Download Exam 2 Key v3 Bio200 Win16

Biology 200, Winter 2016
Exam 2
Name: __________KEY__________________________
Student ID#: ______________ TA: ________________
DO NOT OPEN EXAM UNTIL DIRECTED TO DO SO
• Make sure you have 4 pages of questions and six pages total. Print your
name and information on all pages.
• Please use a pen. Pen is much easier to read, even with extensive
crossing-out. Pencil-written exams are acceptable, but may not receive
full credit to penciled answers on regrades.
• When asked, provide concise and clearly written answers. We may
deduct points if you do not fully answer the question or if your answer is
too vague or too confusing for us to follow.
• Extra information, if incorrect, will lose points.
• Limit your answers to the space provided. If you need extra space, you
can use the bottom of this first page. Indicate “on first page” where
necessary.
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Points Awarded
2 ______ out of 20 points
3 ______ out of 20 points
4 ______ out of 20 points
5 ______ out of 20 points
Total ______ out of 80 points
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Exam 2
Name:_________KEY____________________
[20 points] Transcription and Translation
In the diagram, a snapshot of a single chromosome is shown along with RNA
polymerase and ribosomes. There are RNAs of various sequences, as well as three
different proteins. Use this diagram to answer the questions on pages 2-3.
/5 1a) At the moment of this snapshot shown to the right, which of the following
are likely to be True? Mark all correct answers.
__X__ Proteins are binding to different nucleic acid sequence locations.
__X__ This chromosome contains at least three different genes.
__X__ This species is prokaryotic.
__X__ The hexagonal molecules are likely to move relative to DNA.
_____ Every RNA is currently being used to encode a protein.
/5 1b) Imagine that all tRNAs with the anticodon 5’-CAU-3’ are instantly destroyed.
How many new proteins can be made after this instant?
Explain your answer in 1-2 sentences, max.
This anticodon matches to the start codon.
Without start codons, no new proteins could be made.
Possible alternative: proteins that had already been started could be completed.
/4 1c) All sigma protein molecules are destroyed instantly. How many new proteins
can be made after this instant? Choose the single best answer.
______ An infinite number
___X_ A very large number
______ A small number (less than 10)
______ None
______ A negative number
Explain your answer in 1-2 sentences, max.
The already-existing mRNA can be repeatedly used to make new proteins.
Eventually, these mRNA molecules will also degrade (so infinite is wrong).
/6
1d) On the diagram:
• Indicate with an arrow a place where hydrogen bonds are both made and broken
o Should be at an RNA polymerase molecule.
• Indicate with a star the N-terminus of a protein
o Should be on the leading end of the new proteins being made at ribosomes.
• Indicate with a triangle the 5’ end of a single mRNA
o Should be on the leading end of RNAs being made at an RNA polymerase.
Total: _____
Page 2 of 6
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Exam 2
Name:_________KEY____________________
[20 points] Transcription and translation, continued
/6 1e) Which of the following are changes that happen to a ribosome during termination?
__X__ A protein binds within the ribosome.
_____ A mutation occurs in the ribosome.
_____ Subunits of the ribosome form new bonds with each other.
_____ An anticodon or codon is broken.
__X__ A protein is released from the ribosome.
__X__ A bond between an RNA and an amino acid is broken.
/6 1f) How would you do an experiment to determine which promoters are stronger and which are
weaker? Explain (in general terms) in 2-3 sentences, max.
Answers may vary, but should find a way to test the strength of the binding between sigma and the RNA
polymerase enzyme. Experiments that test protein production do not necessarily tell us about the strength
of the promoter.
/8 1g) You’ve isolated four mutant bacteria in the lab. Based on the descriptions below, judge these
mutants in terms of their likely fitness. Fitnesses should be on a scale from zero to ten, with 7
equaling wild-type fitness and 10 being much more fit than wild-type. A fitness of 0 is instantly and
comprehensively dead. The fitnesses you note should all be relative to each other.
Fitness:
Should be less than 7, but only marginally detrimental (4-6): The gene that encodes Protein
mutation that changed the stop codon into a different codon.
has a
Should be very detrimental (probably 0-2, and less than the others): A mutation in every promoter
moves the -10 and -35 sequences closer to each other in the DNA, but does not change the sequence of
-10 or -35.
Likely neutral or slightly beneficial (7-9): A mutation in the amino-acyl tRNA synthetase enzyme
that increases its rate of reaction.
Should be less than 7, but only marginally detrimental (4-6): Mutant J: A mutation in the
transcription termination signal of 10 different protein-encoding genes that moves this sequence 20
bases farther downstream in the gene.
Total: _____
Page 3 of 6
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Exam 2
Name:_________KEY____________________
[20 points] Replication and Information Flow
Note: For the full version of the exam, the sequence will be filled in and other diagram details may change.
Below is a portion of double stranded DNA from a bacterial chromosome.
The promoter region and the +1 base pair are indicated, as well as the polarity of the two DNA strands.
+1
-10
-35
...5’..CCTTATGGCAAAAGCATAGCGCGCAAACGCTGATAACCGTATGCGCAAATATGCCA..3’....
...3’..GGAATACCGTTTTCGTATCGCGCGTTTGCGACTATTGGCATACGCGTTTATACGGT..5’....
/10 2) What is the translated protein sequence from this gene? Be sure to include the N- and C-termini. You can
abbreviate amino acids using the 3-letter code.
N-Met-Leu-Leu-Pro-C
You’ve been hired by a bioengineering firm to create specific gene structures by mutating bacterial
DNA. For each, pick the mutation that is most likely to consistently create the specific structural
feature that is desired.
Example: Every chromosome has sporadic single breaks between nucleotides in the phosphodiester backbone.
____ Mutation in the gene encoding the helicase enzyme
__X_ Mutation in the start codon of the open reading frame encoding the ligase enzyme
____ Mutation in the origin sequences on a single chromosome
____ Mutation in the promoter regions of all genes encoding DNA polymerase enzymes
5/ 3a) Short chromosomes are copied well. Longer chromosomes are either delayed, or the interior
regions of the DNA are never completely replicated.
_____ Synonymous mutation in the gene encoding the helicase enzyme
_____ Mutation in the promoter regions of all genes encoding DNA polymerase enzymes
_____ Deletion mutation in the area of the chromosome between the genes for ligase and helicase
__X__ Mutation in the start codon of the topoisomerase enzyme
_____ Missense mutation in the open reading frame of the primase enzymes
5/ 3b) Absolutely no DNA is replicated, but when a single-stranded DNA molecule is inserted into
the cell it is replicated.
_____ Synonymous mutation in the gene encoding the topoisomerase enzyme
__X__ Missense mutation in the gene encoding the helicase enzyme
_____ Premature stop mutation in the gene encoding a DNA polymerase enzyme
__X__ Frameshift mutation in the gene encoding ssBP
_____ Large deletion in the promoter region of the gene encoding primase
_____ Large insertion mutation in the gene encoding the topoisomerase enzyme
Explain your answer to 3b in 1-2 sentences, maximum.
The only enzyme in this list whose absence that completely stops replication is the helicase. A
single version of DNA polymerase will not stop all replication, since it could be a specialized
DNA polymerase (like DNA pol I) and not the initial actor.
Total: _____
Page 4 of 6
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Exam 2
[20 points] Bacterial genetic forensics
Name:_________KEY____________________
Researchers at the NIH are trying to use probiotics (helpful bacteria) to cure symptoms in patients suffering
from chronic bowel conditions. A set of four artificial genes is bioengineered into a non-coding region of the
bacterial chromosome as shown in Figure 2 (on Page 6). The problem is that several mutations are decreasing
the effectiveness of that four-gene cluster, and the researchers are having trouble figuring out the cause of
these mutations. They’ve asked you to consult on the problem and try to determine the simplest possible
mechanisms that may have caused these mutations.
/5
4a) There is a mutation in the open reading frame of W. This mutation does not change the
expression or function of W, but it does change the expression of Gene H. Describe the most likely
mutation in 2-3 sentences, maximum.
Answers may vary, but a relatively simple answer is a single synonymous mutation in an area of
the coding region for W that overlaps with the promoter of Gene H. This mutation must change
the binding of sigma to the DNA.
/5
4b) Protein analysis suggests that a few bacteria have a strange ‘combo’ protein. Protein
analysis suggests that the protein has a region that is similar to Protein H and another region that is
similar to protein B. What is the smallest possible mutation that would produce this phenotype?
Choose the single best answer.
______ A loss of a stop codon at the end of Gene H.
______ A loss of a transcription termination signal at the end of Gene H.
__X___ A loss of a transcription termination signal AND a stop codon at the end of Gene H.
______ None of the combinations of mutations might produce this ‘combo’ phenotype.
/5
4c) A mutation occurred that changes the expression of all four genes.
______ This is likely to be a point mutation
______ This is likely to be a large-scale chromosomal inversion
Note: there is a way to make this answer correct with a very specific diagram.
__X___ This is likely to be a large-scale chromosomal deletion
______ This is likely to be a polymerase making many errors
Draw a simple diagram to indicate clearly what has happened.
Answers may vary, but should demonstrate the complete loss of all four coding regions.
/5
4d) Imagine that the region of the chromosome shown in Fig 2 was inserted into the protein
encoding region of a gene for DNA polymerase. Describe the amount of mutations that would result.
Explain your answer in 2-3 sentences, maximum.
Possibility #1: The mutation destroys polymerase function through a single mutation. No other mutations
occur because the organism is likely to die quickly.
Possibility #2: Polymerase function is detrimentally impacted, leading to many mutations as that
polymerase makes mistakes in future cycles of replication.
Total: _____
Page 5 of 6
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Exam 2
Name:_________KEY____________________
Fig 2: Use this reaction diagram for Q4.
Gene W
Gene B
Promoter
for Gene H
om
Pr
Promoter
for Gene F
ote
r fo
en
rG
Promot
e
r for Ge
ne B
eW
Gene H
Gene F
Total: _____
Page 6 of 6
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