2024 ASU BIO 340 {COMBINED EXAMS} LATEST QUESTIONS AND
ANSWERS 100% VERIFIED
What will be the effect on transcription when an inhibitor is in play?
A. Allow transcription B. Prevent transcription - CORRECT ANSWER B Explanation:
An inhibitor prevents activator from binding to DNA; activators are used in
positive regulation (increasing transcription) - thus, the inhibitor prevents the
activator from positive regulating (increasing) transcription.
What will be the effect on transcription when an inducer is in play?
A. Allow transcription B. Prevent transcription - CORRECT ANSWER A Explanation:
When an inducer is in play, it will allow transcription.
For a gene under positive control, which of the following would prevent
transcription?
A. Repressor B. Corepressor C. Inducer D. Effector E. Inhibitor - CORRECT ANSWER
E Explanation: Positive regulation includes activators (affected by effectors and
inhibitors); inhibitors prevent activator from binding to DNA, thus preventing
transcription.
Meiosis - CORRECT ANSWER Cell division that produces reproductive cells in
sexually reproducing organisms
Recombination (crossing over) - CORRECT ANSWER the exchange of genetic
material between homologous chromosomes during meiosis
Meiosis 1: Prophase 1 - CORRECT ANSWER Homologous chromosomes condense,
pair up, and swap segments
Spindle microtubules attach to chromosomes as the nuclear envelope breaks up
chiasmata - CORRECT ANSWER X-shaped regions where crossing over occurred.
crossing over - CORRECT ANSWER the exchange of genes between homologous
chromosomes, resulting in a mixture of parental characteristics in offspring.
Meiosis 1: Prometaphase 1 - CORRECT ANSWER the attachment of the spindle fiber
microtubules to the kinetochore proteins at the centromeres.
,Meiosis 1: Metaphase 1 - CORRECT ANSWER homologous chromosomes are
arranged in the center of the cell with the kinetochores facing opposite poles.
The homologous pairs orient themselves randomly at the equator.
Meiosis 1: Anaphase 1 - CORRECT ANSWER the microtubules pull the linked
chromosomes apart. The sister chromatids remain tightly bound together at the
centromere. The chiasmata are broken in anaphase I as the microtubules
attached to the fused kinetochores pull the homologous chromosomes apart.
Meiosis 1: Telophase 1 and Cytokinesis - CORRECT ANSWER nuclear membranes
form, the cell separates into 2 cells
Meiosis II - CORRECT ANSWER the second phase of meiosis consisting of
chromatids separating, along with the two diploid cells splitting in two
Meiosis produces ________ daughter cells. - CORRECT ANSWER four haploid
If a muscle cell of a typical organism has 32 chromosomes, how many
chromosomes will be in a gamete of that same organism? - CORRECT ANSWER 16
For a gene under negative control, which of the following would you expect to
find bound to the repressor if there is active transcription? A. Repressor B.
Corepressor C. Inducer D. Effector E. Inhibitor - CORRECT ANSWER C Explanation:
In negative regulation, there are repressors (affected by correpressors and
inducers). If there were a corepressor, the repressor would successfully bind to
the operator and prevent transcription. Inducers, however, prevent the binding of
the repressor - thus causing a prevention of it doing its job, and allowing
transcription.
Which molecule serves as the inducer for the lac operon?
A. cAMP B. Glucose C. Lactose D. Allolactose - CORRECT ANSWER D Explanation:
Allolactose is the molecule that binds to repressor protein, making it unable to
bind to the operator.
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 - CORRECT 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 - CORRECT 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 - CORRECT 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 - CORRECT 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 - CORRECT 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 - CORRECT 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 - CORRECT 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 -
CORRECT 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 - CORRECT
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 - CORRECT ANSWER D.
Chromatin remodeling complex
Which regulation of gene expression mechanism is not found in eukaryotes?
