Biology 200
Sample Midterm Exam
Total time: 90 min.
Section: _______________
Family Name: __________________ Given Name: __________________Student #: ___________
(please print)
Instructions:
1. Use black or blue PEN only. Exams that are written in pencil, erasable ink, or that
have corrective tape on anywhere on the exam will not be eligible for re-grading. A
regrade request requires that the exam does not look modified in any way.
2. There are 6 questions on 9 pages. Answer all questions in this exam booklet.
3. You are allowed an 8.5x11 inch, double-sided, hand-written memory aid for this
exam. All memory aids that do not conform to these rules will be taken away.
4. You will have 90 minutes for this exam.
Please write your name and student number on
both the first page AND page 8 of this exam.
Your grade will be written on page 8.
Page 1 of 9
Question 1. (____/5 marks)
When doing research, scientists will often image the same cell using multiple different microscopy
techniques. In this case, the scientists were studying chloroplasts.
A. What type of microscopy was used to generate each of these two images? How do you know?
(3 marks)
Type of microscopy
How do you know?
Image 1:
Bright background, see multiple cells, can’t see
Brightfield
a lot of details.
LOW RESOLUTION
Image 2:
TEM
Can see all the structures of the cell and all the cell
components.
HIGH RESOLUTION
B. The cell in Image 2 is one of the cells seen in Image 1. In image 2, only 6 chloroplasts can be
seen, however in Image 1 there is not a single cell visible with only 6 chloroplasts. Explain how this is
possible. (1 mark)
Have to use cross-section for TEM, you have to slice the sample into thin slices in order to observe the cell.
SINGLE PLANE OF SECTION
C. State one advantage of using the type of microscopy used to generate image 1 over image 2.
(1 mark)
Can use live cells for Brightfield.
Page 2 of 9
Question 2 (6 marks)
For each of the experimental observations below, please state what it tells us about cellular structure
and/or function, and explain why.
A. Nucleoli become heavily radiolabeled when radioactive ribonucleotides are provided to the cell.
It shows that RNA are heavily needed for transcription in the nucleoli - such as mRNA, tRNA, rRNA.
RNA MOVES INTO THE NUCLEUS, ARE INCORPORATED TO rRNA AS IT IS BEING TRANSCRIBED IN THE
NUCLEOLUS
B. When enzymes isolated from cells are heated, they can no longer catalyze a reaction.
Heat denature enzymes so that they can’t catalyze a reaction in the cell.
HEAT DENATURES PROTEINS BY DISRUPTING ALL THE NON-COVALENT BONDS THAN MAINTAIN THE SECONDARY
AND TERTIARY STRUCTURES. THIS MEANS THAT 3D PROTEIN STRUCTURE IS IMPORTANT FOR ENZYME FUNCTION.
C. If the NLS from a nuclear protein is experimentally added to hexokinase (a cytoplasmic protein),
the altered hexokinase is detected in the nucleus.
It shows that the NLS is responsible and necessary for import of protein into the nucleus.
D. When a mutation is introduced at an exon-intron junction of the b-globin gene, this results in the
production of a longer mature mRNA transcript.
This shows that the exon-intron junction has a specific code that allows for snRNP to bind to it and remove the intron.
With mutation, the snRNP cannot recognize the code and therefore would not remove the intron so the mature mRNA would
be longer.
SEQUENCE AT THE JUNCTION IS CRITICAL FOR RECOGNITION BY THE SPLICEOSOME TO REMOVE THE INTRON
OF THE PRE-RNA
E. If mammalian cells are treated with hormones and then imaged using TEM, the results show
increases in the amount of euchromatin relative to heterochromatin in the nucleus.
Euchromatin is the decondensed DNA and it transciptionally active.
This shows that hormones are needed in order for the DNA to be transcribed
The increased euchromatin compared to heterochromatin correlates with increases in
transcriptional activity in response to hormone treatment. This suggests that hormone
treatment stimulates transcription and euchromatin is the transcriptionally active form of
chromatin
F. If cells are treated with SDS (a strong detergent) membrane proteins can be separated form
phospholipids.
A strong detergent is only needed to remove proteins with a strong covalent bond. This shows that membrane proteins
are tightly bonded to the phospholipids by hydrophilic and hydrophobic interactions.
SDS is structurally amphipathic and can form detergent micelles. Detergent miscelles can separate
membrane proteins from membrane phospholipids due to their long hydrophobic tails which associate
with the hydrophobic regions of membrane proteins and the fatty acid tails of phospholipids which
causes the membrane to disassemble.
Page 3 of 9
Question 4 (___ /8 marks)
This is the hydropathy plot of a new membrane protein of the plasma membrane of a eukaryotic
organism.
Further biochemical analysis indicated that this protein is a glycoprotein with a polysaccharide chain
located at its N-terminal end.
A. Predict how the protein will be arranged in the membrane. Draw a diagram and indicate the
plasma membrane, cytosol, extracellular space, N and C termini of the protein. (4 marks)
Only one integral protein in the membrane and rest are in the cytosol or extracellular space.
C and N termini both in the cytosol with N on the left and C on the right.
B. (4 marks) You want to examine a second membrane protein using trypsin digestion followed by
SDS-PAGE. This integral protein is a single membrane-spanning alpha-helix, with its N-terminus on
the extracellular side of the membrane. Predict the results that you would expect to see in this
experiment by drawing protein bands in the gel image for the treatments indicated below. As an
example, Lane 1 has been drawn for you.
Lane 1 (shown): Intact cells, no trypsin treatment
Lane 2: Intact cells treated only with a solution
of high salt concentration
Lane 3: Brief detergent treatment, no trypsin treatment
Lane 4: Trypsin treatment of intact cells
Lane 5: Brief detergent treatment first, followed by trypsin
treatment
1
2
3
4
5
Page 4 of 9
Question 4 (7 marks)
RasGTPase (Ras) is a lipid-linked, plasma membrane-associated protein involved in cell signalling.
Goodwin et al. tagged Ras with GFP and expressed it in cells to examine its mobility by Fluorescence
Recovery After Photobleaching (FRAP). They compared the mobility of Ras in the presence (+) or
absence (-) of a drug called methyl-β-cyclodextrin (MβCD) that removes cholesterol from the
membrane.
Data adapted from Goodwin et.al. 2005. Biophysical Journal. Vol. 89:2 1398-1410
A. Compare the FRAP recovery curves in the upper and lower sets of images. Make sure to indicate
which one is the control (& how you know it's the control) and how fluorescence recovery is
influenced by the test condition. (2 marks)
The one without the drug is the control because it is being added to the cell. With the drug, the recovery is faster whereas
without the drug, the recovery of FRAP is slower.
because treatment with MβCD is the variable, so the
untreated condition provides the basis for comparison to the treated cells
B. Based on the data, draw what you would expect the fluorescence recovery curves for each
condition to look like? Draw and label both curves on the axis below, as well as the x and y axis.
(2 marks)
(-) (+)
(+)
(-)
Fluorescence intensity
Fluorescence recovery
Time
C. Based on this data, explain what you can conclude about how the mobility of Ras within the
plasma membrane may be regulated. (1 mark)
The mobility of Ras within the plasma membrane is regulated (INFLUENCE) by the presence of cholesterol.
In the presence of cholesterol, Ras is less mobile. (CHOLESTEROL RESTRICTS THE MOBILITY OF RAS)
D. Predict how the mobility of Ras might differ in an artificial lipid bilayer (made entirely of
phospholipids) compared to a cellular plasma membrane? Explain. (2 marks)
The mobility of Ras would be higher because there are no cholesterol in an artificial lipid bilayer. Cholesterol is shown to regulated
the mobility of Ras and when it is removed, the FRAP recovery is faster - the mobility become faster. So in an artificial lipid bilayer,
like the experiment where cholesterol is removed, the Ras would move slower.
The mobility of Ras would be faster in an artificial phospholipid membrane because lateral diffusion is not restricted by:
1) lipid composition, presence of cholesterol, lipid rafts
Page 5 of 9
Question 5 (____/7 marks)
Histones can be modified by the addition of acetyl groups (
)
(termed “acetylation”), and this modification affects packaging of
chromatin.
Oppikofer et al. (2011) conducted the following experiment to
examine the effect of acetylation on chromatin structure. In this
experiment, chromatin (with either unmodified or acetylated histones)
was isolated and exposed to micrococcal nuclease for different
periods of time (60 s compared to 120 s).
A. What is the purpose of analyzing chromatin that has not been
exposed to nuclease? (1 mark)
To see if unmodified and acetylated histones will affect the DNA.
It’s a CONTROL that shows how DNA from intact (or undigested)
chromatin runs/looks like in a gel
B. What is the purpose of carrying out nuclease digestion experiments
on DNA with unmodified histones? (1 mark)
So you can compare the affect on DNA with acetylated histones.
Serves as a CONTROL. Tells you what the banding pattern looks like
when DNA HAS NOT BEEN MODIFIED.
C. Describe how the DNA banding patterns change in response to the different treatments. Explain
what has happened to the DNA in each lane, to give this pattern. (3 marks)
Lane 1: DNA has unmodified histones and no nuclease so all DNA is present and it has more than 1200 bp.
Lane 2: DNA has unmodified histones and went through nuclease digestion for 60s. Some parts have been digested
and the rest of the DNA strands have bp ranging from 150 to 1200 with about 200 bp of difference for each strand.
Lane 3: DNA has unmodified histones and went through nuclease digestion for 120s. More parts have been digested
and the rest of the DNA strands have bp ranging from 150 to 800 with bout 200 bp of difference for each strand.
Lane 4: DNA has undergone acetylation and no nuclease. All DNA is present and it has more than 1200 bp.
Lane 5: DNA has undergone acetylation and went through nuclease digestion for 60s. Some parts have been digested
and the rest of the DNA strands have bp ranging from 150 to 1000 with about 200 bp of difference for each strand.
Lane 6: DNA has undergone acetylation and went through nuclease digestion for 120s. Almost all of the DNA has
been digested and only one strand of DNA that is about 150 bp is left.
Lane 1 had no nuclease treatment, so the DNA in the gel is seen as one band at high molecular weight (very long), as
it was in the original intact chromatin (i.e. it is a control).
Lane 2 shows many discreet bands indicating digested DNA fragments of varying sizes. This indicates a repeating
structure within chromatin that protected the DNA from nuclease digestion (i.e. the nucleosome)
Lane 3 shows the same banding pattern as Lane 2, except that the two highest molecular weight DNA bands are
missing. This is due to the longer time of digestion, so the DNA was further digested due to the longer nuclease
exposure.
Lane 4 is the same as Lane 1 (i.e. control)
Lane 5 is similar to Lane 2, but with a missing high-molecular weight DNA bands – indicating that the nuclease is able
to access a greater proportion of DNA more quickly/easily –compared to without histone acetylation
Lane 6 has only the smallest DNA fragment band of the chromatin. This indicates that the DNA from acetylated
chromatin is easily accessed by the nuclease so it can be digested more quickly.
D. Based on this data, what can you conclude about the role of acetylation on histone function and
chromatin structure? Explain. (2 mark
Chromatin is made of DNA wrapped around an octamer of core histones, and further packaging by H1 histone. This
protects and compacts the DNA in the nucleus.
Histone Acetylation loosens chromatin from a compacted, and inaccessible form to one that is more
accessible to nucleases (1 mark).
In the cell, histone acetylation allows for the unpacking of heterochromatin into euchromatin (open
chromatin) to promote transcription or replication (1 mark).
Page 6 of 9
Question 6 – Essay Outline (10 marks)
One of the course level goals of BIOL200 is that you are able to recognize that “all aspects of a
protein’s function are ultimately encoded in its primary amino acid sequence”. Histone H2A is a core
histone protein, involved in the organization and packing of DNA into the nucleus.
Provide a thesis statement and three supporting arguments that describe the importance of the
properties of Histone H2A’s amino acids on its non-covalent interactions with other histones, the DNA
and any other molecules in its environment, that give rise to this proteins’s structure and function.
Thesis statement: (1 sentence recommended, 2 sentences MAX)
•
- some parts are positively charged to interact with the DNA (negatively charged)
- some parts are non-polar to interact with other histones
- the tails of the H2A can have acetylation, methylation, and phosphorylation
Score: _____/1
Argument 1 and evidence: (2 sentences recommended, 3 sentences MAX 1 sentence per bullet)
•
•
•
Score: _____/2
Argument 2 and evidence: (2 sentences recommended, 3 sentences MAX 1 sentence per bullet)
•
•
•
Score: _____/2
Argument 3 on next page
Page 7 of 9
Argument 3 and evidence: (2 sentences recommended, 3 sentences MAX 1 sentence per bullet)
•
•
•
Score: _____/2
Essay Outline Organization: ___/3
Total Essay Outline Score: ______
/ 10
THE SPACE OUTSIDE OF THE BOXES WILL NOT BE MARKED
Family name: ________________ Given name: ___________________ Student #: _____________
For Marking Purposes do not write in this table.
BIOL 200 Midterm Grades
Question
1
2
3
4
5
6
TOTAL
5
6
8
7
7
10
43
Student
Score
Maximum
Score
Page 8 of 9
THIS SPACE WILL NOT BE MARKED – USE IT FOR ROUGH NOTES
Page 9 of 9