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DO NOT OPEN UNTIL TOLD TO START
BIO 312, Section 1, Spring 2011
April 27, 2011 – Exam 3
Name (print neatly)
Instructor
7 digit student ID
INSTRUCTIONS:
1. There are 9 pages to the exam. Make sure you have all of the pages.
2. There are 50 scantron questions.
3. Each question is worth 2 points, for 100 points total.
4. Be sure to provide your student information above and on the scantron.
5. Mark all of your answers clearly on the scantron.
6. If you don’t understand what is being asked, raise your hand to ask for help.
7. You have 50 minutes to complete the exam.
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Scantron portion: problems 1 – 50.
1. Fill in your scantron with your complete name.
2. For each problem #, to get credit you must fill in the appropriate circle on your scantron using
a number 2 pencil.
3. You may write on your exam copy if it helps.
4. When done, turn in to the instructor BOTH your scantron and exam copy.
For problems #1 – 12 the correct answer for each statement is either TRUE or FALSE. For a TRUE
answer fill in the circle underneath the letter A. For a FALSE answer fill in the circle underneath
the letter B.
1. The ground state for prokaryotic gene expression is considered to be “ON”.
True
2. Operator sequence mutations are considered to be “cis-acting”.
True
3. When nutritional resources are abundant bacteriophage λ will enter the lysogenic life cycle.
False
4. A bacteriophage λ with a null mutation in the cro gene will enter the lysogenic life cycle
True
5. The recognition helix of the λ repressor and Cro proteins have the same amino acid
sequences.
False
6. Female and male fruit flies that are both heterozygous mutants for the same maternal-effect
gene (m/+ ♂ crossed to m/+♀) were crossed. If a large number of progeny were obtained, you
would find that approximately ¼ of the progeny exhibit the maternal-effect mutant phenotype.
False
7. Histones isolated from genes that are highly expressed are typically hyperacetylated.
True
8. Synonymous mutations change the codon for an amino acid to a codon for a different amino
acid.
False
9. Nonconservative substitutions alter the codon reading frame for a gene.
False
10. Oncogenes refer to mutant genes that promote a cancer cell phenotype dominantly because
the mutant genes have gained a function.
True
11. In order for enhancers to influence a gene’s expression they must be located next to the
gene’s promoter.
False
12. Human X chromosome dosage compensation involves a complex of proteins interacting
with the male X-chromosome genes resulting in these genes being hypertranscribed.
False
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Problems 13 - 50 are multiple choice questions. Using a #2 pencil, fill in the circle underneath
the letter for the correct answer.
13 and 14. The GAL1 is a gene required for yeast to metabolize galactose as a food source.
13. Mutant yeast cells lacking the GAL80 gene are grown in media with galactose as the only
sugar source. Which of the following choices is correct?
A. Transcription of the GAL1 gene will not occur in these cells.
B. Transcription of the GAL1 gene will occur in these cells.
14. Mutant yeast cells lacking the GAL4 gene are grown in media with galactose as the only
sugar source. Which of the following choices is correct?
A. Transcription of the GAL1 gene will not occur in these cells.
B. Transcription of the GAL1 gene will occur in these cells.
15 - 17. In the figure below the left most images show the points of migration following
electrophoresis for a wild type gene’s mRNA (N) and encoded protein (W). For the problems
below, choose the patterns of migration (select either A, B, C, or D) anticipated for the
following types of mutations to this same gene.
15. Nonsense mutation
B
16. Missense mutation
A
17. Frameshift mutation
C
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18-20. Use the graph below to answer the following problems.
In this experiment two histidine auxotrophic bacteria strains (EC1 and EC2) were given the
chemical compound 11447 at increasing doses in the presence of a mouse liver extract (EC1+LE
and EC2+LE) and without the extract (EC1 and EC2). These bacterial cultures were plated onto
nutritional media lacking histidine and the number of surviving colonies were recorded and
plotted on this graph. The EC1 strain is known to be sensitive to transition mutations and the
EC2 strain is sensitive to indel mutations.
18. This test is known as the
A. Ames
B. Bateson
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C. Morgan
test.
D. Jacob
E. Brenner
19. The mutation events causing the reversion phenotypes result from the ................................
A. removal of DNA sequences from a gene.
B. insertion of DNA sequences into a gene.
C. alteration of a gene’s reading frame.
D. replacement of a base by another of the same chemical category.
E. replacement of a base by another of the other chemical category.
20. What is the significance of the liver extract in this type of experiment?
A. Impairs bacterial cells from being able to correct mutations.
B. Metabolizes non-mutagenic compounds into mutagenic forms.
C. Metabolizes mutagenic compounds into non-mutagenic forms.
D. Liver extract is full of intercalating agents that cause DNA replication errors.
E. Makes bacterial cells permeable to chemical compounds.
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21 – 24. For each problem, select one of the following choices to correctly complete the
statement.
A. processed pseudogenes
B. conventional pseudogenes
C. syntenic genes
D. paralogous genes
E. orthologous genes
21. Genes that originate from an mRNA that was reverse transcribed from mRNA to a cDNA
and inserted somewhere in the genome.
A. processed pseudogenes
22. Genes that acquired disrupting mutations in their coding sequence during the course of
evolution.
B. conventional pseudogenes
23. Genes that are related by gene duplication events that are found in a single genome.
D. paralogous genes
24. Genes that are the same genetic locus in two different species genomes owing to their
inheritance from a common ancestor.
E. orthologous genes
25 - 27. For each problem, choose one of the following choices that correctly completes the
statement.
A. Ultraconserved non-coding elements
B. Expressed Sequence Tags
C. hCONDELs
D. Enhancers
E. Phenocopies
25. Sequencing of a cDNA library generates
can use to find the position of exons in genome a sequence.
B. Expressed Sequence Tags
that bioinformatic approaches
26.
refer to a specific class of identical DNA sequences found by
comparative genomics in the genomes of distantly related species. Some of these sequences
were found to regulate gene transcription, such as for the mouse Isl1 gene.
A. Ultraconserved non-coding elements
27.
are human specific losses of DNA sequences possessed by all other
examined mammal species. One example was the sequence controlling androgen receptor gene
expression in the developing sensory vibrissae of both male mice and chimpanzees.
C. hCONDELs
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28 – 30. Choose one of the following choices that is the correct name for the described
experimental technique.
A. RNAi
B. chromatin immunoprecipitation
C. DNA microarray
D. yeast two-hybrid
E. in situ hybridization
28. Technique that makes visible the locations where a certain mRNA is expressed
E. in situ hybridization
29. This technique uses an antibody specific to a particular protein to purify all of the DNA
sequences in a genome to which the protein is bound.
B. chromatin immunoprecipitation
30. This new technology can be used to monitor the level of transcription for the entire
transcriptome.
C. DNA microarray
31. Which one of the following is not correct about the human genome?
A. Differs in sequence from the chimpanzee genome by 1% due to single nucleotide differences.
B. Most proteins orthologous to chimpanzee proteins either do not differ in amino acid
sequence or differ for just a few amino acids.
C. Human genome has more genes related to color perception than the mouse genome.
D. Compared to the chimpanzee genome, the human genome has lost some genes and has
duplicates of others.
E. Human genome has more genes related to olfaction than the mouse genome.
32 - 35. For each problem, choose one of the following choices that correctly completes the
statement.
A. Trp Repressor
B. AraC
C. Sigma Factor
D. Tryptophan
E. CAP
32.
is a dual function regulatory protein that acts both as an activator and
repressor.
B. AraC
33.
functions to “attenuate” transcription from a certain gene operon in a
metabolic pathway.
D. Tryptophan
34.
interacts with cAMP to positively regulate operon gene expression.
E. CAP
35.
can coordinately control the transcription of a “regulon” of unlinked genes.
C. Sigma Factor
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36 and 37. The below figure is of a mouse chromosome region containing two neighboring
genes. Gene 1 is normally expressed in the developing brain, while Gene 2 is not expressed
in the developing brain. Separately, mutations were created for the sequences A and B. When
the sequence “A” region was mutated, both Gene 1 and Gene 2 are not expressed in the
brain. When the sequence “B” region was mutated, both Gene 1 and Gene 2 are expressed in
the brain.
36. Sequence A functions as ......................
A. a nucleosome. B. an imprinting control region.
D. a coactivator. E. an insulator.
C. an enhancer.
37. Sequence B functions as an ......................
A. a nucleosome. B. an imprinting control region.
D. a coactivator. E. an insulator.
C. an enhancer.
38. After irradiating fruit flies, you notice that a particular male fly has an eye phenotype with
patches having the normal red color and patches that are of a mutant white color. This mutant
white color typically results from mutations that disrupt the gene white. However, you find that
the white gene in this mutant fly has the wild type sequence. What alteration was caused by the
irradiation?
A. Silencing was lost at the MAT locus.
B. The white gene was moved from a region of euchromatin near a region of heterochromatin.
C. The white gene was moved from a region of heterochromatin to a region near euchromatin.
D. The imprinting control region was disrupted for a paternally imprinted gene.
E. The imprinting control region was disrupted for a maternally imprinted gene.
39. The human β-interferon gene remains off until viral infection. Depending on the severity of
the infection the level of β-interferon expression can be adjusted from low to very high levels.
This rheostat-like mechanism of regulation results from the formation of .....................................
A. an enhanceosome.
B. combinatorial regulation.
C. an imprinting control region.
D. a barrier insulator
E. constitutive heterochromatin.
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40. What is the sequential order of action for the classes of toolkit genes during fruit fly
development? Answers below are listed with the earliest acting class on the left and the latest
acting class on the right.
A. gap, maternal-effect, segment-polarity, pair-rule, selector
B. gap, maternal-effect, pair-rule, selector, segment-polarity
C. maternal-effect, gap, selector, pair-rule, segment-polarity
D. maternal-effect, selector, gap, segment-polarity, pair-rule
E. maternal-effect, gap, pair-rule, segment-polarity, selector
41. What class or classes of toolkit genes will be affected in individuals homozygous for a gap
gene null mutation?
A. selector genes
B. pair-rule genes
C. segment-polarity genes
D. all of the above
E. none of the above
42. As you prepare to eat a very ripe banana you notice a funny looking fruit fly. You bring this
fly to Dr. Williams’ lab to inspect it under a microscope. Here you notice that it has seven body
segments rather than the usual 14, and every other body segment is missing. Based on these
findings it can be concluded that this fly has a mutation in a .............
A. selector gene.
B. pair-rule gene.
C. gap gene.
D. segment-polarity gene.
E. Hox gene.
43. Using an antibody that interacts with the protein encoded by the gene hairy, you make
visible the below pattern of embryonic Hairy protein expression. Based on this characteristic
pattern of seven stripes, you can conclude that Hairy is a ...........................................
A. maternal-effect gene
D. gap gene
B. selector gene
E. segment-polarity gene.
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C. pair-rule gene
44. Which of the following in not true about Hox genes?
A. There are 8 Drosophila Hox genes that are all found on the same chromosome.
B. Hox genes are a part of a large class of genes referred to as differentiation genes.
C. Hox genes encode transcription factor proteins.
D. Hox genes are found in a wide diversity of animals.
E. Hox genes all possess a similar gene sequence known as the homeobox.
Problems 45-50.
The map of the lac operon is IPOZY. For the following haploid (45 and 46) and partial diploid
(47-50) E. coli genotypes, use your knowledge of the lactose system to fill in the table as to
whether each protein is produced when there is either No lactose or lactose present. The “+”
sign indicates protein production and the “-“sign indicates no protein production.
β-galactosidase
Permease
Haploid Genotype
No lactose
lactose
No lactose
lactose
45
I+P+O+Z+Y+
1
2
3
4
46
I-P-OCZ+Y+
1
2
3
4
1
2
3
4
Diploid Genotype
47
I+P-O+Z-Y+ /ISP+O+Z+Y-
48
I-P+OCZ-Y+ / I+P+O+Z+Y-
1
2
3
4
49
I+P-OCZ+Y- / ISP+OCZ-Y+
1
2
3
4
50
I+P+OCZ-Y+ / ISP+OCZ+Y-
1
2
3
4
For each problem, choose the one correct answer from the following choices. The order of the
answers below corresponds to 1, 2, 3, 4 in the above table.
45. A. +, +, +, +
B. -, +, -, +
C. -, -, -, -
D. -, -, +, +
E. -, +, +, +
46. A. +, +, +, +
B. -, +, -, +
C. -, -, -, -
D. -, -, +, +
E. -, +, +, +
47. A. +, +, +, +
B. -, +, -, +
C. -, -, -, -
D. -, -, +, +
E. -, +, +, +
48. A. +, +, +, +
B. -, +, -, +
C. -, -, -, -
D. -, -, +, +
E. -, +, +, +
49. A. +, +, +, +
B. -, +, -, +
C. -, -, -, -
D. -, -, +, +
E. -, +, -, -
50. A. +, +, +, +
B. -, +, -, +
C. -, -, -, -
D. +, -, +, -
E. -, +, -, -
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