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Mimg C185A Final Exam Full Solution Study Guide

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Mimg C185A Final Exam Full Solution Study Guide

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MIMG C185A FINAL EXAM FULL SOLUTION
STUDY GUIDE

◉ promotor, filter-binding assay.
Answer: Experiments to determine that *sigma factor* allows for
tight binding to the ________________:
1. Experiment where core vs. holoenzyme was allowed to bind to
radiolabelled DNA, and measured rate at which each jumped onto
nonradioactive DNA
2. Core jumped off radioactive DNA at a much higher rate
3. Doing a ____________________________ saw consistent counts with
holoenzyme, but inconsistent counts with the core


◉ RpoS.
Answer: A sigma subunit that drives the transcription of a lot of
stationary phase genes; presence of these is absent when cells are in
the growing phase.


◉ 28, FlgM, FlgM.
Answer: 1. Sigma _____ drives transcription of late flagellar proteins,
like flagellin and the cap.
2. _______ binds to this when the early flagellar proteins are being
assembled so sigma factor can't go to promoter

,3. When majority of flagella is made, _________ is pumped out and
freed from sigma factor so cap and flagellin can be made


◉ enzymatic.
Answer: Fastest cut off point for regulation of activity is at the
__________________ level (prevent conversion of substrate into product),
which will allow the blocking of regulatory process regardless of
how many proteins and mRNA are already present.


◉ inverted repeats.
Answer: 1. Transcription factors are dimers that bind to
___________________ in the major grooves and along phosphate
backbones.
2. Helix-turn-helix: contains stabilizing helix that dimerizes and
recognition site that binds DNA
3. Leucine finger
4. Zinc finger


◉ trp.
Answer: 1. In absence of __________, repressor is inactive, so operon is
active
2. _________ is a corepressor that binds to inactive repressor, and
makes go conformational change that makes repressor active
3. repressor binds to operator and blocks __________ mRNA synthesis

,4. negative feedback loop


(only one word fits all three spaces)


◉ diauxic growth.
Answer: 1. cells use best carbon source first (glucose)
2. growth on lactose is slower
3. Second level of regulation of the lac operon that activates lac
operon during *high* levels of *cAMP* (*low glucose* levels)


◉ maltose regulon.
Answer: 1. Activator = Mal Ti
2. Maltose binds to MalTi and induces conformation change to active
MalTa (Activator)
3. MalTa binds to promoters and recruits RNA polymerase for
synthesis of mal genes (P, E, K S) (maltose metabolyzing enzymes)
4. MalTa-Polymerase complex binds to four different operons


◉ Attenuation.
Answer: _______________________ of the trp operon:
1. *High Trp*: Leader sequence encodes a short *leader peptide*
with 2 Trp residues is made —> ribosome cruises through
translation —> Transcription *terminates at the 3-4 hairpin*

, 3. *Low Trp* —> ribosome *stalls* because lack of leader peptide —
> terminator sequence is not there because of *2-3 base pairing*
(there is a turn instead) —> transcription of trp structural genes
occurs


◉ two component regulatory systems.
Answer: consist of:
1. sensor kinase that autophosphorylates in response to a signal and
transfer of a phosphate to a response regulator
2. response regulator will bind DNA to initiate transcription of
certain genes or regulate activity of other proteins


◉ osmoregulation, EnvZ, OmpR.
Answer: 1. ___________________ is an example of a two component
system
2. the sensor kinase is ___________
3. response regulator is __________
4. At *low* osmolarity, the system will be *inactivated* because of
*high OmpF* levels (response regulator drives production of more
OmpF)
5. At *high* osmolarity, the system is *activated* because of *high
OmpC* levels (response regulator blocks OmpF production but
drives expression of OmpC)

Información del documento

Subido en
22 de octubre de 2026
Número de páginas
50
Escrito en
2026/2027
Tipo
Examen
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