Download Exam 2 Full KEY v1 Bio200 Sum12

Biology 200, Summer 2012
Exam 2
KEY v1
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Name: _________KEY Version 1____________________
Biology 200
Summer Quarter 2012
Exam #2
DO NOT OPEN EXAM UNTIL DIRECTED TO DO SO
Make sure you have 6 pages, including this one. 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 it is often difficult to give 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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Biology 200, Summer 2012
Exam 2
Name: _________KEY Version 1____________________
1. (18 points) Fill in the blank
For each, write TWO different answers that fit the description given. Be as specific as possible. It is acceptable
to use the same answer for different questions. The full 3 points will be awarded for a correct pair of answers
(+1 for a single correct answer).
Ex)
A type of gene implicated in cancer
i. tumor suppressor
ii. oncogene
progression or mutation
a)
A metabolic energy carrier that is utilized before i. __ATP____________ ii. ___NAD+_________
the linking step and after glucose phosphorylation.
b)
An alternative to aerobic respiration
that occurs in the absence of oxygen.
c)
A transcription factor.
d)
A molecule with lower redox potential than
the electron-carrier Q (ubiquinone).
i. __fermentation_____ ii. __eating sulfide gas_
i. __CAP protein______ ii. ___silencer protein__
i. ___Complex III_____ ii. ___O2___________
e)
Growth media conditions that result in very
i. ___high lactose_____ ii. ____low glucose____
little ß-galactosidase production.
2. [15 points] Organelles
For each organelle, describe the function of an enzyme that you would expect to find at that organelle and not
in most other parts of the cell. The name of the enzyme is not necessary. You should mention a possible
substrate and a reaction that is catalyzed. Answer each in one sentence or less. Be as specific as possible. Also,
fill in the blanks with the number of the correct microenvironment that is created by the organelle. Each can be
used only one time. +2 for each answer enzyme, +1 for each microenvironment.
Microenvironments
__7_Example) Chloroplast: Enzymes change small carbohydrates into
4) dilute & aqueous
5) acidic
larger carbohydrates using the energy from sunlight.
____a) Rough ER:
An enzyme that alters the folding of a protein that is designed to be used in the
hydrophobic layer at the outer membrane of the cell.
____b) Golgi apparatus:
An enzyme that helps to package proteins with similar destinations together
8) basic
10) allows sunlight in
12) rich in lipid building
blocks
13) low in mutating agents
15) rich in traffic-directing
signal proteins
18) rich in motor proteins
19) non-cytoplasmic
protein folding conditions
into small vesicles.
____c) Lysosome:
An enzyme that breaks down lipids into isoprene units and works best in an acidic environment.
____d) Nucleus:
An enzyme that fixes mutations on a single strand of DNA and replaces them with the correct base pair.
____e) Cytoskeleton:
A motor protein that uses ATP to drag a vesicle from the interior of the cell to the external membrane.
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Biology 200, Summer 2012
Exam 2
Name: _________KEY Version 1____________________
3. [21 points] Gene Regulation
a) (6 pts) You are developing an understanding of the similarities and differences between prokaryotes and
eukaryotes. In 3-4 bullet points, describe 3 of the advantages that a single-celled prokaryote has over similar
eukaryotes. Be as specific, clear, concise and thorough as possible. Complete sentences are not necessary.
1. More efficient use of glucose energy with no membranes for NADH to cross
2. Simpler gene regulation requires building fewer transcription factor proteins for greater efficiency
3. Smaller cell size concentrates cellular reactions so no need for organelle construction
100
b) (3 pts) The Silver gene promoter: (Check ALL true answers):
____ Binds basal transcription factors tightly
_X_ Binds basal trancription factors poorly
____ Binds to regulatory transcription factors all of the time
____ Is NOT a double-stranded region of DNA
Black
AMOUNT of mRNA
The data graph shows the total cellular levels of mRNA from 4
eukaryotic genes. Below the graph is a schematic of the transcription
and translation of the “Purple” gene. Each circle in the Purple
protein represents an amino acid.
80
60 Purple
Gold
40
Silver
20
5 min
10 min
15 min
Time
20 min
c) (3 pts) The Black gene is undergoing histone modification at Time 0. What type of change do you predict is
taking place and why? Enter the correct word choice into the following.
This causes [More or Less] acetylation of the histones? _More_____
This change in negative charges on the histone causes DNA to be [More or Less] tightly packed? _Less__
d) (3 pts) Imagine that over time, the Gold protein becomes more often targeted for modification by a ubiquitin
ligase. How would the level of the gold protein change over time and why? Check in front of ALL correct
answers.
___ Gold protein levels stay steady because the mRNA level is steady
___ Gold protein levels increase because Ubiquitin stabilizes proteins
_X_ Gold protein levels decrease because ubiquitin marks proteins for degradation
___ Gold protein levels are unpredictable because of the CAP protein
e) (3 pts) The Purple protein can have either 10 or 13 amino acids. There is no
change to the Purple gene DNA. Explain, including the location of the
mechanism in the cell, in one sentence or less.
DNA of Purple gene
Immature mRNA of Purple
gene
Purple protein
forms
Alternative splicing of the Purple protein into two variants in the cell nucleus (at the spliceosome) creates different
proteins from the same original DNA code.
f) (3 pts) Which statements could explain the Purple data over time? (Check in front of ALL correct answers.)
__X_ At time 1 and 18, the conditions favor the activation of an enhancer protein for this gene
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Biology 200, Summer 2012
Exam 2
Name: _________KEY Version 1____________________
__X_ At time 12, the conditions favor the degradation of an enhancer protein for this gene
____ At time 1 and 18, the conditions favor the binding of a repressor
____ At time 12, the conditions favor the degradation of a silencer protein for this gene
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Biology 200, Summer 2012
Exam 2
Name: _________KEY Version 1____________________
4. (16 points) The Kev Operon
5'
3'
LBS2
LBS3
P
Or1
Or2
ATG KevX TAG ATG KevW TGA ATG
AUG
UAG AUG
UGA AUG
KevP
TAA
3'
5'
UAA
In the prokaryotic dsDNA shown above, Or1
and Or2 are operator sequences that can be bound by the repressor proteins R1 and R2, respectively. R1 and R2
are similar to the E. coli LacI protein. The coding regions for genes KevX, KevW and KevP are shown as well as
the DNA sequences that encode the start and stop codons for each gene.
LBS2 and LBS3 are DNA sequences that can be bound by the proteins T2 and T3, respectively. P is the promoter
for the Kev operon. The proteins KevX and KevW are used to synthesize temporary protective proteins in the
event of an attack by amoeba predators.
a) (4 pts) This organism can recognize when there are no amoebae present. The prokaryote will react efficiently.
When amoeba are gone: (Check in front of ALL correct answers.)
____ KEV X will be bound to LBS 2
____ KEV Y will be bound to OR2
____ T2 will be bound to LBS 2
____ R2 will be bound to LBS 3
_X__ R1 will be bound to OR 1
____ T2 will be bound to OR 2
b) (4 pts) A mutation is analyzed that alters the UGA codon to a CGA codon. What would be the consequence of
that mutation to the protein and to the cell? (Check in front of ALL likely consequences.)
_X__ The Kev mRNAs are produced normally
____ The Kev mRNAs have major errors
____ The Kev proteins are produced normally
____ The cell can fight amoeba normally
_X__ The Kev proteins have major errors
_X__ The cell is more vulnerable to amoeba
c) (4 pts) T3 is a protein that activates transcription of the Kev operon. T2 lowers expression of Kev mRNA. Both
proteins have DNA binding sites that are fairly similar (LBS2 and LBS3 have similar sequences). Besides the
DNA binding region, what is true about the T2 and T3 proteins? (Check in front of ALL correct answers.)
_X_ A surface of T2 protein prohibits RNA polymerase binding
___ A surface of T3 protein prohibits RNA polymerase binding
___ A surface of T2 protein recruits RNA polymerase binding
___ A surface of T3 protein recruits RNA polymerase binding
___ T2 allosterically inhibits T3
___ T3 allosterically inhibits T2
___ T2 is acting like the lac operon CAP protein
_X_ T3 is acting like the lac operon CAP protein
d) (4 pts) The KevP gene encodes a protein that can be released from the prokaryotes and digested by amoeba. If
not digested, this protein will diffuse back into the prokaryote and bind tightly between P and Or1. Why is this
logical for the regulation of Kev gene expression? (Answer in one sentence or less).
The feedback inhibition of this KevP gene keeps the cell efficient by ensuring that when amoeba are absent the higher
concentration of KevP shuts down the unneeded transcription.
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Biology 200, Summer 2012
Exam 2
Name: _________KEY Version 1____________________
5. [21 points] Cellular Respiration
Shown below are several sets of 3
components from aerobic respiration. Put
each set in correct order by labeling the first
of the set with a “1”, second = “2”, 3 = last.
You must have the entire order correct for
credit, but there may be multiple correct
orders for a single set. (3 points each)
Example)_1_Glucose __3_ Krebs Cycle _2_Pyruvate
a)_3_ ATP synthase
_1_ Linking step
_2_ Q electron transporter
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
b) _2_ Acetyl-CoA
_3_ proton motive force
_1_ commitment step
c) _3_ production of water
_2_ FADH2
_1_ pyruvate
d) _1_ cytoplasmic NADH
_2_ production of CO2
_3_ Complex IV
(3 pts)
Eukaryotic
mitochondria
require
a huge amount of ADP in order to create ATP. Where does
e)e)_1_
cutting
a carbohydrate
into two
pieces
all_3_
of athat
cellular
ADP come
from? Answer in 2 sentences or less.
series
of oxidation
reactions
_2_ transport into the mitochondria
Every coupled enzymatic reaction in the mitochondria that uses ATP will generate a Pi and a molecule of ADP,
which are ready for re-binding at the ATP synthase enzyme complex.
f) (3 pts) Choose one of the two statements below. Disagree with the statement, and support your
argument. Make your case in 2 sentences or less. You only need to complete one argument.
Statement #1: The purpose of the Kreb’s cycle is to oxidize carbon-containing molecules.
Statement #2: A eukaryotic organism would rather consume pyruvate than glucose.
I disagree with statement # ____ because:
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Biology 200, Summer 2012
Exam 2
Name: _________KEY Version 1____________________
Take Home: Cancer History
Cancer is a complex and extremely diverse system of related diseases. We know that these diseases are the
result of multiple mutations in cells causing an array of intracellular changes. No single mutation is cancer.
Somehow, the combinations of multiple changes lead to malignant unregulated cell growth.
6. (8 points) Your task is to define a few possible mutations that could contribute to cancer. For each mutation,
you should indicate as specifically as possible how the mutation occurred, where in the cell and in the body
the mutated cell is located, and the mechanism that allows this mutation to lead to cancer. Be creative where
necessary. You should do this in less than one sentence for each mutation (If necessary, you can use two short
sentences). Research outside of Bio200 lectures and labs is not necessary, but is allowed if you want to find
specific examples of parts of this question. Show the diversity of what you know: mutation descriptions
should be as different from each other as is possible while still being specific and correct.
Example) This mutation is in a gene that encodes a signaling molecule to start apoptosis.
A random DNA polymerase III error in a white blood cell’s signal receptor gene causes the loss of social
control so that the cell won’t kill itself when it picks up further DNA damage.
a) This mutation is in a gene that either encodes a cytoskeletal molecule or an enzyme that works on a
cytoskeletal molecule.
The promoter region of an actin polymerase is changed to more efficiently bind RNA polymerase to that
actin filaments continuously grow causing a fatty skin cell to begin to crawl towards blood vessels and
metastasize to other parts of the body.
b) This mutation is in a gene whose product works in or travels through the Golgi apparatus.
A Golgi located enzyme within a neuron packages growth factors to release from the cell too often which
tells other nearby neurons to go through too many rounds of mitosis forming a small brain tumor.
c) This mutation affects the passage of the cell through the cell cycle.
A 4-bp insertion into the Rb gene renders it non-functional in corneal cells in the eye, causing E2F to
increase passage through the S-phase checkpoint too often and leading to a greater chance of corneal tumor
growth.
9. (4 points) Given the mutations you’ve described, imagine a cell with all three mutations. Write a description
of an additional mutation that would be needed for this cell to progress further towards outright malignant
cancer. Again, a single sentence should be enough room to be specific, creative, and correct.
Within this cell, UV-damage causes a thymine dimer to mistakenly recruit enhancer proteins to the
region around the promoter for a telomerase protein which turns telomerase on in the cell allowing it to
survive many rounds of cell division without losing DNA.
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