Name:
TRASK
Zool 3200: Cell Biology
Exam 1
1/30/15
Answer each of the following short answer questions in the space provided; circle the BEST
answer or answers for each multiple choice question. (70 points total)
Regarding the experimental results of Stanley Miller, what supports the conclusion that the
chemicals that comprised earth’s environment 3-4 billion years ago gave rise to life on this
planet? What are the scientific arguments against it? (2 points)
In support: Dr. Miller was able to produce organic molecules that comprise the building blocks of
biological molecules from atoms and small molecules that are predicted to have been available on earth 34 billion years ago.
Arguments against: The molecules that Dr. Miller’s experiment produced are amino acids. Because these
are unable to self-replicate, they are not likely to have been the biological molecules that gave rise to life
on earth
In order for cells to “rise from the primordial soup”, it was necessary that organic/biologic
molecules be contained. Which of the proposed methods of containment mentioned in class
make the most sense to you and why (in other words, compare the advantages/disadvantages or
likelihoods of each)? (2 points)
Two methods were discussed in class: the possibility proposed by Dr. Fox that suggested that proteins,
with turbulence, can form bubbles that provided enclosures to contain other biological molecules AND
the possibility proposed by the same Dr. Miller discussed in the question above—that life arose due to
containment of biological molecules in bubbles in ice. Arguments for the former possibility include that
Dr. Miller was able to produce amino acids in his earlier experiment, supporting the notion that proteins
may have existed to yield the proteinaceous bubbles that Dr. Fox proposed. Arguments for the latter
possibility include that containment could have occurred without additional biological molecules.
However, the colder temperatures necessary to produce ice also limit the rate of molecular movement, so
any contained molecules might be less likely to bump into each other—though if they did, they would
have a relatively consistent shape.
Explain how Fick’s law of diffusion influences the regulation of cell size. (2 points)
All cellular processes depend upon molecules bumping into one another so that chemical reactions can
occur with efficiency. The rate of this ‘bumping’ depends upon molecular diffusion. The factors that
Fick’s law states have influence over this rate include diffusion distance. If diffusion distance is kept
small, chemical reaction occur more efficiently.
Proteins that are destroyed by lysosomal enzymes are synthesized by: (1 point)
a. Ribosomes in the cytoplasm
b. Ribosomes at the endoplasmic reticulum
c. Both a and b
d. Neither a nor b
[Lysosomes degrade proteins originating from outside of the cell; they got to that location because they
were released via the secretory pathway.]
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The matrix of a mitochondrion best corresponds to: (1 point)
a. Nucleoplasm
b. Eukaryotic cytoplasm
c. Prokaryotic cytoplasm
d. Extracellular fluid
Explain your answer to the question above. (1 point)
Because mitochondria are hypothesized to have evolved from ingested bacteria, prokaryotic cytoplasm is
the most reasonable answer. However, mitochondria currently exist in the cytoplasm of eukaryotes, so
eukaryotic cytoplasm (b) is not a bad answer if it is supported.
Which membrane-bound structure(s) would you expect to be more abundant in cells that secrete
testosterone relative to cells that function to secrete a protein hormone like insulin? Explain your
answer. (2 points)
The endoplasmic reticulum (ER), specifically the regions without ribosomes (the ‘smooth’ ER) is the site
of lipid synthesis. Because testosterone is a lid-derived hormone, the smooth ER, or ER without
ribosomes would be expected to be most abundant. A cell secreting a protein hormone like insulin would
most likely have lots of ‘rough’ ER, or ER studded with protein-synthesizing ribosomes.
Which of the three types of cytoskeletal elements would likely be most abundant in an epithelial
cell that makes up the lining of your gastrointestinal tract? (1 point)
Because epithelial cells make up linings, it doesn’t move via flagella, nor does gastrointestinal epithelia
cells have cilia; thus microtubules are not abundant. These cells also don’t crawl or contract, so actin is
also not abundant. However, the structural role of epithelial cells suggests that they have a lot of
scaffolding; this cytoskeletal structure is maintained by intermediate filaments. Those who have taken
anatomy or histology realized that gastrointestinal epithelial cells do have surface microvilli, structures
maintained by polymerized actin. Therefore, if this clarification was indicated in your answer, full credit
was awarded.
Which cytoplasmic structures are involved in the destruction/degradation of molecules, and what
specifically are the molecular targets that each destroys/degrades? (6 points)



Lysosomes degrade molecules, specifically those that enter the cell via endocytosis from the
extracellular space.
Proteasomes degrade molecules, specifically those made by the cell that are no longer needed.
Peroxisomes degrade toxic or harmful molecules such as H2O2.
Did you ever see that ‘Honey, I Shrunk the Kids’ movie? What about ‘Inner Space’? Imagine
that you found yourself in front of the ‘shrinking’ machine that is in each of these movies.
You’ve been minimized to the size of a molecule! Not only that, but you’ve somehow gotten in
the predicament of having been injected into the cytoplasm of a eukaryotic cell! YIKES!
Describe the environment you experience in the cytoplasm and, assuming that you’ll be stuck
inside the cell until you can figure out how the hell to get out, where in the cell will you establish
residence? Why? (3 points)
The cytoplasmic environment is reducing or “harsh”, and oxygen-poor, similar to the environment on the
early earth (the ‘primordial soup’). It is crowded with organelles and macromolecules. Although
establishing residence in any organelle might be an acceptable answer as long as it was appropriately
supported, the most reasonable locations might be the lumen of the ER, the Golgi or a transport vesicle as
these all have internal environments similar to the oxidizing, oxygen-rich environment of the outside
world. The nucleus and mitochondria have environments similar to the cytoplasm, so establishing
residency here would gain you nothing beyond what is already present in the cytoplasm.
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A
B
C
D
In the image of cytoplasm above, proteins are colored blue, nucleic acids are pink, lipids are
yellow and polysaccharides are green. Using the image, answer the following questions:
Which side of the image represents the cell’s exterior surface (left / right)? (1 point)
Left [evident by the green carbohydrates that are only present on the extracellular side of the cell].
Which organelle is depicted in the bracketed region labeled ‘D’? (1 point)
The Golgi apparatus [evident by its three ‘stacks’, representing the cis, medial and trans stacks of this
organelle].
Which monomeric protein makes up the cytoskeletal element identified by ‘A’? (1 point)
Actin. The clues to the identification of this structure as a microfilament include that it is very small and
its location immediately under the plasma membrane.
Which cytoskeletal element is item ‘B’, and what is its primary function in the cytoplasm? (2
points)
This is likely a microtubule, which functions to move things around inside of a cell. It is also possible that
this structure is an intermediate filament; these function to provide cell structure and shape.
What cytosolic components are represented in the item labeled ‘C’ and where are they initially
produced? (2 points)
Because this spherical structure contains a lot of nucleic acid, it is likely a ribosome (containing rRNA).
Ribosomes, or at least their component ‘halves’, are assembled in the nucleolus of the nucleus.
Cell Biology is in the WSU catalog as ZOOL 3200, yet chemistry is a prerequisite for this class.
Explain why an understanding of chemical principles is necessary when trying to understand
cells and cellular processes. (2 points)
Cells are actually a mixture of chemicals that are enclosed and contained by a lipid membrane. All
cellular processes, then, depend upon reactions between the chemicals that comprise cells. Understanding
the foundational principles underlying the chemical reactions that guide, regulate and are integral to
cellular processes will facilitate an understanding of how cells, themselves, function.
The individual monomers that make up biological macromolecules are linked together using a
common chemical reaction. What is this chemical reaction, and why might it, as a reaction
mechanism, have been a favorable one for which evolution selected? (3 points)
The reaction that results in the synthesis of biological polymers is a condensation reaction. This type of
reaction is evolutionarily favorable because it produces water which is both abundant and non-toxic.
Further, the abundance of water makes it relatively easy to reverse the condensation reaction with
hydrolysis.
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‘Model organisms’ are often used in experiments to understand how human cells and humans, in
general, function. If you were interested in studying the processes below, what ‘model
organisms’ might you use to study each process and why? (4 points)
Repair of sunlight-induced DNA mutations:
Generally, ‘model organisms’ are used because they are relatively inexpensive, easily cared for and more
simple to understand. In studying light-induced DNA mutation, then, one might select the simplest
organism that has DNA, as long as it will be damaged by exposure to sunlight. A very reasonable model
organism in this case, then, might be prokaryotic cells like E. coli, or possibly a single-celled eukaryote
like yeast (S. cerevisiae).
The development of a spongiform disease like “mad cow” or Creutzfeldt-Jakob disease:
Because these spongiform diseases affect the brain, use of a single-celled organism as a model organism
in this case is not appropriate. Rather, an inexpensive, easy to care for, simple organism that has a brain,
preferably one organized in a manner similar to those of humans or cows, would be ideal. One possibility,
then, might be a rodent such as a mouse or a rat.
Complete the table below. (4 points)
Chemical Formula Class of biological molecule in which
monomeric polymerization results
C8H16O8
2 Functions of polymeric
molecules
1. Adhesion, structure, energy,
Carbohydrate/polysaccharide
2.communication
NH2CH2COOH
1.Perform work, structure,
Protein
2.communication
C9H16N30O14P3
1.Information storage, perform
Nucleic acids
2.work (RNA), information use
C17H31O2
1.energy, communication,
Fats/lipids
2.containment
A short polypeptide is shown below. Identify and clearly label the amino and carboxyl ends, as well
as each of the peptide bonds. Circle each “R” group. (3 points)
Amino-terminal end
Carboxyl-terminal end
Peptide Bonds
4
Define what is meant by a protein’s tertiary structure. Which type(s) of chemical bonds are most
responsible for maintaining these structures? Provide an example of a condition that might
disrupt a protein’s tertiary structure. (3 points)
A protein’s tertiary structure is its three-dimensional shape in space. The tertiary structure of a protein is
maintained primarily by non-covalent bonds and interactions between the R-groups of the amino acids
that make up the protein. Disruption of these non-covalent bonds can be easily accomplished by altering
environmental conditions such that salt concentrations, pH (hydrogen ion concentrations) or temperature
are significantly changed. Specifically, high salt, low pH and high heat will ‘denature’ or unfold a
protein’s 3D shape.
RNA has been proposed to have been the first nucleic acid to develop on earth due to its
variation in structure and its ability to catalyze chemical reactions. DNA, on the other hand, is
much less variable in its shape and cannot perform cellular work, yet it is currently used—almost
universally—to store and carry genetic information. Though we now know this to be true, it was
a point of contention as recently as 65 years ago. What were the arguments against DNA serving
a hereditary function, and what was the original data that first supported its currently-understood
function? (3 points)
While chromosomes were known to be copied and distributed during the process of cell division, proteins
were known to be comprised of both protein and nucleic acid. The primary reason behind the lack of
acceptance of DNA as the hereditary molecule is that it simply does not have the same degree of
variability as proteins—DNA is comprised of four different building blocks whereas proteins are
assembled from twenty! It seemed unreasonable that something as relatively invariant as DNA might
account for all of the variability seen among live organisms. Furthermore, cells were known to contain
much more protein than DNA—protein was simply more abundant. The data that first supported its
currently-understood function was produced by Avery, McCarty and McLeod. They used Griffiths’
virulent ‘S’ and non-virulent ‘R’ strains of bacteria to assess which of the biological molecules was able
to ‘transform’ non-virulent cells into virulent (deadly) ones. Following separation/purification of each
class of biological molecule from ‘S’ strain bacteria, they mixed each individual biological molecule with
‘R’ strain cells and injected the mixture into mice. Only the DNA (not the protein) was able to ‘transform’
‘R’ bacteria into virulent cells. Further, destruction of the DNA with a nuclease prevented this
transformation. Thus, DNA was the ‘transforming principle’ and the hereditary biological molecule.
To incorporate radioactively-labeled nucleotides into
newly synthesized DNA, researchers use phosphorus32-labeled nucleotides in a DNA synthesis
reaction where the  denotes the position of the
radioactive phosphate—the  phosphate is closest to
the 5’ carbon of deoxyribose, followed by the  and
then the  phosphate as shown in the picture to the left.
Explain why the  position and not the  or  position
is the best to use in these experiments. (2 points)
Only the  phosphorus of the nucleotide triphosphate
shown will be incorporated into a nucleic acid as it is synthesized. The  and  phosphates will be cleaved
from the nucleotide triphosphate and recycled when the phosphate forms a bond with the 3’ –OH of a
polynucleotide chain.
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If the total weight of macromolecules in a cell is 27 picograms, how many picograms of protein
would be predicted to comprise an average bacterial cell? (2 points)
Protein comprises ~15% of a cell’s weight [Per the figure in your textbook, water comprises 70%, RNA 6%,
DNA 1%, small molecules 4%, carbohydrates and lipids each 2% of an average bacterial cell]. Thus, the
macromolecules together comprise 26% if a cell’s total weight; approximately 60% [57.7%] of the weight of
macromolecules is made up of protein. So, if the total weight of a cell’s macromolecules is 27 picgrams
(pcgs), ~60% of this is protein. 57.7% x 27 pcgs (or 0.577 x 27) = ~15.6 pcgs. Full credit was given to any
number between 10-20 pcgs.
Which of the nitrogenous bases in RNA are pyrimidines? (2 points)
Cytosine and Uracil are the two nitrogenous bases in RNA.
Which of the diagrams to the right best represents the form
that DNA has? (1 point)
The image shown in A is the only right-handed helix.
T
C
Label the bases (i.e., A, T, G, C) in the diagram to the left. (2
points)
A
G
The DNA fragment in the figure below is double-stranded at each end, but has a single-stranded region
in the middle in one of the strands. Label the 5’ and 3’ ends of each strand. (2 points)
3’
5’
5’
3’
For a given stretch of DNA with the sequence 5’-CTGAGGCATCGTATCTGATGT-3’, what is the
complementary sequence, written in the 5’ 3’ direction, left to right? (writing DNA sequences in
this 5’ 3’ ‘direction’ is conventional; 1 point)
5’-ACATCAGATACGATGCCTCAG-3’
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Name two chromosomal regions that are associated with heterochromatin in all cells. (2 points)
Telomeres and Centromeres are associated with heterochromatin in all cells.
If the DNA from the nucleus of a human cell were stripped of its proteins, the chromosomes were
straightened and then aligned tightly side-by-side, how wide would the genome be? (2 points)
The DNA double helix is 2 nanometers (nm) in width. There are 23 pairs of chromosomes (46 total
chromosomes) in a human cell. If these 46 chromosomes were stripped of their proteins, straightened, and
aligned side-by-side, the genome would be 2 nm x 46, or 96 nm wide.
DNA is associated with proteins so that it can be packed into the nucleus in an organized manner.
The DNA that will be ‘used’ more frequently is less-tightly packaged and associates with fewer
proteins. This DNA is in the form of
euchromatin
. (1 point)
Isoleucine (Ile) is an essential amino acid, yet too much of it can lead to kidney failure if it is not
properly destroyed. Normal degradation of Ile depends upon an enzyme coded by a gene that is
located on human chromosome 11. Mutations within this gene result in the production of a nonfunctional enzyme; this leads to improper destruction of Ile, which leads to the production of a
harmful byproduct that ultimately affects kidney function. Through an errant recombination event
during meiosis, a mutant sperm cell in which this gene has been translocated (i.e., moved) to the X
chromosome is generated. Although the gene is moved to a different chromosome, it remains intact
and un-mutated. Will simply moving the gene for Ile destruction to the X chromosome have any
effect on its ability to degrade ingested Ile? If so, what effects would you expect in an offspring that
was produced with this mutant sperm? Explain. (3 points)
Even though the translocated gene can function, it may not be able to be ‘turned on’, or transcribed, in all
cells. In a female offspring, which this would be because the sperm used to produce it carried the X
chromosome with the translocated gene for Ile destruction, one of the two X chromosomes will be condensed
into a heterochromatic Barr body from which no transcription will ever occur. This condensation is random,
with a random distribution of cells turning off/condensing the maternal X and a random distribution turning
off/condensing the paternal X. This female offspring will have one copy of the Ile destruction gene on her
maternal chromosome 11, and a second copy of the Ile destruction gene on her paternal X. If the majority
of—or even the correct—cells condense the maternal X, she will still have two functional copies of the Ile
destruction gene that can be transcribed and translated into protein. She’ll be fine. BUT, if the majority (or
particularly detrimental) cells condense the paternal X, the girl will be left with only one copy of the Ile
destruction gene that can be used. She will be able to make the Ile-destroying enzyme, but she’ll only be able
to produce roughly half the normal amount. This might be sufficient to keep her safe and healthy, OR it might
leave her with a build-up of Ile, which would cause kidney damage.
—So, bottom line, she might be fine, but she might not. It just depends upon where and how much of her
paternal X is condensed into a Barr body.
—[For those of you who have already taken genetics, we have to assume that the ‘random distribution’ of
chromosomes during meiosis resulted in the ‘empty’ (non-Ile gene) chromosome 11 to co-distribue with the
X chromosome to which it was translocated—in other words, that the sperm had a chromosome combination
that contained only 1 copy of the Ile destruction gene, and that the one copy was on the X chromosome. For
those of you who have not yet taken genetics, ignore all of this!]
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