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1. bc the signal within the introns can changestrength and allow different selections of exons and splicing of differentintrons out
(also called polyribosome)
string of ribosomes simultaneously translating regions of the same mRNA strand during protein synthesis
Compare the E.coli genome to the human genome
sum of all the nucleotides needed to encode and regulatea RNA or protein
How does the transcription unit (trp operonused) work in PROKARYOTES (ex bacteria)
the dna sits in the genome, the five genes sitnext to each other and have one start site for RNA transcription, so one RNAgets made across the genes, and then that RNA is translated into multipleproteins. the proteins encode enzymes that allow you tosynthesize tryptophan
there are five enzymes that catalyze theconversion from the precursor to Trp. therefore all enzymes encoded are needed tomake tryptophan, which we need in all of our proteins
when rna is get transcribed in bacteria it canbe immediately translated. theres no nucleus, so it comes out of the cytoplasmdoesn’t need to be spliced or anything. can immediately go from RNA to protein
a. what does E.coli’s polycistronic model tell usabout its genome?
i. itis more evolutionarily advanced that human genome. it has streamlined its DNAto get rid of all the “junk” DNA. have selected for smaller genome
1. what are advantages/disadvantages of a small orlarge genome?
i.faster to replicate
ii.no wasted DNA, mutation more likely to hit a gene
i.slower to replicate
ii.less likely that a mutation will hit important gene
1. how does the transcription unit work ineukaryotes?
a. an RNA gets made and chunks of the rna getsspliced out to make the final mRNA that gets translated into protein.
b. must go from pre-mRNA (includes exons andintrons), undergo splicing, becomes mRNA , which is then translated intoproteins
c. premRNA cannot be translated into proteins
1. exons stand for expressed region, intron isintervening sequence
1. slicing machinery occupies a huge amount ofenergy in the nucleus, and is really important why do we keep this mechanism if it is so energyconsuming?
a. each exon has multiple alternatives
b. 38000 proteins can be made from this single gene
1. what happens when alternative splicing creates abad protein? (example growth hormone)
a. it can happen where one exon is partiallyincluded or skipped altogether
b. if exon 3 is skipped, creates a dominantnegative—they may inherit one good, and one bad, but the bad gene dominates,and the more you make of it the shorteryou are
c. a mutation in the growth hormone gene thataffect splicing
a. maybe we make a nonfunctional growth hormone atbirth, and it’s supposed to switch to functioning after birth, and in thoseindividuals who it doesn’t just have a malfunction? this allows the infant tofit out of the birth canal
b. you receive pulses of growth hormone, it tailsoff around the age of 25
c. hard to test whether they are dopin or not bchgh is actually a naturally occurring substance
a. for a given intron we have a 5’ splice site, abranch point and a 3’ splice site. introns always begin with a GU and end withan AG, but surrounding the GU there are some preferred sequences. (consensus)
a. U1 recognizes the 5’ splice site U2 recognizesthe branch point.
thealternative splice sites will not have consensus sequences but the sites thatare used all the time have strong consensus sequences, then you can regulatethe mixing and matching of exons and introns
a. the family has mutations in the splice sitesthat surround exon 3 and it makes their splice sites weaker, the splicingapparatus just misses it and that’s how you get this deficiency
a. in prokaryotes- the dna is not in chromatin, sothe pol is recognizing the dsDNA directly, and does not have to navigate aroundnucleosomes
b. eukaryotes- pol have to navigate around thenucleosomes, RNA has to be transported out of the nucleus into the cytoplasm.there the rna gets translated
b. RNA polymerase
describe the difference in how RNA polymerase knows where the startsite is on a gene in bacteria to humans
a. not as complicated in bacteria bc chromosome issmaller, and the promoters are close together
b. in eukaryotes- theres a large amount of DNA forRNA polymerase to ignore, so something has to recruit it to the start sitebecause we have such an abundance of noncoding DNA
a. transcription is regulated so that only thegenes you need are made, ex a dormant cell does not need flagella if it willdoes not need to be mobile
a. with the sigma subunit positioned and binding tothe DNA’s major groove at the set sequence at -10 and the set sequence at -35,in contact with the bases, some of the atoms on the bases provide template forhydrogen bonding, others are for polymerase to attach.
a.the strength of the promoters at -10 and -35
i.the promoter is bound by sigma
ii.how similar they are to the preferred sequence
iii.the space between the promoters
a. when sigma binds to the -10 and -35 region, thenrecruits other enzymes involved in transcription
beta, beta prime, two alphas, sigma, andsometimes omega, also called RNAPolymerase
a. the core enzyme is holoenzyme without sigma, canbind DNA @ low affinity, just collides and grabs on and surfs along, but onlythe holoenzyme will engage the promoter and start initiation
b. after about 10 nuc, sigma leaves and allows core unit to continue synthesis of transcript
a. sigma’s concentration is limiting and thereforethe sigma subunit is used over and over, the concentration of core enzyme ishigher
a. strength of the sequences at -10 and -35,(thestrength of the promoter), if it matches the consensus you get more frequenttranscription, if not, less
b. differential use of the sigma subunit(what conditionsare the cell under, what genes need to be synthesized at the time, so whatsigma is used)
a. bacteria have other sigma subunits that performthe same task under different conditions
ex. the promoter sequences vary among sigma subunits
What protein is involved in the termination oftranscription in bacteria?
Rho- RNA binding protein
a. the newly formed RNA has elaborate structure,maybe a stem and loop, the polymerase has a hard time making the RNA, maybe assoon as it is made it folds into this structure, so the polymerase slows down,Rho can catch up, and makes polymerase fall off
caused by - built in sequence. or nucleotide deficiency
a. Rna makes a stem loop followed by a strand of Uresidues, causes polymerase to fall off.
1. What are regulators of transcription in bacteria?initiation and attenuation
b.attenuation in tryp operon
i.if the protein being coded for is already present in high amounts, it is not needed, thus RNA synthesis will stop
asthe ribosome translates region 1, it recognizes whether the tryptophan isalready present in high amounts, if it is the ribosome will slow causing a stemloop to form between region 3-4 on the RNA, which will cause transcription toterminate, if it is present in low amount a stem loop will form between regions2-3 but transcription will be continued.
What regulates transcription in Eukaryotes?
a. at the -25 position there is a consensussequence TATA box region, this region is bound by TBP (tata binding protein)functions similar to the (-10,-35 region in bacteria)
TBPbinds to the minor groove, is a key,involved in other polymerases too, but onlywith Pol II does it bind to TATA box, the minor groove is narrow and deep butTBP is able to do this.TBP forcing itself into the minor groove causes the DNAto kink and distort
TBPhas to be directed to the TATA box by a combination of proteins(transcriptionfactors.) that bind to enhancers(sequences on DNA) and promoter proximalelements.
thereare enhancer elements around which are also proteins, these activation domainswill make contact with one another which promotes the assembly of factors thatrecruits TBP to bind to the TATA box in the minor groove causing the DNA tobend, that signals the RNA Pol to be able to find the start site of the geneonthe DNA
1. : 1) the addition of a cap at the 5’ end (7methyl-g cap); 2) poly A tail addition; 3) splicing to get rid of the intronsand join the exons together,
the final mRNA has a 5’cap, a 5’ UTR, the codingregion, 3’ UTR, and poly A tail
a. UAA, UAG, UGA
a. At protein level
a. they enter into the production of RNA, andmutated gene produces mutated RNA that produces a mutated protein
A sequence of three adjacent nucleotides located on one end of transfer RNA. It bounds to the complementary coding triplet of nucleotides in messenger RNA duringtranslation phase of protein synthesis.
1. the trna sythetase charges it with ATP andproofreads itafter the trna has anamino attached to it, trna itself can somehow ensure that the amino acidattached to it matches the anticodon
trna synthetase takes ATP adds it to the aminoacid, then adds that (which is now AMP-amino acid) to the 3’ OH end terminalend of the tRNA.
i.C or U of the mrna can be paired with G of the tRNA
ii.so if Deamination has occurred forming inosine in the 1st position of the anticodon or 3rd position of the codon, it can pair with A, U, or C
1. because we have this flexibility of the thirdposition of the codon, pairing with multiple different possibilities on thetRNA,now we don’t need a full compliment of tRNAs, we can get by with fewer because theres thiswobble position
a. a single mRNA will have multiple ribosomes on itat a given time
this allows genetic material to be amplified
1. the ribosome is located over 3 codon, and astranslation proceeds, there regions of the mRNA serve as sites: E is the entrysite, P is the peptide site, and A is where the charged tRNA is delivered orthe next round of synthesis
i inEukaryotic setting the cap defines which one
ii. inprokaryotic setting the ribosome has the task of determining where to start, soit is much more complicated that eukaryotic setting
joiningtogether of the small subunit, the mRNA and the charged trna over the correctAUG to begin, then recruit the large subunit to start. lots of initiationfactors are involved in this process
i. inbacteria there is a sequence in front of the AUG called the Shine-dalgarnosequence (upstream 5 or 10 nucl)
consensus sequence, determining translation efficiency
the cell theory includes all of the following EXCEPT
What role does the active site Histidine play in thehydrolysis of a peptide bond catalyzed by chymotrypsin?
-Form part of the active site
-Donate a proton to histidine
-Initiatea nucleophilic attack on the carbonyl carbon of the substrate
-Form a transient covalent bond with the substrate
Ser, Thr, Tyr
Water is very abundant and very effective at competingfor the ionized functional groups. Therefore, ionized groups will hydrogen bond with water rather than formionic bonds with each other.
Theenzyme OMP decarboxylase was placed in a solution containing a highconcentration of urea and was found to rapidly lose its enzymaticactivity. Removal of the urearestored the enzyme’s activity.
A. What is urea most likely doing to disrupt the enzyme’s activity?
Causingthe protein to unfold by competing for H-bonds
The enzyme refolded back into its native state.
Muscle (because hexokinaseworks faster at 3.5 mM and therefore more glucose can be trapped in the cellsexpressing hexokinase).
Inglycolysis, the enzyme hexokinase converts glucose to glucose 6-phosphate. What two purposes does theinvestment of 1 ATP at this step serve?
If hexokinase was inhibited,glucose would not be trapped in the cell and stored.
Introduction of double bondsinto the fatty acyl chains of membrane phospholipids.
Human beings have approximately _____ fold more genes than the bacteria E. coli.
This organelle is similar in size and appearance to a bacterial cell.
The phosphoanhydride bonds in ATP are considered “high energy” because
Much more energy is released upon formation of hydrolytic productsthan is needed to break bonds in the reactants.
Alzheimer’s disease may be caused by aggregation of a protein fragment (Aß)extracellularly in large, abnormal structures called _________.
At a pH of 3.0, the product of [H+] x [OH-] will be ______
Which polysaccharide bond cannot be broken by mammalian enzymes that normally digestpolysaccharides?
What is the probability of finding the amino acid sequence VANDYWINS in aprotein?
What property is not a characteristic of enzymes?
For glucose, rotation around carbon 1 as the ring structure opens and closes generates
For a membrane protein where the polypeptide chain spans the thickness of themembrane bilayer (4 – 5 nm), about _____ hydrophobic amino acids are needed whenthe chain is in an α-helix, and about ______ are needed if the chain is in a β-strand.
In the Cn3D program for visualizing the 3D structure of proteins, what happens whenyou go to the Coloring Shortcut and choose “Charge”?
Charged residues turn red (for acidic residues) or blue (for basic residues)
The change in Gibbs free energy is described by the formula _______
∆G = ∆H - T∆S
What level of structure in proteins is held together by intermolecular R group interactions?
Lipid rafts are enriched in the lipids __________.
When human cells die, this lipid gets exposed on the outer leaflet of the plasmamembrane.
The approximately 200 different human cell types in adults are produced from ______cells through the process of ______.
By definition, a kilocalorie (kcal) is the amount of energy required to do what?
Mammals lack the enzyme that hydrolyzes cellulose. Yet many mammals are herbivores andthey eat grass and other plant material for nutrition. How can this be?
Bacteria living the guts of herbivores have enzymes that can degrade cellulose and aid in thedigestion.
The coloration patterns in Siamese Cats arise from a temperature-sensitive mutation. What is happening with the enzyme at themolecular level that explains this?
The enzyme has a mutation that perturbs folding at 37C (or normal bodytemperature), but does not perturb folding at the lower temperature. Tertiaryand quaternary structures are held together by weak bonds (H-bonds, etc.) thatcan be broken by increased temperature. The mutation likely reduces the numberof weak bonds that maintain the proper structure.
Describe how this sugar is metabolized in liver cells and why it is metabolized differently(what enzymes are different?)
Liver cells express glucokinase instead of hexokinase, and glucokinasecannot phosphorylate fructose. A different set of enzymes in the liver willphosphorylate and break fructose down into 3-carbon sugars. Fructose entersglycolysis at DHAP (dihydroxyacetone phosphate) and GAP (glyceraldehyde 3-phosphate).
In one step of glycolysis, glyceraldehyde 3-phosphate dehydrogenase catalyzes the additionof Pi to glyceraldehyde 3-phosphate. This reaction has a positive ∆G of +12 kcal/mol. How doesthe enzyme catalyze this energetically unfavorable reaction?
The addition of Pi is coupled to the oxidation of the aldehyde on glyceraldehyde 3-phosphate and transfer of high energy electrons from the substrate to NAD+ toform NADH. This oxidation reaction has a large negative ∆G of slightly morethan -12 kcal/mol. Therefore, the combined set of reactions catalyzed byglyceraldehyde 3-phosphate dehydrogenase has an overall negative ∆G.
The ∆G for movement of Na+ down itselectrochemical gradient into the cell is -3.1 kcal/mol at typicalintracellular and extracellular Na+ concentrations and a membrane potentialof -70 mV (negative inside, positive outside). If the membrane potential decreased to -20 mV, the ∆G would become
Electron transport complexes pump protons fromthe ______ to the _______?
3 Na+ ions bind and are pumped out of the cell(cytosol -> extracellular space)The terminal phosphate is transferred from ATPto the protein. (also ok… ATP ishydrolyzed leading to phosphorylation of the pump. ATP binding and phosphate transfer to the pumpdrives the E1 to E2 transition.)
2 K+ ions bind from theextracellular side and are pumped into the cytosol (extracellular space -> cytosol).
The phosphate is releasedfrom the pump, allowing it to relax back to the E1 conformation.
It pumps 3 Na+ ions out of the cell and 2 K+into the cell for each ATP hydrolyzed.
The net displacement of +1 charge per ATP hydrolyzedcreates an electrical gradient across the membrane (or a membrane potential). (The pump creates an electrical current bythe net displacement of charge across the membrane.)
To induce rotation, the ring subunits push against subunit ____ as each proton binds.
-It is the most abundant protein on Earth.
-It catalyzes both carbonfixation and photorespiration.
-It is soluble in the stroma ofthe chloroplast
-It is part of the Calvin cycle
What must a protein have in order to leave the cytosol and reside within a membrane-bound organelle?
A. recognizes ER signalsequences
B. causes a pause in translation of secretory proteins
C. facilitates transfer of newly synthesized proteinsto the ER translocon.
D. helps deliver ribosomes to the ER to form the roughER.
A nuclear localization signal of the type found inT-antigen is
1. Fall back toground state & emit heat to the environment.
2.Fall back to ground state & emit a photon of light that is lost to theenvironment. (Fluorescence)
3.Fall back to ground state & emit a photon that is captured by anotherchlorophyll (fluorescence resonance energy transfer - FRET).
4. The excited e- can be transferred to a series of electron acceptors in an e- transport chain to build up a proton gradient/reduce NADP (photooxidation)
(reduction means that an electron is added.)
NADP+ is the final electron acceptor of the electrons transport chain(ETC) Remoting electrons allows their continual flow down the ETC. This generates NADPH which is required as an elctron donor in the calvin cycle.
After P680 absorbs a photon of light and becomes oxidized to P680+, it is reduced backto P680 with an electron extracted from ________
TheRanGEF is localized to the nucleoplasm (by interaction with chromatin).
TheRanGAP is localized to the cytoplasm (because it lacks any sorting signal).
Fusion of this secretory vesicle with its target membranerequires both long fibrous and multisubunit _____ complexes, which are recruited to the correctmembrane by a small GTP binding protein in the ___ family. The final fusion of the membrane bilayers isdriven by a _____ in the vesicle (or VTC) membrane and a ______in the target membrane.
1. By inhibiting some of the KATP channels,sulfonylurea drugs will allow the beta cell membrane to depolarize more easilyas ATP levels rise
2. The depolarized membrane causes a calcium channel to open, increasing the cytosolic concentration of calcium
The mutation likely disrupts the ability of the LDLreceptor to bind to clathrin coat components in clathrin-coated pits. The LDLreceptor normally cycles between the endosomes and plasma membrane many times,removing an LDL particle from the serum every trip. The receptors in thepatient that could not undergo endocytosis would have a greatly diminishedcapacity to remove LDL from the bloodstream.
Dynein (Ciliary dynein) (Outer and inner dynein arms)
The patient would be sterile because he lacksdyneins that load onto the A microtubules. Thus, the cilia will not be able tobeat, and the sperm will be immotile.
The phospholipid bilayer of the red blood cell plasma membrane is supported by peripheralmembrane proteins called ________ that help prevent mechanical rupture.
Ankyrin and spectrin
The Na+/K+ ATPase:
A. pumps 3 Na+ out of the cell per ATP hydrolyzed.B. pumps 2 K+ into the cell per ATP hydrolyzed.
C. generates an electrochemical gradient.
D. can consume as much as 50% of the ATP in a cell.
For regulated secretion in beta cells of the pancreas and in neurons, opening of a______-gated ________ channel is required to drive exocytosis
Which product of the TCA cycle donates electrons to complex II of the electrontransport chain?
Which photosynthetic complex in the thylakoid membrane is most analogous to complexIII in the mitochondrial inner membrane?
Proteins are transported into this organelle using a gated transport mechanism and mosttypically using a T-antigen type of localization signal comprised of ~5 basic amino acids.
The nucleolus is the site of _________ synthesis.
The genome of the bacterium E. coli is considered to be “streamlined”, meaning that nearly all of theDNA encodes genes and there is little extra DNA. In contrast, human genomes are large and have lotsof DNA that does not encode genes, so-called junk DNA. Describe one possible evolutionary advantage for bacterial cells to keep their genome size small.
Describe one possible evolutionary advantage for human cells to having a large genome.
Chance of a single mutation hitting a vital nucleotide or gene is low
Describe one possible evolutionary disadvantage for human cells to maintain a large genome.
Longer time to replicateGreater chance of mutation
Humans inherit one set of genes from their father and one set from their mother therebyexpanding the size of the genome even more, i.e diploid. Describe one advantage to having two sets ofgenes in humans.
Inherited mutations on one set can be offset by normal gene in the other set
When an origin is first recognized and replication begins, two primers are made even thougheventually 4 strands will be synthesized at the two replication forks. Draw a picture of this event and thendescribe why only two primers are immediately made instead of 4?
Since replication can only proceed 5'-3', there is a leading and lagging strand. RNAprimers can be laid down on the leading strands right away but the helix must open upfurther before the lagging strand primers can be laid down.
If the sequence of the RNA in Telomerase is 5’-CCCGGUG-3’, what would be the sequence andpolarity of the telomeric DNA repeat? Would this be a prokaryotic or eukaryotic cell?
hsp70 is a gene that gets turned on during cell stress, particularly after exposure of cells to heat. Broadly, how is the incredible amount of DNA in mammalian cells able to fit within the nucleus?
DNA is packaged into nucleosomes which consist of ~2 loops of DNA wound around a coregroup of 8 histones. The nucleosomes can be further compacted into heterochromatin and areeven more compact during mitosis
hsp70 is a gene that gets turned on during cell stress, particularly after exposure of cells to heat. What is the term used to describe the state of the chromatin around the hsp70 gene in normal,unstressed cells?
hsp70 is a gene that gets turned on during cell stress, particularly after exposure of cells to heat. What is the term used to describe the structure of the chromatin around the hsp70 gene in cellssubjected to stress or heat?
Euchromatin. Activation of the hsp70 gene will cause a rearrangement of the chromatinsurrounding the hsp70 gene allowing gene activation.
hsp70 is a gene that gets turned on during cell stress, particularly after exposure of cells to heat. What controls the change in the chromatin around the hsp70 gene in stressed versus unstressedcells?
The amino terminal tails of the 8 core histones in the nucleosome can be modified byenzymes that either acetylate or methylate the histones.
A chemical mutagen called EMS causes transfer of alkyl groups (carbon chains) to certain bases.Below is a base that contains an alkyl group (R) added by exposure to EMS. What is the base above to which the alkyl group was added?
A chemical mutagen called EMS causes transfer of alkyl groups (carbon chains) to certain bases.Below is a base that contains an alkyl group (R) added by exposure to EMS. How do you expect this modification to affect base pairing?
The oxygen that is now alkylated is normally double bonded making it a good hydrogenacceptor whereas with the addition of the alkyl group, it cannot participate in H-bonding.
Draw the structure of dCTP
For both structures, what base will pair across the DNA helix? Circle the atoms involved inpairing.
Circle three atoms on dCTP and two on the imino form
There are mutant forms of many proteins that are referred to as temperature sensitive mutations. Replication proceeded normally at the permissive temperature but for eachsituation, indicate what would be the specific effect on replication after a shift to the non-permissivetemperature.
A temperature sensitive helicase:
A temperature sensitive 3'-5' exonuclease activity (DNA Polymerase III)
A temperature sensitive 5'-3' exonuclease activity (DNA Polymerase I)
No ability to repair gaps between Okazaki fragments or repair mistakes after replication.
Lots of mutations.
Synthesis and rapid degradation of cyclinproteins are essential for proper cell cycle progression and checkpoint control. What will likely happento the replication status of a cell that suffers a mutation disabling the degradation machinery? Why?
Unregulated passage through the cell cycle due to loss of checkpoints controlled by cyclin-cdk complexes. Rapid proliferation, cancer.
After replication, a portion of one of the daughter chromosomes looks like the following with onemismatch in the middle (a T—G pairing). Assume that the top strand is the parental strand and that thisDNA is from a prokaryotic organism.
Describe the steps needed to repair the mismatch:
A repair endonuclease recognizes the mismatch
The endonuclease cleaves the daughter strand to expose a 3’-OH
DNA Polymerase I begins synthesis from the exposed 3’-OH using its 5’-3’ exonuclease
activity to remove the existing strand and its 5’-3’ polymerase activity to add new DNA
copying the parental strand
DNA ligase rejoins the strand
How would the repair machinery distinguish the parental strand from the new daughter strand?
What property of DNA Polymerase I is essential for this type of repair?
What is the most likely mechanism for how this mismatch arose?
A tautomeric shift that resulted in Tenol:G
A) DNA polymerases proofread newly synthesizedDNA
B) Repair endonucleases nick the DNA to expose3’-OH groups
C) DNA Polymerase I has 5’ to 3’ exonucleaseactivity
D) Theparental DNA is recognized due to methylation
What would youpredict would happen to the error rate of DNA replication if the equilibriumconstant between tautomeric forms was smaller by two orders of magnitude (i.e.100x smaller)? What if the equilibriumconstant was larger by two orders of magnitude?
What would happen toa rapidly dividing cell that lost the main gene that methylates DNA. Why?
No ability to distinguish betweenthe parental and daughter DNA strands so that postreplication repair will havea 50/50 chance of repairing damaged DNA correctly. Thus, more mutations.
Predict the effecton DNA replication for a prokaryoticcell that lost the gene for telomerase.
Why is it important that prokaryotic transcripts terminateat specific positions whereas termination is seemingly random in eukaryotes.
More sequence information contained within the mRNA itselfis required to initiate translation in prokaryotes than in eukaryotes.
What sequences are these and why is more informationneeded in prokaryotes?
1. The Shine-Dalgarno sequence which pairs witha region in 16S rRNA in the small subunit of the ribosome.
2. The start codon which pairs with a methioninetRNA.
Because prokaryoticmRNAs are often polycistronic, the sequence information in the RNA itself is neededto guide the ribosome to the correct AUG start sites.
More sequence information contained within the mRNAitself is required to initiate translation in prokaryotes than in eukaryotes. What allows eukaryotic ribosomes to initiate properly andefficiently without such extensive sequence information in the mRNA?
The Cap structure, a7methyl G residue added to the 5' end of the mRNA via a 5'-5' bond.
Describe the effects on transcription and explain an increase in thedistance between the -10 and -25 elements for sigma.
A mutation thatchanges the promoter from TATA to TGTG
TBP will not bindwith high affinity so the transcription rate will decrease; will still occur at VERY low rates
Effect on transcription of an insertion of100 nucleotides between the TATA box and the normal start site.
No effect, the promoters for tRNA genes arewithin the gene itself.
Atriplet of nucleotides that specifies a single amino acid
How many codons are there in the genetic code?
How can a smaller number of tRNAs pair with all the codons?
7-methyl guanosine triphosphate. This is the modified nucleotide that formsthe cap structure that is added in a 5'-5' bond to the first base of mRNAtranscripts
the creation of alternative reading frames when translating an mRNA.
During transcription of DNA to RNA in prokaryotes:
an RNA polymerase must first bind to a promoter sequence.
Since the two strands of DNA are complementary, for any given gene:
the ribosomal small subunit and tRNAmet are joined by the large subunit over the start codon.
the site where TBP binding leads to recruitment of RNA polymerase II.
Each of the 3 potential reading frames of an mRNA are used to make protein (T/F)
Initiation of translation in prokaryotes involves binding of the sigma factor to a promoter (T/F)
In a prokaryotic ribosome, catalysis of peptide bonds is driven by the ______subunit, whereas identifying the start site is driven by the ___________ subunit.During elongation, each in-coming aminoacyl-tRNA binds to the __-site of the ribosome,whereas the growing peptide chain is held on the tRNA in the _P__-site.
In bacteria, the protein called the _____________ associated with theRNA polymerase is principally responsible for binding to the promoter DNA.
-Occurs between the third base of the codon and the first base of the antibodon
-Is necessary because there are more codons than tRNAs
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