Cell Biology Exam #3-2014
Name__Key____________
Lab Section________________
Multiple choice. Choose the best answer (2 points each)
1.__B___In his transformation experiments, what did Griffith observe?
A) Mutant mice were resistant to bacterial infections.
B) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some
of the living cells into the pathogenic form.
C) Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the
pathogenic strain nonpathogenic.
D) Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains.
E) Mice infected with a pathogenic strain of bacteria can spread the infection to other mice.
2. ___C__ Which of the following investigators was/were responsible for the following discovery?
Chemicals from heat-killed S cells were purified. The chemicals were tested for the ability to transform
live R cells. The transforming agent was found to be DNA.
A) Frederick Griffith
B) Alfred Hershey and Martha Chase
C) Oswald Avery, Maclyn McCarty, and Colin MacLeod
D) Erwin Chargaff
E) Matthew Meselson and Franklin Stahl
3.__B___ In analyzing the number of different bases in a DNA sample, which result would be consistent
with Chargaff’s rules?
A) A = G
B) A + G = C + T
C) A + T = G + T
D) A = C
E) G = T
4. ___E__What is so important about the year 1953 in the history of molecular biology? A) Griffith
completed his transformation experiments; B) Avery McCloud and McCarty obtained evidence that DNA
was the genetic information; C) Hershey and Chase confirmed that DNA was the genetic information;
D ) Messelson and Stahl confirmed that DNA replication was semiconservati ve; E) Watson and Crick
determined the structure of DNA.
5. __C___ Which of the following researchers did not contribute to the first published paper describing
the correct structure of the DNA molecule? A) James Watson; B) Francis Crick; C) Linus Pauling;
D) Rosalind Franklin; E) Maurice Wilkins.
6.__E___DNA polymerase: A) can only make a new strand of DNA in the 5’3’ direction; B) can correct
its own replication errors; C) catalyzes the formation of phosphodiester bonds; D) is found in all living
organisms; E) all of the above
Use the figure below to answer question #7.
7. ___E__ In the late 1950s, Meselson and Stahl grew bacteria in a medium containing "heavy" nitrogen
(15N) and then transferred them to a medium containing 14N. Which of the results (A-E) in the figure
above would be expected after two rounds of DNA replication in the presence of 14N?
8.___C__ What is the function of DNA polymerase III?
A) to unwind the DNA helix during replication; B) to seal together the broken ends of DNA strands; C) to
add nucleotides to the 3' end of a growing DNA strand; D) to degrade damaged DNA molecules; E) to
rejoin the two DNA strands (one new and one old) after replication
9.___A__ You briefly expose bacteria undergoing DNA replication to radioactively labeled nucleotides.
When you centrifuge the DNA isolated from the bacteria, the DNA separates into two classes. One class
of labeled DNA includes very large molecules (thousands or even millions of nucleotides long), and the
other includes short stretches of DNA (several hundred to a few thousand nucleotides in length). These
two classes of DNA probably represent
A) leading strands and Okazaki fragments.
B) lagging strands and Okazaki fragments.
C) Okazaki fragments and RNA primers.
D) leading strands and RNA primers.
E) RNA primers and mitochondrial DNA.
10.__B___ What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?
A) It synthesizes RNA nucleotides to make a primer; B) It joins Okazaki fragments together; C) It unwinds
the parental double helix; D) It stabilizes the unwound parental DNA.
11.__C___ Which of these enzymes is not involved in the process of DNA replication? A) helicase;
B) ligase; C) aminoacyl tRNA synthetase; D) DNA polymerase; E) all of the above are involved.
12. __D___ What does it mean when we say that DNA is antiparallel? A) It has ribose sugars rather than
deoxyribose sugars; B) it contains the nucleotide U instead of T; C) G/C base pairs have 3 hydrogen
bonds and A/T base pairs have 2 hydrogen bonds; D) one strand runs in the 5’ to 3’ direction and its
complement runs in the 3’ to 5’ direction; E) DNA consist of two complementary strands
13.__D___Excision repair of DNA occurs in response to: A) damage to DNA caused by UV radiation; B)
the formation of T-T dimers; C) the formation of Okasaki fragments D) A and B are correct; E) all of the
above are correct.
14.__E___ Suppose you are provided with an actively dividing culture of E. coli bacteria to which
radioactive thymine has been added. What would happen if a cell replicates once in the presence of this
radioactive base?
A) One of the daughter cells, but not the other, would have radioactive DNA.
B) Neither of the two daughter cells would be radioactive.
C) All four bases of the DNA would be radioactive.
D) Radioactive thymine would pair with nonradioactive guanine.
E) DNA in both daughter cells would be radioactive.
15.____B_Which statement did Watson and Crick not make in the paper they published in Nature
regarding the structure of DNA?
A) DNA was a double helix; B) DNA replicates in a semi-conservative ; C) A pairs with T; D) hydrogen
bonding holds the two strands together; E) manner the strands of DNA are antiparallel.
16.__C___This represents an exception to the Central Dogma: A) the fact that the same amino acid has
more than one codon; B) the fact that the same codon can specify a different amino acid in different
organisms; C)HIV virus and their reverse transcriptase enzyme ; D) the one gene one polypeptide
hypothesis; E) none of the above
17. __A___ What is transcription? A) The manufacture of a strand of RNA complementary to a strand of
DNA; B) The manufacture of a new strand of DNA complementary to an old strand of DNA; C) The
manufacture of two new DNA double helices that are identical to an old DNA double helix; D) The
manufacture of a protein based on information carried by RNA. E) The modification of a strand of RNA
prior to the manufacture of a protein.
18.___C__ Which of the following statements is false?
A) RNA polymerase binds to a promoter associated with the gene it transcribes
B) in eukaryotes, transcription factors are required to work with RNA polymerase
C) in eukaryotes, transcription factors are needed to allow the RNA polymerase to bind to the TATA box
which is part of the termination sequence.
D) transcription requires the ribonucleotides A,G,C, and U.
E.) RNA polymerase transcribes the template strand of DNA in the 5’3’ direction.
19. ___B__ A codon:
A) is part of a t-rna molecule; B) is a group of 3 consecutive nucleotides that specifies an amino acid;
C) is a SNRP; D) is a ribozyme; E) is overlapping
20.___D__ A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the
tRNA that binds the mRNA codon is: A) TTT; B) UUA; C) UUU; D) AAA; E) either UAA or TAA
21.___C__Which of the following does not physically interact with m-rna?
A) DNA; B) t-rna; C) amino-acyl t-rna synthetase (charging enzyme); D) ribosomes; E) release factors
22.__E___ Successful expression of a gene in a eukaryotic cell involves which of these processes?
A) transcription; B) translation; C) m-rna processing; D) m-rna transport out of the nucleus; E) all of the
above.
23.___D__ Which of the following is not true of RNA processing? A) Nucleotides may be added at both
ends of the RNA; B) RNA splicing is thought to be catalyzed by spliceosomes; C) A primary transcript is
often much longer than the final RNA molecule that leaves the nucleus; D) Exons are cut out before
mRNA leaves the nucleus.
24. __B___ Introns are significant to biological evolution because A) they are translated into essential
amino acids; B) their presence allows exons to be moved around creating different combinations
(alternative splicing), creating more than one type of m-rna ; C) they correct enzymatic alterations of
DNA bases; D) they protect the mRNA from degeneration; E) they maintain the genetic code by
preventing incorrect DNA base pairings.
25. __B___ Choose the answer that has these events of protein synthesis in the proper sequence.
1. An aminoacyl-tRNA binds to the A site.
2. A peptide bond forms between the new amino acid and a polypeptide chain.
3. tRNA leaves the P site, and the P site remains vacant.
4. A small ribosomal subunit binds with mRNA.
5. tRNA translocates to the P site.
A) 1, 3, 2, 4, 5
B) 4, 1, 2, 5, 3
C) 5, 4, 3, 2, 1
D) 4, 1, 3, 2, 5
E) 2, 4, 5, 1, 3
26. _E____ Accuracy in the translation of mRNA into the primary structure of a protein depends on
specificity in the:
A) shape of the A and P sites of ribosomes. B) binding of ribosomes to mRNA. C) bonding of the
anticodon to the codon. D) attachment of amino acids to tRNAs. E) both C and D
27.___A__ The anticodon of a particular tRNA molecule is: A) complementary to the corresponding
mRNA codon; B) complementary to the corresponding triplet in rRNA; C) the part of tRNA that bonds to
a specific amino acid; D) changeable, depending on the amino acid that attaches to the tRNA; E)
catalytic, making the tRNA a ribozyme.
28.___B__ Which of the following types of mutation, resulting in an error in the mRNA just after the
AUG start codon, is likely to have the most serious effect on the polypeptide product?
A) a deletion of a codon; B) a deletion of 2 nucleotides ; C) a substitution of the third nucleotide in an
ACC codon ; D) a substitution of the first nucleotide of a GGG codon ; E) an insertion of a codon
29. __C___ What is the effect of a nonsense mutation in a gene? A) It changes an amino acid in the
encoded protein; B) It has no effect on the amino acid sequence of the encoded protein; C) It introduces
a premature stop codon into the mRNA; D) It alters the reading frame of the mRNA; E) It prevents
introns from being excised.
30. __D___ The most commonly occurring mutation in people with cystic fibrosis is a deletion of a single
codon. This results in: A) a base-pair substitution; B) a nucleotide mismatch; C) a frameshift mutation;
D) a polypeptide missing an amino acid.; E) a nonsense mutation.
31. (4 points each) Agree or disagree with the following statements. In either case, fully defend your
position.
A. One of the four characteristics of the genetic code was particularly relevant to Eli Lilly’s development
of a technology that allowed them to produce human insulin in bacterial cells.
Agree-If the genetic code were not universal, bacterial cells could not properly translate a human mrna and would not be able to produce the insulin protein.
B. Bacteriophage were a particularly good choice of an experimental model for Hershey and Chase to
use to test the hypothesis that DNA and not protein was the genetic material.
Agree-Because virus are only made of protein and nucleic acid, and these molecules can be selectively
labeled with P32 and S35 respectively, it was easy to determine which molecule, DNA or protein,
entered cells that the virus infected and reprogrammed. Also, when virus infect cells they inject their
genes inside the cell but leave the capsid outside.
C. It is “costly” energetically to make a polypeptide chain during the process of translation.
Agree-Three ATP/GTP are consumed for every amino acid that is incorporated into a
polypeptide chain during the process of translation (one for charging, one for codon
recognition, one for translocation.)
D. Proteins play an important role in both the processes of transcription and translation.
Agree-In transcription both enzymes (RNA polymerase) and in translation charging enzymes as well
proteins that are initiation, elongation, and release factors are required.
32. (10 points) Explain, using text and diagrams, the process of DNA replication referring to all of the
enzymes involved in DNA replication and the following terms: leading strand, lagging strand,
continuous replication, discontinuous replication, Okasaki fragments.
To replicate, DNA must be unwound by helicase, single strand binding proteins attach to the 2 strands
keeping them apart until they replicate, and topoisomerase relieves the “kinks” that form at the
opposite end of the molecule as it unwinds from the other end.
The leading strand is replicated continuously toward the replication form in the 5’3” direction.
Replication begins with RNA primase adding a short RNA primer and then DNA polymerase III lengths
that strand until replication is complete. The RNA primer is cut out using DNA pol I, DNA nucleotides
are added in and then DNA ligase joins them to the continuous strand.
The lagging strand is synthesized in a discontinuous fashion in short fragments (Okasaki fragments)
moving away from the replication fork. Each Okasaki fragment is started with a short RNA primer and
then DNA polymerase III lengths that strand until it reaches another Okasaki fragment. The RNA
primer is cut out using DNA pol I, DNA nucleotides are added in and then DNA ligase joins one Okasaki
fragment to another.
33. (10 points) Using sickle-cell anemia as an example, describe the potential negative and positive
consequences of a base-pair substitution.
Sickle cell anemia is a recessive genetic disorder where all hemoglobin is mutated and this causes red
blood cells to sickle under conditions of even mild exercise. This causes the red cells to become
lodged in capillaries and results in numerous debilitating conditions which may be lethal. You have to
be homozygous for the mutated gene to have sickle cell anemia. The mutated hemoglobin is caused
by a base pair substitution resulting in missense (valine substituted for glutamine).
If you are heterozygous for the condition, your red blood cells only sickle when they become acidic.
This normally doesn’t happen and heterozygotes usually have no negative outcomes. However, if
your red blood cells are invaded with malaria parasites, the red blood cells become acidic (acidic
waste products are produced inside the red blood cells by the malaria parasites). This causes the
parasitized red blood cells to sickle, they are recognized by the immune system as abnormal, and
destroyed (destroying the malarial parasite inside the red cell). Thus heterozygotes for the mutated
gene are at an advantage if malarial parasites are present because they resistant to the parasite.
Individuals that are homozygous for the normal version of the gene never have their red cells sickle,
so they are susceptible to malaria. If they live in an area where malaria is common, they are at a
disadvantage compared to the heterozygotes.