bio 235 exam 4 CORRECT QUESTIONS
& SOLUTIONS(RATED A+)
Translation initiation
-both in ______ &________
- starts with _________ binding to small _____________ subunit - ANSWER
prokaryotes, eukaryotes, mRNA, ribosomal
Translation initiation
- nearest ______ will be the start _____________
- then base pairing with _____ , _______
-recruits large _____________ subunit - ANSWER AUG, codon, fMET, tRNA, ribosomal
Elongation
-The _________ moves ___' to ____' along the ___________ - ANSWER Ribosome, 5,
3, mRNA
Elongation
- _____________ is the____________, it holds everything in place - ANSWER
Ribosome, adopter
Elongation
Newly charged ___________ base pair with codons at ________________ ________.
(a)
-plus their _________ __________ are linked via Peptide bonds at ___________
____________ ______.(p) - ANSWER tRNA's, Aminoacyl site, amino acids, peptidyl
transferase site
The __________________for eukaryotic translation is determined by more complex
sequence rules
-_____________ scans from the ________' end to find the consensus sequence for
optimal start codon (___) recognition
(This is described as part of the Kozak's Rules) - ANSWER start codon, ribosome, 5,
AUG
Euk. Translation
-ribosome scans 5' end to find start codon
-that in mind, the start codon for eukaryotic translation is usually the first AUG
_____________ the 5' Cap! - ANSWER After
End of Elongation
, -_________ is separated from its_________ ____________ and pushed out the
_______ _____________.(e) - ANSWER tRNA, amino acid, exit site
Attenuation - ANSWER Prokaryotic method of genetic regulation, results in the
premature termination of mRNA synthesis due to stem and loop formation in the 5' UTR
polyribose- several ____________ translating a single ____________ - ANSWER
ribosomes, mRNA
termination
-The_______ ____________ is recognized by a ___________ ____________. -
ANSWER stop codon, release factor
Termination
Upon binding by the ____________ ____________, the completed
__________________ ____________ is released, the last ___________dissociates -
ANSWER release factor, polypeptide chain, tRNA
Termination
the large and small subunit s of the ______________ ______________from one-
another and the __________. - ANSWER ribosome dissociate, mRNA
Nonsense-mediated Decay
___________ in the _____________ leaves a mark at the site of _________-
__________ _____________: _____________ of the exon junction complex (EJC). -
ANSWER splicing, nucleus, exon-exon junction, proteins
Nonsense-mediated Decay
-______ _______________ are removed during the _________ __________ of
(pioneer) ________________. - ANSWER EJC proteins, first round, translation
High tryptophan consentration - ANSWER Causes intrinsic termination by stem loop
between 3-4
Low tryptophan concentration - ANSWER Causes stem loop between 2-3, transcription
will continue
Negative regulation - ANSWER Inhibits gene expression
Catabolite repression - ANSWER overall effect of glucose is to prevent lac gene
expression
Positive regulation - ANSWER Stimulates gene expression
& SOLUTIONS(RATED A+)
Translation initiation
-both in ______ &________
- starts with _________ binding to small _____________ subunit - ANSWER
prokaryotes, eukaryotes, mRNA, ribosomal
Translation initiation
- nearest ______ will be the start _____________
- then base pairing with _____ , _______
-recruits large _____________ subunit - ANSWER AUG, codon, fMET, tRNA, ribosomal
Elongation
-The _________ moves ___' to ____' along the ___________ - ANSWER Ribosome, 5,
3, mRNA
Elongation
- _____________ is the____________, it holds everything in place - ANSWER
Ribosome, adopter
Elongation
Newly charged ___________ base pair with codons at ________________ ________.
(a)
-plus their _________ __________ are linked via Peptide bonds at ___________
____________ ______.(p) - ANSWER tRNA's, Aminoacyl site, amino acids, peptidyl
transferase site
The __________________for eukaryotic translation is determined by more complex
sequence rules
-_____________ scans from the ________' end to find the consensus sequence for
optimal start codon (___) recognition
(This is described as part of the Kozak's Rules) - ANSWER start codon, ribosome, 5,
AUG
Euk. Translation
-ribosome scans 5' end to find start codon
-that in mind, the start codon for eukaryotic translation is usually the first AUG
_____________ the 5' Cap! - ANSWER After
End of Elongation
, -_________ is separated from its_________ ____________ and pushed out the
_______ _____________.(e) - ANSWER tRNA, amino acid, exit site
Attenuation - ANSWER Prokaryotic method of genetic regulation, results in the
premature termination of mRNA synthesis due to stem and loop formation in the 5' UTR
polyribose- several ____________ translating a single ____________ - ANSWER
ribosomes, mRNA
termination
-The_______ ____________ is recognized by a ___________ ____________. -
ANSWER stop codon, release factor
Termination
Upon binding by the ____________ ____________, the completed
__________________ ____________ is released, the last ___________dissociates -
ANSWER release factor, polypeptide chain, tRNA
Termination
the large and small subunit s of the ______________ ______________from one-
another and the __________. - ANSWER ribosome dissociate, mRNA
Nonsense-mediated Decay
___________ in the _____________ leaves a mark at the site of _________-
__________ _____________: _____________ of the exon junction complex (EJC). -
ANSWER splicing, nucleus, exon-exon junction, proteins
Nonsense-mediated Decay
-______ _______________ are removed during the _________ __________ of
(pioneer) ________________. - ANSWER EJC proteins, first round, translation
High tryptophan consentration - ANSWER Causes intrinsic termination by stem loop
between 3-4
Low tryptophan concentration - ANSWER Causes stem loop between 2-3, transcription
will continue
Negative regulation - ANSWER Inhibits gene expression
Catabolite repression - ANSWER overall effect of glucose is to prevent lac gene
expression
Positive regulation - ANSWER Stimulates gene expression
