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EBIO 1210: General Biology 1
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Exam 4
July 6, 2012
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1. For this pair of items, choose the option that best describes their relationship.
(A) The probability that amino acids with polar side chains are hydrophobic.
(B) The probability that amino acids with side chains containing an -OH group are
hydrophilic.
A) Item (A) is greater than item (B).
B) Item (A) is less than item (B).
C) Item (A) is exactly or very approximately equal to item (B).
D) Item (A) may stand in more than one of the above relations to item (B).
2. All of the following contain amino acids EXCEPT
A) hemoglobin.
B) cholesterol.
C) antibodies.
D) enzymes.
E) aquaporins.
3. The bonding of two amino acid molecules to form a larger molecule requires which of
the following?
A) removal of a water molecule
B) addition of a water molecule
C) formation of an ionic bond
D) formation of a peptide bond
E) both A and D
4. Upon chemical analysis, a particular polypeptide was found to contain 300 amino acids.
How many mRNA nucleotides were translated to create this protein?
A) 901
B) 900
C) 300
D) 301
E) 100
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5. Which type of interaction stabilizes the alpha helix and the beta pleated sheet structures
of proteins?
A) hydrophobic interactions
B) nonpolar covalent bonds
C) ionic bonds
D) hydrogen bonds
E) peptide bonds
6. The tertiary structure of a protein is the
A) bonding together of several different polypeptide chains by weak bonds.
B) order in which amino acids are joined in a polypeptide chain.
C) unique three-dimensional shape created by bonds between R-groups.
D) organization of a polypeptide chain into an alpha helix or beta pleated sheet.
E) overall protein structure resulting from the aggregation of two or more
polypeptide subunits.
7. A hydrophilic R-group of an amino acid in hemoglobin would NOT be attracted to:
A) the water molecules surrounding hemoglobin.
B) a hydrophobic amino acid R group of hemoglobin.
C) a charged amino acid R group of hemoglobin.
D) a polar amino acid R group of hemoglobin.
E) an amino acid R group of hemoglobin containing an –OH group.
8. Altering which of the following levels of structural organization could change the
function of a protein?
A) primary
B) secondary
C) tertiary
D) quaternary
E) all of the above
9. The structural level of a protein least affected by a disruption in hydrogen bonding is the
A) primary level.
B) secondary level.
C) tertiary level.
D) quaternary level.
E) All structural levels are equally affected.
10. Humans can produce the amino acid alanine from other components in the body, but
valine cannot be produced in the human body. From this you can conclude:
A) Alanine is a non-essential amino acid
B) Valine is a non-essential amino acid
C) Valine is an essential amino acid
D) A and B
E) A and C
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11. In sickle cell anemia, hemoglobin function is affected. WHY?
A) The structure of hemoglobin remains unaffected, but its binding of oxygen is
diminished.
B) One hydrophilic amino acid in the primary structure is replaced by a different
hydrophilic amino acid.
C) One hydrophobic amino acid in the primary structure is replaced by a different
hydrophobic amino acid.
D) The final shape of the protein is altered, as are the shape of red blood cells.
E) The sickle-shaped blood cells aggravate (worsen) the symptoms of malaria.
12. Sickle cell anemia reduces a person’s expected life span by, on average, 25-30 years.
Which of the following is the BEST description of why natural selection has not removed
sickle cell anemia from the human population?
A) Sickle cell anemia is caused by a recessive allele
B) Those with the aa genotype benefit from high survival due to malaria
resistance
C) Those with the AA genotype benefit from high survival due to malaria
resistance
D) Heterozygotes for sickle cell anemia have the highest survival in non-malaria
countries
E) Heterozygotes for sickle cell anemia have the highest survival in malariastricken countries
13. Which of the following is NOT a function of RNA in a normal animal cell?
A) carrying amino acids to ribosomes for protein synthesis.
B) serving as a component of ribosomes.
C) acting as the template to create DNA.
D) splicing of pre-mRNA.
E) all of these are functions of RNA.
14. Which of the following is NOT a difference between DNA and RNA?
A) RNA nucleotides use a different phosphate than DNA nucleotides.
B) One of RNA’s nitrogenous bases is different than DNA’s.
C) DNA nucleotides contain a different sugar than RNA nucleotides.
D) DNA is a double helix, but RNA is usually single-stranded.
E) A and D are correct.
15. A particular triplet of bases in the template strand of DNA is 3' CAT 5'. The
corresponding codon for the mRNA transcribed is
A) 3' CUA 5'.
B) 5' GUA 3'.
C) 5' UTC 3'.
D) 3'GAU 5'.
E) 3’ CUU 5’.
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16. What amino acid sequence will be generated based on the following mRNA codon
sequence?
5' CCAGAUGUCUUCGUCCUUGAAC 3'
A) pro-met-ser-ser-leu-asn
B) pro-asp-val-phe-val-leu-glu
C) met-ser-leu-ser-leu-ser
D) met-ser-ser-ser-leu-asn
E) gin-val-pro-ala-ser-val-asp
17. A polypeptide has the sequence phe-pro-lys-gly-phe-pro. Which of the following
sequences in the template strand of the DNA could code for this polypeptide?
A) 3' AAA-GGG-TTT-CCC-AAA-GGG
B) 3' UUU-CCC-AAA-GGG-UUU-CCC
C) 3' TTT-CCC-AAA-GGG-TTT-CCC
D) 5' GGG-AAA-CCC-AAA-CCC-GGG
E) 5' AAA-TAC-CAT-AAA-CAT-TAC-UGA
18. What is the sequence of a polypeptide based on the following mRNA sequence?
5' . . . CUUUCUUAUUGUCUU 3'
A) leu-cys-tyr-ser-phe
B) cyc-phe-tyr-cys-leu
C) phe-leu-ile-met-val
D) leu-pro-asp-lys-gly
E) leu-ser-tyr-cys-leu
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19. The mRNA sequence for a certain polypeptide is 5’…GAGCCGUAA…3’. If the last A
is changed to a U via a substitution mutation, what effect will this have on the protein?
A) No change
B) The polypeptide will be longer
C) The polypeptide will undergo a frameshift
D) The polypeptide will be shorter
E) The polypeptide will definitely work better
20. Differential gene regulation means that different cell types have the same _______ but
express different _______.
A) proteins, mRNAs
B) genomes, proteins
C) mRNAs, genes
D) genes, chromosomes
E) DNA, genomes
21. Transcription in eukaryotes requires which of the following in addition to RNA
polymerase?
A) snRNA
B) start and stop codons
C) ribosomes and tRNA
D) several transcription factors (TFs)
E) aminoacyl synthetase
22. Which of the following is not involved in mRNA processing?
A) promoter
B) spliceosomes
C) 5' cap and poly (A) tail
D) exons
E) introns
23. What are the coding segments of a stretch of eukaryotic RNA called?
A) anticodons
B) transposons
C) introns
D) exons
E) rightons
24. Which of the following is NOT true about splicing?
A) Splicing involves the removal of non-protein coding sequences
B) Splicing involves the fusion of protein-coding sequences
C) Splicing is done by the ribosome
D) Splicing can generate different mRNAs from the same gene
E) Splicing must occur before translation
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25. Which of the following is TRUE?
A) RNA polymerase uses RNA as a template, and DNA polymerase uses a DNA
template.
B) RNA polymerase requires helicase to unzip the helix for it.
C) RNA polymerase reads 5’ to 3’, but DNA polymerase reads 3’ to 5’.
D) RNA polymerase can initiate RNA synthesis without a primer, but DNA
polymerase requires a primer.
E) RNA polymerase does not need to separate the two strands of DNA in order to
synthesize an RNA copy, whereas DNA polymerase must unwind the double
helix before it can replicate the DNA.
A part of an mRNA molecule with the following sequence is being read by a ribosome: 5'
UGC-GCC 3' (mRNA). The following charged tRNA molecules (with anticodons shown in the
3' to 5' direction) are available. Two of them can correctly match the mRNA to form a
dipeptide.
26. The dipeptide that will form will be
A) cysteine-alanine.
B) proline-threonine.
C) glycine-cysteine.
D) alanine-alanine.
E) threonine-glycine.
27. What is the mRNA codon corresponding to the tRNA to the right?
A) 3’ CCG 5’
B) 5’ CUU 3’
C) 3’ GGC 5’
D) 5’ UUC 3’
E) 5’ UGC 3’
28. What occurs at the top of the tRNA shown in the figure for question 27?
A) The 5' cap of the mRNA will become covalently bound.
B) The codon and anticodon complement one another.
C) The small and large subunits of the ribosome will attach to it
D) The amino acid is attached by aminoacyl tRNA synthetase.
E) The excess nucleotides (ACCA) will be cut off in the ribosome.
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29. A particular triplet of bases in the non-template coding sequence of DNA is GGG. The
anticodon on the tRNA that corresponds to this DNA is
A) CCC.
B) UUA.
C) UUU.
D) GGG.
E) AUG.
30. Put the following translation events in order:
i. elongation of the polypeptide
ii. base pairing of methionine-tRNA to AUG of the mRNA
iii. the larger ribosomal subunit binds to smaller ribosomal subunits
iv. peptide bonding between the first two amino acids
v. the small subunit of the ribosome recognizes and attaches to mRNA
A) i, ii, iii, iv, v
B) v, iv, iii, ii, i
C) v, iii, ii, i, iv
D) v, ii, iv, i, iii
E) v, ii, iii, iv, i
31. As a ribosome moves along an mRNA molecule by one codon, which of the following
occurs?
A) The tRNA that was in the P site moves into the A site
B) The tRNA that was in the A site departs from the ribosome via a tunnel.
C) The tRNA that was in the A site moves into the P site.
D) The tRNA that was in the A site moves to the E site and is released.
E) The polypeptide enters the E site.
32. Which of the following mutations is least likely to be detrimental?
A) A substitution at the beginning of a gene
B) A substitution at the end of a gene
C) A deletion at the end of a gene
D) An insertion at the end of a gene
E) An insertion at the beginning of a gene
33. Each of the following options is a modification of the sentence THECARWASRED.
Which of the following is analogous to a frameshift mutation?
A) CARWASRED
B) THECARWARED
C) THEREDWASCAR
D) THERACWASRED
E) THECARWASDED
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34. Which statement about prokaryotes is NOT true?
A) Prokaryotic translation starts at AUG, which codes for methionine
B) Prokaryotic mRNA receives a 5’ cap before translation
C) In prokaryotes, transcription and translation of an RNA molecule can occur at
the same time
D) Prokaryotic DNA includes a promoter for each gene
E) Prokaryotic ribosomes stop translating at one of three stop codons
35. Which of the following is not a reason cells need to regulate their genes?
A) To produce differentiated tissues and organs
B) In some cases, not all of a cell’s products are needed constantly.
C) Single-celled bacteria often work together in colonies.
D) It is a waste of energy to produce proteins that aren’t needed.
E) Some protein functions are mutually exclusive.
36. What would happen if a scientist experimentally moved the trp repressible operator from
its regular location in a prokaryote’s operon to the end of the gene instead?
A) The trp gene will not be transcribed.
B) RNA polymerase will not be able to bind to the promoter.
C) The repressor will no longer bind to the operator.
D) The enhancer won’t bind to the activators.
E) The trp gene will be transcribed continuously.
37. Spleen cells and kidney cells in one species of animal owe their differences in structure to
A) having different genes expressed.
B) having different genes.
C) having unique ribosomes.
D) having different chromosomes.
E) using different RNA polymerases.
38. The human body regulates blood glucose levels via a negative feedback system. Which of
the following would NOT be consistent with this process?
A) When glucose levels are low, the body releases more glucose into the
bloodstream from storage molecules.
B) When glucose levels are high, the body releases more glucose into the
bloodstream from storage molecules.
C) When glucose levels are high, the body pulls glucose out of the bloodstream
and puts them into storage.
D) When glucose levels are high, the body feels full, triggering us to stop eating.
39. Enhancers are used by ____ to regulate ____.
A) eukaryotes, transcription.
B) eukaryotes, translation.
C) eukaryotes, mRNA splicing.
D) prokaryotes, transcription.
E) prokaryotes, translation.
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40. In eukaryotes, general transcription factors
A) bind to sequences downstream from the promoter during transcription.
B) are required for the expression of specific protein-encoding genes.
C) prevent RNA polymerase from binding to the promoter.
D) help RNA polymerase bind to the promoter.
E) usually lead to a high level of transcription even without additional specific
transcription factors.
41. If you wanted to experimentally turn a human hair follicle cell into a muscle cell, what
would the best way to do this?
A) Replace the follicle cell’s DNA with the muscle cell’s DNA.
B) Replace the follicle cell’s activators with the muscle cell’s activators.
C) Insert a keratin (hair protein) repressor into the follicle cell.
D) Insert an actin (muscle protein) repressor into the follicle cell.
E) Ask it Nicely
42. The bicoid protein is normally concentrated at the head end of an embryo. If large
amounts of the product were injected into the tail end as well, which of the following
would occur?
A) The embryo would grow to an unusually large size.
B) The embryo would grow extra wings and legs.
C) The embryo would probably show no head development and die.
D) Head structures would form in both sides of the embryo.
E) The embryo would develop normally.
43. The hosts a virus can infect are determined by:
A) the proteins on the virus surface and the host cell surface.
B) whether the virus’ nucleic acid is DNA or RNA.
C) the proteins in the host's cytoplasm.
D) the enzymes produced by the virus before it infects the cell.
E) the enzymes carried by the virus.
44. Which of the following is not a part of some or all viruses?
A) nucleic acids
B) a protein coat
C) phospholipids
D) glycoproteins
E) ribosomes
45. What is the name given to viruses with single-stranded RNA that acts as a template for
DNA synthesis?
A) viroids
B) bacteriophages
C) proviruses
D) retroviruses
E) all viruses do this
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46. Which of the following places the events of the viral life cycle in the correct order?
A)
B)
C)
D)
E)
genome replication, capsid assembly, host cell entry
host cell entry, genome replication, capsid assembly
capsid assembly, release from host cell, genome replication
genome replication, host cell entry, host cell release
genome replication, host cell entry, protein synthesis
You isolate an infectious substance that is capable of causing disease in plants, but you do not
know whether the infectious agent is a bacterium or virus. You have four methods at your
disposal that you can use to analyze the substance in order to determine the nature of the
infectious agent.
I.
Treating the substance with nucleases that destroy all nucleic acids and then determining
whether it is still infectious
II.
Filtering the substance to remove all elements smaller than what can be easily seen under
a light microscope
III.
Culturing the substance by itself, away from any plant cells, to see if it can reproduce
IV.
Treating the sample with proteases that digest all proteins and then determining whether
it is still infectious
47. Which treatment would allow you to distinguish between a viral or bacterial disease?
A) I
B) II
C) III
D) IV
E) either II or IV
48. AZT is a nucleotide analog used to treat HIV. It is a nonfunctional thymine (T)
nucleotide. Which step would AZT hamper in the reproductive cycle of the HIV virus?
A) reverse transcription of DNA from RNA
B) entry into the cell
C) transcription of RNA from proviral DNA
D) viral assembly within the cell
49. Which of the following is/are characteristics of HIV?
A) It carries its own reproductive enzymes for reverse transcription.
B) It inserts its genetic material permanently into the host’s genome.
C) Its nucleic acid is DNA
D) A and B
E) All of the above
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50. Which of the following is a correct completion of this chart?
DNA template
DNA non-template
mRNA
3’
A C
G
G
T A
C
C
G A 5’
T
G
A
T
U
A)
DNA template
DNA non-template
mRNA
B)
DNA template
3’ C A C C G G G A A T A T A G A 5’
DNA non-template 3’ G T G G C C C T T A T A T C T 5’
mRNA
3’ C A C C G G G A A U A U A G A 5’
C)
3’ G A C C C G G A T T A T T G A 5’
3’ G A C C C G G A T T A T T G A 5’
3’ G A C C C G G A U U A U U G A 5’
DNA template
3’ C A C C G G G A A T A T A G A 5’
DNA non-template 5’ G T G G C C C T T A T A T C T 3’
mRNA
3’ C A C C G G G A A U A U A G A 5’
D)
DNA template
3’ C A C G G C G T A T A A A G A 5’
DNA non-template 5’ G T G C C G C A T A T T T C T 3’
mRNA
5’ G U G C C G C A U A U U U C U 3’
E)
DNA template
3’ C A C G G C G T A T A A A G A 5’
DNA non-template 5’ G T G C C G C A T A T T T C T 3’
mRNA
5’ G T G C C G C A T A T T T C T 3’
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