Download Exam 4 Review Answers - Iowa State University

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Exam 4 Review
Supplemental Instruction
Iowa State University
Matt C.
Biol/Gen 313
Dr. Myers & Dr. Vollbrecht
04/10/2017
Introduction: Chapters 12, 13, and 14 are covered on this exam.
Multiple Choice
1. After an interrupted mating experiment, you observe the following data
reported as time needed to rescue 50% of the auxotrophic colonies. What
is the order of the genes on the chromosome?
Gene A
Hfr Strain 1 10 min
Hfr Strain 2 4 min
Hfr Strain 3 7 min
a. B, A, C, D, E
b. B, D, E, A, C
c. C, A, B, D, E
Gene B
1 min
2 min
9 min
Gene C
7 min
7 min
4 min
Gene D
3 min
11 min
1 min
Gene E
4 min
10 min
2 min
d. C, E, D, A, B
e. None of the above
To do these questions, look at which genes end up being co-conjugated closest
in time. For example: using Strain 1, we can see that gene E is bounded on
either side by D and C (3 min to 4 min to 7 min) and gene C is bounded by E
and A for similar reasons. You can use the second strain to fill in the gap
between A and B for sure. Additionally, note that the direction doesn’t matter.
This means that C, A, B, D, E is just as correct as C, E, D, B, A since this is just a
circle we’re talking about and we can go in either direction.
2. You discover a new operon in charge of modifying the sugar xylose. When
xylose enters the cell, transcription of this xyl operon begins. A regulator
protein involved in this operon becomes bound to the operator after
binding to xylose. What is the method of regulation described here?
a. Positive inducible
d. Negative repressible
b. Positive repressible
e. Constitutively regulated
c. Negative inducible
The operon begins transcribing when the signal molecule xylose enters the cell
so it is inducible. Additionally, the operon begins transcribing when the
regulator protein binds, so it’s positively regulated.
3. Some genes are under control of the galactose-mediated regulation system
in yeast. In this system, what is the function of Gal80?
a. Gal80 is an activator protein that promotes RNA polymerase binding.
b. Gal80 is an inhibitor that prevents the action of an activator.
c. Gal80 acts to free a separate activator from inhibition.
d. Gal80 is a transcriptional mediator and member of the TFIID
complex.
e. Multiple of the above are correct.
Gal4 is the activator, Gal80 inhibits Gal4, and Gal3 can remove Gal80 in an
inducible fashion. That is what you have to remember about this system.
4. In Arabidopsis thaliana, sequence R is involved in the activation of the
localized sequences K and H. Sequence U exhibits identical response
elements to K and H, but is acted on by the intervening effect of sequence T
and is instead only activated by another sequence R that is further
downstream. What is sequence T an example of?
a. Regulatory gene for an activator
b. Regulatory gene for a repressor
c. Enhancer
d. Insulator
e. Response element
In this question, you should recognize that T is somehow repressing sequence
U which is some form of a gene or response element for a gene. You can’t
really “activate” much beyond that. Since we’re talking about a eukaryote,
then we’re either dealing with a repressor or an insulator. The proof that it’s
an insulator and not a repressor gene or repressor-binding operator is that,
first, the effect of T isn’t diffusible. It affects U, but can’t affect K or H. So, T is
either an insulator or a repressor operator. However, since the downstream R
can reach back and activate U while the upstream R can’t, it’s not a repressor
operator, but an insulator. Note that this was the most involved way I could
word a question identifying an insulator or repressor, so if you followed this
one, you should feel quite confident on this topic.
5. While researching reoccurring gene sequences in Caenorhabditis elegans,
you note that in one C. elegans a sequence surrounded by inverted repeats
is located at 3 different positions in coding regions of the organism, while in
a different sample, the reoccurring element is located in 5 different coding
regions. What is the likely explanation for this?
a. The sequence is the result of duplication.
b. The sequence is the result of non-reciprocal translocation.
c. The sequence is transposable.
d. The sequence is a trinucleotide repeat that develops additional
copies due to slippage.
e. None of the above are good explanations.
Specifically, this is a replicative transposon – the copy & paste form. Since
they’re different coding regions, it’s not the result of trinucleotide repeats and
gene duplication or non-reciprocal translocation occur well too infrequently
for those to be reasonable.
6. What is the purpose of annealing when performing PCR?
a. The lower temperature allows Taq polymerase to renature after the
hotter melting temperature.
b. The lower temperature allows DNA primers to bind to template
strands.
c. The lower temperature allows Taq polymerase to bind to the primertemplate heterodimer.
d. The lower temperature allows dNTPs to more efficiently enter the
polymerase active site and subsequently polymerize.
e. Multiple of the above.
Annealing temperatures are actually kept rather high so that the DNA
molecules are still rather fluid and the primers have every opportunity to
outcompete the larger DNA strand to anneal at their sequence
complement. Taq won’t bind well at these temperatures, but also doesn’t
go through denaturation and renaturation cycles.
7. You observe that a certain transgenic knock-out strain in maize produces an
embryo lethal phenotype. You perform a second transgenic insertion to
reintroduce a wild-type version of the knocked-out gene, but still observe
the lethal phenotype. What most likely explains what has happened,
assuming the second insertion was successful?
a. Complementation has showed that the first transgenic plant’s
phenotype is the result of the location of the first insert, not of the
knock-out.
b. Introduction of the wild-type gene has shown that the genotype
leading to the lethal phenotype is actual a dominant trait.
c. Instead of being a result of the knock-out, chromosomal
rearrangement has occurred in the first plant line that prevents the
formation of gametes.
d. The addition of the new gene has resulted in an enzyme dosage
problem that is also lethal.
e. None of the above are good explanations.
This would be a terribly annoying result experimentally, but it is the best
supported by the data given. This is why reported knock-out strains are
checked via transgenic complementation. I chose this example as it’s a realworld application of complementation. The practice questions and exams
cover more theoretical examples thoroughly.
8. A series of mutations occurs in a strain of E. coli that you’re studying that
results in dramatically overexpressed and super-activated cAMP
phosphodiesterases. These enzymes break down any cAMP that the cell
can manage to make. What is the likely effect of this on the lac operon?
a. The lac operon will be constitutively expressed.
b. The lac operon will produce about half of the usual transcript count.
c. The lac operon will not be transcribed.
d. The lac operon will be unaffected.
e. We can’t tell the effect on the lac operon from this information.
Getting rid of cAMP prevents CAP from forming the CAP-cAMP complex
needed to activate the lac operon. Without CAP-cAMP, the lac operon won’t
be transcribed.
9. Exposure to UV light can damage the DNA. How does it do this?
a. UV light causes spontaneous de-acetylation of histones.
b. UV light causes cytosine methylation.
c. UV light causes a single-stranded break in the DNA.
d. UV light causes adjacent thymines to dimerize.
e. Multiple of the above are methods of action of UV light damage.
Thymine dimers are formed by UV light.
10. You are performing a set of experiments to determine amino acid residues
of structural importance in GAPDH2. You perform a single amino acid
mutation converting a tyrosine to a phenylalanine and perform an enzyme
assay of the mutant enzyme. You note that enzymatic activity matches the
wild-type. What sort of mutation has occurred?
a. Missense
d. Neutral
b. Silent
e. Gain-of-function.
c. Loss-of-function
Unintentionally, both of these are correct answers here. The mutation is
missense since there’s an amino acid change, but it’s also a neutral mutation
since the enzymatic activity isn’t affected.
11. Which of the following methods of expression regulation results in the
shortest time between initiating regulation and observing a change in
expression?
a. Protein modification
b. Histone deacetylation
c. RNA degradation mediated by miRNA
d. Translational control
e. Transcriptional regulation
Modifying a protein has an almost immediate effect on expression since
proteins are usually the actors for any given phenotype. The other regulatory
mechanisms take longer to reach apparent expression, but tend to have a
larger consequence per regulatory event (initiating transcription off of one
gene will have more downstream effects than phosphorylating one protein).
12. While trying to determine the cause of the infection of a human patient,
you isolate reverse transcriptase from infected cells. What is the likely
origin of this enzyme?
a. A bacterial contaminant
b. Part of the phage ghost
c. A dsDNA virus
d. A retrovirus
e. Contamination of human enzymes
Retroviruses (RNA viruses) use reverse transcriptase to convert their RNA
genome to DNA. It’s a reverse transcriptase because you’re moving backwards
from RNA to DNA.
13. What is the function of lacI?
a. It is the response element that is acted on to repress the lac operon.
b. It is a molecule that binds near the lac promoter to physically
impede RNA polymerase.
c. It cleaves lactose into two monosaccharides that can then be
metabolized.
d. It binds to cAMP and represses the lac operon.
e. None of the above.
LacI is the repressor protein, but it doesn’t bind cAMP; it binds allolactose. It’s
repressor activity takes the form of binding to the operator, which is a region
overlapping the end of the lac promoter and beginning of the lac structural
genes. When lacI is bound, it gets in the way of RNA polymerase binding to the
promoter.
14. An intragenic suppressor mutation occurs in gene V that had previously
been affected by a frameshift resulting from a base addition. What could
this mutation be?
a. A base deletion just upstream from the added base.
b. A base addition just downstream from the previously added base.
c. A mutation in a gene Y that covers up the gene V mutation.
d. A crossing over event that pulls away the mutated base.
e. None of the above.
An intragenic suppressor mutation is a suppressor mutation that occurs in the
same gene as the initial mutation. This means that c. is out, though; c. is an
example of intergenic suppression if you want an example of that. As for a
mutation that will undo a base addition frameshift, you can either delete a
base or add two bases. This will get us back to the old reading frame.
15. Which of the below is a cis-acting mutation?
a. LacP+
b. LacIc. LacOC
d. All of the above.
e. Only a. and c.
LacP+ isn’t a mutation, so it’s out. LacI- does produce a polypeptide that
diffuses, but it doesn’t do much of anything. Regardless, we can consider it
trans-acting. LacOC only can act on the operon in its chromosome, so it is cisacting.
16. Which of the following is/are heritable?
a. A silent mutation in the gametophyte.
b. A nonsense mutation in a somatic cell.
c. Increased cytosine methylation.
d. All of the above.
e. Just two of the above.
Answers a. and c. are correct. Cytosine methylation is one of our go-to
examples of an epigenetic change and epigenetic changes are heritable. For a.
and b., we want the mutation to happen in the germ line to be heritable.
Gametophytes make up the germline, so the mutation in a. is heritable, even
though it’s only a silent mutation. Mutations in somatic cells aren’t heritable.
17. A single type of signal molecule enters the nucleus of a eukaryotic cell and
induces transcription in multiple genes. How does this most likely happen?
a. The genes are linked on an operon.
b. Signal integration occurred and expanded the downstream signal.
c. A first gene was activated by the signal directly and was transcribed,
translated, and the resulting product activated the other genes.
d. The activated genes share response elements.
e. None of the above.
In eukaryotes, sharing response elements is analogous to lying on the same
operon in prokaryotes. Sharing response elements puts different genes under
the same regulatory control. In this example, one signal enters and induces
several genes to begin transcribing. To have this effect, those genes must all
have the response element that recognizes that signal. Entering the nucleus,
inducing transcription of an activator gene, processing that transcript,
synthesizing the protein, and shipping the regulator back into the nucleus to
begin affecting transcription is too costly and time-consuming to be likely.
Signal integration wouldn’t accomplish this process.
18. High levels of tryptophan in the cell induce the trp operon regulator to
bind and inhibit transcription. What is the term used to describe
tryptophan in this instance?
a. Attenuator
d. Inducer
b. Activator
e. Co-repressor
c. Repressor
Co-repressor is the name given to signal molecules involved in a repressible
system. Repressors are different; they’re the regulator protein involved in a
negatively regulated system.
19. Which of these structures is a DNA-binding motif in proteins?
a. Beta barrels
b. Coiled coil
c. Zinc fingers
d. Greek key
e. None of the above
Zing fingers, leucine zippers, and helix-turn-helix motifs will bind DNA
sequences. Beta barrels, coiled coils, and Greek keys are unassociated tertiary
structure motifs.
20. Given the following monoploid operons, which would not synthesize
lactose permease under any circumstance?
a. LacI+, LacP+, LacO+, LacZ+, LacY+
b. LacI-, LacP+, LacO+, LacZ-, LacY+
c. LacI+, LacP-, LacOC, LacZ+, LacY+
d. LacIS, LacP+, LacOC, LacZ+, LacY+
e. None of the above
Without a functioning promoter, no polymerase will ever bind and no
transcript will ever be made. The super-repressor in d. cannot repress the
constitutive operator since it doesn’t recognize the changed operator
sequence, just like how wild-type repressor wouldn’t recognize it.
21. Which of the following are epigenetic changes that would affect regulation
of gene expression?
a. Histone acetylation
b. Moving histone complexes along the DNA
c. Cytosine methylation
d. All of the above
e. Just two of the above
Histone (de)acetylation and cytosine (de)methylation are our primary
examples of epigenetic changes that affect regulation, but moving the histones
along the DNA sequence will also have an effect on regulation. There are
questions either the practice exams or practice questions that reference this.
22. Which of the following repeats is most likely to expand during DNA
replication?
a. CAG
d. TAA
b. TAG
e. Multiple of these
c. CGG
Trinucleotide repeat expansion occurs due to DNA slippage. Slippage occurs
when parts of the DNA repeat spontaneously melt a bit and a repeat sequence
towards the front moves back and pairs with an earlier, complementary
sequence. Melting occurs more readily with fewer hydrogen bonds, so T and A
are preferred.
23. You observe the following part of a chromosome with inverted
transposons. Which of the following is most likely to occur?
a. Inversion of the region between the transposons.
b. Excision of the region from one transposon to the other due to
recombination.
c. A mutation in the inverted repeat region that renders both
transposons immobile.
d. All of the above are possible and equally likely.
e. This scenario could never happen.
This occurs as a result of intrachromosomal crossing over. The transposons line
up in the same direction and crossing over occurs. This flips the orientation of
the sequence between the two transposons.
24. Given the following partial diploid chromosome of the lac operon, what is
the mode of regulation?
LacI-, LacP-, LacOC, LacZ+, LacY+
LacIS, LacP+, LacO+, LacZ+, LacY+
a.
b.
c.
d.
e.
Inducible
Repressible
Constitutive transcription
No transcription
This can’t be determined
The top strand doesn’t have a functional repressor or promoter, so it
doesn’t contribute anything to the system. On the bottom strand, a superrepressor is produced and that bottom strand has a normal operator, so
the super-repressor binds permanently and transcription doesn’t occur.