topic 18 Flashcards

1
Q

why is there an RNA translation step in the information flow from DNA (to RNA) to proteins? (3)

A

it provides an amplification which contributes to differential gene expression

since RNA can be degraded (due to its instability in cellular environments), expression of a gene can be stopped/turned off quickly

RNA provides additional opportunities to regulate gene expression through RNA processing, RNA export from the nucleus, etc

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2
Q

what is a consensus sequence? describe it

A

the consensus sequence depicts the most frequent base at each position in a group of functionally related DNA elements. usually, the closer a certain sequence is to the consensus sequence, the better the functionality of the DNA element.

for example, if the sequence is meant to be bound by a certain protein: the more the sequence deviates from the consensus, the less tightly the protein can bind

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3
Q

what is an operon? describe it

A

an operon is a set of bacterial genes that are transcribed from a single promoter expressed as an mRNA. the proteins are then translated from the mRNA which has different coding regions for the different proteins. ***eukaryotes don’t have operons

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4
Q

what is a promoter? what sequences does it include?

A

promoter is the DNA sequence required to initiate transcription of a gene/operon.

the promoter includes the sequences that
1) recognize RNA polymerase and
2) recognize any gene specific regulatory factors.

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5
Q

list in order the sequences/structures that appear on a bacterial operon from the promoter to the terminator

A

promoter

transcriptional start site (+1)

3 protein coding sequences (X,Y,Z)

terminator

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6
Q

describe the promoter structure in bacterial DNA

A

the promoter is usually approximately 100 base pairs. within the bacterial promoter, there are 2 sequence elements called “-35” and “-10 (pribnow box)”. they are named for their approximate location from the +1 start site. the consensus sequences at these sites differ between bacteria.

in between these sites can be any nucleotide as this region does not serve any functional purpose. importance of sequences can be determined with mutations and if it’s still transcribed into RNA.

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7
Q

what is the consensus sequence of -35 and -10 in E coli?

A

in E coli, the consensus sequence for the -35 site is TTGACA and the -10 region is TATAAT.

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8
Q

what is a core enzyme? what is it composed of? how does it recognize the promoter?

A

a core enzyme (bacterial RNA polymerase) is the minimal composition required to transcribe RNA. it consists of 2 alpha, a beta, a beta’, and an omega subunit. the core enzyme can add nucleotides and form a RNA strand using DNA as a template.

it can’t recognize the promoter region. to recognize the promoter region, it needs a sigma subunit. specific sigma subunits can recognize specific promoters.

the core enzyme with a sigma subunit is called the holoenzyme. when bound to the core enzyme, the sigma subunit recognizes the -10 and -35 sequences in the promoter.

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9
Q

what are the steps of transcription? (8) describe them

A
  1. during the 1st step “initiation”, the RNA polymerase holoenzyme binds to the promoter. this is called the closed complex as the DNA is still double stranded.
  2. RNA polymerase unwinds the DNA strands from around the start site (at -10) - it’s now an open complex.
  3. the 1st NTP is brought to the template, base pairing with the base at +1. unlike DNA replication, no primer is required to initiate transcription.
  4. transcription begins and all 4 nucleotides are used as substrates. chain elongation must proceed in the 5’ to 3’ direction, following base pairing rules. the polymerase must follow the DNA in the 3’ to 5’ direction to use it as a template.
  5. after 5-10 nucleotides, the sigma factor isn’t needed for promoter binding anymore and it falls off the holoenzyme. now, transitioning from initiation to elongation.
  6. the transcription bubble moves downstream (5’ to 3’) with the template DNA reannealing behind. elongation is the process of transcribing the rest of the RNA.
  7. RNA synthesis stops when a terminator region is reached and the RNA polymerase falls off.
  8. sigma factor rebinds to the core enzyme (RNAP) and the cycle is repeated
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10
Q

in transcription, why does the RNA polymerase unwind the DNA strands at -10 of the promoter?

A

the consensus sequence here consists of Ts and As, which is easier to unwind compared to GC rich regions, making -10 the ideal point for separating DNA stands.

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11
Q

describe the transcription bubble and the RNA polymerase

A

DNA is unwound for transcription, but it closes again afterwards into the replication bubble. the entire bubble is covered by the RNA polymerase which means no single strand of the DNA is exposed.

in the polymerase, there exists a small rudder which separates the 2 strands, that acts as a channel to bring in new nucleotides and for the RNA to exit

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12
Q

what are the types of termination in bacteria? (2) describe them

A

there are 2 types of termination in bacteria: rho independent (top) and rho dependent (bottom). rho is a small protein that binds to RNA.

in the rho dependent mechanism, rho travels along the RNA. when the polymerase stalls at the terminator sequence, rho catches up with the polymerase and causes it to fall off.

in the rho independent mechanism, termination relies on the polymerase falling off on its own, usually also cause by a terminator sequence in the DNA. the rho independent termination is based on an RNA hairpin formation that destabilizes the transcription bubble.

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13
Q

how are different bacterial promoters transcribed at different levels? (3)

A

all comes down to the consensus sequence and how closely the promoter resembles the optimal promoter sequence.

  1. some -10 and -35 sequences are better than others and are more easily recognized by the sigma subunit. this regulation isn’t dynamic, meaning it can’t react to external triggers and it’s gene specific.
  2. sigma factors are essential to recognize promoters. there are several sigma factors in the cell that each recognize specific sequences. sigma70 is for day to day proteins “house-keeping” gene. sigma54 recognizes the promoter of genes that are required for nitrogen metabolism. sigma38 is for starvation (transfer of nutrients) and sigma 32 is for heat shock.
  3. gene specific regulatory proteins: negative, which represses transcription and positive which activates transcription to be able to regulate transcription, these factors must be able to efficiently bind to DNA. proteins and DNA are highly compatible. 1 alpha helix fits nicely into the major groove of DNA, allowing for a sequence specific interaction. proteins can also interact the DNA backbone, but this isn’t sequence specific. a repressive protein can bind to DNA by binding in the major groove of the DNA.
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14
Q

describe the lac operon’s structure and its surrounding area

A

in the lac operon, 3 lactose metabolizing genes are encoded in the RNA proteins that are beta-galactosidase (protein Z), permease (protein Y), and trans-acetylase (protein A).

in front of the gene, there are 3 regulatory sites: inducer binding site (I), promoter (P), and the operator site (O).

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15
Q

how can transcription be obscured/repressed?

A

transcription can be obscured/repressed. RNA polymerase must first find the promoter region of the gene to initiate transcription, if the region is blocked by another protein (repressor), then there will be little to no expression.

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16
Q

what are the forms of expression control in the lac operon?

A

lac repression (negative control). the repressor bind to the operator site and blocks the RNA polymerase.

activation by the catabolite active protein (positive control). the activator binds to the inducer site and helps the RNA polymerase bind more efficiently.

17
Q

what are the conditions for the lac operon to be turned on?

A

the conditions for the lac operon to be turned on is the presence of lactose and the absence of glucose.

E coli only needs the lac gene when lactose is available. E coli prefers glucose over lactose because it yields more energy for less work. if glucose is present, E coli would stop transcription of the lac gene. this is an example of catabolite repression.

18
Q

what are the key molecules/binding sites in the regulation of the lac operon? describe their roles/processes

A

lac repressor: DNA binding protein

catabolite activator protein (CAP): transcriptional activator protein

lac operator: DNA element that binds lac repressor

CAP binding site: DNA element that binds CAP

the lac repressor binds to the lac operator or the lac repressor binds to a metabolite of lactose called allolactose, preventing the lac repressor binding to the DNA. meaning, that in the presence of lactose, the repressor doesn’t repress the gene expression.

the CAP is a protein that senses glucose metabolism. it can only bind under low glucose conditions. the protein helps the RNA polymerase to bind to the promoter region. when the concentration of glucose in the media is low, the cellular concentration of cAMP rises. at high concentrations of cellular cAMP the E coli, CAP binds cAMP. a conformational change occurs in CAP upon binding cAMP that allows the protein dimer to bind DNA in a site-specific fashion

19
Q

what happens when regions controlling gene expression are mutated?

A

repressor binding site: repressor can’t bind and RNA polymerase will always have access to the promoter.

CAP binding site: CAP can’t bind and RNA polymerase won’t recognize the promoter. therefore, no transcription