cell signaling Flashcards
Which of the following is an example of endocrine signaling?
Epinephrine release by motor neurons at the neuromuscular junction and binding to receptors on adjacent skeletal muscle cells
Antigen stimulation of T lymphocytes, leading to the stimulation and synthesis of a growth factor that drives their own proliferation
Insulin release by β cells in the pancreas, mediating an effect of glucose uptake by muscle cells
None of the above
Insulin release by β cells in the pancreas, mediating an effect of glucose uptake by muscle cells
The term “paracrine signaling” refers to
Question 2 options:
signaling between cells located far from each other.
stimulation of a cell by substances produced by the cell itself.
signaling between cells located close to each other.
signaling between parenchyma cells.
signaling between cells located close to each other.
Which of the following hormones is(are) not synthesized from cholesterol?
Question 3 options:
Testosterone
Progesterone
Corticosteroids
Retinoic acid
Retinoic acid
It is important to recognize that all of the steroid hormones, including testosterone, progesterone, estrogen, glucocorticoids, mineralocorticoids, and even the insect hormone ecdysone, are synthesized from cholesterol.
Which steroid hormone(s) is(are) not secreted by the gonads?
Question 4 options:
Corticosteroids
Progesterone
Estrogen
Testosterone
Corticosteroids, which include glucocorticoids and mineralocorticoids, are
secreted by the adrenal gland
The signaling molecule nitric oxide (NO) functions
Question 5 options:
by binding its cell surface receptor and triggering an intracellular signaling cascade.
by diffusing across the membrane and binding its intracellular receptor, which then activates transcription.
on cells located far from the cells where it was synthesized.
by diffusing across the cell membrane and changing the activities of intracellular enzymes.
by diffusing across the cell membrane and changing the activities of intracellular enzymes.
Nitric oxide does not bind an intracellular receptor.
Nitric oxide does not bind a cell surface receptor.
Unlike most signaling molecules, nitric oxide does not bind a receptor but
rather directly alters the activity of an enzyme.
Nitric oxide is an unstable molecule and thus acts only on cells close to
those that synthesized it
What is the difference between neurotransmitters and neuropeptides?
Question 6 options:
Neuropeptides are generated by neuronal cells but do not transmit signals.
Neurotransmitters are small hydrophilic molecules, and neuropeptides are small proteins.
Neurotransmitters are small protein molecules, and neuropeptides are large ones.
Some neuropeptides can act on distant cells, whereas neurotransmitters cannot.
Neurotransmitters are small hydrophilic molecules, and neuropeptides are small proteins.
Both neurotransmitters and neuropeptides transmit signals, but they are in different chemical classes.
Neurotransmitters are not proteins, and neuropeptides are small, not large, proteins
Both neuropeptides and neurotransmitters can act on distant cells.
Which statement regarding heterotrimeric G proteins in a resting state is true?
Question 7 options:
GDP is bound to the β subunit in a complex with both the α and γ subunits.
GDP is bound to the α subunit in a complex with both the β and γ subunits.
GTP is bound to the α subunit in a complex with both the β and γ subunits.
The proteins are in a complex with G protein-coupled receptors.
Feedback B: Correct! This is the state of G proteins when in a resting state.
G protein-coupled receptors are important molecules involved in signal transduction. Which statement about G protein-coupled receptors is true?
Question 8 options:
They are activated only by steroid hormones.
They bind only guanine nucleotides.
They bind both adenine and guanine nucleotides.
They generally contain seven membrane-spanning α helices.
Feedback A: Incorrect. While steroid hormones do often utilize G protein-coupled receptors, many other cell signaling molecules also use G protein-coupled receptors.
Feedback B: Incorrect. G protein-coupled receptors interact with G proteins, which in turn bind guanine nucleotides.
Feedback C: Incorrect. G protein-coupled receptors interact with G proteins, not with any nucleotides.
Feedback D: Correct! This is a structural characteristic of most G protein-coupled receptors.
Which statement about G protein signaling is false?
Hormone binding induces an interaction of the receptor with the G protein, stimulating the release of GDP and the exchange of GTP on the α subunit.
Once activated, the GTP-bound α subunit dissociates from βγ and interacts with its target.
Activity of the α subunit is terminated by the hydrolysis of GTP to GDP.
The α subunit becomes deactivated when the hormone dissociates from the receptor.
The α subunit becomes deactivated when the hormone dissociates from the receptor.
Feedback A: Incorrect. This is a true statement, but it is only a partially correct answer.
Feedback B: Incorrect. This is a true statement, but it is only a partially correct answer.
Feedback C: Incorrect. This is a true statement, but it is only a partially correct answer. This is a result of the GTPase activity of the α subunit. Once this occurs, the α subunit reassociates with the βγ complex and returns to the resting state.
Feedback D: Correct!
Cyclic AMP (cAMP) is synthesized from ATP by the action of Question 10 options:
phosphodiesterase.
phosphorylase kinase.
adenylyl cyclase.
protein kinase A (PKA).
adenylyl cyclase.
Feedback A: Incorrect. Phosphodiesterase degrades cAMP to AMP.
Feedback B: Incorrect. Phosphorylase kinase is critical to glycogen synthesis.
Feedback C: Correct! This is the key enzyme in cAMP synthesis using ATP as the substrate.
Feedback D: Incorrect. PKA is activated by cAMP.
Most of the effects of cyclic AMP (cAMP) in the cell are mediated by
Question 11 options:
protein kinase A.
ion channels.
protein kinase C.
cAMP phosphodiesterase.
protein kinase A.
Feedback A: Correct! Cyclic AMP activates protein kinase A, which then stimulates the breakdown of glycogen into glucose, and also alters the pattern of gene expression via activation of the transcription factor CREB.
Feedback B: Incorrect. Although cAMP can regulate ion channels such as sodium channels in sensory neurons, it is not the primary means by which it exerts its effects.
Feedback C: Incorrect. Protein kinase C is not activated by cAMP, but rather by diacylglycerol.
Feedback D: Incorrect. cAMP phosphodiesterase is the enzyme that breaks down cAMP to AMP.
Which statement about protein kinase A (PKA) is false?
Question 12 options:
In the inactive state, PKA exists as a tetramer of two regulatory (R) and two catalytic (C) subunits.
PKA binds a total of four molecules of cAMP, one on each of the four subunits.
PKA binds a total of four molecules, two molecules on each of the two regulatory (R) subunits.
Once activated, the catalytic (C) subunits dissociate and activate target molecules.
PKA binds a total of four molecules of cAMP, one on each of the four subunits
Receptor tyrosine kinases represent critical molecules involved in growth and differentiation though phosphorylation of target substrates on tyrosine residues. Which structural feature is not common among all receptor tyrosine kinases?
Question 13 options:
An N-terminal extracellular ligand-binding domain
A single polypeptide
A cytosolic C-terminal domain with tyrosine kinase activity
A single transmembrane α helix
A single polypeptide
Feedback A: Incorrect. They all do share this.
Feedback B: Correct! Many receptor protein-tyrosine kinases are single polypeptides, yet not all are. A notable exception to this would be the insulin receptor.
Feedback C: Incorrect. They all do share this.
Feedback D: Incorrect. They all do share this.
Which of the following is not a commonly observed consequence of the binding of a signaling molecule to its cell surface receptor?
Question 14 options:
Receptor dimerization
Receptor phosphorylation
Conformational changes in the receptor
Increased synthesis of the receptor
Increased synthesis of the receptor
Feedback A: Incorrect. Dimerization of receptors often follows binding by their stimulants.
Feedback B: Incorrect. Receptor phosphorylation often occurs after binding, either by the receptor itself (autophosphorylation), or by associated cytoplasmic kinases.
Feedback C: Incorrect. Conformational changes in the receptor, ultimately resulting in transmission of the signal, do occur following binding.
Feedback D: Correct! The synthesis of the receptor is not typically increased either at the transcriptional or translational level following binding. In fact, sometimes the degradation of the receptor is stimulated as a step toward returning the cell to the resting state.
SH2 domains are
Question 15 options:
protein domains that bind phosphotyrosine-containing peptides.
the domains on receptor tyrosine kinases that contain the phosphorylated tyrosine.
domains that mediate the dimerization of receptor tyrosine kinases.
the domains on receptor tyrosine kinases that possess the kinase activity.
protein domains that bind phosphotyrosine-containing peptides.
Feedback A: Correct! Proteins containing SH2 domains are the first downstream targets of receptor protein-tyrosine kinases.
Feedback B: Incorrect. SH2 domains are not usually found in receptor protein-tyrosine kinases.
Feedback C: Incorrect. SH2 domains are not dimerization domains.
Feedback D: Incorrect. SH2 domains do not have enzymatic activity.
Integrins are transmembrane proteins that connect
Question 16 options:
the nuclear laminae to cytoplasmic kinases.
the extracellular matrix to the cytoskeleton.
focal adhesions to hemidesmosomes.
microtubules to actin filaments.
the extracellular matrix to the cytoskeleton
The MEK kinase (MAP kinase/ERK kinase) is unusual in that it
Question 17 options:
is activated by a kinase.
lies downstream of G protein-coupled receptors.
is a dual-specificity kinase, having the ability to phosphorylate both threonines and tyrosines.
activates a kinase.
is a dual-specificity kinase, having the ability to phosphorylate both threonines and tyrosines.
Most kinases can either phosphorylate serines and threonines or tyrosines, but MEK can phosphorylate both a threonine and a tyrosine residue on ERK2.
Heterotrimeric G proteins are not the only guanine nucleotide-binding proteins. Which of the following represents a family of GTP-binding proteins that act as monomers rather than heterotrimeric compounds?
Question 18 options:
Ras
ERK
Raf
Smad
Ras
The members of the Ras family are often called small GTP-binding
proteins because they are about half the size of the α subunits of heterotrimeric G proteins.
In unstimulated cells, NF-κB proteins are maintained in an inactive state in the cytosol by interactions with
Question 19 options:
Hedgehog.
IκB.
adaptor proteins.
the TNF receptor.
IκB.
An example of signaling by direct cell–cell interactions is the
Question 20 options:
Wingless (Wnt) signaling pathway.
JAK/STAT pathway.
Notch pathway.
pathway leading to vulval development in C. elegans.
Notch pathway.
Notch is a transmembrane protein that is stimulated by other transmembrane proteins (such as Delta) on adjacent cells
Which of the following signaling pathways allows for direct cell–cell signaling by transmembrane proteins?
Question 21 options:
NF-κB
Wnt
Hedgehog
Notch
Notch
In what ways are steroid hormone receptors different from most other types of cellular receptors?
Most cellular receptors span the plasma membrane, where they are ideally positioned to sense extracellular signals and transmit them to the cell’s interior. In contrast, because steroid hormones are small hydrophobic molecules that can slip through membranes, their receptors are located in the interior of the cell, in both the cytoplasm and the nucleus. In addition, once activated, the steroid hormone receptors themselves act as transcription factors, whereas for receptors located in the plasma membrane, the transcription factors usually lie at the end of a signal transduction cascade that includes several components.
In plants, growth can be mediated by what is frequently referred to as “auxin-induced transcription” of genes critical to growth. This terminology was derived from experiments demonstrating that the addition of auxin leads to an increase in the expression of genes responsible for growth. However, since auxin is not a transcription factor and cannot directly activate transcription, how does it regulate gene activity
Many plant growth genes are constitutively or always repressed by proteins that bind DNA(AUX/IAA - repressor proteins) and inhibit transcription by the Auxin Response Factor (ARF). Simply removing them will enhance the rate of transcription. Auxin binds to a receptor with ubiquitin ligase activity ScF TIFI, which leads to the ubiquitination and subsequent proteolytic degradation of the repressor. Thus, even though auxin cannot directly affect gene expression, it can still have dramatic effects on expression by regulating proteins that do directly interact with the genes.
What is the basis for the different responses of nerve cells versus heart muscle cells to acetylcholine
In nerve cells, the acetylcholine receptor doubles as a ligand-gated ion channel composed of five subunits. When bound by acetylcholine, the receptor opens to allow entry of sodium and exit of potassium, thus depolarizing the membrane and triggering an action potential. In contrast, in heart muscle, the acetylcholine receptor has a very different structure—it is a G protein-linked receptor of the seven-transmembrane class. The G protein directly activates a potassium channel, which causes a parasympathetic response in the cardiac muscle. This is a classic example of how a single neurotransmitter can be used for different purposes in different cell types.
Why do enzymes that lie downstream of a cell surface receptor in a signal transduction pathway amplify as well as propagate the signal
Extracellular signaling molecules typically stimulate only a single receptor molecule before being degraded; the receptor, however, can often activate many molecules of the enzyme to which it is directly linked. Since most signal transduction pathways are made up of several enzymes, amplification can occur at each step, resulting in very high amplification by the time the pathway reaches the ultimate component. Thanks to this type of amplification, the cell can be exquisitely sensitive to even very low concentrations of extracellular inducers.
As a cell biologist for a pharmaceutical company, you are charged with developing a drug to inhibit the cAMP response element binding protein (CREB) pathway. Briefly describe the cascade leading to CREB activation and where one might target a drug to block its activation.
In the presence of cAMP, the R subunit is bound by four molecules of cAMP, allowing the catalytic subunits to dissociate, translocate to the nucleus, and then phosphorylate CREB. CREB then homodimerizes, binds to the CRE (cAMP response element) in the promoter of target genes, and activates transcription. Drugs could be designed that would interrupt signaling anywhere along the pathway. For example, a drug could stabilize the R and C subunit interactions of PKA and block translation into the nucleus. A drug could prevent interaction of cAMP with the R subunit and essentially lead to the same result. Similarly, a drug could have a more downstream action that might prevent phosphorylation of CREB, thus not allowing it to become activated. Another option would be to prevent the required dimerization of CREB, thus preventing its ability to bind DNA.
What are the hallmarks of the first steps in signaling through receptor tyrosine kinases?
Receptor protein-tyrosine kinases span the plasma membrane; upon binding by extracellular stimulants, they form dimers with one another and autophosphorylate. The cytoplasmic domain can then bind and activate intracellular proteins containing SH2 domains, which specifically bind peptides containing phosphotyrosine residues. The SH2 domain-containing proteins then go on to activate a variety of different pathways depending upon the identity of the receptor that activated them.
Growth factors and cytokines both lead to tyrosine phosphorylation through receptors, but they do so through different mechanisms. What is the key difference between the receptors for these two classes of ligands in terms of their tyrosine kinase activity
The receptors for growth factors such as EGF are membrane-spanning cell surface receptors in which the cytoplasmic domain of the receptor contains the catalytic activity necessary to phosphorylate proteins on tyrosine residues and often are capable of autophosphorylation. Cytokine receptor activation also leads to tyrosine phosphorylation, but the receptor has no intrinsic catalytic activity. Instead, upon ligand binding, the cytokine receptors interact with a class of proteins called nonreceptor tyrosine kinases, which in turn do have the catalytic activity necessary for tyrosine phosophorylation.
Cytokine receptors frequently form dimers upon ligand binding, and these can be heterodimers or homodimers. Explain how this ability of the receptors to form different combinations of dimers allows three unique cytokines, each with a unique action, to arise from only two cytokine receptors.
If dimerization is required, then the two receptors (R1 and R2) could form homodimers (R1:R1 and R2:R2) and potentially interact with two of the cytokines; the third cytokine could bind to a heterodimer of R1:R2. Given the number of cytokines and receptors and the many combinations such as these that are possible, the complexity of these signaling networks becomes evident.
In the signaling of cells to initiate growth, immediate early genes become activated to initiate the growth process. What would happen to the expression of these immediate early genes if Elk-1 was not able to be dephosphorylated, and why
The immediate early genes would be constitutively activated (i.e., not shut off). Immediate early genes share the serum response element in their promoters that interact with serum response factor (SRF). Following the growth signal pathway ERK phosphorylates and activates Elk-1 that interacts with SRF and activates transcription of the immediate early genes. If it is unable to be de-phosphorylated it would remain active and continually express these genes and with it being a growth signal may well lead to cancer.
What would happen to NF-κB signaling if IκB were unable to be dephosphorylated?
Most likely nothing. When IκB is phosphorylated by IκB kinase, it is targeted for ubiquitination and then degraded by proteasomes. If it could not be phosphorylated, it would still be targeted for ubiquitination and then degraded by proteasomes.
Explain the concept of “feedback loops” in signaling networks
Feedback loops allow for a form of autoregulation of a particular event or activity. In signaling, there are positive feedback loops, in which a downstream element of the pathway activates an upstream element. This leads to a continual activation, even in the absence of the initiating event such as ligand binding to a receptor. The counterpart would be a negative feedback loop, in which a downstream element inhibits the activity of an upstream element, shutting down the pathway. This may occur even in the presence of the initiating event. In the feedforward relay, an upstream element may directly activate not only the next element in the pathway but also an element farther downstream, thus “short circuiting” the elements in between. These concepts have useful applications throughout the study of biology, from the molecular level (as observed here) all the way through organ and systems biology.
Ribosomal RNAs are methylated at specific nucleotide positions. To what extent is the specificity of nucleotide modification achieved through enzyme recognition of sequence and/or other mechanisms?
The sequence specificity of what is modified comes about through guide RNAs, snoRNAs, directing modifications. snoRNAs are part of protein complexes. In the end, it is a protein that catalyzes the reaction. An enzyme—a protein, of course—is specific in terms of the chemical modification that is made: for example, methylation of a base.
Unlike eukaryotic cells, prokaryotic cells do not have a nucleus. What are some possible benefits of the evolution of a nucleus in eukaryotes?
In contrast to prokaryotes, eukaryotic RNA is extensively modified by processes such as splicing and RNA editing, which are critical to the accurate transfer of genetic information from gene to protein. Sequestering the chromosomes in a separate compartment allows RNA-processing events to be more efficient, because the enzymes involved are sequestered in the nucleus and hence have less interference from unrelated enzymes. In addition, because RNA is retained in the nucleus until the processing events are complete, inappropriate translation of unprocessed RNAs does not occur. Furthermore, an additional level of control comes from the entry of regulated transcription factor into the nucleus. This would not be possible if the cytosol were not separated from the nucleus and chromosomes.
How does the clustering and transcriptional pattern for eukaryotic ribosomal RNA genes ensure that there will be equal numbers of copies of the major RNAs available for ribosome assembly?
Ribosomal genes are organized in tandem repeats that contain one copy each of the 18S, 5.8S, and 28S rRNA sequence as a single gene unit. The transcript contains all three sequences, and hence, each is produced after processing of the pre-rRNA in the same amount. All this is located in the nucleolus. Transcription of 5S rRNA occurs outside the nucleolus and may not be as tightly linked in terms of amount.
What is the role of nucleotide sequence in mediating the export of ribosomal RNA from the nucleus?
The recognized export sequences reside on the proteins associated with the ribosomal RNA rather than on the ribosomal RNA itself. For example, a specific exportin, Crm1, recognizes protein sequence features (i.e., amino acids) of proteins bound to the ribosomal RNA rather than the nucleotide sequence of the ribosomal RNA itself
The activity of transcription factors is frequently regulated by means of their localization within eukaryotic cells, such that only upon stimulation is the transcription factor transported into the nucleus, where it can activate its target genes. For a transcription factor that has a nuclear localization signal (NLS), explain two mechanisms by which regulated transport can occur.
Two possible mechanisms involve the masking the NLS such that it is not recognized by importin. This masking can be: (1) intermolecular, whereby another protein binds and obscures the NLS, or (2) intramolecular, whereby the protein itself folds in such a way that the signal is masked. Upon stimulation, the masking protein dissociates or the protein structure is altered such that the NLS is exposed, and transport to the nucleus can occur. Other mechanisms are also possible. For example, the transcription factor may be tethered to another protein that is attached to a cytoplasmic structure, such as a membrane protein or a cytoskeletal protein. In this case, the nuclear localization signal is exposed, but the protein is physically retained in the cytoplasm until the appropriate stimulus liberates it and it is free to make its way into the nucleus.
The nuclear localization signal (NLS) is not removed (cleaved off) after entry of the protein into the nucleus. In contrast, the targeting sequences of proteins destined for organelles such as the mitochondria are located at the N-terminus of the protein and cleaved off once the proteins reach the lumen of the organelle. What are two likely reasons that the nuclear localization signal (NLS) is not cleaved off for nuclear proteins?
During mitosis, the nuclear envelope disintegrates and the nuclear proteins are released into the cytoplasm. Once the envelope re-forms around the daughter nuclei, the nuclear proteins need to be retransported in, thus explaining the need for retention of the NLS. However, in some organisms, such as yeast, the nuclear envelope stays intact during mitosis. Hence, there are likely other reasons for retention of the NLS that may be of equal or greater importance. A second reason is that nuclear localization signals are internal to the polypeptide. Cleaved targeting sequences are typically located at the N-terminus of the protein. Cleavage of an internal sequence has much more consequence for the overall structure of the protein than removal of an N-terminal sequence. In addition, many nuclear proteins shuttle in and out of the nucleus, which would be a third reason for retaining the NLS.
Protein transport into the endoplasmic reticulum (ER) or mitochondria requires unfolding the protein and threading the amino acid chain through a channel or pore into the organelle. In contrast, protein transport into the nucleus does not require an unfolding of the protein. Why is this not necessary for nuclear protein import?
The nuclear pore complex is a huge structure composed of about 30 different pore proteins, and its central channel has a diameter of approximately 10-40 nm, which is large enough for even very large protein complexes. For this reason, it is unnecessary to unfold the protein, which can pass through the pore in its native state
The nuclear pore complex is a huge structure composed of about 30 different pore proteins, and its central channel has a diameter of approximately 10-40 nm, which is large enough for even very large protein complexes. For this reason, it is unnecessary to unfold the protein, which can pass through the pore in its native state
Ran is important to both protein import into and export out of the nucleus. As part of importin recycling to the cytoplasm, Ran in the GTP-bound form complexes with importin inside the nucleus. Formation of this complex results in the release of importin from cargo and the importin/Ran-GTP complex is then recycled back to the cytoplasm. During nuclear protein export in general, GTP-bound Ran binds to exportin inside the nucleus to give rise to a Ran-GFP/exportin/cargo complex that is exported as a unit through the nuclear pore into the cytoplasm. In either case, Ran functions as a GTP-dependent molecular switch. Ran GAP bound to the cytoplasmic filaments of the nuclear pore complex activates the GTPase activity of Ran in the cytoplasm and bound GTP is cleaved to GDP. If Ran GAP were not bound to the cytoplasmic filaments, its effective local concentration would be diluted, and the efficiency of Ran activation would be low. Low affinity would cause a distinct decrease in the Ran GTP gradient across the nuclear pore. This would reduce the efficiency of both import into the nucleus, which is dependent on imporin recycling, and export from the nucleus. Phenotypically, cell growth rates would slow and the defect might be so severe as to be lethal.
What attributes of where replication and transcription occur within the nucleus leads to these sites being designated as “factories”?
DNA replication and its transcription into RNA occur within clustered regions of the nucleus where large complexes of proteins involved in the particular process are concentrated. Because of this clustering into regions and the concentration of protein complexes in these regions, the designation of factory(ies) is appropriate.
In many organisms, the nuclear envelope breaks down during mitosis. As part of this breakdown, the nuclear lamina that underlies the nuclear envelope breaks down. In fact, many investigators think that the breakdown of the lamina initiates envelope breakdown. Propose a reversible mechanism for controlling the polymerization state of the proteins that make up the nuclear lamina.
Reversible phosphorylation/dephosphorylation would provide a mechanism for regulating the polymerization of nuclear lamins. When phosphorylated, negative charge repulsion would lead to dissociation, and after dephosphorylation repolymerization would occur. Specific protein kinases would be involved in the phosphorylation reaction and specific phosphatases in the dephosphorylation reaction. These events could be regulated by various signaling pathways.
Describe the roles of clathrin and associated adaptor proteins in the sorting of proteins to lysosomes.
Proteins destined for lysosomes are recognized in the Golgi and acquire a mannose-6-phosphate “tag,” which then binds to the mannose-6-phosphate receptor, a transmembrane protein. As part of the formation of clathrin-coated vesicles at the trans-Golgi network, an adaptor protein is recruited: Arf/GTP. The adaptor protein has a binding site for interaction with the cytosolic tail of the mannose-6-phosphate receptor and it also serves as a binding site for assembly of the clathrin coat. The clathrin coat forms a lattice resembling a basket around the adaptor proteins, thus inducing the formation of a vesicle that eventually buds off from the trans-Golgi network. Thus, clathrin plays a role in the physical formation of the vesicle, rather than acting as a selectivity factor. The selectivity factor is the adaptor protein.
Binding of SRP to the signal sequence arrests the further translation of the nascent polypeptide chain. How does the arrest of translation promote the subsequent translocation of the polypeptide chain through the Sec61 translocation complex?
SRP, in binding to the nascent polypeptide chain and ribosome, arrests polypeptide elongation. Only when the SRP has become dissociated from the chain and the nascent chain is being pushed through the translocon does polypeptide elongation resume. This results in an extended conformation of the nascent chain that is being pushed through the translocon. Folding occurs in the ER lumen. If SRP bound but did not arrest translation, there would be the need for chaperones in the cytosol to bind to the nascent chain and keep it in an extended conformation as it continued to elongate. In yeast, where posttranslational insertion occurs, this process is fairly common. In mammalian cells, posttranslational insertion is rare, and the pool of cytosolic chaperones is presumably limited.
The viral protein hemagglutinin lies in the membrane that surrounds the influenza virus. Hemagglutinin binds to sialic acid residues on membrane proteins of cells lining the respiratory tract, and the virus is subsequently internalized by endocytosis. What is the likely method by which the viral genome, which is now surrounded by two membranes (its own plus the cellular endosomal membrane), gains access to the proteins within the cytosol that it needs in order to replicate?
Once the virus is internalized, the endocytic vesicle makes its way toward the lysosome via the endosomal compartment, and the compartment becomes increasingly acidic. Once the acidity reaches a certain level, hemagglutinin undergoes a conformational change that causes fusion of the two membranes (viral and endosomal), releasing the viral genome into the cytosol. Thus the cell’s attempt to destroy the virus by transporting it to the lysosome actually leads to its destruction
Suppose you are studying a mutant that you know is impaired in secretion, but you do not know the exact nature of the defect. You examine the sugar content of some of the proteins you know to be N-glycosylated and notice that though they normally contain fucose, galactose, and sialic acid residues, these are absent in proteins purified from the mutant. Instead only N-acetylglucosamine and mannose residues are evident. Where do you think the defect in secretion lies?
There is a defect in transport between the ER and Golgi. In the ER, N-acetylglucosamine, mannose, and glucose residues are added to specific asparagine residues within proteins, but the glucose residues are later removed so that upon reaching the Golgi, they contain only N-acetylglucosamine and mannose residues. Fucose, galactose, and sialic acid residues are added in the Golgi. Thus, an ER-to-Golgi block would explain the lack of these sugar residues in proteins that normally contain them.
The three contiguous membrane domains of the ER—rough ER, smooth ER, and ER exit sites—are all continuous with one another and, depending on cell type, can be present in greatly differing amounts. For example, in cells that synthesize steroid hormones, the smooth ER is much more abundant. In nonsecretory cells, such as cultured human HeLa cells, there is little ER present. Propose a mechanism by which these domains could be generated and maintained
The available data do not offer a clear answer to this question. A plausible working hypothesis is that ER exit sites and rough ER arise as domains within the ER by means of molecular association. In other words, these domains segregate themselves from the smooth ER by self-association. For example, the ribosomes, in binding as polyribosomes to the Sec61 translocon, gather associated components. This ribosome-driven set of associations could lead to others, resulting in rough ER as a distinct domain. Transitional ER is a limited-size domain in which COPII budding occurs. This is a polymerization process that could initiate a set of associations, and with time, key proteins would likely establish association patterns.
In the Golgi apparatus, the enzymes that modify N-linked oligosaccharides are arranged in an assembly-line manner (i.e., in sequential Golgi apparatus subcompartments). In the ER, these enzymes are all present in the same continuous membrane-limited compartment. Propose a reason for this difference.
In the ER, the oligosaccharide precursor to N-linked glycosylation is synthesized and then added to the growing polypeptide cotranslationally. The presence of terminal glucose is used as a “timer” relative to the binding of newly synthesized polypeptide with calnexin during protein folding. The major degradative process is removal of glucose. In the Golgi apparatus or Golgi complex, there is extensive trimming of the oligosaccharide sidechain by glycosidases before the sidechain is rebuilt through glycosyltransferase activity. Physical separation of compartments may lead to better separation of reactions.
A member of the heat-shock class of proteins, BiP, is necessary for the proper folding of proteins in the ER. In terms of function, what do BiP and the heat-shock proteins (HSP) have in common?
The primary function of heat-shock proteins is to prevent damage to the cell upon exposure to conditions that cause protein unfolding; in their absence, unfolded proteins would associate into large aggregates that would be difficult to disentangle, once the cell returned to normal conditions. Like BiP, the inducible heat-shock proteins bind unfolded proteins and assist in their proper folding. In fact, BiP and inducible heat-shock proteins are all chaperone proteins. BiP is an Hsp70 protein-family member.
Imagine a transmembrane molecule that lies in the plasma membrane and serves as a receptor for an extracellular signaling molecule. When the ligand-binding domain is inserted into the ER, will it lie within the lumen of the ER or in the cytoplasm?
The ligand-binding domain will lie within the lumen of the ER, passing through the Golgi in the same orientation. In the final transport step, a vesicle containing the fusion protein will leave the trans Golgi network and fuse with the plasma membrane, thus exposing the lumen of the vesicle to the cell exterior. In addition, any oligosaccharide subunits that may have been added to the protein will be located on the extracellular domain of the protein, since glycosylation takes place in the lumen of the ER and Golgi.
Cargo proteins are transported through the Golgi apparatus to the trans-Golgi network, where they are sorted into various vesicles that are targeted to different destinations. What is the role of vesicles in cargo protein transport through the Golgi apparatus?
Two general models for Golgi function in cargo transport through the organelle are actively being considered: the “stable cisternal model” and the “cisternal maturation model.” In the stable cisternal model, vesicles are the carriers of cargo proteins between cisternae. In the cisternal maturation model, the cisternae are the actual cargo carriers. Vesicles function to retrieve Golgi components and hence to mature the cisternae as they transport cargo. At present, each model explains some, but not all, characteristics of Golgi transport. Further research is needed.
