1-The “hydrophobic effect” is the driving force behind the spontaneous assembly of the lipid bilayer of cell membranes in water. What does that mean exactly?
- It means that the non-polar tails of phospholipids are attracted to one another, forming bonds that hold the membrane in place.
- It means that the polar structures (i.e. water molecules and the head groups of phospholipids) are attracted one another, and their clustering repels non-polar structures away, which pushes non-polar tails of phospholipids into a bilayer.
2-What is a TRUE statement about “trans-unsaturated” fatty acid tails?
- Trans-unsaturated fatty acid chains contain carbon double bonds, while cis-unsaturated fatty acid chains do not.
- Trans unsaturated fatty acid chains contain carbon double bonds, but the chains contain kinks and are difficult to pack tightly together.
- Trans-unsaturated fatty acid chains contain carbon double bonds, but the chains are linear and pack together tightly similarly to how saturated fatty acid tails do.
- Trans unsaturated fatty acids chains do not contain any carbon double bonds. That is why the chains are linear and pack tightly together.
3-What is a TRUE statement about active transport?
a) Active transport is always required to move a solute from an area where it is MORE highly concentrated to an area where it LESS concentrated.
b) lons are so small that they pass easily through lipid bilayers, and therefore do not require active transport.
c) Active transport can be powered by energy from ATP, light, or an electrochemical gradient.
d) Channel proteins carry out active transport.
4-Below is a diagram of a gut epithelial cell. Nat /glucose symporters are located on the edge of the cell facing the gut lumen. Glucose uniporters and Na+/K+ pumps (antiporters) are located on the other edge of the cell. What would happen to the cell if the Na+/K+ pumps stopped functioning?
a)Glucose would build up in the cell, and be unable to exit.
b) The concentration of Na+ would drastically drop in the cell.
c)The cell would be unable to bring glucose into the cell, because the electrochemical gradient of Nat across the membrane from the gut lumen would be lost.
5-How does the protein Adaptin contribute to clathrin-mediated endocytosis?
a) Curved Adaptin proteins bind to each other to form a rounded basket
b) Adaptin proteins bind to cargo protein receptors, and then recruit and bind Clathrin proteins
c) Adaptin protein pinches a thin stretch of membrane to release a coated vesicle into the cytoplasm.
d) Adaptin proteins drive the Clathrin coat to disassemble.
6-What do SNARE proteins do?
a) SNARE proteins tether vesicles to target membranes, but they do mediate the fusion of vesicles with target membranes.
b) SNARE proteins mediate the fusion of vesicles with target membranes, but they can only do this once the vesicle has been tethered to the membrane.
7-What makes misfolded prion proteins so toxic?
a) Hydrophobic interactions between misfolded prion proteins prevent them from binding to one another into a large tertiary structure the cell depends on.
b) Hydrophobic interactions between misfolded prion proteins cause many of them to aggregate together into very large stable structures the cell cannot destroy.
8-What is the function of the E3 protein in the ubiquitin pathway?
a) A chain of E3 proteins is added to proteins that need to be destroyed, and function as a lag for degradation.
b) E3 binds the target protein, thereby bringing it close enough to the E2 protein that E2 can transfer ubiquitin to the protein.
c) E3 binds ubiquitin, thereby activating it.
d) E3 is a large complex of proteins shaped like a cylinder, that proteins tagged for degradation are deposited into to be destroyed.
9-Would you expected gene expression to be high or low in stretches of DNA that are wound around histone tails that have been acetylated or phosphorylated?
a) Gene expression would be higher, because the DNA would be more loosely bound to the histones.
b) Gene expression would be lower, because the DNA would be more tightly bound to the histones.
c) Gene expression would be higher, because the DNA would be more tightly bound to the histones.
d) Gene expression would be lower, because the DNA would be more loosely bound to the histones.
10-How does a nuclear localization signal (NLS) work?
a) An NLS sequence is part of the Importin protein. This enables Importin to bind to proteins destined to be cargo moved from the cytosol into the nucleus.
b) An NLS sequence is a tag on cargo destined to be moved from the cytosol into the nucleus.
c) An NLS sequence is part of Ran GTPase. This enables Ran to bind to and activate Importin, thereby allowing Importin to move into the nucleus.
11)What is a TRUE statement about GTPases?
a) GTPases are in an active conformation (shape) when they are bound to GDP.
b) When GTPases are bound to GTP, GEF proteins can hydrolyze GTP to GDP, which changes the conformation of the GTPase.
c)When GTPases are bound to GDP, GEF proteins can add a phosphate group to the GDP, thereby creating a GTP molecule.
d) When GTPases are bound to GDP, GEF proteins can cause them to drop the GDP, opening up their nucleotide binding site for a new GTP to bind.
12)The cis Golgi network refers to the side that is positioned most closely..
a) To the plasma membrane
b) To the nucleus
13)A single-pass transmembrane protein is created in the endoplasmic reticulum (ER). Its N terminus is exposed to the cytosol, and its C terminus is exposed to the ER lumen. The protein is later packaged into a vesicle that is delivered to the plasma membrane. Once located on the plasma membrane, the N terminus of the transmembrane protein…
a)…continues to be located in the cellular cytosol
b)…is now exposed to the extracellular space
14)Why will future cancer treatments potentially rely on an understanding of glycosylation?
a) Because sugars covering the mitochondria of cancerous cells prevent the release of cytochrome C, thus inhibiting apoptosis
b) Because sugars covering the outer surface of cancerous cells interact with (and sometimes trick) immune cells, allowing cancerous cells to evade detection.
c) Because sugars covering extracellular matrix encourages the attachment of cancerous cells, thereby promoting metastasis.
15)What feature of mitochondria described below is considered to be evidence supporting the
endosymbiont hypothesis?
a) Mitochondria contain nuclei.
b) Mitochondria have their own DNA and ribosomes, which resemble the DNA and ribosomes found in bacteria.
c) Mitochondria have single membranes, rather than double membranes.
d) Mitochondria can survive and reproduce independently, outside of eukaryotic cells.
16)What are cristae?
a) Cristae are proteins that assemble around the middle of a mitochondrion, marking where the mitochondrion will divide into two.
b) Cristae are giant mitochondria found in some fruit fly sperm which are the result of multiple mitochondria fusing together.
c) Cristae are the folds of the mitochondrial inner membrane, which increase the surface area of the inner membrane, thus allowing for greater ATP generation.
d) Cristae are a special type of phospholipid present in mitochondrial and chloroplast membranes that can trap protons (H+’s), thereby making the membranes particularly impermeable.
17)What is TRUE about the pH of the inner compartments of mitochondria?
a) The pH of the intermembrane space must be lower (more acidic) than the pH of the matrix in order for ATP synthesis from cellular respiration to properly occur.
b) The pH of the intermembrane space must be higher (more basic) than the ph of the matrix in order for ATP synthesis from cellular respiration to properly occur.
c) The pH of the intermembrane space must be the same as the pH of the matrix in order for ATP smithesis from cellular respiration to property occur.
18)Which of the following best explain the process of chemiosmosis?
a) High energy electrons are used to pump protons (H+’s) across a membrane, creating a proton gradient.
b) A proton (H+) gradient is used to power ATP synthesis.
c) Glucose is oxidized.
d)Water molecules are oxidized, while carbon dioxide molecules are reduced, resulting in oxygen and glucose as products.
19)The protofilaments of microtubules sometimes curl up, causing the microtubule to fall apart. What do you know to be TRUE about a protoflament that is curled up?
a) The alpha tubulin in the protofilament is bound to GDP,
b) The alpha tubulin in the protoflament is bound to GTP.
c) The beta tubulin in the protofilament is bound to GDP,
d) The beta tubulin in the protoflament is bound to GTP.
20)The “critical concentration” for actin microflament polymerization refers to:
a) The concentration of G actin that must be in solution for them to add onto a microfilament end /either + or -l.rather than fall off of it.
b) The concentration of G actin that always exists in healthy cells.
c) The concentration of G actin that is required in order to activate GTPases.
21)What higher order actin structure does alpha- actinin help support?
a) Tight parallel bundles found in microvilli
b) Contractile bundles of actin found in stress fibers
c) Gel-like actin meshwork
22)What is a FALSE statement about the similarities and differences between kinesin and myosin Il motor proteins?
a) Movement powered by kinesins require that a kinesin head domain is always touching a cytoskeletal fiber, while movement powered by myosin I does not require this
b) The conformational changes required for kinesin function rely on ATP binding and hydrolysis, while the conformational changes required for myosin Il function rely on GTP binding and hydrolysis.
c) The tail domains of Myosin Il associate with one another and form a “thick filament”, while kinesin motor proteins do not do this
d) Kinesin interacts with microtubules, while myosin Il interacts with actin microfilaments.
23)What sort of cellular changes would you expect a sailor sick with scurvy to experience?
a) Their collagen (an extracellular matrix component) would be diminished, resulting in fewer binding sites for integrins. The end result would be less adherent cells.
b) Their collagen (a cytoskeletal fiber) would be diminished, resulting in a loss of contractile motion within the cell, affecting both cell migration and muscle function.
24)In a migrating neutrophil, Rac-GTPase enhances activity at the leading edge, while Rho-GTPase enhances activity at the trailing edge. Normally, ac and Rho antagonize the activity of one another, and this is required for normal neutrophil function. If a mutation occurs, causing Rho to lose its ability to antagonize ac activity in the trailing edge, what do you expect will happen in the neutrophil?
A) Rho-GTPase will be active all over the cell.
b) Lamellopodia will not form properly.
c) Myosin I proteins will bind together into thick filaments in the leading edge, causing the actin meshwork there to contract where it normally does not.
d) WASp will activate the Arp2/3 complex in the trailing edge, causing actin meshwork to form where it normally does not.
25)You are a scientist studying neural crest cell migration, a key developmental event necessary for the formation of the head and face. You are studying a mouse line that has an unknown mutation inhibiting the neural crest cells from detaching from the extracellular matrix (they detach from other cells perfectly fine). You suspect that the mutation may be in a protein key for disassembling:
A) Tight junctions
b) Gap junctions
c) Focal adhesions
d) Desmosomes
26)What type of cell-cell junction depends on Ca2+ to bond identical cadherin proteins on each cell together?
a) Tight junction
b) Adherens junction
c) Focal adhesion
d) Hemidesmosome
27)Insulin is a molecule created by your pancreas that travels extensively, regulating glucose uptake in cells throughout the body. What class of signaling is this?
a) Endocrine signaling
b) Paracrine signaling
c) Contact-dependent signaling
d) Synaptic signaling
28)TRUE or FALSE: All receptors for signaling molecules are transmembrane proteins
a) True
b) False
29)What statement is FALSE about G alpha protein, an important member of the G-protein-coupled receptor signaling pathway?
a) In its inactive form, G-alpha is bound to 2 other proteins and is anchored to the plasma membrane.
b)Like other GTPases, G-alpha is inactive when bound to GDP, and active when bound to GTP.
c) When activated, the G-protein-coupled receptor acts as a GAP for G-alpha.
d) G-alpha proteins regulate the expression of genes promoting cell growth and division.
30)What is a TRUE statement about hormones?
a) Because hormones are hydrophobic, they cannot travel through the bloodstream alone. They must be bound to a carrier protein in order to make this journey.
b) Because hormones are hydrophylic, they cannot cross the plasma membrane alone. They must be bound to a transmembrane receptor protein in order to gain entry into the cell.
31)Which of the following does NOT occur when nitrous oxide (N20) is administered in a dentists office? (Note: Nitrous oxide is converted to nitric oxide (NO)).
Description of pathway:
Acetylcholine released from a nerve terminal bind to a G-protein-coupled receptor on an endothelial
cell of a blood vessel. The receptor then activates a G protein (not pictured), which stimulates IP3 synthesis and release of calcium ions, which activates nitric oxide synthase (NOS). NOS causes endothelial cells to produce nitric oxide from the amino acid arginine. Because nitric oxide is small and hydrophyllic, it can just diffuse across the membrane of the endothelial cell that it is produced in and INTO the neighboring smooth muscle cells where it acts as a signal.
Once inside the smooth muscle cell, nitric oxide binds to and activates guanylyl cyclase, which then stimulates the synthesis of cyclic GMP (cGMP) from GTP. The cyclic GMP triggers a response that causes the smooth muscle cells to relax, increasing blood flow through the vessel.
a) Muscle cells relax, and pain sensation is reduced
b) Increased activation of NO synthase (NOS)
c) Increased production of cyclic GMP (GMP)
d) Increased binding of NO to guanylyl cyclase
32)You use genetic engineering to remove the nuclear localization sequence (NLS) of an estrogen hormone receptor in zebrafish. What do you expect will happen to the cells when they are exposed to estrogen?
a) Increased transcription of genes that are usually highly expressed in the presence of estrogen
b)Decreased transcription of genes that are usually highly expressed in the presence of estrogen.
33)Maria hypothesizes that the transcription factor DEDZ1 promotes apoptosis. She decides to design and conduct an experiment to investigate this hypothesis. Her first experiment is to use genetic engineering to knockout (delete) DEDZ1 from the genome of normal cells, and then measure and compare apoptosis in DEDZ knockout cells versus normal cells by applying Tryphan Blue to each sample.
Examine the graph of her results below. Do the results support Maria’s hypothesis?
a)Yes
b)No
34)TRUE or FALSE: During necrosis, the cell disassembles its cytoskeleton, nuclear envelope and DNA in a self-contained manner. It frequently does this in response to developmental signals.
a) True
b) False
35)What cleaves executioner procaspases, thereby activating them?
a) Initiator caspases
b) Initiator procaspases
c) Bax/Bak proteins
d) Cytochrome C
36)What is a FALSE statement about how Cytochrome C triggers apoptosis?
a) Cytochrome C is normally located within the mitochondrion, but its release promotes apoptosis
b) Cytochrome C must bind to a death receptor to trigger extrinsic apoptosis
c) Several dimers of Cytochrome C bound to APAF-1 coalesce into a structure called the ” apoptosome”
37)What statement is TRUE about Bax and Bak proteins?
a)Bax and Bak are executioner caspases that cleave cytoskeletal proteins and nuclear lamins, among other molecules.
b) Increased Bax and Bak expression is associated with necrosis.
c) When Bax and Bak are activated, they coalesce to form a pore on the mitochondrial membrane.
d) Overexpression of Bax and Bak would probably drive cancerous tumor progression
38)You are a researcher studying yeast cells, and discover a mutant strain of yeast that behave
differently from normal yeast. Specifically, they seem to divide much more quickly than normal yeast cells, and never grow as large as normal yeast cells do before they divide.
It is unknown what gene is mutated in the mutant strain, but a good hypothesis is that:
a)The gene that is mutated normally activates M-Cdk
b)The gene that is mutated normally inhibits M-Cdk.
39)During what phase are cells neither growing nor dividing? (In other words, during what phase are cells senescent/ “paused”)?
a) Mphase
b) GO phase
c) G1 phase
d) S phase
e) G2 phase
40)You are a principal investigator of a laboratory studying the cell cycle, and one of your graduate students asks you how to determine what stage of the cell cycle their cells are in. What do you tell them?
a) Measure the level of different cyclins. The relative levels of these cyclins should tell you which stage the cels are in, because cyclin levels vary throughout the cell cycle.
b) Measure the level of different Cdk’s. The relative levels of these Cak’s should tell you which stage the cels are in, because cyclin levels vary throughout the cell cycle.
c) Examine the cells under a microscope. You will always be able to tell what stage the cells are in by doing this.
41)You have discovered a novel protein BUZZKILL that inhibits the activation of M-Cdk. However, you later discover that M-Cdk directly inhibits the activity of BUZZKILL The result is that once the first few M-Cdks in a cell are activated, those activated M-Cdks will inhibit BUZZKILL proteins, resulting in more rapid activation of other M-Cdks.
Activated M-Cdk’s inhibition of BUZZKILL is an example of:
a) Positive feedback
b) Negative feedback
42)A mouse has one normal E2F allele, and one mutant “null” E2F allele (meaning that a totally useless protein is coded for, instead of a normal E2F allele). How will the pathway leading to the activation of the cell cycle by mitogens be impacted in this mouse, and what will the overall effect be on the mouse’s cells, and on the mouse?
Description of pathway:
The default state of cells in our body is to be senescent (not actively engaged in the cell cycle).
Mitogens are proteins (ligands) that stimulate senescent cells into entering the cell cyde. How this happens: A mitogen binding to its receptor activates as. as activates MAP kinase. These leads to increased transcription of several genes including the transcription factor Myc. Myc promotes transcription of Cyclin D. Cyclin D binds G1-Cdk, activating it.
Activated G1-Cdk inactivates b protein. In its active state, Rb protein binds to and inactivates E2F protein. When b protein is inactivated, it releases and allows activation of E2F protein.
E2F protein promotes transcription of genes required for -phase to occur. This results in the
translation of cyclins associated with S-phase. These cyclins bind to and activate S-Cdk. Activated S-Cok promotes DNA synthesis in a variety of ways.
a)The reactivation of the cell cycle by mitogens will occur more slowly, because it will take longer to build up sufficient active S-Cdk to activate DNA synthesis.
b) The reactivation of the cell cycle by mitogens will occur more quickly, because cyclins associated with S phase will build up more quickly.
c) The reactivation of the cell cycle by mitogens will occur more slowly, because as will be unable to activate MAP kinase as efficiently.
d) The reactivation of the cell cycle by mitogens will occur more quickly, because b protein will constantly be inactive.
43)Name the E3 ligase that ubiquitinates M-cyclins, tagging them for destruction in a proteasome.
a) Securin
b) Separase
c) Anaphase Promoting Complex (APC)
d) Cohesin
44)TRUE or FALSE: If p53 is activated in response to DNA damage, it will drive the cell to self-destruct via apoptosis right away, regardless of the extent of damage or the time it takes to repair the damage.
a) True
b) False
45)Which of the following is NOT one of the ways that M-Cdk promotes mitosis?
a) Phosphorylating Histone H1 and condensing proteins to initiate chromosome condensation
b) Phosphorylating nuclear laminas to promote nuclear envelope breakdown
c) Phosphorylating mitogens to deactivate them, so that cells stay in GO
d) Phosphorylating an inhibitory site on Myosin, so as to prevent early cytokinesis (cell division)