Download Exam 2 Initial Key v2 Bio200 Win17

Biology 200, Winter 2017
Exam 2
Name: _______________KEY______________________
Student ID#: ______________ TA: ________________ Lab Section: ______
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 on regrade requests.
• 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, use
the bottom of the last page. Indicate “on last page” where necessary.
Page
Points Awarded
2 ______ out of 20 points
3 ______ out of 20 points
4 ______ out of 15 points
5 ______ out of 25 points
Total ______ out of 80 points
Codon Table:
Note: The most common start codon is Methionine.
Exam 2
Name:__________KEY___________________
Four mutations are described below. Each mutation is extremely specific and somehow impacts the
transcription or translation of a particular molecule. For each, mark all true answers.
Example: A codon on the coding strand in the coding region of an RNA encoding a region of the sigma protein that contacts the
DNA backbone is changed from 5’GAA3’ to to 5’AAA3’.
______ This increases all transcription.
__X___ This decreases all transcription.
______ This increases translation.
______ This increases translation.
___X__ This likely decreases interactions between the sigma protein and DNA.
/5
1a) In the gene encoding the release factor, base pairs are added downstream of the
DNA that encodes the 1st codon.
______ This increases correct transcription.
___X__ This decreases correct transcription.
______ This increases correct translation.
___X__ This decreases correct translation.
___X__ This impacts the production of all proteins.
/5
1b) For one of the most important protein pieces of the ribosome, the 7th, 8th and 9th
bases in the coding region of the mRNA are changed from 5’-UCA-3’ to 5’-UCG-3’.
______ This increases all transcription.
______ This decreases all transcription.
______ This increases correct translation.
______ This decreases correct translation.
______ This changes the rate at which some (but not all) proteins are produced.
Explain your answer in 1-2 sentences, max.
None of these answers are correct. While the ribosome is important for all proteins, there is no impact
because these mutations combine to form a silent mutation, which does not change the structure of
the encoded protein.
/5
1c) Gene X encodes RNA polymerase, which is made of 122 amino acids. The DNA
encoding the 56th codon of Gene X is changed from 5’-GGA-3’ to 5’-UGA-3’.
______ This increases correct transcription of Gene X.
___X__ This decreases correct transcription of Gene X.
______ This increases correct translation of RNApolymerase
___X__ This decreases correct translation of RNApolymerase.
______ There are likely to be increases in correct transcription of many genes.
Explain your answer in 1-2 sentences, max.
This mutation causes a premature stop codon, which will cause all translated proteins to be very
unlikely to be functional since roughly half of the correct amino acid sequence is missing.
/5
1d) In a prokaryotic genome, an 18-base-pair section of DNA is inserted between the -10
box and the -35 box of Gene Y.
______ This increases transcription of Gene Y.
___X__ This decreases transcription of Gene Y.
______ This increases translation of Protein Y.
___X__ This decreases translation of Protein Y.
______ This is likely to change the tertiary structure of Protein Y.
KEY
Page 2 of 6
Exam 2
Name:__________KEY___________________
/10
2) Draw a diagram of two cells. For each cell, indicate a simplified mixture of DNA, RNA, and
proteins as well as a nucleus. Use your diagram to demonstrate the following difference between
these cells: One cell has a mutation in the gene that encodes Protein G, and the other cell has an error
in a few mRNA molecules that may impact the amount or function of its version of Protein G.
Use your diagram to clearly show the required elements and demonstrate the differences between the cells.
Answers can vary widely. Fully credited diagrams will make clear where the mutation is occurring (in what
nucleotide polymer) and how that impacts the supply of wildtype or mutated proteins in the cytoplasm. The
other cell should show a mix of mutated and non-mutated proteins. The best answers will also make clear the
nature of the mutation.
/6
3) Imagine a cell of Bacteria Q rapidly dividing within your intestine. This Bacteria Q cell needs
to undergo DNA replication as frequently and as quickly as possible. Your analysis of the crystal
structure of a Bacteria Q protein reveals that the active site of helicase is unusually small and unlikely
to bind quickly to DNA. How is Bacteria Q able to overcome this unusual feature of its proteins?
Mark all correct answers:
_______ Bacteria Q must be stealing enzymes from gut cells, because otherwise this difference would
render this necessary protein in Bacteria Q completely non-functional.
_______ Bacteria Q needs no stolen enzymes, since the difference in this protein is unlikely to
impact any important features of the bacterial molecular machinery.
_______ Bacteria Q would survive as long as all DNA molecules were modified by having more
hydrogen bonds to hold them together.
_______ Bacteria Q is likely to outcompete other gut bacteria that are rapidly dividing.
___X__ Bacteria Q is likely to be outcompeted for space by other rapidly dividing gut bacteria.
___X__ Bacteria Q may survive best in very hot environments.
Explanation: This protein has lost some (but not all) function in opening DNA. Stolen proteins might
be useful, but the first two answers each have additional clauses that make them incorrect. More
hydrogen bonds would make helicase more necessary, not less. Bacteria Q will replicate more slowly
and is likely to be outcompeted by other similar cells without this helicase deficiency. Lastly, hot
environments will help helicase function by making the breaking of DNA more likely, so this is
where Bacteria Q may have the best chance.
/4
4) The diagram shown is from the DNA of a mutant cell. The mutation is in one of the genes that
encode enzymes necessary for DNA replication. Using the little information shown in this diagram,
decipher which enzyme is encoded by the mutated gene.
Example diagram:
A short RNA polymer
Two long strands of DNA
The gene mutated in this scenario encodes the replication enzyme known as: DNA pol I or III
(partial credit for “DNA polymerase”)
KEY
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Exam 2
Name:__________KEY___________________
/10 5) Analyze the three mutated sequences below, and then place them in likely order of Most to Least
functional protein produced. Changes have been bolded/underlined where possible.
The wildtype mRNA sequence:
5’-AAGUGAUGGCAGUUAGCAGGGGGGUGAAUUGACGACCCUAUC-3’
Sequence D:
5’-AAGGUGAUGGCAGUUAGCAGGCCCGUGAAUUGACGACCCUAUC-3’
Sequence E:
5’-AAGUGAUGGCUAGUUAGCAGGGGGGUGAAUUGACGACCCUAUC-3’
Sequence F:
5’-AAGUGAUGGCAGUUAGCAGGGGGGUUUGACGACCCUAUC-3’
Most functional protein: D
Middle-est functional protein: F
Deletion of GAA
Least functional protein: E
Explain your reasoning for your choice of Least functional protein in 1-2 sentences, maximum.
The mutation in D impacts a signal amino acid, while the mutation in F impacts two amino acids (this
deletion is 3 base pairs but spans two codons. Mutation E is the least functional because the early
frameshift will produce a premature stop and a tiny, probably-non-functional protein.
Imagine that your research lab is developing a new antibiotic compound. Instead of trying to find a
single molecule that will be poisonous to bacterial cells, you are trying to insert a lethal enzyme.
Colleagues in another country have developed a way to insert the enzyme only into parasitic
bacteria, so human cells and useful bacteria will not be altered.
/5
6a) Which of these enzymatic activities is most likely to quickly decrease the life span of the
average parasitic bacteria? Mark the single most lethal enzyme.
______ Enzyme 1 inserts a ‘5-GGG-3’ after every 5’-UGG-3’ in all mRNAs
______ Enzyme 2 adds an RNA primer to DNA at random locations throughout the genome.
______ Enzyme 3 deletes every 12th occurrence of 5’-GCA-3’ in DNA.
___X__ Enzyme 4 cuts the rRNA responsible for binding to the RBS on mRNAs.
______ Enzyme 5 adds two phosphate groups to each monophosphate nucleotide it encounters.
Explain your answer in 1-2 sentences, maximum.
Enzyme 1 would be detrimental, but only by the addition of 1-2 amino acids in proteins where this
sequence appeared in the coding sequence. Since it impacts all RNAs, though, it is very possible that
this would cause a lethal change in at least one protein.
Enzyme 2 is doing what primase does, just all of the time. This is minimally damaging.
Enzyme 3 is similar to enzyme 1 in scope and impact.
Enzyme 5 is beneficial, as this needs to be done to allow transcription in the first place.
Enzyme 4 is the answer, because it will down-regulate the translation of every protein. This will slow
or stop many necessary processes in the cell and is the most detrimental.
KEY
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Exam 2
Name:__________KEY___________________
/5
Continued as 6b) When treated with the most lethal antibiotic, a few parasitic cells are still able
to survive longer than their kin. These cells are still able to produce enough invasive channel proteins
that steal mRNAs from the cytoplasm of nearby cells to stay alive for a longer period of time.
Specifically, why does this violation of the Central Dogma help these cells that need lots of specific
proteins? Describe how they are able to use this product to survive in 2-3 sentences, max.
Instead of following the standard mechanism of DNA!RNA!protein to do cellular work, this parasitic cell is
taking mRNAs from nearby cells. If they take enough (and get a bit lucky) they will be able to translate proteins
from RNAs that they did not produce, this fulfilling the protein needs of the cell.
/5
7) What is the role of the active site of RNA polymerase? Be sure to indicate the other
molecular components that directly interact with this component, and why those interactions are
important for normal function.
Describe the role of this component in 2-3 sentences, max.
The active site of RNA polymerase pulls apart the two strands of DNA in order to transiently
catalyze phosphodiester bonds on a growing chain of NTPs guided by nucleotide specific hydrogen
bonding. The active site ‘rudder’ component interacts with the DNA (similar to helicase in
replication) and the active site ‘zipper’ structure helps to break DNA/RNA hydrogen bonds so that
the RNA polymer is released.
/5
8) Imagine a mutation or error in the cell such that exit site of the ribosome is non-functional.
What is the impact on translation, and most importantly: why does this mutation cause that impact?
Simply indicating an overall impact will not garner points.
Explain the specific effect of this problem in translation in 2-3 sentences, max.
Because the exit site can not allow the presence of moving tRNAs, tRNAs will be stuck in the P-site and unable
to allow a new tRNA into the ribosome. This will stall the production of proteins at the first two amino-acids,
which will either simply inhabit the ribosome while attached to a tRNA or the ribosome will break apart leaving
only a di-peptide instead of a full protein.
Note: For the full version of the exam, the sequence will be filled in and other diagram details may change.
/10 9) 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
...3’..TACCGAGGTAATGAGGCAGTATTTACATGGCCTATGTTACAGTACTAGATTATA..5’....
...5’..ATGGCTCCATTACTCCGTCATAAATGTACCGGATACAATGTCATGATCTAATAT..3’....
Using the codon table on Page 1, translate the protein sequence from this gene.
Be sure to include the N- and C-termini. You can abbreviate amino acids using the 3-letter code.
Transcript = 5’-ATGACGGAGTAA-3’
Protein = N-Methionine-Threonine-GlutamicAcid-C
KEY
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