MT Cell Bio Flashcards

Question

examples of centromere formation, and comparison of requirements for centromere formation

Answer 1

* In yeaste s.cerevisiae, require 125bp for centromere, require a dozen of proteins and CENP-A variant. * in a contrasting example in more complex organisms like humans, we require thousands of bp for centromere, and do not contain centromere specific DNA sequences, instead we have alpha satelites. However, alpha satelites are also found in other non-centromere chromatins, which suggests that they are not essential for centromere formation. We also have CENP-A variants * extreme example: neocentromeres have been observed to form spontaneously on fragmented chromosomes, some of these positions were originally euchromatin and lacked alpha satelite DNA. * shows the dynamic and that centromere is not defined by one factor

Answer 2

* linking to genome lectures * during evolution, joining of chromosomes have sometimes caused a chromosome to have many centromeres or non-centromeres * using genetic comparison to identify evolutionary history by identifying inactivated multiple centromere regions on one chromosome to show joining of chromosomes

Answer 3

* centromere activity like above need to be inherited to all cell types for correct segregation of cells * de novo centromere formation requires initial seeding event: which is the incorporation of H3-H4 tetramer with the CENP-A variant to form a protein complex * the DNA sequence for centromere often contains alpha satelites, and seeding event happens more often there * the H3-H4 tetramers from each nucleosome are directly inherited by sister DNA helices at replication fork * the presence of one CENP-A recruits more all the way along the chromatin - a propogation event * once a tetramer H3-H4 has CENP-A in it, in the next cell division, it gets inherited again.

Answer 4

* some of the specialized chromatin components are distributed to each sister chromosome after DNA duplication, along with the specially marked nucleosomes that they bind. * After DNA replication, the inherited nucleosomes that are specially modified, acting in concert with the inherited chromatin components, change the pattern of histone modification on the newly formed nucleosomes nearby. * This creates new binding sites for the same chromatin components, which then assemble to complete the structure. The latter process is likely to involve reader– writer–remodeling complexes operating in a manner similar to that previously illustrated

Answer 5

* in enoculated egg of Xenopus, insert nculeus of donor cell which expresses MyoD gene (master transcription regulator for muscles), which is switched off during early embryo stage * when nuclear transferred into an enucleated egg, it starts to produce MyoD (which is not meant to happen, hence showing epigenetic memory) * we know it is epigenetic memory because the mechanism for producing MyoD is due to the activation of its promoter. Promoter is activated if the histones around it is H3.3 has a 4th lysine methylated * when using a mutant MyoD which prevents 4th lysine from being methylated, no histone modification leads to low MyoD production

Answer 6

* crucial to evolution as condensed chromatin packaging allows evolution of eukaryotes * chromatin structure allows for gene expression and hence specialisation of cell types depending on which genes they express * complexed cell memory in comparison to bacteria * local variations of chromatin allow gene expression regulation * chromatin structure/modifications allow for structures like centromere unqiue to eukaryotes allow for inheritence * polycomb group allow for short-term silencing of genes dependent on environment * whereas heterchromatin with HP1 protein and histone variants allow for more long-term persistent effects * some chromatin structures are short lasting, some are more persistent * very dynamic and flexible

Answer 7

* polytene: giant chromosomes found in drosophila: * lampbrush: oocytes, for them to build supplies * polycombs: short change of gene expression

Answer 8

* looped on the top is more highly expressed

Answer 9

* histone variants * post- transcriptional modifications: 'markers' to histone tails/to DNA * methylation of DNA

Answer 10

* Acetylation: acetylation of lysine residues on histone proteins: catalyzed by HATs. reduces postive charge on histones, decreases interaction w DNA, hence more relaxed chromatin structure more accessible DNA to TFs and RNA pol * methylation: Methylation to lysine or arginine, by HMTs. can be mono-di- , tri. Methylation of histone can lead to diff things. H3 at lys 4 trimethylated leads to transcriptional activation, in other places like tri-me at H3 K27 leads to transcriptional repression * phosphorylation: phosphate group to ser, threonine, or tyrosine on histones executred by kinases. Causes chromatin condensation and segregation during mitosis and meisosis. Also play roles in activation of immediate early genes such as response to stress/growth factors. E.g phosphorylation of H3 serine 10 leads to chromatin relaxation and gene activation * ubiquitylation: adds ubiquitin protein to lysine residues on histones by a E1, 2, 3 cascade enzymes. Signal histone removal or serve as marker. i.e H2B ubiquitylation of K12 associated w transcriotional initiation. Very depends of type of signal.

Answer 11

* writers: modify amino acid tags, i.e acetylation...etc * readers: interact w the modified histone modification (inclusing combinations of histone marks), to change local properties of chromatin * reader/writer complxes can recruit more regulatory proteins such as TFs....etc, modification 'read' by specific binding domain modification, and recruit enzymes which alter local chromatin properties * the reader/write complexes and changed chromatin structures can effect biological function * erasers: move modifications

Answer 12

H3K4me3 keeps genes active by recruiting chromatine remodeller NURF. NURF binds to H3K4me3 and remodels chromatin by promoting nucleosome sliding allowing TFs access to binding sequence

Answer 13

* H3K9me3 represses gene activity * this maintenance requires RNA through RNA induced transcriptional silencing as a part of RNAi to silence genes useing RNA molecules and hsitonbe modification 1. RITSC (silencing complex) is directed to the new transcript mRNA by the antisense RNA 2. then a regulator complex CLRC interacts w RITSC, which then causes the Clr4 subunit of the regulator complex to methylate H3K9. Through their interaction of antisense RNA, the CLRC and the RITSC, it targets Clr4 to specific genome site where silencing is required 3. HP1 then recognizes the methylated H3K4me and binds to it, contributing to the formation of heterochromatin

Answer 14

* vertebrates: cytosine can be methylated * usually methylation of DNA causes inactivity (but exeptions present) * DNMT establishes AND maintains cytosine methylation * erasing DNA methylation uses TET enzyme, which oxidizes 5meC

Answer 15

1. CH3 blocks the binding of TFs 2. DNMT recruits trancription inhibiting proteins like HDACs 3. methyl binding domains recruit HDACs too

Answer 16

* when a gene is active, its histone modified is H3K4me3 * this marks the gene and allows it to inactive DNMT, hence stops the DNA methylation. The active gene mark recruits and binds to ADD domain of DNMT which inactivates it * when a gene is inactive, it doesnt have marks which inactivated DNMT * DNMT hence is active to methylate cytosines.

Answer 17

* Maintained and inherited both histone modification and DNA methylations 1. Histone modification is conserved. * histones are recylcles and distributed during DNA replication. Old histones then stay with one of the daughter strands and provides a 'template' for new DNA strands to recruit same histone pattern * histone chaperones help guide and promote modification of histones onto new histones during exchanges * positive feedback: after the initial seeding event, positive feedback mechanism allows them to recruit more histones and add same modifications by enzymes. 2. DNA methylation is conserved * during semi-conserved DNA rep, the mother strand retains DNA methylated patterns. DNMT1 recognizes this and adds same methylation to daughter strand

Answer 18

* DNA replication dilutes histone markers * initially during replications these epigenetic markers may be lost as it uses a recycle of new and old histones. A un modified new histone may be incorporated * however, in post-replication, enzymes which recognize old histones will re-establish this epigenetic pattern * allows gene expression patterns to be maintained in cell line through many divisions

Answer 19

* repression due to H3K37me3, applied by the polycomb repressive complex 2 (PRC2) are diluted because the parental histones are randomely distributed to new strands and combined w new unmodified histones * post- replication: PRC2 complex recognizes the remaining H3K27me3 on the parental histones and binds to them * allosteric activation and spreading: binding of PRC2 to H3K27me3 can allosterically activate the complex. It starts adding more methyls to H3 histones through positive feedback * cohesin holds the 2 sister chromatids tgt during this

Answer 20

* in mammalians, when a parental inherited copy of a gene is activated and the other is inactivated (vice versa) * can unmask mutations that would normally be covered by the other functional copy * in embryo: genes subject to imprinting are methylated * i.e Angelman syndrome, a disease in nervous system that causes speech impairment and mental disability, results due to the expressed mutated gene on one chromosomal homolog and the silencing (imprinting) of the intact gene on the other homolog

Answer 21

* insulin growth factor 2 (igf2) in mouse * cis-regulated * in maternal inherited chromosome, the cis-regulated sequence is blocked due to an insulator element which is activated by CTCF binding and blocks communication between cis-regulatory sequence and Igf2 gene, stopping it from expressed * Because of imprinting, which means that the male gene is methylated. The gene for the insulator protein is methylated, which prevents it from binding to CTCF. which allows cis-regulatory seq to communicate and transcribe the igf2 in male mice, allowing normal growth

Answer 22

* in mammals, to balance out gene product ratio on sex chromosomes, one of the X chromosome of females (XX) is inactivated (X-inactivation) * this allows 2 X chromosomes to be in one cell nucleus without too much gene product, hence careful gene expression

Answer 23

* positive feedback loops allow more stability by buffering against fluctuations of any one Transcription regulator * self-propogation occurs in 2 ways. * cis epigenetic mechanisms: affects only one chromosomal copy * trans epigenetic mechanisms: affects both chromosomes * NOTE: not all epigenetic modifications are long lasting, MOST don't have cell memory, butthose who do likely self propogate and have a positive feedback loop

Answer 24

''the study of changes in gene or chromosome function that are mitotically and/or meiotically heritable and that do not entail a change in DNA sequence”

Answer 25

DNA-RNA-Protein

Answer 26

* semi-conservative * both strands synthesised from a 5' -3' direction * template strand reads 3'-5' * leading strand is continuous from 5'-3', lagging rads from 3'-5', but still synthesises in small DNA fragments (OKazaki fragments) from 5'-3'

Answer 27

* euk: multiple replication forks, Initiated by ORC * proks: 1 replication fork, initiated by DnaA DnaB,

Answer 28

1. origins bound by ORC (initiation) 2. DNA helicase (Mcm2-7) loaded onto ORC inactive form 3. DNA helicase is activated generatingssDNA that can be replicated by polymerases

Answer 29

* short term survival of a cell depends on DNA sequences being conserved, long term survival of a species requires DNA sequences to be changeable over generation to allow adaptation * mutation rates are extremely low in DNA to allow the functions of genes to be preserved. * 1/10^10 nucleotides per cell division * important to conserve genes in somatic cells but more importantly in germ line cells

Answer 30

* late M phase/early G1 phase * known as lincensing * The ORC binds to the origin of the replication form, and cdc6 then combines with the ORC on the DNA * this allows loading of 2 Cdt1/Mcm2-7 complexes

Answer 31

* in late G1 phase * more proteins are recruited and bounf to Mcm2-7 complex including polymerase ε, DDK, CDK, Cdc45

Answer 32

* activation of DNA replication is controlled by 2 protein kinases CDK and DDK * in late G1/S phase: * CDK is recruited by the pre-RC and phosphorylated the pre-loading complex (which contains pol ε) and the sld3 subunit of mcm2-7. This causes the pre-RC to bind to the pre-loading complex * DDK then bounds to the complex and together activated DNA replication by forming the CMG complex/CMG helicase in S phase

Answer 33

1. cdc45, GINS and mcm2-7 bound tgt activates helicase 2. activated helicase generates ssDNA, and opens the replication fork by breaking H bond 3. allows priming by DNA pol a (primers) and pol ε

Answer 34

* replication fork: 2 ss DNA exposed for pairing * DNA pol only capable of synthesising 5'-3'. Which works for the leading strand in a continuous motion * Pol alpha forms a primase complex, and forms a short RNA primer on both leading and lagging strand, to form a RNA primer extended by a short stretch of DNA nucleotides. * leading strand: DNA pol ε takes over alpha and extends a continuous 5'-3' leading strand of DNA. PCNA anchors and allows DNA pol ε to slide along replication fork and follows the helicase as it exposes for ssDNA * lagging strand: DNA pol delta with protein ctf4 synthesizes lagging strand. Since helicase unwinds in a direction away from pol delta, lagging strand is synthesized in small 5'-3' Okazaki fragments. DNA pol delta also guided by PCNA * Lagging strand: primer reomoval and gap filling: As pol delta synthesizes lagging strand, it will end up encoutering the RNA primer from previosu Okazaki fragment and it will displace the primer creating a flap like structure * the flap is removed by FEN1 leaving a nick, which isthen filled in by Pol delta.

Answer 35

DNA pol 3' exonucleases can only add nucleotides on to a 3'OH, which is why Pol a and RNA primase need to add an initial 3'OH end

Answer 36

* lagging strand: DNA pol delta with protein ctf4 synthesizes lagging strand. Since helicase unwinds in a direction away from pol delta, lagging strand is synthesized in small 5'-3' Okazaki fragments. DNA pol delta also guided by PCNA * Lagging strand: primer reomoval and gap filling: As pol delta synthesizes lagging strand, it will end up encoutering the RNA primer from previosu Okazaki fragment and it will displace the primer creating a flap like structure * the flap is removed by FEN1 leaving a nick, which isthen filled in by Pol delta.

Answer 37

* The helicases on each strand pass each other * and replication up to primer on what lagging strand * helicase then switch to encompass both strands * Nicks are filled in by DNA ligase I, * Mcm7 is then ubiquitylated (but not degraded) and removed from chromatin by p97 ATPase

Answer 38

* ORC bind to specific sites, but what these sites are varies among species * In S.cerevisiae: ORC binds to a specific autonomously replicating sequence (ARS) Contain A/T rich element for identification of beginning of replication. Protein Abf1 enhances origin recognition and binding of ORC in s.cerevisiae * In S.pombe, ORC doesnt bind to a strict consensus seq, but binds to a non-specific A/T roch sequence through a domain of its's subunit Orc4. which contains an AT-hook that recognizes these regions * In metazoans replication origin also not determined bys trict sequence. A combination is involved. Epigenetic markers: like H4K20me2 and Acetylation of H4 contribute to this replication origin. Chromatin structure: G4 structures also help determine ORC positionng.

Answer 39

1. avoiding conflict between transcription and replication. Ensure that these 2 mechanisms dont clash, as both replication and transcription involve unwinding DNA 2. Ensuring intervals between each origin is not too larger, so the entire genome can be replicated efficiently and correctly

Answer 40

1. complex development, means that at different developmental stages, the replication may need to be modified in terms of length and duration. To allow for more flexibility, not a specific sequence is used. 2. for example, in embryonic cells, there is a short S phase and reduced transcription. Hence more origin sites are required to replicate short genome in short S phase time. reduced transcription means more origin sites for more replication to occur simultaneously as no risk of clashing into transcription 3. In adult cells however, there is much more transcription going on, and more regulation. Therefore the origin site needs to be more carefully moderated to prevent clashing into transcription, as well as responding to epigenetics. 4. Epgenetic specification gives flexibility to allow variations of S phase length. Epigenetic markers allow better timing adn location of replication origin. This allows fore better response to cellular conditions, and complex regulation depending on cell type and developmental stages

Answer 41

* during cell division (S phase), each origin on the chromosome is fired once only * as cells exit G1 phase and enter S phase, licensing triggers a surge of CDK activity, which activated initiation stage of DNA replication * the CDKs phosphorylate substrates including proteins which form the pre-Replication Complex, which signals the initiation of replication

Answer 42

* In G0 phase (quiescent cells) lack CDKs and DDKs and are not active. Which prevents accidental initiation of DNA rep in wrong timing * Transition from G1 to S phase is a crucial checkpoint where licensing and activation of kinases are required before next processes

Answer 43

1. prevention of early replication or replication when the cell is not ready (i.e not all proteins are synthesized for cell division) 2. E.g in mammalian oocytes, remain in non-dividing stage for extensive periods and dont replicate DNA until developmental cues trigger cell cycle

Answer 44

* In eukaryotes: * Base excision repair (for small base lesions) - DNA glycolusase * Mismatch repair: during DNa rep. MMR proteins recognize and remove mismatch, then resynthesize correct sequence * DNA replication and transcription use topoisomerases: when supercoiling occurs ahead of replication fork, topo breaks DNA strand to relieve supercoiling * Topo I: single stran break, relav negative and positive supercoils * Topo II: DNA gyrase (ONLY FOR BACTERIA VERY important). causes dsbreak in DNA. can manage more complex topological issues

Answer 45

* on a macro scale v.similar mechanism: semi conservative * have the same aim: to ensure DNA is correctly copied and copy passed down to daughter cells * proks and euks have similar mutation rates (proks are slightly higher due to faster replication/not so specific proofreading mechanism), but still both mutation rates are low

Answer 46

* Once CDKs phoshphorylate the proteins in the activation of Mcm2-7 helicase, replication at origin begins and replication fork starts forming

Answer 47

* checkpoints, licensing. Only licensed origins can start replication * after each origin fires once during cell cycle, the origins are not re-licensed, hence preventing it from firing again * After licensing and after pre-RC is formed, the factors forming Pre-RC are degraded to prevent re-replication * Before initiation: all the initiation factors (CDK, DKK) are lacking/inactive, and only when initiation happens they are active

Answer 48

1. after licensing, proteins needed to form a new pre-RC such as cdt1 and cdc6 are degraded or inactivated. 2. in eukaryotes: cdt1 is inhibited by geminin to prevent re-licensing. Level of geminin increases as cell progresses through S phase, ensuring cdt1 cannote re-bind and re-license origins 3. Moreover, the formation of the replication fork leads to cdt1 proteolysis, as it targets cdt1 for ubiquitin mediated degradation, This ensures cdt1 is no longer available to license another origin 4. moreover, the initiation and replication fork formation changes chromatin struicture making it unfavourable for licensing again 5. PCNA also plays signalling role to prevent re-licensing

Answer 49

1. proteolysis of cdc6: low CDK level in G1 permits pre-RC forming, but S phase and G2 phase has high CDK levels which blocks pre-RC to prevent re-replication. This is because high CDK levels phosphorylates Cdc6 and causes it to be recognised and ubiquitylated by SCF for degradation, so no pre-RC formed and no re-replication 2. 2nd function is that it causes nuclear exclusion of mcm2-7: outside S phase, mcm2-7 complex are removed from the nucleusm hence prevents replication of cell in the wrong phase of cell cycle Controls replication timing

Answer 50

* euchromatin regions of origins are replicated first during Early S phase, and heterochromatin origins are replicated in late S phase 1. Resource management: fewer copies of replication proteins required, staggered replication forks allow proteins to be reused. 2. Reduce demand of dNTPs (low dNTPs can lead to genomic instability) 3. timing of replication may influence chromatin modification: epigenetic marks need to be copied to caughter strands, and hence require correct timing to avoid incorrect epigenome

Answer 51

* in diploids during sexual reproudction * produces haploid cells (gametes) carrying a single copy of each chromosome * during sexual reproduction: these gametes join together to form a diploid zygote, which has the potential to form a new individual

Answer 52

* two rounds of chromosome segregation: first round segregates the homologs, and 2nd round * meiosis 2: separates the sister chromatids producing haploids * crossing over and random segregation of homologs, generating genetically different haploid cells * creates genetic variability

Answer 53

* pre-meiotic D phaseL replicated sister chromatids held together throughout the chromosome by cohesin (differs in mitosis in the type of CDKs and cohesin used) * these homologs then randomely line up * meiosis I: homologs recombine in miosis I prophase * homologs separate in through random segregation during meiosis I anaphase * No S phase between meiotic nuclear division * sister chromatids separate in 2nd meiotic anaphase, producing 4 haploids

Answer 54

* in mitosis: the homologs don't pair up, only sister chromatids segregate * in mitosis: only 1 round of division giving 2 daughter cells (vs 2 rounds of segregation giving 4 daughter cells * mitosis produces genetically identical clones of daughter cells, (meiosis produces haploid and genetically diff daughter cells) * meiosis has no S phase between meiotic divisions * meiosis has homologous recombination and random assortment of chromosomes * meiosis: separation of homologs in meiosis I

Answer 55

2^n (where n = number of chromosome pairs * in humans, 23 pairs of chromosomes * hence 2^23 = 8.4 * 10^6 geneticaly diff gametes * and hence 8.4 * 10^6 x2 = 7*10^13 genetically different zygotes * assuming no recombination, recombination will cause even more genetically different children

Answer 56

* in females, oocytes are arrested at meiotic prophase I, and progresses during puberty once ovulation happens * errors in meiotic chromosome segregation can cause trisomy * common one is trisomy 21 causing downe's syndrome, and this error increases with age after 35 more likely to have errors in segregatting chromosome causing trisomy

Answer 57

* prophase I * in order for homologs to recognize each other and pair up and bi-orient on the first meiotic spindle * during prophase I (long period - takes hours in yeatss and weeks in complexed plants) * meiosis cohesin keep the sister chromatids and bivalent homologs closely associated 1. duplicated meiotic prophase chromosomes are long threadlike and closely associated w their sister chromatids 2. Alignment: homologs pair and associate in **pairing** process: through complementary DNA sequences (pairing sites) and causes the homologs to be closely juxtaposed to form a 4 chromatid structure called **bivalent** 3. Recombination: homolog pairs are locked tgt by homologous recombination (crossing over): double strand DNA breaks at sister chromatid. large number of DNA recombination event and cross over of chromatid DNA

Answer 58

* Recombination: * Pre-synaptic alignment: recobination complex (which also assembles double strand break in a chromatids) binds matching sequences on homolog and brings it closer * Synapsis: axial core of a homolog becomes tightly linked to its partner homolog by transverse filamengts to create **synaptonemal complex** which bridges the gap ebtween the homologs to 100nm in gap * crossing over begins before synaptonemal complex assembles but final steps of crossing over occurs when DNA is held in a synaptonemal complex

Answer 59

1. transverse filaments surround the axial cores of the homologs (which have matching DNA seq recognized and brought tgt by pre-synapsis recombination complex) 2. axial cores of each homolog also interact w cohesin to hold together sister chromatids 3. 100nm gap * SC = formed during synapsis

Answer 60

1. leptotene: homologs condense and pair and genetic recombination begins 2. zygotene: synaptonemal complex begins to assemble at sites where homologs are closely associated (pre-synaptic alignment) and where recombination occuring 3. pachytene: assembly of synaptonemal complex complete and homologs are synapsed along entire lengths. (long periods for days - until desynapsis) 4. diplotene: desynapsis occurs: disassembly of synaptonemal complexes along w shortening and condensing of chromosomes. (Dissassemble is complete). This allows an observation of inter-homolog connections: chiasmata (which show sites of recombination by being attached to another cohesin) 5. diakinesis:

Answer 61

* infertility * birth defect

Answer 62

* some species dont need Synaptonemal complex formation for formatyion of chiasmata and recombination * double stran break dependent synapsis: DSB occurs and then this leads to crossing-over, recombination and chiasmata formation

Answer 63

* S.pombe doesnt form SC, but chromosomal movement: ** horsetail nuclear movement**, help align chromosome for recombination * chromosome movement via horsetail nuclear movement and clustering of telomere: causes homologous sequences to come into proximity *

Answer 64

1. nucleasecauses double strand break with 3' overhanging 2. hence causing either crossingover or joining of DNA strands again

Answer 65

* homolog separation unique to meiosis, does not occur in mitosis and is avoided 1. in metaphase I: sister kinetochores of one homolog must attach to same spindle pole (this is AVOIDED in mitosis). Cohesin links at recombination sites (chiasmata) between sister chromatids of homolog and all along the sister chromatids to allow the homologs to be close together 2. Metaphase I: the 2 sister kinetochores of 1 homolog are fused into a single microtubule binding unit (syntelic attachment) and is attached to a single pole 3. fusion of sister kinetochore: achieved by monopolis that is localized at kinetochores and is removed after meiosis I to allow sister chromatid separation (allow bioriented on spindle during meiosis II - similar to mitosis) * at this stage (metaphase I) the spindles have attached correctly to sister chromatid kinetocore, but not yet pulled apart

Answer 66

* connect homologs and keep all 4 sister chromatids together in close proximity through cohesin links

Answer 67

* meiosis I: 4 kinetochores * when incorrect attachment of meiotic spindle to sister kinetochores

Answer 68

* meiosis I (before anaphase I): cohesin throughout the sister chromatids and hold bivalent structure of homolog pair * meiosis I (anaphase I): cohesin lost in sister chromatid arms to allow segregation of homologs * mitosis: kept until anaphase to allow division of sister chromatids

Answer 69

* cannot lose cohesion completely as this would also split the sister chromatids in meiosis I 1. anaphase I: cohesin retained around centromere and lost from arms. The shugoshin protein protect cohesin around centromere from being cleaved off by separase. Separase is actiavted in APC/C activation, which leads to destruction of securin, and activation of separase 2. Shugoshin recruite PP2A phosphatase to remove phosphates from cetromeric cohesins. 3. separases can only cleave off phosphorylated cohesins, hence cannot cleave non-phosphorylated centromeric cohesins 4. this allows sister chromatids to be linked for bi-orientation at equator for meiosis II. 5. shugoshin is inactivated after meiosis I.

Answer 70

* meiosis I: anaphase I: needs to keep centromeric cohesin, but remove arm cohesin in sister chromatids * The shugoshin protein protect cohesin around centromere from being cleaved off by separase. 1. Shugoshin recruite PP2A phosphatase to remove phosphates from cetromeric cohesins. 2. separases can only cleave off phosphorylated cohesins, hence cannot cleave non-phosphorylated centromeric cohesins 3. this allows sister chromatids to be linked for bi-orientation at equator for meiosis II. 4. shugoshin is inactivated after meiosis I.

Answer 71

1. Anaphase II: sister chromatids are aligned bi-oriented at the equator and are to be segregated by spindle fibres. 2. APC/C activation once again triggers Separase to cleave off cohesin 3. without shugoshin, sister chromatids are separated due to lack of cohesin and spindle fibres attach to each sister'chromatids kinetochore and splits the sister chromatids to opposite poles 4. in subsequent cytokinese they are split off to form 4 haploid daughter cells (gametes)

Answer 72

1. securin in bound to separase and inhibits its function 2. when APOC present, securin is degraded and separase becomes activated 3. active separase cleaves cohesin off sister chromatids

Answer 73

1. low CDK required to load ORC (mcm2-7), once loaded, CDK level increases and BLOCKS ORC loading/mcm2-7 loading 2. low CDK activity which initiates licensing in DNA replication is initiated by cyclin B proteolysis 3. in Meiosis I and II, there is limited cyclin B proteolysis, hence CDK level is not low, and this stops the licensing and loading of ORC and prevents formation of Pre-RC. Hence blocks DNA replication

Answer 74

* Inject Frog oocyte with mRNA of WeeI (CDK inhibitor) during Meiosis I and II interval * this causes CDK to be inactivated in these oocytes, and this led to DNA replication and loading of mcm helicase (licensing) * In normal conditions, CDK present and activel and no DNA rep in interval of MI and MII

Answer 75

* Nucleotide addition cycle: 1. Translocation: RNAP moves along DNA template in 3'-5' direction, moving the growing RNA chain from pre-translocated to post-translocated state 2. NTP binding/releasing: A NTP conplementary to DNA template base pairs w DNA in active site of RNAP 3. Catalysis: RNAP catalyzes formation of phosphodiester bond of correct NTP. this adds a nucleotide to the 3' end of RNA molecule. Energy used for this comes from hydrolysis of NTP (releasing PPi) 4. PPi Release/binding: PPi released from active site of RNAP and completes addition

Answer 76

* RNA Pol is dual-functionality, it catalyzes the addition of nucleotides and the excision of incorrect nucleotides thru a backtrack reaction via hydrolysis

Answer 77

* RNAPs in eukaryotes and prokaryotes have similar structure, and hence thought to all derive from a common ancestor * for eukaryotes: 1. Initiation (recruitment of RNAP): TFB recruits TBP binds to TATA box and recruits RNAP to form a complex 2. Initiation (closed complex): DNA helix still in tact 3. Initiation: open complex: RNAP unwraps DNA exposing template strand 4. Abortive Initiation: RNAP synthesizes abortive RNA fragments, before successfully escaping the promoter to transition into elongation phase 5. Elongation: RNAP associated with elongation factors and adds nucleotides to RNA transcript 6. Termination: RNAP dissociates from DNA template and releases the full length RNA molecule. U-rich tracts signal RNAP to stop transcription

Answer 78

* relating back to the RNA world hypothesis * likely that RNA Polymerases in euks and proks originated from same common ancestor * RNAPs are diverse in tree of life, from more simple structure in prokaryotes, to more complex and more diverse RNAPs in eukaryotes * Fe-S cluster is common in all, and is an important catalytic structure that allows transcription to occur * Mg+ active site * function of RNAPs are conserved and is universal, despite their different structures * Single subunit phage RNAPs can carry out transcription w/o additional factors, showing that the other factors are not essential for transcription, and that RNAP is essential * as RNA features of all 3 domains are conserved, suggests that transcription (or RNA repliacation mechanism) had occured prior to division of three domains

Answer 79

* 6 units * sigma factor binds specifically to the promoter region on bacterial DNA * addition of sigma factor tgt is called the DNA polymerase holoenzyme * sigma unit only binds with the rest of the RNA polymerase when in comes into contact w DNA (binds weakly)

Answer 80

* in both euks + proks 1. initiation 2. elongation 3. termination

Answer 81

1. sigma factor binds to RNAP to form RNAP holoenzyme and quickly slides through DNA until it reaches a promoter, which then the polymerase binds tightly (because sigma factor makes specific contacts) 2. RNAP opens up the DNA helix and expose short stretch of DNA forming transcription bubble. bubble is stabilized as RNAP binds to non-template DNA strand 3. Template strand exposed and forms mRNA by scrunching mechanism (RNAP still bound to promoter and pulls upstream DNA into the active site and expands transcription bubble) causes stress and causes release of abortive RNA fragments (abortive initiation) 4. then scrunching causes core-enzyme of RNAP to break its interaction w promote and sigma factor is disgarded 5. RNAP moves down DNA elongating it - expanding transcription bubble 6. chain elongation continues at 20nucleotides/sec for bacterial RNAP 7. RNAP encounters terminator signal. Template strand has string of ATs when transcribed forms into a hairpin strcuture and causes RNA transcript to dissociate and causes release of the DNA template strand.

Answer 82

* very complicated steps * requires holoenzyme (RNAP + sigma factor) * abortive initiation through creating stress abortive RNAs * if RNAP released prematurely mist start over again

Answer 83

* how RNA searches a specific site for initiation (finding the promoter) * includes 1D hopping, 1D diffusion (sliding), and intersegmental transfer

Answer 84

1. 1D Hopping: RNAP binds to DNA and moves by dissociating and reassociating with the DNA, effectively "hopping" along it. This allows RNAP to cover short distances rapidly, but it's a local mode of movement. 2. 1D Diffusion (Sliding): RNAP binds to DNA and then slides along it in a linear fashion. This "sliding" or "1D diffusion" means the enzyme stays in contact with the DNA while moving along its length. It's an effective way to search for promoter sites without fully detaching from the DNA molecule. 3. Intersegmental Transfer: This is when RNAP transfers from one segment of DNA to another non-contiguous segment. DNA looping can bring distant regions into close proximity, allowing RNAP to jump or transfer between these regions. It's a way for RNAP to efficiently sample different areas of the genome. 4. 3D Diffusion (Jumping): Apart from moving along the DNA, RNAP can also completely dissociate from the DNA and diffuse through the nucleoplasm in three dimensions, before reassociating with the DNA elsewhere. This allows RNAP to search for promoter regions that are not in its immediate vicinity.

Answer 85

A consensus sequence is a way of representing the most frequent nucleotides at each position in a collection of aligned sequences. - after aligning multiple sequences, finding the most common base for each position

Answer 86

-35 Region: This hexameric sequence is located approximately 35 nucleotides upstream of the transcription start site. It's recognized by the σ factor, which is a part of the RNA polymerase holoenzyme, and is essential for the binding of the enzyme to the DNA. -10 Region (Pribnow box): Another hexameric sequence found about 10 nucleotides upstream of the transcription start site. It is also critical for RNA polymerase binding and initiation of transcription. The consensus sequences for these regions in E. coli are typically: -35 Region: TTGACA -10 Region: TATAAT

Answer 87

* sigma factor has Helix-Turn-Helix (HTH) motifs which are structural motifs for the sigma factor to bind to DNA at specific sequences. These motifs insert into the major groove of DNA helix and make sequence specific contact that determine where RNAP bind * Once a sigma factor engages w the DNA, the N-terminal fo the sigma factor initially blocks the DNA binding domains. Only when the core RNAP binds, it causes conformational change which allows sigma factor to bind to the promoter seq * once sigma factor's DNA-binding domains recogniza and bind to the DNA at -35 and -10 regions of promoter, it positions RNAP to be at the correct site for transcription initiation

Answer 88

* sigma and anti-sigma factors are involved in gene expression in bacteria * sigma factors bind to RNAP holoenzyme and direct it to specific promoter regions on DNA where transcription begins * diff sigma factor recognize fdiff promoter sequences, hence regulates gene expression * For example, Sigma 70 recognizes the consensus sequence TTGACA followed by 16-18 base pairs and then TATAAT. This sequence is commonly found in promoters targeted by Sigma 70 in E. coli. * antisigma factors are proteins that bind to sigma factors to inhibit their function. * switches of genes/regulatory mechanism

Answer 89

* prokaryotes: rho-dependent, rho-independant (intrinsic termination method) * eukaryotes: cleavage and polyadenylation + RNA pol II

Answer 90

* both bacteria transcription termination mechanism 1. rho-dependent termination: rho protein (ATP-dependent helicase). Rho-protein recognizes specific sequences (rut sequence, C-rich) on the new RNA transcript and binds to it. Moves along the RNA towards RNA polymerase, unwinds RNA-DNA hybrid within transcription bubble, leading to release of RNA transcript. Rho has 2 domains, N-terminal RNA binding domain, and C-terminal ATPase domain, uses ATP to dissociate 2. Intrinsic termination: Relies on specific sequence in RNA transcript or DNA sequence. A GC-rich region of the RNA forms a hairpi loop followed by a series of U bases. Hairpinstructure causes RNAP to pause and week bonds between U bases and A in DNA template strand causes detachment of RNA and DNA. HOWEVER: in some conditions, mRNA can also form anti-terminator hairpins to prevent termination of transcription

Answer 91

* eukaryotes have 3 RNA polymerase, share similar seq and structural domains: RNA polymerase I transcribes 18S/28S rRNA. * RNA polymerase II transcribes mRNA and some small RNAs. * RNA polymerase III transcribes tRNAs, 5S ribosomal RNA, and also some other small RNA * prokaryotes only have 1 RNAP:

Answer 92

* Initiation: euks have 3 diff types of RNAP, proks only has 1 RNAP * euks' TBF +PolII recognize TATA box promoter sequence, proks require -10 and -35 promoter sequence recognized by sigma factor * termination: proks: rho dependent/independent, euks: polyA + cleavage + splicing

Answer 93

Plants also evolved two additional types of RNA polymerases from RNA polymerase II: RNAP IV and V. They are involved in the production of non-coding RNAs that regulate gene expression - evolved latest from Pol II

Answer 94

Eukaryotic transcription initiation must take place on DNA that is pack- aged into nucleosomes and higher-order forms of chromatin structure features that are absent from bacterial chromo- somes. In GTF, the TFIID and TFIIH are very important. 1. TFIID binds to TATA binding protein 2. Then the TFIID w/ TATA binds to DNA TATA box, and TFIIA binds to this complex to stabilize it. 3. TFIIB then binds and recruits RNA pol 2 4. TFIIF and TFIIE recruits the important TFIIH which completes the large complex and can act as a DNA helicase/kinase, by phosphorylating polII, as it recruitrs the capping enzyme 5. Pol II then goes on to transcribe nucleotides (72 nucleotides/sec) 6. Capped polymerase stops it from getting degraded. d. Promoter region can increase of decrease rate of transcription for a gene i. Or repressors can bind to DNA motifs which prevent DNA transcription

Answer 95

a. PAS (polyadenykation sequence) is a termination sequence b. And pol II transcribes the PAS and continues c. CAP complex binds to pol II in RNA d. This leaves the RNA end after the PASwhich is naked and not capped/protected, hence CAP binds to it and stops its from being hydrolysed e. Poly A tail is then added to the 3’ end to prevent endonucleation

Answer 96

RNA polymerase I is recruited by a basal transcription factor SL1 that is placed at a TATA element by factors that bind the upstream promoter element .

Answer 97

RNA polymerase III can bind three different types of promoters, which consist of conserved elements arranged in alternative configurations.

Answer 98

Escape from the Initiating Complex: Once the RNA polymerase (same for I, II, III) is bound to the promoter, forming the pre-initiation complex, it must transition to the elongation phase of transcription. This is often referred to as "promoter escape," where the RNA polymerase leaves the promoter region to begin synthesizing RNA along the DNA template. Recruitment of New Complexes: After the RNA polymerase has cleared the promoter, the transcription initiation complex that remains bound to the promoter can then recruit another RNA polymerase molecule. This allows for multiple rounds of transcription initiation, enabling the gene to be transcribed repeatedly, which is essential for producing the required amounts of RNA.

Answer 99

1. Cleavage and Polyadenylation: As RNA polymerase II transcribes the DNA into RNA, it eventually transcribes a sequence that signals the end of the gene, known as the polyadenylation signal (typically AAUAAA in humans). Two key multi-protein complexes, Cleavage and Polyadenylation Specificity Factor (CPSF) and Cleavage stimulation factor (CstF), recognize this signal and bind to the RNA. 2. Pre-mRNA Cleavage: CPSF and CstF together direct the endonucleolytic cleavage of the pre-mRNA about 10-30 nucleotides downstream of the polyadenylation signal. 3. Poly(A) Tail Addition: Poly(A) Polymerase (PAP) adds a string of adenine nucleotides (the poly(A) tail) to the newly created 3' end of the RNA. This polyadenylation is important for mRNA stability and nuclear export. 4. RNA Processing and Export: Guanylyl transferase adds a 7-methylguanosine cap to the 5' end of the pre-mRNA, which occurs co-transcriptionally (as the RNA emerges from RNA polymerase II). This capping is crucial for RNA stability and later for initiation of translation. Methyltransferase then adds a methyl group to the guanine cap, forming the 7-methylguanylate cap (m7G). 5. Binding of Poly(A) Binding Protein (PABP): PABP binds to the poly(A) tail, contributing to the stability and regulation of the mRNA. 6. Transcription Termination: Tafter cleavage, the RNA polymerase II continues to transcribe for a short distance before it dissociates from the DNA. This could be facilitated by the still-transcribing polymerase encountering the cleaved, polyadenylated RNA, or by other factors that signal the polymerase to disengage. Endoribonuclease may be involved in degrading the RNA downstream of the cleavage site, which may help to dislodge RNA Pol II from the DNA. 7. Torpedo Model: A proposed mechanism for the actual termination step is the "torpedo model," in which an exonuclease (like the 5’ to 3’ exoribonuclease Xrn2 in humans) degrades the downstream, uncapped RNA following cleavage and chases down the RNA polymerase. Once it catches up to the polymerase, the degradation of the RNA induces the polymerase to terminate transcription.

Answer 100

1. RNA pol II tail (aka C-terminal domain) gets phosphorylated to be detatched from initiation proteins and start transcription elongation 2. As it detaches from initiation proteins it also allows new proteins to become attached to the CTD which function in RNA elongation and RNA processing steps 3. the CTD is essentially a scaffold which holds a variety of proteins close by the RNA pol II, so that it can be transferred and used on the nascent RNA when required 4. This allows the RNA pol II to not only elongate and transcribe nascent RNA, but also allow its proteins to process the nascent RNA as it is formed 5. allows overall rate to be sped up

Answer 101

1. 5' Cap (7-mehylguanosine) 2. 3' Poly A tail 3. splicing

Answer 102

* prokaryotic mRNA can be transcribed in an operon (multiple related genes controlled by one promoter i.e lac Operon) * Eukaryotic mRNA is always 1 protein controlled by 1 promoter (however alternative splicing can create more that one protein)

Answer 103

- Polycistronic transcription is specific to prokaryotes and some viruses, where a single mRNA transcript can encode multiple proteins, because it contains several open reading frames. Each of these frames corresponds to a different protein, which is a common feature in bacterial DNA. - i guess it is sort of bacteria's way to get one gene to code for multiple proteins (like alternative splicing in eukaryotes)

Answer 104

1. capping proteins are attached onto ser5 of CTD of RNA pol II once it has been phosphorylated by TFIIF during transcription initiation 2. Once RNA pol II starts transcription and produces aroun 25 nucleotides on nascent RNA, the RNA capping proteins add a 5' cap to nascent RNA prevent it from being hydrolyzed by exonucleases 3. CTD phosphorylated at Ser2 positions, and eventually dephosphorylated at Ser5, anddissociates w DNA to allow reinitiation 4. this eventually leads to the attraction of 3'-processing proteins and splicing proteins

Answer 105

1. we can compare RNA pol II with RNA pol I and III transcripts 2. RNA pol I and III also produced RNA transcript, but they are uncapped at 5' because of a lack of CTD 3. The 5ʹ-methyl cap also has important role in the translation of mRNAs in the cytosol, and helps w mRNA further processing and exporting 4. hence the 5' cap mRNA helops distinguish it from other types of RNA

Answer 106

1. RNA pol II's CTD during initiation is phosphorylated by TFIIF at ser5, and hence dissociates from initiation proteins 2. 3 capping enzymes acting is succession is bound to CTD (as CTD is phosphorylated at Ser5): a phosphatase, a guanyl transferase, a methyl transferase 3. once RNA pol II produces about 25 nulceotides of nascent RNA transcript, these 3 enzymes start capping process in succession 4. phosphatase removes a phosphate from 5' end of nascent RNA 5. Guanyl transferase adds a GMP to the 5' end via reverse linkage (5'-5' link instead of 5'-3' link) 6. then methyl transferase adds a methyl group to the Guanosine

Answer 107

1. RNA pol II CTD contains some components which later form the spliceosome 2. during elongation, the nascent RNA is marked by components of spliceosome or additional SR proteins to mark downstream and upstream positions of where a splicing event should take place. This reduces exon skipping and cryptic splicing errors during splicing, and increases accuracy, to allow splicing to be made at correct location 3. spliceosome in a snRNP containing RNA and proteins to capture, splice and release RNA through forming a lariat and 2 phosphoryl-transferase reactions 4. 3 locations on the intron must be defined for spliceosome to recognize and splice: the 5' splice site, the 3' splice site and the branch pount in the intron. (recognized by spliceosome by consensus sequences, and other additional steps) 5. after splicing and joining of the 2 exon sequences, the exonjunction complex (EJC) will be bound to the location of the splicing event to mark location of splicing event and to determine mRNA fate 6. IMPORTANT: splicing labelling and marking occurs during elongation, but the actual splicing reaction doesnt take place until the transcription is complete!

Answer 108

1. each splicing event removes 1 intron 2. 2 sequential phosphoryl-transfer reactions (transesterification) 3. joins 2 exons together well removing th eintron between as a lariat

Answer 109

* catalyzes pre-mRNA splicing * is a complex of 5 additional RNA molecules and several hundred proteins * hydrolyzes many ATPs while being flexible enough to deal with a huge avriety of introns * key steps in RNA splicing performed by RNA * these RNA recognize consensus sequences in the introns * spliceosome contains the RNA molecules (snRNA): U1, U2, U4, U5, U6, each is complexed with 7+ protein subunits to form snRNP (which is the core of the spliceosome) * recognizes 5' splice junction, 3' splice junction, branch point site, consensus sequence in introns (due to base pairing)

Answer 110

1. ensure accuracy (last step of transcription) acts as a checkpoint 2. allows flexibility to deal with huge variety of introns found in euakryotic cell

Answer 111

* differ hugely in size (some introns are 100 nucleotides some are 10,000) * although they all have consensus sequences but still may be difficult to determine (maybe due to cryptic splicing) * majority of a gene is intron (up to 95%+)

Answer 112

* suggest: exon-intron arrangement may facilitate emergence on new proteins over evolutionary time scales * presence of numerous introns in DNA allow genetic recombination to allow combining of exons from diff genes to emerge ne proteins * many cells have composed a common set of protein domains * alternative splicing increases coding potential of genomes

Answer 113

* 2 phosphoryl-transfer reactions (transesterification) 1. A specific adenine nucleotide in the intron sequence (the branch point) attacks the 5' splice site and cuts the sugar phosphate backbone of RNA at 5' splice point (first phosphoryl transfer reaction) 2. the cut 5' site become covalently linked to adenine nucleotide there by creating a loop in RNA 3. the released 3'OH end of exon sequence then reacts w the start of the nect exon sequence 3' splice site 4. that joins the 2 exons tgt and releases lariat (2nd phosphoryl transfer) 5. releases lariat is broken down to single nucleotides and recycles

Answer 114

1. 5' splice site: starting with GU---- 2. branch point: the Adenine 3. 3' splice site: ends with AG * Apart from starting GU and ending AG, the rets of the consensus sequence is variable * distance along RNA between 3 splicing sites are highly variable depending on length of intron * base pairing between RNA transcript and spliceosome help identify these regions

Answer 115

* few components of the spliceosome assemble on pre-mRNA as it emerges to prevent error in splicing as the RNA transcript emerges (marks the positions to ensure correct splicing but the actual splicing doesnt occur until transcript completed) * snRNA that is carried by phosphorylated CTD on RNA Pol II. As RNA transcript emerges, the 5' splice site will only be labelled with the 3' splice site that emerged form the polymerase by snRNP. Sites are labelled as they emergy, to prevent exon skipping * Exon definition; as RNA synthesis occurs, group of SR proteins assemble on 5' and 3' splice sites, then recruit U1 snRNA which marke the downstream and U2 to mark upstream of exons (prevents cryptic sites and exon skipping)

Answer 116

1. the U1 snRNP forms base pairs with 5'spice junction (due to marking of SR proteins). The BBP (branch point binding protein) and U2 auxilliary factor recognize and bind to branch point site 2. U2 snRNP recruited and replaces BBP and U2 auxiliary factor and forms base pairs with branch point consensus sequence 3. the U4/U6* U5 triple snRNP brings together the U1 and U2, and sebsequent rearrangements abd reactions break apart U4/U6, causing U6 to siplace U1 at 5' junction. This change in arrangement in spliceosome allows creation of active site which catalyzes the first phosphoryl transfer reaction 4. additional RNA-RNA rearrangements create the active site for 2nd phosphoryl transferase reaction, and completes the splice. This 3' splice site cleavage, and joins the 2 exon sequences, while putting a EJC protein to mark splice event

Answer 117

* ATP not required during the 2 transesterification reactions as the breaking of bonds during this releases high energy phosphate bonds * however, ATP hydrolysis required for assembly and reassembly of spliceosome; to break RNA-RNA interactions between U4/U6 to allow formation of new ones and new rearrangements

Answer 118

1. allow splicing signals on the pre-RNA to be examined and checked by snRNPs multiple times prior splicing. E.g U1 snRNP recognizes 5' splice thru base pairing and SR proteins, and as splicing proceeds U1 is replaced by U6 which also requires base pairing. This ensures base-pairing signals are correct and not mis-paired Increases overall accuracy 2. rearrangements create active sites for the 2 transesterifications. Only after splicing signals are checked, the 2 active sites are sequetially created. Orderly progression, with checks between each active site creation, along with specificity of active site ensures rare errors 3. Once splicing complete, snRNPS remain bound to lariat, disassembly of snRNP from RNA and each other require RNA-RNA rearrangements that also require ATP hydrolysis (return to original configuration to be used again)

Answer 119

* caalytic sites formed both by protein and RNA

Answer 120

* the rearrangements through base pairing and creation of active sites prevent errors * 2 types of errors: 1. exon skipping: skipped an exon 2. Cryptic splice site selection: Cryptic splicing signals are nucleotide sequences of RNA that closely resemble true splicing signals and are sometimes mistakenly used by the spliceosome. * there are 2 ways to prevent this 1. Coupling transcription with splicing: as transcription proceeds, the phosphorylated CTD of RNA Pol II carries several components of spliceosome (U1 and U2), which are transferred to RNA transcript as it emerged. The snRNPs are only bound to 5' splice site once the 3' splice site emerged. Coordination of labelling splice sites as they evolve prevent exon skipping 2. exon definition: exon sizes are much uniform than introns (avg 150 across eukaryotes) hence allows better labelling. SR proteins assemble on exon sequence and makr teh 5' 3' splice sites. SR proteins then recruit U1 (downstream mark) and U2 (upstream mark). SR proteins bind to specific RNA seqs on EXONS (splicing enhancers). These sites can be created without affecting amino acid sequence due to wobbling base (redundancy in amino acid code)

Answer 121

1. Nucleosomes tend to be positioned over exons, hence can determine the speed of RNA transcript emergence and hence allow proteins responsible for exon definitions to assemble on RNA transcript as it forms 2. bcuz splicing and transcription is coupled, hence the rate of transcription will effect rate and accuracy of splicing. If transcription occurs slowly, it prevents exon skipping, as the spliceosome will formbefore the next splice site emerged. Nucleosome in condensed chromatin will cause RNA pol II to pause or take longer to transcribe 3. mechanism unknown: chromatin structure/histone modification can attract components of spileosome and these components can then be transferred to nascent RNA

Answer 122

1. adapting to mutations: Mutations in splicing sequence often lead to new splicing patterns.Common outcomes include exon skipping or the use of cryptic splice sites.This flexibility suggests that RNA splicing's adaptability has been crucial in evolutionary processes. Mutations affecting splicing can lead to severe diseases, such as β-thalassemia, cystic fibrosis 2. alternative splicing: Alternative splicing allows for different proteins to be produced from the same gene, enhancing the genome's coding potential.

Answer 123

1. lariat mechanism involving snRNPs is used rather than a simpler nuclease cleavage 2. evidence for RNA world theory: Early cells likely used RNA as both genetic storage and as catalysts, predating the use of DNA and proteins for these roles. RNA-catalyzed splicing was crucial in these early cells, supporting the evolutionary precedence of RNA in cellular functions. 3. Certain RNA introns can splice themselves without proteins, evident in organisms like the ciliate Tetrahymena, some bacteriophage T4 genes, and in mitochondrial and chloroplast genes, hence eukaryotic splicing probably evolved from that

Answer 124

1. 5' capping occurs as soon as 25 nucleotides are made 2. splicing components are transferred from CTD to RNA transcript to mark splicing sites, as RNA nascent transcript emerges 3. as RNA polymerase reaches a gene, 3' poly A tail added

Answer 125

1. transcription termination signal encoded in genome 2. these signals are transcribed into RNA and are recognized by 2 enzymes: CstF (cleavage stimulation fcator) and CPSF (cleavage and polyadenylation specificicty factor); these enzymes travel on CTD RNA Pol II tail and transferred on to the 3'end and the signal is transcribes 3. CstF and CPSF bind to termination signal on nascent RNA and create the 3' end of mRNA by first cleaving it from RNA pol II. Then 3 additional processing steps occur * Poly-A Polymerase (PAP) adds one at a time 200 A. nucleotides to 3' produced by cleavage (added via norma 5'-3' linkage), These nucloeitde precursor in ATP and doesnt have a template, hence poly-A-tail is not encoded in genome * Then, poly-A binding proteins assemble on to it and determine final length of Poly-A -tail 5. after mRNA being cleaved, the RNA Pol II continues transcribing (as it is still bound to DNA), it produces hundred of nucleotides but as these nucleotides lack 5' cap, they are quickly degraded. Continued RNA degradation causes RNA pol to dissociate and terminate transcription

Answer 126

1. enable protection from exonuclease 2. form complex that facilitate ribosome assembly to initiate translation

Answer 127

1. a splicing mechanism that forms circRNAs 2. Black splicing mechanism: downstream of one exon is linked to the upstream of the same exon (or another upstream exon), this creates a loop containing the introns to be removed 3. this process is catalyzed by exon definition complexes and in backsplicing they cause a circular linkage forming circRNAs 4. circRNAs dont have 5' or 3' ends as they r a closed loop 5. circular property makes them resistant to degradation by exonuclear, and becomes more stable 6. circular RNAS have functional roles within cell due to stability: act as molecular sponges for microRNAs and affetc they gene expression, regulate protein synthesis

Answer 128

* an alternative splicing mechanism involving U11/U12 instead of U1/U2, which also forms a lariat and has the U4,U5,U6 triple complex * dedicated to minor class of introns

Answer 129

* bacterial and organelle pre-mRNA contail self splicing introns * this catalytic property is conseved by secondary intron structures, which are absent in nulcear introns

Answer 130

* a first quality check control system prior to translation 1. Initiation of mRNA Transport: * The process begins as the mRNA molecule moves from the nucleus to the cytosol. * As the 5ʹ end of the mRNA exits the nuclear pore, it encounters a ribosome that starts translating it. 2. EJCs are bound to the mRNA at each splice site. * As the ribosome translates the mRNA, it displaces these EJCs as it moves along. 3. Normal Termination of Translation: * In a properly spliced mRNA, the normal stop codon is located in the last exon. * By the time the ribosome reaches this stop codon, all EJCs should be displaced. * If no EJCs remain when the ribosome stalls at the stop codon, the mRNA is considered correct and continues to full translation in the cytosol. 4. Nonsense-Mediated Decay: If a stop codon is encountered prematurely while EJCs are still bound to the mRNA, it indicates a potential error in the mRNA. * This early stop (nonsense codon) triggers the rapid degradation of the mRNA. * This mechanism allows the cell to "inspect" each mRNA for errors during its first round of translation as it exits the nucleus.

Answer 131

1. properly processed mRNA lack certain proteins like snRNP, if snRNP present in mRNA after processing then aberrant splicing occured, and this signals nuclear exosome to degrade it 2. other RNA debris, improperly processed mRNAs is not exported into cytosol and remain in nucleus and eventually degraded 3. hnRNPs (heterogeneous nuclear ribonuclear proteins) are abundant on emerging pre-mRNA molecules whichhelp unwind hairpin RNA to make splicing signals readable 4. mRNA export through nuclear pore complexes (NPCs) through nuclear transporters, and macromolecules like mRNA protein complexes require active transport 5. export - ready mRNA hovers at pore entrance and goes through final checks like nonsense mediated mRNA decay before being transported through NPC 6. REF protein is a part of EJC protein of readily transported mRNAs 7. TF TAP/Mex bind to REF and actively transports mRNA through nuclear pore 8. TAP/Mex dissociates from mRNA once in cytoplasm

Answer 132

1. redundancy 2. universal (except mitochondria and candida albicans) 3. non-overlapping (with the exception of some polycistronic RNAs which have multiple ORFs)

Answer 133

* ribosomes (either 80s or 70s) are made out proteins and rRNA and are assembled in small (40s or 30s) and large subunits (60s or 50s) * these subunits are initially formed within nucleolus and then exported separatelt through nuclear pores into cytoplasm * ribosomal proteins after being synthesized in cytoplasm are transported by to nucleus to combine to rRNA in nucleolus. After assembly, these subunits (not fully activated) are transported back out to the cytoplasm * transort mech: go through NPC via nuclear export signals which are recognized by karyopherins/exportins) * in cytoplasm, the large and small subunits come to gether and bind to translate mRNA

Answer 134

* prokaryotes/organelles: 70s ribosomes: small subunit (30s=16s rRNA) large subunit (50s=23 rRNA + 5s rRNA) * eukaryotes: 80s ribosome: small subunit (40s= 18s rRNA) large subunit (60s = 28S rRNA, 5.8s rRNA, 5s rRNA)

Answer 135

* 80% of RNAs in cells are rRNAs (noncoding) * RNA synthesis non-coding: euks have multiple RNA Pols instead of bacteria which only has 1 * RNA Pol I is specialized for synthesizing rRNAs in euks - this does not have a CTD tail, which is why rRNA transcripts are not capped or polyadenylated * rRNAs are in high demand to comstruct enough ribosomes, hence cells have multiple copies of rRNA genes: humans have 200 copies per haploid genome across 5 chromosomes * eukaryotic rRNAs include 18s, 5.8s, 28s and 5s * extensive modification occru in pre-rRNA including 100 methylations and 100 isomeriations to pseudouridine which aid in rRNA folding/assembly/fucnction * snoRNAs (small nucleolar RNAs) guide modifications and cleavage of pre-rRNAs into mature forms * these guide snoRNAs brings modifying enzymes to pre-rRNA via base pairing * snoRNAs are encoded by introns in other genes (especially introns that code ribosomal proteins) and are synthesized by RNA POl II

Answer 136

1. euks have Untranslated Regions (UTRs) on 5' end (5' G-Cap) and 3' end (poly-A tail), also have UTRs for initiation and termination 2. translated region = ORFs, read in 3nts (each by 1 codon) 3. start codon = AUG coding for Met (determines reading frame); stop codon (UAA, UAG, UGA) 4. Initiation in eukaryotes: eukaryotic mRNA = monocistronic, ribosome bind to 5' Cap and scan to find AUG start codon 5.ORF important to locate because wrong ORF = nonsense and frameshift

Answer 137

1. proks dont have 5'cap but does have a 3' polyAtail (j shorter) 2. proks also have additional UTR regions other than PolyA and 5' cap 3. ORFs and start codon = AUG (met) and same stop codons as euks 4. In proks the shine dalgarno sequence (SD) is followed by the AUG start codon in mRNA 5. Proks have operons where a series of related proteins are transcribed into 1 mRNA uner control of 1 promoter sequence

Answer 138

1. silent mutations (1/3) 2. missense mutations (2/3) 3. nonsense mutations - move the ORF or stop codon (<5%) 4. Frameshift by indels

Answer 139

* tRNA has acceptor stem where the 3'CCA end can carru and activates charged amino acid * middle loop = location of anticodon and recognizes codons * 1 tRNA = recognize several codons due to wobble base theory * tRNA contains modified RNA bases like Inosine * the 3 loopps can interact to fold into L shape (3D) * allows 3D shape to have anticodon and acceptor stem attached to the activated aa on the other end

Answer 140

1. formation of tRNA - biogenesis 2. tRNA precursors are first transcribed from tRNA gene 3. RNAse P will process 5' end of pre-tRNA and RNAse D process 3' end by removing extra seqs 4. introns removed by endonucleases 5. acceptor stem 3' CCA added by tRNA nucleotidyl transferase (Post Translational modifications) 6. base modifications by snoRNAs - tRNA base modification by enzymes is required for stabilization of tRNA structure and recognition by aminoacyl-tRNA-synthetase

Answer 141

1. amino acid need to be first charged before it can become active to join tRNA - this is bcuz the formation of the peptide bond is not theromodynamically feasible 2. amino acid becomes charged by ATP, amino acid first adenylated and releases a pyrophosphate from original ATP 3. the adenylated aa will then be transferred and attached to 3'OH end of tRNA by amino-acid-tRNA-synthetase (aaRS) and releases an AMP to form amino acyl tRNA 4. this forms a tRNA which is bound to its corresponding amino acid = charged tRNA (or amino acyl tRNA)

Answer 142

1. after amino acid is charged (amino acyl) 2. the recognition of tRNA to the correct amino acid is carried out by the specific corresponding aminoacyl-tRNA-synthetase (aaRS) 3. 20 aa and 20 aaRS (each aaRS recognizes one aa) but 31-41 tRNAs bcuz some aaRS can recognize multiple tRNAs 4. the specific aaRS catalyzes covalent attachement between tRNA and aa through ATP hydrolysis 5. correct aa chosen due to highest affinity with specific synthetase over the remaining 19 synthetase 6. the same synthetase will also recognize the specific anticodon sequence of tRNA

Answer 143

1. form covalent bonds between correct amino acid and its specific tRNA 2. proofreading and editing function: modify incorrect covalent linakge: aaRS can carry out hydrolysis to break the covalent linkage within the aminoacyl-tRNA

Answer 144

* during translation: multiple ribosomes translate on one mRNA: multiple copies of proteins can be made at once * occur in both euks and proks

Answer 145

* only rRNA in ribosome has catalytic properties (RNA world hypothesis) * proteins in ribosome only act as a structural supporting role * hence argued that translation is dependent on mostly RNAs: with rRNAs catalyzing translation, mRNA and tRNA catalyze bond formation and polypeptide formation

Answer 146

* ribosome structure is different * prok transcription + translation all occur in cytoplasm, and occur simultaneously (synchronized process); allows immediate translation as no/little post transcriptional modification occurs (no physical barrieri) * euks: cannot synchronize (asynchronized process) as there is a membrane bound nucleus between cytoplasm

Answer 147

1. Initiation at strat codon AUG: N terminus to C terminus: mRNA needs to first be linked to ribosome 2. in both euks and proks the starting aa met must be a fMet (formylated methionine), but only the first one 3. in proks, all methionine is fMet, but for proks only the first methionine is fMet 4. tRNA caryying fMet attaches to 40s ribosome, then as a complex they scan mRNA to find first AUG and form codon-anticodon interaction, which causes large subunit to finally join and form initiation complex 5. fMet can then be removed later after 10aa is synthesized 6. existing Initiation factors help binding. of mRNA to ribosome by joining mRNA to small subunit before it binds to large subunit Elongation: 1. E (exit), P (polypeptide channel), A site 2. 3 processes of elongation: codon recognition, peptide bond formation, and translocation 3. After initiation the A and E sites are still vacant whilst the P site is occupied by the fMet-tRNA 4. Codon recognition: * The next aa-tRNA is brought depending on codon * It is brough to A site by the elongation factor: EF-Tu which is also a G protein and binds to aa-tRNA * protects the aminoacyl-tRNA from being hydrolysed * EF-Tu ensures that the anticodon-codon bonding is correct, as the EF-Tu complex only leaves the ribosome if bonding is correct. * During this process GTP is hydrolysed into GDP. * EF-Tu factor also helps align the aminoacyl-tRNA in P site to be ready for peptide bond formation in a process called ‘Accommodation’. The GDP can then be phosphorylated to form GTP again using EF-Ts 5. peptide formed by peptidyl transferase centre located next to P site: formation due to aa-tRNA attacks ester linkage on tRNA and fMet 6. Translocation: EF-G hydrolyses GTP to provide energy to move mRNA by 3 nucleotides from A-P to E Termination 1. stop codon: UAA, UGA and UAG are reached: Release factor RFs bind to ribosome instead of tRNA, which modify peptidyl transferase to promote water molecules attack on ester bond between tRNA an dpolypeptide chain 2. tRNA detaches and leaved ribosome through P channel 3. tRNA and mRNA remain bound to ribosome until Ribosome release factor (RRF) and EF-G bind which hydrolyses GTP and causes dissociation of tRNA and mRNA from ribosome

Answer 148

* 10 million ribosomes per cell * Each ribosome makes 1-2 proteins per minute * Proteins translated in the cytoplasm can be directed to the nucleus, mitochondria, chloroplasts or peroxisomes * 30% of all proteins are directed to the ER and secreted

Answer 149

1. pulse chase analysis was used to understand secretory pathway of proteins: tracks progression of a labelle substance through a cellular pathway over time. I.e 1974 Nobel Price for tracking pancreatic acinar cells 2. Steps: 1. first expose cell to radioactive aminoa cids for 3 minutes (the pulse). Proteins synthesized in these 3 minutes will incorporate the labelled aa. 2. then add a large amount of unlabelled aa (the chase) to dilute the labelled one to ensure after 3 minutes, no more labelled aa are incorporated into proteins 3. thren prepare cells forexamination with and EM at various time points fater the change Results: * 0 minutes: right after pulse: radiocative proteins are seen in ER as small dots * 10 mins: labeled protein moves to Golgi apparatus, where proteins are further processed * 20mins: label now in condensing vesicles (CV) where proteins are packaged * 40mins: label found in secretory vesicles ready for secretion conclusion: * demonstrates how newly synthesized proteins in pancreatic acinar cells are processed and trafficked through cell organelles to be secreted

Answer 150

* ER is very structurally diverse and specialised to accomodate for more diver/flexible functions: different cells possess diff types of ER, some have more rough ER some have more smooth ER * rER is involved in protein synthesis. Most proteins are imported to the ER co-translationally: meaning they are in the process of polypeptide synthesis as they are transported and moved onto the rER * Cotranslational: in co-translational import, ribosomes bind to the ER membrane during co-translational translocation. ribosome binds to membrane and one end of the protein is translocated into the ER while the rest of the polypeptide chain is being synthesized * in contrast: post-translational import: import of proteins into mitochondria/chloroplast doesnt require ribosome to bind to ER, and polypeptide enters organelle/ER for modification after it is released from ribosome

Answer 151

* transitional ER: is sER regions which bud off as transport vesicles carrying proteins and lipid to golgi * sER is more abundunt in cells that specialise in lipid metabolism such as those synthesizing steroid hormones from cholesterol, a larger sER in these cells allow enzymes for cholesterol modification and hormone production * sER in hepatocytes (liver cells) have a v larger sER. Here iot produces lipoprotein particles which transport lipids in bloodtsream; contains enzymes for synthesizing lipids and detoxifying lipid soluble drugs. * cytochrom P450 enzymes in sER help make water insoluble substances -> water soluble for secretion * Ca2+ also stored in sER and rER: i.e in muscle cells contain sER (sarcoplasmic reticulum) which is responsible for releasing and uptaking Ca2+ which is crucial for muscle contraction and relaxation

Answer 152

- ER is very structurally diverse and specialised to accomodate for more diver/flexible functions: different cells possess diff types of ER, some have more rough ER some have more smooth ER - sER and rER all have their specialized functions - calcium storage in ER: stores Ca2+ from cytosol; Ca2+ pumps transport Ca2+ into ER lumen which is aided by high conc of Ca2+ binding proteins in ER for storage - specific regions of ER may b specialized for C2+ storage which can contribute to rapid cellular response to signals - i.e in muscle cells contain sER (sarcoplasmic reticulum) which is responsible for releasing and uptaking Ca2+ which is crucial for muscle contraction and relaxation

Answer 153

* a type of inhibitor of protein synthesis in both eukaryotes and prokaryotes * Causes the premature release of nascent polypeptide chains by its addition to the growing chain end * can be used in experiments to test protein synthesis

Answer 154

* reconstitution experiment is by breaking up/homogenizing ER to form little ER vesicles called microsomes which serve as the functional representation of ER; maintaining capabilities like protein translocation, glycosylation, Ca2+ handling, and lipid synthesis * rER -> rough microsomes have ribosomes attached so are denser, and can be separated from sER by centrifuge * sER/Golgi -> smooth microsomes: often from liver/muscle cells In reconstitution experiments: rER microsomes are treated w radio labelled puromycin took label into lumen * to answer questions such as: Are rER associated ribosomes diff to cytosolic ribosomes, and is the signal for secretion in mRNA or protein? FIndings: * Cytosolic and ER-bound ribosomes are identical and inter-changeable in vitro * Discovery of signal peptides in secreted proteins - 1972 * Early Research and Signal Hypothesis: * Discovered in the 1970s, signal sequences were noted in secreted proteins undergoing translocation across the ER membrane. * In vitro experiments showed that these proteins were larger without microsomes but correctly sized with microsomes, indicating the presence and cleavage of a signal sequence.

Answer 155

* through reconstitution experiments using labelled microsomes * discovered signal peptides/sequences in secreted proteins * they were first identified in proteins which were imported into the rER * signal peptides in proteins target them to ER * signal sequence in the polypeptide chain in/on ER is cleaved by the signal peptidase on ER membrane

Answer 156

* Function of the ER in Protein Capturing: The ER captures specific proteins from the cytosol during their synthesis. include transmembrane proteins that embed in the ER membrane and water-soluble proteins that fully enter the ER lumen. * Destinations of ER-Imported Proteins: Transmembrane proteins may function within the ER, become part of the plasma membrane, or reside in other organelle membranes. Water-soluble proteins are either secreted outside the cell or stay within the lumen of the ER or other organelles. * ER Signal Sequence: All proteins targeted to the ER have an ER signal sequence. This sequence directs proteins to the ER membrane, initiating their translocation.

Answer 157

* original proposed protein translocation from cytosol to ER * when ER signal sequence is emerged in the polypeptide still attached to the ribosome, it directs ribosome to a translocator on ER membrane which forms a pore in the membrane to where in the ER the polypeptide should translocate to * a signal peptidase is closely associated with the translocator and clips off the signal sequence during translation -> mature protein * the mature protein is released into the lumen of the ER immediately after its synthesis is completed * The translocator is closed until the ribosome has bound, so that the permeability barrier of the er membrane is maintained at all times.

Answer 158

* by Milstein and Brownlee * notes that IgG light chains (a type of antibody) are initially translated into a form that has a slight heavier molecular weight than the mature final product * these initially heavier molecular weight of IgG was found to have an extension at the N terminus (which is the begining part of protein as it is synthesized) * after the protein is fully synthesized, these N terminal extensions are enzymatically removed * removeal of extension to become normal weight of IgG were only found in microsomes, and when not in microsome this extra sequence would not be removed * demonstrated that the N-terminal extensions were signal sequences to allow differentiation between proteins which should be secreted vs proteins which should remain in cell

Answer 159

* Signal Recognition Particles (SRP) bind signal peptide as it emerged from ribosome and pauses translation * SRP bound ribosome attaches to SRP receptor in ER membrane * Translation continues and translocation begins: SRP and SRP receptor displaved and recycled * translocation of SRP to a specific ER channel/protein translocater where the Signal peptide is cleaved off

Answer 160

1. ER signal sequence on a protein is guided to rER by SRP and SRP receptor 2. Components of SRP: in animal cells SRP complex includes 6 polypeptide chains and 1 small RNA molecue; across tree of life contains homologous SRP components which suggest evolutionary conservation of this targeting mechanism 3. signal sequence specificity: Er signal sequences are diverse but share common central region of non-polar amino acids; SRPs can bind various signal sequences thanks to a flexible hydrophobic pocket lined with methionines 4. SRP has rodlike shape and wraps around ribosome binding the emerging signal sequence which temporarily blocks translation; this allows ribosome SRP complex to bind to ER membranes and precent premature release of protein into cytosol 5. Importance of Co-translational import: pausing translation ensures proteins meant for secretion are not degraded in cytosol and ont fold prematurely before ER membrane 6. vs post-translational translocation: in mitochondria imports requires chaperones to keep protein unfolded 7. SRP and SRP receptor interaction: SEP exposes binding site for SRP receptor on ER membrane when a signal sequence is bound; SRP-receptor interaction docks ribosome to a protein translocator in the ER membrane; post binding SRP is released and protein continues to be synthesiszed directly into ER

Answer 161

1. 2 types of ribosome population: - membrane bound ribosomes (synthesize proteins entering ER); - free ribosomes (synthesize other ceullar protein) 2. polyribosomes and ER attachment: multiple ribosomes can attach to single mRNA forming polyribosome;If the mRNA encodes an ER-targeted protein, the entire polyribosome attaches to the ER membrane, directed by the signal sequences on the nascent polypeptides. Ribosomes detach and rejoin the cytosolic pool after completing protein synthesis, while the mRNA remains associated with the ER membrane through a changing set of ribosomes.

Answer 162

* SRP-54: for signal recognition: also has affinity for non tranlsating ribosomes and ribosome nascent chain complexes that lack a signal peptide; SRP 54 binds to the 20ss hydrophobic sequence on signal peptide * SRP 9 and SRP 14: translational arrest/pause * SRP-68 and SRP-72: ER docking * GTP hydrolysys on SRP 54 and SRP receptor are required to complete the docking and release the SRP

Answer 163

* the Alu domain of the SRO stops protein biosynthesis by competition with the elongation factor binding on the ribosome

Answer 164

* recognition of signal peptides by SRP is enhanced by competition iwith the NAC which improves accuracy of protein sorting * Role of NAC: NAC binds to nascent polypeptides as they emergy during translation process; it has higher affinity and binds more preferentially to other hydroPHILLIC sequences over hydroPHOBIC signal peptides * henceall emerging non-signal peptide or hydrophillic sequences will be bound to NAC, leaving the SRP to bind to signal peptide * this binding is part of quality control to ensure proteins are correctly sorted * mutations in NAC lead to errors in protein sorting: i.e mito proteins can be secreted out of cell instead to to mito

Answer 165

* debated whether polypeptides crossed ER via lipid bilayer or protein channel * discovered translocator ER channel confirmed protein channel involved * Translocator structure: idenitified by Sec61 complex, forming a water filled channel in ER membrane for polypeptide traversal * Sec 61 is highly conserved across bacteria and euks * Sec 61 complex structure: composed of 3 subunits with alpha helices forming a central channel; * channel gating: channel within sec61 complex is gated by a short alpha helix; helix keeps channel closed when no translocation occurs and opens when polypeptide translocation occurs * Ion permeability: when not in use channel remains closed to prevent ions like ca2+ from leaking out of ER; hydrophobic aa side chains form a seal during translocation to prevent ion leakd * opening mechanism: sec61 pore can open along a seam on the side, allowing lateral access to membrane's hydrophobic core; side openeing allow integration of transmembrane proteins into the bilayer and the release of signal peptides * Sec61 complex in euk cells: 4 sec 61 complexes form a larger translocator assembly to help grow and modify the polypeptide; includes oligosaccaride transferase and signal peptidase; this complex assembly = translocon

Answer 166

* Translocator structure: idenitified by Sec61 complex, forming a water filled channel in ER membrane for polypeptide traversal * Sec 61 complex structure: composed of 3 subunits (alpha beta gamma) with alpha helices forming a central channel; * channel gating: channel within sec61 complex is gated by a short alpha helix; helix keeps channel closed when no translocation occurs and opens when polypeptide translocation occurs * Sec61 complex in euk cells: 4 sec 61 complexes form a larger translocator assembly to help grow and modify the polypeptide; includes oligosaccaride transferase and signal peptidase; this complex assembly = translocon

Answer 167

* negatively charged residues at opening of translocator meets positive charged residues on signal peptide causes opening * opening mechanism: sec61 pore can open along a seam on the side, allowing lateral access to membrane's hydrophobic core; side openeing allow integration of transmembrane proteins into the bilayer and the release of signal peptides * Ion permeability: when not in use channel remains closed to prevent ions like ca2+ from leaking out of ER; hydrophobic aa side chains form a seal during translocation to prevent ion leakd

Answer 168

1. it explains why and how ribosomes are bound to ER

Answer 169

1. fully synthesized proteins can also be translocated across ER (post translationally) 2. common in yeast ER and bacterial plasma membranes 3. ER translocator requires accessory proteins to help feed the complete polypeptide into channel to facilitate translocation 4. In bacteria, the SecA ATPase acts as a motor, using ATP hydrolysis to induce conformational changes and push the polypeptide through the translocator. 5. Eukaryotic cells have different accessory proteins associated with the Sec61 complex. These proteins help to recruit an hsp70-like chaperone protein (BiP) on the ER lumen side to pull the polypeptide chain into the ER. 6. Role of BiP: BiP binding and release, powered by ATP, drive the unidirectional translocation of the polypeptide into the ER. 7. Chaperons binding in cytoplasm: Proteins destined for post-translational import into the ER are kept unfolded in the cytosol by chaperones until they're transported, similar to mitochondrial and chloroplast proteins

Answer 170

1. In bacteria, the SecA ATPase acts as a motor, using ATP hydrolysis to induce conformational changes and push the polypeptide through the translocator. 2. Eukaryotic cells have different accessory proteins associated with the Sec61 complex. These proteins help to recruit an hsp70-like chaperone protein (BiP) on the ER lumen side to pull the polypeptide chain into the ER.

Answer 171

* used when post-translational translocation of polypeptide across ER membrane in eukaryotes like yeast * functions as accessory proteins associated with sec61 complex to pull polypeptide chain into ER * BiP located on ER lumen side and pull polypeptide chain into ER: does this by binding and release, powered by ATP, drive the unidirectional translocation of the polypeptide into the ER.

Answer 172

1. signal sequence initiates opening of Sec61 translocator pore by binding to an specific site on the pore 2. this signal sequence recognized twice: 1. by SRP in cytosol; 2. within the pore as a start transfer signal 3. dual recognition of the signal sequence helps ensure correct entry of proteins into the ER lumen 4. signal sequence contacts both sec61 complex and hydrophobic core of lipid bilayer during translocation 5. after cleavage by ER signal peptidase, the signal seq is released into membrane and degraded 6. the translocator opens laterally to release hydrophobic portions of protein into lipid bilayer 7. in single pass transmembrane proteins: a stop transfer sequence anchors protein in membrane. This created a membrane spanning alpha helic w N-Terminus in ER lumena dn C-terminus in cytosol 8. there can also be the case where the SP start signal is internal rather than on the N-terminal end of protein where the orientation can be either direction

Answer 173

1. in this case the internal signal sequence acts as start transfer signal to initiate translocation until it reaches stop transfer signal which causes polypeptide to be released into bilayer until it encounters stop transfer sequence 2. in multipass transmembrane proteins there is more than one stop and start transfer signal by threading it across the membrane 3. proteins inserted into ER from cytosolic side to ensure consistent orientation, resulting in asymmetrical ER membrane w diff protein domains exposed on either side

Answer 174

1. The Sec61 channel in eukaryotes and the SecY channel in prokaryotes are universally conserved structures that facilitate protein translocation and membrane integration. 2. These channels are formed by a single heterotrimeric complex, creating a narrow pore that allows polypeptides to pass through in an extended conformation. 3. Secretory proteins are translocated when a polypeptide loop inserts into the channel, displacing the channel's plug domain. 4. Co-translational and post-translational translocations occur through different mechanisms, with ribosomal feeding in the former and chaperone BiP ratcheting or SecA ATPase pushing in the latter, while SecD/F proteins aid in prokaryotes. 5. The channel maintains the membrane barrier, sealing off ions and small molecules with pore ring amino acids and a plug domain in its resting state, and forming a seal around the polypeptide during active translocation. 6. Transmembrane (TM) segments of membrane proteins are integrated into the lipid bilayer by moving from the channel interior to the lipid phase through a lateral gate.

Answer 175

* How do proteins exit the ER? * Bulk Flow or Signal Mediated? * Experimental evidence for Bulk Flow * Signals for retention * Calnexin and the glycocode * Degradation of misfolded proteins through the ERAD pathway * Evidence for receptor mediated exit * Receptors, coats and packaging

Answer 176

* endocytic (endocytosis) * The secretory pathway leads outward from the endoplasmic reticulum (ER) toward the Golgi apparatus and cell surface, with a side route leading to lysosomes, * while the endocytic path- way leads inward from the plasma membrane. * Retrieval pathway: is to balance out the flow of membrane in the opposite direction, and can act as a quality control to ensure that incorrect proteins which left a compartment through secretory pathways can be brought back

Answer 177

* if 2 proteins in one compartment have different fates, then there must be signalled events involved 1. Bulk Flow Hypothesis: This idea posits that proteins are exported from the ER without any specific signals—export is the default pathway. Proteins that need to stay in the ER require a retention signal. The prediction here is that proteins leave the ER in proportion to their abundance within the ER, and all proteins exit at the same rate (bulk-flow rate). 2. Signal Mediated Export Hypothesis: Contrary to the Bulk Flow Hypothesis, this suggests that a specific signal is essential for proteins to be exported from the ER. The default state for proteins is to remain in the ER unless they have an export signal. The efficiency of the signal determines the proportion of protein exported from the ER.

Answer 178

* in 1980 * Bacterial proteins can be targeted to ER of animals, plants, and fungi with addition of signal peptide * This signal peptide is a short stretch of amino acids that directs the protein to the ER membrane where it can be imported into the ER lumen. * Once inside ER, proteins typically follow secretory pathway and are exported out of ER to go to various destinations like golgi, lysosomes or cell surface * Evidence for bulk flow: prokaryotic proteins (like bacteria lack ER) dont have signal peptide but are exported by default - hence aligning w bulk flow hypothesis

Answer 179

* not all proteins are secreted at the same rate, hence indicates that theres some specificity in how proteins are exported from ER * Study by Lodish 1983: showed rate determining step of secretion is the export from ER and not the abundance of proteins within ER * suggested that active export signals may be present in proteins that are secreted faster whilst slower secretion rates may indicate a default bulk flow pathway * when bacterial proteinss are introduces into euk cells, they are exported at same rate of slowest native proteins, which suggest that lack of specialised export signal results in deafult bulk flow rate

Answer 180

* couldn't find a candidate for these signals * early candidate proposed to be N-glycans: N-linked glycans are a type of carb modification to N-atom of aa to act as potential export signal * proposed because these are common in proteins which were secreted from ER, and the N-glycan modification are made as proteins enter ER * Use of Tunicamycin to test this idea. Tunicamycin inhibits addition of N-glycans to proteins in ER. so if N-glycan was an export signal, then when we add tunicamycin the abundance of secreted proteins would decrease. * Result of tunicamycin: it only blocked the export of SOME glycoproteins (which hald supported this idea taht N-glycans were export signals), however NOT ALL glycoproteins export were blocked, some non-glycosylated proteins were still secreted -> showed inconsistency and incomplete inhibition, hence not as simple as N-glycan = export signal Reason for rejection of N-glycan: * Not all secreted proteins are glycosylated, which means some proteins are secreted without these sugar modifications. Some proteins that remain in the ER are glycosylated, suggesting that glycosylation does not necessarily serve as an exclusive export signal. Tunicamycin does not block the secretion of all glycoproteins, indicating that glycosylation is not the sole determinant for the export process. Another candidate: peptide sequences * specific amino acid sequences in proteins might serve as export signals. Rejection of Peptide sequences: * No obvious uniquely conserved sequence motifs (patterns) were identified that could serve as export signals for all proteins. * Mutational studies of secreted proteins failed to identify any common necessary sequence. * When scientists made mutations in these proteins, they couldn't find a single mutation or a small group of mutations that consistently stopped secretion. * Many mutations found inhibited secretion, but they were too numerous and scattered across different proteins, making it difficult to define any clear export signal(s).

Answer 181

* it was mainly about the rate of proteins, and that the rate of protein export was dependent on specific signals * all agreed that Export = default

Answer 182

1. Premise = N-glycan are NOT export signals, neither are any other carbohydrate moeity nor glycosylation motif (but this was later turned over, as glycan was not a direct signal, but did subsequently contribute to signal) 2. Experiment: used a synthesized radiolabelled small tripeptide and tracked the time peptide time it took to exit from the cell 3. as this tripeptide does not have any export signals, and is not large enough to be actively transported or retained, it must exit via bulkt transport, hence should reflect the bulk flow rate 4. question of is the rate of bulk flow equal to the fastest or slowest export of native proteins 5. answer = bulk flow is the fastest rate of export, indicating that default bulk flow secretes proteins quicker, however is less efficient (overall slower as it did it in v small quantities) 6. HOWEVER: quicker secretion,, but secretion quantity is very low 7. hence maybe proteins w export signals are secreted slower but in more abundance 8. ALSO IMPLIED: FOR RESIDANCY AND RETAINED PROTEINS IN ER there needs to be a signal

Answer 183

* Synthetic glycosylated tripeptides exported at maximum rate * Bulk flow can account for maximum rates of export * No need for export signals * Implies a signal for ER residency: * Late 80s Hugh Pelham and colleagues discover K/HDEL retrieval signal and receptor system for ER resident proteins

Answer 184

* Protein folding ‘Quality Control’ (QC) * export rates are determined by folding rates * only correctly folded/assembled proteins are exported * Mutant proteins retained e.g. CFTR DF508, et al. * Explains tunicamycin data: absence of glycan increases probability of misfolding * Explains distributed nature of export mutations – their effect is on folding

Answer 185

* quality control of proteins and protein folding is due to ER chaperons * ER chaperons assist in correct protein folding * ER chaperons involved in protein folding in ER = Calnexin (CNX), and Calreticulin (CRT) is its soluble form in the lumen * CNX and CRT specifically recognize and associates w proteins w processed high mannose glycans (a type of oligosaccaride during glycosylation) * Glycan processing: glycan is added and modified to the proteins for recognition in ER lumen. glycoproteins are synthesized in ER w glycan precursor that is then transferred to protein. Glycan is then modified by glucosidase I, II (trim specific sugars from glycan), an UGT enzyme which adds a glucose molecule back to glycan * function of calnexin (CNX) interacts w glycoprotein after trimming by ER mannosidase but before UGT/ It assists in folding by binding to a glycoprotein that has a single glucose left on their glycan after initial triming. The single glucose acts as a signal for CNX to bind * after CNX binds, UGT can potentially add a glucose back if protein was misfolded, indicating a signal for protein to be retained in ER for additional attempts of refoldiong (this is a QC method to ensure only correctly folded proteins are exported) * CRT and CNX are highly specific lectins - Lectins = protein which binds to specific carb structures * CNX and CRT binds specifically to Glc1 glycan (glycan w 1 glucose left after trimming) * UGT adds glucose back to misfolded proteins, which cause them to reassociate w CNX after another round of folding, and retain them in ER

Answer 186

1. glycoproteins are synthesized in ER with a glycan precursor that is transferred to nascent protein 2. glycan is then modified by series of reaction by enzymes such as glucosidase i, II and ER mannosidase (which trim specific sugars from the glycan) 3. THEN enzyme UGT adds a glucose molecule back to the glycan 4. CNX calnexin binds to glycoprotein after trimming and before UGT 5. it binds to glycoprotein which only have a single glucose (glc1 glycan) as a signal for CNX to bind 6. IGT can then add a glucose back tothe protein if protein was misfolded as a signal to retain protein in ER 7. this is a quality control check, to ensure no misfolded proteins leave ER

Answer 187

1. UGT can add a glucose back to the protein after ER chaperon CNX associates with it 2. this addition of 1 glucose signals that this protein is misfolded and needs to be retained in ER for further re-folding and correction

Answer 188

1. Oligosaccaride transferase: attaches a high mannose tree to asparagine residues on nascent protein being synthesized in ER (N-linked glycosylation) 2. Initial Trimming: after glycosylation: Glucosidase I, II sequentially remove the 2 terminal glucose residues from the newly added glycan 3. CNX then bind to the trimmed glucose end. These chaperons help protein to fold properly as they recognize the single glucose left on glycan after trimming 4. Further trimming: as protein starts to fold, glucosidase II removed the last glucose (glu1) on glycan which prompts release of glycoprotein from CNX. Correctly folded protein will have no more glucose, and are ready for secretory pathway 5. the rate of protein dissocaited from CNX is critical as it foverns the rate of protein exported to golgi, the faster they fold and detach from chaperons, the faster they can move on

Answer 189

* oligosaccaride transferase: step 1 in protein folding in ER: adds high mannose tree to asparagine residues on nascent protein as it is synthesized in ER * Glucosidase I, II are involved in INITIAL trimming: to sequentially remove the 2 terminal glucose from newly added glycan * glucosidase II also involved in further trimming: it removes the last glu1 and prompts release of glycoprotein from CNX for secretory pathway * UGT: enzyme that selectively re-glucosylates unfolded proteins in ER after trimming- adds a glucose back onto incorrectly folded proteins * ER Mannosidase I: ER mannosidase I is an enzyme that removes a mannose sugar, set time limit for protein before export

Answer 190

when Oligosaccharide transferase attaches a high mannose tree to asparagine residues on a nascent protein as it is being synthesized. This process is known as N-linked glycosylation because the sugar tree is linked to the nitrogen of the asparagine side chain.

Answer 191

1. Role of UGT: enzyme that selectively re-glucosylates unfolded proteins in ER - adds a glucose back onto incorrectly folded proteins 2. discrimination of unfolded proteins: can distinguish incorrect folding by * exposure to hydrophobic sequences: parts of protein that should be inside a correctly folded protein is exposed when misfolded * accessibility to glycan core: specific structure of glycan which is exposed only when protein in incorrectly folded 3. Retention in ER: if UGT adds a glucose back to a misfolded protein, it allows protein to bind again to CNX, which can help protein fold properly again. 4. This retains incorrect folded proteins in ER acting as a QC

Answer 192

* process of adding and removing glucose residues * allow glycoprotein to have multiple opportunities to refold to correct shape * It’s a cyclic process where a protein can be continuously assessed for proper folding, only when UGT checks it as corre t and unglycosylated, then it will exit ER

Answer 193

1. If hydrophobic sequences which are meant to be inside and not exposed is exposed in proteins 2. if acceessible to glycan core

Answer 194

1. ER chaperones like CNX 2. cytosolic HSP70 3. Protein disulfide isomerase 4. ER mannosidase I

Answer 195

1. further processing of their attached glycans' mannose required before exiting 2. mannose trimming: trimming mannose core from Man9 to Man8, prepares for export to golgi apparatus where further glycan processing occurs 3. ER Mannosidase I is an enzyme that removes a mannose sugar and acts slowly. ensures that only proteins that have had adequate time to fold properly (and have been subject to quality control processes) are exported. 4. Reduction of chaperon interaction: The conversion of a glycan from Man9 to Man8 reduces the protein’s interaction with the chaperones CRT/CNX and the enzyme UGT. As they are trimmed to Man8, they no longer have the single glucose that is recognized by CNX/CRT for refolding, which means they will not be re-glucosylated by UGT and are thus released from the chaperone-mediated folding cycle. 5. Export from ER

Answer 196

1. a quality control mechanism that disposes proteins which fail to fold correctly 2. ER mannosidase I also works on unfolded proteins by converting their glycans from Man9 to Man8, and reduces its intercations w chaperons and UGT (juts like for correctly folded protein) 3. if protein fails to fold after ER quality control cycle, then the unfolded Man8 protein become substrate for ER mannosidase II, and further trims Man8 to Man7 4. EDEM (ER degradation enhancing mannosidase like Proteins) are involved to help recognize misfolded/unfolded proteins and target them for degradation 5. Proteins w Man7 are recognized by ERAD lectins wich tag these proteins for removal 6. Misfolded proteins are retranslocated back through ER membrane through channel complex Sec61 7. once identified misfolded and tagged ERAD proteins, the proteins are ubiquitinated 8. ubiquitinated proteins signals destruction by proteasome 9. ubiquitinated protein is then targeted to the 26S proteasome, which recognizes the ubiquitin tags and proceeds to degrade the protein into its amino acid components.

Answer 197

goes to ERAD pathway and degraded eventually

Answer 198

1. bulk flow theory assumes that proteins exported from ER to golgi in non-selective manner, without sorting concentration mechanism, hence assumes the conc and distribution of these cargo proteins should stay the same 2. however Golgi has a higher conc of cargo that ER 3. the retrival pathway accounts for a part of this difference in concentration, (through using COPI recycling mechanism from golgi to ER) 4. More importantly the discovery of Exit sites on ER through COP II vesicles show that cargo proteins are a more regulated process, and not just bulk flow

Answer 199

1. COPI recycling from golgi 2. selectively retruns excaped ER residents and membrane 3. remaining cargo after retrieval and recycling is hence more concentrated at the golgi

Answer 200

1. his experiment showed that there were specific 'exit sites' on the ER which had high concentrations of cargo proteins, before its exported 2. the conbcentration at exit sites on ER showed that there existed a selective mechanism instead of a non selective bulk flow 3. in mammals these specialized regions of exit site on ER, are linked to the packaging of cargo proteins into COPII vesicles, hence showing specificity and selectiveness

Answer 201

1. balch argues that it is a ctive process (but others argue it is passive later): the maintenance of proteins in the ER is an active process, involving the retrieval and retention pathways 2. reticuplasmins = residence proteins in ER, they remain in ER and are not transported to golgi 3. these ER residence protein interact through forming calcium cross bridges through their calcium binding domains 4. ER residences such as CNX use this to interact and to maintain in Er rather than diffusing into COPII vesicles 5. COP II vesicles transport protein from ER to golgi 6. K/HDEL is a signal seq that tags certain proteins for retention in ER/ If these protein accidentally get exported to golgi, the COP I vesicles recognize this signal and retrieve them back to ER 7. CRT without K/HDEL secreted much rapidly, hence shows retention isgnlas actively keep proteins within cell

Answer 202

1. exclusion model suggest cargo prorein are concentrated and exported out of ER via a passive way of being SQUEEZED OUT from areas densely packed with resident ER proteins 2. The accumulation of cargo proteins at the ER exit sites happens passively because these are the areas less populated by reticuloplasmins. It is a consequence of cargo proteins being excluded from areas where reticuloplasmins are prevalent. 3. The concentration of cargo proteins is further enhanced by the retrieval of escaped ER resident proteins by COPI-coated vesicles. This process can inadvertently increase the concentration of cargo proteins by removing mislocalized resident proteins.

Answer 203

* Passive: means proteins move from ER to golgi without being selective * active concentration mode: requires signal on cargo proteins to designate them for transport, receptors need to identify transport signals on cargo proteins, active recruitment into COP II vesicles * in conclusion: they found in 1994, that COP II vesicles has mechanisms whichselectively packaged cargo proteins (hence active)

Answer 204

1. the COPII Sec24p receptor system 2. COPII coat assembly: a complex of proteins assemble on ER membrane to form a vesicle to transport cargo proteins 3. Sar1p activation: Sar1p is GTPase which initiates COP II vesicle formation 4. Sec12p an ER membrane protein activates Sar1p by GDP -> GTP and activates Sar1p to embed into Er membrane. 5. Sar1p exposes its hydrophobic domain to anchor it inplace 6. Recruitment of Sec24p/Sec23p: Sar1p-GTP then recruits Sec24p and Sec23p. Sec23p serves as a GAP (GTPase-activating protein) for Sar1p, eventually triggering GTP hydrolysis to disassemble the coat after vesicle formation. Sec24p is the component that directly interacts with cargo proteins. It has cargo-binding sites that recognize specific signals on cargo proteins, ensuring that only proteins meant to be exported from the ER are packaged into the vesicle. 7. Membrane-bound cargo proteins bind to Sec24p via their cytoplasmic tails that contain export signals. These signals are specific sequences or structures that are recognized by Sec24p. 8. The Sec13p-Sec31p complex is another component of the COPII coat. It binds to the Sec24p/Sec23p complex to complete the formation of the COPII vesicle. 9. nteraction of these proteins leads to the concentration of cargo within a defined region of the ER membrane and the subsequent budding off of a COPII vesicle that encapsulates the cargo for delivery to the Golgi.

Answer 205

1. COPII vesicles selectively transport cargo proteins from the ER to the Golgi apparatus. 2. The cytoplasmic tails of cargo proteins interact directly with Sec24p, a component of the COPII vesicle coat. This interaction is crucial for the recognition and selection of cargo proteins for transport. 3. By analyzing protein sequences, overrepresented motifs were identified as potential export signals. 4. Multiple export motifs were found, such as di-acidic and di-hydrophobic motifs, which are signals for export. These motifs are used by Sec24p to differentiate cargo proteins from resident ER proteins. 5. Deletion and transplant experiments were conducted: Deleting these motifs from cargo proteins led to a reduction in export rates. Transplanting these motifs to other proteins increased their export rates. 6. Specific binding sites for these export signals have been identified on the Sec24p protein.

Answer 206

1. soluble cargo receptors are ER membrane proteins which recognize soluble cargo proteins for export 2. These cargo recptors on ER membrane are large and can protrude the ER lumen which allows them to bind to cargo proteins 3. cytolsolic tails on these receptors contain export signals (di-acidic, di-hydrophobic motifs) and are recognized by COPII components and they package the cargo protein along w cargo receptor into COPII vesicle 4. These receptor have KKxx motif which is a signal for retrieval, whicha llows these receptors to be returned to ER if they are captured by COP II vesicle into golgi 5 there is no common motif signals for all cargo proteins bcuz many diff receptors involved, each recognizing diff motifs in diff cargos

Answer 207

1. It contains an export signal that facilitates the packaging of the receptor-cargo complex into COPII vesicles. 2. It contains a retrieval signal (KKxx motif) that ensures the receptor is returned to the ER from the Golgi if necessary.

Answer 208

1. study of loss of function mutants for cargo receptors in yeast 2. these mutants have specific secretion defects, and only export certain proteins 3. debate about whether cargo receptors simply bind cargo to promote proper folding (quality control, QC) or whether they are directly involved in the rapid export of the cargo. 4. Mutation for ppa and Erv1p leads to defects in Er export process 5. A specific example given is the erv29 mutant, which displays a selective loss of export for the pre-pro-alpha factor (ppα), suggesting that the Erv29p receptor is specific to ppα. 6. It has been shown that Erv1p contains both export signals for leaving the ER and retrieval motifs for returning to the ER in its cytoplasmic tail, allowing it to cycle between the ER and the Golgi as needed. 7. Erv1p has a luminal domain that binds to ppα, indicating a direct interaction between the cargo receptor and its cargo.

Answer 209

* misfolded proteins detected by UGT and ERAD lectins, leading to retential and destruction * folded proteins (export receptors will recognize) and lead to faster packaging and secretion

Answer 210

1. Phosphorylation: Addition of a phosphate group to serine, threonine, or tyrosine residues. It often controls the activity of proteins and signaling pathways. 2. Ubiquitination: Attachment of ubiquitin proteins to a lysine residue on a target protein, which can signal for protein degradation via the proteasome or alter protein activity or localization. 3. Acetylation: Addition of an acetyl group, typically at lysine residues, affecting gene expression by modifying chromatin structure and protein stability. 4. Methylation: Addition of methyl groups to arginine or lysine residues on histones, influencing gene expression and protein interactions. 5. Glycosylation: Attachment of sugar moieties to asparagine or serine/threonine residues, crucial for protein folding, stability, and cell-cell interactions. 6. Sumoylation: Similar to ubiquitination, it involves adding SUMO proteins (small ubiquitin-like modifier) to lysine residues, affecting nuclear-cytosolic transport, transcriptional regulation, and protein stability. 7. Sulfation: Addition of sulfate groups to tyrosine residues, which is important for protein-protein interactions, especially in growth factors and receptors. 8. Lipidation: Addition of lipid groups, such as farnesyl and geranylgeranyl, to cysteine residues, which helps anchor proteins to membranes. 9. Nitrosylation: Addition of a nitric oxide group to cysteine residues, which can regulate enzyme activity and protein function.

Answer 211

1. range of building blocks 2. nature of glycosidic bond 3. branching of polysaccharide chains

Answer 212

* Glycosyl transferases (GTs) produce glycans * Catalyse glycosidic bond formation with specific monosaccharides * Integral membrane proteins in ER/Golgi * Make glycans on lumen or cytoplasmic side of the membrane

Answer 213

1. Glycosyltransferases are enzymes that transfer sugar monosaccarides from activated donor molecules to specific acceptor molecules (proteins or other sugars). 2. They are highly specific to both the sugar donor and the acceptor molecule. 3. Responsible for the formation of glycosidic bonds and the diversity of glycan structures.

Answer 214

1. Monosaccharides donors are activated by linking them to a nucleotide triphosphate, forming nucleotide sugars (e.g., glucose to UDP glucose then to dolichol phosphate glucose). 2. variety of nucleotide sugar donors are made in cytoplasm 3. in the example of glucose to UDT glucose to Dolichol phosphate glucose, the second step is done by linking via glycosidic bond to anomeric carbon

Answer 215

* They are transported from cytoplasm to lumen of ER/golgi in 2 ways 1. Through antiporter mechanism: UDP-Glucose is transported to lumen. GT then reacts w it to release UMP and the glucose is added to glycan, UMP is then transported back from lumen to cytoplasm 2. Through Flippase Mechanism: once initial sugars are attached to dolichol P on the cytoplasmic side, a flippase enzyme flips the dolichol and its attached sugars onto ER membrane and lumen side

Answer 216

* mentioned also in quality control, it is important in protein folding, and subsequentoy acts as an export signal * found common in eukaryotes and archaea, rare in prokaryotes * Bond ype: attachment of oligosaccaride to the nitrogen atom of an Asn side chain on a protein/polypeptide * This bond is typically formed in the endoplasmic reticulum (ER).

Answer 217

* On Secreted Proteins: N-glycosylation occurs on asparagine residues of proteins destined for secretion. * On Transmembrane Proteins: It also affects asparagine residues within the extracellular domains of transmembrane proteins.

Answer 218

* sequon = specific amino acid sequence where N-glycosylation occurs

Answer 219

* Oligosaccharyltransferase (OST): This is an 8-mer membrane protein complex in the rough ER that recognizes the sequon on nascent proteins as they are being synthesized. * OST cotranslationally transfers a preassembled glycan precursor to the asparagine residue within the sequon via an N-glycosidic bond.

Answer 220

1. initial trimming and transfer of glycans are all modification on a N-glycan 2. UGT recognizes unfolded protein by reglycosylating them on the N glycan

Answer 221

* further triming through glycosyl hydrolases trim N glycans

Answer 222

In Human Plasma: N-glycan structures found in human plasma are extremely diverse, creating a complex pattern of glycosylation that can vary due to genetic, environmental, and physiological factors. Between Species: N-glycan structures vary significantly between species, which is particularly noticeable when comparing humans to invertebrates, as referenced in parasitology studies. In Different Tissues and Cells: Tissue and cell-specific glycosylation patterns exist, affecting the function and interaction of glycoproteins.

Answer 223

1. Heterologous Expression: The heterogeneity of N-glycans is a critical consideration when producing vaccines in different host systems, such as yeast or insect cells, which may glycosylate proteins differently than human cells. 2. SARS-CoV-2 Spike Protein: The spike protein of SARS-CoV-2 has 22 sequons for N-glycosylation. When this protein is expressed in human HEK cells, it is predominantly N-glycosylated but exhibits a high degree of heterogeneity

Answer 224

1. O-glycosidic bond on Ser and Thr (bound to oxygen on Ser or Thr and some others) 2. prokaryotes also use O glysosylase transferase (OGTs) 3. they are extracellular and cytoplasmic 4. highly diverse glycans

Answer 225

1. Pro-hydroxylation by prolyl hydroxylases (requires Vit. C), e.g. primary component in collagen 2. Lys-hydroxylation by lysyl hydroxylases (requires Vit. C), e.g. another component in collagen 3. Tyr-hydroxylation e.g. glycogenin (primer for glycogen)

Answer 226

1. GalNac initiated in Golgi by ~20 types of GalNac transferases, each recognising S/T in motifs. Further GTs generate 8 core glycans, which are further modified. * Example: O-glycans on hinge region of IgM1 and IgG3 protects against proteases 2. Galactose; Example: different GT expression causes O-glycan polymorphism on glycophorin, recognised by IgM during blood transfusion 3. Mannose: Initiated in ER, completed in Golgi * Example: a-dystroglycan in muscle – links exoskeleton to cytoskeleton 4. Xylose initiated in ER, completed in Golgi; * Example: proteoglycans in animal extracellular matrix 5. Fucose initiated in ER, completed in Golgi. * Example: cell-cell recognition by Notch1 receptor during neuron development in fruitfly 6. Arabinose (only in plants) On hydroxyproline in SPPP repeat * Example: extensins – cell wall expansion 7. GlcNac: Single, unmodified unit – reversible, cytonuclear Added by GlcNac transferases (OGT) on Ser/Thr of undefined motifs. Removal by O-GlcNacase OGA; crosstalk with phosphorylation

Answer 227

1. glypiation 2. prenylation 3. myristoylation 4. palmitoylation

Answer 228

* Transport is carried out by glycosyl transferases, they catalyse protein glycosylation in 2 ways: 1. Build glycans directly onto protein, and extends the glycan 2. other way is to build a oligosaccaride precursor on a dolichol-phosphate, then the precursor is added to a protein by oligosaccaride transferase (OST)

Answer 229

* glypiation is GPI anchor attachment: it attaches proteins to cell membrane via a GPI anchor * it is a way for cells to anchor proteinw to cell membrane without protein having a transmembrane domain * occurs in eukaryotes, some archaea but not prokaryotes * proteins are attached to GPI anchor and anchored to membrane via C terminus * it occurs in ER lumen co-translationally

Answer 230

1. occurs in ER lumen co-translationally 2. catalysed by GPI transamidase 3. C-terminal of protein has GPI signal but no consensus sequence

Answer 231

* Phosphatidylinositol + Glucosamine + 3 x mannose-phospho-ethanolamine * first 2 steps occur in cytoplasm, and then fliped to ER lumen

Answer 232

* Phospholipase C (PLC) can hydrolyse lipid double tail from GPI proteins

Answer 233

* Matrix Metallo-Proteases (MMPs) * Extracellular, PM-tethered proteases Regulate the extracellular matrix High expression in invasive cancers

Answer 234

1. Reversible: S-palmitoylation 2. irreversible: N-myristolatuon 3. irreversible: Prenylation

Answer 235

* palmitoylation refers to the addition of palmitate (palmitic acid) a 16 carbon saturated fatty acid * ot occurs mostly on cystein residues through a thioester bond (altho also on serine, lysein through forming oxyester and amide linkages)

Answer 236

1. Addition of Palmitate (Palmitoylation): 2. Enzymatic Catalysis: Palmitoylation is typically catalyzed by a family of enzymes known as DHHC-type palmitoyl acyltransferases (PATs). 3. Enzyme Recognition: These enzymes recognize specific protein substrates and transfer palmitate from palmitoyl-CoA to the sulphur atom on the side chain of cysteine residues. 4. Catalytic Domain: The DHHC refers to an Asp-His-His-Cys amino acid motif in the catalytic domain of the enzyme PAT. PATs are transmembrane enzymes 5. Location: Palmitoylation can occur in various cellular compartments, including the Golgi apparatus, endoplasmic reticulum, and plasma membrane.

Answer 237

* Removal of Palmitate (Depalmitoylation): 1. Thioesterase Enzymes: The reversal of palmitoylation, known as depalmitoylation, is catalyzed by acyl protein thioesterases (APTs) and palmitoyl protein thioesterases (PPTs). 2. Regulation of Protein Function: This dynamic cycle of palmitoylation and depalmitoylation can regulate protein function and localization within the cell. 3. Subcellular Localization: Depalmitoylation can cause proteins to return to their original location within the cell or to move to different cellular compartments.

Answer 238

* Protein transport across the Golgi controlled by palmitoylation PATs are enriched in the Golgi * Palmitoylation concentrates membrane cargo at the cisternal rim * as it is reversible, it regulates signalling capabilities and protein interaction

Answer 239

* Fatty Acid: Myristic acid is linked to proteins through an amide bond. * Target Site: It most commonly occurs at a glycine residue at the N-terminus of proteins after the N-terminal methionine is cleaved off.

Answer 240

* most common time * During Protein Synthesis: As proteins are being synthesized by ribosomes, the N-terminal methionine is cleaved by a methionine aminopeptidase, exposing the glycine residue. Enzyme Attachment: A myristoyl-CoA molecule, which is myristic acid linked to coenzyme A, is used as the donor for the fatty acid chain. Catalysis: The enzyme N-myristoyltransferase (NMT) then transfers the myristoyl group from myristoyl-CoA to the exposed glycine residue, forming an amide linkage.

Answer 241

* less common * Recognition of Internal Glycine: This can occur when an internal glycine residue is exposed after proteolytic cleavage or during the rearrangement of a protein, which is less frequent. * Enzymatic Activity: It is assumed that enzyme N-myristoyltransferase (NMT) would also catalyze this reaction, but the specifics of recognition and transfer in a post-translational context are less well-understood than the co-translational mechanism.

Answer 242

1. Protein-Membrane Association: The myristoyl group allows proteins to attach reversibly to cell membranes, anchoring them to the lipid bilayer. 2. Protein-Protein Interactions: It also facilitates protein-protein interactions within the membrane, influencing signal transduction and other cellular communication processes. 3. Subcellular Localization: The modification can affect the protein’s localization within the cell and is often involved in targeting proteins to specific membranes, like the plasma membrane or Golgi apparatus.

Answer 243

1. Irreversible Process: Unlike palmitoylation, myristoylation is generally considered an irreversible modification, so it’s a more permanent change to the protein’s characteristics. 2. Co-Translational Advantage: The co-translational nature of myristoylation ensures that newly synthesized proteins can be immediately directed to membranes without requiring additional regulation or modification.

Answer 244

1. intrinsic pathway of apoptosis 2. where cleavage and myristoylation of BID (pro-apoptotic protein) by capase 8 and NMT causes BID to bind to mito membrane 3. BID binds to BAK and causes BAK oligomiseration 4. this initiates cytochrome c release, triggering apoptosis programmed cell death (PCD) in animals

Answer 245

* Prenyl Groups: The prenyl groups are 15-carbon farnesyl or 20-carbon geranylgeranyl isoprenoids. * Target Amino Acids: Prenylation typically occurs at the C-terminus of the protein at a cysteine residue within a specific motif called the ‘CAAX box’, where ‘C’ is cysteine, ‘A’ is usually an aliphatic amino acid, and ‘X’ is any amino acid.

Answer 246

* Mechanism of Prenylation 1. Enzymatic Catalysis: * Farnesyltransferase (FTase) or Geranylgeranyltransferase (GGTase): These enzymes catalyze the transfer of farnesyl or geranylgeranyl groups from their respective isoprenoid pyrophosphate donors (farnesyl pyrophosphate or geranylgeranyl pyrophosphate) to the sulfur atom of the cysteine residue within the CAAX box motif of the target protein. 2. CAAX Box Processing: * Recognition: FTase or GGTase recognizes the CAAX motif at the C-terminus of the target protein. * Prenylation: The isoprenoid group is attached to the cysteine via a thioether bond. * Proteolytic Cleavage: After prenylation, the three amino acids ‘AAX’ are usually cleaved off by specific proteases. * Carboxymethylation: Additionally, the now C-terminal cysteine may be further modified by carboxymethylation.

Answer 247

1. Membrane Association: Prenylation helps anchor proteins to the cell membrane. 2. Protein-Protein Interactions: It can also facilitate interactions with other membrane-bound proteins and contribute to the assembly of signaling complexes. 3. Subcellular Localization: Prenylation affects the trafficking of proteins to different organelles and membrane compartments.

Answer 248

1. the farnesylation/prenylation of phodopsin kinase GRK1 phosphorylates rhodopsin in retinal membrane in eye 2. this initiates phototransduction cascade 3. GRK1 dysfunction causes night blindness

Answer 249

6 in total 1. Glycosylations: N and O glycosylations (both extracellular on ER membrane) 2. Lipidation: Glypiation: on plasma membrane but extracellular towards cytoplasm; Myristoylation, palmitoylation, prenylation all inside plasma membrane 3. palimitoylation is reversible, myristoylation irreversible and is co- and post- translational, prenylation is also irreversible

Answer 250

1. phosphorylation 2. methylation 3. acetylation

Answer 251

* Small, rapid covalent modifications which alter protein function * Usually reversible

Answer 252

* Kinase enzymes catalyse addition of phosphate group * Phosphatase enzymes catalyse removal of phosphate group * Major regulatory PTM - >50% proteins are phosphorylated * Usually occurs in cytoplasm or nucleus

Answer 253

* phosphorylation usually activates or switched on or off a gene (changes it from on to off, or vice versa) * Radical change in charge – causes conformational change or altered interactions * Can activate or inactivate * Quick and transient

Answer 254

* Common targets are polar (-OH groups in side chains); * most common is Ser (79%) * Phosphorylation catalysed by target-specific kinases; specificity includes sequence surrounding target residue * change of charge from 0 to -2

Answer 255

* Less common targets are basic residues (His, Lys, Arg), acidic residues (Glu, Asp) and cysteine. * these are more difficult to detect because basic they can be more unstable, hence more difficult to maintain when prepping for mass spec analysis

Answer 256

* human kinome is a complete set of protein kinases present in human cell ( they r enymes which transfer phosphate groups from ATP to protein substrates via phosphprylation) * 518 kinase enzymes in human kinome = diversity and specificity of kinase substrate reaction * 3 common categories of kinases 1. Tyrosine Kinases: These are the kinases that specifically phosphorylate tyrosine residues on proteins. EG: the Epidermal Growth Factor Receptor (EGFR), which is important for cell growth and differentiation. 2. Ser/Thr Kinases: kinases that phosphorylate Ser and Thr. E.g Cyclin-dependent kinases (CDKs), which play a key role in cell cycle regulation. 3. MAPK (Mitogen-Activated Protein Kinases): involved in responding to extracellular signals and is crucial for processes like inflammation, cell growth, and apoptosis (programmed cell death).

Answer 257

* Kinase inhibitors are a class of drugs that block the kinase activity, preventing phosphorylation, and thus, can disrupt cancer cell growth and survival. * Tyrosine Kinase Group: This includes drugs that inhibit EGFR/HER (important for cell growth), VEGFR (involved in blood vessel formation), and ABL (associated with chronic myeloid leukemia). * Ser/Thr Kinase Group: This includes inhibitors of BRAF (associated with melanoma) and MEK1/2 (involved in the MAPK/ERK pathway). * Many of these drugs are analogues of ATP, which means they are structurally similar to ATP and can compete with ATP for the binding site on the kinase. This inhibits the kinase's ability to transfer phosphate groups to its substrates. These inhibitors are often used to target mutant kinases that are overactive in cancer cells.

Answer 258

1. Protein kinases are enzymes that modify other proteins by chemically adding phosphate groups to them (phosphorylation). 2. Structure + function: protein kinases share the same fold (structure), mechanism, and regulation, but their functions can vary depending on the cellular environment 3. Example Function of PKA: Role of PKA in glucose release from glycogen during the adrenaline-mediated fight-or-flight response. * PKA (protein Kinase A) phosphorylates Phosphorylase Kinase (PhK) * PhK phosphorylates Glycogen Phosphorylase (PG) * PG catalyses release of glucose-1-phosphate from Glycogen

Answer 259

1. Kinase domain: responsible for enzymatic catalytic activity of PKA 2. N-lobe: beta strand and binds to ATP 3. C-lobe: alpha helices which binds to substrate

Answer 260

* Mechanism of ATP Binding: 1. When ATP binds to the N-lobe of PKA, a glycine-rich loop within the N-lobe moves to "close" around ATP, ensuring that ATP is held in the correct position for catalysis. 2. A structural change occurs where the N-lobe shifts to accommodate the ATP, allowing the correct orientation for the transfer of the gamma phosphate (the outermost phosphate in ATP) to substrate protein 3. this causes the PKA to enclose ATP upon binding * Mechanism of substrate protein binding: 1. The substrate-binding groove is located between the N-lobe and the C-lobe, and it is designed to position the substrate so that the specific amino acid residue that needs to be phosphorylated (serine or threonine) is aligned with the gamma phosphate of ATP. 2. PKA recognizes the substrate: hydrophobic pockets within the C-lobe interact with hydrophobic residues on the substrate, and charged pockets interact with charged residues.

Answer 261

1. PKA catalyzes the transfer of the gamma phosphate from ATP to the hydroxyl group of a serine or threonine residue on the substrate protein. 2, The residues K72, N171, D166, and K168 in PKA interact with the gamma phosphate and are crucial for proper positioning of ATP and the substrate. 3. Magnesium ions (Mg²⁺) are also important as they help to stabilize the negative charges of the phosphate groups in ATP. 4. Once everything is correctly aligned, the catalytic residues within PKA facilitate the transfer of the gamma phosphate from ATP to the substrate, resulting in a phosphorylated protein. 5. The transfer of the gamma phosphate to the substrate protein is a key regulatory step in many cellular processes. The phosphate group changes the structural conformation of the substrate protein, thereby changing its activity.

Answer 262

1. Binding of ATP: ATP binds to PKA with the help of magnesium ions (Mg²⁺). The Mg²⁺ ions stabilize the negative charges on the phosphate groups of ATP, specifically the beta (β) and gamma (γ) phosphates. This stabilization is crucial for the phosphorylation reaction to occur. 2. Positioning of Serine Residue: The substrate protein, which contains the amino acid serine (Ser) that is to be phosphorylated, is positioned such that the hydroxyl group (OH) of the serine side chain is in close proximity to the gamma phosphate of ATP. 3. Catalysis: The catalytic residues within PKA facilitate the reaction. Aspartate (D166) acts as a base, removing a proton from the serine's hydroxyl group. This deprotonation makes the oxygen atom of the serine more nucleophilic, meaning it is ready to attack the gamma phosphate of ATP. 4. Transition State: As the serine oxygen attacks the gamma phosphate, a transition state is formed. This is a high-energy, unstable state where the bonds are in the process of breaking and forming. The energy of this state is reduced by the stabilization provided by the surrounding amino acids and Mg²⁺ ions. 5. Phosphate Transfer: As the transition state resolves, the bond between the gamma phosphate and the beta phosphate of ATP is broken. The gamma phosphate is then covalently attached to the serine residue, converting it into phosphoserine (pSer). 6. Product Release: The newly formed phosphoserine is released from PKA, and the enzyme is ready for another cycle of catalysis.

Answer 263

* PKA is involved in signalling cascade using GPCR receptor protein * when a glucagon of adrenaline binds to GPCR: it causes hydrolysis of GTP to GDP, which causes release of alpha subunit on G protein to bind and activate Adenylate cyclase * activated adenylate cyclase converts TAP to cAMP * cAMP then causes the PKA regulatory inhibitor to bind to cAMP instead of PKA, which causes inactivated PKA to become active * PKA is then phosphorylated by PDK1 (bound to PIP3 on cell membrane) by hydrolysis of ATP to ADP

Answer 264

1. to deactivate PKA, signal causes cAMP to stop producing 2. ebzymes like phosphodiesterases breakdonw cAMP into AMP, reducing cAMP conc in cell 3. Hence cAMP molecules disscoaite and are not bound to PKA regulatory units, which cause the regulatory subunits to bind and inhibit PKA again causing PKA to be inactivates

Answer 265

* phosphatases cause hydrolysis reactions * phosphatases can be specific or general

Answer 266

* Pseudomonas syringae (tomato leaf spot disease) recognition by FLS2 * receptor kinases like FLS2 recognize specific bacterial components like flagellin fragments * this recognition by FLS2 triggers signalling cascade and phosphorylation of other proteins BAK1 and BIK1 * this leads to the activation of both the MAPK pathway and CDKs * phsophorylation and activation of CDK causes phosphorylation of RBOHD -> oxidative burst which limit pathogen spread * MAPKinases phosphorylate another signalling causing transcriptional reprogramming

Answer 267

1. Upon flagellin recognition, FLS2 interacts with BAK1 and activates BIK1. 2. This complex then activates downstream MAP kinases (like MEKK1, MKK5, MPK3) and CDKs, which then phosphorylate various transcription factors. 3. This phosphorylation modulates gene expression to bolster plant defense mechanisms, including the production of reactive oxygen species (ROS) for an oxidative burst to limit pathogenic spread

Answer 268

1. CDKs phosphorylate RBOHD, a component responsible for the production of reactive oxygen species (ROS). 2. The ROS serve as a defense mechanism to create a hostile environment to limit pathogen growth and signal to other parts of the plant that there is an infection.

Answer 269

1. Pseudomonas syringae injects type 3 effectors (T3Es) into the host plant cells. 2. These effectors include AvrPto and AvrPtoB, which inhibit kinase signaling, and others like AvrPphB, HopB1, HopF2, and HopAI1, which target various components of the immune response pathway to suppress plant defense mechanisms.

Answer 270

1. AvrPto: Kinase inhibitor that blocks FLS2/BIK1 signaling. 2. AvrPtoB: E3 ligase that leads to ubiquitination and degradation of FLS2/BAK1. 3. AvrPphB: Cysteine protease that cleaves BIK1. 4. HopB1: Serine protease that cleaves phosphorylated BIK1. 5. HopF2: ADP ribosylase, inactivates MKK5. 6. HopAI1: Phosphatase, inactivates MPK3.

Answer 271

1. covalent modification of protein with methyl 2. methyl group comes from donor SAM (S-adenosylmethionine) to SAH (S-adenosyl-homocysteine) 3. mono/di/tri methylation can occur on same residue 4. methylation increases hydrophobicity 5. methyltransferases add -CH3 6. demethylases remove CH3

Answer 272

* commonly occur on 7 different residues * split into N-methylation and O methylation

Answer 273

* most methyltransferases contain SET domains * these domains help catalyse the transfer of methyl groups from SAM donor to amino asid lysine residues on histone tails * hence can also regulate gene expression via chromatin structure, epigenetic marks...etc

Answer 274

* SET9 methyltransferases specifically methylates lysine 372 (K372) on the p53 protein, * p53 protein is a crucial transcription factor that regulates cell cycle, apoptosis, and DNA repair. * Methylation of p53 by SET9 can influence p53's stability and activity. * Notably, the p53 protein is found mutated in about 50% of human cancers, indicating its vital role in preventing cancer development.

Answer 275

1. it has a substrate binding groove which recogizes P53 peptide 2. has a cofactor pocket which opens to bind to SAM/SAH

Answer 276

1. SET9 has a substrate binding groove that specifically recognizes the p53 peptide. 2. This specificity ensures that the methyl group is transferred to the correct lysine residue (K372) on p53, which is critical for the proper regulation of p53's role in the cell.

Answer 277

1. The open cofactor pocket of SET9 is responsible for binding SAM, the methyl donor, and SAH, the product after the methyl transfer. 2. This pocket is essential for the catalytic activity of SET9, allowing it to carry out the transfer of methyl groups to target lysine residues.

Answer 278

1. N-terminal methylation is a form of protein modification that occurs after the removal of the initiator methionine (iMet) from a protein. 2. This process is catalyzed by N-terminal methyltransferases (NTMTs) and often targets proteins with a consensus sequence of Met-X-Pro-Lys, where X can be alanine, proline, or serine.

Answer 279

1. after prenylation and processing of C-terminal by ICMT in ER membrane 2. improves association of prteins with plasma membrane association by removing negative charges

Answer 280

1. NTMTs typically catalyze methylation on large protein complexes at the N-terminus following the specific consensus sequence. 2. This methylation is important for the proper functioning and stabilization of protein complexes. 3. consensus sequence of Met-X-Pro-Lys, where X can be alanine, proline, or serine.

Answer 281

1. RCC1 (Regulator of chromatin condensation-1) requires N-terminal methylation by the enzyme PRMT6 to bind DNA, condense chromatin, and facilitate normal mitosis. 2. This modification helps RCC1 to engage with chromatin effectively and influences cell cycle progression.

Answer 282

1. Low PRMT6 levels lead to decreased methylation and increased degradation of RCC1, which results in less chromatin-bound RCC1, mitotic defects, cell cycle arrest, low self-renewal, low tumorigenicity, and increased radiation sensitivity. 2. High PRMT6 levels increase RCC1 methylation and promote chromatin binding of RCC1, which correlates with proper mitosis, cell cycle progression, high self-renewal, high tumorigenicity, and radiation resistance.

Answer 283

1. PRMT6 can be inhibited using specific inhibitors like EPZ020411 and CX-4945. 2. These inhibitors can potentially modulate RCC1 methylation status, thereby impacting the chromatin-binding ability of RCC1 and influencing the cell cycle and cancer progression.

Answer 284

1. His73 methylation on b-actin by methyl transferase SETD3 Found in most eukaryotes, not yeast. 2. Methylation decreases rate of ATP hydrolysis in actin; stabilises actin filaments. 3. Mutation/Knock out of SETD3 methyltrasferase in mice led to reduced progeny because of birth defects and problems

Answer 285

1. Flagella of salmonella are methylated by methyl transferase FliB on thei lysine residue 2. Methylated flagella are more hydrophobic and interact with epithelial cells: promotes bacterial adhesion and host cell invasion

Answer 286

1. Asymmetric and symmetric Arg methylation is catalysed by different protein Arg (R) methyl transferases (PRMTs, SET proteins). 2. is involved in histone methylation

Answer 287

1. Methylation of Lys (K) and Arg (R) residues by specific histone methyl trasferases (HMTs, SET domain proteins); removal by various histone demethylases (HDMs) * H3K4me: active genes * H3K9me: constitutive heterochromatin * H3K27me: facultative heterochromatin

Answer 288

1. O-methyl removal via hydrolysis, releasing methanol 2. N-methyl removal via reduction (e.g. FAD co-factor) then hydrolysis

Answer 289

1. Covalent modification of protein with acetyl (-COCH3) 2. General (at N-terminus) and regulatory (at Lys) 3. Acetyltransferases add acetyl group 3. Deacetylases remove acetyl group 4. donor for acetyl is Acetyl CoA -> CoA

Answer 290

* Transfer of acetyl (-COCH3) from Acetyl CoA * by acetyl transferase

Answer 291

* Acetylation of N-termini (Nt) and lysine residues (K) * Removes positive charge on residues

Answer 292

1. N-terminal acetylation is a common modification in the cytoplasm of all organisms, where approximately 85% of human proteins have acetylated N-termini. 2. This modification can influence protein stability, localization, and function.

Answer 293

1. N-terminal acetyltransferases (NATs), which all contain a Gcn5-related N-acetyltransferase (GNAT) domain, catalyze N-terminal acetylation. 2. Bacteria typically have 3 NATs, while eukaryotes have 5-8 NATs. 3. These enzymes are often associated with ribosomes and recognize N-terminal dipeptides, although their full roles are not completely understood.

Answer 294

1. Lysine acetylation is catalyzed by lysine acetyltransferases (KATs) and histone acetyltransferases (HATs), which can be GNAT-based enzymes like NATs. 2. Deacetylation, the removal of acetyl groups, is catalyzed by lysine deacetylases (KDACs), also known as histone deacetylases (HDACs).

Answer 295

1. Acetylation on histone 3 at lysine 27 (H3K27Ac) generally activates transcription, 2. HOWEVER: acetylation of p53 protein at K120/164 prevents p53 from binding to DNA. 3. This modification is crucial for regulating transcription and protein-DNA interactions, and it can have significant implications in processes such as cell cycle regulation and tumor suppression.

Answer 296

1. GNAT (Gcn5-related N-acetyltransferase) domains are present in enzymes like GCN5 2. are responsible for transferring an acetyl group from Acetyl Coenzyme A (AcCoA) to target lysine residues on proteins, notably histones.

Answer 297

1. GCN5 is a lysine acetyltransferase (KAT) that acetylates the 14th lysine (K14) of histone-3 (H3), forming H3K14Ac. 2. This modification is crucial for transcriptional activation. 3. GCN5 specifically recognizes and acetylates sequences with the pattern G-K-x-P, where 'x' can be alanine, proline, or serine.

Answer 298

1. Substrate (histone-3 N-terminus) and cofactor (AcetylCoA) bound to GCN5 enzyme (structure not shown)

Answer 299

1. Acetylation of lysine 40 on α-tubulin by α-tubulin acetyltransferase (αTAT) increases the half-life of tubulin and is associated with stabilized microtubules. 2. Microtubules that are acetylated at this site are less likely to undergo breakage 3. can contribute to microtubule repair, and resists bending 4. important for cellular integrity and dynamics.

Answer 300

* 3 types of small PTMs: Phosphorylation, methylation, acetylation * Cofactor donating the PTM * Enzyme catalysing the PTM addition * Subcellular localisation of PTM addition * Residue receiving the PTM * Enzyme catalysing PTM removal * General effect of PTM on protein function * Examples of proteins affected by this PTM

Answer 301

* Phosphorylation, methylation and acetylation regulate most proteins! * Small, reversible PTMs affecting local charge/polarity of proteins. * Addition/removal of PTMs is catalysed by specific enzymes. * Phosphate/acetyl/methyl groups donated by nucleotide co-factors. * Transferases (kinases, SETs, GNATs) are large gene families recognising different substrates. * Removal occurs via smaller, diverse families, often hydrolases. * Well understood examples include PTMs on PKA, histone-3, actin, tubulin and flagellin.

Answer 302

* N-Glycosylation * O-glycosylation * Glypiation * Palmitoylation * Myristoylation * Prenylation

Answer 303

* The nucleus and nuclear envelope are believed to have originated through two main hypotheses related to eukaryotic evolution: 1. Extension of the Hydrogen Hypothesis: * a symbiotic relationship between an archaeal host and alpha proteobacteria, leading to increased surface area for substance exchange, internalization of the bacteria, and eventually the formation of the nuclear envelope to separate the archaeal genome from bacterial processes and to protect nuclear genome from ROS * metabolic syntropy between methase (produced by archaea) and CO2 O2 produced by bacteria (exchange in metabolites)problem: assumption of 100mya environment, and unsure if the assumption for 100bya for origin of euks is correct 2. Autogeneous Hypothesis: * Suggests that the nuclear envelope developed to separate the splicing of genes from translation due to the interference of group II introns with gene expression, leading to the emergence of spliceosomes, nuclear envelope, and nuclear pores. * suggested that the archaea hpst and proteobacteria shared 2 gene expression system * Group II introns are mobile ribozymes that self-splice from precursor RNAs to yield excised intron lariat RNAs, which then invade new genomic DNA sites by reverse splicing. * Gene expression was stopped/impeded by translation of introns/unspliced transcripts in the chromosome * This allowed separation of splicing from translation. Which solves the translation problem with introns

Answer 304

1. Eukaryotic DNA is enclosed by a nuclear envelope, contains approximately 1000 times more DNA than prokaryotes, 2. Euk DNA is packaged with histones in linear chromosomes within sub-nuclear compartments like the nucleolus. 3. In contrast, bacteria typically have DNA that is not enclosed by a nuclear envelope and is less complex in terms of packaging and quantity.

Answer 305

* The 3D organization of the genome is established through the spatial arrangement of chromatin within the nucleus, facilitating the regulation of gene expression. * This involves higher-level DNA packing with histones, the formation of chromatin fibers, and their further organization into chromosomes and specific nuclear territories.

Answer 306

1. Yes, the 3D organization of the genome is closely related to its function. 2. The spatial arrangement of genes and regulatory elements within the nucleus can influence gene expression patterns, 3. allowing for efficient coordination of cellular processes and metabolic demands.

Answer 307

* Higher-level organization of the genome serves several functions: 1. Facilitates efficient gene regulation and expression. 2. Enables compartmentalization of different processes, like separating transcription and translation. 3. Protects the genome from damage, such as from reactive oxygen species (ROS).

Answer 308

1. General principles include the conservation of genome integrity, 2. the regulation of gene expression through spatial positioning, 3. and the evolution of cellular mechanisms to resolve conflicts, such as those posed by the presence of introns.

Answer 309

1. The "Inside Out" hypothesis suggests that an eocyte bacterium, a hypothetical ancestral prokaryotic group, extended blebs towards an epibiotic bacterium (future mitochondrion). 2. These blebs engulfed the epibiotic bacterium, leading to the formation of half nuclear pores. 3. As blebbing continued, full nuclear pores formed, the S layer disintegrated, and the mitochondria escaped into the cytoplasm. 4. This process supports the origin of the nuclear pore complex, indicated by the homology with coat proteins like tubulin and actin.

Answer 310

1. Ecological constraints provide a realistic framework for evolutionary events, such as the availability of hydrogen and methane, and the spatial relationships between organisms. 2. For example, in the hydrogen hypothesis, it only makes sense if hydrogen and methane were present in the environment, suggesting the need for a sensible spatial ecology for the endosymbiotic relationship to evolve.

Answer 311

1. The E3 model stands for Entangle, Engulf, Endogenize and is the only model with experimental evidence from lab cultures. 2. It suggests a symbiotic relationship between Asgard archaea and other microorganisms like the Halodesulfovibrio bacterium, and Methanogenium archaeon. 3. lab experiment showed that they could only culture asgard achaea along w a mixed culture of another bactera/archaeon 4. Suggesting syntrophy between Asgard archaea and Halodesulfovibrio bacterium (85%), Methanogenium archaeon (2%)

Answer 312

1. Extension of hydrogen hypothesis 2. Autogeneous development of the nuclear envelope 3. Inside out hypothesis 4. E3 model hypothesis

Answer 313

1. The eukaryotic genome is packaged in an extremely compact manner, with DNA wrapped around histone proteins to form nucleosomes. 2. These nucleosomes further fold to create chromatin, which condenses into chromosomes during cell division. 3. This complex packaging allows eukaryotes to control and express the large amount of DNA within the nucleus efficiently.

Answer 314

1. The Lane and Martin Hypothesis suggests that the endosymbiosis of mitochondria led to an increase in bioenergetic membranes coinciding with mitochondrial genome reduction. 2. This allowed a massive increase in the number of genes that can be expressed in eukaryotes and the innovation of new protein folds, contributing to cellular complexity.

Answer 315

One hypothesis posits that mitochondria evolved before the nucleus, based on the idea that mitochondria are a prerequisite for the complexity seen in eukaryotic DNA, the nucleus, and the cell as a whole. This suggests that the original host of the mitochondrion had a prokaryotic genome organization (which could not be achieved without mitochondria), and the nucleus developed much later.

Answer 316

* Chromosome condensation during eukaryotic mitosis involves: 1. Phosphorylation of histone proteins by cyclin-dependent kinases (CDKs). 2. The action of SMC proteins like condensin and cohesin, which are crucial for condensing and wrapping up DNA during mitosis. 3. Cohesin holds sister chromatids together, while condensin helps in organizing chromosomes into their compact mitotic structure.

Answer 317

1. Bacterial chromosomes are typically circular and less than 10Mb in DNA content. 2. Unlike eukaryotes, bacterial DNA does not associate with histones. 3. Instead, the nucleoid is organized as a series of interwound supercoiled loops emanating from a dense core, a structure facilitated by bacterial SMC proteins.

Answer 318

1. Yes, archaea also have histones with DNA wrapped around them, similar to eukaryotes, and possess SMC proteins. 2. Given that SMC proteins are essential for packaging DNA in bacteria, archaea, and eukaryotes, they must have evolved very early in the history of life and are vital across all three domains of life for organizing chromosomes.

Answer 319

1. Fluorescence in-situ hybridization (FISH) is a technique used to visualize the spatial relationship and distance of specific DNA sequences within the cell. 2. It can range from a gene level to a chromosomal scale, indicating the precise location of these sequences on chromosomes in different cells.

Answer 320

1. 3C (Chromosome Conformation Capture) and Hi-C are molecular biology techniques that interrogate chromatin interactions based on physical proximity within the nucleus. 2. They are used to study the three-dimensional structure of chromatin and reveal how different regions of the chromosome interact with each other.

Answer 321

1. Chromatin Immunoprecipitation (ChIP) is a method used to map the chromatin state and binding of regulatory proteins. 2. By using antibodies against covalent histone modifications, site-specific transcription factors, or transcriptional cofactors, ChIP provides insight into the regulatory mechanisms of gene expression.

Answer 322

1. DNAse hypersensitivity testing identifies accessible regions of DNA that are not protected by nucleosomes or DNA-binding proteins. 2. These regions are often indicative of active regulatory elements involved in gene expression.

Answer 323

1. During interphase, each chromosome is housed within a distinct region of the nucleus known as a chromosome territory (CT). 2. The positioning of these territories within the nucleus can vary between cell types and under different conditions, playing a role in gene regulation.

Answer 324

1. The interchromatin compartment (IC) is the space between chromosome territories, filled with various non-chromatin domains. 2. The IC contains factors essential for nuclear processes like transcription, splicing, DNA replication, and DNA repair.

Answer 325

1. he location of a gene within its chromosome territory can influence its transcriptional activity. 2. Active genes are typically found closer to the edges of their territories, near the IC, facilitating easier access to transcription machinery, 3. while less active genes might be situated deeper within the territories.

Answer 326

1. The spatial organization of chromosomes within the nuclear environment can significantly influence gene expression and cell function. 2. This organization changes in response to various signals or conditions, and disruptions to typical nuclear/chromosome structure are often associated with diseases, such as cancer.

Answer 327

* Chromosome Conformation Capture (3C) and its derivative Hi-C are methods to probe the spatial organization of chromosomes. Steps include: 1. Crosslink DNA with formaldehyde to preserve spatial organization. 2. Cut DNA with restriction enzymes like HindIII or NheI. 3. Fill ends and mark with biotin for easy pull-down later. 4. Ligate loose ends to connect interacting DNA pieces. 5. Purify, shear DNA, and pull down biotin-labeled fragments. 6. Sequence using paired-end sequencing to identify proximal DNA sequences.

Answer 328

1. Level 1: Chromosome territories at the nuclear scale, where individual chromosomes occupy distinct regions. 2. Level 2: Compartments at the chromosome scale segregate chromatin by activity. 3. Level 3: Topologically Associated Domains (TADs) within compartments where DNA sequences within the same TAD interact more frequently. 4. Level 4: Loops at the megabase scale within TADs, essential for gene regulation.

Answer 329

1. Yes, TADs are a conserved feature found in various species, from bacteria to mammals, highlighting their fundamental role in genome organization and gene regulation. 2. In bacteria, similar structures are referred to as Chromosome Interacting Domains (CIDs), emphasizing the evolutionary significance of these domains. 1.

Answer 330

topologically associating domain is a self-interacting genomic region, meaning that DNA sequences within a TAD physically interact with each other more frequently than with sequences outside the TAD.

Answer 331

* LADs and TADs are both crucial for chromatin organization: 1. TADs are about intra-chromosomal interactions and compartmentalization. 2. LADs involve the genome's interaction with the nuclear lamina, often linked to gene repression. 3. TADs bring promoters close to regulatory elements, aiding gene regulation, 4. while LADs contribute to the spatial organization and the 3D nuclear architecture

Answer 332

1. SMC proteins, like SMC1/SMC3, form rings that hold sister chromatids together during S phase. 2. SMC2/SMC4 stabilize the 300nm fiber, a level of chromosomal compaction. 3. These rings open during the metaphase/anaphase transition in mitosis to allow chromatid separation.

Answer 333

* Condensin in yeast binds to DNA and uses ATP hydrolysis to actively form DNA loops, extruding them at a rate of 1kb per second. * The loop formation is asymmetrical, with one side longer than the other.

Answer 334

1. Human condensin I and II can compact nucleosome-bound DNA. 2. They also stabilize the structure during decondensation of DN, maintaining the nucleosomes' position during spontaneous structural reversals.

Answer 335

1. Chromatin loop sizes are regulated by CTCF, which recognizes specific DNA motifs, and cohesin, which assists in loop formation. 2. Cohesin's ATP-dependent movement along chromatin extrudes loops, while convergent CTCF sites can halt this movement. 3. Absence of WAPL or PDS5 leads to extended loops or altered structures.

Answer 336

1. TADs form ~1MB loops that are conserved across species and are marked by CTCF binding sites. 2. They bring regulatory elements close together for coordinated gene regulation. 3. LADs, associated with the nuclear lamina, are usually transcriptionally silent and enriched in methylated histones, indicating repressive states.

Answer 337

1. Active polymerase clusters and transcription factors are key architectural features of genome spatial organization. 2. Genes located near transcription factories have increased activation frequency. 3. This spatial organization suggests potential functions and behaviors of regulatory elements like enhancers and super-enhancers.

Answer 338

1. Architectural proteins like cohesin and CTCF. 2. Chromosome territories, compartments, domains, and loops. 3. A transcription-driven organization that aligns the genome's structure with its functional expression.

Answer 339

* The nuclear envelope consists of two parallel membrane bilayers: 1. the inner nuclear membrane (INM) and 2. the outer nuclear membrane (ONM), which is continuous with the endoplasmic reticulum (ER) and studded with ribosomes. 3. Between them is the perinuclear space, leading into the ER. 4. The nuclear lamina, composed of intermediate filaments called lamins, is attached to the INM and provides structural support.

Answer 340

1. The nuclear envelope serves to separate the genome from the cytoplasm, organizing chromatin and regulating gene expression through its interactions with microtubules and heterochromatin. 2. It facilitates the localization and migration of the nucleus, and protein complexes spanning the envelope link to the cytoskeleton to assist in cellular movement.

Answer 341

1. The NPC is a large protein complex (~100 MDa) embedded in the nuclear envelope, acting as a semipermeable gate. 2. It's assembled from ~30 nucleoporins (NUPs) that form a symmetric core with 8 rotational symmetry and an asymmetrical structure on both the nuclear and cytoplasmic sides. 3. FG repeats within NUPs create a selective barrier for active transport of macromolecules.

Answer 342

* The symmetric core is responsible for bidirectional transport, featuring an inner ring spanning the INM and ONM, with cytoplasmic and nucleoplasmic rings at each end. * Scaffold NUPs form the NPC's structural foundation, with Y complexes contributing to the lattice structure. * Transmembrane NUPs anchor the complex to the membranes, while vesicle coat and adaptin-related NUPs stabilize the membrane curvature.

Answer 343

* The central pore of the NPC is the main passageway for molecules moving between the nucleus and cytoplasm. * It contains FG repeat proteins, which are anchored on one end and extend into the center, forming a dynamic meshwork with a hydrophobic nature that selectively filters macromolecules requiring active transport.

Answer 344

1. The nuclear basket on the nuclear side and the cytoplasmic fibrils on the cytoplasmic side constitute the NPC's asymmetrical structures. 2. They provide docking sites for transport protein complexes and control the disassembly of these complexes, facilitating the directionality of macromolecule transport.

Answer 345

* NPCs are highly conserved across eukaryotes, with structural flexibility that allows the central pore diameter to accommodate larger macromolecules. * The number and distribution of NPCs vary depending on cell type, age, cell cycle stage, and metabolic state, which might contribute to pore-to-pore heterogeneity.

Answer 346

* Proteins that shuttle in and out of the nucleus, such as karyopherins, RAN GTPases, MAPKs, and transcription factors (TFs), utilize active transport. * Karyopherins mediate the transport of molecules through the nuclear pore complexes (NPCs) by interacting with nucleoporins, and this process is regulated by the Ran GTPase gradient. * Karyopherins = are proteins involved in transporting molecules between the cytoplasm and the nucleus of a eukaryotic cell

Answer 347

* RNA molecules including mRNA, miRNA, tRNA, and rRNA are exported from the nucleus through various pathways. * Bulk export occurs for most RNA types, while alternative pathways may be used for specific RNAs or under certain conditions.

Answer 348

* Nuclear proteins such as histones, lamins, polymerases, transcription factors, and those involved in RNA editing, degradation, and DNA repair are imported into the nucleus. * This process is guided by nuclear localization signals (NLS) that are recognized by importin receptors.

Answer 349

* Transport receptors are essential for nuclear transport events. * They bind to cargo to form a complex, interact with FG NUPs, and facilitate passage through the NPC. * Protein Transport receptors: 1. Karyopherins transport receptors such as Importin β and exportins) and RNAs (NXT1/NTF2 heterodimers for mRNA). Used for both RNA and Protein transport RNA transport receptors: 1. Karyopherins 2. NXT1/NTF2 heterodimer mRNA

Answer 350

1. Importins, like Importin β, are HEAT REPEAT proteins that bind cargo proteins either directly or via adaptor proteins like Importin α. 2. Nuclear Localization Signals (NLS) are sorting signals on proteins that direct them to the nucleus, either as short sequences rich in lysines and arginines (monopartite or bipartite) or as PY-NLS with specific terminal sequences.

Answer 351

1. The receptor-cargo complex cannot enter the nucleus directly due to the hydrophobic environment of the NPC. 2. Cargo binding to the importin receptor exposes the hydrophobic side of HEAT repeats, allowing interaction with FG repeats. 3. Multivalent binding of FG repeats enhances stability, though the mechanism is not fully understood.

Answer 352

* Ran GTPase acts as a molecular switch, existing in either a GTP- or GDP-bound form. * It regulates the interaction between importin and cargo, with Ran GAP (GTPase activating proteins) and Ran GEF (guanine nucleotide exchange factor) serving as regulators. * RanGTP induces the dissociation of the importin-cargo complex inside the nucleus, ensuring unidirectional transport.

Answer 353

1. The importin-cargo complex enters the nucleus through the NPC. 2. Importin β binds to Ran-GTP, which causes cargo dissociation. 3. The Importin-Ran-GTP complex exits the nucleus. 4. Cytoplasmic RanGAP promotes GTP hydrolysis, leading to complex dissociation. 5. Ran-GDP returns to the nucleus via NTF2. 6. Nuclear RanGEF converts Ran-GDP back to Ran-GTP.

Answer 354

1. Yes, while the movement of the receptor-cargo complex through the NPC doesn't require energy, the RanGTPase-mediated control of import does. 2. This energy is vital for the dissociation and directionality of transport, with spatial separation of Ran regulators maintaining an asymmetric distribution of RanGDP and RanGTP across the nuclear envelope.

Answer 355

1. Protein export utilizes karyopherins and the Ran system. 2. Sorting signals for export are called nuclear export signals (NES), typically a hydrophobic motif. 3. Exportin receptors cooperatively bind to Ran-GTP and cargo, and GTP hydrolysis leads to disassembly of the complex. Exportin1 (Xpo1) in humans and CRM1 in yeast are examples.

Answer 356

1. Exportin binds to Ran-GTP, increasing its affinity for the cargo, forming the exportin-cargo-RanGTP complex. 2. The complex moves through the NPC. 3. Cytoplasmic RanGAP triggers GTP hydrolysis, disassembling the complex. 4. Exportin returns to the nucleus. 5. Ran-GDP is transported into the nucleus by NTF2. 6. Nuclear RanGEF induces the formation of RanGTP.

Answer 357

1. Bulk mRNA export, constituting about 95% of mRNA transport, is Ran-independent, relying on the heterodimeric receptor complex Nxf1-Nxt1 in humans or Mex67-Mtr2 in yeast. 2. This pathway involves the assembly of an export-competent mRNP complex, receptor binding and NPC translocation, and dissociation of the receptor from the mRNP.

Answer 358

1. Ran-dependent export is used for selective transport of various RNA types and involves karyopherin-exportins like Exportin 1 for mRNA and snRNA, Exportin-t for tRNA, and Exportin 5 for miRNA. 2. This system allows for sequence-specific RNA transport and is essential for mRNAs related to cell cycle and proliferation.

Answer 359

1, Assembly of an export‐competent mRNA-containing ribonucleoprotein (mRNP) complex 2, Receptor binding and translocation of mRNP through the NPC 3, Dissociation of receptor from the translocated mRNP complex

Answer 360

* in protein export Sorting signals are nuclear export signal or NES * Karyopherin receptors called exportins * Exportin binds cooperatively to RanGTPs and cargo (NES) * GTP hydrolysis will result in disassembly of the receptor-cargo-Ran complex * exportin1 (Xpo1) - human, CRM1- yeast

Answer 361

* Both rely on RanGTP-RanGDP system * Upon cargo binding karyopherin receptors undergo conformational change to increase hydrophobic surfaces * RanGTP- receptor complex forms in the nucleus and exits the nucleus * Importin βs exclusive RanGTP binding (importin either binds Ran or cargo) * Exportins cooperative RanGTP binding (exportin needs to bind to ran and cargo to transport)

Answer 362

1. Ran-independent * Bulk mRNA export ( ~95%) * Heterodimeric receptor complex * Nxf1-Nxt1 - human * Mex62-Mtr2- yeast 2. Ran-dependent/karyopherin dependent * Selected mRNA, tRNA, rRNA, miRNA, snRNA export * Karyopherin- Exportins * Exportin 1 (XPO1, CRM1): mRNA, snRNA, rRNA Exportin 5 :miRNA * Exportin-t : tRNA

Answer 363

1. 5' methyl guanosine cap binds CBP (cap binding protein) complex; CBPs promote splicing 2. After splicing Exon junction complex (EJC) and Serine-Arginine (SR) proteins are deposited in the coding region. First EJC and CBC together recruit TREX and Dbp5 DEAD-box RNA helicase. 3. Exon interactome (green) helps folding and packaging the coding region, including ATP-dependent DEAD-box proteins. 4. After 3' UTR cleavage and polyA synthesis, PolyA tail binding Protein complex ( PABPC) forms

MT Cell Bio Flashcards

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