final-115 Flashcards
(5.1) Crick’s adapter hypothesis, which posited the existence of small RNA molecules (tRNAs) to explain the decoding of messenger RNA, assumed the existence of an unknown class of enzymes. Name this enzyme class and describe the reaction(s) that these enzymes must catalyze
The enzyme class required for this process is aminoacyl-tRNA synthetase. There are 20 different aminoacyl-tRNA synthetase for each corresponding to 1 amino acid in the genetic code. The enzymes catalyzes the following, first, by transfer of AMP from ATP to the specific amino acid, forming an aminoacyl-AMP, via mixed anhydride bond formation. Secondly, the transfer of “activated” amino acid from aminoacyl-AMP to tRNA, causing an exchange between the mix anhydride bond for a ester bond. Most aminoacyl-tRNA synthetase also catalyze a third reaction, the hydrolysis reaction of either(or both) aminoacyl-AMP or aminoacyl-tRNA, to ensure fidelity in protein synthesis.
(5.2) Which step of the reaction cycle of aminoacyl-tRNA synthetases transfers free energy from a chemical work reservoir to the reaction cycle, and which step dissipates this free energy for kinetic proofreading? Answer by reformulating the suggested answer to question (4.2)
The formation of aminoacyl-AMP (via ATP and amino acid) introduces free energy into the system, by driving the reaction (aminoacyl-AMP hydrolysis) away from equilibrium, which lies far on the side of amino acid and AMP. This free energy is later dissipated during the hydrolysis of aminoacyl-AMP back into amino acid and AMP, effectively making the proofreading step irreversible
(5.3) If protein synthesis would proceed from the carboxy to the amino terminus of the protein, how would a plot of specific radioactivity of hemoglobin oligopeptides on the y-axis against the order of oligopeptides from N (left) to C (right) on the x-axis have looked like in Howard Dintzis experiment (cf. Lecture 8)?
If protein synthesis proceeds in the opposite direction (C to N), the radioactivity would be highest at the C-terminal end and decrease towards the N-terminal end. This means the graph would be monotonically declining , with a negative slope.
(5.4) What is the speed of the large subunit of the eukaryotic ribosome in a centrifuge force field with an acceleration of 10^6m/s (i.e., 100,000 × g)?
The speed is given by
v = 60S × 100^-13 (s/S)× 10^6 (m/s^2) =
6 × 100^-6 (m/s)
Therefore the ribosomes move at 6 micrometers per second.
(5.5) The treatment of tRNAs with periodate prevents the charging of tRNAs with amino acids. Where, therefore, must the amino acid be linked to the tRNA? Explain your answer.
The amino acid has to be linked to either the 2’- or 3’- hydroxyl group of this nucleotide because periodate oxidizes vicinal diols and the only vicinal diols present in tRNA molecules is at the 3’-terminal nucleotide.
Oxidation at this site prevents the tRNA from attaching the amino acid, blocking the charging of tRNAs.
(5.6) The catalytic activity of the ribosome is reduced by 6 orders of magnitude upon replacement of the 3’-terminal nucleotide of tRNAs with a 2’-deoxynucleotide. Explain this observation. (How does peptide bond formation proceed?)
Peptide bond formation occurs when the amino group of aa-tRNA attacks the carbonyl carbon of peptidyl-tRNA, both of which are linked to tRNA via ester bonds at the 3’-hydroxyl group. The tRNA of the peptidyl-tRNA leaves as a 3’-alkoxide ion, creating a high-energy transition state. This state is stabilized when the 2’-hydroxyl group of peptidyl-tRNA donates a proton to the leaving group.
The 6-fold reduction in catalytic activity occurs because replacing the 3’-terminal nucleotide with 2’-deoxyribose removes the 2’-hydroxyl group, which is essential for proton donation. Without it, the transition state remains unstable, slowing the reaction.
(5.7) Conversion of eEF1A∙GTP to eEF1A∙GDP by GTP hydrolysis frees aa-tRNAs for peptide bond formation. What prevents the rebinding of free aa-tRNAs to the ribosome?
To prevent free aminoacyl-tRNA from rebinding to the ribosome, free aa-tRNAs preferentially bind to eEF1A∙GTP, forming a ternary complex (eEF1A∙GTP∙aa-tRNA). This ensures that only tRNAs correctly bound to eEF1A∙GTP can enter the ribosome, preventing free aa-tRNAs from interfering with translation.
(5.8) For every peptide bond formed at the elongating ribosome, eEF1A hydrolyzes one GTP. How is the free energy of GTP hydrolysis transferred to the elongating ribosome, including auxiliary factors, and how is this energy eventually dissipated and for what purpose?
The hydrolysis of GTP to GDP of eEF1A pushes the reaction away from equilibrium, favoring the formation of eEF1A-GDP. This free energy is stored and later released when eEF1A-GDP interacts with eEF1B, catalyzing an exchange of GDP to GTP, regenerating eEF1A-GTP. Ensuring eEF1A-GTP is available to accurately deliver free aminoacyl-tRNA to the ribosome (forming a ternary complex).
Free energy is dissipated through GTP hydrolysis occurring when eEF1A-GTP binds to ribosome ensuring it is available to deliver tRNA to the ribosome.
(5.9) The elongating ribosome plus auxiliary factors increase the specificity of amino acid selection via three distinct mechanisms. Name the three mechanisms and state in one sentence the principle of by which selectivity is improved.
Helix recognition: increases the energy difference between correct and incorrectly matched aminoacyl-tRNAs.
Accommodation: Ribosome slows down peptide bond formation, allowing incorrect aa-tRNAs to dissociate before being incorporated into the peptide bond formation.
Kinetic proofreading: kinetically distinguishes between correct and incorrect aa-tRNAs twice before and after eEF1A∙GDP dissociation at the expense of free energy dissipation (GTP hydrolysis)
(5.10) Peptide bond formation requires the conversion of one ATP to AMP and two orthophosphates. What is the role of this conversion in peptide bond formation? Name the enzymes that convert ATP to AMP and orthophosphate for peptide bond formation.
The formation of a peptide bond via condensation of amino acids is an endergonic reaction. To drive this process, Aminoacyl-tRNA synthase activates amino acids by linking them to their respective tRNA via ester bond formation. This requires two ATPs: conversion of ATP to AMP, and pyrophosphate (splits into two orthophosphates). The ribosome then couples peptide bond formation to ester cleavage, an exergonic reaction (peptides are more stable than ester.
(5.11) The discovery that the ribosome is a ribozyme solved a fundamental problem in molecular biology. What problem?
The problem was, If all enzymes are proteins, including the enzyme that catalyzes protein synthesis, the ribosome, then how did proteins ever come into being? This could be solved through the catalytic activity of the ribosome brought up by RNA and not protein.
(5.12) It is generally thought that codon recognition by eRF1 does not involve kinetic proofreading. Argue why this might be true (cf. Slide 35, Lecture 8).
Ribosome and eRF1 do not require free energy input to reach the catalytically active complex. For this reason if eRF1 binds to an incorrectly non-stop codon, it could only be corrected once by reversing the initial binding reaction between eRF1 and ribosome. Unlike in the case of kinetic proofreading, where it can be corrected twice, disproving eRF1 involves kinetic proofreading.
(5.13) The mechanism of translation requires GTP hydrolysis for ribosomal movement along the mRNA, and ester bond cleavage for peptide bond formation. What would it 4 take to reverse this process and make the ribosome (preferentially) move in the 3′ → 5′ direction (and thus transfer amino acids from protein to tRNA in an mRNA directed manner)?
In the case of translation, the process is strictly not irreversible by mechanical design (it would violate the 2nd law of thermodynamics) but rather due to the large dissipation of free energy. Translation favors the forward direction (5’ -> 3’) because it is an exergonic process that is driven: by GTP hydrolysis ( 2 GTP converted in GDP) as well as ATP hydrolysis (two ATPs are converted into ADP) and protein folding (exergonic process). In order to reverse the process it will require: protein unfolding, altering GTP to GDP concentration ratio of phosphates and a very large concentration of tRNA, in order for the process to go against it’s natural (favored) direction.
(6.1) Most prokaryotic transcripts are polycistronic whereas almost all eukaryotic
transcripts are not. How is this difference mechanistically explained?
In bacteria, translation initiation relies on Shine-Delgarno (SD) sequences, which are complementary to 16S ribosomal RNA. Since multiple SD squences can exist on a single mRNA, translation on start at multiple sites, making it be transcript polycistronic. In contrast, eukaryotic ribosome scans from the 5’ end to find the start codon, restricting the number of the number of start site to one, resulting in most eurkaryotic transcripts be monocistronic.
(6.2) It was thought that initiator met-tRNA is special in the sense that it is the only aatRNA that binds to the P-site of the full ribosome.
(i) Which experimental observation
disproved this notion?
(ii) In which sense is the initiator met-tRNA special according to Nomura?
(i) Experiment with high magnesium concentration and in a cell-free translation system, revealed it could still translate RNA, even those who lack AUG. Thus, implies any aminoacyl-tRNAs can bind to the P site under certian conditions.
(ii) However, according to Nomura, initator Met-tRNA is special because its ability to not bind to the A site, instead directly involved in the ribosome assembly.
(6.3) It may be argued that “strong” Shine Dalgarno (SD) sequences might place any codon in the P-site of the ribosome, which would allow for the initiation of translation at any codon (provided it is properly positioned with regard to the SD sequence. Indeed, AUG is used by only 83% of E. coli genes, the rest employ GUG, UUG, and to a lesser extent also AUU. The latter do not encode for methionine, yet the synthesis of all proteins begins with methionine. Explain this observation by employing Nomura’s hypothesis for translational initiation!
A start codon is linked to ribosome assembly which requires the initiator tRNA, one specific aa-tRNA charged with a methionine.
Thus, regardless of the start codon used, the first amino acid incorported is always methione.
(6.4) What is the most significant chemical difference between RNA and DNA?
Provide examples for two distinct important roles of this difference.
The most significant chemical difference among DNA and RNA is the presence of a 2’ hydroxyl (-OH group) in the sugar ribose of RNA. Wheras DNA contains a deoxyribose, lacking this hydroxyl group. Two examples are the 2’ hydroxyl group causes 5’-phosphester bond more likely to undergo hydroysis. In addition the 2’ hydroxyl group plays a catalytic role in peptide bond formation.
(6.5) Picornaviruses inhibit translation of host cell proteins while their own RNA
genome is still translated. Explain the underlying mechanism!
Picornaviruses encodes a protease that cleaves elF4G, preventing the recuriment of the host mRNA to the 43S preinitiation complex. Therefore leads to inihibiting host cell protein translation. However, the viral RNA does not rely on the 5’ cap dependent scanning instead it initiates translation through an IRES (Internal ribosome entry site).
(6.6) Various kinases inhibits protein biosynthesis via phosphorylation of eIF2. Explain the underlying mechanism of inhibition!
Phosphorylation of elF2 pervents the exchange of GDP for GTP by inhibiting elF2B, functions as a guanine nucleotide exchange factor (GEF) for elF2. As a result the phosphorylated elF2 remains in its GDP-bound inactive state, preventing the formation of elF2-GTP. Since translation initiation required the ternary complex (elF2-GTP-initiator Met-tRNA) the loss of elF2-GTP blocks the initiation of translation.
(6.7) eIF4E-binding proteins inhibit translation by competing with eIF4G for eIF4Ebinding. How is this mechanism employed in the developing Drosophila embryo to inhibit the translation of specific mRNAs and not all mRNAs?
In the developing Drosophila embryo, this inhibition of elF4E-binding protien CUP is tethered to the specific mRNA via the sequence-specific RNA-binding protein Bruno. This tethering increases relative to concentration of CUP / elF4E, thus suggesting that only translation specific mRNA are affected via Bruno an CUP interaction.
The elF4E-binding protien CUP inhibits translation by competing with elF4G for binding to elF4E, inhibiting ribosome recruitment to the mRNA.
blocking elF4G from interactiong with elF4E
(6.8) What are the functions of the bacterial sigma factor?
Which eukaryotic factors serve the same functions instead?
The sigma functions are
(i) core promoter recognition;
(ii) open complex formation;
(iii) coupling promoter escape of RNA polymerase to RNA synthesis.
In eukaryotes, function
(i) is fulfilled by TFIID and TFIIB;
function (ii) by TFIIH and TFIIF;
and function (iii) by TFIIB.
In bacteria, sigma factor is a specialized subunit of RNA Polymerase, essential for transcription initiation, that includes 3 major functions: Core promoter recongition (helps RNA Poly identify/bind to the promoter region of DNA), Open complex formation (faciliates the unwinding of DNA at the promoter, allowing RNA poly to access the templete strand for transcription) Coupling promoter escape of RNA poly (once transcription begin, sigma factor helps RNA poly transition from initiation to elongation by releasing the promoter interaction.)
In the case of Eukaryotes, transcription initiation requires multiple transcription factors instea of one. In comparison to the bacteria, eukaryote includes: TFllD and TFllB, simiarly to the core promoter recongition, ensures RNA Poly bind to correct DNA sequnce. Similar to the open complex, TFllH and TFllF helps by unwinding DNA and stabilization, as for the coupling promoter, TFllB, helps RNA Polymerase transition from initiation to elongation.
(6.9) Promoter escape by RNA polymerase II is coupled to RNA synthesis. Describe the mechanism of this coupling!
RNA Poly II, is anchored to the promoter through its interaction TFllB. This in done through TFllB’ B-finger domain that is linked on the RNA exit channel of the RNA Poly II. As, RNA synthesis begins, the B-finger domain gets displaced, due to the RNA chain starting to grow. Resulting in a dissruption between the interaction of TBIIB and RNA Poly II. Resulting in RNA Poly II to lose its attachment to the promoter, thus initiate transcription elongation phase.
(6.10) Transcriptional activators encompass two functionally and structurally distinct domains.
Which are these domains and what are their functions?
Transcriptional activator contians two distinct domains, DNA binding domain and activation domain. The DNA binding domain is responsible for sequence-specific DNA recognition. In the case of activation domain,its role is to recurit cofactors that help activation of transcription. An example could be SAGA, modifies chromatin structure.
(6.11) Let
T + D ⇌ TD
be a binding reaction between a transcription factor T and a DNA molecule D. The
occupancy (w) of D by T is defined as the equilibrium fraction of D bound by T, i.e.,
w ≡
[TD]/
[D] + [TD]
where [∙] is the equilibrium concentration of ∙. Show that
w = [TD]
[D] + [TD],
where Kd, is the equilibrium dissociation constant of the reaction.
How does (w) change as you gradually increase the concentration of the transcription factor?
Consider the implications of your answer for the specificity of transcriptional regulation.
The second equation follows from first by multiplying the numeator and denominator with [T]/[T][D] and also using the defintion of equilibrum dissociation constant Kd= [T][D]/[TD]. As the concentration of transcription factor [T] increases toward inifinity, the occupancy (w) apporached 1, no matter the value of Kd.
Thus, indicates that at a high concentration transcription factor, T, will bind to DNA non specifically. In order to maintain specifity in transcription regulation, the concentration of transcription factor must be kept low.
(6.12) Use the result of the previous problem to explain how the Kd, of a DNA binding protein may be measured in an electrophoretic mobility shift experiment; suppose you know the total protein concentration but not the free concentration.
EMSA naked and bound DNA can be separated. According to the equation
above, at w = 0.5, the free protein concentration equals the kd,. The free protein
concentration is approximately equal to the total protein concentration as long as the
DNA concentration remains sufficiently below kd.
In an electrophoretic mobility shift assay (EMSA), free DNA and DNA bound to a protein can be separated
based on their mobility in a gel. To measure the dissociation constant Kd of a DNA-binding protein, the protein
concentration at which half of the DNA is bound ( ω = 0.5) is determined. At this point, the free protein
concentration [ T ] equals Kd . If the DNA concentration is kept sufficiently low compared to Kd , the free
protein concentration is approximately equal to the total protein concentration. By identifying the total protein
concentration that results in half-maximal DNA binding ( ω = 0.5), K d can be directly estimated.
(6.14) It was thought that transcriptional activators activate transcription by
recruitment of TBP via Tafs, i.e., Tafs serve as adapter proteins for TBP recruitment.
What experimental results argue against this conjecture?
Although all gene require TBP for transcription, not all require TFIID, the
Taf-independent genes. Thus, not all activators work by recruiting TFIID to promoters.
It may be argued that activators of Taf-dependent genes do. But this is also false: The combination of UAS from Taf-dependent genes with core promoter sequences from Tafindependent genes is Taf-independent. Second, Mediator but not TFIID has antisquelching activity.
While all genes require TBP for transcription, not all require TFIID, which contain TBP (TATA-binding protein) through TAFs (TBP-associated factor). Some genes are TAF-independent, meaning they can initiate transcription without TFIID. This proves not all activator recruit TFIID to promoters, contradicting the idea of TAFs being essential for TBF recruitment. Additionally, experiment shows using mixture of UAS from a TAF-dependent gene with a TAF-independent core promoter result in TAF-independent transcription, Proving that activator do not nesssarily rely on TAFs. Lastly, rather than TFIID, Mediator prevent squelching (gene regulation), resulting in antisquelching activity, further supporting that TAFs are not always required for activiator function.
TBP is a subunit of both TFIID and SAGA. In both complexes other subunits
interact with TBP such that TBP cannot bind DNA.
What might be the purpose of this inhibition?
TBP is required for the transcription of all genes, i.e., it must be present in large
concentration. Specific and many nonspecific binding sites would be occupied all the time. Inhibition of TBP prevents such interactions, until activators facilitate deinhibition.
TBP (TATA binding protein) serves as a subunit for TFIID and SAGA, however in both complex, other subunit will interact with TBP and inhibit it from binding onto DNA. TBP is required for transcription of all genes, therefore TBP has a high concentration presence. Without this inhibition regulation, TBP will be capable of binding to either specific or non-specific site. Thus, leads to uncontrolled transcription and gene misregualtion. Though this inhibition of TBP it will ensure TBP binds to DNA only when activators signal and prevent unwanted interaction.
Transcriptional activators encompass two functionally and structurally distinct
domains. What are these domains and what are their functions?
DNA binding domain and activation domain. The first is required for
sequence-specific DNA recognition, the second for recruitment of cofactors for
activation (e.g., SAGA, Mediator).
(It was thought that Mediator solely functions as an adaptor for the recruitment of RNA pol II by activators to promoters. What experimental results argue against this conjecture?
First, the loss of Mediator could only be overcome by Mediator and not an
increase in concentration of any other GTF or RNA polymerase. Second, loss of the
Mediator tail abolishes Mediator association with UAS elements, but not core
promoters.
Mediator stimulates the phosphorylation of the carboxy-terminal domain (CTD) of
RNA Pol II by Kin28 a TFIIH subunit. CTD phosphorylation is coupled to promoter
release. Which experimental observation argued against the notion that Kin28
stimulation is the sole (or even most important) function of Mediator?
The ability to stimulate Kin28 and to activate transcription can be uncoupled;
loss of Srb4 activity does not interfere with CTD phosphorylation, but prevents
transcriptional activation.
In an in vitro transcription reaction a high concentration of Gal4 is used to inhibit
(squelch) transcription from a promoter that is activated by the transcriptional activator
Gcn4. You have isolated a yet unknown activity M that has anti-squelching activity, i.e.
addition of M alleviates squelching of Gcn4-activated transcription. You reason that M
might be Mediator, an activity is required for activator-dependent transcription which,
therefore, is the subject of competition between activators. However, there is an
alternative hypothesis (H): Suppose M is an inhibitor of Gal4 (H), and not Mediator.
Propose a simple in vitro transcription experiment that allows you to test H. Describe the expected outcomes of your experiment for both hypotheses ¾ if H is true, and if M is Mediator.
Set up a transcription reaction in vitro with a Gal4-dependent promoter and
Gal4. If M is an inhibitor of Gal4, it will inhibit transcription. Mediator, however, would
instead promote transcription.
What are the diameter and height of the nucleosome core particle (in nanometers)?
How much DNA is bound to the histone octamer (in base pairs)? How often is the DNA spooled around the histone octamer and what is the handedness of the spooling. By which factor, approximately, is the size of RNA polymerase smaller or larger than that
of the nucleosome?
Diameter is 10 nm, height 6.5 nm, 147 base pairs wrap around the histone
octamer in a left-handed helical path of 1.65 turns. RNA Pol II is ~1.5-fold larger than
the nucleosome.
With 140 non-covalent bonds of an energy per bond of, say, 2.5 KbT, we would
expect that it takes 350 KbT to unspool a nucleosome. Magnetic tweezer experiments on single nucleosomes however indicate that it only takes about 45 KbT.
How may this discrepancy be explained?
Energy is required to bend DNA. The nucleosome is “spring-loaded.”
Although histones make contacts with the sugar phosphate backbone of DNA,
rather than the bases, nucleosomes are more stable on some DNA sequences than
others.
What is the characteristic of such sequences?
What is the physical property of
such sequences that renders them more suitable for nucleosome formation?
Nucleosomes preferentially form at sequences with a 10 base pair periodicity
of A/T-rich dinucleotides. Such sequences bend more easily bend around the histone
octamer.
Argue why the regulation of transcription in most if not all eukaryotes requires DNA looping? Why might yeast be an exception?
Eukaryotic activators are thought to stimulate transcription by recruiting TBP
(SAGA, TFIID) and holoenzyme (Mediator plus RNA pol II) to enhancers. These factors
must be handed over to core promoters, which requires the looping of the intervening DNA when the distance (in bps) between enhancers and core promoter is large. In yeast, the distances between UASs and core promoters are generally quite short
What is a topologically associating domain (TAD)? By what mechanism may
TADs confine the interaction of core promoters with enhancers?
A TAD is a chromosome domain whose segments are more likely to interact
with segments within the TAD than outside the TAD. TAD boundaries, CTCF binding
sites, may confine loop extrusion by SMC proteins. Enhancer-core promoter interactions that are mediated by SMC complexes will then be confined to interactions within the same TAD.
