Exam 3: Chp 26 and Partial 27 Flashcards

1
Q

What is Transcription?

A

Transcription is the first step in gene expression, in which information from a gene is used to construct a functional product such as a protein. The goal of transcription is to make a RNA copy of a gene’s DNA sequence. For a protein-coding gene, the RNA copy, or transcript, carries the information needed to build a polypeptide (protein or protein subunit). Eukaryotic transcripts need to go through some processing steps before translation into proteins.

The main enzyme involved in transcription is RNA polymerase, which uses a single-stranded DNA template to synthesize a complementary strand of RNA. Specifically, RNA polymerase builds an RNA strand in the 5’ to 3’ direction, adding each new nucleotide to the 3’ end of the strand.

Transcription has three stages: initiation, elongation, and termination.

DNA contains the information needed to make all of our proteins, and that RNA is a messenger that carries this information to the ribosomes (RNA–protein complexes)

This copy, called messenger RNA (mRNA), carries the gene’s protein information encoded in DNA. In humans and other complex organisms, mRNA moves from the cell nucleus to the cell cytoplasm (watery interior), where it is used for synthesizing the encoded protein.

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2
Q

How does RNAP recognize the correct DNA strand and initiate RNA synthesis at the beginning of a gene (or operon)?

A

RNAP binds to its initiation sites through base sequences known as promoters that are recognized by the corresponding σ70 factor.

The core enzyme initiates transcription only at promoter because of the σ (σ70 is predominant in E. coli).

the σ subunit does not stay closely associated with the core enzyme (αββ’ω) except when helping to initiate transcription.

Once RNA polymerases are in the right place to start copying DNA, they just begin making RNA by joining together RNA nucleotides complementary to the DNA template.

Initiation of transcription requires binding of RNA polymerase to a small single strand portion in the promoter of a gene.

RNA synthesis is normally initiated at specific sites on the DNA template.

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3
Q

What is a promoter? Where is the “+1 position” in a promoter?

A

A promoter is a non-coding sequence and a region of DNA upstream of a gene where relevant proteins (such as RNA polymerase and transcription factors) bind to initiate transcription of that gene. The resulting transcription produces an RNA molecule (such as mRNA)
The holoenzyme forms tight complexes with promoters.
In prokaryotes, the promoter consists of two short sequences at -10 and -35 positions upstream from the transcription start site.

+1 position is the position where the first nucleotide is synthesized during transcription.

A base pair in a promoter region is assigned a negative or positive number that indicates its position, upstream or downstream in the direction of RNAP travel, from the first nucleotide that is transcribed to RNA; this start site is +1 and there is no 0. Because RNA is synthesized in the 5′ → 3′ direction (see below), the promoter is said to lie upstream of the RNA’s starting nucleotide.

upstream DNA is the DNA which occurs towards the 5’ end from a particular point on the DNA strand this DNA mainly contain the regulatory elements of a gene, including the promoter site and the transcription binding sites whereas the downstream DNA is the DNA which occurs towards the 3’ end The downstream DNA of a gene contains the protein-coding region of the gene. In eukaryotes, it contains exons and introns. The protein-coding region undergoes transcription to produce a functional RNA molecule

The purpose of the promoter is to bind transcription factors that control the initiation of transcription.

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4
Q

What are the consensus sequences in the -10 region—Pribnow box and -35 region in E. coli?

A

In prokaryotes, the promoter consists of two short sequences at -10 and -35 positions upstream from the transcription start site.

The sequence at -10 is called the Pribnow box, or the -10 element, and usually consists of the six nucleotides TATAAT. The Pribnow box is absolutely essential to start transcription in prokaryotes.
The other sequence at -35 (the -35 element) usually consists of the six nucleotides TTGACA. Its presence allows a very high transcription rate.

To get a better sense of how a promoter works, let’s look an example from bacteria. A typical bacterial promoter contains two important DNA sequences, the - 10 and -35 elements.
RNA polymerase recognizes and binds directly to these sequences.
Once the RNA polymerase has bound, it can open up the DNA and get to work. DNA opening occurs at the -10 element, where the strands are easy to separate due to the many As and Ts (which bind to each other using just two hydrogen bonds, rather than the three hydrogen bonds of Gs and Cs).

They come 35 and 10 nucleotides before the initiation site. The minus signs just mean that they are before, not after, the initiation site.
The footprint of RNAP holoenzyme indicates that it contacts the promoter primarily at its –10 and –35 regions.

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5
Q

which regions of the σ factor bind to the -10 and -35 regions, respectively?

A
  • σ70 factor can be divided into four regions: region 1, 2, 3 and 4.
  • Region 4 possesses a common DNA binding motif called a helix-turn-helix. Region 4 binds to the -35 element through this motif. One of these helices inserts into the major groove and interacts with bases in -35; the other lies on the top of the groove and contacts with the backbone.
  • Region 2 forms a α helix that binds to -10 region where the DNA melting is initiated during the transition from the closed to open complex. Thus, the region 2 does more than simply DNA binding—DNA melting, ssDNA stabilization.
  • Region 3 interacts with extended -10 where present.
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6
Q

From which nucleotide is transcription mostly initiated?

A

The initiating (+1) nucleotide is mostly A or G.

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7
Q

How many strands of DNA are copied in transcription? What is the template and non template strand?

A

1,In contrast to replication, which requires that both strands of the chromosome be entirely copied, the regulated expression of genetic information involves the copying of much smaller, single-strand portions of the genome.

The DNA strand that serves as a template during transcription is known as the antisense or noncoding strand because its sequence is complementary to that of the RNA. The other DNA strand, the nontemplate strand, which has the same sequence as the transcribed RNA (except for the replacement of U with T), is known as the sense or coding strand.

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8
Q

Most eukaryotic genes are transcribed individually. However, prokaryotic genes are frequently transcribed together. These genetic units are called?

A

operons, typically contain genes with related functions

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9
Q

Overview of initiation using terms, closed and open complex and bubbles

A

RNA polymerase, together with initiation factors in many cases, binds to a specific region of the promoter of a gene (-10 and -35 region in E. coli), forming a close complex. The DNA around the point where transcription will start unwinds and the base pairs are melted, producing a “bubble” of a single-stranded DNA and forming a open complex. DNA melting occurs between positions -9 and +2 relative to the transcription start site.

• Unlike DNA replication, transcription is initiated by RNA polymerase without the need for a primer.

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10
Q

How do alternative σ factors direct RNA polymerase to alternative promoters?

A

???

Gene Expression Is Controlled by Different σ Factors:

In bacteriophage: The bacterial phage uses three sigma factors in succession to regulate expression of its genome. This ensures that viral genes are expressed in the order in which they are needed.

In E. coli: heat shock induces the amount of a new sigma factor, sigma 32, which displaces sigma 70 form a proportion of RNA ploymerases, and direct those enzymes to transcribe genes whose products protect the cell from the effects of heat shock

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11
Q

In what direction does the RNA chain grow?

A

The RNA Chain Grows from the 5′ to 3′ End

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12
Q

which regions of the σ factor bind to the -10 and -35 regions, respectively?

A

σ Factor Mediates Binding of Polymerase to the Promoter:

  • σ70 factor can be divided into four regions: region 1, 2, 3 and 4.
  • Region 4 possesses a common DNA binding motif called a helix-turn-helix. Region 4 binds to the -35 element through this motif. One of these helices inserts into the major groove and interacts with bases in -35; the other lies on the top of the groove and contacts with the backbone. * Region 2 forms a α helix that binds to -10 region where the DNA melting is initiated during the transition from the closed to open complex. Thus, the region 2 does more than simply DNA binding—DNA melting, ssDNA stabilization.
  • Region 3 interacts with extended -10 where present.
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13
Q

What happens before entering the elongation phase for E.coli?
When does regulation of gene expression occur?

A

RNA polymerase synthesizes several short RNAs before entering the elongation phase caused by Strain that builds up in the DNA template. This period is called abortive initiation.
releases after only ∼9 to 11 nt have been polymerized,

The regulation of gene expression generally occurs in the step of transcription initiation.

In abortive initiation, the RNAP fails to escape the promoter and instead relieves the conformational tension by releasing the newly synthesized RNA fragment, thereby letting the transcription bubble relax to its normal size. The RNAP then reinitiates transcription from the +1 position.

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14
Q

Close complex and open complex

A

RNA polymerase, together with initiation factors in many cases, binds to a specific region of the promoter of a gene (-10 and -35 region in E. coli), forming a close complex.

The DNA around the point where transcription will start unwinds and the base pairs are melted, producing a “bubble” of a single-stranded DNA and forming a open complex. DNA melting occurs between positions -9 and +2 relative to the transcription start site.

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15
Q

Describe Elongation in Transcription in prokaryotes

A
  • Elongation: once the RNA polymerase has synthesized a short stretch of RNA (~10 bases), the transcription shifts into elongation phase. This transition requires further conformational changes in polymerases that lead it to grip template more firmly.
  • RNA polymerase is processive during chain elongation.
  • Like replication, transcription always occurs in a 5’ to 3’ direction. i.e. the ribonucleotide is added to the 3’ end of the growing chain.
  • Transcription is rapid. The in vivo rate of transcription is 20 to 50 nucleotides per second at 37oC.
  • During elongation, the enzyme complex performs additional tasks besides catalysis of RNA synthesis.
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16
Q

What are the functions of RNA polymerase besides catalyzing RNA synthesis?

A
  • During elongation, the enzyme complex performs additional tasks besides catalysis of RNA synthesis including unwinding the DNA in front and re-annealling it behind, dissociating the growing RNA chain from the template as it moves along, and proofreading the synthesized RNA.
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17
Q

What is the significance of RNA association with the template only in few base pairs?

A

• Only a few bases are paired between RNA and its template DNA.

• The displacement is critical for the RNA to be translated to produce its protein products in prokaryotes.

• This association ensures the gene can be transcribed by multiple RNA polymerase at the same time. Thus, a cell can synthesize large number of transcripts from single gene in a short time.

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18
Q

What is it indicated that a actively transcribing gene shows a arrowhead-like shape under electron microscope?

A

The synthesis of RNAs that are needed in large quantities, rRNAs, for example, is initiated as often as is sterically possible, about once per second. This gives rise to an arrowhead appearance of the transcribed DNA

The “arrowhead” structures result from the increasing lengths of the nascent RNA chains as the RNA polymerases synthesizing them move from the initiation site on the DNA to the termination site.

Electron micrograph indicates that DNA contains specific sites at which transcription is terminated.

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19
Q

Where does transcription stop in prokaryotes?

A

At specific sites which require Rho factors to terminate transcription

once the polymerase has transcribed the length of the gene, it must stop and release the RNA product. This process requires transcription termination sequences—the terminators and often proteins binding to the terminators. Some terminators are wellcharacterized but others are less clear.

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20
Q

What is termination in transcription in prokaryotes?

A

Termination: once the polymerase has transcribed the length of the gene, it must stop and release the RNA product. This process requires transcription termination sequences—the terminators and often proteins binding to the terminators. Some terminators are wellcharacterized but others are less clear.

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21
Q

What is a terminator and what are the two types?

A
  • Terminator: a specific sequence downstream of the gene coding region functioning for transcription termination.
  • Two types of terminator: Rho-independent and Rho-dependent.
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22
Q

Mechanisms of Rho-independent and Rho-dependent terminators.

A

Transcription Termination by Rho-independent Terminator—
* The hairpin forms in the RNA as soon as this region has been transcribed.

  • The hairpin structure disrupts RNA polymerase just as it is transcribing the AT rich stretch.
  • The combination of the hairpin structure and the weak interaction between Us and As conspires to dissociate RNA from the DNA template.

Rho-dependent Terminator—
* Rho-dependent terminators are less well-characterized.

  • Rho (ρ) is a ring-shaped protein with six identical subunits, binds to single-stranded RNA when RNA exits from the polymerase.
  • Rho is directed to a RNA by binding to the so-called rut (for Rho utilization) site consisting of 80-100 nucleotides that do not fold into a secondary structure and are rich in C residues but poor in G.
  • Rho only binds to those transcripts still being transcribed beyond the end of a gene, but not the transcripts being translated.
  • Rho is a helicase that unwinds RNA-DNA and RNA-RNA double helices by translocating along a single strand of RNA in 5’-3’ direction. This process uses the energy derived from NTP hydrolysis activity from Rho.
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23
Q

Difference between prokaryotes and eukaryotes transcription polymerases?

A

eukaryotic transcription is distinguished by having 3 RNAPs and by much more complicated control sequences. prokaryotes only have 1

Bacteria require only one additional initiation factor (σ), whereas several initiation factors, called the general transcription factors (GTFs), are required for efficient and promoter-specific initiation in eukaryotes

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24
Q

Pol I, II, and III are responsible for the synthesis of what different types of RNA?

A
  1. RNA polymerase I (RNAP I), which is located in the nucleoli and synthesizes the precursors of most rRNAs.
  2. RNA polymerase II (RNAP II), which occurs in the nucleoplasm, synthesizes the mRNA precursors.
  3. RNA polymerase III (RNAP III), which also occurs in the nucleoplasm, synthesizes the precursors of 5S rRNA, the tRNAs, and a variety of other small nuclear and cytosolic RNAs.
  • In addition, eukaryotic cells contain separated mitochondrial and (in plants) chloroplast RNA polymerases responsible for transcription of genes in these organelles.
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25
Q

Mechanisms of rifampicin, actinomycin D, and amatoxin to inhibit transcription.

A

rifamycin B,- specifically inhibit transcription by prokaryotic but not eukaryotic RNA polymerases. The selectivity and high potency of rifampicin (2 × 10 −8 M results in 50% inhibition of bacterial RNA polymerase) make it a medically useful bactericidal agent. Rifamycins inhibit neither the binding of RNA polymerase to the promoter nor the formation of the first phosphodiester bonds, but they prevent further chain elongation. The inactivated RNA polymerase remains bound to the promoter, thereby blocking initiation by uninhibited enzyme. Once RNA chain elongation has occurred, however, rifamycins have no effect on the subsequent elongation process.

Actinomycin D (right), a useful antineoplastic (anticancer) agent produced by Streptomyces antibioticus, tightly binds to duplex DNA
and, in doing so, strongly inhibits both transcription and DNA replication, presumably by interfering with the passage of RNA and DNA polymerases.

The poisonous mushroom Amanita phalloides (death cap), which is responsible for the majority of fatal mushroom poisonings in Europe, contains several types of toxic substances, including a series of unusual bicyclic octapeptides known as amatoxins. which is representative of the amatoxins, forms a tight 1:1 complex with RNAP II (K = 10 −8 M) and a looser one with RNAP III (K = 10 −6 M) so as to specifically block their elongation steps.

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26
Q

The structure of Pol II core promoter.

A
  • Core promoter: the minimal set of sequence elements required for accurate transcription initiation by the pol II machinery.
  • A core promoter is typically about 40-50 nucleotides long, extending either upstream or downstream of the transcription start site. * Five elements were found in Pol II core promoter, i.e. TFIIB recognition element (BRE), the TATA box, the initiator (Inr), the Motif ten element (MTE), and the downstream promoter element (DPE). 2/3 of the promoters lack TATA box, about half have Inr. Core promoters without TATA box used MTE or DPE for initiation
  • Typically, a promoter includes only two or three of these four elements.
  • Promoter elements can also be located at -50 to -110 region, for example, CCAAT Box is a common element found in many eukaryotic promoters
27
Q

Describe initiation in eukaryotes. General transcription factors (GTFs) and their binding sites on the core Pol II promoter.

A

The basic mechanism for initiating transcription of structural genes in eukaryotes requires protein factors binding selectively to the promoter regions.

• In eukaryotes, a set of at least six general transcription factors (GTF) (see table below) operate as a formal equivalent of a prokaryotic σ factor.

  • The complete set of general transcription factors and polymerase, bound together at the promoter and poised for initiation, is called pre-initiation complex (PIC).
  • GTFs help polymerase bind to promoter and melt the DNA (comparable to the transition from closed to open complex in bacteria). They also help polymerase escape from the promoter to enter elongation phase.

Functions of General Transcription Factors

  • TFIID— consists of TBP (TATA-binding protein) and TAFs (TBP Associated Factors). TBP recognizes the TATA box sequence and specifically binds to this sequence in the Pol II core promoter. Two TAFs bind DNA elements (Inr and DPE) at the promoter. Another TAF can regulate the binding of TBP to TATA box by using its inhibitory flat to bind to the DNA-binding surface of TBP.
  • TFIIA—probably stabilizes the TBP-TATA box interaction.
  • TFIIB—a protein with single polypeptides chain, binds to BRE through the interactions with major groove and to downstream sequence of the TATA by interactions with minor groove. TFIIB appears to bridge TATA-bound TBP and Pol II.
  • TFIIF—binding of TFIIF-Pol II stabilizes DNA-TBP-TFIIB complex and helps recruit TFIIE and TFIIH.
  • TFIIE—helps recruit and regulates TFIIH.
  • TFIIH—9 subunits, controls the ATP-dependent transition of the pre-initiation complex to the open complex. Two subunits function as ATPase and another, a protein kinase has roles in promoter melting and escape. TFIIH is also involved in DNA mismatch repair.
28
Q

Which component in a general transcription factor binds to the TATA box?

A

Binding of TBP to the TATA box

29
Q

Which transcription factor is responsible for the melting of DNA in the core promoter?

A

TFIIH

30
Q

The unusual Protein-DNA interaction between TBP and the TATA sequence.

A

Binding of TBP to the TATA box

  • TBP is a saddle-shaped molecule that consists of two structurally similar domains arranged with pseudo-twofold symmetry.
  • TBP binds to DNA using β sheets and distorts DNA by inserting β sheets into the minor groove, causing the minor groove to be widened to a almost flat conformation and bending of the DNA by a angle of ~80 o . The interaction between TBP and DNA involves only a limited number of hydrogen bonds between the protein and the edge of the base pairs in the minor groove. Instead, much of the specificity of TBP-TATA interaction is imposed by two pairs of phenylalanine side chains that intercalate between the base pairs at either end of the recognition sequence and drive the strong bend of the DNA. Therefore, to select the TATA sequence, TBP relies on the ability of the TATA sequence to undergo a specific structural distortion.
31
Q

The role of CTD phosphorylation in transcription.

A

A new set of factors stimulate Pol II elongation and RNA proofreading

• Transition from initiation to elongation involves the Pol II shedding most of its initiation factors, e.g. GTFs and mediators. A new set of factors, including elongation factors (e.g. TFIIS and hSPT5) that stimulate elongation and other factors required for RNA processing (i.e. 5’ capping, splicing, and 3’ polyadenylation). These factors are recruited to the CTD of Pol II.
* These factors favor the phosphorylated form of the CTD. Thus, phosphorylation of the CTD facilitates both shedding the initiation factors and recruiting elongation and RNA processing factors.
• Different enzymes are recruited depending on the phosphorylation state of the tail.

• TFIIS promotes elongation and also contributes to proofreading by stimulating RNAse activity in Pol II

32
Q

What are the three major posttranscriptional processing of mRNA in eukaryotes?

A

The molecule that’s directly made by transcription in one of your (eukaryotic) cells is called a pre-mRNA, reflecting that it needs to go through a few more steps to become an actual messenger RNA (mRNA). These are:

I. 5’ Capping of mRNA
II. 3’ Polyadenylation of mRNA
III. RNA Splicing

Addition of a 5’ cap to the beginning of the RNA
Addition of a poly-A tail (tail of A nucleotides) to the end of the RNA
Chopping out of introns, or “junk” sequences, and pasting together of the remaining, good sequences (exons)

Once it’s completed these steps, the RNA is a mature mRNA. It can travel out of the nucleus and be used to make a protein.

33
Q

The 5’ cap structure in eukaryotic mRNA.

A
  • Eukaryotic mRNAs have a cap structure consisting of a 7-methylguanosine residue joined to the transcript’s initial (5’) nucleotide via a unusual 5’-5’ triphosphate bridge.
  • The RNA is capped when it is still only 20-40 nucleotides long—when the transcription cycle has progressed only to the transition between the initiation and elongation phases.
  • The addition of the guanosine residue is catalyzed by a RNA triphosphatase to remove the leading phosphate in mRNA, then by a specific guanylyl transferase, and then the guanosine residue is methylated by a methyl transferase
  • Capped mRNAs are bound by cap-binding complex (CBC) and are resistant to 5’exonucleolytic degradation

The 5’ cap is added to the first nucleotide in the transcript during transcription. The cap is a modified guanine (G) nucleotide, and it protects the transcript from being broken down.

34
Q

In which stage does 5’ capping occur?

A

the transition between the initiation and elongation phases or Post-transcriptional processing ???

35
Q

Which polymerase is used for polyadenylation and what is the characteristic of this polymerase?

A

proteins and leading to RNA cleavage and polyadenylation.

  • Poly(A) tail is unique to transcripts made by Pol II, a feature that allows experimental isolation of protein coding mRNAs by affinity chromatography.
  • It is not clear what determines the length of the polyA tail, but the polyA binding protein is involved in this process.
  • PolyA tail bound by PABP may prevent mRNAs from digestion.
36
Q

Exons and introns, and their arrangement in a gene.

A

Eukaryotic Genes Consist of Alternating Exons and Introns

  • The coding sequence of most eukaryotic genes are interspersed with unexpressed regions.
  • The primary transcripts, called pre-mRNAs or heterogenous nuclear RNAs (hnRNAs), consist of expressed sequences (exons) and non-expressed intervening sequences (introns). * Exon and intron sequences are alternately arranged in a gene.
  • The length of introns in vertebrate genes ranges from ~65 to 200,000 nt.
  • Introns are to be removed from pre-mRNA and exons are joined, or spliced, together.

RNA splicing, specific parts of the pre-mRNA, called introns are recognized and removed by a protein-and-RNA complex called the spliceosome. Introns can be viewed as “junk” sequences that must be cut out so the “good parts version” of the RNA molecule can be assembled.
What are the “good parts”? The pieces of the RNA that are not chopped out are called exons. The exons are pasted together by the spliceosome to make the final, mature mRNA that is shipped out of the nucleus.

37
Q

What is pre-mRNA or hnRNA?

A

The primary transcripts, called pre-mRNAs or heterogenous nuclear RNAs (hnRNAs),

38
Q

The consensus sequences at exon-intron junction.

A

The Exon-Intron Junctions of Eukaryotic Pre-mRNA Contain Consensus Sequences

  • Most of exon-intron junctions have a invariant GU at the intron’s 5’ boundary.
  • Most of exon-intron junctions have a invariant AG at the intron’s 3’ boundary.
  • These consensus sequences are necessary and sufficient to define a splice junction.
39
Q

The chemical reactions and the lariat structure formation during mRNA splicing.

A

Exons Are Spliced in a Two-Stage Reaction The splicing reaction occurs via two transesterification reactions, which does not require free energy input:

  1. A 2’,5’-phosphodiesterbond forms between a intron adenosine residue and the intron 5’terminal phosphate group, resulting in the formation of a lariat structure.
  2. The free 3’-OH group of the 5’ exon attacks the 5’-phosphate of the 5’-terminal residue of the 3’ exon, forming a 3’,5’-phosphodiester bond, thereby splicing the two exons together and eliminating the intron in its lariat form. In vivo, the intron in lariat form is rapidly degraded.
40
Q

Splicosome-mediated splicing.

snRNPs and their functions.

A

Splicing Is Mediated by snRNPs

  • The transesterification reactions during splicing are mediated by a huge molecular complex called the spliceosome, which made up of snRNPs and other proteins.
  • snRNAs: small nuclear RNAs, highly conserved 60 to 300 nt RNAs.
  • snRNPs: small nuclear ribonucleoproteins, protein complexes consisting of snRNAs and proteins. Five snRNPs—U1, U2, U4, U5, and U6 are involved in the spliceosomemediated splicing. snRNPs have three roles in splicing: recognize the 5’ splicing site and the branch site through base pairing between snRNAs in the snRNPs and the splicing or branch sites (figure a and b below), bring those sites together through RNA-RNA interaction between snRNPs (figure c below), and catalyze the RNA cleavage and joining reactions.
41
Q

snRNPs and their functions.

A

snRNPs have three roles in splicing: recognize the 5’ splicing site and the branch site through base pairing between snRNAs in the snRNPs and the splicing or branch sites (figure a and b below), bring those sites together through RNA-RNA interaction between snRNPs (figure c below), and catalyze the RNA cleavage and joining reactions.

snRNP core proteins
➢ All four snRNPs, U1, U2, U4, and U5, contain the same so-called snRNP core proteins, which consists of seven Sm proteins named B, D1, D2, D3, E, F and G

➢ The Sm proteins share a common fold with a N-terminal helix followed by five-stranded antiparallel sheet

➢ These Sm proteins form a heptameric ring, and the funnel-shaped central hole of the Sm ring is lined with positive charges and interacts with a conserved sequence AAUUUGUGG known as Sm RNA motif present in all four snRNAs

42
Q

What are the difference between mRNA splicing and mRNA editing?

A
  • Like mRNA splicing, mRNA editing can also change the sequence of a RNA after it has been transcribed. Thus, the protein produced upon translation is different from that predicted from the gene sequence.

The term RNA editing describes those molecular processes in which the information content is altered in an RNA molecule
trons are removed by RNA splicing as RNA matures, meaning that they are not expressed in the final messenger RNA (mRNA) product, while exons go on to be covalently bonded to one another in order to create mature mRNA.

  • There are two mechanisms that mediate editing: site-specific deamination and guide RNA-directed uridine insertion or deletion.

mRNA splicing: rearrangement of mRNA
mRNA editing: Individual bases are either inserted, deleted, or changed

43
Q

What enzymes are used for rRNA processing in prokaryotic cells?

A

I. Prokaryotic rRNA Processing

  • E. coli has three types of rRNAs, the 5S, 16S, and 23S rRNAs. Seven operons transcribe these rRNAs and each operon contains one nearly identical copy of each of the three rRNA genes.
  • The polycistronic primary transcripts of these operons contain, in addition to rRNAs, as many as four tRNAs.
  • The primary transcripts are processed through endonucleolytic cleavages by RNases.
44
Q

Group I intron and group II intron Self-splicing

A

Some Eukaryotic rRNA Genes Contain Introns and Their Transcripts Are Self-splicing

Self-splicing RNAs are known as group I introns.

Discovery of self-splicing demonstrated that RNA can act as an enzyme—ribozyme.

The chemistry of splicing and the RNA intermediates produced during splicing of group II introns are the same as for nuclear pre-mRNAs, but the reactions are catalyzed by RNA enzyme encoded by intron (ribozyme)

  • In both group I and group II introns, although these introns can splice themselves out of

RNA molecules unaided by proteins in vitro, in vivo they typically do require protein

components to stimulate the reaction.

45
Q

What is RNA translation? Where does it occur?

A

Process by which the sequence of nucleotides in a message RNA molecule directs the incorporation of amino acids into protein. It occurs on a ribosome

46
Q

The characteristics of genetic codon.

A

A polypeptide chain contains up to 20 types of amino acids.
Only 4 types of nucleotide residues in mRNAs.
In fact, a codon consisting of three nucleotides specifies a single amino acid, proving sufficient combinations (43=64) to specify all the amino acids.
Codon: sequence of three nucleotides in a DNA or mRNA that represents the instruction for incorporation of a specific amino acid into a growing polypeptide chain.
The genetic code is non-overlapping, degenerate, triplet code.

  • The genetic code is based on sets of three-nucleotide codons that are read sequentially.
  • Each codon represents one amino acid or a Stop signal.
  • The genetic code is degenerate, nonrandom, and nearly universal.

One of the most fascinating puzzles in molecular biology is how a sequence of nucleotides composed of only four types of residues can specify the sequence of up to 20 types of amino acids in a polypeptide chain. Clearly, a one-to-one correspondence between nucleotides and amino acids is not possible. A group of several bases, termed a codon, is necessary to specify a single amino acid. A triplet code—that is, one with 3 bases per codon—is more than sufficient to specify all the amino acids since there are 4 3 = 64 different triplets of bases. A doublet code with 2 bases per codon (4 2 = 16 possible doublets) would be inadequate. The triplet code allows many amino acids to be specified by more than one codon. Such a code, in a term borrowed from mathematics, is said to be degenerate.
1.The code is highly degenerate.
2. The arrangement of the code table is nonrandom.
3.UAG, UAA, and UGA are Stop codons.
4. AUG and GUG are chain initiation codons. (start codons)

47
Q

Reading frame and frameshift.

A

Reading frame is the way the letters are divided into codons.

A. Codons Are Triplets That Are Read Sequentially
Crick and Brenner’s experiments.
Reading frame: The phase in which nucleotides are read in sets of three to encode a protein. A message RNA molecule can be read in any one of three reading frames, but generally only one of which will give the required protein. Deletions or insertions of nucleotides can cause frameshift mutation.

Brenner concluded that the genetic code is read in a sequential manner starting from a fixed point in the gene. The insertion or deletion of a nucleotide shifts the reading frame (grouping) in which succeeding nucleotides are read as codons. Insertions or deletions of nucleotides are therefore known as frameshift mutations.

48
Q

What is a nonsense codon?

A

UAG, UAA, and UGA are Stop codons. These three codons (also known as nonsense codons) do not specify amino acids but signal the ribosome to terminate polypeptide chain elongation.

49
Q

An mRNA does not directly recognize amino acids. Rather..?

A

it specifically binds tRNAs that each carry a corresponding amino acid.

50
Q

The common features and structure of tRNA.

A

A. tRNA Structure

Almost all known tRNAs can be schematically arranged in the so-called cloverleaf secondary structure.
Starting from the 5′ end, they have the following common features:
The common features of tRNAs:-
-A 5’-terminal phosphate group.
-A 7-bp stem that includes the 5’-terminal nucleotide and that may contain non-Watson-Crick base pairs such as G:U. This assembly is known as the acceptor or amino acid stem.
-A 3- or 4-bp stem ending in a 5-to 7-nt loop that frequently contains the modified base dihydrouridine (D), called D arm.
-A 5-bp stem ending in a loop that contains the anticodon, called anticodon arm.
-A 5-bp stem ending in a loop that usually contains the TΨC, called TΨC or T arm.
-A 3’-CCA sequence with a free 3’-OH group. The –CCA may be genetically specified or enzymatically appended depending on the species.

They vary in length from 54 to 100 nucleotides (18–28 kD) although most have ∼76 nucleotides.

51
Q

The tertiary structure of tRNAs.

A

tRNA Has a Complex Tertiary Structure:
-tRNA molecules have an L-shape in which the acceptor and T stems form one leg and the D and anticodon stems form the other.
-A tRNA’s complex tertiary structure is maintained by extensive stacking interactions and base pairing within and between the helical stems.

52
Q

Why do all tRNAs have similar tertiary structure?

A

Maintained by extensive stacking interactions and base pairing within and between the helical stems

53
Q

What are the two critical recognition steps during translation process?

A

Ribosomes, as we shall see, cannot distinguish as to whether a correct or incorrect amino acid group is linked to a tRNA. Therefore, accurate translation requires two equally important recognition steps:

  1. The correct amino acid must be selected for covalent attachment to a tRNA by an aminoacyl–tRNA synthetase (
  2. The correct aminoacyl–tRNA (aa–tRNA) must pair with an mRNA codon at the ribosome

-Covalent attachment of a correct amino acid to a tRNA by aminoacyl-tRNA synthetase
-Codon-anticodon recognition between aminoacyl-tRNA and an mRNA

54
Q

How is aminoacyl-tRNA formed from aminoacyl-tRNA synthetase (aaRS) ?

A

An amino acid-specific aminoacyl-tRNA synthetase (aaRS) appends an amino acid to the 3’-terminal ribose residue of its cognate tRNA at either its 3’-OH group or 2’-OH group to form an aminoacyl-tRNA.

55
Q

Two-step tRNA charging by aminoacyl-tRNA synthetase.

A

Aminoacyl tRNA Synthetases Charge tRNAs in Two Steps:

Step 1: Adenylation. Amino acid reacts with ATP to become adenylated with the concomitant release of pyrophosphate. The principal driving force for the adenylation reaction is the subsequent hydrolysis of pyrophosphate by pyrophosphatase.
Step 2: tRNA charging. Adenylated amino acid, which remains tightly bound to the synthetase, reacts with tRNA to form aminoacyl-tRNA and release AMP. Amino acid is appended to tRNA via either 3’-OH or 2’-OH.
-Aminoacyl-tRNA is a “high energy” compound; for this reason, the amino acid is said to be “activated” and the tRNA is said to be “charged”.

1.The amino acid is first “activated” by its reaction with ATP to form an aminoacyl–adenylate, which, with all but three aaRSs, can occur in the absence of tRNA. Indeed, this intermediate can be isolated, although it normally remains tightly bound to the enzyme.
2. The mixed anhydride then reacts with tRNA to form the aa–tRNA:

56
Q

Differences between two classes of aaRS.

A

Differences Between Class I and Class II Aminoacyl-tRNA Synthetases

  1. Structural motifs. The Class I enzymes share two homologous polypeptide segments that have the consensus sequences His-Ile-Gly-His (HIGH) and Lys-Met-Ser-Lys-Ser (KMSKS), which are components of a dinucleotide-binding fold (Rossman fold). The Class II synthetases lack these consensus motifs but have three other sequences in common, which consists of a seven-stranded antiparallel β sheet with three flanking helices and that forms the core of their catalytic domains.
  2. Anticodon recognition. Many Class I aaRSs must recognize the anticodon to aminoacylate their cognate tRNAs. In contrast, several Class II enzymes do not interact with their bound tRNA’s anticodon.
  3. Site of aminoacylation. All Class I enzymes aminoacylate their bound tRNA’s 3’-terminal 2’-OH group, whereas Class II enzymes charge the 3’-OH group.
  4. Amino acid specificity. The amino acids for which the Class I synthetases are specific tend to be larger and more hydrophobic than those for Class II synthetases.
57
Q

What is isoaccepting tRNA?

A

Aminoacyl-tRNA Synthetases Recognize Unique Structural Features of tRNA :
Isoaccepting tRNAs: different tRNAs that carry the same amino acid (because the genetic code is degenerate).

58
Q

Which structural portions of tRNA are important for aaRS-tRNA interaction?

A

The acceptor stem is critical for the enzyme-tRNA interaction, not necessary including the anticodon.

The most common synthetases contact sites on tRNA occur on the inner face of the L-shaped tRNA.

59
Q

What structural difference at the aaRS-tRNA interaction site could account for the difference on usage of OH group by class I and class II aaRSs?

A

Different Modes of tRNA Binding by GlnRS and AspRS
The Class I synthetase GlnRS binds tRNAGln from the minor groove side of its acceptor stem so as to bend its 3’ end into a hairpin conformation.
The Class II synthetase AspRS binds tRNAAsp from the major groove side of its acceptor stem so that its 3’ end continues its helical path on entering the active site.
These structural difference account for the observation that Class I and Class II enzymes aminoacylate different OH groups on the 3’ terminal ribose of tRNA.

60
Q

What are the two reasons for high accuracy of aminoacyl-tRNA formation?

A

The combined selectivities of the aminoacylation and editing steps are responsible for the high fidelity of translation, a phenomenon that occurs at the expense of ATP hydrolysis.

Aminoacyl-tRNA Formation Is Very Accurate
The charging of a tRNA with its cognate amino acid is a remarkably accurate process. For example, isoleucine and valine differ by only a single methylene group. However, IleRS transfers ~40,000 isoleucines to tRNAIle for every valine it so transfers.
Isoleucyl-tRNA synthetase is able to discriminate against valine twice: in the initial binding and adenylation of the amino acid, and then in the editing of the charged tRNA.
Although both amino acids will fit into the synthetase amino acid binding site, interaction with a extra methylene group on isoleucine will provide extra free energy. Even this relatively small difference in free energy will make binding to isoleucine approximately 100-fold more likely than binding to valine.
In addition to its catalytic pocket (for adenylation), the isoleucyl-tRNA synthetase has a nearby editing pocket (a deep cleft in the enzyme) that allows it to proof read the product of the adenylation reaction, i.e. AMP-valine. AMP-valine can fit into this editing pocket, where is hydrolyzed and released as free valine and AMP. In contrast, AMP-isoleucine is too large to enter the editing pocket and hence is not subjected to hydrolysis. Therefore, the editing pocket is a molecular sieve that excludes AMP-isoleucine but not AMP-valine.

61
Q

On which position of an anticodon does non-watson-crick base pairing commonly occur?

A

It seems that non-Watson-Crick base pairing can occur at the third codon-anticodon position. The third (5’) anticodon position commonly contains a modified base such as Gm or I.

62
Q

Codon Usage

A

Studies in yeast Saccharomyces cerevisiae has revealed a remarkable bias in their codon usage. Only 25 of the 61 coding triplets are commonly used.
The preferred codons are those that are most nearly complementary, in the Watson-Crick sense, to the anticodons in the most abundant species in each set of isoaccepting tRNAs.
The degree with which the preferred codons occur in a given gene is strongly correlated with the gene’s level of expression in both yeast and E. coli.

63
Q

Which codon can code for selenocysteine?

A

Incorporation of this amino acid requires some variation in the standard translation machinery, including a unique tRNA and a UGA stop codon that is reinterpreted as a Sec (selenocysteine) codon.