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BIOL 226 > Transcription/Translation > Flashcards

Transcription/Translation Flashcards

1
Q

Transcription is the first step in ___

A

gene regulation

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2
Q

what is responsible for RNA synthesis

A

RNA polymerase

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3
Q

What is the difference between RNA pol and DNA pol

A

RNA pol doesn’t require a primer sequence to initiate RNA synthesis

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4
Q

RNA polymerase I (its products and location)

A

Products - 5.8S, 18S and 28S (ribosomal RNAs) structural component- excluding 5S rRNA

Location - nucleolus

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5
Q

RNA polymerase II (its products and location)

A

Products - Nuclear pre - mRNA

Location - nucleus

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6
Q

RNA polymerase III (its products + function and location)

A

Products - transfer RNA (tRNA)

The function of the product - serve as adaptor molecules that align amino acids during mRNA template

Location - Nucleus

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7
Q

Genome size increases therefore, there is a ____ protein-coding sequence and ______ non - coding sequences

A

a decrease in protein-coding sequences

Increase non - coding sequences

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8
Q

What are transcription factors

A

proteins required for RNA polymerase II to initiate transcription

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9
Q

There are 2 categories of transcription factors:

A
  1. General transcription factors - involves in transcription from all polymerase II promoters
  2. Gene-specific transcription factors - bind to the promoter and an enhancer element
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10
Q

What is the function of the general transcription factors

A

Recruits RNA polymerase II

Formation of the preinitiation complex

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11
Q

What if the function of gene-specific transcription factors

A

bind to promoter + enhancer element

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12
Q

Steps of Transcription: (9 steps)

A
  1. TATA box (regulatory seuqunece) transcribed by RNA pol. II
  2. TBP binds to the TATA box
  3. TAFs are proteins associated with TBP in the genral transcription factor, TFIID
  4. TFIIB binds with TFFID at core pormoter
  5. RNA pol. II + TFIIIF are recruited using TFIIB to get to TFIID + promoter
  6. C- terminal domain of RNA pol. II requires phosphorylation prior to initiation of transcription
  7. TFIIE + TFIIH complete the formation of the initiation complex
  8. TFIIH has both a kinase activity that phosphorylates CTD and a helicase that unwinds the DNA
  9. RNA pol. II starts synthesis and leaves the promoter and TFIID behind
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13
Q

Steps in Elongation:

Translation

A
  1. tRNA/ribosome complex is startingat AUG start site of mRNA (P site of the ribosome)
  2. The next charge tRNA comes to the A site
  3. Peptide bond formation occurs between the amino acids
  4. Ribosome moves downstream - now the A site is ready for the next charged tRNA
  5. Repeat steps 1-4
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14
Q

What is different about a mature mRNA versus a regular mRNA

A

A mature mRNA consists of a 3’poly AA tail, PABP and 5’m7G cap eIFs

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15
Q

Steps in Initiation - Translation

A
  1. met tRNA + small ribosomal subunit + initiation factors
    * Charged tRNA binds to 40S subunit before 40S binds to the mRNA
  2. the 40S/tRNA/initiation factor complex meets up with a mature mRNA + binds to the 5’m7Gcap
  3. The entire complex sites on 5’end of the mRNA + slides along the mRNA with it finds AUG start codon.
    - This requires energy
    - There is not a 60S ribosomal unit yet in step 3!
  4. Most of the complex (except for 40S subunit + tRNA) dissociate form start site 60S subunit attaches to 40S + tRNA
    - GTP hydrolysis is required to release the initiation factors.
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16
Q

What is PABP

A

PABP is a Poly A binding protein

  • It helps recruit enzymes for polyadenylation and remain bond to the poly-a-tail until translation
  • Protects poly-A tail fro degradation
  • Binds mRNA to translation initiation factors to facilitate binding to the small ribosomal subunit
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17
Q

Splicing - what is it and when/where does it occur?

A

Splicing - introns are removed and exons are joined together to form a mature mRNA

  • Occurs in the nucleus before the RNA migrates to the cytoplasm to be translated into proteins
  • Only the pre mRNA transcript contains both introns and exons. Mature mRNA only contain uninterrupted coding information
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18
Q

Name for coding and non - coding sequences

A

Intorns = non - coding sequences

Exons = coding sequences

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19
Q

Do Prokaryotes and Eukaryotes have introns?

A

No, only eukaryotes

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20
Q

Two mechanisms of splicing

A
  1. Self - splicing (enzymatic activity - ribozyme) no energy required
  2. RNA/protein complex-mediated splicing - spliceosome
21
Q

which snRNAs are required for splicesome - dependent splicing

A

U1, U2, U4, U5, and U6

22
Q

How does the spliceosome function?

A

snRNA in the SNRPs bind the transcript through nucleotide interactions with the conserved splicing sites to bring the exons in close proximity for the reactions to occur

23
Q

Order of events in each exon/intron/exon (splieosome)

A
  1. U1 binds 5’ site, U2 binds branch site
  2. U4/U5/U6 bind to U1/U2
  3. Axon - axon boundaries brought closer
  4. U4/U6/U1 leave the complex
  5. Transesterification reactions take place
  6. Intron sequences are removed with the formation of a loop (lariat)
24
Q

Poly A polymerization

A

The 3’ poly-A tail - the poly-A polymerases recognize processed transcripts as templates to add the poly-A tail

25
Q

What is the purpose of the poly(A) tail

A

the enhance mRNA stabilization in the cytoplasm

26
Q

True or False:

The complexity of the transcriptome mirrors an increase in the genome

A

False

27
Q

Transcription in Prokaryotes

- RNA chain Initiation

A
  1. RNA polymerase (holoenzyme) binds directly to a promoter in DNA
  2. Unwinds the two strand of DNA by RNA polymerase to provide a single-stranded template
  3. phosphodiester bonds between ribonucleotides occur
  4. RNA polymerase binding is guided by the recognition of cis-elements in the promoter of genes
  5. +1 site is where transcription states (between promoter regions and ATG)
  6. RNA chain elongation
28
Q

Transcription in Prokaryotes - RNA chain Termination

A
  1. RNA chain grows from 5’ to 3’
  2. RNA poly (helicase) unwinds DNA
  3. DNA rewinding happens by reforming hydrogen bonds between the two DNA strands
  4. RNA poly detached from DNA template + RNA strand is released
29
Q

What is the difference in RNA pol in prokaryotes and Eukaryotes

A

Prokaryotes - 1 RNA pol
Eukaryotes - 5 RNA pol - all RNA pol in eukaryotes requires the activity of trasncription facotrs to intiiate polymerization

30
Q

True or False - Both prokaryotes and Eukaryotes experience post-transcriptional modifications

A

False - only Eukaryotes experience post-transcriptional modifications like intron splicing/5’ capping and 3’ polyadenylation

31
Q

where does transcription and translation occur in both eukaryotes and prokaryotes?

A

Eukaryotes - transcription occurs in the nucleus and translation occurs in the cytoplasm
Prokaryotes - both occur simultaneously in the cytoplasm

32
Q

RNA pol IV (its products and location)

A

Products - Small interfering RNAs

Location - Nucleus (plant)

33
Q

RNA pol V (its products and location)

A

Products - some siRNAs plus noncoding (antisense) transcripts of siRNA target genes

Location - Nucleus (plant)

34
Q

Which RNA pol is found in a different location than the others

A

RNA pol I - it is found in the nucleolus whereas the others are found in the nucleus

35
Q

What assists RNA pol II and why

A

Transcription factors - to help recognize and initiate transcription at the promoter

36
Q

Translation invovles which RNA polymerases

A

RNA pol I - all rRNAS but 5S rRNA

RNA pol II - mRNA

RNA pol II - tRNA, 5S rRNA

37
Q

Describe the components of the Prokaryotic 70S ribosome

A

Large 50S subunit (28S and 5S rRNAs)+ 30S subunit (16S rRNA)

38
Q

Describe the components of the Eukaryotic 80S ribosome

A

Large 60S subunit (28S, 5.8S and 5S rRNAs) + 40S subunit (18S)

39
Q

tRNA

A

small adapters between amino acids and the codons in mRNA molecules

tRNA contains modified nucleotides to avoid mispairing with the codon

40
Q

Aminoacyl

A

tRNA synthetase attaches an amino acid to its specific tRNA

41
Q

What would cocur if the information and anticodon did not match a tRNA?

A
  1. The tRNA would not be able to catalyze the addition of the incorrect amino acid in the peptide chain
  2. would be blind to the incompatibility and would normally catalyze the incorrect amino acid.
42
Q

The genetic code is composed of ____ and is non - _____ and is comma _____

A

nucleotide sequence and is non - reversible and is comma free

43
Q

Start and Stop codons

A

Start- AUG

Stop - UAA, UAG, and UGA

44
Q

What is Hachioji DNA

A

8 nucleotides instead of 4 (A-t, G-C, Z-P, S-B)

can be replicated and transcribed

45
Q

What is the Wobble Rules

A

Wobble Rules identify base-pair interactions between mRNA (3’ end of codon) and tRNA (5’ end of anticodon) (do not follow Watson and Crick Rules

46
Q

The wobble excpetion is when…

A

there is a degeneracy of the third nucleotide of the codon only happens AFTER the first tow nucleotides form stable interactions with the corresponding anticodon

47
Q

Amber (UAG) mutations (what occurs + the result)

A

mutations that introduce a premature stop codon in the transcript will cause a short version of the protein

Result - leads to a lack of function and a mutant phenotype

48
Q

What is the effect of a nonsense mutation

A

a stop codon therefore, it prevents the amber mutant from terminating their polypeotide prematurely

BIOL 226 (6 decks)

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