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2024 ASU BIO 340 EXAM 3 LATEST QUESTIONS AND ANSWERS 100% VERIFIED WITH RATIONALES
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2024 ASU BIO 340 EXAM 3 LATEST QUESTIONS AND ANSWERS 100% VERIFIED WITH RATIONALES
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2024 ASU BIO 340 EXAM 3 LATEST 2024-2025 QUESTIONS AND ANSWERS 100%
VERIFIED WITH RATIONALES
Which of the following conditions will result in the greatest levels of
transcription of the lac operon?
A. Lactose absent, glucose absent B. Lactose absent, glucose present
C. Lactose present, glucose present D. Lactose present, glucose absen
- ANSWER-D Explanation: From the perspective of the cell, it is most
efficient to repress expression of the genes allowing for metabolism
of lactose when it's absent, to allow it when present, but only to
prioritize it when glucose (the preferred food source) is absent. In C,
allolactose will also be present, allowing transcription, but positive
regulation (i.e. increase in transcription) only happens as in D, when
glucose is absent.
Which of the following mutations will result in the lowest level of
transcription of the lac operon when lactose is present?
A. An operator mutation that prevents binding of the lac repressor B.
A mutation in the lac repressor that prevents it from binding to the
operator C. A mutation in CAP that prevents it from binding to the
CAP binding site D. A mutation in the lac repressor that prevents it
from binding allolactose - ANSWER-D Explanation: A - if repressor
,can't bind, the gene will be permanently turned on, there will be
transcription. B - exactly the same effect as A. C - "tempting but
wrong"; CAP can't bind, will cause lower but not total prevention of
transcription. D - correct answer; if the inducer is totally ineffective,
then the repressor will permanently bind to operator and totally
prevent transcription.
The regulatory elements in the lac operon act to:
A. Repress transcription when there is no lactose present B. Repress
transcription when there is lactose present C. Activate transcription
when there is no glucose present D. Activate transcription when
there is glucose present E. A and C - ANSWER-E Explanation: The cell
wants to both repress transcription when there is no lactose present
and activate it when glucose is absent (C "slightly less good" answer
than A, but still true). Similar to Clicker Question 6
A bacterium needs to make a certain lipid that it cannot obtain from
its environment. That same lipid is involved in regulating the
expression of the enzymes that make it. Which of the following is the
most likely regulatory function of the lipid?
A. Effector B. Activator C. Inducer D. Co-repressor E. Operator -
ANSWER-D Explanation: If you have that lipid, you would want it to
not make any more. So, if you have the lipid in the cell, you would
want to turn off the transcription, so it would likely be involved in
negative regulation. Effectors and activators are both involved in
positive regulation. An inducer is under negative regulation but its
job is to turn on transcription. A co-repressor will only turn the gene
off if there is a signal the cell can sense. Operator is a piece of DNA.
,A bacterium is unable to transport lactose into the cell to be broken
down. Which gene is likely mutated in this bacterium?
A. lacZ B. lacY C. lacI D. lacP E. lacO - ANSWER-B Explanation: If you
can't transport lactose, then it means permease isn't working. So lacY
is likely mutated since that is the gene that encodes permease.
Under which conditions will E. coli make the least amount of beta-
galactosidase?
A. I+ O+ Z+, glucose absent, lactose present B. Is Oc Z+, glucose
present, lactose absent C. I- O+ Z+, glucose absent, lactose absent D.
Is O+ Z+, glucose absent, lactose present - ANSWER-D Explanation: A
is wild type, so the expression level would be high, B has a super
repressor, meaning is not able to bind allolactose, but since we have
constitutive operators, the gene is going to turn "on". For C, the
repressor can't bind the operator, so the gene will not undergo
negative regulation. For D, the super repressor bound to the wild-
type operator will keep the gene off no matter what, even though
lactose is present.
In terms of lac operon regulation, what happens when an I+, O+, P+,
Z- mutant E. coli is grown in medium containing very low glucose and
high lactose?
A. CAP is bound to the DNA but the lac repressor is not, E. coli grows
B. CAP is bound to the DNA but the lac repressor is not, E. coli does
not grow C. Lac repressor is bound to the DNA but CAP is not, E. coli
grows D. Lac repressor is bound to the DNA but CAP is not, E. coli
does not grow E. Both CAP and lac repressor are bound to DNA, E.
coli does not grow - ANSWER-E Explanation: Three parts, think about
whether CAP is bound, Lac repressor is bound, and does the cell
grow. In this case, the regulation is normal, but there is no β-gal. No
, β-gal means that allolactose can not be made, so the lac repressor is
always bound. There is low glucose, so cAMP is being made and CAP
will bind the DNA (this is independent of the lac operon function).
Since lactose can't be used and there is low glucose, the e.coli will
not grow.
Which of the following tends to loosen the interaction between DNA
and histones? A. Acetylation of histone tails B. Deacetylation of
histone tails - ANSWER-A. Acetylation of histone tails
Which of the following regulatory sequences may be located tens of
thousands of nucleotides away from the genes they regulate?
A. Core promoters B. Enhancers C. Proximal elements D. TATA boxes -
ANSWER-B. Enhancers
Which of the following enzymes can slide a histone octamer to a
different position on a eukaryotic chromosome?
A. Chromatin modification complex B. Histone acetyltransferase C.
Topoisomerase D. Chromatin remodeling complex - ANSWER-D.
Chromatin remodeling complex
Which regulation of gene expression mechanism is not found in
eukaryotes?
A. Different strength promoters B. Activator binding to DNA C.
Repressor binding to operator D. Alternative splicing - ANSWER-C
Explanation: Repressors bind to silencers in eukaryotes (operators
specific to prokaryotes)
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