cell signaling

Question 1

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

Answer 1

Insulin release by β cells in the pancreas, mediating an effect of glucose uptake by muscle cells

Question 2

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.

Answer 2

signaling between cells located close to each other.

Question 3

Which of the following hormones is(are) not synthesized from cholesterol?
Question 3 options:

Testosterone

Progesterone

Corticosteroids

Retinoic acid

Answer 3

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.

Question 4

Which steroid hormone(s) is(are) not secreted by the gonads?
Question 4 options:

Corticosteroids

Progesterone

Estrogen

Testosterone

Answer 4

Corticosteroids, which include glucocorticoids and mineralocorticoids, are
secreted by the adrenal gland

Question 5

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.

Answer 5

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

Question 6

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.

Answer 6

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.

Question 7

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.

Answer 7

Feedback B: Correct! This is the state of G proteins when in a resting state.

Question 8

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.

Answer 8

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.

Question 9

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.

Answer 9

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!

Question 10

Cyclic AMP (cAMP) is synthesized from ATP by the action of
Question 10 options:

phosphodiesterase.

phosphorylase kinase.

adenylyl cyclase.

protein kinase A (PKA).

Answer 10

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.

Question 11

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.

Answer 11

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.

Question 12

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.

Answer 12

PKA binds a total of four molecules of cAMP, one on each of the four subunits

Question 13

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

Answer 13

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.

Question 14

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

Answer 14

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.

Question 15

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.

Answer 15

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.

Question 16

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.

Answer 16

the extracellular matrix to the cytoskeleton

Question 17

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.

Answer 17

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.

Question 18

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

Answer 18

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.

Question 19

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.

Answer 19

IκB.

Question 20

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.

Answer 20

Notch pathway.

Notch is a transmembrane protein that is stimulated by other transmembrane proteins (such as Delta) on adjacent cells

Question 21

Which of the following signaling pathways allows for direct cell–cell signaling by transmembrane proteins?
Question 21 options:

NF-κB

Wnt

Hedgehog

Notch

Answer 21

Notch

Question 22

In what ways are steroid hormone receptors different from most other types of cellular receptors?

Answer 22

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.

Question 23

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

Answer 23

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.

Question 24

What is the basis for the different responses of nerve cells versus heart muscle cells to acetylcholine

Answer 24

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.

Question 25

Why do enzymes that lie downstream of a cell surface receptor in a signal transduction pathway amplify as well as propagate the signal

Answer 25

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.

Question 26

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.

Answer 26

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.

Question 27

What are the hallmarks of the first steps in signaling through receptor tyrosine kinases?

Answer 27

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.

Question 28

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

Answer 28

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.

Question 29

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.

Answer 29

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.

Question 30

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

Answer 30

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.

Question 31

What would happen to NF-κB signaling if IκB were unable to be dephosphorylated?

Answer 31

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.

Question 32

Explain the concept of "feedback loops" in signaling networks

Answer 32

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.

Question 33

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?

Answer 33

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.

Question 34

Unlike eukaryotic cells, prokaryotic cells do not have a nucleus. What are some possible benefits of the evolution of a nucleus in eukaryotes?

Answer 34

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.

Question 35

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?

Answer 35

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.

Question 36

What is the role of nucleotide sequence in mediating the export of ribosomal RNA from the nucleus?

Answer 36

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

Question 37

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.

Answer 37

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.

Question 38

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?

Answer 38

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.

Question 39

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?

Answer 39

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

Question 40

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

Answer 40

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.

Question 41

What attributes of where replication and transcription occur within the nucleus leads to these sites being designated as "factories"?

Answer 41

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.

Question 42

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.

Answer 42

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.

Question 43

Describe the roles of clathrin and associated adaptor proteins in the sorting of proteins to lysosomes.

Answer 43

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.

Question 44

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?

Answer 44

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.

Question 45

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?

Answer 45

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

Question 46

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?

Answer 46

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.

Question 47

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

Answer 47

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.

Question 48

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.

Answer 48

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.

Question 49

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?

Answer 49

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.

Question 50

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?

Answer 50

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.

Question 51

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?

Answer 51

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.

Question 52

Suppose that a student you are tutoring comes to you carrying two centrifuge tubes, each with a pellet at the bottom. She knows that one tube contains a pellet of mitochondria and the other a pellet of chloroplasts, but she cannot remember which is which. How would you help her?

Answer 52

Chloroplasts are full of chlorophyll and thus are green, whereas mitochondria are colorless.

Question 53

At low pH, the chemical 2,4-dinitrophenol (DNP) is neutral and can diffuse freely across membranes, including those of mitochondria. At high pH, it gives off a proton, becomes negatively charged, and can no longer diffuse across membranes. What effect would DNP have on the proton gradient between the mitochondrial intermembrane space and matrix?

Answer 53

DNP would have the effect of dissipating the proton gradient across the inner membrane. It would be neutral at pH 7 in the intermembrane space and thus pass freely into the mitochondrial lumen, where it would encounter a high pH environment and release its proton. This would neutralize the proton gradient and drastically decrease the overall ATP production of the cell as well.

Question 54

What are "porins"?

Answer 54

Porins are transmembrane proteins that form large pores and are found in chloroplast and mitochondrial outer membranes, as well as in the outer membranes of Gram-negative bacteria. They allow the passage of molecules smaller than 6,000 daltons, and thus the space between the inner and outer membranes is equivalent to the cytosol in its concentration of ions and small molecules. The inner membrane does not contain porins, and thus ions and small molecules can cross it only in a regulated way via channels and transporters. Porins are also found in the outer membranes of bacteria.

Question 55

What is the similar shared function of mitochondrial matrix processing peptidase (MPP) and chloroplast stromal processing peptidase (SPP)?

Answer 55

First, remember that a peptidase cleaves peptide bonds in polypeptides and proteins. But in both mitochondrial and chloroplast transport there are N-terminal amino acid sequences that target polypepides and proteins to the Tom and Toc complexes in mitochondria and chloroplast outer membrane and subsequently to the Tim and Tic complexes of the inner membranes. In mitochondria, this is the 15-55 amino acid N-terminal presequences, and in chloroplasts, it is the 30-100 amino acid N-terminal transit peptides. Once in the inner membrane, MMP cleaves the presequence in mitochondria and SPP cleaves the transit peptide in chloroplasts.

Question 56

Mitochondrial mRNAs have short poly-A sequences at their 3′ end. Poly-A is generally considered to be a feature of eukaryotic, not bacterial, mRNA. How can this observation be reconciled with an endosymbiotic origin for mitochondria?

Answer 56

The mitochondria of today are not the mitochondria of endosymbiotic origin. Present-day mitochondria have evolved from their original status of newly introduced endosymbionts. In reality, the mitochondrial genome shows a mix of bacterial and eukaryotic traits, with the poly-A sequence a eukaryotic-like trait of mitochondrial mRNA. Presumably this trait originated after the introduction of mitochondria into eukaryotic cells.

Question 57

Zellweger syndrome is caused by defects in genes coding for peroxisomal protein import. Why are defects in such genes more likely to be lethal than a defect in a gene encoding a single enzyme present in the peroxisomal lumen?

Answer 57

A defect in import machinery affects the import of several proteins into the lumen of the peroxisome. A mutation in a single peroxisomal enzyme affects only that enzyme. Therefore, the import machinery mutation should have a greater effect and likelihood of being lethal.

Question 58

Explain the role of endosymbiosis in the evolution of mitochondria and chloroplasts.

Answer 58

Because of the prokaryotic nature of the organelle and the striking similarity between the genomes of mitochondria and some bacteria (most notably Rickettsia prowazekii), it has been hypothesized that mitochondria evolved from an endocytic event in which a bacterium was endocytosed by a eukaryotic cell. The eukaryotic cell would have provided the bacterium with protection from the outside world, and the eukaryotic cell would have benefited from the bacterium's oxidative phosphorylation system for energy production. The evolution of chloroplasts is hypothesized to have occurred in much the same way but at a somewhat later date.

Question 59

A human hereditary disease has two seemingly unrelated symptoms: an inability to clear mucus from the respiratory system and male sterility. What could be the cause of this disease?

Answer 59

The failure of cilia to beat would disable the clearing of the respiratory tract, and the failure of flagella to beat would render sperm immobile—one source of male sterility. This disease is caused by a failure to produce dynein, the motor protein that drives ciliary and flagellar motion. The basis of a third symptom, the occasional placement of the heart to the right side of the thorax instead of to the left, is still unknow

Question 60

Describe the genetic defect that causes Duchenne's and Becker's muscular dystrophy and the molecular processes that characterize the diseases.

Answer 60

The diseases are caused by a mutation in the gene for dystrophin, a spectrin-related gene that links the actin network beneath the cell surface (the cortex) to transmembrane proteins in the plasma membrane. The transmembrane proteins, in turn, are bound to components of the extracellular matrix, and thus dystrophin plays a role in linking the cortex to the extracellular matrix. This firm anchoring to the extracellular matrix stabilizes muscle cells; without it, the constant stress of contraction results in their destruction and consequently in the loss of muscle tissue that characterizes the diseases.

Question 61

One approach to testing whether keratin intermediate filaments are dynamic is to inject biotin-labeled keratin into living fibroblasts and then to compare, at different times, the distribution of biotin-labeled keratin with that of endogenous keratin intermediate filaments. Assuming that keratin intermediate filaments turn over dynamically with a half-time of one hour, predict the comparative distribution of biotin-labeled keratin and keratin intermediate filaments 10 minutes and four hours after microinjection.

Answer 61

Any incorporation of the biotin-labeled keratin into keratin intermediate filaments takes time. After 10 minutes, the injected biotin-labeled keratin should be diffusely distributed in the cell. After four hours, individual keratin intermediate filaments will have turned over (with a one-hour half-time). The biotin-labeled keratin should now be incorporated into the keratin intermediate filaments, and the distribution of the biotin-labeled keratin and of endogenous keratin in the intermediate filaments should be the same.

Question 62

Actin filaments are a major element of the cytoskeleton. The cytoskeleton provides a framework for the cell and acts as a scaffold that both determines cell shape and positions organelles within cells. For example, the cytoskeleton provides the tracks along which organelles move. At the same time, the cytoskeleton is subject to cleavage by proteins like cofilin. What is the apparent function of cofilin?

Answer 62

Cofilin is an actin filament; it severs protein that binds to actin freed during filament severing. Actin filaments must be dynamic for the cell to able to move. Hence, the ability of cofilin to sever actin filaments creates a dynamic actin cytoskeleton.

Question 63

Why might colchicine, a tubulin-binding drug, be used to treat cancer?

Answer 63

Colchicine, an alkaloid derived from plants, binds tightly to tubulin and inhibits its polymerization, thereby preventing the assembly of the mitotic spindle. The drug acts very quickly and inhibits cell division within a few minutes, hence its usefulness as an anticancer drug.

Question 64

In vitro, at tubulin concentrations intermediate between the critical concentration for assembly at the plus and minus ends, microtubules treadmill. As a consequence of treadmilling, the microtubules move. In what direction do the microtubules move and why?

Answer 64

Tubulin addition is faster at the plus end than at the minus end of microtubules. There will be an intermediate concentration of free tubulin dimer at which there is net growth at the plus end and net loss of tubulin at the minus end—this is treadmilling. As a result of treadmilling, individual microtubules move in the direction of their plus ends.

Question 65

The positioning of various organelles—for example, the Golgi apparatus in cells—is the outcome of a balance between dynein and kinesin. What is the expected distribution of the Golgi apparatus in cells in which the dynein function is inhibited?

Answer 65

Disruption of activity would mean there is no minus-end-directed motor tugging at the Golgi apparatus. Kinesin, a plus-end-directed motor, continues to be active, and the Golgi apparatus (or fragments thereof) will be pulled toward the plus end of microtubules (i.e., the cell periphery)

Question 66

How does Ca2+ molecularly regulate the activity of myosin motors

Answer 66

In striated muscle, actin-myosin contraction is regulated by the binding of Ca2+ to troponin. Troponin binds to actin in the absence of Ca2+ and blocks the binding of myosin to actin. In the presence of Ca2+, troponin changes shape, and myosin can bind to actin. In nonmuscle and smooth muscle cells, myosin activity is regulated by phosphorylation. Myosin light-chain kinase (MLCK) is the responsible enzyme. MLCK activity is regulated by the binding of Ca2+ to calmodulin, which in turn binds to MLCK.

Question 67

The kinetic properties of myosin have been optimized such that it moves very rapidly along actin filaments. Specifically, this movement is accomplished by myosin's very brief attachments to the microfilament along which it moves. In contrast, kinesin moves along microtubules more slowly, with longer periods of attachment between each "step." How might these differences be explained in evolutionary terms?

Answer 67

Because myosin works in conjunction with so many other myosin molecules in muscle contraction, it is unlikely that hundreds of myosin molecules will all let go at once. Therefore, myosin can afford to sacrifice attachment for speed. In contrast, because vesicles and organelles are moved along a microtubule by just a few molecules of kinesin, the chances that they will all let go at once are higher. Therefore, kinesin takes shorter "steps" along the microtubule with longer attachment periods. This increases the chances that the cargo will make it to its final destination without falling off the microtubule.

Question 68

What is one way in which the more rapid growth of actin filaments at one end of the cell (the plus end) compared to the other end (the minus end) is advantageous to the cell?

Answer 68

Since the plus end grows 5 to 10 times more rapidly than the minus end, microfilaments essentially grow in one direction. Thus, correct orientation of the plus end will cause a cell to move toward an attractant without any counteracting force in the opposite direction.

Question 69

Hemidesmosomes link intermediate filaments inside intestinal epithelial cells to the basal lamina, an organized zone of extracellular matrix that underlies the intestinal epithelium. What is the advantage of this to a flexible tissue?

Answer 69

Two forms of contacts hold the intestinal epithelium together. One is cell-cell adhesions, and the other is hemidesmosomes. Hemidesmosomes link integrins as transmembrane proteins to adaptors and intermediate filaments inside the cell and to laminins and the basal lamina outside the cell. Broadly, the desmosome contributes to the strength of the cell layer.

Question 70

Integrins are dimeric transmembrane proteins of the plasma membrane. According to the fluid mosaic model of membrane structure (see Chapter 14), transmembrane proteins should be free to diffuse within the lipid bilayer. Yet, as shown by Tamkun and colleagues in 1986, integrins have a very restricted distribution in the plasma membrane of fibroblasts. The distribution often appears as a series of tracks across the cell surface. How can this distribution be explained?

Answer 70

Integrins interact with various intracellular proteins that bridge to actin filaments or intermediate filaments. These cytoskeletal elements are arranged in a linear manner and create a linear arrangement of integrin, as revealed by immunofluorescence staining.

Question 71

A standard approach to detaching tissue culture cells from substratum is to incubate the culture briefly in saline containing a divalent cation chelator and trypsin. How does this procedure detach the cells from the surface

Answer 71

Both integrins and cadherins require divalent cations for their adhesive properties. Removal of divalent cations by chelators will greatly decrease the adhesive properties of these molecules. Cell adhesion molecules are also cell-surface proteins. Trypsin will digest these and also lead to a decrease in cell adhesiveness: hence cell detachment occurs.

Question 72

In vivo, epithelial cells exhibit basolateral and apical surfaces. The tight junctional complex separates these two distinct cell surfaces from each other and joins the cells into an epithelial cell layer. In vivo cells showing such properties are resistant to ion flow across the cell layer. In vitro, many cultured epithelial cell systems show high electrical resistance but fail to exhibit a polarized distribution of proteins and lipids between the apical and basolateral surfaces. Present a working hypothesis to explain this outcome.

Answer 72

According to this experimental outcome, the existence of tight junction complexes between cells appears insufficient to cause polarized protein and lipid distribution. The presence of tight junctions seems, at most, to permit the generation of polarized protein distributions. Other processes must be important in creating the asymmetry. Targeted membrane trafficking that delivers proteins selectively is one possible hypothesis, although this would not explain why the trafficking is targeted in the first place. Small differences in cell-cell or cell-basal lamina interactions might be important.

Question 73

A given protein is regulated by a protein kinase and a protein phosphatase, and is active when phosphorylated. Which would you expect to INCREASE the protein's activity?

Answer 73

Expression of a mutant, nonfunctional phosphatase that blocks the activity of the cell's normal phosphatases (a dominant-negative phosphatase)

Question 74

Which statement about phosphorylation is correct?

Answer 74

b. The negative charge of the phosphate groups can attract positively charged amino acid side chains of the protein and therefore alter the overall shape of the protein and its binding sites. c. The addition and removal of phosphate groups from proteins occurs in response to signals that specify a change in the state of a cell.

Question 75

Ras is a small G protein that regulates cell proliferation in response to growth factors. When Ras is in its active form, cells proliferate. The activity of Ras is carefully regulated by two other proteins: Ras GEF (guanine nucleotide exchange factor) which stimulates binding of GTP by Ras, and Ras GAP (GTPase activating protein) which stimulates GTP hydrolysis by Ras. The activities of these regulatory proteins are, in turn, also regulated. Ras activity stimulates cell proliferation. Which of the following changes in GAP and GEF proteins might cause a cell to increase proliferation?

Answer 75

a. A nonfunctional GAP. | d. A permanently active GEF

Question 76

Ran's role in regulating nucleocytoplasmic transport is based on a mechanism in which the cell maintains a ______ nuclear concentration and a very low cytoplasmic concentration of ______.

Answer 76

a. low Ran-GDP, Ran-GTP | c. high Ran-GTP, Ran-GTP

Question 77

Of these five steps listed, what is the FOURTH step in making a nuclear protein?

Answer 77

d. The protein folds.

Question 78

Which of the following is FALSE about nuclear localization signal and nuclear export signal?

Answer 78

The NES and NLS have similar sequences and function and work in the same direction.

Question 79

Which of the following is not a function of the nuclear lamina?

Answer 79

it is a point of attachment for the enzymes that synthesize histone. It binds to ribosomes that are conducting protein synthesis.

Question 80

You isolated a protein and tagged it with a fluorescent probe. You microinjected the protein into the cytoplasm of a Xenopus oocyte and observed the localization of the protein as shown in the diagram. The stippled (dotted) region is the area that appears fluorescent. What is the correct interpretation of the experiment?

Answer 80

a. Tail is necessary but not sufficient for the protein to be transported into the nucleus. d. Neither tail nor head alone is sufficient for import into the nucleus.

Question 81

Which of the following things can be moved either in, out, or in both directions through nuclear pores?

Answer 81

a. mRNAs b. snRNAs c. ribosomal subunits d. tRNAs

Question 82

What happens to Ran-GTP after it has been transported to the cytoplasm bound to an importin?

Answer 82

Bound GTP is hydrolyzed to GDP. | Ran-GDP is released from the importin.

Question 83

After transport into the nuclear compartment, what interacts with the importin-cargo complex and what effect does it have on the complex?

Answer 83

Ran-GTP, causes the importin-cargo complex to disassemble

Question 84

Which protein serves as an accessory protein that typically resides in the cytoplasm where it promotes the conversion via hydrolysis of Ran-GTP to Ran-GDP?

Answer 84

Ran-GAP

Question 85

Which of the following correctly describes components that interact to successfully complete the import of cargo proteins from the cytoplasm to the nucleus?

Answer 85

cargo—importin—Ran-GTP complex

Question 86

What would you expect to happen if you massively overexpressed a very active RanGEF in the cytoplasm, such that the normal activity of RanGAP was completely overridden?

Answer 86

Ran would not be recycled back to the nucleus by Ntf2. Any exported Cargo/Exportin/RanGTP complexes would abnormally persist in the cytoplasm due to the excess RanGTP there Importins returning to the cytoplasm could not bind cargo proteins because they would remain bound to the excess RanGTP present there

Question 87

Of these five steps listed, what is the FOURTH step in making a nuclear protein?

Answer 87

An importin binds to the NLS and moves the protein through the nuclear pore into the nucleus.

Question 88

Competition experiments have been used to determine that there are different kinds of exportins for each species of RNA. You are studying importins for proteins. You translate in vitro two different nls-containing proteins. Protein A is fused to GFP, while the other (protein B) is left unlabeled. Which statement below correctly describes results and interpretations that you might obtain from observing the behavior of these proteins injected into the cytoplasm of oocytes?

Answer 88

b. Injecting an excess of protein B would decrease import of Protein A into the nucleus if both use the same importin. c. Injecting an excess of Protein B would not alter import of Protein A into the nucleus if both use different importins.

Question 89

f gold particles of varying size are injected into the cell, they can be seen to enter what structure? What technique was used to visualize these events?

Answer 89

nuclear pores, electron microscopy

Question 90

Your investigations discover a protein that is normally present in the cytoplasm, but when you treat cells with a hormone, it moves to the nucleus. What is the most likely explanation?

Answer 90

The nuclear localization signal is initially buried in the protein until, after hormone treatment, the protein changes its conformation thereby exposing the NLS and allowing transport to the nucleus

Question 91

You take the normal NLS and fuse it to a non-nuclear protein like ovalbumin. Next, you fuse GFP in-frame with this new protein and inject it into the cytoplasm of the cell and use microscopy to monitor what happens to the protein. What would you expect to see?

Answer 91

GFP would be visualized in the nucleus of the cell.

Question 92

What happens to Ran-GTP after it has been transported back to the cytoplasm following its release of cargo from the importin α/β molecule?

Answer 92

It is hydrolyzed to Ran-GDP. | The Ran-GDP is released from the importin β subunit.

Question 93

How can an importin subunit be transported back to the cytoplasm?

Answer 93

a. Importin β subunit can diffuse back through the nuclear pore complex bound to RanGTP. b. Importin a can be transported back to the cytoplasm by Cas, an exportin, in a trimeric complex that included Ran-GTP.

Question 94

Your undergraduate honors project is to isolate a new protein from the nucleus of the cell. In order to test that you got the right protein, you decide to inject it into the cytoplasm of frog eggs and see if it is transported into the nucleus. You are relieved to find that this experiment works, and your advisor breaks out the champagne (which is probably a violation of some university policy). Next, you decide to isolate the domain containing the targeting signal by taking 5 equal sized pieces of the protein and testing them individually. You find that no matter which piece you test, none are transported into the nucleus. However, when you take pieces 2 and 3 and mix them, both pieces are imported. No other combination works except those containing pieces 2 and 3. Your conclusion is:

Answer 94

2 and 3 are both necessary for targeting. | Neither 2 nor 3 alone are sufficient for targeting

Question 95

Once a protein-exportin-RanGTP complex has reached the cytoplasm, all of the following occur except:

Answer 95

Ran-GDP is directly rephosphorylated to GTP on the nuclear side. (Explanation: RanGEF exchanges GTP for GDP on Ran once RanGDP enters in nucleus; there is no phosphorylation of RanGDP; the production of GTP is a separate set of biochemical reactions)

Question 96

Which of the following is FALSE about the Ran G-protein

Answer 96

RanGTP does not encounter GAP when it returns to the cytoplasm.

Question 97

If a nuclear localization signal were added to the gene encoding a lysosmal protein, where would the protein likely be found?

Answer 97

Lysosomes

Question 98

Which of the following is true regarding the Endoplasmic Reticulum

Answer 98

a. site of co-translational glycosylation of secretory proteins b. site of phospholipid synthesis c. site of GPI lipid anchoring of membrane proteins d. site of synthesis of oligosaccharides for co-translational attachment to proteins

Question 99

Which one of the following is not a function of the ER?

Answer 99

Site of degradation of phagocytosed bacteria.

Question 100

Which of the proteins below is(are) not made on the RER or typically glycosylated and thus converted to glycoproteins?

Answer 100

mitochondrial enzymes

Question 101

Which step listed is not true about protein synthesis on the rough ER?

Answer 101

The signal peptide binds to the hydrophobic site on the ribosome.

Question 102

All of the following are essential components of the complex that directs a nascent protein into the lumen of the RER except:

Answer 102

Cis golgi

Question 103

What are the two sites within a cell at which protein synthesis is generally thought to occur?

Answer 103

cytosolic surface of RER and free ribosomes

Question 104

What effect does the docking of the SRP/ribosome/growing polypeptide chain complex to the SRP receptor have on protein synthesis?

Answer 104

The SRP will release from the ribosome. | Protein synthesis resumes

Question 105

All of the following hold the SRP-ribosome-nascent polypeptide complex onto its binding site on the RER membrane except:

Answer 105

Its interaction with sphingomyelin.

Question 106

Hydrophobic alpha-helices that become transmembrane domains, and that interact with the translocon machinery to determine their orientation in the membrane are also known as:

Answer 106

topogenic sequence

Question 107

What poorly understood event occurs during the docking of the ribosome to the translocon and the subsequent release of SRP and SRP receptor from the ribosome and translocon?

Answer 107

GTP binding and hydrolysis by both the SRP and SRP receptor

Question 108

Many integral membrane proteins have a single segment in the nascent chain that serves as both a signal sequence for binding SRP and a sequence that codes for insertion into the lipid bilayer. What is it called?

Answer 108

internal start transfer or signal-anchor sequence

Question 109

Stop-transfer sequences typically include ___________.

Answer 109

at least 15 continuous hydrophobic or uncharged amino acids

Question 110

How is the orientation of membrane proteins in the membrane thought to be accomplished?

Answer 110

During synthesis, the translocon inner lining orients the nascent polypeptide so the more positive end of the TM domain faces the cytosol.

Question 111

What would be the effect on synthesis of an ER-targeted protein if you overexpressed a mutant SRP that was able to bind to the signal sequence and the ribosome, but could not interact properly with the SRP receptor?

Answer 111

Synthesis of the protein would be inhibited

Question 112

What would be the effect on synthesis of an ER-targeted protein if you overexpressed a mutant SRP that was unable to bind to, and interact properly with, the signal sequence and ribosome?

Answer 112

The protein would end up a cytosolic protein.

Question 113

If free ribosomes in the presence of RER vesicles with a mutant, nonfunctional SRP receptor that cannot bind the SRP, are placed in a test tube with mRNA for a protein with an n-terminal signal sequence, and everything else needed synthesize proteins, along with the addition of SRP, where are the proteins found after their production?

Answer 113

There are no full-length proteins made

Question 114

You isolated cellular components to study the transport of proteins in a cell. After a lot of hard work, you take a vacation, but you forget to write down how far along in the purification you progressed. If you put some of your preparation in a test tube, and then add mRNA for a secretory protein, you find that the mRNA is completely translated and the new protein is sensitive to protease. Your preparation has:

Answer 114

Ribosomes and microsomes, but no SRP.

Question 115

Which of the following could prevent a secretory protein from being secreted normally?

Answer 115

a. A mutation that causes addition of mannose-6-phosphate to the oligosaccharide structure. b. A mutation that prevents addition of N-linked oligosaccharides in the ER. c. A mutation that results in deletion of the start transfer sequence. d. Addition of an ER-retrieval sequence

Question 116

Which of the proteins below is (are) NOT made on the RER or typically glycosylated and thus converted to glycoproteins?

Answer 116

Nuclear proteins.

Question 117

Nuclear proteins.

Answer 117

soluble proteins found inside lysosomes | secretory proteins

Question 118

You are studying two proteins, with the goal of understanding how they might be targeted to the ER. Protein A has the same molecular weight when synthesized in vitro without rough ER (RER) vesicles as in the presence of RER vesicles. Protein B is apparently 10-15 amino acids shorter when synthesized in the presence of vesicles. What statement about the results is true?

Answer 118

Protein B uses an n-terminal signal sequence.

Question 119

The following are involved with synthesis of a secreted protein. Which is the FOURTH step of the five listed?

Answer 119

The SRP dissociates from the ribosome and start transfer sequence

Question 120

One of the hallmarks of co-translational import of proteins into the ER is the insensitivity of proteins to protease if translation occurs in the presence of SRP, SRP receptor and RER vesicles. You are studying synthesis of what you think is a protein that resides in the extracellular matrix (secreted from the cell). You do the classic protease experiment and find that the protein is not degraded at all by protease. Which statement below is correct?

Answer 120

The protein uses an n-terminal signal sequence.

Question 121

In those cases where post-translational transport of proteins into the ER lumen occurs, how do proteins enter the ER?

Answer 121

They go through the same channels as the proteins that enter co-translationally. They are held in an unfolded conformation by chaperones so they may pass through translocons.

Question 122

If free ribosomes in the presence of RER vesicles are placed in a test tube with mRNA for a protein with an n-terminal signal sequence, and everything else needed to synthesize proteins, together with a mutant, nonfunctional SRP that cannot bind the signal sequence, where would the proteins be found after their production?

Answer 122

Floating free in the aqueous solution in the test tube.

Question 123

What is the purpose of molecular chaperones like BiP and calnexin?

Answer 123

They associate with unfolded and misfolded proteins and help them regain their native structure.

Question 124

Where are the signal peptidase and oligosaccharyltransferase located?

Answer 124

Embedded in the RER membrane facing the lumen. | Associated with the translocation channel.

Question 125

If you engineered a new protein by taking the gene for a single pass transmembrane protein that has an n-terminal signal sequence and remove the stop transfer sequence from the middle of the gene, what is a possible destination of the new protein

Answer 125

None is a possible destination.

Question 126

If you inserted into a cell a gene that codes for a protein that normally enters the nucleus because it has a c-terminal NLS, but you added an N-terminal ER signal sequence to it, what would you expect the FINAL destination of the protein to be?

Answer 126

Secreted from the cell.

Question 127

Which of the following is not a therapeutic effect of aspirin?

Answer 127

Feedback A: Correct! A substance stored in blood platelets, platelet-derived growth factor (PDGF), has this effect. Feedback B: Incorrect. By inhibiting the synthesis of prostaglandins, aspirin reduces inflammation and pain. Feedback C: Incorrect. By inhibiting the synthesis of thromboxane, aspirin inhibits platelet aggregation and blood clotting. Feedback D: Incorrect. By inhibiting the synthesis of prostaglandins that cause cell proliferation, aspirin reduces the incidence of colon cancer

Question 128

Integrins are transmembrane proteins that connect

Answer 128

Feedback A: Incorrect. Integrins are not located in the nuclear membrane. Feedback B: Correct! Integrins span the plasma membrane and provide a communication link between the extracellular matrix and the cytoskeleton within the cell. Feedback C: Incorrect. Focal adhesions and hemidesmosomes are two kinds of cell matrix junctions, and they are not connected to one another via integrins. Feedback D: Incorrect. These are both components of the cytoskeleton within the cell, and integrins are transmembrane proteins that connect components on opposite sides of a membrane

Question 129

What are some examples of endocrine, autocrine, and paracrine signaling?

Answer 129

Autocrine signaling is when a cell sends signals to itself, paracrine signaling is when neighboring cells signal to each other, and endocrine signaling is when cells, via hormones, relay messages throughout the body.