FRNSC 420 - EXAM 4 PRACTICE QUESTIONS AND ANSWERS (100% PASS)
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Course
FRNSC421W /FRNSC 420
Institution
FRNSC421W /FRNSC 420
FRNSC 420 - EXAM 4 PRACTICE QUESTIONS AND
ANSWERS (100% PASS)
positive allosteric regulation - Answer️️ -binding of a regulatory molecule to an
enzyme induces a conformation change, increasing the enzyme's affinity for a
substrate
- proteins are activators = enhance RNAP binding and/or ind...
FRNSC 420 - EXAM 4 PRACTICE QUESTIONS AND
ANSWERS (100% PASS)
positive allosteric regulation - Answer✔️✔️-binding of a regulatory molecule to an
enzyme induces a conformation change, increasing the enzyme's affinity for a
substrate
- proteins are activators = enhance RNAP binding and/or induce conformational
change, can change binding constant (KB) or rate constant (kf), lowers activation
energies
negative allosteric regulation - Answer✔️✔️-bound regulatory molecule must
release to induce the conformation change necessary for substrate to bind an
enzyme
- proteins are repressors = block RNAP binding or escape
allosteric regulation of Cdk - Answer✔️✔️-replication initiation in eukaryotes
- Cdk is inactive when not bound to cyclin, with helices blocking its active site;
when Cdk is inactive, helicases are readily loaded onto DNA
- cyclin binds to Cdk to induce a conformational change that exposes its active site
- active site is phosphorylated by kinase CAK; when activated, Cdk will
phosphorylate Cdc6/Cdt1 in the eukaryotic PIC
- phosphorylation activates loaded helicases and prevents others from loading,
allowing for the unwinding of DNA and subsequent initiation of replication
effects of binding constants on promoters - Answer✔️✔️-promoter/RNAP
interaction strength can be determined by the amount of the holoenzyme needed to
occupy ½ of the promoter
- KB = [RP]/[R][P]
- equilibrium in bound & unbound RNAP is [RP]/[R][P] = 1
- from that, you get KB = 1/[R], shows KB when ½ of the promoter is occupied
- when KB is large, less holoenzyme is needed to bind ½ of the promoter =
stronger RNAP/promoter interaction
- when KB is smaller, more holoenzyme needed = weaker
- also dependent on rate constant k, where larger k = quicker conversion
allosteric regulation of glnA - Answer✔️✔️-RNAP is pre-bound to glnA (glutamine
synthetase gene, involved nitrogen metabolism) in the closed complex
- RNAP has a σ⁵⁴ factor = unable to unwind DNA, must be activated
- activator is NtrC, binds 4 sites at ~150 bps upstream from promoter
- IHF contains pseudo-loops which bind DNA and bend it 180⁰, facilitating
interaction between NtrC and the σ⁵⁴ factor
- NtrC utilizes ATPase activity; energy from ATP hydrolysis induces a
conformational change in RNAP to 'open' it
- process is regulated by environmental conditions in the cell; NtrC can only bind
to DNA in low nitrogen levels
- NtrC is phosphorylated by kinase NtrB, which induces a conformational change
in NtrC to expose its DNA-binding domain
- only then it can bind DNA and activate transcription
promoter spacing in merT - Answer✔️✔️-protein MerR controls the expression of
the merT gene, expression is activated in the presence of mercury (Hg)
- MerR binds sequence between -35 and -10 regions of the merT promoter opposite
to RNAP (contains σ⁷⁰) for simultaneous binding
- spacing between the -35 and -10 elements is ~19 bps, much larger than the usual
15-17 bps; elements are unaligned on DNA, facing opposite sides of the helix
- without Hg, MerR binds to lock this conformation in position; structural
distortion prevents RNAP from transcribing the gene
- when bound to Hg, MerR undergoes a conformational change that allows for
DNA to twist at the center of the promoter
- new configuration allows for the -35 and -10 elements to align with one another
catabolite repression - Answer✔️✔️-in E. coli, presence of glucose causes the down
regulation of operons whose products participate in the metabolism of other carbon
sources
- glucose is preferred carbon source in E. coli
- when lactose is absent, the lac repressor will remain bound to the lac operator,
only dissociating when lactose is present = basal transcription
- through induction (activation), transcription of the lac operon is further activated
in the presence of lactose when glucose is absent entirely
operons - Answer✔️✔️-cluster of coordinated, regulated genes; contain:
- structural genes that encode enzymes
- regulatory genes that encode repressors or activators
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