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BIO 325_Genetics > Chapter 12_Gene Transcription And RNA Modification > Flashcards

Chapter 12_Gene Transcription And RNA Modification Flashcards

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1
Q

What is the definition of a gene at the molecular level?

A

A segment of DNA that is used to make a functional product, either an RNA molecule or a polypeptide.

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2
Q

Transcription

A

Produces an RNA copy of a gene from a DNA sequence.

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3
Q

Structural Genes

A

Encode the amino acid sequence of a polypeptide.

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4
Q

Describe the central dogma of genetics

A

DNA replication –> Transcription –> Translation –> Polypeptide

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5
Q

Messenger RNA (mRNA)

A

When a structural gene is transcribed, the first product is an RNA molecule, mRNA.

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6
Q

Translation

A

Polypeptide synthesis. The sequence of nucleotides within the mRNA determines the sequence of amino acids in a polypeptide.

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7
Q

Promoter

A

Site for RNA polymerase

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8
Q

Terminator

A

Specifies the end of transcription.

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9
Q

Regulatory Sequences

A

Site for the binding of regulatory proteins; the role of regulatory proteins is to influence the rate of transcription. Regulatory sequences can be found in a variety of locations.

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10
Q

DNA vs. mRNA:

A
  • DNA: Deals with transcription. regulatory sequences, promoter, terminator
  • mRNA: Deals with translation. ribosomal binding site, start codon, codons, stop codon
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11
Q

Ribosomal Binding Site

A

Site for ribosome binding; translation begins near this site in the mRNA. In eukaryotes, the ribosome scans the mRNA for a start codon.

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12
Q

Start Codon

A

Specifies the first amino acid in a polypeptide sequence, usually a formylmethionine (in bacteria) or a methionine (in eukaryotes).

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13
Q

Codons

A

3-nucleotide sequences within the mRNA that specify particular amino acids. The sequence of codons within mRNA determines the sequence of amino acids within a polypeptide.

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14
Q

Stop Codon

A

Specifies the end of polypeptide synthesis.

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15
Q

Can bacterial mRNA be polycistronic?

A

Yes, which means it encodes two or more polypeptides.

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16
Q

Transcription Factors

A

Recognizes base sequences in the DNA and controls transcription.

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17
Q

The three stages of transcription:

A
  • Initiation: The promoter functions as a recognition site for transcription factors. The transcription factor(s) enables RNA polymerase to bind to the promoter. Following binding, the DNA is denatured into a bubble known as the open complex.
  • Elongation: (Synthesis of the RNA transcript) RNA polymerase slides along the DNA in an open complex to synthesize RNA.
  • Termination: A terminator is reached that causes RNA polymerase and the RNA transcript to dissociate from the DNA.
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18
Q

Transcriptional Start Site

A

This site is the first base used as a template for RNA transcription and is denoted +1. The bases preceding this site are numbered in a negative direction. No base is numbered zero.

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19
Q

Sequence Elements

A

Particularly critical for promoter recognition.

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20
Q

What are the important sequence elements in E. coli and other bacteria?

A
  • The sequence in the coding strand at the -35 region is 5’-TTGACA-3’
  • The sequence in the coding strand at the -10 region is 5’TATAAT-3’, which is called the Pribnow box
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21
Q

Consensus sequence

A

The most commonly occurring bases within a sequence element. This sequence is efficiently recognized by proteins that initiate transcription.

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22
Q

The enzyme that catalyzes the synthesis of RNA is

A

RNA polymerase.

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23
Q

Core Enzyme (RNA Polymerase)

A

Composed of five subunits, two alphas, Beta, Beta Prime, and w.

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24
Q

Holoenzyme

A

RNA polymerase with a sixth subunit, the sigma factor.

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25
Q

How is a promoter identified?

A
  • When the holoenzyme encounters a promoter sequence, sigma factor protein contains a structure called a helix-turn-helix motif that can bind tightly to these regions.
  • Alpha helices within the protein fit into the major groove of the DNA double helix and form hydrogen bonds with the bases. Hydrogen bonding occurs between nucleotides in the -35 and -10 regions of the promoter and amino acid side chains in the helix-turn-helix structure of sigma factor.
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26
Q

How is the process of transcription initiated?

A

When sigma factor within the holoenzyme has bound to the promoter region to form the closed complex. For transcription to begin, the double stranded DNA must then be unwound into an open complex.

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27
Q

Where does the unwinding occur to make an open complex in double stranded DNA?

A

At the -100 region, which contains only AT base pairs. It’s easier because AT base pairs have only 2 hydrogen bonds, while GC base pairs have 3.

28
Q

How does elongation begin?

A

A short strand of RNA is made within the open complex, and then sigma factor is released from the core enzyme. The release of sigma factor marks the transition to the elongation phase of transcription. The core enzyme may now slide down the DNA to synthesize a strand of RNA.

29
Q

RNA polymerase moves along the template strand in a 3’ to 5’ direction, why?

A

If it moves in a 3’ to 5’ direction, the RNA is synthesized in a 5’ to 3’ direction.

30
Q

How does termination begin?

A

The hydrogen bonding between the DNA and RNA within the open complex is of central importance in preventing dissociation of RNA polymerase from the template strand. Termination occurs when this short RNA-DNA hybrid region is forced to separate, thereby releasing RNA polymerase as well as the newly made RNA transcript.

31
Q

In E. coli, there are two different mechanisms for termination have been identified:

A
  • Rho dependent and independent formation.
32
Q

Rho-Dependent Termination

A
  • The termination process requires two components.
    • First, a sequence upstream from the terminator, called the rut site for rho utilization site, acts as a recognition site for the binding of the rho protein. After the rut site is synthesized in the RNA, rho protein binds to the RNA and moves in the direction of RNA polymerase.
    • Second, the site where termination actually takes place. At this terminator site, the DNA encodes an RNA sequence containing several GC base pairs that for a stem-loop structure. This loop forms almost immediately after the RNA sequence is synthesized and quickly binds to RNA polymerase. This binding results in a conformational change that causes RNA polymerase to pause in its synthesis of RNA. The pause allows rho protein to catch up to the stem loop, pass through it, and break the hydrogen bonds between the DNA and RNA within the open complex.
33
Q

How does the rho protein facilitate termination?

A

The rho protein functions as a helicase, an enzyme that can separate RNA-DNA hybrid regions.

34
Q

Rho-Independent Termination

A
  • The terminator is composed of two adjacent nucleotide sequences that function within the RNA. One is a uracil rich sequence located at the 3’ end of the RNA. The second is adjacent to the uracil rich sequence and promotes the formation of a stem loop structure.
  • The stem loop causes the RNA polymerase to pause. This pausing is stabilized by other proteins that bind to RNA polymerase.
  • ## At the precise time of the pause, the U rich sequence in the RNA transcript is bound to the DNA template strand. (This hydrogen bonding of RNA to DNA keeps RNA polymerase clamped onto the DNA). However, the binding of this U rich sequence to the DNA template strand is very weak, causing the RNA transcript to spontaneously dissociate from the DNA and cease further transcription.
35
Q

Intrinsic Termination

A

Termination that does not require a protein to physically remove the RNA transcript from the DNA. In E coli, half of the genes show intrinsic termination, the other half are terminated by rho protein.

36
Q

The genetic material within the nucleus of a eukaryotic cell is transcribed by…

A

…three different RNA polymerase enzymes, designated RNA polymerase I, II, and III.

37
Q

Describe the role of each RNA polymerase in eukaryotic cells.

A
  • One: Transcribes all of the genes that encode ribosomal RNA except for the 5S rRNA.
  • Two: Plays a major role in cellular transcription because it transcribes all of the structural genes. It is responsible for the synthesis of all mRNA and also transcribes certain snRNA genes, which are needed for pre-mRNA splicing.
  • Three: Transcribes all tRNA genes and the 5s rRNA gene.
38
Q

For structural genes, at least three features are found in most promoters:

A

Regulatory elements, a TATA box, and a transcriptional start site.

39
Q

Core Promoter

A

A relatively short DNA sequence that is necessary for transcription to take place. IT consists of a TATAAA sequence called the TATA box and the transcriptional start site, where transcription begins.

40
Q

Basal Transcription

A

A low level of transcription, pertains to the core promoter which by itself produces a low level of transcription.

41
Q

TATA Box

A

About 25 bp upstream from a transcriptional start site. It is important in determining the precise starting point for transcription.

42
Q

There are two categories of regulatory elements:

A
  • Enhancers: Needed to stimulate transcription (also called activating sequences).
  • Silencers: DNA sequences that are recognized by transcription factors that inhibit transcription.

A common location of regulatory elements is the -50 to -100 region. However, the locations of regulatory elements vary considerably among different eukaryotic genes.

43
Q

cis-acting genes

A

Always found within the same chromosome as the genes they regulate.
- EX: DNA sequences such as the TATA box, enhancers, and silencers, and they exert their effects only over a particular gene.

44
Q

Trans-acting factors

A

The regulatory transcription factors that bind to said elements (cis-acting genes).

45
Q

Transcription of Eukaryotic Structural Genes is initiated when…

A

…RNA polymerase II and General transcription factors bind to a promoter sequence.

46
Q

Three categories of proteins are needed for basal transcription at the core promoter:

A

RNA polymerase II, general transcription factors, and mediator.

47
Q

Describe in VIVID detail the steps leading to the formation of the open complex

A
  • TFIID binds to the TATA box. TFIID is a complex of proteins that includes the TATA-binding protein (TPB) and several TPB-associated factors (TAFs).
  • TFIIB binds to TFIID.
  • TFIIB acts as a bridge to bind RNA polymerase II and TFIIF.
  • TFIIE and TFIIH bind to RNA polymerase II to form a preinitiation or closed complex.
  • TFIIH acts as a helicase to form an open complex. TFIIH also phosphorylates the CTD domain of RNA polymerase II. CTD phosphorylation breaks the contact between TFIIB and RNA polymerase II. TFIIB, TFIIE, and TFIIH are released.
48
Q

the colinearity of gene expression

A

The one to one correspondence between the sequence of codons in the DNA coding strand and the amino acid sequence of the polypeptide.

49
Q

Exons

A

Regions that are contained within mature RNA where coding sequences are found.

50
Q

Introns

A

(intervening sequences) Sequences that are found between the exons.

51
Q

RNA splicing

A

During transcription, an RNA is made corresponding to the ENTIRE gene sequence. Subsequently, as it matures, the sequences in the RNA that correspond to the introns are removed and the exons are connected, or spliced together.

52
Q

Possible mechanisms for transcriptional termination of RNA polymerase II

A

RNA polymerase II transcribes a gene past the polyA signal sequence. The RNA is cleaved just past the polyA signal sequence. RNA polymerase continues transcribing the DNA.

  • Allosteric Model: After passing the polyA signal sequence, RNA polymerase II is destabilized due to the release of elongation factors or the binding of termination factors. Termination occurs.
  • Torpedo Model: An exonuclease binds to the 5’ end of the RNA that is still being transcribed and degrades it in a 5’ to 3’ direction. Exonuclease catches up to RNA polymerase II and causes termination.
53
Q

Nucleolus

A

In eukaryotes, where the cleavage of 45S rRNA into smaller rRNAs and the assembly of ribosomal subunits occurs.

54
Q

Exonuclease vs. Endonuclease

A
  • Exonuclease: A type of enzyme that cleaves a covalent bond between two nucleotides at one end of a strand. Starting at one end, an exonuclease can digest a strand, one nucleotide at a time. Some go from 3’ to 5’, others go 5’ to 3’.
  • Endonuclease: Can cleave the bond between two adjacent nucleotides within a strand.
55
Q

What happens when mRNA is bound to DNA

A

It prevents the DNA from forming a double helix. When they bind to each other, it is called hybridization. This new molecule then forms a loop, called an R loop (RNA displacement loop).

56
Q

cDNA

A

complementary DNA. It is a copy of DNA made by mRNA. However, because of RNA splicing, the cDNA would be missing base sequences if it had any introns.

57
Q

Self-splicing

A
  • Splicing that does not require the aid of other catalysts. Instead, the RNA functions as its own ribozyme.
  • The splicing among group I and group II introns occurs this way.
58
Q

Maturases

A

Proteins that enhance the rate of splicing of group I and group II introns.

59
Q

Spliceosome

A
  • Needed for pre-mRNA splicing.
60
Q

A spliceosome is a large complex composed of…

A

…several subuints known as snRNPs. Each contains small nuclear RNA and a set of proteins.

61
Q

During splicing, the subunits of a spliceosome carry out several functions:

A
  • First, spliceosome subunits bind to an intron sequence and precisely recognize the intron exon boundaries.
  • In addition, the spliceosome must hold the pre mRNA in the correct configuration to ensure the splicing together of the exons
  • Finally, the spliceosome catalyzes the chemical reactions that cause the introns to be removed and the exons to be covalently linked.
62
Q

Capping

A

At their 5’ end, most mature mRNAs have a 7-methylguanosine covalently attached. It occurs while the pre-mRNA is being made by RNA polymerase II, usually when the transcript is only 20 to 25 nucleotides in length.

63
Q

Describe the process of capping

A
  • First, an enzyme called RNA 5’ - triphosphatase removes one of the phosphates, and then a second enzyme guanylyltransferase, uses guanosine triphosphate (GTP) to attach a guanosine monophosphate (GMP) to the 5’ end.
  • Finally, a methyltransferase attaches a methyl group to the guanine base.
64
Q

What are the functions of the 7-methylguanosine cap?

A
  • The cap structure is recognized by cap binding proteins
  • Recognized by initiation factors that are needed during the early stages of translation
  • May be important in the efficient splicing of introns, particularly the first intron located nearest the 5’ end.
65
Q

polyA tail

A

A string of adenine nucleotides on mRNAs. It is important for mRNA stability and in the synthesis of polypeptides. It is not encoded in the gene sequence.

66
Q

Polyadenylation

A

Because the polyA tail is not encoded in the gene sequence, it is added enzymatically after the pre-mRNA has been completely transcribed.

BIO 325_Genetics (17 decks)

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