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Exam 1 Review
Supplemental Instruction
Iowa State University
Matt C.
Biol/Gen 313
Dr. Myers & Dr. Vollbrecht
01/29/2017
Introduction: The chapters to be covered by this exam are 1, 8, 9, and 10. This review will
largely focus on material from the later 3 chapters, but there are several example questions from
Chapter 1 on Blackboard.
Your exam will be all multiple choice, but there are some short answer questions included on
this worksheet to make sure you really know the material. This should help you study better.
Multiple Choice
1. Arabidopsis thaliana is a commonly-used model organism for C3 plants. It is widely used
for all of the following reasons EXCEPT for that:
a. It grows quickly in the lab.
b. It is inexpensive to grow.
c. It produces a specific, limited number of offspring.
d. Its genome is small.
e. All of these are reasons why A. thaliana is a model system.
2. Heterozygous inheritance of one copy of the sickle-cell hemoglobin gene helps to confer
greater resistance to malaria. Research groups interested in this phenomenon track the
number of heterozygous individuals in different parts of Africa as related to malaria
infection rates. What type of genetic research would this be considered?
a. Evolutionary genetics.
b. Population genetics.
c. Molecular genetics.
d. Transmission genetics.
e. None of the above answers.
3. Below is the heterocyclic base adenine. Which of the following heterocyclic bases will
hydrogen bond with adenine in double-stranded DNA?
a.
Adenine
b.
c.
d.
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4. Griffith’s experiment in 1928 was used to identify the chemical nature of the
“transforming principle” that could pass one bacterial strain’s virulence phenotype onto a
non-virulent strain. This transforming principle was DNA and Griffith’s experiment
helped prove DNA’s function as genetic material. To come to this conclusion, Griffith
heat-killed three virulent bacterial samples and treated each sample with one of protease,
RNase, or DNase. He then tested to see which samples could still pass on virulence. What
would his results have looked like if protein was the transforming principle?
a. Only the samples treated with DNase and RNase would cause
transformation.
b. Only the samples treated with protease and DNase would cause transformation.
c. Only the samples treated with protease and RNase would cause transformation.
d. Only samples treated with protease would cause transformation.
e. None of the above would be the case.
5. You come across an unknown nucleic acid and measure the concentration of uracil to be
27% of the total nitrogenous bases and the concentration of cytosine to be 24%. What is
the most likely conclusion you can come to about the nucleic acid?
a.
b.
c.
d.
e.
The molecule is DNA with [A]=27% and [G]=24%.
The molecule is RNA with [A]=27% and [G]=24%.
The molecule is single-stranded DNA.
The molecule is single-stranded RNA.
You do not have enough information.
6. Which of the following single-stranded RNA sequences will form a hairpin structure?
a. -AGAGACCCCCCCAGAGAb. -AGAGACCCCCCCTCTCTc. -AAAAAGGGGGGAAAAAd. -AGAGAGGGGGGCTCTCe. RNA doesn’t form secondary structures like hairpins.
First note, the Ts should be replaced by Us because it is RNA. That was my mistake. Second,
while the sequence in B will match up better, the sequence in D could indeed form a small
hairpin.
7. Which of the following is known to be TRUE about centromeres?
a. Centromere sequences help prevent chromosome shortening in eukaryotes.
b. Centromere sequences are added just prior to metaphase I so that spindle
microtubules may attach to the chromosome.
c. Centromeres have a specific consensus sequence that is identified during cellular
replication and to which the kinetochore binds.
d. Centromeres are made up of primarily euchromatin.
e. None of the above are true.
8. A mutation knocks out expression of DNA gyrase in E. coli. What is the likely effect this
would have?
a. DNA replication would be strongly inhibited or stopped due to supercoiling.
b. RNA transcription would be strongly inhibited or stopped due to supercoiling.
c. DNA replication would be strongly inhibited or stopped due to the lack of ability
to split hydrogen bonding between complementary bases.
d. RNA transcription would be strongly inhibited or stopped due to the lack of
ability to split hydrogen bonding between complementary bases.
e. None of the above would happen.
9. A certain molecule of chromatin is found to contain 10 molecules of histone H1. How
many molecules of histone H4 would you expect there to be in this sample?
a. 5
b. 10
c. 15
d. 20
e. None of the above would be expected.
Match the given terms with their representation marked below. Use each term only once.
A.
B.
C.
D.
10. Leading strand. B
11. Parent strand. D. A also works, but A is used later.
12. Okazaki fragment. E
13. 3’-end. A
14. Discontinuous replication. C
E.
15. Telomerase activity helps solve which problem?
a. Telomere disassociation.
b. Base mismatch during DNA synthesis.
c. Eukaryotic end-replication issues.
d. Binding of the spindle apparatus during cell division.
e. None of the above.
16. A eukaryotic organism doesn’t add licensing factors during the G1 phase, but skips over
the G1 checkpoint to begin DNA replication in S phase. What is likely to occur?
a. The genome will be over-replicated.
b. No DNA replication will occur.
c. DNA replication will be completed, but with many more mutations than usual.
d. Chromosomes will shorten following replication.
e. None of the above.
17. What is the primary function of sigma factors?
a. To identify promoter regions in bacteria and assist in transcription initiation.
b. To improve transcription elongation speed.
c. To provide a proof-reading mechanism to the transcription holoenzyme.
d. To put together the other components of the RNA polymerase complex.
e. None of the above.
18. Poly-adenylation accomplishes what?
a. Lengthens telomere sequences.
b. Creates a tail on the 3’ end of mRNAs.
c. Protects the 5’ end of DNA from degradation.
d. Modifies histone complexes.
e. Multiple of the above are true.
19. Below is an image of a DNA and mature mRNA hybrid. How many introns are contained
in the DNA gene sequence?
a.
b.
c.
d.
e.
2
3
4
5
This cannot be determined.
20. Several mutations occur in the terminator region of a bacterial gene following the
inverted repeats that switches out A bases for T bases. What is the likely effect of this
change?
a. If the gene is rho-dependent, it won’t be transcribed.
b. If the gene is rho-independent, it won’t be transcribed.
c. The gene won’t be transcribed regardless of its rho-dependency.
d. The gene will be transcribed, but it will continue past the terminator until it
reaches the next terminator.
e. Gene transcription will not be affected significantly.
21. Which of the following positions has a partial positive charge?
a. A.
A.
b. B.
B.
c. C.
C.
d. D.
e. None of the above.
D.
22. What is the approximate size of the human diploid genome in base pairs?
a.
b.
c.
d.
e.
46
200,000
6,000,000
3,000,000,000
None of the above.
23. Which of the following is not an activity of DNA polymerase I?
a. 5’-3’ DNA polymerase.
b. 3’-5’ DNA polymerase.
c. 5’-3’ exonuclease.
d. 3’-5’ exonuclease.
e. All of the above are activities of DNA polymerase I.
24. All of the following are requirements for transcription EXCEPT:
a. Priming with RNA primase.
b. Sufficient NTPs.
c. A DNA template.
d. The RNA polymerase complex.
e. All of these are requirements for transcription.
25. Studies have shown that the knockout of telomerase in eukaryotic organisms leads to
premature aging and death, however; telomerase isn’t synthesized at all in bacteria. What
is the best explanation for this?
a. Bacteria only have one chromosome.
b. Bacteria have a circular chromosome.
c. Bacteria only have one origin of replication.
d. Bacteria don’t age regardless of telomere length.
e. Bacteria actually do synthesize telomerase to deal with telomere shortening.
26. Which type of RNA listed below is useful due to its catalytic activity.
a. mRNA
b. rRNA
c. tRNA
d. CRISPR RNA
e. None of the above are catalytic molecules.
27. A research group investigates the mechanism that produces a recessive phenotype in C.
elegans. What type of genetic research is this an example of?
a. Evolutionary genetics.
b. Population genetics.
c. Molecular genetics.
d. Transmission genetics.
e. None of the above answers.
28. Edwin Chargaff proposed the pattern for base pairing using his experimental results on
the concentrations of each nitrogenous base in a double-stranded DNA sample. If base
pairing instead occurred between A and C or G and T, what would his results be?
a. [A]=[G]
b. [C]=[G]
c. [A]+[T]=[C]+[G]
d. [T]=[G]
e. None of the above.
Another one I screwed up. Both C. and D. are 100% right.
29. Below is a ribose sugar with numerically-labelled carbon atoms. To which carbon would
the phosphate be attached in a singular nucleotide?
5.
4.
a.
b.
c.
d.
e.
1.
3.
2.
Carbon 1
Carbon 2
Carbon 3
Carbon 4
Carbon 5
30. Which of the following is an epigenetic change?
a. Methylation of cytosine bases to repress expression.
b. Acetylation of positively-charged sidechains on histone proteins to increase gene
expression.
c. Repressor proteins bind permanently to a gene promoter, preventing its
expression.
d. A prion infection that passes between dividing cells.
e. All of these are possible examples of heritable epigenetic changes.
31. Which of the following heterocyclic bases is cytosine?
a.
c.
b.
d.
e. None of the above.
32. Which of these proteins has to act first during DNA replication at one origin?
a. DNA primase.
b. DNA helicase.
c. DNA polymerase III.
d. DNA ligase.
e. Multiple of the above act at the same time at one origin.
33. Why does acetylation weaken histone binding to DNA strands?
a. Acetylation removes positive charges that interact with DNA.
b. Acetylation removes negative charges that interact with DNA.
c. Acetylation changes the preferred interactions between histone proteins.
d. Acetylation allows the histone H1 protein to clamp onto the nucleosome complex.
e. None of the above.
34. DNA polymerases are quite accurate with just their 5’-3’ polymerase activity, but also
have a proofreading mechanism to improve that accuracy further. What is the name given
to this mechanism?
a. 3’-5’ polymerase activity.
b. 2’-hydroxyl phosphodiester bond cleavage.
c. 5’-3’ endonuclease activity.
d. 3’-5’ endonuclease activity.
e. None of these are involved in proofreading.
35. For a certain gene, the sequence of the non-template strand is -ATTGACCTG-. What is
the mRNA sequence without processing?
a. -ATTGACCTGb. -TAACTGGACc. -AUUGACCUGd. -UAACUGGACe. You cannot know from this information.
The non-template strand is also known as the coding strand. Its sequence is the same as the
mRNA sequence from the template strand, but with switched U and T bases.
Match the given terms with their representation marked below. You can reuse terms.
B.
A.
D.
F.
G.
C.
E.
36. Template strand. C.
37. Coding strand. B.
38. Promoter. A.
39. Terminator. E.
40. 5’- end. G. Or B. if you want to make that argument.
D. is the RNA coding region. It’s possible that you’ll be provided with a diagram with even less
information on the test, so be prepared for that. The other practice exams have examples of that.
41. An RNA that normally associates with telomerase has sequence UAUCCCCUAUCCCC.
A mutation occurs that converts the sequence to UAUCCCCUAACCCC. What will
likely happen as a result of the mutation?
a. Proteins that normally associate with the telomere after synthesis won’t be able to
recognize the new sequence.
b. Telomerase will not be able to reattach after one round of synthesis.
c. Telomerase won’t be able to bind at all. No new synthesis occurs.
d. Telomerase will target the wrong DNA strand.
e. None of these will happen. Telomerase activity would proceed as normal.
Telomerase works because its RNA sequence is just a series of repeats leading to that G-rich
telomere sequence. It has this repeat so that it can attach to the gap left by the lagging strand and
synthesize new DNA onto the old, parent strand. This new DNA on the old strand will have the
telomere repeat sequence, so telomerase can just re-associate onto the end of the strand to
continue synthesizing new DNA onto the parent strand. Eventually, the extension to the parent
strand will be long enough that the rest of the DNA replication machinery may return and
synthesize a normal lagging strand Okazaki fragment.
In this case, the telomerase repeat has been ruined in the mutation, so it won’t reattach. However,
it will still be able to bind initially and add the one pattern more of DNA.
42. A certain gene sequence in the DNA template has three exon regions with two introns in
between. Alternative splicing occurs and connects exon 1 to exon 3. What happens to
exon 2?
a. Exon 2 will form its own mRNA and be transcribed.
b. Exon 2 will attach to the end of exon 3.
c. Exon 2 will be degraded with the two introns.
d. Exon 2 will be degraded separately from the two introns since it doesn’t have a
branch point site.
e. None of the above since this splicing can’t occur.
The lariat structure formed during splicing marks introns for degradation. Since exon 2 will be
contained in this lariat, it will be degraded along with the other two introns.
43. What is wrong with the picture below describing the formation of a new phosphodiester
bond?
a.
b.
c.
d.
The nucleophilic attack is occurring in the wrong direction.
The phosphodiester bond is being formed on the wrong carbon.
Water is not generated in this reaction.
RNA doesn’t polymerize with a condensation reaction.
e. None of the above. The diagram is accurate.
This diagram is saying that the oxygen on the phosphate attacks the 3’ carbon and displaces the
hydroxyl group there. This isn’t how the chemistry actually occurs and Dr. Myers wants you to
know that it is nucleophilic attack by the 3’ hydroxyl onto the phosphorous of the phosphate that
is actually happening.
44. Which of the following positions has tetrahedral (pyramid-like) geometry?
A.
C.
B.
D.
a.
b.
c.
d.
e.
A.
B.
C.
D.
None of the above.
Tetrahedral geometry occurs at sp3 hybridized positions. Position D. with the nitrogen is the only
position marked that lacks a double bond. It will be sp3 hybridized (more or less – O-chem is
complicated).
45. The Hershey-Chase blender experiment determined the chemical nature of the genetic
material of viruses. What would conclusion would you draw if the chemical with
radioactive phosphorus was found in the bacterial pellet?
a. That protein is the viral genetic material.
b. That DNA is the viral genetic material.
c. That both DNA and protein are involved in passing on genetic information.
d. That neither DNA nor protein is involved in passing on genetic information.
e. That the experiment was performed incorrectly.
Radioactive phosphorus labeled the DNA components. Radioactive sulfur labeled the protein
part. Whichever part ends up in the bacterial pellet you conclude to be inside the bacteria itself,
and also the genetic material of the virus. Radioactive phosphorus ending up in the pellet was
what actually occurred in the experiment – quite in opposition to what I usually force you all to
answer.
46. What is accomplished by RNAi?
a. Alternative splicing.
b. Origin licensing during the G1 phase.
c. Inhibition of the expression of a target gene.
d. Catalysis of self-splicing in tRNAs and rRNAs.
e. Targeting proteins for secretion to the Golgi bodies.
47. Which of the following sequences would you expect to disassociate from its complement
at the lowest temperature?
a. -AATGCATb. -ATTGCGTc. -GGCATGCd. -GCCATACe. Not enough information is provided.
Temperature of dissociation is related to the strength of the hydrogen bonds involved. The more
A-T pairs you have, the fewer hydrogen bonds and the lower the temperature of dissociation.