Exam 3

Question 1

L15. List and briefly explain four mechanisms by which an organism/species could acquire “new” protein coding genes.

Answer 1

1. Mutation: single base changes, insertions, or deletions in the protein coding region of a gene.
2. Gene duplication (and divergence by mutation): the duplication of a protein-encoding gene through
   unequal-crossing over, and subsequent divergence through the accumulation of genetic changes
   (mutations).
3. Exon/domain duplication: the duplication of a one or more exons (with intevening introns) in a
   protein-encoding gene through unequal-crossing over, and subsequent divergence through the
   accumulation of genetic changes (mutations).
4. Exon/domain shuffling: the shuffling of one or more exons from one gene to the next through the
   mobilization of transposons.
5. Horizontal gene transfer: the transfer of a protein coding region from one organism to another

Question 2

L15. What is a “pseudogene,” and how might pseudogenes arise? What features of a normal gene would you expect a pseudogene to include? What distinguishes a pseudogene from
a “real” gene?

Answer 2

A pseudogene is a copy of a functional gene (ie protein coding gene) that no longer codes for a functional product, usually because it contains multiple stop-codons, insertions, and or
deletions. If generated by a gene duplication/divergence, a pseudogene might be expected to
contain a similar array of exons and introns as found in its functional counterpart. It may also
include any upstream/downstream control elements of sequences (promoters, polyA signals, etc)…
but it may not, depending on the extent of the duplication.

So, why do pseudogenes contain multiple genetic lesions, while the “normal” gene does not?
What keeps stop codons from accumulating in the “normal” functional copy of the gene?

Question 3

L15. Some pseudogenes in eukaryotes include introns, just as their normal, functional counterparts.
However, many (most?!) mammalian pseudogenes lack introns found in their related functional
genes, and most of these include a poly A region near their 3’ end (sense strand). Propose a
mechanism by which such a “processed” pseudogene might be created. Would you expect
processed pseudogenes to be expressed in the same cells as its “normal” counterpart? Why or
why not?

Answer 3

“Processed” pseudogenes are thought to have arisen by reverse-transcription of a mRNA…
and subsequent insertion of the “cDNA” into the cellular genome, in a manner very similar to the
transposition of a retrotransposon. Thus, the processed pseudogene would lack introns, and might
include a “poly-A tail” (which is not normally coded by the genome).
It is unlikely that processed pseudogenes would be expressed in a normal manner, since the
retrotransposition process would not include the promoter and other control elements. It is possible
that a processed pseudogene might be expressed in some tissues, if it was inserted downstream of
another promoter

Question 4

What is the difference between “germ-line” and “somatic cells?

Answer 4

The term “somatic” or “somatic cells” refer to cells of the body, while “germ-line” cells are
those that will contribute to the next generation: ie, reproductive cells such as sperm, eggs, and
their precursors.

Question 5

L15. Transposable elements such as Alu, L1, and SVA, are scattered throughout the human
genome, and recent studies suggest that transposition occurs as frequently as ~1/20 live births. If
a transposable element jumped into an important gene in one of your cells when you were a baby
and caused a disease, is it likely that YOUR child would also have the disease? Briefly explain.

Answer 5

No, unless the transposon “hopped” into a gene in your germ-line [which likely would not
cause a phenotypic change (“disease”) in you].

Question 6

L15.
Protein domains related to EGF, serine proteases, and containing
kingle domains, are found in a wide variety of animal proteins,
and it is not uncommon to find all three domains within a single
polypeptide. What genetic mechanism could account for the
combination of these domains found in proteins such as urokinase
and clotting factor IX

Answer 6

Exon/Domain shuffling through the mobilization of transposons.

Question 7

L15. Hypothet gene encoding proteins A. Propose a model for the origin of gene 1 and gene 2. What evidence supports the model?

Answer 7

pose a model for the origin of HRP-1 and HRG-2. What evidence supports your model?
The simplest model for the origin of HRP-1 and HRG-2 is gene duplication and divergence, as
evident by: (1) sequence similarity, (2) similarity in gene structure (introns/exons), and (3)
clustering with the hypothetin gene on chromosome 1 (which argues against whole genome
duplication)

Question 8

L15. Hypothet Genes B
You were surprised to find that HRG-2 and HRG-3 have single base changes (*) introducing
multiple stop codons in their coding regions. What would you conclude about HRG-2 and HRG3?

Answer 8

The accumulation of stop codons indicates they cannot make functional product. Both are
“pseudogenes.”

Question 9

L15. Hypothet Genes C
Examination reveals that HRG-3 has no upstream promoter, lacks the intron found in the the
other genes, and has a stretch of ~100 A’s follows exon 2 (in the sense strand). Propose a model
for the origin of HRG-3?

Answer 9

All of these characteristics (polyA, introns removed, no promoter) are
found in “processed” pseudogenes, in which a cellular mRNA is reverse transcribed to make a cDNA
and then inserted (more or less randomly) into the genome.

Question 10

L15. Hypothet Genes D
You made knock-outs of the Hypothetin and HRP-1 genes, and introduced them individually
into hypothets. Both were found to be essential (knocking them out was lethal). What would you
conclude about the function(s) of Hypothetin and HRP-1.

Answer 10

The products of both the hypothetin
gene and HRP-1 gene must have essential functions, and the functions are DIFFERENT. If the two
products had the same/similar functions, they would be redundant and individual knockouts would
not be lethal (you could knock out one, and the other would compensate).

Question 11

L16. Start Codons?

Answer 11

AUG

Question 12

L16. Stop Codons

Answer 12

UAA
UAG
UGA

Question 13

L16. Which reading window would be found in the miiddle of a polypeptide?

Answer 13

The one without multiple stop codons inserted.

Question 14

L16. What determines which product was actually formed during translation?

Answer 14

During translation, the reading frame, and thus the sequence of the protein product, is
determined by the “start codon (AUG),” which encodes methionine. With few exceptions, protein
synthesis begins with methionine (formyl-methionine, or fmet, in prokaryotes). Often, the initial
methionine is cleaved post-translationally

Question 15

L15. How might you explain the tandem organization of Ig domains in IgG
heavy chains? What cellular mechanism might be involved?

Answer 15

Domain duplication by unequal crossing over (homologous

recombination), followed by divergence

Question 16

L15. How would you explain the obvious structural similarities of IgG heavy and light chains?
What cellular mechanism might be involved?

Answer 16

Gene duplication by unequal crossing over (homologous recombination), followed by
divergence.

Question 17

L16. Sketch cloverleaf tRNA

Include, acceptor system/sequence, correct conjugation amino acid and various loops.

Answer 17

DIAGRAM

Question 18

L16. In prokaryotes, transcription and translation can
occur simultaneously: ribosomes can begin translating a
mRNA before transcription of the mRNA is complete.
Why can’t this happen in eukaryotes?

Answer 18

can’t this happen in eukaryotes?
In prokaryotes, both transcription and translation
occur in a common compartment (the cytoplasm). Thus,
ribosomes have access to the 5’-end of nascent mRNA. In
eukaryotes, transcription and translation occur in separate compartments (the nucleus and
cytoplasm, respectively). Nascent transcripts have to be processed (capping, splicing,
polyadenylation) before they are transported to the cytoplasm for translation.

Question 19

L16. How does the process of translation initiation differ in prokaryotes and eukaryotes?

Answer 19

Prokaryotes translation begins with formyl-methionine (fmet). In addition, polycistronic
prokaryotic mRNAs include conserved “ribosome binding sites” (= “Shine-Delgarno” sequences)
upstream from each start codon (AUG).
In eukaryotes, the small ribosomal subunit with tRNAi
met and eIFs is loaded at the 5’ end of
the mRNA (facilitated by 5’ cap and poly A tail) and scans to find the start codon (AUG).
[Though we didn’t discuss it in lecture, the order of assembly of the initiation complexes also
differs in prokaryotes and eukaryotes].

Question 20

L16. Many antibiotics act by specifically inhibiting the function of prokaryotic ribosomes, without
affecting the ribosomes found in the cytoplasm of eukaryotic cells. Yet, many of these antibiotics
have side effects that limit their usefulness. Explain.

Answer 20

Eukaryotic mitochondria contain ribosomes that are very similar to those found in their
prokaryotic ancestors. Thus, they share some of the same sensitivities to antibiotics. Poisoning of
mitochondrial ribosomes (and thus mito translation) can cause side effects seen with some
antibiotics.

Question 21

L16. Outline 3 steps in the elongation of a polypeptide by a ribosome.

Answer 21

(1) The next charged AA-tRNA binds to A site of ribosome, as a complex with EF-1-GTP. GTP
hydrolysis allows release of EF-1-GDP.
(2) The new peptide bond is formed by breaking the ester bond linking the growing polypeptide to
the tRNA in P site, and transferring the polypeptide to the free amino group of the AA-tRNA in the
A-site. The energy for peptide bond formation comes from that stored in the peptidyl-tRNA bond
that is broken. The rxn is catalyzed by the peptidyl transferase activity of the ribosome,
associated with the rRNA. Coincident with peptide bond formation, the large ribosomal subunit is
translocated 1 codon (3 bases) in the 3’ direction.
(3) EF-2-GTP binds to ribosome. GTP hydrolysis releases EF-2-GDP, and small subunit is
translocated 1 codon (3 bases) in the 3’ direction. The empty tRNA from step 2 is released from
the E site of the ribosome.

Question 22

L16. How does GTP hydrolysis contribute to the fidelity of translation?

Answer 22

GTP hydrolysis by EF-1 (EF-Tu in prkaryotes) provides the energy for “proof reading” of the
aminoacyl-tRNA loaded into the A-site of the ribosome. Peptide bond formation cannot occur until
EF-1 hydrolyses its bound GTP to GDP, which releases EF-1 from the aa-tRNA and ribosome. If the
wrong aa-tRNA is bound, it cannot base pair with the codon, and is released before EF-1 can
hydrolyze GTP and be released. This allows another AA-tRNA to enter the A-site…

Question 23

L16. What provides the energy for peptide bond formation?

Answer 23

The energy stored in the ester bond linking the AA to the tRNA… the equivalent of two high
energy bonds used during charging

Question 24

L16. What provides the energy for translocation of the ribosomal subunits?

Answer 24

The energy of charging powers peptide bond formation and translocation of the large
ribosomal subunit. The GTP hydrolyzed by EF-2 (EF-G in prokaryotes) powers translocation of the
small subunit.

Question 25

L16. How is translation terminated?

Answer 25

In a normal “wild type” cell, there are no AA-tRNAs which can base-pair with stop codons
(UAG, UAA, UGA). When the ribosome reaches a stop codon, it (the stop codon) is recognized by
one of the “release factors.” The release factor binds in the A-site and causes the ribosome to
hydrolyse (break by adding water) the ester bond linking the polypeptide to the tRNA in the P-site.
The polypeptide is released… the ribosome subunits dissociate from the mRNA to be re-used.

Question 26

L16. E. coli cells containing the Amber suppressor gene “read through” the UAG stop codon by
inserting tyr in the protein instead of stopping translation. Propose a molecular
mechanism/explanation for the amber suppressor

How to test hypothesis?

What precautions might cells evolve to prevent such suppressor activities from being
detrimental?

Answer 26

Cells bearing the amber suppressor mutation express a mutated tRNAtyr , in which the first
position of the anti-codon has been changed to C, allowing it to base pair with the UAG (stop) codon
instead of the UAC or UAU codons for tyr

B:
Clone and sequence the gene encoding the tRNA for tyr

C:
The coding region of most mRNAs are followed by multiple stop codons (if one is
“missed,” it is followed by another)

Question 27

L16. Compare and/or contrast the cost (in high energy phosphate bonds) of adding a single amino
acid to a protein with the cost of adding 3 nucleotides (= 1 codon) to a mRNA. How would you
reconcile this comparison with the statement made in lecture that translation was “energetically
expensive,” accounting for as much as 80% of the energy used by cells?

Answer 27

It takes six high energy bonds to add one codon to an mRNA: two per nucleotide added. Yet,
it takes “only” four high energy bonds to add an amino acid (two for charging that also power
peptide bond synthesis, one for binding the aa-tRNA to the A-site, and one to translocate the
ribosome). Why, then, is protein synthesis more demanding of energy? Because a single mRNA
molecule can be translated many, many times, making many copies of the same protein.

Question 28

L16. Eukaryotic heat shock proteins were first identified in the fruit fly Drosphila melanogaster,
where they were expressed in response to brief treatments at elevated temperatures. Expression
of the HSPs conferred resistance to subsequent, more prolonged treatment at higher
temperatures. Based on your understanding of the cellular roles of HSPs, how might they help
cells resist or recover from high temperature?

Answer 28

High temperatures “denature” (unfold) proteins. This inactivates the protein… but it also
can result in formation of large protein precipitates, as the exposed hydrophobic regions of
proteins (usually buried in the center, now exposed) aggregate. Both protein denaturation and the
formation of these protein aggregates can be toxic to the cell. “Heat shock proteins” like those of
the HSP70 family can bind the hydrophobic regions of unfolded proteins, preventing formation of
protein aggregates. Along with HSP60s, they may also help refold some proteins. Both HSP70s and
HSP60s use cycles of ATP hydrolysis during the binding/release/folding process

Question 29

L16. Briefly explain the “fate” of mis-folded or damaged proteins? B. How are they “marked”
for disposal?

Answer 29

Damaged or mis-folded proteins are marked for degradation by addition of
polyubiquitin. They are then targeted to the proteosome fordegradation/recycling.

Question 30

L16. Why can mis-folded proteins be so damaging to cells?

Answer 30

Mis-folded or unfolded proteins often
have hydrophobic regions exposed… these exposed hydrophobic patches can lead to formation of
large aggregates of unfolded or mis-folded protein in the cytoplasm. Accumulation of these
agrgegates can be toxic (particularly to neurons), and are the cause of several human diseases

Question 31

L16. The largest protein known, up to 1 m in length, is Titin. Titin is found in myofibrils (the
contractile machinery of muscle cells) and functions as a “molecular spring” that provides much
of the intrinsic elasticity of muscles (wikipedia). Titin contains 244 folded protein domains
connected by unstructured regions. The folded domains unfold under tension, and refold when
the tension is released. In ECB4 question 7-16B, you calculated the probability of making an
error-free Titin molecule (what was that probability… what fraction of Titin molecules are error
free?). Given that small fraction of error-free Titin molecules made… how can it function in
muscle? How might muscle cells tolerate such an error rate?
Just speculating…

Answer 31

If the sequence error causes the Titin molecule to fold incorrectly, it should be tagged with
ubiquitin and degraded in the proteosome.
However, much of Titin’s 3-D structure is unstructured, without any defined secondary
structure or folding. Amino acid substitutions in these regions (particularly if they are conservative
substitutions) might have little effect on the structure or function of the protein molecule.
Likewise, conservative substitutions in the structured domains also might yield a functional protein.

Question 32

L17. Which would you expect to diffuse into/out of the nucleus by simple
diffusion? Why?

Answer 32

H2O and dNTPs

Both are sufficiently small to diffuse through the disordered mesh of protein in the central channel of the nuclear.

Question 33

L17. Draw simple diagrams comparing the
molecular mechanisms of nuclear import
and nuclear export,

Answer 33

DIAGRAM

Question 34

L17. What physical structure or
feature of cargo proteins signal for
nuclear transport?

Answer 34

NLS: A sequence
of basic amino acids: lysines and
arginines. NES: A leucine-rich
sequence.

Question 35

L17. What protein(s) act as receptors

for that signal?

Answer 35

Karyopherins:”

Importins and Exportins

Question 36

L17. How is the cargo transported

across the nuclear envelope?

Answer 36

explained in DIAGRAM

Question 37

L17. How is GTP hydrolysis and
exchange coupled to import/export
reactions? Why is nuclear transport
blocked by non-hydrolyzable
analogs of GTP?

Answer 37

GTP hydrolysis is
required for Ran to release
transportins. In the presence of the
non-hydrolyzable GTP analogs, all of
the transportins become
bound/sequestered in complexes
with Ran, and nuclear transport will
come to a halt...

Question 38

L17. Based upon your understanding of the mechanisms nuclear targeting, where would you
expect to find proteins containing (A) a sequence consisting of multiple basic amino acids
(KKKRK)? (B) a leucine-rich sequence of amino acids (LLPPLERLTL)?
(C) both basic (KKKRK) and leucine-rich (LLPPLERLTL) amino acid
sequences?

Answer 38

A) Nucleus (an NLS).
B) (Cytoplasm (an NES).
C) Shuttling in and out of the nucleus. Where it is the most abundant at steady state depends on the strength of the two signals.

Question 39

L17. Compare and/or contrast the mechanisms/paths for importing proteins into the
mitochondrial matrix and chloroplast stroma.

Answer 39

Proteins are targeted to the mitochondrial matrix and chloroplast stroma by sequence
signals (usually) found at their N-termini. In both cases, the targeting signals (mitochondrial “signal
sequences” and chloroplast “transit peptides” ) are “amphipathic” -helices, in which polar and
charged amino acids are clustered on one side. Although both signals are amphipathic helices, the
signals are specific, and there is no “cross talk” between the mito and chloroplast targeting import
systems.
In both cases, signals
are recognized by receptors
in the outer membrane, which
then pass them to
translocators in the outer
and inner membranes (“TOM”
and “TIM” in mitos; “TOC”
and “TIC” in chloroplasts).
The signals are cleaved in the
matrix or stroma

Question 40

L17. What is the energy source

for importing proteins into the mitochondrial matrix?

Answer 40

(1) ATP (and cytoplasmic HSP70) is required to unfold proteins prior to import into the matrix;
(2) The electrochemical proton gradient is required (for targeting the signal to TIM?);
(3) ATP and a matrix HSP70 are required for import and to refold the protein in the matrix.

Question 41

L17. How are proteins targeted to other mitochondrial compartments and membranes, such as the
intermembrane space or inner membrane?

Answer 41

There are actually at least two paths to the IMS/inner
membrane:
(1) A “stop transfer” signal halts translocation through TIM (the “direct route”)… DIAGRAM
or (2) After transport into the matrix, cleavage of the matrix targeting signal exposes a second
signal which engages machinery to transport the protein back across the inner membrane (the “in and
out route”) DIAGRAM

Question 42

L18. How to make a Type I protein (N in ER, C in Cytoplasm)

Answer 42

an N-terminal “start transfer” signal or
“signal peptide” starts translocation of the N-terminal domain into the ER, and is then cleaved off
by signal peptidase and degraded. A “stop transfer” domain serves as the transmembrane (TM)
domain, and the C-terminus is completed in the cytoplasm.

Question 43

L18. How to make a type 2 protein (N in cytoplasm, C in ER)

Answer 43

To make “type 2” membrane proteins, translation of the N-terminal domain occurs in the
cytoplasm. An internal hydrophobic domain acts as an uncleaved “start transfer” sequence and
becomes the transmembrane domain. As translation continues, the C-terminal domain is transferred
into the

Question 44

L18. How to make a type 3 (multipass) protein with N and C both in ER

Answer 44

To make a multipass (type 3) transmembrane protein with its N-terminus in the ER lumen, a
cleavable signal sequence (“signal peptide”) starts translocation of the first ER domain before being
cleaved. The signal peptide is removed by signal peptidase and degraded. A stop transfer sequence
stops translocation and serves as the first transmembrane domain, and the next domain is made in
the cytoplasm. Start and stop transfer sequences then alternate and serve as the additional
transmembrane domains. If the last “topogenic sequence” is a start transfer (as in this example),the C-terminus will be in the ER lumen. If the last topogenic sequence is a stop transfer sequence,
the C-terminus will be in the cytoplasm

Question 45

L18. How to make a type 3 (multipass) protein with both the N and C terminals in the cytoplasm

Answer 45

To make a multipass transmembrane protein with its N-terminus in the cytoplasm, the first
domain is made in the cytoplasm. An internal, uncleaved start transfer sequence serves as the first
transmembrane domain and begins translocation of the first ER domain. Stop and start transfer
sequences then alternate and serve as the additional transmembrane domains. If the last “topogenic
sequence” is a stop transfer (as in this example), the C-terminus will be made in the cytoplasm. If
the last “topogenic sequence is a start transfer sequence, the C-terminus will be in the ER lumen.

Question 46

L18. What features of “start-transfer” and “stop-transfer” sequences allow them to serve as
membrane spanning domains?

Answer 46

Both start and stop transfer domains are non-polar sequences of
~15-25 amino acids predicted to form -helices.
Although similar in character, “start transfer” and “stop transfer” sequences are not
entirely interchangeable. Often, there are polar are charged amino acids flanking the topogenic
sequence, which give it directionality. For example, highly charged lysine side chains often anchor
the cytoplasmic side of the topogenic sequence. Thus, a topogenic sequence immediately preceded
by one or more lysines typically acts as a “start transfer” sequence, while a sequence followed
closely by one or more lysines often acts as a “stop transfer” (draw it out, to see how the lysines
end up in the cytoplasm in each case).

Question 47

L18. Sketch the expected orientation of LDL-R in the membranes of the
RER, the Golgi apparatus, and the endosome.

Answer 47

In all of the above compartments, the C-terminus is in the
cytoplasm and the N-terminus is in the lumen. The topology of the
protein (with respect to membranes) is established during synthesis,
and does not chamge during vesicle trafficking..

Question 48

L18. Based on your knowledge of the mechanisms for targeting membrane proteins,
predict/diagram the organization of “topogenic” sequences required to insert the nascent LDL-R
into the membrane of the endoplasmic reticulum. Point out structural features important to the
function of these “topogenic” sequences.

Answer 48

LDL-R is a type I membrane protein, so the map of topogenic sequences would look similar to
that in Q1A (above), with a cleavable signal peptide and an internal stop transfer sequence:

Question 49

L18. Outline the mechanism by which the LDL-R is inserted into the membrane of the ER,
indicating the role of the “topogenic” sequences and identify any important components of the
targeting/translocation machinery.

Answer 49

Translation begins with assembly of the ribosome on the LDL-R mRNA (see L16 notes), then…
1. SRP binds the signal peptide on the N-terminus of LDL-R, once it is exposed outside the
ribosome… Binding of SRP transiently arrests translation and targets/docks the complex to
the SRP-R and translocon in the RER membrane.
2. SRP is released, translation is resumed, and translocation of LDL-R into the ER lumen
commences. At some point, signal peptidase cleaves the signal peptide from the N-terminus,
which is then degraded (cleavage can occur before translation/translocation is completed).
3. When the stop transfer sequence is translated, is exposed outside the ribosome, and enters
the translocon, it signals the translocon to open laterally, and release LDL-R. The stop
translocation sequence functions as the transmembrane domain.

Question 50

L18. In addition to the “topogenic sequences,” what three components of the translocation
machinery are needed for targeting and inserting EMP into the membrane of the RER? List the
components in order of their action, and briefly describe the function(s) attributed to each
component

Answer 50

1. Signal recognition particle (SRP).
   a. Binds “start transfer” sequence (signal sequence; domain B in this example)
   b. Transiently arrests translation
   c. Targets nascent polypeptide/ribosome/mRNA complex to RER membrane by binding to
   SRP-receptor.
2. SRP-receptor (an integral membrane protein in RER).
   Binds SRP bound to nascent polypeptide. Hands protein/ribosome complex to translocon.
3. Translocon (composed of sec61 ).
   An aqueous channel in the RER membrane. Ribosome docks to translocon. Start transfer (B
   in this example) anchored in translocon and will become transmembrane domain. First
   lumenal domain (C in this example) of nascent polypeptide is translocated thru the
   translocon into ER lumen.

Question 51

L18. Describe the mechanism by which prolactin is translocated across the ER membrane,
including the identity of any required cellular factors.

Answer 51

Translation begins with assembly of the ribosome on the prolactin mRNA (see L16 notes), then…
1. SRP binds the signal peptide on the N-terminus of preprolactin/prolactin, once it is exposed
outside the ribosome…
2. Binding of SRP transiently arrests translation and targets the complex to the SRP-R and
translocon in the RER membrane.
3. SRP is released, translation is resumed, and translocation of LDL-R into the ER lumen
commences. At some point, signal peptidase cleaves the signal peptide from the N-terminus,
which is then degraded (cleavage can occur before translation/translocation is completed).
4. Completion of translation releases prolactin into the lumen of the RER. It will be
transported to the Golgi in transport vesicles, and then to the plasma membrane, where it
will be secreted.

Question 52

L18. What is the functional significance of the difference in molecular mass (~2 kDa) between
preprolactin and prolactin? What is the structural nature of the region of polypeptide removed.

Answer 52

The extra mass corresponds to the mass of the signal peptide that is cleaved off in the ER.
The signal peptide would be a non-polar (hydrophobic) -helix ~15-25 amino acids long

Question 53

L18. Compare and interpret the results obtained with samples A-C. What do these results tell you
about the timing of targeting and translocation to the ER? What two indicators demonstrate that
the protein product has been translocated into microsomes

Answer 53

In the absence of SRP or RER membranes (rxn A), the mRNA is translated to make
preprolactin (PPL; the 25 kDa product containing the N-terminal signal peptide). It cannot be
targeted, so does not get processed to the mature 23 kDa form (prolactin, or PL)… and it is
susceptible to protease digestion.
When translated in the presence of RER microsomes (rxn C; which contain loosely bound
SRP), the signal peptide targets the nascent PPL (during synthesis) to the RER membrane, where it
can be translocated into the RER microsomes. In the RER microsomes, the signal peptide is
proteolytically removed to make 23 kDa PL… evidence that the product is in the RER. Translocation
into the ER can be confirmed by showing the 23 kDa product is now inaccessible to the added
protease, because it is inside the RER When translation is completed in the absence of RER microsomes (rxn B), and they are
added later (reaction B), the nascent 25 kDa PPL is NOT targeted and is NOT translocated and
processed. It is this result that tells us that targeting and translocation is COTRANSLATIONAL

Question 54

L18. What do the results of samples D and E tell you about the functions of SRP

Answer 54

In rxn D, SRP is added in the absence of RER microsomes. No product is detected. This tells
us that SRP arrests translation.
In rxn E, RER is added to the SRP-arrested reaction from rxn D… and now a product is made
that is targeted to the RER, processed to mature PL, and is protease resistant. In this rxn, addition
of the RER microsomes has relieved the SRP-induced arrest and restored targeting. This suggests
that SRP functions to arrest translation, providing time for the nascent protein/ribosome complex
to be targeted to the RER membrane

Question 55

L18. Why can’t a classical ER signal sequence located at the C-terminus function target a protein
for insertion into the RER

Answer 55

For an ER signal sequence to bind SRP and function in targeting,
it must be exposed outside the ribosome. The channel thru which the nascent polypeptide
exits the ribosome spans about 70 amino acids of the nascent polypeptide. Thus, the Cterminus
of the nascent polypeptide is never exposed to SRP and the targeting machinery
until translation is completed. There is a separate mechanism for targeting C_terminally
anchored membrane proteins to the RER

Question 56

L18. Based on the function of

SNAREs, how would tetanus toxin block neuronal transmission?

Answer 56

Cleavage/destruction of VAMP

would block docking and fusion of synaptic vesicles, and thus release of neurotransmitter

Question 57

L18. Arf and Sar1 are a small
GTPases, related to Ras, that are
required for transport by COPcoated
vesicles. Diagram and
briefly explain the role of Arf/Sar1
in the assembly and disassembly of
the COP (coatomer) coats.

Answer 57

1. A GEF in the donor (ER) membrane
   induces the “coat recruitment
   GTPase” (Arf or Sar1) to release GDP
   and bind GTP. Active GTPase extends its hydrophobic “tail,” which inserts into the donor membrane (see inset).
2. GTPase recruits COPs, which assemble and drive vesicle budding.
3. GTP hydrolysis releases causes the GTPase to retract its “tail”, releasing it from the membrane,
   and resulting in coat disassembly.

Question 58

L18. Brefeldin A (BFA) is a small fungal toxin that is thought to act by inactivating the Arfguanine
nucleotide release protein (GEF/GNRP). Predict the effect of BFA on transport by COPcoated
vesicles, indicating at which step the pathway would be blocked.

Answer 58

BFA blocks activation of the coat recruitment GTPase, and thus assembly of the coat.
Interestingly, it specifically disrupts FORWARD transport, while allowing retrograde transport to
continue… thus the Golgi apparatus becomes fragmented into many small vesicles and many Golgi
proteins end up recycled back to the ER.

Question 59

L18. GTPS is a GTP analogue, in which the oxygen linking the - and -phosphates has been
replaced with a sulfur atom. GTPS is not readily hydrolyzed by cellular GTPases, such as Arf.
Predict the effect of GTPS on transport by COP-coated vesicles, indicating at which step the
pathway would be blocked.

Answer 59

When added to an ER/Golgi transport reaction in vitro,GTPS will activate the coat
recruitment GTPase. Since ARF/Sar1 cannot hydrolyse GTPS, the vesicles will be unable to uncoat.
Though they will dock to the target membrane compartments, they will be unable to fuse and
transport will be blocked

Question 60

L18. Discuss the following statement: “Secretion is the default pathway (followed by soluble
proteins in the ER).”

Answer 60

The “default” pathway is that taken by “something” (in this case, a soluble secretory
protein) in the absence of any other signals. So, what this says is that, without needing any
additional signals, a protein translocated into the lumen of the ER will be transported through the
Golgi and released outside of the cell. Secretion appears to be the “default” for many proteins
translocated into the ER. However, more recent studies suggest that some proteins require ER exit
signals to direct them to the Golgi.

Question 61

L18. How could you
demonstrate that the conserved KDEL sequence is:
A. …necessary for retention of BiP in the ER?

Answer 61

Use recombinant DNA technology to clone a BIP
cDNA and delete, modify, or mask (by adding additional amino acids) the KDEL sequence. Clone the
modified cDNA into an expression vector, transfect it into cells, and use antibodies (or fuse to GFP
and look directly) to determine whether BIP is “retained” in the ER or secreted

Question 62

L18. How could you
demonstrate that the conserved KDEL sequence is:
B:… sufficient for retention of a protein in the ER

Answer 62

Use recombinant DNA technology to add KDEL

to a protein that is normally secreted, and demonstrate that the protein is now “retained” in the ER

Question 63

L18. How might you demonstrate that the KDEL sequence functions as an ER “retrieval signal,”
rather than a “retention” signal (hint: how does retrieval differ from retention?)

Answer 63

Look for
evidence of Golgi processing in proteins in the ER. Pelham et al. did this by adding a KDEL “ER
retrieval” signal to a lysosomal protein (a protease called cathepsin D). Lysosomal proteins are
modified by addition of M6P (the lysosomal targeting signal) in the Golgi… although the KDELtagged
cathepsin was located in the ER, it was “marked” with the M6P lysosomal targeting signal
indicating it had been sent to and retrieved from the Golgi.

Question 64

L18. How does the lysosomal targeting

signal differ from other signals we have discussed in lecture?

Answer 64

The signal for lysosomal targeting is mannose-6-phosphate added to the carbohydrate of
lysosomal proteins… the “immediate” signal is in the carbohydrate, not the protein (however,
lysosomal proteins must be recognized and modified by M6P addition… though the recognition
mechanism is not yet clear).
The mutation described changes the NXS glycosylation signal in GPX to SXS… since the M6P
targeting signal is attached to te carbohydrate of lysosomal proteins, the mutation prevents
correct targeting, and GPX is secreted, by default.

Question 65

L18. In which cellular compartment would you expect to find a protein containing (briefly
explain your reasoning; two points each):
A. …both a mitochondrial signal sequence at its N-terminus and an internal, uncleaved ER
signal sequence

Answer 65

Plasma membrane. Protein will be co-translationally targeted to the ER
and inserted into the ER membrane. If secretion is default, protein will then be transported
through the Golgi to the plasma membrane. Mito import signal will never be seen by the mito
import machinery

Question 66

L18. In which cellular compartment would you expect to find an internal, uncleaved ER signal sequence near the middle of the protein and a stretch
of six basic amino acids near (but not at) the C-terminus

Answer 66

Plasma membrane. Protein will be
co-translationally targeted to the ER and inserted into the ER membrane, with its Cterminal
domain in the lumen. If secretion is default, protein will then be transported
through the Golgi to the plasma membrane. Tho’ a stretch of basic amino acids is
characteristic of an NLS, in this case, they were co-translationally translocated into the
lumen of the RER, and were never accessible to the nuclear targeting machinery.

Question 67

L18. IN which cellular compartment would you expect to find an N-terminal signal peptide, internal “stop transfer” sequence, and C-terminal LysLys-X-X
(KKXX) sequence?

Answer 67

ER membrane. Signal sequence will co-translationally target to
the ER membrane. Stop transfer will stop translocation, resulting in a membrane protein.
Membrane protein will be transported to Golgi for processing, but KKXX sequence at Cterminus
will target back to the ER.

Question 68

L18. Summary Outline or diagram (i.e. a flow chart or
simple map) the pathways(s) followed by these
three proteins en route to their final target
compartment, including or indicating (1) any
intermediate compartments involved in
targeting/trafficking; (2) where the targeting
pathways diverge; (3) steps in the processing
of the protein precursors along the way; and (4)
key targeting signals or sequences

Answer 68

DIAGRAM +
The signal sequence (non-polar -helix) is
cleaved in the RER lumen.

2. Glycosylation begins in the RER.
3. Carbohydrate is processed in the RER and
   Golgi.
4. M6P is added in the cis-Golgi

Question 69

L18. What targeting signal is required for ALL THREE of these proteins to reach their proper
destinations? Briefly explain

Answer 69

An ER signal sequence (non-polar -helix)! All three of the proteins
must first be targeted to the ER

Question 70

Look at wobble pairing

Answer 70

slides