA from the histone structure. which allows for transcription. deacetylation leads to a condensed structure.

Understand the role and function of suppressor tRNAs

there are 3 classes of suppressors, one for each termination codon. f. Suppressor tRNAs have mutated anticodons i. Changes the codon that is recognized ii. Extends polypeptide chain beyond initial stop codon nonsense mutation - converts a codon (in mRNA) that specifies an AA to one of the three stop codons. 1. Suppression of nonsense mutations a mutation in the tRNA anticodon creates an aminoacyl-tRNA that can recognize the termination codon. 2. Read-through of termination codons iii. Missense suppression missense mutations change a codon representing one AA into a codon representing another AA 1. Changes in tRNA cause the original AA or a different amino acid to be incorporated as long as it restores function, it is a missence suppressor iv. Suppressor tRNAs exist for every stop codon v. Compete with wt tRNAs that have the same anticodon

Understand the structural/function relationship of tRNA molecules

vi. tRNA molecules are synthesized from larger precursor molecules and processed following transcription 1. on the 3' end, an endonuclease cleaves the precursor and an exonuclease trims in the 3'-5' direction. 3'-CCA trinucleotide used for AA attachment is added after synthesis by tRNA nucleotidyltransferase. this enzyme adds the CCA one at a time with no template. 2. ribonuclease P Enzyme recognizes the global L-shaped tRNA structure and specifically hydrolyzes the phosphodiester linkage that forms the mature 5' end of the molecule, leaving a 5'-phosphate group. Modificaitons located in close proximity to anti-codon loop affect tRNAs ability to bp with mRNA a. May function by constraining available motion of anticodon b. Modifications to bases within anticodon loop affect base pairing precision c. isolucine at nucleotide 34