molecular bio

Question 1

DNA synthesis

Answer 1

it’s catalyzed by DNA polymerases: they make sure that the incoming nucleotide is the right match for the template, and if there’s a perfect match the DNA polymerase catalyzes the reaction. it monitors the ability of the incoming nucleotide to form an AT or GC base pair, rather than detecting the exact nucleotide that enters the active site. DNA polymerases can distinguish between ribo and deoxyribo triphosphates: this discrimination is mediated by the steric exclusion of rNTP from the active site. two metal ions bound to DNA pol catalyze nucleotide addition: the palm domain contains the primary elements of the catalytic site and it binds two metal ions (Mg or Zn) that alter the chemical environment around the correctly base-paired dNTP and the 3’OH of the primer. it also monitors the base pairing. the fingers bind to the incoming dNTP and once a correct pair is formed the fingers enclose the dNTP. the thumb interacts with DNA that has been sythesized recently, maintaining a strong association between the substrate and the poly. the incoming nucleotide base pair with the next available template base. this interaction causes the fingers of the pol to close around the dNTP. the metal ions catalyze the formation of the next phosphodiester bond. attachment of the base paired nucleotide to the primer leads to the reopening of the fingers and the movement of the primer:template junction by one base pair. the degree of processivity is defined as the average number of nucleotides added each time the enzyme binds a primer:template junction.

Question 2

proofreading exonuclease

Answer 2

the removal of mismatched nucleotides is facilitated by the reduces ability of DNA pol to add a nucleotide ajacent to an incorrect pairing. the 3’OH group of the wrong nucleotide is not on the active site anymore so there’s a conformational change in the enzyme and it gets exposed to another region, the exonuclease active site. exonuclease > an enzyme that cleaves nucleotides one at a time from the end of a polynucleotide chain.

Question 3

the replication fork

Answer 3

the junction between the newly separated template strands and the unreplicated duplex DNA. the antiparallel nature of DNA creates a complication for the simultaneous replication of the two exposed templates at the replication fork. Because DNA is synthesized only by elongating a 3’ end, only one of the two exposed templates can be replicated continuously as the replication fork moves. On this template strand, the polymerase chases the replication fork. The newly synthesized DNA strand is the leading strand. On the other strand, the template directs the DNA polymerase to move in the opposite direction of the replication fork. The new DNA strand directed by this template is known as the lagging strand the resulting short fragments of new DNA formed on the lagging strand are called Okazaki fragments. They get covalently joined together to generate a continuous strand of DNA shortly after being synthesized.

Question 4

DNA replication

Answer 4

primase is dedicated to making short RNA primers on an ssDNA template; it adds ribonucleotides and not deoxy. RNA primers must be removed to complete the DNA replication. primase activity is increased when it associates with another protein that acts at the replication fork > helicase. this protein unwinds the DNA. removal of primers can be thought as a DNA repair event. to replace the RNA primers with DNA; RNase H recognizes and removed moth of each primer. removal leaves a gap perfect for DNA pol: it fills this gap until every nucleotide is base paired, leabing a DNAmolecule that is complete except for a break in the phosphodiester backbone. this nick can be repaired by DNA ligase. ligase is a 3 step process. 1> binds to ATP = release of energy; 2 > charged enzyme transfers the AMP to the 5’ = pyrophosphate bond; 3 > nucleophilic attack

Question 5

DNA pol

Answer 5

E.Coli > 5 = DNA pol III is the primary enzyme and is generally found to be part of a holoenzyme. DNA pol I is specialized for the removal of the RNA primers
eukaryotes > three are essential: δ, ε and α. DNA pol α initiates synthesis of new DNA strands. It consists of a two-subunit DNA pol α and a two-subunit primase. After the primase synthesizes an RNA primer, the resulting RNA primer:template junction is handed off to the associated DNA pol α to initiate DNA synthesis. DNA pol α is rapidly replaced by the highly processive DNA pol δ and pol ε (which do not have primase activity). DNA pol ε synthesized the leading strands and DNA pol δ the lagging strand.

Question 6

initiator and replicator

Answer 6

initiator > a six protein complex, ORC (origin recognition complex), which recognizes a conserved sequence, the A element, as well as a less conserved one, the B element. binding of the ORC does not lead to strand separation. all initiator proteins select the sites that will become origins of replication. the initiator protein is the only protein involved in the initiation of replication.
replicator > includes a binding site for the initiator protein that nucleates the assembly of the replication initiation machinery, including a stretch of AT-rich DNA. the initiator recognizes a DNA element in the replicator and activates the initiation.

Question 7

initiation

Answer 7

loaded helicases are activated by protein kinases CDK and DDK. the regulation is tightly coupled to the functions of CDKs: activate loaded helicase and inhibit helicase loading. CDK levels are low in G1 (the helicases are being loaded but are not active). as the cell cycle moves forward and we enter S, CDK can activate the helicases and inhibit at the same time the loading of new ones. when a cell divides, modifications have to be passed on. H3, H4 tetramers (they stay together) and H2A, H2B dimers are composed of either all new or all old histones. The tetramers are randomly transferred to one of the two daughter cells and will form the base(s) of the new nucleosome. H2A and H2B are expelled with newly synthesized dimers to a soluble pool.

Question 8

finishing replication

Answer 8

for a circular chromosome, the machinery replicates the entire molecule, but the daughter molecules are topologically linked to each other. the two circular DNA molecules must be disengaged from each other or decatenated. this separation is accomplished by the action of type II topoisomerases. replication problem > although shortening of the molecule would only occur on one of the strands, every round would result in the shortening of one of the two daughter DNA molecules. most eukaryotic cells have telomeres, generally composed of head-to-tail repeats of a TG-rich DNA sequence. although many of these repeats are double stranded, the 3’ end extends beyond the 5’ end. this structure acts as a novel origin of replication that compensates. this origin does not interact with the same proteins as other origins, but recruits telomerase. telomerase acts to extend the 3’ end of its DNA substrate. unlike other polymerases, telomerase uses its RNA component as the template for adding the telomeric sequence to the 3’ terminus at the end of the chromosome. by providing an extended 3’ end, telomerase provides an additional template for the lagging-strand replication machinery. by synthesizing and extending RNA primers using the telomerase extended 3’ as a template, the cell can increase the lenght of the 5’ end of the chromosome

Question 9

mutations

Answer 9

permanent changes in the DNA sequence. the simplest are switches: transitions (pyr to pyr/pur to pur) and transversions (pyr to pur/pur to pyr). somatic mutations occur in somatic cells and are not passed from one generation to the other one. germline mutattions affect gametes. changes within genes are called point mutations. DNA lesions are sites of damage in the base pairing of DNA structure and are classified in: abasic site (a base is missing while the sugar backbone is intact), mismatch (caused by replication errors), modified bases (these lesions are caused by changes to the bases themselves), single stranded breaks (caused bt a nick in the backbone), double stranded breaks (both backbones are broken) and interstrand crosslink(the two strands become covalently linked to each other and cannot open). the three major sources of mutations are inaccuracy in DNA replication, chemical damage to the genetic material and transposons. microsatellites are hot spots for mutations. they have some sequences that are repeated many times.

Question 10

excision repair

Answer 10

base and nucleotide.
BER > glycosylase recognizes and removes the damaged base by hydrolyzing the glycosidic bond. the resulting abasic sugar is removed from the DNA backbone in a further endonucleolytic step. after the damaged nucleotide has been removed, a repair DNA pol and ligase restore an intact strand using the undamaged strand as a template
NER works best by recognizing distortions to the shape of the double helic. such distortions trigger a chain of events that lead to the removal of a short single stranded segment that includes the lesion. this removal creates a single strand gap in the DNA, which is filled in by DNA polymerase using the undamaged strand as a template

Question 11

DSB repair systems

Answer 11

NHEJ (NON HOMOLOGOUS END JOINING) > Ku70 and 80 detect the double strand and they bind at the end. because they bind to the ends, they protect them from further damage. when they bind they recruit a kinase, DNA PKcs which forms a complex with Artemis and cuts the DNA to clean up the damage. NHEJ is mutagenic. the original sequence across the break is not faithfully restored.
homologous recombination > after the two molecules are aligned, breaks are introduced. once the breaks are formed the ends are further processed to generate regions of single stranded DNA. initial short regions of base pairing are formed: this pairing occurs when a single stranded region of DNA pairs with its complementary strand in the homologous duplex.
after strand invasion, the two DNA molecules become connected by crossing DNA strands to form a Holliday junction. this junction can move along the DNA by the repeated melting and formation of base pairs. each time the junction moves, base pairs are broken in the parental DNA molecules while identical base pairs are formed in the recombination intermediate. this process is called branch migration.
the process to regenerate DNA molecules and finish genetic exchange is called resolution and can be achieved either by cleavage of the holliday junction or by a process of dissolution

Question 12

RecBCD pathway

Answer 12

major DSB repair pathway in E.Coli. The RecBCD enzyme binds at the broken DNA end and rapidly unwinds the duplex in an ATP-dependent manner. As unwinding proceeds, the 3’ terminated strand is continuously degraded by an endonuclease activity. This unwinding and digestion continues until the Chi sequence is encountered, at which point the enzyme pauses, and the nuclease switches to the other DNA strand. The enzyme then continues unwinding the duplex, digesting the 5’ terminated strand and loading the RecA strand-exchange protein onto the 3’ strand.

Question 13

gene conversion

Answer 13

mechanism 1 > if the A gene is very close to the site of the DSB, the 3’ ssDNA tails invade the duplexes and may copy the a information, which could replace the A info.
mechanism 2 > invokves the repair of base pair mismatches that occur in the recombination intermediates. if either strand invasion or branch migration includes the A/a gene, a segment of heteroduplex DNA carrying the A sequence on one strand and the a on the other would be formed. this region of DNA carrying base-pair mismatches could be recognized and acted on by the cellular mismatch repair enzymes. when these enzymes detect a mismatched base pair, they cut a short stretch of DNA from one strand and the repair DNA pol fills the gap. therefore, after their action, both strands will carry the sequence encoding either the A or a info, and gene conversion will be observed.

Question 14

site specific

Answer 14

recombinases recognize specific sequences where recombination will occur and bring the ends physically close to each other in a synaptic complex. between the recombination sites there’s the crossover region. the side chain of a serine residue within the active site attacks a specific phosphodiester bond in the recombination site. during recombination four single strands of dna must be cleaved and rejoined. the serine recombinases cleave all four strands prior to strand exchange. the dna strands can be rejoined simply by reversing the cleavage reaction. the mechanism of action of the tyrosine recombinases is very similar to that of the serine recombinases except a tyrosine side chain is involved in the cleavage reaction. the tyrosine recombinases cleave and rejoin two dna strands first generating a holiday junction.they rejoin the other two dna strands resolving the holiday junction. a given recombination site always has a defined polarity: if both recombination sites are on the same dna molecule they can be a direct repeat with respect to one another or inverted with respect to one another. recombination between a pair of recombination sites organized as direct repeats will result in deletion of the dna segment between the two sites; in contrast recombination between a pair of inverted recombination sites will invert the dna segment

Question 15

biological function of CSSR (conservative site specific recombination)

Answer 15

cells and viruses use it for a wide variety of biological functions. many phages insert their DNA into the host chromosome during infection - to alter gene expression - to help maintain the structural integrity of circular DNA molecles during ccles of DNA replication, homologous recombination and cell division.

Question 16

salmonella hin recombinase

Answer 16

inverts a segment of the bacterial chromosome to allow expression of two alternative sets of genes. hin recombination is an example of a class of recombination reactions known as programmed rearrangements. the genes that are controlled by the inversion process encode two alternative forms of flagellin (H1 and H2 forms), the protein component of flagellar filament. flagella are a common target for the immune sustem. by using Hin to switch between H1 and H2 some individuals in the bacterial population can avoid recognition of this surface structure by the immune system.

Question 17

transposition

Answer 17

a transposon is a sequence of DNA which can move from a place to another.
virus like retrotransposons > have the recombination sites within the long terminal repeats. they insert into new sites in the genome of the host cell. a cycle of transposition starts with transcription of the retrotransposon DNA sequence into RNA by a cellular RNA polymerase. transcritpion initiates at the promoter sequence and continues until it generates a nearly complete RNA copy of the element’s DNA. the RNA is then reverse-transcribed to create a double stranded DNA molecule.
polyA retrotransposons > they have two regions, the UTR and the polyA tail. they move using an RNA intermediate and target-site primed reverse transcription. the first step is the transcription of the DNA. this newly synthesized RNA is exported to the cytoplasm and translated to generate the ORF1 and ORF2 proteins. the protein-RNA complex re-enters the nucleus and associates with the cellular DNA. a nick is introduced in the chromosomal DNA and the 3’OH serves as the primer for reverse transcription of the element RNA: the ORF2 proteins also catalyze this DNA synthesis. the remaining steps of transposition include synthesis of the second DNA strand, repair of gaps at the insertion site and sealing of the DNA strands.

Question 18

cut and paste, intermediate, copy and paste

Answer 18

cut and paste > to initiate recombination, the transposase binds to the terminal inverted repeats. it brings the two ends of the transposon DNA together to generate the synaptic complex or transpososome. the transposase cleaves one DNA strand at each end of the transposon generating two free 3’OH groups which attack the DNA phosphodiester bonds at the site of the new insertion. this DNA segment is called the target DNA. the transposon DNA is covalently joined to the DNA at the target site. this reaction occurs by a one-step transesterification reaction called DNA strand transfer.
intermediate > generated after DNA strand transfer, has the 3’ ends of the transposon attached to the target DNA. the fact that the two sites of DNA strand transfer on the two strands are separated by a few nucleotides results in short single stranded DNA gaps. these gaps are filled by a DNA repair polymerase encoded by the host cell.
mechanism 1 > an enzyme can be used to cleave the second strand ;
mechanism 2 > Tn5 and 10 cleave the non-transferred strand by generating a DNA hairpin
copy and paste > the synaptic complex forms, the ends of the transposon get cleaved, the DNA replication proteins from the host cell assemble at one of the forks

Question 19

transcription vs replication

Answer 19

the main difference is the end product: the first one has a double strand as a result, while the second one a single strand of DNA. the RNA polymerase does not need a primer and does not stay attached to the DNA. transcription is less accurate than replication (1 error in 10.000 vs 1 in 10.000.000). any mistake that arises during replication can be catastrophic since it becomes permanent in the genome while transcription produces only transient copies and normally several from each transcribed region. transcription selectively copies only certain parts of the genome while replication must copy the whole strand.

Question 20

transcription in bacteria

Answer 20

in E.Coli, the predominant σ factor is σ70: its promoters have two conserved sequences, -35 and -10. The σ70 factor can be divided into four regions 1 through 4. The regions that recognize the -10 and -35 elements are region 2 and 4. Two helices within region 4 form a helix-turn-helix. One of these helices inserts into the major groove and interacts with bases in the -35 region; the other lies across the top of the groove, making contacts with the DNA backbone. The -10 region is also recognized by an α helix; whereas the -35 region simply provides binding energy to secure polymerase to the promoter, within the -10 region DNA melting is initiated. The initial binding of RNA polymerase to the promoter DNA in the closed complex leaves the DNA in double-stranded form. The next stage in initiation requires the enzyme to become more intimately engaged with the promoter.
If the bacterial enzyme bearing σ70 isomerization is the result of a spontaneous conformational change in the DNA, isomerization is essentially irreversible and typically guarantees that transcription will subsequently initiate.

Question 21

transcription

Answer 21

a promoter is the DNA sequence that initially binds the RNA polymerase. the first step is the initial binding of polymerase to a promoter to form the closed complex. in the second step, the complex undergoes a transition to the open complex. the first two ribonucleotides are brought into the active site are aligned on the template and joined together. once an enzyme makes a transcript longer that 10 nucleotides, it is said to have escaped the promoter. the σ subunit eventually is released using energy accumulated during the scrunching phase and the release of the σ subunit causes conformational changes in the RNA Pol that facilitate the promoter escape and thus the transition into the elongation phase. During elongation, the RNA Pol adds one nucleotide at a time to the growing RNA transcript. During this phase, the RNA Pol uses a step mechanism and advances in a single step a distance equivalent to a base pair for every nucleotide it adds to the growing RNA chain. Therefore the site of the bubble remains constant. There are two proof-reading processes: pyrophosphorolytic editing (the RNA Pol uses its active site, in a simple back reaction, to catalyze the removal of an incorrectly inserted ribonucleotide, by reincorporation of PPI; the enzyme can then incorporate another ribonucleotide in its place in the growing RNA chain) and hydrolytic editing (the RNA Pol backtracks by one or more nucleotides and cleaves the RNA product, removing the error containing sequence). At the end of a gene there are sequences called terminators that trigger the RNA Pol to dissociate from the DNA and release the RNA chain it has made.

Question 22

RHO termination

Answer 22

a protein made of six identical subunits that form its characteristic ring shaped structure. bacteria have two types of termination: rho dependent and rho independent. rho depedent terminatrs have ill defined RNA elements, the rut sites, and for them to work requires the action of the rho factor, which binds to single stranded RNA as it exits the polymerase. the protein also has an ATPase activity, and once attached to the transcript uses the enrgy to induce termination. rho pushes polymerase forward and pulls RNA out of the polymerase. rho independent terminators consist of two sequence elements: a short inverted repeat followed by a stretch of about eight AT base pairs. these elements affect the polymerase only after being transcribed. when polymerase transcribes an inverted repeat sequence, the RNA can form a stem-loop structure. the hairpin induces termination by pishing the poly forward or inducing a change in it.

Question 23

transcription in eukaryotes

Answer 23

the process is identical, the machinery is not. all eukaryotes have at least three different RNA pols (vs 1) + a sigma factor (vs GTF). the TATA binding protein (TBP) binds the TATA element on DNA and starts the formation of the transcriptional pre initiation complex. TFIIB binds the DNA and TBP. doing so it forms a molecular binding platform that is recognized by RNA pol. TBP binds to the TATA sequence and provides a platform to recruit other GTFs and the poly itself to the promoter > promoter melting.

Question 24

TRCF

Answer 24

RNA poly can become arrested and need removing. the transcription repair coupling factor removes the stalled RNA pol and repairs the DNA. TRCF has an ATPase activity, it binds double stranded DNA upstream of the poly and uses the ATPase motor to translocate along the DNA until it encounters the stalled RNA poly. the collision pushes RNA pol forwards, either allowing it to restart elongation or causing dissociation of the ternary complex.

Question 25

capping and polyA

Answer 25

processing transcribed RNA includes capping and polyA. these modifications allow the cell to assess whether both ends of an mRNA molecule are present before translation. capping and splicing happen during the transcription while the polyA has to happen at the end. when an mRNA is transcribed its called pre mRNA and its made of introns and exons. through splicing, all introns get removed. the capping involves the addition of a methylated guanine to the 5' end of the RNA. once the poly has reached the end of the gene, it encounbters the polyA sequence signal, which triggers CPSF (cleavage and polyA specificity factor) and CSTG (cleavage stimulation factor). polyA is mediated by an enzyme called polyA poly, which adds approximately 200 adenines to 3'. this enzyme uses ATP as a precursor and adds the nucleotides without a template.

Question 26

torpedo model

Answer 26

after cleavage and polyA, the RNA pol continues to move along the DNA template, generating a second RNA molecule. the free end of the second RNA is uncapped and thus can be distinguished from genuine transcripts. this new RNA is recognized by an RNase called Rat1 in yeast or Xrn2 in humans. pol I is required for the expression of only one gene that’s encoding the rRNA precursor. The promoter for the rRNA gene comprises two parts: the core element and the UCE (upstream control element). In addition to Pol I, Initiation requires two other factors: SL1 and UBF. Pol III transcribes genes encoding tRNAs and 5S rRNA. Pol III promoters are located downstream from the transcription start site; the promoters for tRNA genes consist of two regions, called Box A and Box B, separated by a short element (while the Pol III promoter for the 5S rRNA gene contains Box A and Box C). Other promoters contain a TATA element like those of Pol II.

Question 27

open reading frame

Answer 27

the protein coding region(s) of each mRNA is composed of a continuous, non-overlapping string of codon called ORF. each ORF specifies a single protein and starts and ends at itnernal sites within the mRNA: the first and last codons of an ORF are known as the start and stop codons. the start codon specifies the first amino acid to be incorporated into the growing polypeptide cham. mRNAs with multiple ORFs are plycistronic, and those with a single one are monocistronic. to facilitate binding by a ribosome, many prokaryotic ORFs contain the ribosome binding site (shine dalgarno). some prokaryotic ORFs lack a strong RBS but are nonetheless actively translated. these ORFs are not the first ones but come just after another ORF in a polycistronic message. the start codon of the downstram ORF often overlaps the 3' end of the upstream ORF. this phenomenon of linked translation between overlapping ORFs is known as translational coupling. eukaryotic mRNAs recruit ribosomes using a specific chemical modification called the 5' cap. the guanine nucleotide of the 5' cap is connected to the 5' end of the mRNA through three phosphate groups. the resulting 5' cap is required to recruit the ribosome to the mRNA. once bound to the mRNA, the ribosome moves in a 5' to 3' direction until it encounters a 5' AUG 3' start codon ( scanning ).

Question 28

transfer tRNA

Answer 28

there are many tupes of tRNA molecules, but each is attached to a specific amino acid, and each recognizes a particular codon, or codons, in the mRNA. a striking aspect of tRNAs is the presence of several unusual bases in their primary structure. pseudouridine is derived from uridine isomerization, while dihydrouridine by the enzymatic reduction of uridine. these modified bases are not essential for tRNA function, but cells lacking these modified bases show reduced rates of growth. the principal features of the tRNA cloverlead are 1) the acceptor stem (so-named because it’s the site of attachment of the amino acid, is formed by pairing between the 5’ and 3’ ends of the tRNA molecule); 2) the ΨU loop; 3) the D loop; 4) the anticodon loop, which contains the anticodon, a three-nucleotide-long sequence that is responsible for recognizing the codon by base pairing with the mRNA; 5) the variable loop.

Question 29

attachment of amino acids to tRNA

Answer 29

tRNA molecules to which an amino acid is attached are said to be charged, and tRNAs that lack an amino acid are said to be uncharged. Charging requires an acyl linkage between the carboxyl groups of the amino acid and the hydroxyl group. The energy released when this acyl bond is broken is coupled to the formation of the peptide bonds that link amino acids to each other in polypeptide chains. Step one is adenylylation, in which the amino acid reacts with ATP to become adenylylated with the concomitant release of pyrophosphate. Adenylylation refert to transfer of AMP. Step two is tRNA charging in which the adenylylated amino acid reacts with tRNA.

Question 30

introns

Answer 30

group 1 > they use a free guanine that binds in a pocket made by a secondary structure of the intron itself. instead of a branchpoint A residue, they use a free guanine nucleotide or nucleoside. this G is bound by the RNA, and its 3' OH is presented to the 5' splice site, initiating a transesterification reaction that fuses the G to the 5' . groups 2 > the sequence is such that they can make several secondary structures called domains that resemble the structures made by snRNPs. the adenine at the branchpoint is a free one. the sequences in the domains are complementary to each other.

Question 31

splicing errors

Answer 31

exon skipping > happens if the spliceosome components bound at the 5' splice site of one exn interact with spliceosome components bound at the 3' splice site of not the nexr exon but one beyond cryptic splice site selection > happens when the spliceosome recognizes pseudo splice sites, sequences that resemble legitimate splice sites exon definition > serine arginine rich proteins bind to sequences called exonic splicing enhancers within the exons and recruit the U2AF proteins to the 3' splice site and U1 snRNP to the 5' site.

Question 32

nonsense-mediated decay

Answer 32

nonsense-mediated decay: this mechanism ensures that only messages that have one or another exon survive. it results from the fact that including both exons produces an mRNA that contains a premature termination codon. proteins that regulate splicing bind to specific sites called exonic or intronic splicing enhancers (ESE or ISE) and exonic or intronic silencers (ESS and ISS). the presence or activity of a given protein can determine whether a particular splice site is used in a particular cell type or at a particular stage of development. activators bind to the pre-mRNA and U2AF to facilitate the assembly of the spliceosome.

Question 33

rna editing

Answer 33

editing may include insertion, deletion and base substitution of nucleotides within the edited RNA molecule. it's an essential mechanism of gene regulation: it increases the number of proteins. site-specific deamination > a regulated process in which targeted nucleotide residues within mRNA are deaminated. another example of mRNA editing by enzymatic deamination includes adenosine deamination. this reaction is performed by the enzyme ADAR (adenosine deaminase acting on RNA) and produces inosine. inosine is detected as guanine during translation. insertion/deletion of uridines > occurs mainly in mitochondria of trypanosomes. in this case, multiple Us are inserted or removed from specific regions of mRNAs after transcription. it is found in the RNA transcripts that encode proteins in the mitochondria of trypanosomes. multiple Us can be inserted or deleted from the RNA by so-called guide RNAs. these additions cause a change in the reading frame, a shift that is required to generate the correct protein.

Question 34

isoaccepting tRNA

Answer 34

because one amino acid can be specified by more than one codon, one synthetase may recognize and charge more than one tRNA: the synthetases must recognize the proper amino acid and the proper tRNA to charge. for other synthetases, the nucleotide sequence of the amino acid accepting arm is the key recognition determinant. in some cases, changing a single base in the acceptor stem (discriminator) is sufficient to convert the recognition specificity of a tRNA from one synthetase to another. the correct amino acid has the highest affinity for the active site pocket of its sunthetase and is therefore favored over the other 19. a second discrimination step is called hydrolytic edition and occurs after the amino acid has been covalently linked to AMP. when tRNA binds, the synthetase tries to force the adenylated amino acid into a second editing pocket in the enzyme. the precise dimensions of this pocket exclude the correct amino acid, while allowing access by closely related amino acids

Question 35

the ribosome

Answer 35

macromolecular machine that directs the synthesis of proteins. the large subunit has the peptidyl transfer center which is where the two amino acids are bound to each other. the small subunit is the decoding center in which charged tRNA reads and decodes the codon units of the mRNA. the ribosome has three binding sites for tRNA: A (amino acylated tRNA), P (peptidyl tRNA) and E (for the tRNA that is released after the growing polypeptide chain has been transferred to the aminoacyl tRNA. the mRNA interacts with the small subunit: enters and exits the decoding center through two narrow channels in the small subunit. a region that allows the tRNA to enter the core of the ribosome is found in the large subunit. a second channel through the large subunit allows the amino acid to come out. polyribosomes: an mRNA bearing multiple ribosomes is known as a polyribosome or a poltsome. the result is the simultaneous synthesis of multiple polypeptided. in prokaryotic cells translation starts when transcription has not ended yet.

Question 36

the peptide bond

Answer 36

the growing chain is attached to a peptidyl tRNA and the incoming amino acid is attached to an aminoacyl-tRNA. the bond between the aminoacyl-tRNA and the amino acid is not broken during the formation of the next peptide bond. the bond between the peptidyl-tRNA and the growinf polypeptide chain is broken as the growing chain is attached to the amino group of the amino acid attached to the aminoacyl tRNA to form a new peptide bond. this method requires that the amino terminus of the protein be synthesized before the carboxyl terminus. the growing polypeptide chain is transferred from the peptidyl tRNA to the aminoacyl tRNA

Question 37

initiation

Answer 37

the association of the small subunit with the mRNA is mediated by base pairing interactions. translation initiation is the only time a tRNA binds to the P-site without previously occupying the A site. after the initiator is charged with a methionine, a formyl group is added by Met-tRNA transformylase. the initiator is coupled to N formyl methionine and is referred to as tRNAfMet. deformylase removes the formyl groups from the amino terminus during or after the synthesis of the polypeptide chain. the initiation is catalyzed by IF1, IF2 and IF3. 1 prevents tRNAs from binding to the portion of the small subunit that will become part of the A site; 2 is a GTPase and interacts with three key components of the initiation machinery, facilitating the association of tRNA N-formylmethionine with the small subunit and prevents other charged tRNAs from associating with the small subunit. 3 binds to the small subunit and blocks it from reassociating with the large subunit. only the P-site can bind a tRNA. with all three initiation factors bound, the small subunit is prepared to bind to the mRNA and the initiator tRNA forming 30S. the small subunit undergoes a change in conformation, which results in the release of IF3. the large subunit is free to bind to the small subunit, which causes the hydrolysis of GTP. release of IF2-GDP as well as IF1. the initiation complex 70S is formed at the start site of the mRNA with initiator tRNA in the P site and an empty A site.

Question 38

initiation in eukaryotes

Answer 38

step 1 >selection of the tRNA by elF2. ternary comples > leF2, GTP, Met tRNA + 40S complex = 43S. step 2 > elF4G+ elF4E + mRNA + elF4B activates the RNA helicase activity of elF4A. step 3 > the small subunit and its associated factors move along the mRNA in a 5' to 3' direction in search of the start codon. correct base pairing changes the conformation of the 48S complex, leading to the release of elF1 and a change in conformation of elF5. these events stimulate elF2 to hydrolyze its associated GTP. elF2 is released. step 4 > elF5B + initiation tRNA = 60S subunit > release of the remaining initiation factors. the tRNA Met is placed in the P site of the resulting 80s initiation complex.

Question 39

elongation

Answer 39

methionine-carrying tRNA starts out in the P site. a fresh codon is exposed in the A site. once the matching tRNA has landed in the A site, it's time for the formation of the peptide bond that connects one amino acid to another. this step transfers the methionine from the first tRNA onto the amino acid of the second tRNA in the A site. the methionine forms the N-terminus of the polypeptide, and the other amino acid is the C-terminus.

Question 40

selection against incorrect aminoacyl tRNAs

Answer 40

at the heart of the selection of the correct aminoacyl tRNA is the base pairing between the carhged tRNA and the codon displayes in the A site of the ribosome. these are three different mechanisms: 1 > additional hydrogen bonds are formed between two adenine residues of the rRNA and the minor groove of the anticodon-codon pair only when the first two bases of the anticodon-codon pair form correct watson crick base pairs. 2 > correct codon anticodon base pairing facilitates EF-Tu bonding to the aminoacyl-tRNA to interact with the factor binding center inducing GTP hydrolysis and EF-Tu release. 3 > only correctly base paired aminoacyl tRNAs remain associated with the ribosome as they rotate into the correct position for peptide bond formation (tRNA accommodation). the ribosome cannot control if the amino acid that's coming is the right one but the charged tRNA. thus the translation machinery relies on the high fidelity of the aminoacyl-tRNA synthetases to ensure the accurate decoding of each mRNA.

Question 41

translating the mRNA molecule

Answer 41

this reaction is catalyzed by the 23S rRNA component of the large subunit. once the peptidyl transferase reaction has occurred, the tRNA in the P site is no longer attached to an amino acid, and the growing polypeptide chain is linked to the tRNA in the A site. for a new round of peptide chain elongation to occur, the P site tRNA must move to the E site and the A site tRNA must move to the P site. at the same time the mRNA must move by three nucleotides to expose the next codon (translocation). the translocation is initiated in the large subunit, and before the small subunit translocation as well, the tRNAs are said to be in hybrid states. EF-G binds and stabilizes the ribosome in the hybrid state. when EF-G/GTP binds, it contacts the factor binding center of the large subunit, which stimulates GTP hydrolysis. EF-G/GDP binds to the A site of the decoding center and triggers conformational changes within the ribosome to complete the translocation. after GTP hydrolysis, both EF-Tu and EF-G must release their bound GDP and bind a new molecule of GTP. for EF-G, this is a simple process, because GDP has a lower affinity for EF-G than does GTP. for EF-Tu a second protein is required to exchange GDP for GTP. the elongation factor EF-T acts as a GTP exchange factor for EF-Tu.

Question 42

release factors

Answer 42

``` class 1 > they bind to the specific termination codons and hydrolyze the polypeptide tRNA linkage in the P site. class 2 > they stimulate the dissociation of the class 1 factors from the rubosome after release. prokaryotes have two class 1 release factors: RF1 (recognizes the stop codon UAG and UAA) and RF2 (recognizes UGA and UAA) eukaryotes have a single class 1 release factor called eRF1 ```

Question 43

ribosome recycling factors

Answer 43

ribosome recycling is the complete dissociation of tRNAs, mRNAs, small and large subunits. in prokaryotic cells, the rybosome recycling factor cooperates with EF-G and IF3 to recycle ribosomes after polypeptide release.

Question 44

regulation

Answer 44

protein or RNA binding near the ribosome binding site negatively regulates bacterial translation initiation. the primary target of regulators of bacterial initiation is to interfere with the recognition of the RBS by the 30S subunit. inhibitory proteins associate with sequences near the RBS and physically inhibit base pairing between the RBS and the 16SrRNA. this regulation occurs for proteins that need it immediately. gene encoding ribosomal proteins are organized in operons (sequences of DNA that contain a cluster of genes under the control of a single promoter). ribosomal proteins are translational repressors of their own synthesis. while operons are regulated at the level of RNA synthesis, the most important control is at the level of translation of the mRNA. the control is the result of autorepression. for each ribosomal protein operon, a complex of two of the encoded ribosomal proteins binds that operon's mRNA near the translation initiation sequence.

Question 45

iron deficiency

Answer 45

blood lacks healthy red cells, reduce physical and cognitive development in children, increases child mortality, tiredness and lack of energy, shortness of breath, noticeable heartbeats (heart palpitations), pale skin

Question 46

iron excess

Answer 46

increased cellular oxidation liver diseases heart problems diabetes and increased cancer risk.

Question 47

ferritin

Answer 47

it's the major regulator of iron levels in the human body. the levels of ferritin must respond rapidly to the levels of free iron in the body. ferritin translation is regulated by iron binding proteins called iron regulatory proteins (IRPs). IRPs recognuze a specific hairpin structure formed at the 5' end of the ferritin mRNA called the iron regulatory element (IRE). the ability of IRPs to recognize the IRE is controlled by the leves of iron in the cell. when iron levels are low, IRPs can bind IRE and block ferritin translation

Question 48

quality control systems

Answer 48

1 > nonsense-mediated RNA decay: exon junction complexes are assembled on the mRNA because of splicing and are located at each exon-exon boundary. the exon junction complexes and the eRF3 that is bound to the prematurely terminatinf ribosome lead to the activation of multiple enzymes that cleave the mRNA, remove the 5' cap and/or remove the polyA tail 2 > nonstop-mediated decay: when an mRNA lacking a stop codon is translated, the ribosome translates through the polyA tail. the stalled ribosome is bound by two proteins that stimulate dissociation of the translation complex. exonucleases and endonucleases degrate the non-stop mRNA. 3 > no-go decay: recognizes ribosomes that are stalled on an mRNA. DOm34 and Hbs1 bind to the stalled ribosome and stimulate its dissociation into the large and small subunits.

Question 49

constitutive expression

Answer 49

at many promoters, in the absence of regulatory proteins, RNA pol binds only weakly. when the RNA pol does occasionally binds, it undergoes a transition to the open complex and initiates transcription. this gives a low level of constitutive expression called the basal level: in this case, binding of RNA pol is the rate limiting step. to activate transcription an activator can just help the pol bind the promoter. the activator uses one surface to bind to a site on the DNA near the promoter; whith aother surface, it interacts with RNA pol, bringing the enzyme to the promoter. this mechanism, often called recruitment, is an example of cooperative binding of proteins to DNA: at this promoter, an activator must stimulate the transition from a closed to an open complex. this mechanisme is an example of allostery.

Question 50

the LAC operon

Answer 50

E.Coli can grow in media which contain lactose as their sole carbon source. to metabolize this sugar, it must produce two enzymes: ꞵ-galactosidase (cleaves the bond joining the two monosaccharides) and lac permease (allows the lactose to enter the cell). three lactose related genes: lacZ (codes for the galactosidase), lacY (permease) and lacA (transacelutase). all three genes are next to each other on its genome and together are the lac operon. the lac promoter directs transcription of all three genes as a single mRNA (polycistronic message). these genes are expressed at high levels only when lactose is availabe, and glucose is not. two regulatory proteins are involved: one is an activator CAP (CRP) and the lac repressor. under basal conditions, there are only a few copies of the repressor. the CAP has two binding domains, one for DNA and one for RNA pol. the allosteric effector for CAP is cAMP: only when CAP is complexed with it does the protein adopt a conformation that binds DNA. E.COli encodes several other sigma subunits that can replace sigma70 under certain circumstances: one of these is sigma 32, which is a heat shock factor: when E.COli is subject to heat shock, the amount of sigma32 in the cell increases. sigma32 displaces sigma70 from a proportion of RNA pol and directs those enzymes to transcribe genes whose products protect the cell from the effects of heat shock.

Question 51

activators that induce allosteric changes

Answer 51

NtrC controls expression of genes involved in nitrogen metabolism by inducing a conformational change in a pre-bound RNA pol, triggering transition to the open complex. it binds DNA only in the presence of low nitrogen levels: when nitrogen is low, NtrC is phosphorylated and activated by a specific protein kinase. once active it binds DNA sequences located approximately 150 bp upstream of the promoter for the glnA gene. the form of RNA pol that transcribes this geme contains the sigma54 subunit. this RNA pol binds to the glnA promoter in a stable closed complex in the absence of NtrC: once active, NtrC interacts directly with the RNA pol/DNA complex.

Question 52

bacteriophage λ

Answer 52

bacteriophage lambda has most genes encoding coat proteins, proteins involved in DNA replication, recombination and lysis. the region of interest contains two genes (cl and cro) and three promoters (PR, PL and PRM). PRM transcribes only the cl gene. PR and PL are strong constitutive promoters; they bind RNA pol efficiently and direct transcription without help from an activator. the cl gene encodes the lambda repressor. the lambda repressor can both activate and repress transcription. as a repressor, it binds to sites that overlap the promoter and excludes the RNA pol. cro (control of repressor and others) only represses transcription. cro binds OR3 with the highest affinity. the lambda repressor binds DNA cooperatively and cannot simultaneously make contact with a third dimer at that adjacent site. during lysogeny, PR and PL are off. for lytic growth, a single cro dimer is bound to OR3, repressing PRM. PR binds the RNA pol and directs transcription of lytic genes.

Question 53

RecA

Answer 53

if DNA damage is present, E.Coli activates a protein called RecA, which is involved in DNA repair. activated RecA stimulates autocleavage of LexA, a protein that represses genes encoding DNA repair enzymes. the lambda repressor resembles LexA and thus is cleaved by RecA upon activation of the SOS response pathway. RecA removes the carboxy domain of the repressor and so dimerization and cooperativity are lost. the two key events are cII and cIII. cII is on the right of cI and is transcribed from PR; cIII is on the left and transcribed from PL. the cII protein is a transcriptional activator as it stimulates transcription of the cI (repressor) gene from that promoter. the repressor gene can be transcribed from two different promoters (PR and PRM). only when sufficient repressor has been made from PRE can that repressor bind to OR1 and OR2 and direct its own synthesis from PRM. repressor synthesis is established by transcription from one promoter (stimulated by one activator) and then maintained by transcription from another promoter (positive autoreg). upon infection, transcription is initiated. PR directs synthesis of cro and cII. once cro reaches a certain level, it will bind OR3 and block PRM, thereby favoring lytic development. cII expression on the other hand, favors lysogenic growth by directing transcription of the repressor gene. for successful lysogeny, the repressor must then bind to OR1 and OR2 and activate PRM.

Question 54

multiplicity of infection

Answer 54

moi > a measure of how many phage particles infect a given bacterial cell. if the average number is <1 particles per cell = lysis, otherwise lysogeny.

Question 55

reporter genes

Answer 55

researchers attach them to a regulatory sequence of another gene of interest in bacteria, cell culture, animals or plants. such genes are called reporters because the characteristics they confer to organisms expressing them are easily identified and measured, or because they are selectable markers. reporter genes are often used as an indication of whether a certain gene has been taken up or expressed in the cell or organism population.

Question 56

activators have separate binding and activating functions

Answer 56

eukaryotic activators rarely bind directly to the RNA pol: recruitment of nucleosome modifiers and remodelers to open the promoter; recruitment of general transcription factors as well as protein complexes. the HSP70 gene in drosophila is controlled by two activators working together. the GAGA binding factor recruits enough of the transcription machinery to the promoter for initiation of transcription.

Question 57

synergistic activation

Answer 57

two or more activators wor together on the same promoter. 1 > multiple activators each recruiting a different component of the transcriptional machinery 2 > multiple activators helping each other bind 3 > multiple activators recruiting a single component of the transcriptional machinery.

Question 58

repressors

Answer 58

they work by binding to sites that overlap with an activator, binding to a site on DNA beside an activator and interacting with that activator, occluding its activating region, binding to a site upstream of a gene (inhibit transcription initiation), recruiting nucleosome modifiers that compact the chromatin or remove groups recognized by the machinery

Question 59

dna methylation

Answer 59

typically the methylated nucleotide is a cytosine, which is followed by a guanine. CpG islands are CpG rich regions in the genome, which are often located in the promoter regions of genes and some regulatory elements. cytosine is unmethylated in its natural state and becomes methylated only after it's incorporated into double-stranded DNA. methylated DNA sequences are recognized by DNA binding proteins that recruit histone deacetylases and methylases, which then modify nearby chromatin. thus, DNA methylation can mark sites where heterochromatin subsequently forms.

Question 60

imprinting

Answer 60

the H19 gene on the maternal chromosome is expressed, whereas the copy received from the paternal chromosome is off. for lgf2 the opposite is true. the enhancer can activate either of the two genes. the enhancer cannot activate the maternal lgf2 gene because on that chromosome, the lCR binds and insulator: CTCF. on the paternal chromosome in contrast the ICR element and the H19 promoter are methylated. in that state, the transcriptional machinery and CTCF cannot bind the H19 promoter and the ICR. thus the enhancer now activates the Igf2 gene. the paternal H19 gene is further repressed beacuse its methylated DNA is recognized by a specific protein that recruits histone deacetylases and histone methylases, which further prevent its expression.

Question 61

riboswitches

Answer 61

regulatory segment of a mRNA that binds a small molecule, resulting in a change in production of proteins. an mRNA that contains a riboswitch is directly involved in regulating its own activity. Riboswitches are composed of two domains: the aptamer domain and the expression platform The aptamer domain acts as a receptor that specifically binds a ligand. The expression platform acts directly on gene expression. Common to both domains is something called the switching sequence, and its placement dictates the expression outcome of the mRNA. T BOX > an RNA-based regulatory mechanism that modulates expression of a wide variety of amino acid-related genes

Question 62

miRNA

Answer 62

small single stranded non coding RNA molecules that bind target mRNA to prevent protein production. miRNAs are transcribed by RNA pol II and III. they can originate from introns of protein-coding genes or from intergenic regions. the regulatory functions are accomplished through the RNA-induced silencing complex (RISC). the degree and nature of the complementarity between the miRNA and mRNA target determine the gene silencing mechanism: slicer dependent degradation or independent translation inhibition. pre-miRNA is a single strand that makes a loop in itself, a loop the dicer then cuts. the guide strand is complementary to the target RNA. if there's a perfect match, the mRNA will be cleaved. if there's no perfect match bt a partial one, the RISC complex does not degrate the mRNA but prevent its translation. activated RISC binds target mRNAs through watson crick base pairing between the guide strand and the target mRNA. target recognition relies heavily on base-pairing between the seed of the miRNA and the target mRNA: a near perfect extensive base pairing between the miRNA and the mRNA leads to the degradation of the mRNA.

Question 63

AGO dependent gene silencing

Answer 63

Ago 2 is a member of the Argonaute protein family. its the catalytic subunit that performs the initial mRNA cleavage; for this reason, Argonaute is often called Slicer.

Question 64

miRNA dependent inhibition of translation

Answer 64

multiple complementary sites with imperfect base pairing create bulges in the RNA duplex inhibiting the slicer activity. miRNA can inhibit translation at the initiation and elongation steps.

Question 65

small interfering rna (siRNA)

Answer 65

once inside cells, siRNAs are processed similarly to miRNAs. the sequence of siRNA is highly complementary to the target mRNA: while miRNAs work like trans acting regulators, siRNAs are typically generated by transcripts of the regions on which they act and are thus described as working in cis.

Question 66

piwi interacting rna (piRNA)

Answer 66

a type of non coding RNA that interacts with PIPI proteins, members of the argonaute family. in animals, piRNAs maintain genomic integrity by repressing the mobilization of transposable elements, regulate the expression of target genes via transcriptional or post transcriptional mechanisms and help fight virus infections. involved in the regulation of other genetic elements in germ line cells. a decrease or absence of PIWI gene expression is correlated with an increased expression of transposons.

Question 67

Y RNA

Answer 67

humans have four (hY1, 3, 4 and 5): they are clustered together in the same locus on chromosome 7. they're essential for initiation of DNA replication: this function is mediated by the upper stem domain.

Question 68

lncRNAs

Answer 68

they're involved in numerous important biological phenomena such as imprinting genomic loci, shaping chromosome conformation and allosterically regulating enzymatic activity. their overexpression, deficiency and mutation has been implicated in numerous disorders. they can be generally divided into three categories: non functional (result of transcriptional noice), non necessary (their act alone is sufficient but the transcript itself is not necessary) and functional (act in cis and or in trans).

Question 69

xist regulates chromosome X inactivation

Answer 69

female mammals have two X chromosomes, whereas males have only one X and a Y chromosome; any gene encoded by the X chromosome would be expressed at twice the level in females as in males. avoiding such problems required what is called dosage compensation, in which females inactivate one of their two X chromosomes. xist is encoded from a locus of the X chromosome known as Xic (X inactivation center). Xist RNA coats the X chromosome. Xist recruits other factors that modify and condense chromatin and methylate the DNA.

Question 70

circular RNAs

Answer 70

broadly expressed in mammalian cells in a tissue specific manner. they are not translated into proteins. the most well known physiological function for them is that of regulators of miRNA function. they also form complexes with proteins to regulate the cell cycle and translation. their expression is increased in many cancer cells.

Question 71

miR-7

Answer 71

1 > as a direct regulator of alpha-synuclein (proposed a role in parkinson's disease) 2 > its inhibition in pancreatic B cells increases mTOR signaling 3 > miR7 directly targets and downregulates central oncogenic factors in cancer-associated signaling pathways (EGF receptor and IRS-1)

Question 72

CRISPR/Cas9

Answer 72

a technique used to edit genes. it stands for clustered regularly interspaced short palindromic sequences. Cas9 is an RNA guided nuclease whose sequence specificity arises from pairing between its guide RNA and the target DNA site. Cas9 can be programmed to target new sites just by changing its guide RNA sequence. the CRISPR system is an adaptive immune mechanism present in many bacteria

Question 73

mediated immunity

Answer 73

step 1 > Cas proteins form a complex which cleaves foreign viral DNA. the foreign DNA is incorporated into the bacterial CRISPR loci as a repeat-spacer unit. the newest viral sequence is always integrated at the 5' end of the CRISPR locus. each locus carries a temporal record of prior infections. step 2 > processed and bound to the Cas protein. the CRISPR locus is transcribed to produce pre-cRNA: the pre-cRNA is then cleaved into shorter CRISPR RNA. step 3 > crRNAs/Cas complexes seek out complementaty viral DNA sequences and cleaves them.

Question 74

PCR (polymerase chain reaction)

Answer 74

a laboratory technique used to amplify DNA sequences. the method involves using short DNA sequences called primers to select the portion of the genome to be amplified. the temperature of the sample is repeatedly raised and lowered to help a DNA replication enzyme copy the target DNA sequence. the technique can produce a billion copies of the target sequence.

Question 75

primer selection

Answer 75

primers are oligonucleotide sequences of single stranded nucleic acids. for example, there's 1/4 chances of finding an A C G or T in any sequence, a 1/16 of finding any dinucleotide sequence, 1/256 a given 4 base sequence.

Question 76

mice

Answer 76

GENETICALLY MODIFIED MICE > 3 categories: transgenic mice, knockout mice and knock-in mice. TRANSGENIC MICE: mutant animals carrying experimentally introduced foreign genetic elements in all their cells, including the germline. KNOCK-OUT MICE: targeted mutation designed to delete or inactivate a specific gene. KNOCK-IN MICE: targeted insertion of a mutation, gene or DNA fragment. TRANSGENESIS > the stable integration of foreign DNA into a host genome. EARLY TRANSGENIC MICE > 1980-81: several groups reported the development of transgenic mice by microinjecting genes into the pronucleus of a fertilized egg. 1982: the first visible genotype was shown by Dr. Palmiter and colleagues in mice. ADVANTAGE OF MICE OVER WORMS AND FLIES: they’re a mammalian model with organ systems, tissues, physiology highly comparable to humans. Most genes are well-conserved between humans and mice. ADVANTAGE OF MICE OVER OTHER MAMMALIAN MODELS > the availability of genetically homogeneous inbred strains, easy to breed (short generation time), easy to house given their small size, availability of embryonic stem cells and easy of gene modification in the germline.