BCH 211

Transcription initiation and elongation Basal transcription for a TATA containing Pol II promoter:

TBP (TATA-binding protein) associates with the TATA box slightly upstream of the initiator/promoter on the DNA.

TBP is associated with several TAFs (TBP associated factors); TBP plus TAFs are known as TFIID.

TBP binds the minor groove of the DNA and flattens it; has two specific Phenylalanine residues which insert into the TATA box base pairs.

Next, TFIIA comes in and stabilizes TBP-DNA and TAF-DNA interactions. Not essential for basal transcription but necessary for regulated. Also has anti-repressor function against TAF250.

Next, TFIIB is recruited. Contacts DNA upstream and downstream of the TATA box; serves to stabilize TBP-DNA interaction. It is repressed by NC2 via competitive binding.

Next, TFIIF and RNA polymerase II are recruited. TFIIF has high affinity for dsDNA and contacts the promoter directly; holds polymerase II back in order to inhibit promoter escape. TFIIF also remains associated with Pol II during elongation in order to reduce pausing.

Next, TFIIE and TFIIH are recruited, though it is possible that they arrive pre-associated with TFIIF and Pol II. TFIIE simply has stimulatory function for TFIIH. TFIIH has two helicase subunits (XPB and XPD) which use ATP to melt/unwind dsDNA around the promoter. The CDK7 subunit of TFIIH also has kinase activity; phosphorylates the CTD of Pol II at serine-5 in order to stimulate elongation. During elongation TFIIH is involved in nucleotide excision repair.

Promoter escape: Pol II transcribed about 9 base pairs and then dissociates from TFIID. Escape is held back (until the proper time) by TFIIF which is tightly bound to dsDNA at promoter; TFIIF also decreases frequency of abortion by increasing nucleotide addition so transcript reaches the 9 nucleotide threshold quickly. TFIIH helicase activity disrupts TFIIF’s binding to DNA.

Elongation involves many periods of pausing and backsliding. The 3’OH of the nascent transcript must be aligned with the DNA template and the catalytic site of RNA pol II. Pausing occurs when Pol II modestly and reversibely backslides on the template; in this case SII activates an endonuclease which cleaves the RNA that is out of line, allowing Pol II to leave paused state and continue. Elongation arrests when the enzyme severely backslides.

Proteins with anti-pause function: elongin and PTEFB (phosphorylates CTD at serine 12)

Proteins with pro-pause function: DSIF and NELF (associate with non-phosphorylated CTD)

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Nucleosomes Core histone protein consists of H2A, H2B, H3, and H4. Two of each, forming an

octamer. (2A and 2B form dimers while H3 and H4 form a tetramer) Histone Lysines can be methylated, acetylated, or ubiquitinated. The positive charge normally increases the interaction between the protein and

negatively charged phosphates on the DNA backbone. Methylation does not alter charge but protects against further modification. Acetylation cancels out the charge. Histones are acetylated by specific histone

acetyltransferases (HATS). This modification reduces association of DNA with the nucleosome, as well as signaling for other proteins to bind in order for more transcription activating modifications to occur.

RNA polymerase overcomes the presence of nucleosomes through Brownian motion: DNA strand is not firmly fixed on each histone; the strands fluctuate and “breathe”. RNA Pol II takes advantage of these natural fluctuations in order to slide one or a few base pairs forward. Tighter nucleosomes thus induce more pausing/backsliding.

Pausing is evolutionarily conserved because splicing sites are chosen as they emerge from the polymerase; thus pausing at one region increases the likelihood of that site being “chosen”. This promotes alternative splicing of some genes.

Hyperacetylated histones can signal for the SETII complex to recruit machinery to rapidly deacetylate/methylate histones, resulting in closed chromatin structure.

Regulated transcription Mediator is one of the largest co-activators. It interacts with many activators

during regulated transcription. Mediator also has similar (synergistic) function with many TAFs. It associates with the non-phosphorylated CTD of pol II (pre-initiation, as TFIIH phosphorylates the CTD when it is time for elongation) Mediator promotes the activity of TFIIH, has HAT activity on H3 and H4, and dissociates from the pol II CTD as soon as the CTD is phosphorylated by TFIIH.

SAGA also has HAT activity and is made up of many of the same TAFs as TFIID. Has varied subunits which contact activators, acetylate histones, and contact the core promoter/TBP.

SWI/SNF is an ATP-dependent remodeling complex which physically slides nucleosomes out of the way in order to activate transcription.

Chaperones such as FACT (act on H2A, H2B) and ASFI (act on H3, H4) promote dissociation of histone subunits in front of the elongating polymerase while other factors such as Spt6 reassemble nucleosomes behind it.

SIN3/NURD are co-repressors with histone deacetylase (HDAC) activity. HDAC activity, unlike HAT activity, is nonspecific and these co-repressors can deacetylate many different acetylated histones. NURD has two HDAC subunits as well as a remodeling subunit which uses ATP to slide nucleosomes around in order to repress transcription.

Other repressors include ModI and Not which can interact with TBP and prevent it from binding the TATA box. NC2 interacts with TBP to prevent the binding of TFIIB. Srb10/11 can be part of the mediator complex but are repressors: prematurely phosphorylate Pol II CTD at serine 2 and serine 5. This makes the

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Pol II act like an elongating polymerase so it is unable to be recruited to the promoter.

Nuclear cis splicing Most introns have several conserved sequences: the first two nucleotides of the

intron are almost always a GU; this is the 5’ splice site. The intron ends in AG; this is the 3’ splice site. There is also a highly conserved branch point (and a polypyrimidine tract) in the body of the intron. The branch point always contains a specific A which will eventually be “bulged” out.

The entire splicing process is catalyzed by snRNPs (small nuclear ribonuclear proteins): U1, U2, U4, U5, U6.

U1 associates with the 5’ splice site, a downstream sequence, and the 3’ splice site. (contacts the first and last nucleotides of intron)

U2 interacts with the branch point; base pairs with every nucleotide of the branch point except for a specific A which ends up bulging out due to lack of base pairing.

U4, U5, and U6 enter together as a complex. U4 is initially paired extensively with U6 but eventually rearranges and leaves once U6 begins interacting with U2. The 3’ end of U6 first base pairs with the 5’ end of U2; this is helix II. Once U4 leaves, U6 base pairs extensively with U2; forming helix I. Helix I is the catalytic center of the first transesterification (positions the branch point A for attack).

U5 has a loop sequence which interacts with the last few nucleotides of the 5’ exon; preventing it from floating away after the first step.

Actual reaction: two step transesterification: 2’ hydroxyl of branch point A nucleophilically attacks the 5’ splice site at the G, forming a 2’-5’ bond. This releases exon 1 (which now has a free 3’OH). The free hydroxyl of the 5’ exon then nucleophilically attacks the 3’ splice site. This releases the intrn in lariat form and ligates the two exons together.

Drosophila sex determination U2 can easily bind the yeast branch point (it is long enough) but in higher

eukaryotes the branch point is significantly shorter and not as conserved. U2 thus requires accessory factors (U2AF) in order to stably bind.

The nearby polypyrimidine tract is recognized by U2AF65, while the AG of the 3’ splice site is recognized and bound by U2AF35. The branch point is usually quite close to the 3’ splice site.

AG-dependent introns have short PPTs so U2AF65 binding is not very stable; cooperative binding of U2AF35 to the 3’ splice site is also required.

AG-independent introns have large PPTs so U2AF65 is bound stably enough to recruit U2 without requiring U2AF35.

In Drosophila, the msl-2 gene must be spliced in order to specifiy a male. The PPT is too far away from the 3’ splice site and the PPT is quite short; thus

both U2AF35 and U2AF65 cannot bind very stably to the intron. In females, SXL (sex lethal) binds to the PPT and easily outcompetes U2AF from binding. Thus, there is no splicing event in females. In males, there is no SXL and U2AF is able

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to bind eventually due to lack of competition; splicing results in determination of males.

hnRNP proteins can also bind nonspecifically to 5’ splice sites and block them from recognition by SR proteins. This can lead to selection of alternate splice sites.

Group I trans splicing Trans splicing involves the splicing of two separately transcribed RNAs into a

single product. Lower animals often have polycistronic mRNAs. Outrons are intervening

sequences between two cistrons (cistrons contain both introns and exons). The leader sequence, from a separate RNA, must be trans spliced onto the front of the operon promoter.

Outrons are similar to introns but lack 5’ splice sites. They still have 3’ splice sites and branch points.

Process: First U2AF binds to the polypyrimidine tract of the outron. All snRNPs are involved except for U1, since there is no 5’ splice site for it to bind to. U2 then binds to the branch point, U5 binds to the 5’ cistron and U6 binds to the 3’ splice site. The 2’OH of the bulged branch point A attacks the 5’ splice site of the splice leader RNA (a separate RNA). This forms a 2’-5’ covalent bond and a Y structure, as well as a free 3’OH on the liberated SL sequence. This free 3’hydroxyl of the SL RNA then attacks the 3’ splice site of the pre-mRNA, ligates to the cistron and releases the outron as a Y structure.

The SL RNA has associated proteins, including an Sm binding site. SL1 trans splices the first cistron of a polycistronic mRNA while SL2 is trans spliced onto every subsequent cistron. This is position dependent, based on the presence of polyadenylation sequences between cistrons, not sequence dependent.

RNA editing Mitochondrial genome of trypanosomes is composed of 20-50 maxicircles which

encode normal mitochondrial genes, and 10,000+ concatenated minicircles which encode guide RNAs.

Mitochondrial transcripts (mRNAs) require extensive addition and/or deletion of U’s in order to express functional proteins.

Guide RNAs consist of three regions: an anchor region which base pairs perfectly with the target mRNA, an editing region which is not perfectly base paired, and a poly-U tail which binds to the 5’ half of the mRNA after cleavage in order to prevent it from floating away.

Mechanism: There is always a mismatch between the guide RNA and the mRNA; the first step is endonucleolytic cleavage at this location, followed by either addition or deletion of U’s based on the complementary sequence of the editing region of the guide RNA (in order to base pair with either A’s or G’s). U’s are added the the 3’ end of the 5’ half of the mRNA. Guide RNAs act sequentially and relatively specifically; multiple guide RNAs edit the same mRNA strand at different regions.

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The entire process is catalyzed by the 20S editosome complex. This complex contains a terminal UTP transferase (a TUTase) which is responsible for the addition of U’s to the target mRNA. There is a helicase to unwind the guide strand from the target strand, there is an exoUase which is responsible for the deletion of U’s, there is RNase activity for the initial endonucleolytic cleavage, and there are two ligase subunits which reseal the cleaved mRNA after editing is complete.

Apolipoprotein B mRNA Apolipoprotein B has two main isoforms: the unedited full protein in the liver

(ApoB-100) functions to transport tryglycerides while the truncated form in the intenstines (ApoB-48) is used for absorption of lipids.

The truncated protein is created due to the emergence of a premature stop codon due to deamination of a C to form a U. This results in a UAA codon which terminates translation early; this occurs only in the intestine.

Mechanism: Acf (APOBEC complementation factor) binds to an upstream AU rich element and then recruits APOBEC (ApoB mRNA editing cytodine deaminase), which is a dimer. APOBEC is then positioned over the specific CAA codon and removes an amine group from the C in order to convert it into a U. This results in a premature stop codon on the ApoB mRNA and a smaller protein in the intestine.

Group I introns Group I introns are self-splicing; no snRNPs are required. The 5’ end of the intron

base pairs with the 5’ exon forming a double-stranded internal guide sequence. The MMM sequence of the intron can also base pair with the 5’ end of the 3’ exon. This results in folding so that the 5’ and 3’ splice sites are right next to each other. The intron has a guanosine binding pocket which contains an exonuclear (free) guanosine with a free 3’OH. This free hydroxyl attacks the 5’ splice site and then the 3’OH of the 5’ splice site attacks the 3’ splice site. The last nucleotide of the upstream exon is a U which forms a G-U wobble base pair (with the external G) When the 3’OH of the 5’ exon attacks the 3’ splice site, the intron is released and the two exons are ligated together.

Ribozymes RNase P processes pre-tRNA to mature tRNA by cleaving the 5’ end of the

tRNA. It is also responsible for cleaving the 4.5S RNA in E.Coli. RNase P is composed of one molecule of catalytic RNA and 1-10 associated proteins based on the species. The RNA portion of RNase P binds to the the tRNA stem and uses Mg2+ to cleave the 5’ end of the tRNA.

Hammerhead ribozymes are structures found during viroid replication. The viroid genome does not encode any proteins and replicates using a rolling circle mechanism, forming a long strand with many complementary repeats of the genome (concatemers). These individual genome sequences fold themselves out into hammerhead ribozyme structures and cleave out, requiring only Mg2+. The hammerhead structure has three stem regions, as well as ssRNA and dsRNA

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regions with lots of non Watson-Crick base pairing. An unusual cyclic 2’-3’ bond is formed during the cleavage event; there is also a specific C which is popped out into a catalytic pocket.

Telomerase is a catalytic protein with an RNA component and reverse transcriptase activity. The RNA template is complementary to telomeric repeats (has one and half copies of a telomeric repeat). The extra one half is used to base pair to the complementary repeat while the other whole copy is used as a template to reverse transcribe DNA, using the 3’ end of the DNA (chromosome) as a primer. Germ cells have long telomeres and high telomerase activity, stem cells have medium, while somatic cells have short telomeres and little to none telomerase activity.

5’ cap addition The 5’ cap is an inverted, methylated GTP (methylated on carbon 7) and bound to

the front of the mRNA through a 5’-5’ bond. The addition of the cap is catalyzed by enzymes associated with the CTD of Pol II, co-transcriptionally.

The first nucleotide of the RNA strand keeps 3 phosphate groups normally; an RNA triphosphatase then clips off the terminal phosphate of the 5’ nucleotide. RNA guanyltransferase then hydrolyzes a pyrophosphate off of a GTP, resulting in GMP. The remaining phosphate of this GMP then attaches to the 5’ end of the RNA. Next, methyltransferases methylate the cap structure; some are monomethylated while others are trimethylated.

CBC (cap binding complex) associates with the 7-methyl guanosine at the cap. CBC in the nucleus is later replaced by eiF4e in the cytoplasm. CBC promotes splicing by facilitating the association of the U1 snRNP with the closest 5’ splice site. CBC also promotes transport out of the nucleus.

Polyadenylation The hexonucleotide sequence AAUAAA signals for polyadenylation. Slightly

downstream of this is a CA polyA cleavage site, further downstream is a G/U rich sequence, and upstream of the polyadenylation sequence is a U rich sequence.

CPSF (cleavage and polyadenylation specificity factor) binds to the hexonucleotide sequence and has five subunits: 160, 100, 73, 30, FIP.

The downstream G/U rich sequence is bound by CstF (cleavage stimulation factor) which has a 64 kDA subunit which contacts the RNA and a 77 kDA subunit which contacts CPSF 160. CPSF and CstF have cooperative binding.

CFI and CFII (cleavage factors) also bind the RNA and are required for cleavage. Mechanism: CPSF and CstF are initially associated with the CTD of Pol II. After

proper signal (transcription of the polyA sequence), they dissociate and assemble onto the nascent mRNA. CPSF-73 is an endonuclease which cleaves the mRNA after the A on the downstream CA sequence. CPSF-73 requires proper binding and association with CstF and CFI, II in order to properly cleave. Poly A polymerase (PAP) is initially associated with the upstream U rich sequence. It has very low specificity (adds As slowly to almost anything) but once it is stimulated and directed by CPSF, it will slowly use ATP to add As and form the polyA tail after the cleavage site. After about 12 As are added, PAB (poly A binding protein)

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binds to the tail. This stimulates and speeds up PAP and also signals it to stop at the proper time. (PAB II is bound at this time, replaced by PAB I in the cytosol).

Cytoplasmic polyadenylation in Xenopus oocytes: transcription in oocytes occurs early and then halts until the 4,000 cell mid-blastula stage. mRNAs are sequestered in the cytosol; since there is no transcription there is no transcriptional control. Regulated instead by translation and cytoplasmic polyadenylation. RNA is first polyadenylated and then sent into the cytoplasm where the tail is shortened; not translated unless the tail is relengthened. Only mRNAs with cytoplasmic polyadenylation elements (CPEs) can undergo cytoplasmic polyadenylation and therefore become translated. The CPE (an AU rich element) is bound by CPEB (cytoplasmic polyadenylation element binding protein), which interacts with Maskin which interacts with EIF4E at the cap and prevents translation by preventing 4E from interacting with 4G. When it comes time for translation, however, CPEB gets phosphorylated and CPSF is recruited to the polyadenylation sequence. This recruits a cytoplasmic Poly A polymerase which lengthens the polyA tail. PAB I’s bind and interact with EIF4G, which interacts with 4E, which recruits the 40S ribsome. 4G+4E are now able to displace Maskin.

RNA transport Exportin-t binds to tRNAs and exports them out of the nucleus. Part of the

exportin superfamily. snRNPs have a monomethylated CAP bound by CPB, which interacts with

PHAX, which interacts with the exportin CRM1 and RanGTP. This exports the whole complex to the cytoplasm; following this RanGTP hydrolyzes its GTP to dissociate the complex. CBC then binds to importin alpha and beta to re-enter the nucleus. The reason snRNPs go to the cytoplasm is to “pick up” associated proteins. When Sm proteins bind to the Sm binding sites on snRNPs, the snRNP cap becomes hypermethylated. This signals for snRPortin-1 to bind which interacts with importin B, which results in reimport to nucleus. The only exception is U6, which is transcribed by pol III and never leaves the nucleus.

5S rRNA is present in the large subunit of the ribosome. It is exported differentially in somatic cells and in oocytes and it is transcribed by pol III.The 5S rRNA initially associates transiently with La. In oocytes, the 5S rRNA is transported into the cytoplasm, even though ribosome assembly occurs in the nucleolus. In oocytes, the 5S rRNA is sent into the cytoplasm for temporary storage. TFIIIA helps to bring in Pol III during transcription; TFIIIA also associates with the 5S rRNA to form the 7S RNP. This complex can interact with the CRM1 recepter to be transported into the cytoplasm. This 7S RNP is stored into cytoplasm until enough L5 (another ribosomal subunit) is translated. Once there is enough L5, L5 displaces TFIIIA and binds to the 5S rRNA. This forms an import complex which takes the 5S and L5 into the nucleolus for ribosomal assembly.

In somatic cells, however, L5 is already high and L5 in the nucleus can directly bind 5S rRNA before export and ship it to the nucleolus.

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mRNA export: most do not use a member of the exportin family. As RNA is spliced, EJCs (exon junction complex) are deposited 20-24 nucleotides upstream of where exons are ligated. Thus, there is one EJC for every intron that was removed. Aly/REF associates with the EJC. TAP(p15) then associates with Aly/REF. TAP targets the mRNA to the nuclear pore for export.

Some mRNAs do use exportins, however. AU rich element containing RNAs (ARE) have unstable AU elements in their 3’ untranslated regions. These serve as binding sites for HuR which can interact with Trn2 (an exportin). After export, secretory protein encoding RNAs are sent to the ER while others are translated by free ribosomes.

Deadenylation dependent decay The 5’ UTR or leader sequence is the region between the cap and the initiator

codon on an mRNA, and the 3’ UTR is between the stop codon and the poly-A tail. Deadenylation dependent decay requires that the poly-A tail be shortened by deadenylases.

The major deadenylase in yeast is the CCR4-Not complex while the mammalian deadenylase is PARN.

PAB, when bound to the poly-A tail, inhibits CCR4. However, natural fluctuations of the mRNA eventually result in one or two nucleotides being exposed. These exposed nucleotides are “nibbled” away by the deadenylase, until the PAB falls off. (Each PAB requires at least ~12 nucleotides to bind stably). Eventually, all of the PABs will fall off after a critical short length (when the poly A tail is shorter than 12 nts). This triggers decay: first ribosomes stop loading onto the RNA which signals Pat I to recruit DHH1 (an RNA helicase), which recruits LSM1-7, a 7 member ring complex, which recruits the decapping complex. The decapping complex is composed of DCP-1 and DCP-2, 1 is a regulatory subunit. DCP is assisted by EDC 1, 2, 3 (enhancers of decapping), it then removes the 5’ cap, rendering the mRNA susceptible to degradation by Xrn1. Xrn1 is a very powerful 5’-3’ exoribonuclease. Another possibility is that after the last PAB falls off, the exosome (a complex with many 3’-5’ exoribonucleases) degrades the mRNA from the end. DCP-S then cleaves off the cap of the small fragment leftover by the exosome complex.

P-bodies (processing bodies) are clusters of RNA and degradation machinery found throughout the cell. Size of P-bodies depends on the amount of RNA in an untranslated state. If translation is inhibited, then P-bodies grow larger as more mRNAs are being degraded. RNAs destined for P-bodies must be void of ribosomes.

The presence of PAB inhibits the decapping complex because PAB interacts with EIF4G which interacts with EIF4E which interacts with the cap. This stabilizes 4E’s binding to the cap, preventing decapping. The presence of PAB also indicates that there are active transcribing ribosomes, making degradation impossible.

Deadenylation independent decay

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Nonsense mediated decay: since there are lots of splicing events, the possibility for error is relatively high. A premature stop codon means that the mRNA must be degraded because a truncated protein can be deleterious.

Every removed intron is marked with an exon junction complex. EJC consists of a core of four proteins (EIF4A3, MLN-51, Y14, magoh) Upf2 and Upf3 are also part of the EJC (Upf2 joins in the cytoplasm, part of the cEJC)

During translation, ribosomes knock off EJCs. Normally all EJCs are knocked off, but in the presence of a premature stop codon the ribosome will stall and be unable to knock off downstream EJCs.

During termination, ERF1 and 3 are release factors. Normally, ERF3 interacts with PAB (because last stop codon is usually close to the poly A tail) and this promotes efficient termination.

A PTC however, results in inefficient termination because PAB is too far away. ERF3 instead interacts with Upf-1, Smg1 kinase, and ERF1; forming the SURF complex. Upf2 and 3 are on the (downstream) EJCs; in the case of a PTC, Upf1 on the SURF complex interacts with Upf2 and 3 on one of the remaining EJCs. This signals Smg1 kinase to phosphorylate Upf1 which triggers the recruitment of additional Smg factors which results in mRNA decay.

Yeast do not have EJCs, Upf 2 and 3 are instead free floating but the mechanism is the same: ERF3 can either bind to PAB (normal termination) or Upf1 which recruits Upf2 and 3 resulting in inefficient termination, recruitment of decapping enzymes, and decay.

Other forms of deadenylation independent decay : Rps28B RNA, endonucleolytic cleavage, no-go decay (ribosomes stall at

secondary structure), and non-stop decay (no stop codon, ribosomes stall at the end and knock off PABs, Ski7 tRNA mimic)

Regulated mRNA stability ARE RNAs and stuff. Synchronous (distributive) vs desynchronous (processive)

deadenylation. If there is a whole question dedicated to this then I will punch a kitten.

Cell cycle regulated histones Each histone protein is encoded by two separate genes: a normal one and a cell-

cycle regulated one. During S-phase there is a high demand for histone protein to bind to all of the newly synthesized DNA; during this time the half-life for cell cycle regulated histone mRNA jumps from 10 minutes to 40 minutes.

These cell-cycle regulated histone mRNAs have no poly-A tail, instead they have a 3’ stem loop which is highly conserved and is 20-40 nucleotides downstream of the stop codon.

Mechanism for stability increase: U7 snRNP has a region which base pairs to the histone downstream element (HDE) which is 9-14 nucleotides upstream of the stem loop. SLBP (stem loop binding protein) is bound to the stem loop and interacts with U7 through ZFP-100. U7 is also bound to a 7 member Sm ring complex which stabilizes binding to ZFP.

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Upstream of the loop is a sequence which interacts with the U2 snRNP (brought in via protein-RNA contacts, not RNA-RNA like in splicing). U2 interacts with the SF3B complex. U2 ends up recruiting CPSF (directly to CPSF-100) and also recruits CstF and Symplekin. CPSF-73 cleaves the mRNA downstream of the stem loop, just upstream of where U7 was bound. Then an exonuclease trims off a bit. This shortened 3’ UTR results in a twofold increase in stability.

At the end of S phase, a TUTase adds a few Us to the 3’ end of the RNA which recruits LSM which results in decapping, or it can also recruit the exosome which has a high affinity for RNA with long stretches of Us.

Translation initiation Prokaryotes: mRNA is often polycistronic. Ribosome has 30s and 50s subunits.

The 30s subunit can “see” the RNA in two dimensions and can therefore bind to any of the cistrons. It does not have to start with the first cistron. The 30s subunit initiates at the proper AUG start codon due to its 16s rRNA component. The 16s rRNA has a sequence at its 3’ end which is complementary to the Shine-Delgarno sequence (AGGAGG) which is upstream of every start codon. The rest of the 16s rRNA is full of stem loops and base pairing, but the 3’ end is single stranded and free to bind to the S-D sequence. The strength of a particular start codon is determined by its distance from the S-D sequence and how close the S-D sequence is to the consensus AGGAGG. Varying distances (usually 5-10 nucleotides) between S-D sequence and start codon can result in more translation for one cistron vs another.

Eukaryotes: mRNAs are almost always monocistronic. There is also no Shine-Delgarno consensus sequence. Instead there is a conserved Kozzak sequence:

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 +1 +2 +3 +4A/G A/G C C A/G C C A U G G/A

Mutations of the A/Gs have the most effect. Better context initiation codons are closest to the Kozzak consensus sequence. However, the 40S ribosome does not interact with the RNA through base pairing. The 18s rRNA (the eukaryotic homolog for the 16s rRNA) is missing the 3’ region which would normally bind a consensus sequence. Instead, the 40s ribosome binds at the cap structure and scans down the 5’UTR in one dimension until it reaches an AUG with good context. The presence of secondary structure near a start codon can also aid its selection by causing the 40s ribosome to pause over the codon.

Initiation mechanism: The 40S ribosome is sensitive to secondary structure and can often get stuck. Initially, eIF4F (which consists of eIF4E bound to the cap, the larger eIF4G, and eIF4A) interacts with the mRNA. eIF4G brings in 4A; eIF4A is a helicase which melts secondary structure in the 5’UTR. 4A is stimulated by eIF4B.

Meanwhile, the 40s ribosome is associated with eIF3 and eIF1a. eIF3 binding to 40s prevents the 40s from binding to the 60s and forming the full 80s complex prematurely. Next, a ternary complex composed of eIF2, GTP, and Met-tRNA

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bind to the 40s subunit. The 40s ribosome with eIF2, eIF3, GTP, Met-tRNA, and eIF1a is also known as the 43s complex.

The 43s complex is recruited to the mRNA through interaction between eIF3 and eIF4G. This recruitment only happens after eIF4A melts the secondary structures. Thus, the 40s ribosome subunit is almost always recruited to wherever eIF4G happens to be, and since 4G interacts with 4E at the cap, this means that the 40s ribosome is almost always recruited directly to the 5’ end of the RNA.

eIF5 also interacts with eIF2 and eIF3, stabilizing them. The 40s ribosome will then scan down the mRNA and once an initiation codon is recognized, eIF5 triggers eIF2 to hydrolyze its GTP to GDP. eIF1a and eIF1 are important in recognizing the proper AUG (the met-tRNA is also very important for this). eIF2’s GTP hydrolysis triggers dissociation of all of the factors. eIF5B is a GTPase which promotes recruitment of the 60s subunit; this forms the 80s ribosome. GTP is hydrolyzed after the full 80s ribosome is formed. Now eIF2B recycles the GTP attached to eIF2 so that it may bind another met-tRNA to form a new ternary complex for another round of initiation.

Translation elongation The ribosome has 3 sites: P (peptidyl), A (acceptor), and E (exit). The initial

tRNA with the peptide chain is in the P site. eEF1a is bound to a charged tRNA and a GTP. It brings tRNAs into the A site. If the anticodon on the tRNA fits (is complementary to) the codon on the RNA in the A site of the ribosome, then eEF1a will hydrolyze its GTP and leave the tRNA in the A site. eEF1b will then recycle the GTP so that eEF1a can bind a new charged tRNA.

The next step is peptidyl transfer; the nascent peptide from the P site tRNA is completely transferred over onto the amino acid on the new tRNA in the A site. This increases the length of the nascent peptide by one amino acid. This entire step is catalyzed by the 23s rRNA (part of the large subunit of the ribosome). The 3’ ends of the A site and the P site tRNAs are positioned very close to one another due to the ribosome structure. The nitrogen of the A-site amino acid terminal amino group is positioned close to the carbonyl carbon of the first P-site amino acid. Adenosine2951 of the 23s rRNA, due to its environment, is particularly basic and is close to the A-site amino acid amino group. This A then pulls off a proton from the amino group on the A-site amino acid, resulting in a negatively charged nitrogen which nucleophilically attacks the nearby carbonyl carbon of the P-site amino acid. This breaks the bond between the nascent peptide and the P-site tRNA and attaches it onto the A-site amino acid. Thus, the ribosome functions as a ribozyme.

Following peptidyl transfer, the ribosome translocates down the mRNA by 3 bases, this is catalyzed by eEF2 which hydrolyzes GTP in order to do this. This translocation results in the A-site tRNA (with the newly added peptide) to be moved to the P-site, and the old (now uncharged) P-site tRNA is in the exit site. This frees up the A-site for a new charged tRNA to be brought in by eEF1a.

Termination: There is no tRNA which has an anticodon complementary to a stop codon. Instead, eRF1 (a tRNA mimic) binds to the A site. eRF3 hydrolyzes GTP and interacts with eRF1, eventually transferring the nascent peptide from the P

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site tRNA onto water. This releases the peptide. eRF1 can physically recognize any of the three stop codons; eRF1, eRF3, and GTP form a ternary complex.

Global translation regulation Can occur due to starvation, heat shock, etc. The two key points where global

translational regulation can occur are eIF4E and its binding to the cap and eIF2 and its GTP hydrolysis and ability to bring in met-tRNA.

eIF4G has three main domains: CTD, NTD, and middle domain. The NTD is bound by eIF4E and PAB. eIF3 (and therefore the 40s ribosome) interacts with the middle domain, and eIF4A interacts with the middle domain and the CTD. The CTD also interacts with Mnk kinase. Mnk phosphorylates the eIF4E which is bound to the same 4G on serine-209, this tightens the binding of 4E to the cap and therefore increases translation. 4E is thus only phosphorylated when it is associated with the rest of the 4F complex. This phosphorylation is heavily regulated; Mnk is regulated by many signal transduction pathways.

One method of repressing translation is 4E-BP which (when 4E-BP is unphosphorylated) binds to eIF4E and prevents it from binding to 4G. When 4E-BP is phosphorylated by kinases such as Frap/mTOR then it cannot inhibit 4E. Frap/mTOR can also phosphorylate the NTD of eIF4G, opening up the structure and activating translation further.

mRNAs compete for eIF4E, since it is one of the rarest translation factors. Poor competing mRNAs evolved that way in order to control amount of protein expression; not all mRNAs need to always be translated at the maximum rate. Growth factor RNAs, for example, have long leader sequences with lots of secondary structure. This reduces the ability of 4E to bind and decreases the amount of growth factor proteins.

PAB can also interact with the NTD of 4G. PAB binding also serves to increase the binding of 4E to the cap, resulting in mutual stabilization. PAB can exist in multiple phosphorylation states: phosphorylated PAB prefers to interact with 4G while non-phosphorylated PAB prefers to interact with eIF4B.

Picornaviruses Viruses in this family (including poliovirus and rhinoviruses) have a plus sense

ssRNA genome. The genomic RNA (which functions as an mRNA) has a polyA tail but no 5’cap. Instead there is a viral protein (VpG) bound to the 5’ end of the RNA, but this is removed upon viral entry into the cell, resulting in just a U at the 5’ end. The 5’ UTR is also very large (800 nucleotides compared to 50-100 for normal ones) and contains extensive secondary structure as well as many premature AUGs, some of which have good context. This completely prevents any type of normal translation initiation.

Since there is no 5’cap, eIF4E cannot bind and even if it could, there is too much secondary structure for eIF4A to melt so the 40s ribosome could never reach the ORF. Instead, there is an IRES (internal ribosome entry site) in the 5’UTR which directly recruits eIF4G through its middle domain. eIF4G can then recruit eIF3 which is attached to the 40s ribosome. This process ends up recruiting the 40s

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ribosome right on top of (or directly upstream of) the actual start codon for the ORF, resulting in translation.

This translation results in the synthesis of a single viral polyprotein, which is cleaved out by viral proteases 2a and 3c. 2a can also target eIF4G, cleaving it between the NTD and the middle domain. This relieves almost all competition between the viral RNA and cellular mRNAs. This is because host mRNAs recruit eIF4G through its NTD interaction with eIF4E, therefore at this point host mRNAs will all recruit a useless 4G NTD. Meanwhile, the viral IRES will recruit the rest of the eIF4G protein, including the middle domain which is necessary to recruit eIF2 and the 40s ribosome. Picornaviruses infection can also result in dephosphorylation of eIF4E-BP, resulting in sequestering of 4E away from host mRNAs so that they don’t even get a chance to recruit the useless 4G NTD.

The host cell evolved a defense mechanism in the form of PKR. PKR is a kinase which autophosphorylates and activates upon binding to dsRNA (the replication intermediate for the viral genome is double stranded RNA). Upon activation, PKR phosphorylates eIF2 on serine-51 of the alpha subunit, this regulates eIF2 by “trapping” it together with eIF2B (which normally recycles eIF2’s GTP) to form a nonproductive complex. Eventually all of the useful eIF2 is depleted and protein synthesis is completely shut down (including viral protein synthesis which does rely on eIF2).

Viruses evolved a counter-strategy in the form of binding PKR and sequestering it with viral proteins. Rhinoviruses can also create small pieces of dsRNA which clog up PKR and prevent it from being activated. Mammals still haven’t evolved a counter-measure because we’re weak as hell so instead we get vaccinated or just die.

Cricket paralysis virus mRNA has a polyA tail and a 5’cap. The mRNA is also bicistronic, the first

cistron encodes viral replicase (RNA dependent RNA polymerase) and the second ORF encodes coat proteins.

Initially, only the first ORF is translated. The synthesized viral replicase then quickly makes copies of the genomic RNA before PKR becomes activated in response to the dsRNA intermediates. Once PKR is activated, eIF2 becomes phosphorylated and trapped so that translation is globally repressed.

During this time, the second ORF gets expressed. This is possible because there is an IRES in between the two open reading frames. The IRES folds up to resemble a tRNA and even has an “anticodon” stem loop (GGA) which base pairs with a downstream CCU in the intercistronic sequence. The ribosome is directly recruited and this CCU and the “tRNA” are now in the P-site. The open A-site is then bound by a real charged tRNA (almost always alanine because the next codon is usually GCU) After this, peptidyl transfer and elongation proceeds as normal and coat proteins are synthesized. These coat proteins and the previously replicated viral ssRNA genomes are then assembled into virion particles.

Regulation of iron homeostasis

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Levels of iron in the cell must be tightly regulated in order to maintain proper conditions. Transferrin receptors are embedded in the plasma membrane in order to intake transferrin (which is bound by two molecules of ferrous iron) when the cell requires iron. When the cell has excess iron, ferritin protein is expressed in order to bind the extra iron and sequester it.

Ferritin RNA has a conserved stem loop in the 5’ UTR. This is known as an IRE (iron response element). IRE-BP is a protein which binds to IRE’s only when there is very low iron. This is because when IRE-BP has iron in its core, it functions as aconitase. However, when iron is low, it loses its iron core and the 4th

domain folds out from the other three domains via its hinge domain. This opens up the IRE-BP structure and allows it to bind to IRE stem loops. Thus, when iron is very low, IRE-BP binds to the stem loop IRE in the 5’ UTR of ferritin mRNA. This stabilizes the stem loop structure and prevents the 40s ribosome from being recruited and therefore prevents translation of ferritin. Once iron levels increase, however, IRE-BP dissociates from the stem loop and translation proceeds as normal. Thus, ferritin is only expressed under conditions of high iron; which makes sense because ferritin is used to sequester excess cellular iron.

Transferrin receptor mRNA, on the other hand, is only translated under iron starvation conditions. The 3’ UTR has multiple IRE stem loops. Normally there is an exposed cleavage site which results in endonucleolytic cleavage and degradation of the mRNA. However, under low iron conditions, these IRE’s are bound by IRE-BPs. These proteins hold the stem loops in a specific conformation which stabilizes the mRNA and hides the cleavage site so that the transferrin mRNA ends up being translated. If cellular iron levels increase then the IRE-BPs will fall off and cleavage will occur. Thus, transferrin receptor protein is only expressed during conditions of low iron, which makes sense because transferrin receptors are used to intake extracellular iron (attached to transferrin proteins).

GCN4 During normal conditions, amino acid biosynthesis in yeast is repressed.

However, under starvation conditions, this pathway must be activated for survival. GCN4 is a transcriptional activator which activates amino acid biosynthesis.

The GCN4 mRNA has four small open reading frames upstream of the actual ORF encoding the protein. These small ORFs do not encode any functional protein.

Under non-starvation conditions: Translation initiates as normal at the 5’ end of the mRNA. The 40s ribosome scans down the leader sequence until it encounters the first small ORF. eIF2 (which brought in the met-tRNA) hydrolyzes its GTP and dissociates once its met-tRNA base pairs with the start codon. The 60s ribosome is recruited and the first small ORF is translated. Following translation of the first small ORF, all of the 60s subunits and half of the 40s subunits dissociate from the mRNA. The remaining 40s subunits continue scanning down “blindly” since they do not have an eIF2 or met-tRNA/GTP bound to provide specificity. Therefore, the 40s ribosomes scan past the second and third small ORFs. However, under non-starvation conditions a new eIF2 ternary complex

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almost always binds by the time the 4th ORF is reached. The 4th ORF is translated normally, but this time all of the ribosomes completely dissociate due to a downstream termination sequence. Therefore under non-starvation conditions, the actual GCN4 ORF is never reached and the protein is never synthesized.

Under starvation conditions: GCN2 (a kinase) has a histidine amino acyl tRNA synthetase-like domain which allows it to bind uncharged tRNAs and activate. Under amino acid starvation conditions, there is a buildup of uncharged tRNAs so GCN2 is activated in large quantities. GCN2 phosphorylates eIF2 on the alpha subunit at serine-51 (just like PKR); this causes eIF2 and eIF2B to be “trapped” together in a nonfunctional complex. This serves to greatly reduce the amount of available eIF2 in the cell. So under starvation conditions the first ORF of GCN4 is translated normally and 50% of the 40s ribosomes dissociate while the rest continue scanning blindly, waiting for an eIF2 ternary complex to bind. However this time a new eIF2 does not bind until much later; the ribosome continues to scan way past the 4th small ORF and past the termination sequence. A new eIF2 ternary complex does not bind until the 40s ribosome reaches the start codon for the actual ORF encoding GCN4. Thus, GCN4 is expressed and goes on to activate amino acid biosynthesis.

RNAi siRNAs are formed after dsRNA is cleaved by a protein called Dicer. This

processesing results in a 21 nucleotide long dsRNA duplex with a 3’OH and two unbased nucleotides (overhangs) at each 3’ end, as well as free 5’phosphates. Dicer can use any dsRNA as a substrate. Dicer is also coupled with an RNA binding protein called R2D2. Dicer has a PAZ domain with a cleft to fit the 2 base overhangs of the dsRNA, as well as two RNase III-like domains which are offset from one another by a distance of 2 nucleotides to create the overhangs, as well as its own RNA binding domain (separate from R2D2). The processed siRNA duplex is delivered/loaded onto the RISC complex by Dicer and R2D2, which themselves become part of the complex. Only one of the original siRNA strands remains associated with the complex; usually the one with the most unstable 5’ end. This is usually the strand with the 5’ end not buried in the Dicer PAZ domain. The passenger strand then dissociates and the remaining strand becomes the guide strand, which provides specificity to the RISC complex. The passenger strand is separated out by Dicer’s internal helicase activity.

The RISC holoenzyme also contains Argonaute (previously known as Slicer) which cleaves RNA which base pairs with the newly incorporated guide strand. Once the guide strand binds with the target mRNA, Argonaute uses its RNase-H like domain to cleave the target strand but not the guide strand.

miRNAs, unlike siRNAs, are actual gene products encoded by the cell. They are Pol II transcripts. The primary transcript (pri-miRNA) folds up into many secondary structures with a stem loop at one end and many bulges. These pri-miRNAs serve as substrates for dicer-like proteins (DCL1 in plants or Drosha in fruit flies). The first cleavage event cleaves off one end of the pri-miRNA resulting in a pre-miRNA. The pre-miRNA then binds to exportin-5 in order to be transported to the cytoplasm. There a cytoplasmic dicer cleaves the stem loop end,

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resulting in a mature miRNA/miRNA* duplex. (in plants both processing events occur in the nucleus and the mature duplex is shipped out into the cytoplasm by binding Tasty, an exportin) A helicase unwinds the two strands and the guide strand is incorporated into the RISC complex. Once incorporated, the miRNA strand functions in much the same way as siRNAs; leading the RISC complex to specific RNAs due to complementary base pairing. The main difference, however, is that miRNAs often do not have perfect base pairing with their target mRNAs, while siRNAs do. This results in translational repression (blocks ribosomes) instead of cleavage; although this is not always the case.

Short RNAs (miRNA or siRNA) can also direct the RISC complex to chromatin and bind genomic DNA in order to recruit methyltransferases to methylate histones and cause transcriptional repression. This is often used to silence transposons. Both siRNAs and miRNAs are methylated at 2’OHs by Hen1 in order to protect against degradation.

Natural siRNAs are formed when, during regular transcription of a gene, the opposite strand can also be transcribed. This results in short RNA strands which are complementary to the functional mRNA, resulting in regions of dsRNA. This is processed by Dicer to form siRNAs. RNA dependent RNA polymerases (RDRPs) can also elongate the initial mRNA/nat siRNA duplex longer so more siRNAs are formed by Dicer.

Trans acting siRNAs (Tas) are Pol II transcripts which are sent into the cytoplasm to be base paired by miRNAs. This results in cleavage of the Tas RNA into small fragments; these small fragments are used as substrate by RDRPs to form dsRNAs. These are processed by Dicer into multiple different siRNAs with diverse mRNA targets.

Repeat-associated siRNAs are formed when methylated DNA is transcribed and the transcript becomes dsRNA due to the action of an RDRP. This is processed by Dicer and incorporated into RISC to go back and retarget the original DNA in order to recruit more methyltransferases to make sure that transcriptional repression is maintained.

Protein targeting and transport Proteins can be targeted to the nucleus, mitochondria, chloroplasts, ER,

peroxisomes, or secreted out of the cell. Nucleus: Proteins targeted to the nucleus require an NLS (nuclear localization

sequence). The NLS is a short stretch of basic amino acids which is recognized and bound by importin alpha, which is also bound by importin beta. Importin B can then interact with cytoplasmic filaments of the nuclear pore and cause the protein to be “dunked” through the pore into the nucleoplasm. Upon arrival, the protein dissociates from the importins. Ran is important for directionality; RCC (Ran-GEF) is in the nucleus and regenerates Ran-GTP so there is high Ran-GTP in the nucleus. Ran-GAP in the cytosol promotes GTP hydrolysis so that there is high Ran-GDP in the cytoplasm.

Endoplasmic reticulum/secretory pathway : the luminal space of the ER membrane system is in contact with the nuclear envelope. There is also indirect interaction between the ER and the Golgi via trafficking for the secretory pathway. All

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secretory proteins are synthesized by ribosomes on the ER (the “rough ER”). This is because secreted proteins often require disulfide bridges and the luminal space of the ER has an oxidizing environment which promotes such structures. Proteins targeted to the ER have an N-terminal signal peptide which is cleaved off only once the protein is transported/translated into the ER. Proteins can then have additional signal sequences which promote transport to other subcellular locations such as vacuoles or lysosomes; however all secretory pathway proteins, regardless of where they end up first go to the ER. Mechanism: First, RNAs start being translated on free floating ribosomes. Once the ribosome synthesizes about 60-75 amino acids, about 25 amino acids are exposed from the ribosome. These 25 amino acids are the most N-terminal residues, and they make up the signal peptide. This signal peptide contains targeting information; it is recognized by the signal recognition particle (SRP). The signal peptide interacts with SRP54 as well as SRPs RNA component (the 7S RNA). SRP binding to the signal peptide causes translational arrest; this is due to the action of the 7S RNA and the SRP9/14 protein subunits. SRP then docks on to the SRP receptor on the endoplasmic reticulum membrane. SRP receptor has an alpha subunit which is exposed to the cytoplasm and a beta subunit which is embedded in the membrane. The SRP receptor then delivers SRP (along with the arrested ribosome) to a translocon on the ER membrane. The translocon is a pore system (also known as Sec61 complex, with 3 subunits: alpha, beta, gamma). This causes the large subunit of the ribosome to dock on the cytosolic face of the translocon. Next, the signal peptide which was bound to SRP is transferred to bind to the translocon. The signal peptide is released into the pore and SRP and the SRP receptor all hydrolyze GTP and dissociate. This allows the ribosome to resume translation and the nascent peptide is transported through the translocon into the ER as it is synthesized. As the peptide emerges in the luminal space, a signal peptidase cleaves off the signal peptide.

Some proteins (rare) are translated completely in the cytoplasm and then sent into the ER as mature proteins. These proteins must associate with molecular chaperones such as hsp70 which help to unfold the protein and thread it through the ER translocon. The luminal space has a different environment from the cytosol so translocons must be sealed in order to prevent leakage. Normally, the ribosome seals off the cytosolic side so this is not an issue, but in the case of post-translational translocation the proteins must associate with the Sec63 complex (a different translocon) and be threaded through by ATP-dependent chaperones on the luminal side which associate with the NTD of the protein.

If proteins in the ER cannot fold/refold properly then they must be sent out of the ER for degradation since there is no proteolytic degradation machinery in the ER. These proteins are threaded through translocon into the cytosol where Cue1 will ubiquinate the NTD as it emerges, recruiting the 26s proteasome and causing complete degradation.

Vesicles are formed through the action of coat proteins which surround/associate with a section of membrane (ER membrane, Golgi membrane, plasma membrane, etc). These coat proteins interact with adaptors which interact with membrane embedded receptors. Coat proteins are recruited by GTPases such as ARF (ARF

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only recruits when it is bound by GTP). The coat proteins then form a shell around the membrane portion, which is then “pinched” off the membrane. The coat proteins then dissociate very quickly. The vesicle membrane has important targeting proteins called vSNAREs (vesicle SNAREs) which can interact with complementry tSNAREs (target SNAREs) on the target membrane. Only the appropriate vSNARE will interact with the appropriate tSNARE to form a complex. NSF (N-sensitive factor) and SNAPs also associate with the vSNARE/tSNARE complex. NSF is an ATPase which hydrolyzes ATP to dissociate the complex after cargo is dumped. Rab is another GTP binding protein which interacts with the vesicle as it is budding off. Set-1 is associated with the tSNARE and helps keep vSNAREs away; Set-1 is overcome by Rab. This ensures that only the appropriate vSNARE and tSNARE interact.

Targeting the mitochondria: The mitochondria has a double membrane; the inner membrane is larger so it has invaginations in order to fit. There are contact points between the two membranes which contain pores for transport. Out membrane pores are Toms (transport outer membrane) and inner membrane channel proteins are Tims (transport inner membrane). Proteins destined for the mitochondria often interact with chaperones such as hsp70 to keep them unfolded. The NTD of the protein has targeting info, it interacts with Tom20/22 which deliver the protein to the proper channel on the outer membrane to be threaded through. Some proteins however, require a mitochondrial import stimulation factor (MSF). MSF associates with these proteins and delivers them to Tom70/37 which delivers them to Tom20/22. Tom 22 has an acidic face which allows it to interact with the positive charged NTD signal sequence on the protein. The MSF pathway requires ATP hydrolysis to release MSF from the protein, while the MSF-independent pathway does not require ATP. After the protein is threaded through into the mitochondrial matrix, there is a signal peptidase which cleaves off the NTD.

Targeting the chloroplast: The chloroplast is also double membraned. Outer membrane channels are Tocs, inner membrane channels are Tics. Proteins interact with chaperones such as hsp70 or 14-3-3 and are threaded through Tocs and Tics. There is also a signal peptidase in the stroma which cleaves off the NTD signal peptide. Proteins destined for thylakods have additional targeting information.

Protein splicing Inteins are cleaved out to leave exteins, similar to how introns are cleaved out to

leave exons in RNA splicing. However, inteins do not have many conserved sequence elements. Inteins are generally 134-500 amino acids long. The first amino acid of the intein is always a cysteine or a serine while the last two amino acids are always histidine followed by asparagine. There is also a threonine and histidine motif conserved near the N-terminal end of the intein. The first amino acid of the C extein is always serine, cysteine, or threonine.

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Inteins, along with Group I introns, are products of selfish genetic elements. Group I introns (after being spliced out) can actually have an open reading frame which encodes an endonuclease. Spliced out inteins can also refold to become functional endonucleases.

This ability is found often in single celled organisms which conjugate in order to reproduce. For example, if the recipient cells gains two copies of the same gene, one with an intron/intein and the other without (if the donor cell had the copy with the intron while the recipient cell copy did not): the endonuclease encoded by the intein/intron can cleave the DNA of the gene from which it originated but only if the gene does not have the intron/intein. This is because the intron/intein masks the cleavage site. After cleaving the intronless gene copy, there is a double stranded break and the “in tact” gene copy (which contains an intron or intein sequence) is used as a template to repair the break. This results in both copies of the gene ending up with the group I intron or intein sequence. This is how these selfish genetic elements end up spreading.

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The first step is an N-O shift. The OH of the first serine of the intein (or the SH if the first residue is a cysteine) nucleophilically attacks the carbonyl carbon of the last amino acid of the N-extein. This shifts the N-extein to an ester linkage on a serine side chain instead of being attached via a normal peptide bond.

The second step is transesterification. The first serine of the C-extein attacks the carbonyl carbon of the ester formed by the first step.

In the third step the last amino acid of the intein, asparagine, cyclizes. This occurs because the terminal NH2 of the asparagine side chain attacks its own carbonyl carbon in order to break the peptide bond. This releases the intein with a branched asparagine ring structure at the C-terminal end and leaves the two exteins covalently attached to each other via an ester bond.

In the final step, an O-N shift occurs and the ester linkage formed in the first step is restored to a peptide bond, the serine R group is restored, and the two exteins are now ligated together and functional.

Trans protein splicing: two separately translated exteins can be fused/spliced together by attaching half of an intein to each extein. These inteins will come together and splice out, ligating the two separately translated exteins together to form a functional protein. This can be useful for protein domain “shuffling”, resulting in alternate protein formation.

Prion disease Lesions occur in brain due to aggregation of prion protein. Native protein

structure is 4 alpha helices, 2 in front and 2 in back. However, the scrapy conformation of the prion protein has two of the alpha helices unfolded to become Beta sheets. These sheets cause aggregation of multiple proteins. PrpSC can bind to a normal PrpC and promote unfolding so that the PrpC becomes a PrpSC (scrapy conformation)

Apoptosis Programmed cell death which begins with condensation of the nucleus due to

DNA fragmentation, followed by complete fragmentation of the nucleus, followed by fragmentation of the cell through vesicles, and ultimately complete degradation/necrosis of the cell.

The entire process is mediated by caspases (cysteine containing aspartic acid proteases) which can cleave many targets required for apoptosis, including nuclear proteins and endonucleases which become activated. Caspase precursors (procaspases) contain long prodomains at the NTD which is cleaved out to form a mature caspase when it is time for apoptosis. Procaspases also have a cleavage site between their large and small subunits, ultimately resulting in functional heterotetramers with two copies of each subunit. Divided into initiator caspases (8, 9, maybe 2 and 10) and effector caspases.

Mitochondrial activated apoptosis: insults to the mitochondria result in swelling or open channels, resulting in release of cytochrome c. Cytochrome c then binds to Apaf-1 (apoptotic protease activating factor) on its A domain which contains WD40 repeats. This changes the conformation of Apaf-1 so it can use ATP to dimerize through its NOD (nucleotide binding dimerization domain). Apaf-1 also has a 3rd domain called the CARD (caspase recruitment domain) which can interact with the CARD region of the prodomain of procaspase 9. Multiple molecules of procaspase 9 are recruited. These procaspases are not completely inactive; when two are brought together in close proximity by Apaf-1, their weak proteolytic activity is enough for the procaspases to cleave and process each other, resulting in formation of mature caspase 9 proteins. These initiator caspases then go on to activate effector caspases to induce apoptosis.

Receptor mediated apoptosis: The external side of the plasma membrane has Fas and TNF receptors. When the proper ligand (Fas or TNF) binds to the receptor, the receptor will oligerimize. The cytosolic side of the receptor has a death domain which can interact with the death domain of the FADD adaptor protein (fas associated death domain). FADD has a separate death effector domain which can interact with the death effector domain in the prodomain of procaspase 8. Multiple procaspase 8s assemble through association with membrane bound

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FADD adaptors. The procaspases cleave and process each other and result in active caspase 8 proteins.

Initiator procaspases have very long prodomains which are necessary to interact with adaptor proteins. Effector procaspases have short prodomains which are simply cleaved out by activated initiator caspases. The active effector caspases can then go on to begin the execution phase.

There is cross-talk between the two pathways. Caspase 8 can activate/cleave a protein called Bid. Bid can then bind to the mitochondrion and stimulate the release of cytochrome c. There are pro-apoptotic and pro-survival members of the Bid family.

DNA cleavage occurs due to activation of CAD (DFF40) which is an endonuclease. CAD is normally inhibited by ICAD, but caspase 10 can cleave ICAD and relieve this inhibition. This results in activation of CAD which goes on to cleave genomic DNA to result in 180 bp fragments (cleaved around histones). eIF4G is also cleaved by caspases between the NTD and MD. This prevents translation of most mRNAs but promotes translation of cellular mRNAs which contain an IRES.

BCH 212Traugh:

cAMP-dependent protein kinase (PKA) The general kinase reaction (for all kinases): the protein kinase cleaves a

phosphate group off of ATP (or possibly GTP in the case of Casein kinase 2) and forms a covalent ester linkage between the freed phosphate and the hydroxyl group of serine, threonine, or tyrosine side chain on the substrate protein. This introduces a site specific negative charge and site specific additional mass which almost always alters the activity of the substrate in some meaningful way.

Structure of PKA: R2C2 – two regulatory subunits which are similar but separate gene products. Each R subunit contains a cAMP binding site, a dimerization domain, and a C binding domain. The two catalytic subunits are separate and inactive when bound by regulatory subunits.

Activation: B-adrenergic receptors, when bound by ligand (hormones) activate receptor associated G-proteins (Gs in this case). G proteins have an alpha domain which binds GTP/GDP and a By subunit which integrates with the membrane. Upon stimulation, the Ga loses its GDP and gains GTP, resulting in activation (the receptor acts as a GEF for Ga). Ga dissociates from the By subunits and translocates (while still on the membrane) towards adenylate cyclase. AC is then activated and begins forming cyclic AMP. These newly formed cAMP molecules can then bind to the PKA regulatory subunits (two cAMPs per subunit), reducing the regulatory subunits affinity for the C subunits. This results in dissociation of the regulatory subunits and the catalytic subunits are free.

Regulation: a-adrenergic receptors, when bound by ligand, activate receptor associated Gi proteins, which translocate to AC and inhibit it, reducing cAMP

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levels and ultimately repressing PKA. The free catalytic subunits can also be bound and inhibited by PKI, which acts as a pseudosubstrate.

Subdomain I of PKA has a phosphate anchor region and the catalytic loop has a specific catalytic aspartate in subdomain VI. Subdomain VII of the activation loop also contains DFG which binds to Mg2+ which binds to ATP (the phosphate groups from ATP are used to phosphorylate targets). Thr 197 in the activation loop must be phosphorylated by another kinase for activation; this brings Thr197 closer to Arg165 (both of which are bound to a phosphate group). This ultimately brings the catalytic Asp166 closer to the ATP for cleavage.

Protein Kinase C Structure: One polypeptide with a catalytic and a regulatory domain. The

regulatory domain has two cysteine rich regions: C1 which binds to DAG or phorbol esters and C2 which binds Ca2+.

Activation: alpha1 adrenergic receptors can activate Gq which activates phospholipase C. PLC cleaves PIP2 (phosphatidyl inositol 4,5 bisphosphate) into IP3 (inositol triphosphate) which travels to the ER and causes release of calcium and DAG (diacylglycerol) which stays integrated into the membrane. PKC complexes with calcium, DAG, and phospholipids in order to activate. Binding of Calcium to C2 of the regulatory domain of PKC increases its affinity for binding anionic phospholipids, resulting in conformational change which attracts PKC to the membrane where it can also bind DAG. Once associated with DAG/Calcium/phospholipids, PKC is phosphorylated on an activation loop threonine by PDK-1. Next, PKC autophosphorylates at two sites on the C-terminus resulting in a fully catalytically correct conformation.

Regulation: PKC can be downregulated by proteolysis by calpain I (under low levels of Ca2+) which results in a regulatory fragment being stuck in the membrane. TPA (PMA, or phorbol esters) mimics the structure of DAG and binds to PKC, integrating it further into the membrane than usual and holding it in the “on” position. Calpain I is more likely to cleave PKC when it is associated with TPA as well.

Target: PKC phosphorylates basic residues on proteins involved in growth, differentiation, and apoptosis.

Phosphatidyl inositol compounds (Pdnlns) IP3 is an example of one of these compounds. PKC phosphorylation

downregulates this pathway to reduce Ca2+ levels back to normal (once PKC has activated and done its job) since high levels of Ca2+ in the cytosol can be harmful to the cell. PKC also regulates itself by stimulating degradation of IP3 to IP2, pumping Ca2+ out of the cell, and inhibiting hydrolysis of PIP2.

Phosphatases PP1 is a serine/threonine phosphatase with a catalytic subunit and one inhibitory

subunit (either I2 or I1). It can be regulated via phosphorylation of I2 by glycogen synthase kinase 3 (GSK-3) which results in dissociation of the inhibitory subunit and activation of PP1. I1 inhibits PP1 only when I1 is phosphorylated by PKA.

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PP1 can also bind to targeting subunits to be sent to specific locations (G to glycogen, M to actomyosin, etc)

PP2A is also a serine/threonine phosphatase. It is trimeric with catalytic (C), regulatory (B), and scaffold (A) subunits. The activity is dependent on subunit composition and PP2A is also the major phosphatase in the cell.

PP2B is also known as calcinurin and it is also a ser/thr phosphatase. It has one catalytic and one regulatory subunit, as well as a Fe/Zn center. PP2B requires Ca2+ and Mg2+ for activation. The regulatory subunit binds up to 4 Ca2+ in order to activate; activity is further stimulated by binding of Calmodulin/Ca2+ to the catalytic subunit. PP2B has limited substrate specificity compared to PP1 and PP2A.

PP2C is also a ser/thr phosphatase which is a monomer. It requires Mg2+ for activation.

Phosphotyrosine phosphatases: ensure growth related sites are dephosphorylated in the absence of growth signals. PTPs cleave at a rate 100,000 times faster than serine/threonine phosphatases. VH1 and Cdc25 (specific to Cdc2) are dual specificity phosphatases.

Calmodulin Small protein with four Ca2+ binding sites; two or more calciums are required for

activation. CAM kinases are activated in response to binding extracellular calcium.

A Kinase anchoring protein (AKAP) PKA can bind to different AKAPs in order to be localized to alternate subcellular

locations. AKAPs bind to the dimeric regulatory subunits of PKA and direct it. AKAP 75/79 can bind both PKA and PP2B which results in rapid phosphate turnover.

Lipid-bound proteins (myristilated, for example) are targeted to the cytosolic side of the plasma membrane.

Casein Kinase 2 Structure: Two alpha catalytic subunits and two beta regulatory subunits. There

are alternate isoforms of the a subunits, resulting in different combinations of subunits. These subunits can only be separated by denaturation. CK2 is constitutively active and does not require any messengers to function. The free alpha subunits are also constitutively active, but have weaker function.

Activity: CK2 is very non-specific and found in all eukaryotes. It is also unique because it can utilize both GTP and ATP. CK2 has diverse substrates (RNA polymerase, mRNA, glycogen synthase). It recognizes acidic residues and phosphorylation by CK2 is stable and not readily phosphorylated. Thus, CK2 substrates are almost always phosphorylated. Mutations in CK2 are lethal; it has important functions in growth and metabolism. The beta regulatory subunits bind to the substrates while the alpha catalytic subunits phosphorylate; however the beta subunits can also function independently and cause conformational change in substrates without resulting in phosphorylation.

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Regulation: CK2 requires ATP (or GTP) and Mg2+. It is activated by polyamine binding to acidic loops in the B subunits and it is inhibited in red blood cells by acidic regulatory compounds such as 2, 3-DPG and drugs such as Heparin.

Cell cycle and Cyclin-dependent protein kinase I CDK-1 is also known as cdc2. Cyclin dependent kinases form a 1:1 complex

with cyclins, which regulate the activity of cdks. CDK-1 is inactive during most of the cell cycle because tyrosine 15 is phosphorylated by Wee1 and thr161 is phosphorylated by Cak (cdk activating kinase). Just prior to mitosis (during the G2/M interface) the tyrosine is dephosphorylated and CDK-1 (cdc2) is active. Nearly all cdks are regulated in a similar manner.

p21 and p27 are inhibitors which bind CDKs. CDK levels change reciprocally as cells progress through the cell cycle.

CDKs associate with cyclins; cdc2 (cdk-1) binds to cdc13 (a cyclin) to form the maturation promotion factor, MPF.

Late in mitosis, cdc2 is inactivated by dephosphorylation of thr161 by cdc25 which results in inactivation of cdc2. The associated cyclin is then destroyed by proteolysis.

Stress-induced cytostasis and apoptosis A cell will enter a cytostatic state in response to moderate stress. Under these

conditions, cell division is inhibited until the damage (usually to DNA) is repaired or the cytotoxic components are removed.

Apoptotic stress, on the other hand, results in programmed cell death. Examples of apoptotic stress stimuli include severe UV radiation, oxidative stress, and response to a FAS or TNF ligand. Cells undergoing apoptosis start by displaying chromatin aggregation and then “blobbing” of the nucleus, then fragmentation of the entire cell (forming apoptotic bodies), followed by immune cell cleanup or necrosis.

Apoptosis is mediated by caspases (cysteine-containing aspartate specific proteases) which initially exist as zymogens (procaspases) which must be activated by proteolysis. Initiator caspases activate other caspases, while effector caspases actually perform the “execution”. Other caspases are involved in inflammation.

Activation of initiator caspases can occur due to binding of hormones such as TNF or Fas to receptors on cell membranes, which cause aggregation of procaspases and cross-activation by proteolysis. Another method of initiator caspase activation is via mitochondrial swelling which results in release of cytochrome c into the cytosol. Cyt c can then bind to Apaf-1, which dimerizes and also promotes aggregation of procaspases and activation.

Anti-apoptotic proteins such as Bcl-2 can also prevent the onset of apoptosis. p21-activated protein kinases (PAK) exist in three major forms: alpha-PAK

(Pak1), beta-PAK (Pak3) and gamma-PAK (Pak2). Pak1 is involved in cell growth, Pak2 is ubiquitous and promotes stress response, while Pak3 is involved in memory storage in neural cells.

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Pak2 is a major effector of the cell stress response: transiently activated by Cdc42 in response to moderate stress. Once activated, Pak2 shuts down cell growth by phosphorylating growth related proteins. During apoptosis, Pak2 is cleaved and made constitutively active by caspase 3, in which case it goes on to phosphorylate extra protein sites in order to promote apoptosis instead of cytostatic stress. Pak2 requires 8 different autophosphorylations in order to be fully active.

Liu

JAK-STAT pathway The cytokine receptor is embedded in the membrane as a hetero or homodimer

(sometimes as a trimer). JAK is a tyrosine kinase which is associated with the cytosolic portion of the receptor. JAK has a tyrosine kinase domain, a pseudokinase domain, an SH2 domain in order to bind phosphotyrosines, and a FERM domain which allows it to bind to the receptor. The receptor itself does not have any intrinsic kinase activity.

First step: The two JAKs associated with the receptors (one each) trans-autophosphorylate each other. This increases their activity.

Second step: The phosphorylated JAKs now phosphorylate the receptor (on the cytosolic side).

Third step: STATs (signal transducer and activator of transcription) have an SH2 domain, a DNA binding domain, a linker domain, and a conserved tyrosine which allows for dimerization only when phosphorylated. One STAT binds to each phosphorylated receptor (via the STAT SH2 domain). This brings each STAT is close proximity to a membrane-associated JAK, which phosphorylate the STATs. These tyrosine phosphorylated STATs are then able to dimerize. The STAT dimmers are then able to translocate into the nucleus and bind to DNA to act as activators of transcription.

Regulation of JAKs: JAKs can be negatively regulated by suppressors of cytokine signaling (SOCs) proteins as well as by protein tyrosine phosphatases. SOCs can either inhibit kinase activity of JAKs via its SH2 domain or inhibit the association of the receptor and STATs also via its SH2 domain. When bound to JAK via its SH2 domain, Socs can also recruit ubiquitination machinery via its Socs box domain, resulting in ubuiqutilyation and subsequent degradation of any JAK associated with the same Socs protein. Protein tyrosine phosphatases also have SH2 domains which allow them to associate with activated/phosphorylated JAKs and dephosphorylate them.

Regulation of STATS: PTPs in the cytoplasm and nucleus can also inactivate STATs by dephosphorylating the critical tyrosine and thus preventing dimerization. PIAS proteins can also bind to STATs in the nucleus and cause the STAT to inhibit transcription instead of activating it.

Lipophilic hormone signaling Lipophilic hormones act through a conserved set of nuclear receptor proteins.

Nuclear receptor proteins have a DNA binding domain, a ligand binding domain (binds lipophilic hormones), a hinge domain, two transcription activating domains, and a dimerization sequence within the DBD or LBD. These NRs bind

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to specific DNA sequences in promoters of hormone-regulated genes, usually as dimmers. They repress transcription until bound by agonist.

Examples of lipophlic hormones are steroids, retinoids, thyroids, and vitamin D3. They bind to carrier proteins in blood and interact with cytosolic or nuclear receptors; they can also traverse plasma membranes.

Type I nuclear receptors exist free-floating in the nucleus or cytoplasm and only bind to the hormone response element (HRE, inverted repeats) upon ligand binding.

Type II nuclear receptors are already pre-bound to the DNA as well as co-repressor proteins. Ligand binding changes the conformation of the heterodimer and dissociates the corepressor and recruits coactivator proteins. Type II NRs bind direct repeats rather than inverted repeats like type I.

Upon activation/ligand binding the NR recruits co-activators such as Swi/Snf/Brg which use ATP to remodel chromatin and slide nucleosomes out of the way to activate transcription. NRs can also recruit HATs such as SAGA and the TRAP/DRAP/ARC complex which mediates the interaction between nuclear receptors and RNA polymerase II.

Nuclear receptors can also be initially bound to co-repressors such as N-CoR (bound to unliganded NR) or SMRT which is a silencing mediator which recruits HDACs to repress transcription.

Proteolysis There are four classical types of endopeptidases: serine/threonine proteases with a

S/T involved in covalent binding, cysteine proteases with a thiol group involved in a covalent intermediate, aspartic proteases with two asps involved in catalysis, and metalloproteases which have metal ions as a part of their structure.

Proteases are often localized in order to prevent inappropriate/uncontrolled degradation of proteins. Some are stored extracellularly (trypsin, coagulation cascade), some are stored in the lysosome (active only in acidic pH), some are membrane-bound (cleave substrates as they pass through membranes)

PARs (protease activated receptors) are embedded in the membrane and have an external tethered ligand. The ligand can only interact with the receptor after a proteolytic cleavage event which cleaves off the extra peptide sequence after the ligand, exposing the ligands binding domain. The PAR can then be disabled by completely cleaving off the tethered ligand.

SREBP precursor protein is sent to the Golgi when cellular sterol levels drop, where it is cleaved by two proteases. This releases the transcription factor domain bHLH which travels to the nucleus to direct transcription of target genes.

Cytoplasmic proteases are highly regulated, usually bound by inhibitor. The major source of cytoplasmic proteolysis is the 26S proteasome.

Ubiquitylation pathway: free ubiquitin is first activated in an ATP-dependent manner by the formation of a thiol-ester linkage between E1 and the carboxy-terminal glycine of ubiquitin. In the next step, the Ub is transferred from E1 to one of many E2s (still attached via a thiol-ester linkage). The third step differs based on which class of E3s is recruited. E3 always has a substrate already bound; in RING E3s the ubiquitin is transferred directly from E2 to the substrate while in

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HECT domain E3s th ubiquitin is first transferred from E2 to the HECT domain of E3 and then finally to the substrate. There are more E2s than E1s and more E3s than E2s; thus E3 determines specificity. Ubiquitins are added to either the N-terminal end of the protein or to the amino group of lysine side chains and attached via an isopeptide linkage. A chain of at least four Ubs are required in order to signal for degradation. Four Ubs on four separate lysine residues would not be sufficient; they actually have to be in a chain (Ubs can be linked/added on to each other via isopeptide linkages)

The 20S core proteasome is 700 kDa and consists of an a7b7b7a7 ring structure which consists of multiple proteolytic activities. The eukaryotic a and b rings are composed of distinct subunits while the bacterial subunits are all the same (ie, all alphas are the same and all betas are the same). The B subunits have catalytic sites. The 26s complete proteasome specifically recognizes target (ubiquitinated) proteins. The 19S regulator (lid and base) add onto the 20S core in an ATP-dependent manner to form the full 26S proteasome which is 2000 kDa. The 19s regulator is responsible for recognizing ubiquitination/possible substrates and also opens up an orifice in the ring which allows substrates to enter.

Rpn1 and Rpn2 are scaffold proteins and Rpn10 and Rpn13 are ubiquitin receptors.

The N-terminal amino acid of the substrate is important in ubiquitin-mediated proteolysis and dissociation/association of the 26s is regulated by phosphorylation.

Tumor suppressor p53 p53 has a transactivation domain, an SH3 domain (binds polyproline stretches), a

central DNA-binding domain, an NLS, a tetramerization domain, and a C-terminal regulatory domain. p53 is a transcription factor which is essential for checkpoint control that arrests cells with damaged DNA in G1 (this checkpoint control can also lead to apoptosis if there is severe DNA damage).

DNA damage activates PTM which activates p53, resulting in transcriptional regulation: activation of cell-cycle arrest and senescence (aging, halts DNA division), and in some cases repression of apoptosis inhibitors and activation of apoptosis accelerators.

Normal cell growth, on the other hand, results in active MDM2. MDM2 is an E3 ubiquitin ligase specific to p53. When not needed, MDM2 is sequestered in the nucleolus and bound by p14ARF.

p53 has a short half-life and is maintained at low levels normally due to degradation promoted by MDM2.

Signals for p53 activation include DNA damage, abberant growth singals, and multiple signal transduction pathways. p53 is activated via protein stability, subcellular localization, DNA binding, or coactivator recruitment.

p53 also has many possible phosphorylation sites. Phosphorylation on serine-15 by DNA-PK results in dissociation from MDM2 (increases half-life) while phosphorylation at thr-55 by TAF1 causes p53 to be associated with MDM2 and sequestered away from the nucleus. This is because phosphorylation at thr-55 promotes the association between p53 and CRM-1, which is an exportin.

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p53 can also be acetylated by p300/CBP and P/CAF in response to DNA damage; this inhibits ubiquitination and promotes DNA binding, co-activator recruitment, and also stabilizes the p53 tetramer.

DeFea

Tyrosine kinases 2/3rds of known tyrosine kinases are receptor tyrosine kinases, the other 1/3rd are

non-receptor tks. None have been found in lower eukaryotes. The first tyrosine kinase discovered was the NRTK Src, isolated from the Rous

Sarcoma virus. It was discovered when the antibody used to immunoprecipitate out the putative kinase actually ended up getting phosphorylated. After using SDS-PAGE, acid hydrolysis, and then phosphoamino acid analysis, it was found that the IgG heavy chain was phosphorylated on a tyrosine.

The next TK discovered was the RTK EGFR (epidermal growth factor receptor). Flourescently tagged EGF ligand was used to determine that the receptor had intrinsic tyrosine kinase activity.

Can experimentally determine the requirement for tyrosine phosphorylation in a pathway by substituting tyrosine with phenylalanine. Phenylalanine is similar enough in structure to prevent unintended changes in protein behavior, while at the same time it lacks the hydroxyl group so it cannot be phosphorylated.

Receptor tyrosine kinases RTKs are single membrane spanning proteins with variable extracellular domains.

All of them have cytosolic kinase domains; however some have the kinase domain split in two by a small “kinase insert”.

RTK activation involves two main processes: an increase in intrinsic catalytic activity and creation of binding sites to downstream components (with SH2 domains). Both of these processes are fulfilled by trans-autophosphorylation (two RTKs of the same time phosphorylate each other). This usually requires dimerization. Some of the phosphotyrosines in the activation loop will stimulate kinase activity while others will generate docking sites for signaling proteins which bind phosphotyrosines with their SH2 domains.

Agonist/ligand binding to the RTK creates a favorable configuration which promotes oligerimization of monomeric RTKs or structural rearrangement of dimeric receptors (such as Insulin receptor). This facilitates trans autophosphorylation by bringing the phosphor-acceptor site of one receptor in proximity to the kinase domain of the other.

Methods of dimerization: Ligand ligation; in this case the dimeric ligand simply causes the receptors to dimerize. ECM ligation; receptors dimerize upon binding to a ligand bound to the extracellular matrix. The last method is cell/cell ligation; the ligands (ephrins) are clustered in the membrane of an adjacent cell and attach to the receptors.

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EGFR: Ligand binding to individual receptors induces a conformational change exposing dimerization interface, it can also promote tetramerization of dimeric receptors.

Insulin receptor: IR already exists as a dimer. Upon insulin binding to the receptor, the two alpha subunits are ligated together. This brings the two beta subunits close enough together to promote trans-autophosphorylation. Some tyrosines have negative regulatory effect: Y1162 in the activation loop hydrogen bonds with a catalytic Asp, phosphorylation of Y1162 prevents this interaction by adding a negative charge and causes the phosphorylation lip to flip out and be exposed (relieves inhibition). Phosphorylation on Y1163 and Y1158 positively regulate kinase activity when phosphorylated by stabilizing the active conformation.

Non-receptor tyrosine kinases Src has an SH1 catalytic domain in the C-terminal half (with key phosphorylation

sites at Y416 in activation loop and Y527 in C-tail), an SH2 domain to bind phosphotyrosines, an SH3 domain to bind polyproline stretches, a polyproline-rich linker region, and N-terminal myristoylation which allows for membrane anchoring. Most NRTKs still have methods of attaching to the membrane.

SH2 domains are sequences of about 100 amino acids with an anti-parallel beta-sheet surrounded by two alpha-helices. This structure allows SH2 domains to bind to phosphotyrosines; though not every SH2 domain can bind to every pY: there is additional specificity contributed by the surrounding amino acids.

SH3 domains are B-barrels composed of 5-6 antiparallel B-strands. SH3 domains bind to polyproline stretches; this is possible because proline residues are “stiff” and usually relatively exposed.

When Src is inactive the helix next to its active site is mis-aligned so ATP cannot bind. Y416 is dephosphorylated while Y527 is phosphorylated; this pY interacts with Srcs own SH2 domain. This interaction also causes Src’s polyproline linker to interact with and block its own SH3 domain.

Activation of Src requires a phosphatase to dephosphorylate Src on Y527; this prevents the SH2 interaction as well as the PPII/SH3 interaction. The next step in Src activation is tyrosine phosphorylation of Y416 which stabilizes the active state, aligning the helices within the activation loop to permit ATP binding. When Src binds to the membrane via its myristoylation domain, the conformation changes to favor phosphorylation of Y416 and dephosphorylation of Y527 (favors activation).

Mutations which disrupt the SH3/PPII interaction result in constitutive activation and possible cancer. Viral src (vSrc) lacks the regulatory Y527 (the one that must normally be dephosphorylated) so it is constitutively active.

NRTK activation is very similar to RTK activation; there is trans autophorphorylation which results in increased intrinsic tyrosine kinase activity. The only difference is that phosphorylation on some NRTK sites can decrease activity (like Y527 on Src).

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Tyrosine kinase downstream signaling Grb2 is a common downstream binding/adaptor protein. It has two SH3 domains

and an SH2 domain. It can bind directly to pYs on RTKs, or indirectly to RTKs through Shc. Grb2 can then bind downstream signaling proteins on their polyproline stretches via its SH3 domain.

Shc is another adaptor protein with a C-terminal SH2 domain. It binds directly to RTK and NRTK pY’s and is also phosphorylated by the TKs which it binds to.

NRTKs and RTKs often signal to mitogenic pathways such as the MAP kinase cascade, via SH2 and SH3 adaptor proteins.

Examples of downstream signaling molecules with SH2 domains: GEFs (guanine exchange factors) which can activate monomeric GTPases, GAPs (GTPase accelerating proteins) which turn off monomeric GTPases, phospholipases, and PI3K. GTPases of the Ras family promote proliferation through the MAPK cascade while GTPases of the Rho family are important for cytoskeletal reorganization.

EGFR signaling: EGF ligand binds to EGFR, EGFR dimers tetramerize and trans autophosphorylate. Grb2 binds to pYs on EGFR via its SH2 domain, then binds to polyprolines on Sos (son of sevenless) via its SH3 domain. Sos is a GEF for Ras, which goes on to activate the MAPK cascade. The pYs on EGFR can also be bound by PLC, which cleaves PIP2 to IP3 and DAG in order to activate PKC. The pY’s can also be bound by PI3K which activates cell survival cascades and cytoskeletal formation.

PDGF signaling: Platelet derived growth factor (PDGF) ligand binds to the receptor, promoting dimerization and trans-autophosphorylation. PLC and PI3K can dock on to these pYs in the same way as on EGFR. Ras-GAP can also bind to the pY’s, however in this case binding results in decreased activity for the GAP. This allows Ras to stay associated with GTP (and thus remain active) for longer and thus promote the MAPK cascade and cell proliferation.

Insulin receptor signaling: InsR is a preformed dimer and insulin binding promotes trans autophosphorylation. The pY’s on InsR can then serve as docking sites for Shc or Grb2 and promote the mitogenic effects of insulin. The pY’s on InsR can also be bound by IRS (insulin receptor substrate) which gets phosphorylated by the IR after binding. The activated IRS can then recruit PI3K in order to mediate the metabolic effects of insulin.

Src signaling: The SH2 and SH3 domains of Src can also bind to PYK2, FAK, and EGFR. PYK2 and FAK are tyrosine kinases which further activate Src. CSK is another kinase which can phosphorylate the C-terminus of Src and activate it to prevent cancer. FAK (focal adhesion kinase) interacts with integrins and has a FERM domain which allows cross-talk between RTKs and NRTKs. Phosphorylation sites of FAK also allow for docking by SH2 proteins and proline-rich sequences allow docking of SH3 proteins.

MAP Kinases MAP kinases are all serine/threonine kinases which are activated by

phosphorylation on a Thr and a Tyr (TXY) within the activation loop. They are activated by dual specificity kinases (MEK, or MAPKK). MAP kinases do not

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have regulatory subunits and they are proline-directed kinases; they recognize substrate variations of the motif P-X-S/T-P.

Each MAP kinase has one or two very specific MAP kinase kinases. For example, ERK is phosphorylated by MEK. MEK is a dual specificity kinase; it phosphorylates ERK on a tyrosine and on a threonine. MAP kinase kinase kinases (MAPKKKs, Raf in the ERK pathway) are slightly less specific. There is an equal amount of ERK and MEK, while there is significantly less Raf. This is because one Raf can phosphorylate many MEKs; this is the amplification step. MEK, on the other hand, will only phosphorylate one ERK (no amplification).

In the classic ERK1/2 cascade an RTK is activated by ligand binding. The pYs on the RTK generate docking sites for Grb2 or Shc (which can then interact with Grb2 via its polyproline stretch and Grb2’s SH3 domain). Grb2 then interacts with the polyproline stretch on Sos via its SH3 domain. Sos is then activated and acts as a GEF for ras, which now gains a GTP and becomes active. The activated Ras will then bind to Raf and bring it to the membrane, bring it to upstream kinases, or simply change its conformation in order to activate it. Thr and Ser phosphorylation by other kinases such as PKC can also partially activate Raf. Raf is initially in an autoinhibitory state but Ras binding/membrane association opens up the structure to allow for phosphorylation/activation. Raf will then phosphorylate multiple MEKs, and each MEK will phosphorylate one ERK on a thr and a tyr. This allows ERK to dimerize and then enter the nucleus in order to phosphorylate downstream targets, resulting in activation of genes involved in growth and proliferation.

Src can also be involved by activating Shc, which activates Grb2, which activates sos, which activates ras, etc.

Integrins can also signal the cascade and prevent cells from proliferating if they are not attached to the ECM (for example, tumor cells).

PKC can also activate Pyk2 which activates Src. PKC can also inactivate Ras-GAP, further increasing the signal cascade.

In white blood cells, PKA can also activate Rap which can activate B-Raf, which can then activate MEK.

Nuclear targets of ERK1/2 include C-myc, Rsk, MSK1/2, ELK1. Involved in cell-cycle and serum response.

Cytosolic targets of ERK include MAP (microtubule associated protein), myosin light chain kinase (cytoskeletal reorganization), B-arrestin (destabilizes its binding to receptors), IRS-1 (desensitizes insulin signaling), MNK1/2 (activates phospholipase A) and most importantly ERK can also phosphorylate MEK and Raf in an inhibitory way in order to turn off its own pathway (negative feedback).

ERK requirement in a pathway is determined by using specific inhibitors of MEK1 activation or phosphorylation of protein of interest with a recombinant constitutively active ERK. ERK1/2 activation can be studied by using phosphospecific antibodies recognizing activated ERK as well as immunoprecipitation of ERK followed by an in vitro kinase assay.

JNK (another MAP kinase) cascade: An RTK can activate PI3K (which is a GEF) which activates RAC1 or CDC42 (rho family GTPases). Stress can also directly activate Rac1/Cdc42. These then activate PAK which phosphorylates and

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activates MEKK1/4, which activates multiple MEK4/7s, which then go on to activate JNK1/2 which translocates into the nucleus to phosphorylate the transcription factor c-JUN. Another possible method to get to Jnk is via TAK1 and Fas and a whole bunch of other stuff that we don’t have to know. Downstream effects of JNK include cell cycle and DNA repair, although it can also activate p53 which results in inhibition of cell cycle progression.

P38 MAPK mediates response to osmotic shock. ERK5 shares many downstream targets with ERK1/2 and is activated in the same

way. It is unique because it can be phosphorylated in the cytosol or nucleus because MEKK2 can shuttle in and out of the nucleus without being bound to ERK5. ERK5 can also promote cell survival by activating anti-apoptotic proteins such as BAD.

ERK activation mechanics MEKs phosphorylate ERKs on T and Y in the TXY motif. The X is important for

specificity. MEK and ERK interact very tightly compared to other kinases; they can often be coimmunoprecipitated together.

Activation sequence: first ERK is phosphorylated on the tyrosine. This provides a more favorable conformation for threonine phosphorylation. When both are phosphorylated ERK is activated. Tyrosine phosphorylation by itself has almost no effect on activity, and the threonine is never phosphorylated alone.

Catalytic core of ERK2 is formed by N and C-terminal lobes which contains a phosphorylation lip that is opposite of the ATP binding site. Catalytic activity involves ATP binding to residues in the N-terminus. A critical Lys at codon 52 is also required for activity. The N terminus and C terminus both make up the kinase domain.

Dominant negative ERK can be used as a research tool: mutating Lys52 allows recombinant ERK to bind and sequester substrate but never phosphorylate it. This can be used instead of MEK inhibitors to determine if ERK is required in a pathway.

ERK activation domain: phosphorylation on Y185 stabilizes a salt bridge which may facilitate T183 phosphorylation. pT induces conformational changes required for activity.

Substrate recognition site is P-X-S/T-P. Catalytic core has a small pocket which only allows small hydrophilic residues to fit; proline only fits because the side chain faces away from the kinase surface. ERK can also dock to other proteins; the ERK docking domain is positioned opposite the catalytic core.

Crystal structure of ERK suggests a dimerization interface. ERK has an NLS but no NES; MEK has an NES so prolonged association between MEK and ERK allows ERK to leave the nucleus after it is done.

Scaffolds such as Ste5 can hold multiple components of the MAPK cascade together for sequential activation. In fact, Ste5 is required for most in vivo MAPK activation in yeast.

KSR (kinase suppressor of Ras) is a mammalian Ste5 like scaffold which holds MEK1/2. It has a kinase domain but it doesn’t really phosphorylate anything.

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MP1 is a MEK binding partner which facilitates MEK phosphorylation of ERK. Also localizes late endosomes to the small adaptor protein p14.

B-arrestins normally bind phosphorylated GPCRs in order to diminish G-protein coupling to the receptor, and they also facilitate receptor endocytosis. However, B-arrestins can also act as scaffolds which are required for activation of ERK downstream of some GPCRs and RTKs. B arrestins can recruit Raf, MEK, and ERK into a complex which determines its subcellular localization: the scaffold prevents ERK from entering the nucleus and instead directs it to the cytoskeleton to promote cytoskeletal reorganization and motility.

G-protein coupled receptors GPCRs are the largest receptor superfamily. They are heptahelical (7

transmembrane helices) and they couple to heterotrimeric G-proteins. GPCRs transmit signal from extracellular activators to intracellular proteins.

Heterotrimeric G-proteins: 3 subunits; alpha, beta, and gamma. The alpha subunit binds and hydrolyzes GTP and binds downstream components (such as adenylate cyclase in B-adrenergic signaling). The By subunit is associated with the membrane and can also have intrinsic activity. They inhibit Ga while in a heterotrimer. The GPCR acts as a guanine exchange factor (GEF) for Ga.

GPCR activation of G-proteins: ligand induced activation of the GPCR promomtes association of the receptor with the heterotrimeric G-protien and exchange of GTP for GDP. The Ga now has a bound GTP and is able to dissociate from the receptor and By in order to transducer downstream effects. The free By subunit can also (when unbound by alpha) transducer separate downstream effects. One of the downstream signals is usually the activation of kinases that phosphorylate the receptor and promote dissociation of the G-protein. The separated G-protein remains bound to its target protein and signals independently of the GPCR until an RGS (regulator of G-protein signal) binds to Ga and promotes GTP hydrolysis. Ga-GDP then reassociates with the bg subunits and the receptor. GTP hydrolysis is thus an internal “clock” that controls G-protein signaling.

Discovery of G proteins: It was known that glucagon increased cAMP production and caused activation of PKA, but the mechanism was unknown. It was also known that glucagon increased GTPase activity in the membrane and that non-hydrolyzable GTP was better at activating AC (Galpha would stay on, but they didn’t know that yet). Used purified Cyc- cells which could not make cAMP (were actually missing Gs); found that normal cell membranes could rescue cAMP production. This is how Gas was first purified. Cyc- cells were also resistant to cholera toxin; this is because the toxin causes ADP ribosylation of Gas and causes constitutive activation. Both of these factors led to discovery of G-proteins.

GPCRS were discovered by co-purifying B-adrenergic ligand with the receptor. This was difficult because GPCRs are not very common in the membrane. Activity was measured by using artificial vesicles with purified B2AR and mixing these lipid vesicles (with GPCRs) with erythrocyte membranes to promote fusion. The next step was simply measuring if AC exhibited increased activity.

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GPCR features: phosphorylation results in desensitization. There are also multiple disulfide bonds which are involved in the structure and ligand binding. Palmitoylation promotes G-protein interaction and N-linked glycosylation promotes ligand binding.

Different types of alpha subunits: Gas binds to and activates AC and can also activate ion channels in some cells (neurons). Gai binds and inhibits AC. Gaq binds and activates PLCB wihle Ga12/13 binds and activates p115RhoGEF and tissue specific effectors.

G-protein By subunits inhibit Gas function when bound. However, once released from Gai the By subunits can activate Rho-GEFs as well as ion channels, PLC, as well as scaffold proteins involved in MAPK signaling.

The slow response to cAMP/PKA: In this case PKA activates cAMP specific genes by activating the transcription factor CREB which binds to target genes and initiate their expression. This is a “slow” response because it takes time for transcription/translation of the target factors.

The fast response to cAMP/PKA: low blood glucose stimulates glucagon release and this glucagon binds to Gas coupled receptors in the liver and eventually activates PKA. PKA can then activate glycogen phosphorylase kinase which activates glycogen phosphorylase which uses phosphorolysis to break down glycogen and release free glucose. The epinephrine (fight or flight) response uses this same concept in skeletal muscle. These are much more rapid effects.

Gaq activation of PLC results in cleavage of PIP2 into IP3 and DAG, resulting in Ca2+ mobilization from the ER and eventual activation of PKC.

Pertussis toxin is useful because it ADP ribosylates Gai and prevents it from binding adenylate cyclase and inhibiting it. This is useful to determine if your protein relies on or is Gai. This has a similar effect with cholera toxin, which ADP ribosylates Gas and constitutively activates it. Both end up activating cAMP production.

BARK (Beta adrenergic receptor kinase) phosphorylates GPCRs and causes them to uncouple from Gs and instead couple to Gi.

B-arrestins are usually involved in terminating signals but they can also bind to GPCRs and activate signaling in a manner independent of G-proteins. However, B-arrestin and G-protein pathways can also antagonize each other.

Monomeric GTPases Monomeric GTPases bind and activate other proteins in signaling cascades. They

are activated by exchange of GTP for GDP and their activity is determined by GTP hydrolysis which acts as an intrinsic timer. Monomeric GTPases can also attach to membrane lipid modifications; membrane association is necessary for activity.

GAPs are GTPase-activating proteins which stimulate GTP hydrolysis and thus inactivate the GTPase. GEFs are guanine nucleotide exchange factors which facilitate exchange of GDP for GTP, activating the monomeric GTPase. GEFs achieve this by inducing a conformational change causing the GTPase to simply release its GDP; a new GTP spontaneously binds because there is 10 times more GTP than GDP in cellular conditions. GDIs are guanine nucleotide dissociation

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inhibitors which make it harder for a GEF to exchange the GTPase guanine, thus rendering the protein less able to be activated.

Ras superfamily: 21kD GTPases with varied functions. Ras-GAPS often contain SH3 domains, SH2 domains, as well as PH domains which grant high affinity for specific phospholipids. Sos (son of sevenless) is the primary GEF for Ras; it contains polyproline stretches which allows it to interact with upstream Grb2 (Grb2 has a SH3 domain).

GTPase mutations: constitutively active mutations are common in cancer; they cannot hydrolyze GT and are thus permanently active. Dominant negative mutants cannot bind GTP and thus block activation of endogenous proteins. Most common Ras mutations in cancer are mis-sense resulting in a one amino acid change (usually Gly12 to Ala or Gly13 to Asp) which renders the protein consitutively active because it cannot hydrolyze GTP.

Ras family GTPases are often modified via farnesylation, which is essential for lipid targeting. Membrane localization of Ras is crucial for the GEF induced conformation change that allows it to activate Raf.

Ras is important for development but is not as essential for adults because cell division is not as prominent.

Rho family GTPases have three main subclasses: Rho (RhoA, involved in actin stress fibers and cytoskeletal stabilization), Cdc42 (involved in filopodia for cell migration), and Rac (involved in membrane ruffling and lamellipodia).

Non-muscle cell motility Actin is an abundant 43kD ATPase conserved in all eukaryotes. It polymerizes to

form filaments; it is a polar protein with a barbed end and a pointed end. The barbed end contains the ATP binding site and ATPase activity. The barbed end dominates filament assembly kinetics.

Actin assembly: polymerization occurs at both ends but the fast growing barbed end dominates growth (10 monomers per second). Growing of filaments occurs at barbed ends. Only ATP-actin is added and ATPase activity is not required for addition. However, ATP is hydrolyzed immediately after addition and the cleaved Pi is not released until about 70s after hydrolysis. Thus, the newly added monomers in the growing end have either ATP or ADP+Pi while the “older” actin monomers only have ADP. This forms a “cap” of ATP and ADP+Pi which serves to add polarity to the filament. Leading edges for movement are formed when actin filaments extend, branch, and push the membrane out. Branching requires polymerization to occur from the side of a filament.

Spontaneous assembly: the initial binding of two monomers is very slow. However, adding a third monomer to a nucleus of two actins is faster, and the addition process gets faster and faster as more monomers come together. The initial nucleation is the reason for the “lag phase”.

Actin must be reorganized in response to signal (for the cell to move) and this is done by either increasing the number of exposed barbed ends, chopping up existing filaments, disassembling focal contacts/pulling off CapZ from barbed ends, or most importantly by helping speed up actin nucleation (which is the rate limiting step).

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The Arp 2/3 complex is a seven subunit complex found in all actin containing organisms. It is involved in pointed end caping activity, side-binding (important for branching) and most importantly for de novo actin assembly. Arp2/3 promotes nucleation. Arp3 binds to the pointed end of a monomer or the side of a filament; Arp 2 and 3 together “mimic” an Actin dimer. The cell sees these proteins as a stable nucleus and begins adding actin onto the Arp nucleus. This overcomes the rate limiting step and promotes rapid filament elongation.

Arp2/3 must be activated by WAS family proteins which are also called nucleation promoting factors (NPFs). NPFs cause a conformational change in the complex which brings Arp2/3 together.

WASP family proteins (NPFs) have an acidic domain which binds Arp2/3, a central domain, a WH2 domain which binds actin monomers, a polyproline stretch which binds profilin-actin monomers and a CRIB domain which binds GTPases of the Rho family.

These NPFs must also be activated. There are many different methods for NPF activation: RhoA GTPases can cause a conformational change. PIP2 can recruit the NPF to the membrane. SH3 domain containing proteins can link cytoskeletal NRTKs to NPFs via its polyproline domain. Tyrosine phosphorylation can also stabilize NPFs; as well as Ser/Thr phosphorylation of WAVE by ERK and N-WASP by CK2.

In the inactive state, Y291 of WASP cannot be phosphorylated. However, Cdc42 (a GTPase) can bind WASP and open up its conformation as well as recruiting it to the membrane where it binds PIP2. This facilitates tyrosine phosphorylation by Lyn and Btk which in turn stabilizes the open form. It can also be opened up and activated without any phosphorylation. Once activated, the WASP can then activate Arp2/3 which can then promote actin nucleation.

The Nck adaptor protein has SH2 and SH3 domains and allows WASPs to interact with EGFR, PDGFR and other TK pathways via SH2 domains.

F-BAR proteins help to activate NPFs without any GTPase activity. Wave/Scar are NPFs with no CRIB motif and cannot bind to GTPases. Instead

they are bound by repressor proteins which are removed by the Rac-1 scaffold. Actin polymerization is regulated by specialized actin binding proteins. Profilin

and thymosin sequester actin monomers, CapZ binds to barbed ends to prevent new monomers from adding, Gelsolin severs filaments and also caps the barbed end (activated by calcium). Cofilin severs filaments which produces free monomers and “actin seeds”. Arp2/3 also cap the pointed end in order to prevent depolymerization from the back.

Receptor signaling to NPFs: chemoattractant signal results in activation of a receptor which activates a GEF for Cdc42. Cdc42 can then activate a WASp which then activates Arp2/3, resulting in rapid actin polymerization and migration toward the chemoattractant.

In wound healing, PDGF binds to PDGFR (an RTK) which stimulates proliferation and cell migration at the same time; cells migrate to the center of the wound and then divide.

Cofilin/ADF binds to the sides of filaments and twists them, weakening the contact between subunits. Cofilin binds preferentially to ADP actin so it

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disassembles “older” filaments. It also does not cap the barbed end so it results in exposed barbed ends. Phosphorylation by LIM kinase causes cofilin to dissociate from filaments while dephosphorylation by SLINGSHOT activates it. Dephosphorylated (activated) cofilin can also be suppressed by PIP2. Basically if the cell gets a signal to move, SLINGSHOT is activated for a few seconds and cofilin is activated. It chops up existing old filaments while WASps cause nucleation at the other end (using the actin subunits freed by cofilin cleavage). The new filaments won’t be cleaved because they are bound to ATP or ADP+Pi. By the time the phosphate dissociates, RhoA activates LIM kinase and cofilin is phosphorylated and inactivated.

The membrane is pushed forward by treadmilling caused by spatially restrained cofilin activity combined with nucleation and elongation.

Focal contacts are attachment between actin cytoskeleton and the extracellular matrix. Protein comlexes hook the actin filaments to integrins at the plasma membrane, these integrins then signal to NRTKs which aid in focal adhesion turnover. CAPZ is also found on barbed ends of actin filaments at focal contacts.

Desensitization and Down Regulation Methods of down regulating a signal: receptor modification can desensitize it,

sequestration or endocytosis can result in lysosomal degradation. Specific kinases and phosphatases can turn off previously activated kinases or relay proteins, or simply an inhibitory pathway can be turned on that counteracts the signal.

Receptor downregulation: the first step is usually desensitization via a modification (often downstream of the receptor itself) which renders the receptor to be unable to bind any more ligand or respond to ligand binding. Homologous desensitization is when the agonist renders the receptor immune to activation by additional agonist addition (the same agonist). This usually involves phosphorylation. Homologous desensitization is an example of feedback inhibition because it usually involves a downstream kinase or phosphatase. Heterologous desensitization occurs when the activation of an inhibitory pathway renders the receptor immune to activation by additional agonist addition.

The receptor can also be downregulated via internalization or degradation/recycling (via proteasome or lysosome).

GRKs are G-protein receptor kinases. GRK2 binds Ga and acts as a GAP, it can also bind to Gby subunits and bring them back to Ga. Most importantly it binds and phosphorylates GPCRs to desensitize them.

GRKs and B-arrestins can also be involved; after a GRK phosphorylates a receptor at multiple sites, a B-arrestin can bind to the phosphorylated receptor and prevent signaling.

During endocytosis, an activated but desensitized receptor buds off from the membrane into an early endosome, these vesicles bud off and can either return to the membrane or be sent to the proteasome or lysosome for degradation. In clathirin-dependent endocytosis, clathrin (a 2 subunit protein) associates in groups of three to form a triskelion. This forms a cage around the membrane, protecting it and targeting it to specific locations.

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Arrestins “arrest” GPCR signaling by facilitating interaction with clathrin coated pits. Arrestins can also serve as scaffolds for signaling cascades. Arrestins bind to phosphorylated receptors (though they can also bind unphosphorylated ones). B-arrestins can inhibit PI3K, Akt, LIMK, NFkB and they can romote activity of localized MAPK, Cofilin, RhoA, and Src.

Downregulation of RTK signaling can also occur via dephosphorylation by protein tyrosine phosphatases (PTPs). Feedback inhibition can also occur; phosphorylation of the receptor by a downstream kinase at a specific location can inhibit further Y phosphorylation via steric inhibition. Clathrin is required for internalization of RTKs.

PTPs contain SH2 domains (in order to bind phosphotyrosines). The SHP family of PTPs dephosphorylate some RTKs including PDGFR and they can be regulated by S/T phosphorylation themselves. Receptor-like PTPs also exist but most do not have specific ligands. CD45 is an example.

BCH 184

Study Questions – Crystallography

1. Why do you need a crystalline array of protein in order to obtain a 3-D molecularmodel of that protein by X-ray crystallography?

2. Why is X-ray crystallography conducted with X-rays rather than with light rays?X-rays are used because they have much smaller possible wavelengths than visible light rays. This allows for a resolution of 1-1.54A, allowing visualization of atomic structures.

3. What is a structure factor?A structure factor is a mathematical description of how the crystal scatters/diffracts X-ray radiation. Each reflection is an individual structure factor while each X-ray diffracted to the detector is a sum of contributions of all atoms in a unit cell.

4. What is a unit cell?A parallelepiped of minimum volume enclosing a full compliment of asymmetric units which reflects symmetry properties of the entire group. Basically, the crystal is made up of large quantities of identical unit cells.

5. The reflections observed in a diffraction pattern represent X-rays that diffract offperiodically repeating ______________________________________________.

6. What are Miller Indices?Miller indices are families of planes making intercepts with the unit cell edges, identified by three integers h, k, and l.

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7. What are Bragg Planes?Bragg’s Law states that reflections are only observed on film when the diffractions of waves of all planes constructively interfere. When Bragg’s law is satisfied then nλ = 2d sinΘ. D is the distance between planes of atoms, n is any integer, and λ is wavelength.

8. What information is provided by the intensity of an individual structure factor?The intensity of a reflected wave depends on the electron density at and between the planes. The intensity of a structure factor provides the amplitude; i.e., how displaced atoms are. This can be used to calculate electron density.

9. What information is provided by the phase of an individual structure factor?The phase of a structure factor is the time/distance required for the wave crest to reach a defined point from the origin. The phase of an individual structure factor provides information about electron locations and eventually the position of electron density between hkl (Bragg) planes.

10 What is a Fourier transform?A Fourier sum is used to describe a complicated periodic wave as a sum of simple waves. The Fourier transform is the precice mathematical description of diffraction. It is used to transform structure factor data back and forth between electron density data.

11. What is a Friedel Pair?

The couple of reflections h, k, l and is called a Friedel pair. Their intensities are equal if there is no absorption, but differ otherwise. Friedel’s law then does not hold. Generally speaking, when absorption is present, equivalent reflections generated by the symmetry elements in the crystal have intensities different from those of equivalent reflections generated by the introduction of an additional inversion centre in normal scattering. Friedel pairs are used in the resolution of the phase problem for the solution of crystal structures.

12. When none of the atoms in the unit cell absorb the incident X-ray radiation, thetwo members of a Friedel Pair have the same intensity. Why?Friedel’s law only holds when there is no anomalous scattering.

13. What are some consequences of eliminating higher frequency reflections fromthe Fourier Sum used to calculate a structural model?Eliminating higher frequency reflections results in loss of high resolution data.

14. What is a Patterson Map? What information can be obtained from it?A Patterson map is used to locate the heavy atom in a unit cell when using MIR. The Patterson map is basically a difference map of the diffraction of the protein with the heavy atom minus the protein without a heavy atom in order to determine the diffraction pattern of just the heavy atom.

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15. How does the symmetry of asymmetric units in the unit cell help you determinethe (x,y,z) coordinates of a heavy atom in the unit cell?Relating symmetry from the 2D unit cell vector combined over the crystal provides all three coordinates of the heavy atom (u, v, w on Patterson map, and x, y, z on the unit cell).

16. In Multiple Isomorphous Replacement (MIR), what does “isomorphous” refer to?Isomorphous refers to the addition of a heavy atom isotope to the protein.

17. Briefly describe the sequence of steps you must carry out when using MultipleIsomorphous Replacement (MIR) to obtain the phases of structure factors.First obtain native crystal data (Fp). Then soak the crystal with a heavy atom and determine Fph. Next, calculate the difference in diffraction to obtain Fh. Then use a Patterson map to determine the location of the heavy atoms. Finally, use Harker diagrams in order to obtain structure factors.

18. Briefly describe how SIRAS differs from MIR.In MIR, the multiple vector equations are provided by two different Fh vectors (which correspond to multiple heavy atoms at different sites). In SIRAS however, the multiple vector equations are provided by anomalous scattering from a single heavy atom that provides the phase Fph, then Fph and Fh together can define Fp. Atoms which do not absorb X-rays have their electrons oscillate at the same frequency; this acts as another source of radiation but with a 180 phase shift. This is coherent scattering. Anomalous scattering occurs when atoms which do absorb X-rays have their electrons excited to a higher quantum state or ejected; a phase change then occurs from scattering dependent on atomic number (not 180 degrees). Friedel’s law holds only when there is no anomalous scattering.

19. Briefly describe how Multiwavelength Anomalous Diffraction (MAD) differs fromMultiple Isomorphous Replacement (MIR) as a means of obtaining the phases ofstructure factors.In MAD, the diffraction data is recorded at three or more wavelengths (at, below, and above the metal ion absorption edge) yielding 3 sets of ΔFr and ΔFi values. Data sets from a single heavy atom derivative at different wavelengths provide similar data to that obtained from using distinct heavy atoms (like with MIR). Therefore in MAD one can obtain sufficient phasing information from a single crystal.

20. Briefly describe the sequence of steps you must carry out when using MolecularReplacement to obtain the phases of structure factors.First obtain the phasing model for the protein of interest from pdb. Next, collect data (the diffraction pattern) from your protein. After that superimpose the model from pdb onto the data obtained from your protein (using a Patterson map). Figure out the best orientation between model and protein and finally compute the R-factor (residual index).

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21. Why is the initial electron density map calculated for new structure oftendisappointing?The initial F0 map is usually disappointing due to noisy structure; it is difficult to distinguish protein from solvents especially if the protein is not fully crystalline.

22. What is solvent flattening?Assigning average electron density values to regions based on if they are apparently in the protein or in the solvent. The average electron density values should be higher for regions in the protein and lower for regions in the solvent. This smoothes out the map to reveal small random fluctuations, resulting in an improved map: p’(x,y,z)

23. What is a Fo – Fc map? Why does a crystallographer calculate such a map?What do the peaks in a Fo – Fc map represent?This map is produced by a Fourier synthesis and it is used to minimize model bias. Electron density in this map has peaks: positive means the real unit cell has more electron density at that (x, y, z) location than implied by the molecular model, while a negative peak means that the real unit cell (from your data) has less density at that location. The map emphasizes errors in the current model or differences between your crystallized protein and the model on pdb.

24. What is a 2Fo – Fc map? Why does a crystallographer calculate such a map?What does a strong positive peak in such a map tell you? What does a weakpeak that bulges in a particular direction tell you?This map is much less loisy and does not have any negative peaks. Strong peaks indicate that your protein of interest has electron density in the correct position. A weak peak indicates low density and a bulge in one direction indicates that the correct position is towards the location that the bulge is facing.

25. What is “Energy Refinement”?Least squares analyses are conducted on the overall energy model (bond, angle, conformational, non-covalent interactions such as H-bonds) in an attempt to find the lowest-energy structure that resembles the current model.

26. Define R-factor. What should the R-factor be for a protein refined to a resolutionof 2.5 Å? Approximately what is the R-factor if random phases are used forcalculating a model?

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The R-factor is the residual index, used to converge Fobs and Fcalc. Each Fobs corresponds to a measured reflection intensity and Fcalc is the amplitude of the corresponding structure factor calculated from the model.

If refined to 2.5A, R should be <0.2; an excellent model has R of .12-.14. If completely random phases are used (ie, the model doesn’t match your measured protein at all) then R = 0.6.

27. Describe the difference between R and Rfree. How is Rfree calculated? Why isRfree considered to be “a more demanding and revealing criterion of modelquality” than R?Rfree is computed from only a small set of randomly chosen reflections (about 5%) which were set aside from the beginning and not used during refinement. Since Rfree measures how well the current model predicts a subset of intensities not included in refinement, Rfree is better and less-biased. Rfree is generally larger than R.

Debus Study Questions – NMR

1. Write an expression for the Boltzmann distribution and explain what it tells you about energy level spacings in NMR versus those in optical absorption or vibrational spectroscopy. nB/na = e^(-ΔE/RT)In NMR there is a very small energy gap (1/25,000) between the ground state and the excited state and most atoms are in the ground (up spin) state. In optical absorption only the ground state is populated at equilibrium and there is a large energy gap.

2. What is a transition dipole moment and why is it important in spectroscopy? A transition dipole moment is a measure of charge redistribution that accompanies a transition. A transition is only active if significant dipolar charge redistribution occurs.

3. What causes 1H chemical shifts in NMR? The environment: either J coupling, electron shielding, or the presence of 13C or 15N.

4. What is “J coupling” (also known as “spin-spin coupling” or “indirect dipole dipole coupling”) and what causes it? J-coupling occurs when magnetic nuclei interact through intervening electrons in covalent bonds. It is a weak interaction which can be sensed through at most 2-3 bonds.

5. What is the primary advantage of Fourier-Transform NMR over continuous-wave NMR? FT (Pulse) NMR improves signal averaging and allows the recording of 100 free induction decays in the time it takes one continuous wave spectrum to record. There is also a 10x increase in signal/noise ratio.

6. Why is it possible to excite 13C or 15N nuclei independently of 1H nuclei? 13C and 15N nuclei transitions appear on different regions of the electromagnetic spectrum; therefore they absorb different frequences from 1H and can be excited independently.

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7. In optical absorption spectroscopy, a molecule that is promoted to an excited state spontaneously relaxes to the ground state in tens of nanoseconds. In NMR spectroscopy, what is required for relaxation to occur? In your answer write a simple expression for the relaxation rate constant, label its primary components, and explain where the distance dependence comes from.

Spontaneous relaxation takes too long with NMR, therefore it must utilize spin-lattice relaxation. Spin-lattice relaxation requires a fluctuating magnetic field which occurs during random tumbling of molecules in solution.

8. How does NOE relaxation differ from normal spin-lattice relaxation? In NOE relaxation, one irradiated proton will transfer its energy into the lattice, which will then enhance the energy of another proton transition if the protons are dipolar coupled.

9. Why are 1H nuclei so much more effective at causing relaxation than 2H or 13C nuclei? 1H nuclei have a massive gyromagnetic (charge/mass) ratio compared to other nucleui. Spin lattice relaxation is therefore dominated by dipolar interactions between 1H nuclei.

10. What is the primary advantage of 2D-NMR over 1D-NMR? 2D NMR allows one to perform a Fourier transform of the 1D data; it also provides the amplitude of the frequency over time. This results in all positive peaks, further refinement, and the potential to create a molecular model.

11. What is the pulse sequence for a 2D-COSY experiment? Explain how the 2nd dimension is produced. 1. 90 excitation pulse2. Wait t13. 90 detection pulse4. Wait t2 for results.The second dimension is produced by varying the intial wait time (t1) and then performing a Fourier transform to go from time data to frequency data.

12. What specific information does an off-diagonal peak in a 2D-COSY spectrum provide? J-coupling.

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13. What is the pulse sequence for a 2D-NOESY experiment? Explain how the 2nd dimension is produced. 1. 90 excitation pulse2. Wait t13. 90 excitation pulse4. Wait tm (mixing time)5. 90 detection pulseThe 2nd dimension is produced by varying the mixing time and performing a Fourier transform.

14. What is the significance of the “mixing time” in a 2D-NOESY pulse sequence? Mixing time allows for relaxation. If mixing time is too short, COSY peaks will appear in the NOESY spectrum. If it is too long then the entire system will relax and there will be no signal. An intermediate mixing time allows efficient NOE relaxation and allows the nuclei to disperse energy into the lattice and then into neighboring nuclei.

15. What specific information does an off-diagonal peak in a 2D-NOESY spectrum provide? Dipolar coupling.

16. Under what circumstances do COSY peaks appear in NOESY spectra? When mixing time is too short, COSY peaks (indicating J-coupling, which happens earlier) will appear in NOESY spectra.

17. What are two reasons that a NOESY peak might NOT be seen between two particular protons? If mixing time is too long or if the protons are more than 5A apart. (5A is the maximum distance for dipolar coupling to occur).

18. Why does unusually rapid motion in a protein domain diminish the rate of NOE relaxation? Unusually rapid motion results in many individual frequences but low spectral density and weak relaxation.

19. What are “3JHNα” couplings? How do their magnitudes help you determine elements of secondary structure in proteins by NMR? They are J-couplings (when one nuclei senses the magnetic field of another nuclei through intervening electrons) between an alpha carbon and an amide nitrogen (3 bonds apart).

20. How would you identify α-helical domains in a protein using NMR [i.e., what specific NOE or other couplings (or “connectivities”) would you examine and what are their strengths in helical domains]? For an alpha helical domain, dnn is strong, dan is weak, dan(i, i+3) will be moderate, and dab(i, i+3) is also moderate. This is because the helix turns, so interactions between hydrogens even several residues apart are possible because they are physically close. dnn

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refers to interactions between two neighbouring amide hydrogens, dan is between the hydrogen on the alpha carbon and the hydrogen on the closest amino hydrogen, dan(i, i+3) is between the alpha carbon hydrogen and the amino hydrogen 3 residues away, and dab(i, i+3) is between the alpha carbon hydrogen and an R group hydrogen 3 residues apart. This all refers to J-coupling. Dipolar coupling refers to the direct interaction between two magnetic dipoles (nuclei).

21. How would you identify β-sheet domains in a protein using NMR (i.e., what specific NOE or other couplings (or “connectivities”) would you examine and what are their strengths)? In a Beta sheet, dnn is weak, dan is strong, dan(i, i+3) will be “zero”, and dab(i, i+3) is also “zero”. This is because B-sheets are relatively straight, so residues far apart will not be able to interact.

22. Describe how you would use COSY and NOESY spectra to obtain a structural model for a small protein (e.g., one having a MW of about 12 kDa). In your answer, explain how you would identify individual peaks in the spectra. Start with peak assignment, then look at J-coupling (COSY), and then look at dipolar coupling (NOESY). This will provide distance constraints (strong is close, medium is further, and weak is the furthest H-H distance). This allows for calculation of accurate 3D models because there are so many data points.

23. Briefly explain how 3D-NMR spectroscopy differs from 2D-NMR spectroscopy.3D NMR uses proton peaks and chemical shifts from either the amide nitrogens (15N) or aliphatic carbons (13C) which the protons will interact with.

24. What is the primary advantage of 3D-NMR over 2D-NMR?Due to the addition of chemical shifts, the data is spread over a 3rd dimension which allows for refinement of larger molecules.

25. How does 4D-NMR differ from 3D-NMR? 4D-NMR uses 13C and 15N chemical shifts, while 3D-NMR uses one or the other.

26. Figures depicting NMR-derived structural models of macromolecules often show many overlapping structural models. What do these individual structural models represent?Structural models represent flexibility of the molecule in solution. Additional constraints remove degrees of freedom and increase refinement. More constraints reduce the degrees of freedom which results in more refinement. For example, 16 constraints would result in a 2A resolution.

Electron Crystallography and EPR 1. In electron crystallography, how is phase information obtained? The phase information is obtained by viewing electron micrographs in different angles of tilt and then using the Fourier transform of an electron microscopy image.

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2. Why are the diffraction patterns and images recorded in electron crystallography usually “noisy“? The diffraction pattern is noisy because of low electron density and the fact that all non symmetric data points are considered “noise”.

3. To obtain sufficient data to calculate a 3-D molecular model of a membrane protein, diffraction data and images are recorded at different angles of tilt (i.e., the sample stage is tilted in the electron beam). Why is it necessary to record data at different angles of tilt? The crystalline array structure is in 2D so the image must be tilted in order to reconstruct the entire 3D image.

4. Approximately what percentage of structural models in the protein data base corresponds to unique membrane proteins? 1%

5. Membrane proteins make up approximately what percentage of the proteins found in any organism? 20-30%

6. Describe the physical basis for the differences between EPR and NMR.EPR NMRElectrons Protons-1/2 is stable +1/2 is stableSpin+orbital angular momentum Only spin angular momentumMicrowaves Radiofrequency wavesMassive gyromagnetic ratio

7. Why is the typical EPR magnet so much smaller (provides a much smaller magnetic field) than the typical NMR magnet? Electrons have a 2000x higher gyromagnetic ratio so a smaller magnet is sufficient.

8. Describe how an EPR spectrum is recorded. What does the detector actually measure? The detector measures microwaves reflected by the sample (which contains unpaired electrons). First, load samples into the cavity and then adjust so that no microwaves go back to the detector, then scan the magnetic field.

9. What is meant by the terms “isotropic”, “axial” and “rhombic” in reference to EPR spectra? They refer to the symmetry of EPR spectra. Isotropic is one peak, axial is two, and rhombic is three peaks.

10. What is the effect of a 1H nucleus (I=1/2) on the EPR spectrum of an unpaired electron? Of a 14N nucleus (I=1)?

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An H nucleus (I=1/2) will either increase or decrease the magnetic field felt by the unpaired electron. This splits the EPR resonance in two. A 15N nuclei (I=1) will cause three ossible orientations.

11. Why are EPR spectra usually so anisotropic? Most electron orbitals are anisotropic (not spherically symmetric) so the EPR spectra is also.

12. The EPR spectra of tyrosine radicals in different proteins differ markedly. Why? Tyrosines can have different angles between methylene protons and the aromatic ring.

13. What is ENDOR spectroscopy? What type of information can be obtained from it? Electron nuclear double resonance; it involves saturating one EPR line and then applying a radio-frequency field of steadily increasing frequency. This yields individual hyperfine couplings. The position of the ENDOR transition provides the identity of nuclei giving rise to hyperfine interactions.

14. What is ESEEM spectroscopy? What type of information can be obtained from it? Electron spin echo envelope modulation; a 180 pulse causes electrons to precess toward one another in order to amplify the signal. ESEEM spectroscopy detects electrons coupled to 14N or 1H nuclei and senses the new magnetic moment and frequency of precession changes by hyperfine interactions.

15. What is the difference between hyperfine and superhyperfine interactions? Hyperfine interactions are between the metal ion nucleus and unpaired electrons while superhyperfine interactions are between the ligand of the metal ion with unpaired electrons.

16. Describe how spin-labels can be used to study biological systems.Spin labels are organic molecules with unpaired electrons. They can be used to probe protein binding sites and have their motion hindered by other groups in the binding site. The spectrum will then provide info on the dimensions and rigidity of the binding site.

17. For proteins containing transition metal ions, what type of information can be obtained with EPR spectroscopy? EPR spectroscopy can provide information about octahedral vs tetrahedral orbitals, high vs low spin states, and the superhyperfine splitting can determine electron and ligand interactions.

18. To what do the terms “high spin” and “low spin” refer? In high spin there is a small energy gap between orbitals; the most stable configuration is with all electrons unpaired. Low spin refers to a large energy gap with the most stable configuration having all electrons paired.

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19. What is Zero Field Splitting? What causes it? Zero-field splittings occur when spin-spin interactions between unpaired electrons produce splittings of the electron spin states in the absence of a magnetic field. ZFS magnitude depends on the symmetry of the metal ion and is anisotropic.

20. Why are EPR spectra of metal ions frequently only observable at very low temperatures (below 150 K and often below 10 K)?Relaxation normally occurs too fast due to strong spin-orbital coupling; a lower temperature is required to slow relaxation enough for observation.

BCH 184 2009PRACTICE EXAM QUESTIONS IN OPTICAL SPECTROSCOPY (Dunn’sLectures, PART 1)(1) (12 Points). On the axes given below, draw and label three energy wellsrepresenting the ground electronic state, S0, the singlet excited state, S1, and thetriplet excited state, T1 for a molecule. Then label the diagram to identify thefollowing:(a) Vibrational energy levels in each state.(b) Rotational energy levels in the S0 state.(c) An electronic transition for absorption giving a singlet excited state.(d) An electronic transition for intersystem crossing.(e) An electronic transition for fluorescence.(f) An electronic transition for phosphorescence.

(2) (5 Points). Explain why fluorescence is considered to be a kinetic phenomenon.Fluorescence is considered to be a kinetif phenomenon because it is only observed when it occurs more quickly than non-radiative processes. (Non-radiative processes are when the single excited state returns to the ground state without emitting a photon by transfer of energy to other molecules as heat)

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(3) Why does the electronic absorption spectrum of a compound always appear with a maximum at an energy that is higher than the energy of the fluorescence maximum?Fluorescence emission is always a lower energy than the electronic absorption spectrum because energy is always lost to non-radiative processes (decay to lowest vibrational energy level of es and decay of vibrationally excited gs to lowest vibrational state of gs)

(4) (10 Points). Briefly explain why the fluorescence of Trp 59 in horse heart Cyt cis quenched in the native state of the protein, but not in the denatured state.Trp59’s emission spectrum overlaps with the absorbance spectrum of the nearby (H-bonded) heme group. Normally the heme group steals the emission energy of Trp59 and quenches its fuorescence so only the heme emission at 500 nm is seen. This is an example of Foerster energy transfer (FRET). Trp emission is still seen but is weak. However, when the protein is denatured, the Trp is far away from the heme and a strong fluorescence from Trp 59 is seen (around 350 nm).

This diagram illustrates the Franck-Condon principle which states that at the instant the electron enters the new orbital, the molecule will still have the same bond lengths and angles as the ground state. This is because nuclear vibrations happen later as the molecule moves towards the lowest vibrational energy. It is only after this that energy is released as heat and bond lengths change. Vertical transitions corresponding to no nuclear displacement during an electronic transition are highly favored.

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When uncleaved, the emission peak of Trp at 350nm is quenched by the absorption peak (350-370nm) by Ser-ANS, so only Ser-ANS’s fluorescence emission peak at 500 nm is seen. Cleavage will prevent this FRET and Trp will absorb the 280nm excitation energy and Trp’s normal fluorescence emission spectra at 350 nm will be seen.

In the case of a polar excited state, a polar solvent will stabilize the excited state and therefore reduce the energy of the electronic transition. Thus the answer is C.

(8) (2 Points). Spectroscopy is the observation of quantum mechanics.

(9) (2 Points). Quantum Mechanically, light has particle properties, and the theseparticles are called photons.

(10) (2 Points). Increasing the number of conjugated double bonds in an organicmolecule causes the spectrum to shift to a longer wavelength. (30nm per bond)

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(11) (4 Points). Name the electronic transitions that acetone can undergo, andclassify these transitions with regard to the relative energies of the transitions andthe expected intensities.Pi to pi star has high intensity and energy and n to pi star (weaker, forbidden) requires a rotation and displacement of charge.(12) (4 Points). Draw a structural diagram depicting a a π,π* transition forethylene.

(13) (2 Points). Define Absorbance (use equations and illustrate your answer with adiagram).Absorbance is the amount of energy absorbed by the sample. A = log[psolvent/psolution]

(14) (2 Points). Write out Beer’s Law and explain what effect light scattering willhave on a plot of Absorbance versus chromophore concentration.A = ebc (e is the molar absorbance coefficient, b in the path length of the absorbing bilayer, and c is the concentration of the absorbing species) Light scattering will increase the apparent absorbance because the scattered light does not reach the detector (the machine operates on the assumption that any light that does not reach the detector was absorbed by the chromophore).

(15) (10 Points). Draw simple diagrams (indicating essential mechanical/opticalelements) for the following:(a) A scanning spectrophotometer that uses a single photomultiplier detector.(b) A scanning spectrophotometer that uses a photodiode array detector.

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(16) (2 Points). Your sample is expected to absorb between 200 and 260 nm. Whattype of cuvette will you need to use in order to measure the absorption spectrum?(a) Glass(b) Quartz(c) PlasticGlass and plastic cuvettes can only measure absorption spectra between 350-2000 nm. Quartz can measure far below 350nm (ultraviolet). Tungsten lamps are used for visible light (350-800nm) while an H2 or D2 lamp is used for UV (160-350nm)(17) (2 Points). which occurs more quickly, a π,π∗ transition or a molecularvibration?A π,π∗ transition occurs 16x faster than an inter-molecular vibration.

(18) (6 Points). Sketch the absorption spectrum for a peptide that contains one eachof the following amino acid residues:L-Trp, L-Phe, L-Glu, and Gly.The amide peptide bond transitions for n, pi star are at around 210 nm while the pi, pi star transitions are at 190 nm. The side chain peaks for His, Arg, Glu, Gln, and Aspartate are at 210 nm, which means they are buried under the envelope of the peptide band transitions. Tyrosine has an intense pi, pi star transition at 270 nm, phenylalanine at 260 nm, and tryptophan at 280 nm. Cysteine also has a weak absorption spectrum at 250 nm.

(19) (7 Points). Explain why the spectrum of Cyt c changes upon changing thesolvent from aqueous pH 7.0 at 25o C to 9 M urea at pH 2.0 and 25o C.Denaturation prevents quenching of Trp59’s emission spectra by the nearby heme group because after denaturation the heme group is too far away.

(20) (3 Points). In the tryptophan synthase bienzyme complex, the commonmetabolite, indole is transferred from the α-subunit to the β-subunit via achannel. This process is called substrate channeling.

(21) (4 Points). The catalytic cycles of the α- and β-sites of tryptophan synthase aresynchronized by switching the α-site open and close in response to thechemical events occurring at the β-site.

(22) (2 Points). The escape of the common metabolite indole is prevented by theLid alpha - the sites between alpha and beta closed conformation.

(23) (4 Points). In the rapid mixing experiment where indoline is reacted with the α-aminoacrylate intermediate, E(A-A), what effect does the α-site ligand, glycerolphosphate, have on the reaction rate? (Explain the glycerol phosphate effect.)

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Glycerol phosphate slows down the reaction by promoting the a-lid closing. Due to this slower speed, it is possible to observe intermediates. Important for studying the mechanism.

(24) (2 Points). In fluorescence spectroscopy, what is the Stokes Shift and whatcauses the shift.The Stokes Shift is the change/difference between the absorption spectra and the emission spectra due to lost energy from the non-radiative decay processes.

(25) (4 Points). Sketch the essential mechanical/optical elements of a recordingspectorfluorometer.

BCH 184 2009PRACTICE EXAM QUESTIONS IN OPTICAL SPECTROSCOPY (Dunn’sLectures, PART 2)

(1) (2 Points). In fluorescence spectroscopy, what is a Dynamic Stokes Shift andwhat causes the shift?The dynamic Stokes shift is a wavelength increase/shift caused by solvent/molecular reorganization around the chromophore excited state resulting in an emission shift to longer wavelength during the excited state lifteim. Increased molecular flexibility results in increased time in the excited state and thus a longer shift.

(2) (6 Points). Describe the processes that result in the green bioluminescence of thejellyfish, Aequorea Victoria. What does FRET have to do with this?Foerster energy transfer (FRET) is when the absorption spectrum of a nearby chromophore overlaps the emission spectrum of the fluorescent molecule. The energy of the es can be transferred to the chromophore through a dipole-induced dipole mechanism. The system must be in resonance (energy separations must match). In the jellyfish, the absorption spectrum of GFP overlaps the emission spectrum of the enzyme product so GFP steals the energy and emits fluorescence at 508nm (the protein chromophore absorbs at 395nm or 475nm)

(3) (6 Points). The fluorescent protein variant from Discosoma coral, mPlumexhibits strong Dynamic Stokes Shift behavior. The wild-type fluorescent proteinfrom which mPlum was derived, DsRed, does not show a Dynamic Stokes Shift.Explain why.In the wild type protein the fluorophore is buried and rigidly bound so the protein microenvironment around the fluorophore cannot reorganize to stabilize the es and shift to a longer wavelength; therefore there is no red color.

(4) (6 Points). Name light scattering artifacts which interfere with the measurement of fluorescence emission spectra. What molecular processes give rise to these? How

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are these artifacts related to the wavelength of the excitation beam?Elastic scattering also known as Rayleigh-Tyndall scattering. Incident radiation is not only absorbed or transmitted by the sample but also scattered in all directions. This results in inaccurate results but does not change the emission wavelength. Inelastic scattering is also known as Raman scattering. Some of the incident scattered energy from Rayleigh scattering is abstracted and converted into vibrational and rotational energy. The resulting scattered energy is of lower energy and longer wavelength than the incident radiation. The result is a weak emission band separated from the incident radiation.

(5) (2 Points). Why is the sky blue and the grass green?The sky is blue due to Rayleigh scattering of light by particles in the atmosphere which are much smaller than the wavelength of sunlight. The scattering is inversely proportion to wavelength^4 so shorter wavelength of blue light scatters more. Grass is green because chlorophyll has an absorption peak which corresponds to red light and has emission spectra that corresponds to the complementary color green.

(6) (2 Points). What is the inner filter effect? How can it be minimized?The inner filter effect occurs at high concentrations of protein; more light is emitted from the front than the back of the sample cuvette. Thus, the intensity of light emitted from the front is greater. This effect can be minimized by keeping the chromophore concentration low or using a microcuvette.

(7) (6 Points). How does time-resolved fluorescence depolarization work? What is acommon physical biochemical application of time-resolved fluorescencedepolarization to the study of proteins?A short pulse of vertically polarized light is directed at the sample. The light is then absorbed, promoting the subset of molecules with transition dipole vectors aligned with the polarized light to an excited singlet state. Excited molecules are held in a rigid matrix (gel or crystal) so the return to ground state will result in emitted light that is also vertically polarized. Rotation during the interval between absorption and emission will decrease the polarization with time at a rate which reflects the rate at which the molecules rotate (rigidly bound means less rotation). This is useful for determining the dynamics of protein motion and interaction.

(8) (4 Points). Given a UV-Vis absorption spectrum, what part of the spectrumprovides a measure of the probability of the transition to the excited state and whatpart provides a measure of the energies associated with the excitation?The probability of a transition is the area under the curve while the measured energies correspond to the maxima.

BCH 110A

Lecture 1

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Henderson Hasselbach: pH = pKA + log[(base)/(acid)] pKa = -logKa = log1/Ka Higher Ka = strong acid, higher pKa = weak acid (less likely to give up its H+) Non-covalent interactions: van der Waals, ionic, H-bonds, hydrophobic

interactions Ionic interactions: dielectric constant is higher for polar solvents. F = Q1Q2/er^2,

r is distance, Q is charges, and F is the energy of interaction. Ionic interactions are stronger in polar solvents.

Van der waals interactions between two atoms are strongest at about 4-5A. Interactions are individually very weak but the sum of many can be important.

Buffers maintain homeostasis by resisting changes in pH when small amounts of base or acid are added. Buffer systems involve an aqueous solution of a weak acid and its conjugae base; buffer range is usually +- 1 pH unit from its pKa.

Physiologically important buffer systems are inorganic phosphate (phosphoric acid) and organic phosphates (phosphomonoesters).

Lectures 2-3: Biological thermodynamics and ligand binding ΔG = ΔH – TΔS Actual free energy change (ΔG) depends on two parameters: the standard free

energy change for that reaction (ΔGo) and the actual mass action ratio. Thus ΔG’ = ΔGo + RTln(actual mass action ratio) Catalysts increase the rate of biological reaction but do not change Keq (position

of equilibrium is not altered) or ΔG. The catalyst merely increases the rate at which the reaction goes toward equilibrium.

Ligand binding: equilibrium dissocation constant is Kd = {[P][L]}/{[PL]}. A lower Kd indicates a higher affinity between the enzyme and the ligand and tighter binding.

θ = fractional saturation; the fraction of total binding sites on the protein that are occupied by ligand. θ = [L]bound/[P]total = [P.L]/{[P]+[P.L]} Ranges from 0-1.

Kd is the ligand concentration when fractional saturation is at 0.5; it is thus the concentration of ligand needed to half-saturate the binding sites.

ATP has high ΔGo of hydrolysis but energy is still required to break the bond. The large free energy is due to many interactions, including relief of charge repulsion, resonance stabilization of the phosphate (increased entropy), ionization (release of H+ at pH 7) as well as greater solvation of products.

Lectures 3-4: Amino acids and peptide bonds Almost all natural amino acids are L-isomers.

Group Amino Acid Typical pKaTerminal a-carboxyl All peptides 3.1Side chain carboxyl Glutamate, Aspartate 4.1Imidazole group Histidine 6.0Guanidino group Arginine 12.5Aromatic hydroxyl Tyrosine 10.9Thiol Cysteine 8.3Epsilon amino Lysine 10.8

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Terminal a-amino All peptides 8.0

pKas are found at half equivalent points on titration curves. Equilibrium for peptide bonds lies far in the direction of hydrolysis; however this

hydrolysis is slow on a biological timescale. Thus, peptide bonds are metastable. Protein hydroxylation requires vitamin C, lack of vitamin C results in scurvy. Carboxylation requires vitamin K, lack causes blood clotting deficiency.

Lecture 5: Peptide bond properties and secondary structure Peptide bond has partial double bond character: planarity, rigidity, cis-trans

isomers are possible, and dihedral angles.

Phi (Φ) is the angle between the alpha carbon and the amide nitrogen while psi is the angle between the alpha carbon and the carbonyl carbon.

The alpha helix is usually right handed in proteins and involves intrachain H-bonds between peptide bond elements 4 residues apart. 3.6 residues per turn, R groups are protruding outside, amide nitrogens are all pointed back and carbonyl oxygens are all pointing downward. Proline and glycine are rarely found in alpha helices. A-keratin has two right-handed a-helices coiled around each other in a left handed direction.

B-sheets have nearly fully extended backbones in a regular, repetitive pattern (all residues have the same phi and psi angles). Side chains point in alternate directions for adjacent residues in chain. Real B-sheets are not ideal/flat, they are twisted due to the chiral effect (twists to the right due to L-amino acids). B-turns usually involve Gly, Asn, Ser (small hydrophilic residues) or Proline.

Fibrous proteins have polypeptide chains assembled into elongated rope-like or sheet-like structures. They are usually water insoluble.

a-keratins have two right handed helices coiled together to form a superhelix (coiled coil). Rich in ala, val, leu, ile, met, and phe. Stabilized by hydrophobic interactons.

Collagen makes up ¼ of all protein in the body. Its primary structure is rich in Proline, hydroxyproline, and glycine. Has tripeptide repeats Gly-X-Y where X is

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often Pro and Y is hydroxyproline. Collagen helix is unique; left handed with 3 residues per turn and 3-stranded coiled coil as supersecondary structure. Triple helix is very rigid; every third residue must be a glycine because nothing larger will fit. Stabilized by kinks in Pro and Hyp rings, not intra-chain H-bonds. Hydroxylation of some tyrosines is important for structure; requires vitamin C (ascorbic acid).

Silk fibronin has lots of Gly, Ala, and Ser (small residues). Made up of many stacked antiparallel B-sheets. The sheets are fully extended so silk is flexible (sheets can slide because sheet-sheet van der Waals interactions are weak).

Lecture 6: Protein Tertiary Structures Folding of globular proteins requires minimization of solvent-accessible surface

area (burying hydrophobic side chains and exposing hydrophilic side chains), maximizing hydrogen bonds within the protein (bury the backbone and h-bond polar R-groups to each other), and the chiral effect (L-amino acids cause backbone to be right-handed).

Myoglobin is a water-soluble globular protein. Binds O2 in muscle cells for storage and is mostly (70%) a-helical; the rest is mostly turns and loops. The helices are amphipathic with many charged residues on the surface.

Triose phosphate isomerase is an a/B barrel protein with a parallel 8 stranded B-barrel on interior surrounded by a-helices).

Immunoglobulins have domains consisting of 2 antiparallel B sheets each, with loops between the B strands.

Porins are large B-barrels (large antiparallel B sheet) with hydrophobic residues on outer surface to interact with hydrophobic lipid core of membrane with hydrophilic residues in the water-filled inner pore.

Lecture 7: Protein Quaternary Structure and folding The folded form of a protein has a slightly lower free energy and is therefore

favorable. This is because internal interactions reduce ΔH and the hydrophobic effect redues –TΔS (favors folding) while conformational entropy (loss of entropy due to less solvation etc) increases –TΔS and favors unfolding.

RNase P experiment: denatured with urea and B-mercaptoethanol, but was able to reform on its own. Showed that native tertiary structure is determined entirely by the primary structure (aa sequence).

Many diseases result from defects in protein folding. Cross-beta structures are very thermodynamically stable misfolded proteins .

Alzheimer’s disease: intracellular aggregates of a protein called tau also form. Amyloid-B peptide is normally in extracellular plaques but can also enter cell membranes. Specific proteases can cleave off intracellular and extracellular domains of the AP precursor, the isolated peptide loses a-helical structure and forms extended 2-layer B sheets. This will form an aggregation cascade and damage cells.

Lecture 8 and 9: Ligand Binding and Allosteric Regulation

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Myoglobin binds O2 delivered to tissue by blood and stores it until needed as the terminal e- acceptor for energy metabolism. Myoglobin has 8 a helices and a globin fold. The interior is densely packed with most hydrophobic residues being buried. The heme prosthetic group allows it to bind O2. Myoglobin is monomeric; a single polypeptide chain with 153 residues.

Hemoglobin is a heterotetramer (a2B2). The a and B subunits are homologous to each other and to Mb. The globin fold is conserved and the quaternary structure is two aB heterodimers with strong interactions at the interfaces within aB protomers. R state in the longs = high O2 affinity, T state in the tissues = low O2 affinity. Big changes at interfaces between protomers when going between R and T states.

Heme: four N atoms in heme are coordinated to 4 positions on the protein. Proximal His is attached to heme Fe2+, distal His is part of E helix and becomes hydrogen bonded to the distal O atom (of the O2 substrate). The O2 binds between Fe2+ and the distal His. 2 other hydrophobic residues (Val and Phe) help keep Fe2+ from becoming oxidized to Fe3+.

Binding of O2 to myoglobin is hyperbolic, but binding of O2 to hemoglobin is sigmoid (reduced affinity due to possibility of T state).

Tertiary structure changes when hemoglobin goes from T to R state. The T state has ion pairs and a much larger central cavity.

Hill plot shows degree of cooperativity; nh is the Hill coefficient. The maximum value is the number of binding sites (4 for hemoglobin). If nh > 1 then there is cooperative binding, nh = 1 means no cooperativity or there is only one subunit, and nh < 1 means negative cooperativity (binding of ligand to one site reduces the affinity for the other sites).

Structural basis of cooperativity: binding of O2 brings Fe2+ into plane with heme porphyrin ring, pulling the proximal His with it: this moves the F helix. This causes sstructural changes in interface between a1B1 and a2B2 protomers, triggering quaternary change in the whole tetramer. The R state loses 2 salt links due to loss of ion pairs.

Allosteric inhibitors of O2 binding: 2, 3-BPG, protons, and CO2. Higher [H+] or lower pH in tissues stimulates release/delivery of O2. Binding of CO2 in tissues also promotes release of O2 because CO2 needs to bind to the T state Hb to be delivered back to the lungs to be exhaled.

2, 3-bisphosphoglycerate is the main allosteric inhibitor of human Hb. It binds to the central cavity of the T state Hb and stabilizies it; BPG is highly negative and requires several positive residues in the central cavity to bind.

Fetal hemoglobin: serine instead of a histidine in the central cavity, reduces affinity for BPG. This increases affinity for O2 and allows fetus to pick up oxygen from maternal Hb.

Sickle cell: Glu to Val mutation, Val sticks to hydrophobic patches on other tetramers and forms aggregates/fibers when Hb is in the T state.

Lecture 10: Enzymes Enzymes do not alter Keq (concentrations of substrate/product at equilibrium) or

ΔG. They do however increase rate constants and thus increase rates of reactions.

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An enzyme that increases kF (forward rate) by a factor of 100 must also increase kR (reverse rate) by a factor of 100. Enzymes decrease ΔG‡ which is the free energy of activation. It is the free energy barrier which must be overcome; the transition state has the highest free energy and an enzyme serves to reduce that.

Enzymes reduce ΔG‡ by changing the pathway of the reaction and tightly binding transition states. This increases the speed of the slowest (rate-limiting) step.

Lecture 11: Enzyme Kinetics First order reaction is substrate to product. The rate = velocity = Kf[S] Second order reactoion has R + S leading to P + Q. The rate = Kf[R][S] Enzyme catalyzed reactions are zero order at high [S], the rate becomes

independent of substrate concentration. Steady state assumption is that the concentration of ES is not changing. The rate

of formation of ES = rate of breakdown of ES. Kcat = Vmax/[Et] where Et is the concentration of total enzyme active sites. Kcat

is the turnover number. Kcat/Km provides the catalytic efficiency of the enzyme, cannot be greater than

the limit imposed by diffusion control. Most enzymes operate near ½ Vmax. Vo/Vmax is fractional saturation.

Lecture 12: Bisubstrate reactions and enzyme inhibitors Lineweaver Burk, 1/Vo vs 1/S C is the factor by which Ks is changed due to inhibitor. Infinite = competitive

inhibition, 1 = noncompetitive, 0 = uncompetitive. Ks is the rate of E+S to ES and cKs is the rate of EI + S to ESI.

Two types of bisubstrate reactions: sequential (single displacement, ordered or random) vs ping-pong (double displacement).

Reversible inhibitors: competitive increases Km but has no effect on Vmax. Uncompetetive only bids ES complex and prevents product formation, decreases both Km and Vmax by the same factor. Pure noncompetitive decreases Vmax but has no effect on Km. Pure noncompetitive is mixed inhibition where the inhibitor has the same affinity to both E and ES.

Irreversible inhibitors cause covalent modification of the enzyme. Includes chemical modifications, substrate analogs, and suicide substrates that “kill” the active site that they bind to.

Lecture 13: Chemical mechanisms of catalysis General catalytic mechanisms are proximity + orientation (enzymes bring

reactants close together in the proper position), desolvation (reactants move out of H2O, increases electrostatic interactions due to loss of polarity), tight transition state binding, induced fit, general acid-base catalysis, covalent catalysis, metal ion catalysis, and electrostatic effects.

There are four classes of proteases: serine proteases such as chymotrypsin which have a serine hydroxyl perform a nucleophilic attack, cysteine proteases such as papain which uses a cysteine thiol group in the same way, aspartate proteases such as HIV proteases which uses H2O as a nucleophile

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assisted by two Asps, and finally metalloproteases which use H2O as the nucloephile and metal ions.

Chymotrypsin mechanism: Chymotrypsin has a binding pocket for the substrate, cleaves after an aromatic residue.The catalytic triad consists of Ser195, His57, and Asp102. First the serine becomes a potent nucleophile with assistance of a nearby His which acts as a base (accepts the proton from Ser). The Serine then performs a nucleophilic attack on the substrate, forming an acyl enzyme covalent intermediate. The oxyanion hole holds this transition state tightly. In the second step, water attacks the carbonyl carbon of the acyl enzyme using Histidine again as a base. The ester bond is hydrolyzed to regenerate an alcohol group while the rest of the peptide is released. Aspartate maintains the orientation of the Serine and Histidine.

Lecture 15: Enzyme Regulation 5 principle methods of enzyme regulation: allosteric control, multiple forms of

enzymes (isozymes), interaction with regulatory proteins, reversible covalent modification, and proteolytic activation.

Aspartate transcarbamoylase (ATCase) catalyzes the first committed step in the metabolic pathway for synthesis of pyrimidine nucleotides. The products are carbamoyl phosphate and aspartate and the product is N-carbamoylaspratate which undergoes a few more reactions to make a cytidine base.

PALA is a bisubstrate analog which resembles the reaction intermediate and prevents formation of product.

ATCase has 12 subunits with cooperative binding. 6 catalytic subunits and 6 regulatory subunits which can bind CTP and ATP. Also has R state and T state; the T state predominates by a factor of 200 and has a very high Km for Asp (takes a lot more Asp to reach ½ Vmax) Cooperative substrate binding occurs; Asp binding to active site on one subunit increases Asp binding affinity for the other subunits. CTP is an allosteric inhibitor (it is a product, so this is feedback inhibition). ATP is an activator for two reasons; indicates cell has extra energy for biosynthesis reactions, and also indicates that there are more purines than pyrimidines and pyrimidine synthesis is required to balance things out. CTP binds to and stabilizes T state while ATP binds to and stabilizes the R state.

Different isozymes of hexokinase: muscle hexokinase operates near Vmax while glucokinase in liver operates slower and is regulated heavily based on changes in concentrations of blood glucose.

Calmodulin: binding of Calcium exposes hydrophobic patches which can serve as docking regions for target proteins such as PP2B.

Lecture 16: Enzyme Regulation 2 PKA, when activated, phosphorylates both glycogen synthase and glycogen

phosphorylase. This activates phosphorylase and inactivates synthase, resulting in breakdown of glycogen and release of free glucose.

Zymogens are inactive precursors which are activated by cleavage. Chymotrypsinogen is an example.

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Blood clotting is a cascade of proteolytic events. Fibrinogen is cleaved by thrombin, resulting in soluble fibrin monomers. These monomers self-associate into protofibrils (soft clots) and covalent cross-linking of these results in the final clot. Vitamin K is required for activity of glutamate carboxylase, which hydoxylates glutamates; these hydroxyglutamates can bind calcium which is attached to phospholipids on plateley surfaces. This holds the platelet complex and clotting factors in the right location. Termination of the clotting cascade involves removal of clotting factors via protease degradation, inhibitors, and dilution by blood flow.

Lecture 17: Lipids and membrane bilayers Glycerophospholipids have a glycerol backbone and two fatty acyl “tails” in an

ester linkage plus a polar “head group”. Phospholipases cleave glycerophospholipids, generating important products such as signals (DAG and IP3) or arachidonic acid (precursor of eicosanoids like prostaglandins).

Sphingolipids have a sphingosine backbone and a fatty acid chain in amide linkage and either a carbohydrate or phosphate ester of another alcohol like choline or ethanolamine.

Cholesterol has a structure of 4 fused hydrocarbon rings. 3 have 6 carbons and 1 has 5 carbons. Planar, rigid, neutral, and amphipathic. Influences fluidity of membranes.

Membrane fluidity: longer fatty acid chainlength results in more packing of tails, less fluidity, and higher transition temperature. Fewer double bonds also adds more packing and less fluidity. Cholesterol serves as a fluidity “buffer”.

Lateral diffusion is rapid for both proteins and lipids within the plane of the membrane (flow around). Transverse diffusion (flip-flop from cytosolic side to extracellular side etc) of both proteins and lipids is extremely slow unless mediated by protein flippases. Lipid and protein composition in the two leaflets of bilayer is asymmetric.

Lecture 18: Membrane proteins and transport Integral membrane proteins: the polar backbone of the transported peptide goes

through the hydrophobic membrane core and has extensive H-bonding by forming secondary structures.

Glycophorin A has a large molecular weight since it is 60% carbohydrate. These carbohydrates can be blood group antigens. The NTD is on the outside of the cell exposed to water (hydrophobic residues) while the CTD is on the cytosolic side. 19 hydrophobic amino acids go through the membrane as an alpha helix.

Bacteriorhodobsin has 7 transmembrane helices, helps maintain transmembrane proton gradient.

Hydropathy plots are used to plot free energy of transfer of side chain from a hydrophobic to a hydrophilic solvent. More hydrophilic side chains makes this favorable while more hydrophobic side chains makes this unfavorable (positive free energy).

Porins are found in the membrane, the backbone is fully H-bonded in the B-barrel. The hydrophilic residues are on the inside and form a water-filled pore.

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Cyclooxygenase (PGH2 synthase, COX) uses arachidonic acid as a substrate. The water insoluble substrate enters the enzyme active site via a hydrophobic channel. NSAIDs are drugs which block the channel and prevent prostaglandin synthesis. This reduces inflammation. (Aspirin and ibuprofen are NSAIDs).

ΔGt = RTln C2/C1 where C1 = concentration at the origin and C2 = concentration and destination. Uncharged solutes move from region of higher concentration to regions of lower concentration. For charged solutes, the electrical potential also comes into play: ΔGt = RTln C2/C1 + ZFΔpsi with Δpsi being the charge gradient, Z being the absolute value of the ions electrical charge and F being the Faraday constant.

Acetylcholine receptor: Ach is a neurotransmitter, binds to the receptor and induces an action potential. Ach receptor is a ligand gated channel; ligand binding promotes the open conformation so ions can pass through.

GLUT1 glucose transporter: protein mediated transport, not an ion channel. Facilitates passive transport of glucose down its concentration gradient. It has a polar pocket which is specific for glucose and uses the rocking-banana model. Only one side is open at a time. One conformation is open to the outside to pick up glucose, the other conformation is open to inside the cell to release it.

Active transport is driven by ATP or some solute/ion gradient (secondary active transport). Primary active transport generally involves ATPases which can be used for uphill ion transport. SERCA is a membrane spanning ATPase which uses a specific Asp residue on the cytosolic side which gets phosphorylated and allows calcium ions to bind in the transmembrane domain.

Lecture 19: Signaling (PKA, PKC, insulin receptor RTK) I’ve seen this somewhere before.

BCH 110BZiegler

Lecture 12: Pentose Phosphate Pathway

• What two products of the pentose phosphate pathway are needed by cells?Ribose-5-phosphate and NADPH

• In what part of the cell does the pentose phosphate pathway occur?PPP occurs in the cytosol.

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• Write out the reactions of the OXIDATIVE phase of the pentosephosphate pathway. How is the oxidative phase regulated?G-6-P to lactone to 6-phosphogluconate to ribulose-5-phosphate and then to ribose-5-phosphate. Generates CO2 (loss of one carbon) and two molecules of NADPH. The pathway is regulated by the rate of Glucose-6-phosphate dehydrogenase and by NADP/NADPH ratio. High NADP+ activates the pathway while high NADPH inhibits it.

• Outline the nonoxidative stage, with names and structures of the starting and ending compounds (3 moles of 2 different pentose phosphates at one "end", glycolytic intermediates at the other "end"), and names of the enzymes (including those that interconvert the various pentose phosphates) and necessary cofactors, but not structures of intermediate compounds in the transketolase and transaldolasereactions.The non-oxidative phase is reversible. First, one ribulose-5-phosphate becomes ribose-5-phosphate via phosphopentose isomerase. And another one becomes xylulose-5-phosphate via phosphopentose epimerase. Then there are 3 carbon shuffling reactions: First there is a 2-C transfer by transketolase (TPP), then a 3-C transfer by transaldolase (Schiff base), and then another 2-C transfer by transketolase. The end results are fructose-6-phosphate (which can go through glycolysis or gluconeogenesis) and glyceraldehyde-3-phosphate (same).

– What type of "fragment" is transferred by transketolase, what coenzyme is involved, and what other enzymes have you previously studied with similar mechanisms?CH2OHC=OPyruvate decarboxylase and pyruvate dehydrogenase also has similar activity.

– What type of "fragment" is transferred by transaldolase, and what glycolytic enzyme has a similar mechanism?CH2OHC=OCHOHAldolase

– For each enzyme, what is the nucleophile on the donor fragment that is stabilized by the enzyme, that then attacks the carbonyl C of thealdose acceptor?The carbanion.

Which phase of the pentose phosphate pathway (oxidative, or nonoxidative)is reversible in the cell?The non-oxidative phase is reversible.

• Explain the metabolic role of the pentose phosphate pathway. Showthe metabolic fate of glucose-6-phosphate in each of the followingsituations:

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1. The needs for ribose-5-phosphate and NADPH are balanced.(Glucose-6-phosphate is oxidatively decarboxylated and theribulose-5-phosphate is converted to ribose-5-phosphate.)2. Much more NADPH than ribose-5-phosphate is required.(Glucose-6-phosphate is completely oxidized to CO2 by runningthe glycolytic products of the nonoxidative reactions back toglucose-6-phosphate for repeated "rounds" of the cycle.)3. Much more NADPH than ribose-5-phosphate is required, andenergy (ATP) is also needed. (Glucose-6-phosphate is oxidativelydecarboxylated, and the 5-C products are cycled back to glycolyticintermediates and catabolized to pyruvate and beyond.)4. Much more ribose-5-phosphate than NADPH is required.(Glucose-6-phosphate is converted by glycolytic reactions tofructose-6-phosphate and glyceraldehyde-3-phosphate, and thenonoxidative reactions are run "backwards" to generate pentosephosphates.)

Lecture 13: Mitochondrial Electron Transport

• Define/explain: respiration, oxidative phosphorylation, cristae, dehydrogenase, oxidoreductase.Respiration: any metabolic process that leads to uptake of O2 and release of CO2.Oxidative phosphorylation: enzymatic phosphorylation of ADP to make ATP, coupled to electron transfer from a substrate to O2.Cristae: foldings of the mitochondrial inner membrane which provide huge surface area.Dehydrogenase/oxireductase enzymes collect electrons from nutrients such as NAD(P)H or FADH2.

• List 4 ways (forms) in which electrons can be transferred from one molecule to another.As electrons, as H atoms, as hydride ions, and through direct addition of O to C, forming an O-C bond.

• Explain how values of standard reduction potentials relate to relativeaffinities for electrons under standard conditions, and be able tointerconvert ΔG°' and ΔE'o.ΔG°' = -nF ΔE°', where E = standard reduction potential and n = # of electrons transferred.

• What two structurally very different water-soluble coenzymes are usedby cells as electron carriers?NADH and FADH2.

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• (review from earlier in course) Explain the function of NAD+ and NADP+,including the structures of the coenzymes (especially the oxidized andreduced forms of the nicotinamide moiety), the number of electronstransferred and the form in which they are transferred, the half-reactionfor the reduction of NAD(P)+, and a balanced chemical equation for ageneral reaction describing oxidation of a substrate AH2 by NAD(P)+.• (review): Explain the function of FAD (and FMN), including recognitionof the structures of the coenzymes, in particular the “business end”(oxidized and reduced forms of the flavin isoalloxazine ring system), thenumber of electrons transferred and the form in which they are transferred, the half-reaction for 2-electron reduction of either FMN orFAD, and a balanced chemical equation for a general reaction describingoxidation of a substrate AH2 by FAD.• Compare NAD(P)+ with the flavin coenzymes with regard to how tightlythe coenzyme is generally bound to the enzyme that uses it.

• Characterize oxidative phosphorylation in terms of driving force for ATPsynthesis, cellular location, and ultimate acceptor of electrons (acceptorhalf reaction).Oxidative phosphorylation uses an H+ gradient to drive ATP synthesis. It takes place in the inner mitochondrial membrane and uses O2 as the ultimate electron acceptor.

• List 5 types of electron carriers participating in the mitochondrial electrontransport chain and explain whether they transfer 1 electron only, 1 or 2electrons, or always 2 electrons at a time.Flavoproteins transfer 1 or 2, ubiquinone transfers 1 or 2, cytochromes transfer only 1, iron-sulfur proteins transfer only 1, and dehydrogenase enzymes transfer two at a time from NADH and one at a time from FADH2.

• Explain what happens to the electron carriers before and after the blockin an electron transport chain when an inhibitor that blocks a certain stepis added.Carriers after the block accumulate in the oxidized form while everything before is reduced.

• List the 4 electron carrying complexes in mitochondrial electron transfer by Roman numeral (I - IV) and by name, the overall reaction catalyzed by each, and what the redox components are (NAD+, flavin, heme Fe, Fe-S, quinone, etc.) for each, and how they fit (in order) into the electron transport pathway in mitochondria.Complex I is NADH dehydrogenase. Two electrons are transferred from NADH to Q and 4 protons are pumped into the intermembrane space. Complex II is succinate dehydrogenase, two electrons are transferred (one at a time) from succinate to FAD to Q (which becomes QH2) and no protons are pumped. Complex III is cytochrome bc1, QH2 transfers its electrons to cytochrome C and 4 protons are pumped into the intermembrane

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space. Complex IV is cytochrome c oxidase, cytochrome c (reduced) transfers its electrons to oxygen to form water and release 4 protons into the intermembrane space.

• What is the role of Q and of cytochrome c with respect to the large integral membrane protein complexes as components of the chains?Why is the mobility of the electron-carrying quinones and of cytochrome c in the membrane important in mitochondrial electron transport?

Q (ubiquinone) is freely diffusible inside the bilayer. QH2 must be mobile in order to shuttle electrons from complexes I and II to complex III. Cytochrome c is a small water-soluble peripheral membrane protein which shuttles electrons from complex III to IV.

• Which of the large protein complexes in the ETC carries out net oxidation of just 1 QH2 but increases the number of protons pumped into the intermembrane space (the P side of the inner mitrochondrial membrane)from just the 2 H+ from that one QH2, to 4 H+ by carrying out the “Q cycle”?Complex III carries out the Q cycle in order to pump out extra protons.

• Which of the mitochondrial ETC complexes pump protons, and in what direction/orientation?Complex I, III, and IV pump protons from the mitochondrial matrix into the intermembrane space.

• Quantitatively, how does the free energy conserved from the redox reactions of the ETC relate to the concentration gradient of protons across the inner mitochondrial membrane and the charge separation across that membrane?The exergonic reactions of ETC maintain the proton gradient; about 200 kJ/mol of the 220 kJ/mol released by NADH oxidation is conserved in the electrochemical gradient.

• Write the overall reaction for the vectorial process in the mitochondrial inner membrane of transfer of 2 electrons from NADH to oxygen.NADH + 11H+

N + 1/2O2 > NAD+ + 10H+P + H2O

Lecture 14: Oxidative Phosphorylation• Define/explain: oxidative phosphorylation, chemiosmotic model, electrochemical potential, protonmotive force, ΔpH, binding change mechanism of ATP synthase, uncouplerNo.

• Characterize oxidative phosphorylation in terms of driving force for ATPsynthesis, cellular location, the ultimate acceptor of electrons (the acceptorhalf reaction.

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The driving force for ATP synthesis is the proton gradient, it takes place in the inner mitochondrial membrane, and the ultimate acceptor of electrons is oxygen.

• Which of the mitochondrial ETS complexes pump protons, and in what direction/orientation?Complex I, III, and IV.

• How is electron transport in mitochondria coupled to ATP synthesis?Where does the ATP synthesis occur? Describe the chemiosmotic theory, including the name of its originator. Explain the term "protonmotive force"in terms of ΔG and the 2 terms contributing to it. (Review the ΔG equationfrom active transport, the "electrochemical potential", with a chemical andan electrical gradient term; for protonmotive force, be able to point out how ΔpH fits into the ΔG equation.)Electron transport is coupled to ATP synthesis via the proton gradient and ATP synthase. The F1 component of ATP synthase on the matrix side of the membrane is responsible for ATP synthesis while Fo is the pore. The proton-motive force is the proton gradient described as ΔPH (concentration gradient) and Δpsi (charge gradient).

• How do CN–, oligomycin, and dinitrophenol (DNP) inhibit ATP synthesis, and does each of them inhibit both electron transport and ATP synthesis, or just ATP synthesis? Why does prevention of proton flow through the Fo component of ATP synthase by oligomycin inhibit the electron transfer reactions (oxygen consumption)? Why does DNP inhibit ATP synthesis but not electron transfer (oxygen consumption)?CN- inhibits electron transport from cytochrome c oxidase to oxygen; this blocks electron transport and eventually ATP synthesis because electron flow stops and no more protons are pumped. Oligomycin inhibits the Fo pore, preventing electron transport for some reason. DNP is an uncoupler, electron transport still works, but DNP makes extra pores in the inner membrane, allowing H+ to flow back through and completely removing the H+ gradient and preventing ATP synthesis.

• Describe the general structure of the FoF1 ATP synthase, and explainhow its structure relates to the binding change mechanism, couplingthe proton gradient across the mitochondrial inner membrane with ATPsynthesis.Binding change mechanism: ATP synthase normally rapidly intercoverts ADP + Pi to ATP, but it binds ATP much more tightly. This makes release of ATP into solution very unfavorable. Proton flow is required to change the conformation. ATP synthase has 3 B catalytic subunits which are identical but in different conformations. One binds to ATP, one to ADP, and another one is empty. Gamma (another subunit) interacts with one Beta subunit at a time and rotates it in 120 increments, driven by H+ flow. Flow through the Fo pore subunit drives rotation of C10 which drives rotation of gamma, which drives one B subunit at a time into an “empty” conformation, then to ADP binding, then to tight ATP binding, and then again to empty to release the ATP. This is rotational catalysis and it’s pretty cool.

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• What is the biological purpose of thermogenin, and what does it do?Thermogenin is an uncoupler found in brown adipose tissue which causes the proton gradient to dissipate and form heat instead of ATP. It is used for maintenance of body temperature.

• Briefly explain (name of transporter, direction of transport, cotransport --sym or anti -- with what other component) how inorganic phosphate istransported from the cytosol into the mitochondrial matrix, and how ADP3– is transported in and ATP4– out of the mitochondrial matrix.Phosphate translocase uses symport transport (H2PO4- and H+) to transport phosphate from the cytosol to the mitochondrial matrix; it is needed in the matrix for ATP synthesis.

• Describe (with diagrams, reactions and names of enzymes, most of which you have learned before) 2 shuttle systems used for moving electrons from cytosolic NADH into the mitochondrion for reoxidation by the electron transport chain (ETC), thus regenerating cytosolic NAD+, and explain whether those electrons enter the ETC at the "level" of NADH or of FADH2, and what P/O ratio results and why.The glycerol phosphate shuttle transfers electrons from NADH in the cytosol to DHAP, which becomes glycerol-3-phosphate (via g-3-p dehydrogenase) which then goes back to DHAP and gives the lost electrons to FADH2, which transfers its electrons to quinone in complex II. This skips one proton pumping step (since FADH2 enters at complex II instead of I) so there is less ATP formed in the end. The malate-aspartate shuttle transfers electrons to NAD+ as malate is shipped into the mitochondrial matrix from the intermembrane space. Malate is reduced to OAA which undergoes a transamination to form Aspartate, which is shipped into the intermembrane space to regenerate OAA and then Malate to repeat the cycle. Since electrons go directly to NAD+, more ATP is formed. 2.5 P/O ratio compared to 1.5 P/O ratio for FADH2.

• Explain how the concentrations of [ADP] and [ATP] regulate the rates of ATP synthase, and of glycolysis, PDH, and the TCA cycle. Which enzymes are inhibited by ATP and activated by ADP and/or AMP?ATP inhibits PDH, glycolytic PFK-1 and pyruvate kinase, as well as TCA citrate synthase, isocitrate dehydrogenase, and a-ketoglutarate dehydrogenase. ADP and AMP pretty much activate all of these.

Lecture 15: Photosynthesis• Define/explain: thylakoid membranes, grana, stroma, photosystem, photochemical reaction center, antenna pigments, P680, P700, accessory pigments, chlorophylls, oxygen-evolving/water-splitting complex (and Mn complex), cytochrome b6f complex, plastocyanin, ferredoxin, cyclic vs.non-cyclic electron flow.• Describe the "anatomy" of a chloroplast: outer membrane, inner membrane, thylakoids, stroma, thylakoid lumen.

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• Describe how absorption of a photon by an antenna pigment leads (by exciton transfer) to a separation of charge in the reaction center of a photosystem.A photon of light excites an antenna molecule (such as chlorophyll or an accessory pigment) which raises an electron to a higher energy level. The excited antenna molecule then passes this energy to a neighboring chlorophyll molecule (resonance energy transfer) exciting it. The energy is then transferred to a reaction-center chlorophyll, this then passes electrons to an electron acceptor and the electron hole in the reaction center hole is filled by an electron from an electron donor.

• Explain how separation of charge in the reaction center leads to electron flow in a photosystem, and to acceptance of an electron to neutralize the radical cation of the reaction center chlorophyll, enabling it to return to ground state.The excited reaction center chlorophylls are P680 (photosystem II) and P700 (photosystem I). The electron donor fixes the electron hole (cation); the electron donor is the water splitting complex for P680 and plastocyanin for P700.

• Describe the "Z-scheme" for the light-driven electron transport reactions of photosynthesis, including the order of components in electron transfer reactions, the tandem functioning of PSII and PSI, and the role of photons. Which complex carries out Q cycle, like function of mito cyt bc1 complex?Which components carry out 1-e– and which carry out 2-e– transfers?The water splitting (O2 evolving) complex breaks down water into ½ O2 (while protons are sent to lumen to create a gradient) and electrons which are passed to P680 of photosystem II. A photon of light then activates an electron and raises P680 to a higher energy level. The excited electrons are then passed through several carriers including a reduced quinone and then to the Cytochrome b6f complex, which passes them to plastocyanin. Cytochrome b6f can also oxidize PQH2 in a Q cycle, pumping protons and passing electrons one at a time to plastocyanin. Plastocyanin is the electron donor for P700 of photosystem I. Another photon of light activates these electrons, raising P700 to an excited state. These electrons are then passed down to quinones, then to an Fe-S protein and finally to Ferredoxin. Fd then passes electrons to NADP+ via ferrodoxin:NADP oxidoreductase, which is a flavoprotein that picks up 1 electron at a time and passes two to NADP+.) NADP is thus the final electron acceptor for the light reactions.

• Why is the mobility of the electron-carrying quinones in the membrane important both in mitochondrial electron transport and in photosynthetic electron transport? (role of quinones with respect to other e– carriers)Mobility is required to shuttle electrons between complexes.

• On which side of the thylakoid membrane (stroma or lumen) is plastocyanin? ferredoxin? Where is NADPH made? Where is ATP made? Regarding the electrochemical gradient, which side is more negative (N) and which side more positive

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(P)? What are the 2 mechanisms (system components) that establish and maintain the proton gradient in chloroplasts?Plastocyanin is on the thylakod lumen (P) side. Ferredoxin is peripherally on the stroma side of the membrane; NADPH is made wherever Fd is. ATP is synthesized by ATP synthase on the thylakoid membrane on the stromal side. The stromal side is more negative (protons go from lumen to stroma) and protons are pumped out by the cytb6f Q cycle and by the water splitting complex.

• How many photons must be absorbed by PSII for every 2 e –ʼs from oxidation of 1 H2O → 1/2 O2 used to reduce PQB to PQBH2?Two photons per PQb reduced (2 photons per H2O)

• How many photons must be absorbed by PSI for every 2 e–ʼs from oxidation of 2 plastocyanins (received from PSII) used to reduce 1 NADP+ to NADPH?Two photons per NADPH produced (one for each electron)

• Write a balanced equation for the Z scheme reduction of NADP+ by H2O inthe light reactions of photosynthesis. How many electrons and how manyphotons total for the 2 photosystems are involved/required in oxidizingtwo (2) H2O to O2 and producing 2 NADPH?2H2O + 2NADP + 8 photons > O2 + 2NADPH + 2H+8 photons and 4 electrons are required to produce O2 and NADPH.

• Explain the sequence of events in the water-splitting (oxygen-evolving)complex, including the way electrons flow, the roles of the Mn4 cluster andthe TyrZ residue of the protein, and what the water-splitting complex doesrelative to P680+.Water-splitting activity of O2-evolving complex electrons pass one at a time to P680; these electrons are removed from H2O in “packets” of 4 from oxidation of 2H2O by the Mn cluster (Mn4+ to 4e- > Mn) which passes them one at a time to a Tyrosine radical, which passes them one at a time to P680. This releases the 4 protons into the lumen.

• Describe the structure and orientation of CFoF1 ATP synthase in thylakoidmembranes, the energy source driving condensation of ADP with Pi, andmechanism coupling of the energy source to ATP synthesis (same as thatof mitochondrial ATP synthase).Protons from the thylakoid lumen must pass through the ATP synthase Fo pore to return to the more negative chloroplast stroma. The mechanism is similar to the ETC ATP synthase (rotational catalysis).

• Per 4 e– (i.e., per 8 hν or per O2 evolved), about how many ATP aresynthesized?3 ATP are formed for every O2.

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• Compare the electron transport chain and oxidative phosphorylation inmitochondria with photosynthetic electron transport andphotophosphorylation in chloroplasts. Be sure to include the difference inthe source(s) (donors) of electrons in the two cases, and the ultimateelectron acceptors in the two cases, as well as the approximatestoichimetry of number of ATP synthesized per 2 electrons (depends onsource of e–ʼs in mitochondrial system) or per 4 electrons transferred.

Mitochondrial respiration Photosynthesis light reactionsElectron donors are NADH, FADH2 Electron donor is H2OFinal acceptor is O2 Final acceptor is NADP2 ATP per NADH (2 electrons)1.5 ATP per FADH2 (2 electrons)

3 ATP per H2O (4 electrons)

Cyclic electron flow is when electrons fall back to photosystem II in order to create more ATP, but no NADPH is produced!

Lecture 16: Photosynthesis Calvin Cycle• Define/explain: CO2 fixation, C assimilation, chloroplast stroma,mesophyll cell, bundle sheath cell, C4 plant, CAM plant.

• Outline the stages of CO2 fixation by the Calvin Cycle: fixation, phosphorylation and reduction, and regeneration. In what cellular compartment in plants do these reactions take place?Stage 1 is ribulose 1,5 bisphosphate + CO2 becoming two molecules of 3-phosphoglycerate via ribulose 1,5 bisphosphate carboxylase/oxygenase (rubisco). In the second stage, 3-PG becomes 1,3 bisphosphoglycerate via 3-PG kinase, and then this becomes glyceraldehyde-3-phosphate via glyceraldehyde-3-phosphate dehydrogenase.The third stage is regeneration of ribulose-1-5-bisphosphate. Five of these three carbon sugars (glyceraldehyde 3 phosphate and dihydroxyacetone phosphate) are shuffled via several trasketolase and transaldolase and other reactions to regenerate three 5-carbon sugars. This entire process occurs in the chloroplast stroma.

• Stage 1: What does “rubisco” stand for, including the “c”and the “o”? What compound serves as the “acceptor” for CO2 that is being fixed? In the chemical mechanism of the rubisco (carboxylase) reaction, what arethe general roles of the carbamyl-Lys residue and the Mg2+ ion? What is the name of the intermediate that actually reacts with CO2? What are the substrates and products of the carboxylase reaction? Where are the genes that encode the large and small subunits of rubisco?Rubisco stands for ribulose 1,5 bisphosphate carboxylase/oxygenase. Ribulose 1, 5 bisphosphate is the carbon acceptor. The carbomoylated Lysine residue helps to coordinate the Mg2+ which polarizes the CO2 substrate for nucleophilic attack by

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enediolate of Ribulose-1,5 bisphosphate. 8 large subunits of rubisco are encoded in chloroplast DNA while the 8 small subunits are nuclear encoded genes.

• Write out the reactions by which the products of the rubisco reaction areconverted to glyceraldehyde 3-phosphate (GA3P), including names ofenzymes, names and structures of reactants/products, and coenzymesinvolved. What is the overall stoichiometry of Stages 1 + 2, giving netfixation of 3 carbon atoms (enough for net production of 1 GA3Pmolecule), from the fixation reaction (rubisco) to glyceraldehyde 3-phosphate, including number of NADPH, number of ATP, and products?Ribulose 1,5 bisphosphate + 3CO2 > 3-PG > 3-PG + 6 ATP + 6 NADPH > glyceraldehyde-3-phosphate

• Outline 4 possible fates of the glyceraldehyde 3-phosphate thus produced,including the cellular compartments in which these “fates” are pursued and the 3-carbon compound(s) which cross(es) the membrane if transport to adifferent compartment is required. Review the metabolic pathways bywhich 2 GA3P produced in the Calvin Cycle can be converted into glucoseor can be degraded to pyruvate (Dr. Luben’s lectures, week 1).Glyceraldehyde-3-phosphate can either go to step 3 for carbon shuffling to form 5-C sugars or be transported to the cytosol as dihydroxyacetone phosphate and used in cytosol for glycolysis, or it can be converted in the cytosol to sucrose for transport to other tissues, or converted in stroma to starch for storage in chloroplasts.

• Summarize what is accomplished in Stage 3, the “regeneration” phase ofthe Calvin Cycle, starting with the fates of 6 GA3P produced in Stage 2.What is the overall stoichiometry of the reactions for conversion of 5molecules of GA3P to 3 molecules of RuBP (why 3 RuBP?), includingother reactants needed and products produced?6 glyceraldehyde-3-phosphates produced in stage 2; five of them go into the third stage for regeneration while the other one is a net gain of carbons.

• Review the nonoxidative reactions of the pentose phosphate pathway(Lecture 12, Lehn. chapter 14, pp. 560-564), including the transaldolaseand transketolase mechanisms. What kind of group is transferred fromwhat type of donor compound to what type of acceptor compound for eachof these enzymes (including enzyme specificity for configuration of C3 ondonor compound)? What coenzyme is used by transketolase? Reviewthe chemical mechanism of this coenzyme in transferring the activatedgroup, with reference to the Calvin Cycle reactions.Transketolase uses TPP, 2-C transfer. Transaldolase uses Schiff Base, 3-C transfer.

• The 5-C sugars generated by the aldolase and transketolase reactions inthe Calvin Cycle are (2) xylulose 5-P and (1) ribose 5-P. How are thesecompounds converted to ribulose 5-phosphate (names of enzymes), and

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what enzyme converts ribulose 5-P to ribulose 1,5-bisphosphate (RuBP)?The penultimate products of the regeneration reactions (step 3) are xylulose-5-phosphate and ribose-5-phosphate. X5P becomes ribulose-5-phosphate via ribulose 5 p epimerase and ribose-5-phosphate becomes ribulose-5-phosphate via ribulose-5-phosphate epimerase. Then ribulose-5-phosphate kinase uses ATP to create ribulose-1,5-bisphosphate.

• Explain the stoichiometry of ATP and NADPH production by the light reactions of photosynthesis and the stoichiometry of use of ATP and NADPH in the Calvin Cycle, including the issue of Pi shortage in the chloroplast stroma, and how that problem is resolved.Dark reactions: 3CO2 + 9ATP + 6NADPH + 5H2O > 1C3H6O2-P + 9ADP + 8Pi + 6NADP + 6H+This results in a shortage of Pi in the stroma (since 9 ATPs are broken down but only 8 free phosphates are released) so Pi triose antiport system must be used; pumps DHAP into the cytosol and phosphate into the stroma.

• How can the Pi-triose phosphate antiporter, with the enzymes triosephosphate isomerase, glyceraldehyde 3-phosphate dehydrogenase, and phosphoglycerate kinase, isozymes in stroma and in cytosol, be used to carry electrons from stromal NADPH into cytosolic NADH, and“transport” ATP from the stroma to the cytosol?

• Describe the activation of C assimilation enzymes by light, making themless active in the dark, more active in the light? Specifically,1) How does light help to activate rubisco?2) How does light result in reduction of thioredoxin, and how does reducedthioredoxin activate several of the Calvin Cycle enzymes? (You donʼt need to memorize the list of specific enzymes regulated in this way.)Light causes the H+ gradient so there is a higher pH in the stroma, which activates rubisco (specific Lysine becomes carbamoylated) There is also more Mg2+ in the stroma due to the charge, which further activates rubisco.

• Outline the process of photorespiration, including the cycle salvaging phosphoglycolate by converting it back to 3-phosphoglycerate. What activity of what enzyme is responsible for photorespiration? Why is photorespiration detrimental to a plantʼs efficiency of C fixation (what enzyme in the salvage produces CO2), and what factors control the relative partitioning of RuBP into the Calvin Cycle vs. the photorespirationpathway?In photorespiration, rubisco fixes oxygen instead of CO2. In this case ribulose-1, 5-bisphosphate is broken down to form only one molecule of glyceraldheyde-3-phosphate and one molecule of phosphoglyolate (instead of two molecules of G-3-P) The phosphoglyolate is salvaged by losing its phoshphate and being sent to the peroxisome where it is oxidized to glyoxylate, then transaminated to glycine, then sent to mitochondria where the carboxylate is released as CO2, the amino group is salvaged by

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glutamine synthetase, and the alpha carbon goes to THF which makes glyceraldyhyde 3 phosphate using ATP. Basically wastes a whole bunch of shit because rubisco screwed up.

• How do C4 plants avoid/reduce photorespiration?Specialized leaf structure to separate O2 from rubisco; CO2 is carried to mesophyll cells as OAA and malate.

• How do CAM plants avoid/reduce photorespiration?CAM plants open stomata at night and close them during the day. Carbon is trapped as OAA and malate and then used during the daytime by rubisco.

• In what compartment of a plant cell is starch synthesized? In what compartment is sucrose synthesized? What starting compounds can cross the chloroplast inner membrane between compartments?Starch is synthesized in chloroplast stroma and sucrose is synthesized in the cytosol of leaf cells. Triose phosphates from the Calvin cycle in the stroma goes to the cytosol.

• What kind of coenzyme is involved in activation of carbohydrate monomers for formation of glycosidic bonds? What is the “activated” form of the sugar monomers for substrates for starch synthase in plants? For sucrose synthesis in plants? For glycogen synthase in animals?

UDP-Glucose is used for sucrose synthesis.

Lecture 17: Nitrogen Metabolism• What kind of organisms (general terms) have the ability to reduce N2 to NH3? What organisms have the ability to convert NH4+ into organic nitrogen-containing compounds?Bacteria (usually in roots of legumes) reduce N2 into NH3+, and then all organisms can convert NH4+ into organic nitrogen-containing compounds.

• Describe the reactions involved in conversion of N2 to NH4+ (or NH3), including name of the overall process, names of enzymes, substrates and products, "who" is getting reduced and "who" is getting oxidized in the process, and where the electrons are coming from. The process is very favorable thermodynamically (ΔG°' ≈ –33.3 kJ/mol), so why hasn't nearly all the N2 in the atmosphere long since been converted to NH3?N2 + 10H + 8e + 16 ATP > 2NH4 + 16ADP + 16Pi + H2Dinitrogenase reductase provides electrons with high reducing power. It binds to dinitrogenase and transfers 8 electrons one at a time (has a 4Fe-4S center) to reduce dinitrogenase. Electrons originally came from the oxidation of 4CoA and 4 pyruvate to form 4CO2 and 4 acetyl-CoA (pyruvate dehydrogenase reaction).

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• Why do nitrogen-fixing bacteria in root nodules of legumes have to protect their nitrogenase complexes from O2, and how do they do that? What do plant host root cells do for the bacteria? What do the bacteria do for plants?O2 deactivates the nitrogenase complex so heme groups from the plant root cells bind O2 and keep it away from the bacterial nitrogenase complex.

• What reaction do all organisms have for conversion of NH4+ to an organicform of N (enzyme, substrates and products, with structures)?Glutamine synthetase: Glutamate becomes Glutamine using NH4+ and ATP, releasing ADP+Pi.

• What reaction do bacteria and plants use to transfer the amido group ofglutamine to α-ketoglutarate to produce glutamate (substrates, products,structures, coenzymes, etc.). Why is NADPH involved in this reaction?Plants and bacteria use glutamate synthase which transfers an amino group from glutamine to a-ketoglutarate, resulting in two glutamates. NADPH is used as an electron donor.

• Write the sum of the reactions in the two previous objectives, in bacteria and plants (incorporation of NH4+ into an amino acid used in transaminations)?• Why is one enzyme called a “synthetase” and the other enzyme a“synthase”? (You may want to look that up in text glossary.)

• Describe the allosteric regulation of glutamine synthetase; explain thecovalent regulation system [effects of adenylylation/deadenylylation of GSby the Adenylyl Transferase (AT) and the uridylylation/deuridylylation of theregulatory protein P by the Uridylyl Transferase (UT), and how high[glutamine] results in inhibition of GS].Glutamine synthetase, when it has an AMP attached is inactive. Adenylyl transferase attached AMP when AT itself does not have a UMP on it. When AT has an attached UMP, it removes the AMP from glutamine synthetase. Uridyl transferase attaches UMP to adenylyl transferase when there is low glutamine; indicating that glutamine synthetase should be activated to create more. High glutamine removes this UMP, causing AT to be active and attach AMP to glutamine synthetase, inactivating it.

• Name 2 different coenzymes involved in 1-carbon transfer reactions, and state what oxidation states of those 1-C units are usually transferred by which coenzyme.1-C transfers are usually done by THF or H4 folate, usually at aldehyde or alcohol oxidation levels.

• (review from chapter 18) What coenzyme is involved in transaminationreactions? Write out a reaction as an example of a transamination reactionwith glutamate as the donor of the amino group.

• Explain the general structure and mechanism of glutamine amidotransferases.

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Glutamine amidotransferases have two domains: one for glutamine, the other for the acceptor. An NH3 group is removed from glutamine and passed via a channel to the acceptor domain for amination.

• What are the 10 “essential” amino acids in humans? (What does “essential”amino acid mean?) What 10 AAs can humans make de novo?The essential AA’s are methionine, threonine, lysine, valine, leucine, isoleucine, trp, phenylalanine, tyrosine, and histidine. Essential means that they must be obtained through the diet as we lack biosynthetic pathways to synthesize these amino acids.

• What are the 6 “families” of amino acid biosynthetic pathways (i.e., what arethe metabolic biosynthetic precursors that define those families)?

• What is the metabolic precursor in some amino acid and nucleotidebiosynthetic pathways that is made from ribose 5-phosphate? Write thereaction for its formation, with name of the enzyme catalyzing the reaction.Where in metabolism did the ribose 5-P come from?PRPP (5-phosphoribosyl-1-pyrophosphate)

Ribose-5-phosphate + ATP > AMP + PRPP

Lecture 18: Amino Acid Biosynthesis• What are the 6 “families” of amino acids based on biosynthetic precursors,i.e. what are those precursors?The 6 precursors are a-ketoglutarate (glutamate, glutamine, proline, arginine), 3-phosphoglycerate (serine, cysteine, glycine), oxaloacetate (aspratate, asparagine, methionine, threonine, lysine), pyruvate (alanine, valine, leucine, isoleucine), phosphoenolpyruvate + erythose 4-phosphate (tryptophan, phenylalanine, tyrosine) and ribose-5-phosphate (histidine).• What 2 citric acid cycle intermediates serve as precursors for amino acids,and what amino acids come from each of them?• What is the precursor/intermediate that brings ribose-5-phosphate to thereactions involved in Trp, His, purine and pyrimidine biosynthesis? Writethe reaction producing that precursor from ribose-5-P (N Metabolism 1)PRPP?

• What steps are required to convert glutamate into proline? (No enzymenames required) Hint: look at the structures and think what needs to bedone to Glu to get to Pro.Glutamate must be phosphorylated and then dephoshorylated (coupled with reduction) and then the alpha amino group attacks the carbonyl of the new aldehyde, resulting in cyclization.

• Why is Glu acetylated before the rest of Arg biosynthesis begins? Hint:the next 2 steps after acetylation are the same as the first 2 steps in Pro

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biosynthesis. What happens next in Pro pathway, when the amino group isNOT acetylated?The acetylation prevents the alpha amino group from attacking the carbonyl; thus preventing cyclization like with proline. Arginine also requires a transamination at the end from glutamate (which becomes a-kg)

• Outline the steps in conversion of 3-phosphoglycerate to Ser and then Gly.Where does the Ser N (and thus the glycine and cysteine N) come from?Where did the hydroxymethyl group from Ser go when Gly was generated?The serine N comes from glutamate amidotransferase and hydroxymethyl group from Serine goes to methylene H4 folate.

• Outline the conversion of serine to cysteine (2 steps) in mammals. Wheredid the S atom come from, its immediates. The source and its “ultimate” (original)source for humans?TH4 folate takes a methylene group from serine and the sulfur comes from diet methionine; all combine to form homocysteine.

• What amino acid biosynthetic reaction is slowed down by a mammalian enzyme deficiency that causes build-up of homocysteine in blood serum?The formation of cysteine is slowed down.

• Write the reactions for conversion of pyruvate to alanine and conversion ofoxaloacetate to aspartate. Where did the N come from for Ala and Asp?Pyruvate goes to alanine and OAA goes to aspartate. The N came from transamination by glutamate.

• Write the reaction for conversion of aspartate to asparagine, with enzyme name.The enzyme is asparagine synthetase, requires amidotransfer from Glutamine.

• What amino acids are made from PEP and erythrose 4-phosphate, and whatis the common intermediate from which the pathways for synthesis of thoseamino acids all diverge?– What herbicide inhibits the step in which a second molecule of PEPenters the pathway?– Where did the amino group come from for Phe and Tyr?– How is Tyr generated from Phe in animals (name of enzyme)?– What compounds are the sources for the structure of Trp?– What is the enzyme catalyzing the final 2 steps of Trp biosynthesis?What is “intermediate channeling” (substrate channeling)?Chorismate is a common intermediate for amino acids coming from PEP + erythose-4-phosphate. The amino group for Phe and Tyr comes from transamination (Glutamate) and conversion of Phenylalanine to Tyrosine requires Phe hydroxylase in animals. Tryptophans 6 member ring comes from chorismate and the N comes from glutamate. Two extra carbons come from PRPP. The enzyme that catalyzes the final two steps is tryptophan synthase. Substrate channeling is when the indole intermediate doesn’t

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actually leave the enzyme; it instead passes through a channel from the alpha to the beta subunit.

• What 2 amino acids have PRPP contributing to their biosyntheses?Tryptophan and Histidine.

• Given the structure of imidazole acetol 3-phosphate (so imidazole ring isalready formed), how would you convert it to histidine in 3 steps?First a transamination (gluatamate becomes a-kg) in order to generate the alpha amino group. Then a phosphatase removes the terminal phosphate, and finally a dehydrogenase performs an oxidation to generate the alpha carboxyl group and Histidine is complete.

Lecture 19: Regulation of AA biosynthesis and Nucleotide Biosynthesis• Explain/define: allosteric regulation, feedback inhibition, concertedinhibition, enzyme multiplicity, sequential feedback inhibition, substratechannelingEnzyme multiplicity is when multiple isozymes exist, each controlled by different allosteric modulators. Sequential feedback inhibition: one endproduct can inhibit multiple enzymes in its own biosynthesis.

• (Review) Explain the allosteric and covalent regulation of glutaminesynthetase.

• What compound accumulates and causes jaundice (yellowing of skin andeyeballs) when the liver malfunctions or the bile duct is blocked?Billirubin accumulates in blood and causes jaundice.

• What type of enzyme/reaction converts some amino acids to biologicallyimportant primary amines, and what kind of coenzyme is involved in thereactions?Decarboxylases (PLP coenzyme) make amino acids into hormones/neurotransmitters.

• Draw the structures of dopa (dihydroxyphenylalanine) and of dopamine, andname the enzyme that converts tyrosine to dopamine. Aromatic aa decarboxylase (Tyrosine goes to dopa, which goes to dopamine after another carboxyl group is removed).

• What are epinephrine and norephinephrine (structure and very generalfunctions), and how are they synthesized? (no enzyme names, but startingcompounds, and which is made from the other)They are hormones/neurotransmitters which are involved in flight or flight response. Dopamine is their precursor.

• What is histamine, and how is it synthesized (starting from what compound)?

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Decarboxylated histidine. Compound involved in allergic response.

• What is serotonin, and how is it synthesized (starting from what compound)?Decarboxylated tryptophan, it is a neurotransmitter.

• What is the structure of glutathione, what is its general function in cells, and(overview) how is it synthesized?Glutathione is gammaGlutamate- Cysteine – Glutamate. It is a reducing agent which maintains a reducing environment; prevents structures such as disulfide bridges from forming in the cytosol.

• Where do the methyl group and phosphate come from in synthesis ofcreatine phosphate, and what is the general function of creatine phosphate?The phosphate comes from ATP and the methyl comes from adoMethionine. Creatine phosphate is used as temporary energy storage.

• What are the two different general strategies/pathways used for de novosynthesis of purines vs. pyrimidines?Purines are synthesized with the sugar first (PRPP) and then the bae last. Pyrimidines are synthesized with the base ring first, and then PRPP is added to form the sugar backbone.

• What reaction represents the first committed step in purine biosynthesis?The first commited step in purine biosynthesis is PRPP to 5-phospho-B-a-ribosylamine, catalyzed by Glutamine amidotransferase, converting Glutamine to Glutamate and also releasing a pyrophosphate.

• In purine biosynthesis, what compound is the last intermediate before the 2-step branches producing GMP and AMP, respectively?The last intermediate for purine biosynthesis before the commited steps is IMP – inositate monophosphate.

• Describe the 4 regulatory mechanisms determining the rates of biosynthesisof AMP and GMP.Allosteric inhibition by endproducts (IMP, GMP, AMP) Concerted inhibition by AMP and GMP.GMP inhibits GMP from IMP production and AMP inhibits AMP production. Production of AMP requires GTP and vice versa, results in balanced nucleotide levels. PRPP synthesis is also inhibited by ADP and GDP.

• For the cytosolic isozyme of carbamoyl phosphate synthetase that makescarbamoyl-phosphate for pyrimidine biosynthesis, where does the nitrogencome from? For the isozyme in liver mitochondrial matrix (urea cycle),where does nitrogen come from?

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The cytosolic form of carbamoyl phosphate synthetase uses glutamine for ammonia, while the urea cycle isozyme uses inorganic NH4+.

• What enzyme catalyzes the first committed step in pyrimidine biosynthesis inE. coli, and how is that enzyme regulated? (Review from BCH 110A, chapter 6, pp. 220-222 as well as chapter 22, pp. 886-887.)ATCase catalyzes the first commited step, it is inhibited by CTP and activated by ATP via allosteric regulation.

• How are nucleoside monophosphates converted to the corresponding di andto triphosphates?High energy phosphate transfers among nucleotides.

Lecture 20: Deoxynucleotides

1. Explain how ribonucleotides are converted to deoxyribonucleotides, including the name of the enzyme. From what protein does the enzyme get the electrons it uses to reduce the ribonucleotides, and what coenzyme is the original electron donor?

Ribonucleotide reductase first reduces the 2’CHOH of ribose to CH2. Reducing equivalents are provided by NADPH. Next a buried Tyr radical takes an electron from a Cysteine thiol to initiation the reaction by generating a cysteine thiyl radical (S-) in the active site. This radical then abstracts a hydrogen atom from the ribose C-3’ to generate C(3). Two other cysteine residues are oxidized to S-S. Ribose is reduced to deoxyribose and the disulfide is re-reduced by NADPH via thioredoxin.

2. Describe the role of the stable radical and the highly reactive thiyl radical in the initial steps of the mechanism of ribonucleotide reductase. When a ribonucleotide is reduced, what component of the enzyme is oxidized?

Two cysteines are oxidized when the ribonucleotide is reduced. (Disulfide bond)

3. Briefly describe the regulation of ribonucleotide reductase at the primary regulation sites and at the substrate specificity sites. How do different endproducts alter the substrate specificity to keep the levels of the different deoxynucleotides balanced?

Specific regulation: ATP, dATP, dGTP, dTTP inhibit only reduction of themselves. Primary regulation is by ATP and dATP; this keeps the nucleotides balanced.

4. Outline the steps by which deoxythymidylate is made from dCDP or dUDP, including names of enzymes.

CDP goes through ribonucleotide reductase to become dCDP, which then goes through nucleoside diphosphate kinase to become dCTP, which then goes through deaminase to become dUTP, which goes through dUTPase to become dUMP (lol) which then goes through thymidylate synthase to become dTMP.

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5. What are the substrates for thymidylate synthase and what other enzymes/reactions are required for it to continue working? How is thetrahydrofolate regenerated from dihiydrofolate after the thymidylate synthase reaction and how is H4 folate “recharged” with a 1-carbon unit? Why are the enzymes in this cycle such attractive targets for cancer chemotherapeutic agents? Name 2 such drugs and explain molecular basis for their effects.

Thymidylate synthase transfers a 1-C unit at CH2OH oxidation level from Methylene THF and then reduces the 1-C to the methyl state with electrons from THF oxidizing THF to DHF, while converting dUMP to dTMP. NADPH re-reduces DHF to THF via dihydrofolate reductase, serine recharges THF with a 1-C group via Serine hydroxymethyl transferase to regenerate Methylene-THF.

6. In purine degradation, what is the enzyme that converts adenosine to inosine? What condition is caused by a hereditary lack of this enzyme?

Adenosine deaminase converts adenosine to inosine, deficiency results in SCID immunodeficiency which pretty much means you’re dead.

7. What is the final waste product of purine catabolism in primates?The final waste product of purine catabolism in primates is uric acid.

8. What metabolic condition is triggered by excess purines and build-up of the purine breakdown product? What drug is commonly used to treat this condition, and what enzyme does it inhibit?

Buildup of uric acid can result in gout (urate in joints). Xanthine oxidase is the enzyme which converts xanthine into uric acid; inhibiting it with drugs such as allopurinol can be used to treat gout. NSAIDs are also useful for treating gout.

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9. What is the general reaction by which bases from nucleic acid degradation are salvaged? Where does the ribose come from?

Salvage generally involves reacting a base with PRPP to form a nucleotide monophosphate and a pyrophosphate. These salvage pathways recycle free purines and pyrimidine bases to recreate nucleotides.

10. What is HGPRT, and what hereditary disease is caused by a complete lack of this enzyme?

HGPRT is hypoxanthine guanine phosphoribosyltransferase (salvages hypoxanthine to IMP and guanine to GMP). Deficiency results in excessive uric acid production and causes crazy ass neurological abnormalities like aggressiveness, destructive behavior, and compulsion toward self-mutilation. These symptoms can also be caused by studying for the comp exam, listening to Ke$ha, or hearing a million vuvuzelas blasting at once for 90 god damn minutes

Luben

Lecture 21: Glycogen

Glycogen consists of glucose subunits linked by a(1-4) glycosidic bonds with branch points where a(1-6) bonds form. Starch has the same chemical structure but the branches are longer.

Glucose residues are phosphorylzed from nonreducing end of the glycogen polymer by glycogen phosphorylase, resulting in the release of glucose-1-phosphate. G-1-P is rapidly converted to G-6-P by phosphoglucomutase.

Glycogen debranching enzyme transfers residues from short branches to nonreducing ends and hydrolyzes the 1-6 bonds at branch points. When branch points are hydrolyzed, free glucose is released, not G-1-P.

For biosynthesis of glycogen, G-1-P is activated by addition of UDP to form UDP-glucose, carried out by nucleotide diphosphate-sugar phosphorylase. To synthesize glycogen, UDP-glucose is condensed with the nonreducing end of an already existing glycogen molecule by glycogen synthase. No ATP is necessary. Glycogenin can create “de novo” glycogen chains which can act as primers which can be further lengthened by glycogen synthase.

Epinephrine or glucagon activate AC which creates cAMP which activates PKA which phosphorylates both glycogen synthase and phosphorylase. Synthase is inactivated and phosphorylase is activated. Glucose causes a conformational change that makes glycogen phosphorylase more likey to be dephosphorylated; so it is inactive when there is already a lot of free glucose.

Glycogen synthase kinase 3 (GSK-3) also phosphorylates and represses glycogen synthase. PP1 removes this phosphorylation.

Lecture 22: Biosynthesis of Fatty Acids and Triacylglycerols Fatty acids are synthesized by a stepwise addition of two-carbon subunits to a

fatty acyl hydrocarbon chain, which is attached to the enzyme fatty acid synthase. The immediate source of the two-carbon subunits is malonyl-CoA, whose three

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carbon functional group is synthesized from acetyl-CoA by the rate-limiting acetyl-CoA carboxylase. Transfers a carboxyl group from bicarbonate to biotin, which then reacts with acetyl-CoA to produce malonyl-CoA.

Fatty acid synthase is a large complex that has at least six enzymatic activites arranged in a highly organized manner.

Mechanism: First, transfer of an acetyl group from acetyl-CoA to the thiol group of a Cys residue at one of the active sites of FAS. A malonyl group is then transferred from malonyl CoA to the thiol group of ACP (part of FAS). Next is condensation of the acetyl group with the malonyl group, with the loss of CO2. This lengthens the chain. Then there is a reduction of the B-ketoacyl chain to a B-hydroxyacyl form. The next step is dehydration to a double-bond and finally a reduction to form a saturated fatty-acyl-ACP complex. Finally, the saturated fatty-acyl chain is transferred from ACP to the fatty acid synthase SH group where the initial acetyl group had been located. The complex is then ready to repeat in order to add another two carbons.

Triacylglycerol synthesis involves the condensation of fatty acids (in the CoA activated form) with derivatives of glycerol-3-phosphate.

Lecture 23: Cholesterol biosynthesis A bunch of acetyl-CoAs go through some shit to become acetoacyl CoAs and

then react with HMG-CoA reductase to form HMG-CoA which goes through some other shit to become activated isoprene subunits. A bunch of isoprenes come together and form 30-C squalene which cyclizes and does some other shit to eventually form the cholesterol rings.

Cholesterol metabolism is regulated by HMG-CoA reductase, the synthesis of LDL receptors, and regulating the rate of esterification (cholesterol removal). HMG-CoA reductase is activated by insulin, inhibited by glucagon.

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