Download Fe2+ is absorbed from the lumen of the gut (in the small intestine) by

C2006/F2402 ’11
Exam #2 -- Key
See your returned exam for complete versions of questions 1-4. (There is no question 5.)
For Q 1 to 4, each answer is worth 1 pt and each explanation 2 pts, unless it says otherwise.
1. A-1. In the alternative processing of DMT RNA, the two cases differ in the (5’ donor splice site(s) used) .
A-2. Each copy of the gene for DMT contains (both – exon 16 & 16a).
Explanation (4 pts): The DNA must contain both, as it contains all sequences. The picture indicates that exons
16 and 16a overlap – 16a includes 16. (It was okay if you circled 16a and explained this.) The entire section is
transcribed, left to right, and then the RNA is spliced to include all of 16a or just 16 (which corresponds to the 5’
part of 16a). The part of 16a between the 3’ end of 16 and the 3’end of 16a is part of an intron in one case and an
exon in another. It all depends on where the 5’ (donor) end of intron 16 is – at the end of exon 16 or the end of
exon 16a. The 3’ (acceptor) end of the intron 16 is always the same – at the start (5’ end) of exon 17.
We know the poly A addition site is the same, at the 3’ end of exon 17, in both cases. (If it were earlier, exon 17
wouldn’t always be included in the mRNA.) Note that the 5’ donor site refers to the 5’ end of the intron being
removed, not to the 5’ end of the exon being included.
B. One form of DMT mRNA contains an IRE. (Each ans. 2 pts.)
B-1. The DMT mRNA with an IRE should have the IRE near the (3’) (5’) (either) end.
B-2. If the iron regulatory protein binds to the DMT mRNA (with the IRE) the binding is most likely to
(inhibit degradation of the mRNA) .
Explanation: B-1. The gene is always written with the sense strand (and the direction of transcription) going left
to right, 5’ to 3’. The IRE must be encoded in exon 16a, which is near the 3’ end of the gene and mRNA.
B-2. When the iron supply is low, the body needs more DMT to absorb iron. Therefore binding of the iron reg.
protein must lead to more DMT production, not less. Binding to the 3’ end of the mRNA is more likely to affect
degradation of the message; binding at the 5’ end is more likely to affect initiation of translation.
B-3. In people with HCC, the body accumulates excess iron. Why?
(a) The primary problem in HCC is probably improper regulation of (degradation of DMT mRNA)
(b). The process(es) you have chosen should be (lower than normal) .
(c) In HCC, binding of regulatory protein to IREs should occur at levels of iron that are (higher than normal) .
Explanation (4 pts): (a). There is too much mRNA with the IRE, but normal amounts w/o the IRE. Therefore the
problem is not transcription, but some later step that affects one mRNA and not the other. The only thing that
fits is inhibiting degradation of the mRNA with the IRE (without affecting the mRNA w/o the IRE). This makes
sense, given that the iron reg. protein binds to the IRE and inhibits degradation, as explained in part B-2.
(b). You have too much mRNA, not too little. There is too little degradation, and the mRNA (w/ the IRE) builds
up.
(c). In normal people, the reg. protein does not bind when iron is abundant. Here it continues to bind, and allows
DMT mRNA (w/ the IRE) to build up, even when the iron is high and there is enough DMT. (In real life it is
likely that other mechanisms besides binding of the reg. protein allow stabilization of DMT mRNA (w/ IRE) at
high levels of iron.)
C. Section S of DMT is between TM domains 3 and 4; Q is between TM domains 4 & 5. (Ans 2 pts each).
C-1. In endosomes, which domain(s) could contain glycosylation? (S).
C-2. Relative to the side with section S, iron is transported (the same way in both membranes).
Explanation: C-1. From the information given, S will be extracellular and Q will be in the cytoplasm. That
means S was inside the lumen of the ER or Golgi when the protein was made, and that’s where glycosylation
occurs. Note that glycosylation does not occur on the outside of the cell, but domains on the extracellular side of
the plasma membrane are usually glycosylated. The extracellular domains were glycosylated while inside the
lumen of the EMS (before the proteins got to the cell surface). Domains in the cytoplasm are never glycosylated.
(This did not need to be explained, but if it was stated incorrectly, 1 pt. was deducted.)
C-2. When DMT is inserted in to the ER membrane, or transported in a vesicle to the Golgi, domain S is in the
lumen. When vesicles form (from the Golgi) and fuse with the plasma membrane, domain S is extracellular.
When endocytosis occurs and DMT goes to endosomes, domain S will be in the lumen again. So iron is always
transported from the side with domain S (lumen or extracellular side) to the other side, which is in the
cytoplasm.
2. Ferroportin, or FPN, is an iron transporter that allows iron to exit the epithelial cells of the intestine and enter
the blood stream. Hepcidin is a peptide made by the liver that binds to receptors on the epithelial cells and
causes FPN to be degraded inside the cell.
A-1. Hepcidin is best described as a (endocrine) .
A-2. In normal people, you would expect hepcidin binding to be highest when the body iron is (high).
A-3. HHC can be caused by levels of HCC that are (lower than normal).
Explanation: A-1. Hepcidin travels through the blood from the liver to the intestine. Therefore it is an endocrine
by definition. The liver and intestine may be reasonably close in the body, but their cells are not right next to
each other, so hepcidin cannot diffuse over to the intestinal cells and be a paracrine. (This did not need to be
explained.)
A-2. Hepcidin cuts back on iron transport to the blood and body in general. Therefore it binds when there is
already enough iron. If iron is low it doesn’t help to accumulate it in epithelial cells, because all the cells need it.
The iron has to get to the blood to do any good. (The iron can then be carried to all parts of the body, including
epithelial cells.)
A-3. If there isn’t enough hepcidin, FPN will remain in the membrane and allow iron to pass into the blood even
when it isn’t needed. You need hepcidin to cut back on FPN and iron transport once there is enough iron in the
body. Excess hepcidin would lead to low levels of body iron, not to high levels.
B-1. In epithelial cells, FPN should be located (on the BL surface). (2 pts each answer.)
B-2. In response to hepcidin, FPN is most likely to be degraded in (lysosomes).
Explanation: B-1. To absorb iron from the lumen, you need DMT on the apical or lumen side to get the iron
into epithelial cells and FPN on the BL side to get the iron out of the cells. (This did not need to be explained.)
B-2. If you remove a PL protein it will be part of a vesicle formed by endocytosis. It won’t be in the cytosol, so
it won’t be able to get to proteasomes. Endocytic vesicles containing material to be destroyed usually deliver
their contents to endosomes and thence to lysosomes.
C-1. Which of the following should bind to the SRP when the protein is being made? (Both FPN & hepcidin)
C-2. Which of the two proteins should have more hydrophobic sequences (at least 25 amino acids long)? (FPN).
Explanation (3 pts). Both proteins are made on ribosomes attached to the ER, so both must have signal peptides
that bind the SRP. FPN is a TM protein, probably a multipass, so it must have at least one hydrophobic section
to anchor it in the membrane. It could have one TM section only, doubling as an SP, but it is likely to have both
an SP and additional hydrophobic sections. Hepcidin is a secreted protein, so it should not have a hydrobphobic
section to anchor it in the ER membrane. It must have an SP, but it should not have any other long hydrophobic
sections – and the SP must be removed so the protein can pass all the way into the ER lumen. The mature
protein must be soluble in order to be secreted.
3. In your organism, the transport vesicles in the Golgi always move anterograde
A-1. Before you add your drug, the vesicles should be moving toward the side of the Golgi (farthest from the
ER) .
A-2. After you add your drug, you expect a change of movement through the Golgi of (cargo).
(modifying enzymes) (both) (neither).
A-3. The ‘medial’ enzyme should be located in (medial) part(s) of the Golgi.
Note: If you thought ‘anterograde’ meant ‘retrograde’ or trans to cis, this is the cisternal maturation model, and
answers to A-2 & A-3 are ‘modifying enzymes’ and ‘trans.’ You were given full credit for A-2 & A-3 for this
(& for A-4) if and only if it was clearly explained, and there were no contradictions in your answers or
explanations.)
Explanation: Anterograde means ‘forward’ or from cis to trans. (This did not need to be explained.) If the
vesicles move cis to trans, this is the stationary model – Golgi sacs do not move, and neither do the enzymes
inside them. The transport vesicles carry cargo (newly made proteins) from sac to sac, and appropriate
modifications occur in each sac. After you add the drug, the cargo is stuck and cannot move forward, and the
enzymes remain (as always) in their normal position. (If vesicles move the other way, the drug will not affect
cargo progression, but the modifying enzymes will be moved on to the ‘next’ part of the Golgi.)
Note: This problem is not about vesicles carrying cargo from ER to Golgi, but about vesicles carrying material
from one sac to another.
Answer to Problem 3, cont.
A-4. (2 pts each ans.) If vesicles move anterograde, or cis to trans, and you add the drug, the answer is both -cargo should not be normally modified & cargo should accumulate in part of the Golgi.
Explanation: After you add the drug, the cargo is stuck and cannot move forward, and the enzymes remain (as
always) in their normal position. Since the cargo cannot move to the next part of the Golgi with the next set of
modifying enzymes, the cargo won’t be properly modified. Cargo will build up in the cis side as vesicles from
the ER deliver more cargo, and this cargo won’t get to the medial or trans Golgi with the corresponding
modifying enzymes.
Note: If vesicles move trans to cis, cargo should progress normally as sacs mature, but cargo probably would not
be normally modified (as all the modifying enzymes would be lost or accumulate in the trans part).
4. A. (2 pts each ans.) In yeast, the DNA in nucleosome cores is 147 BP long, and the linkers are 18 BP long
A-1. The length of the DNA in band # 1 in the ‘ladder’ should be (longer in humans).
A-2. The length of the DNA in the band after prolonged MN treatment should be (the same length in both)*.
A-3. Per fixed length of DNA, the number of molecules of H4 should be (higher in yeast).
Explanation: A-1. When the chromatin is digested for a short time, the shortest band in the ‘ladder’ should
contain DNA from one nucleosome = the length of the core + linker. The enzyme cuts preferentially once in
each linker, so the shortest band should contain the DNA from one core plus part of the linker on each side. The
other bands in the ‘ladder’ should contain multiples of the length found in the shortest band. For humans that
length of DNA in one nucleosome is about 200 BP; for yeast it should be 147 + 18 or 165 BP.
A-2. When the chromatin is digested for a long time, the linker sequences are completely digested leaving the
DNA in the core intact. The difference in length between yeast and human nucleosome DNA is largely in the
linkers – the length of the DNA in the cores is about the same.*
A-3. If the length of the DNA in each nucleosome is shorter, as in yeast, there are more nucleosomes per length
of DNA. (This did not need to be explained.)
*Note: If you said the band was longer in yeast because human cores are 147 BP and yeast cores are 145 BP that
was ok. However it is important to note that the structure of nucleosome cores is strongly preserved in all
eukaryotes, so that the similarities are much more striking than the differences.
B. (2 pts each answer.) The paper says: “(The DNA has) a ~140 bp nucleosome-free promoter region, which
was found at ~95% of all 5761 genes, and a transcription start site that is ~13 bp inside the upstream border of
the #1 nucleosome (core).”
B-1. This core of nucleosome #1 is positioned (overlapping the start of transcription).
B-2. If the chromatin was digested with MN, the protected areas should be (-13 to -1) (+1 to +13) (+13 to +26).
Explanation (3 pts): The core should be located around the start of transcription, so as to protect the DNA from
aprox. -13 to +134. (In other words, the DNA from -13 to +134 or 147 BP total should be inside the core.) The
DNA from about +134 to +152 should not be protected because it is in a linker. The second nucleosome core
should start at +152. (Picture should look more or less like the picture below, where circles are nucleosome
cores – DNA is wrapped twice around the histone octamers in each core.) Many students drew the 1st
nucleosome on one side or the other of the start of transcription (usually on the 3’ or downstream side). Partial
credit was given if the picture fit your answers, and showed that (wherever you put the nucleosome) the DNA in
the core was protected from digestion, but the DNA in the linker was not protected.
Start of Transcription
-13
+134
(18 BP)
+152
Answer to problem 4, cont.
C. In a human cell, you would expect a nucleosome-free promoter region for (<95% of all euchromatic genes).
Explanation: In any given cell of a multicellular eukaryote (not dividing), most of the genes are not transcribed.
Therefore you would expect most of the genes to be euchromatic, but not loose enough to transcribe. You would
expect the promoters of most genes to be in nucleosomes, just like the rest of the DNA. Only those promoters
that belong to active genes should be nucleosome free, that is, hypersensitive to DNase, because they have
bound TFs in lieu of histones.
D. (2 pts each answer). Suppose you can make a recombinant yeast gene that contains the first 50 codons for a
secreted protein followed by the entire coding region for histone 4.
D-1. The resulting recombinant protein should have both an NLS and a SP.
D-2. The recombinant protein will probably end up (outside the cell).
Explanation (3 pts). A secreted protein has to be made on ribosomes attached to the ER and end up entirely in
the ER lumen. Therefore it must have a signal peptide (SP) that is removed, and no additional hydrophobic
sections that can anchor it in the membrane. When the peptide chain is complete, the secretory protein should be
a soluble protein in the lumen. The protein then travels inside vesicles, first to the Golgi and then to the plasma
membrane. When the secretory or default vesicle fuses with the plasma membrane, the protein ends up secreted,
that is, outside the cell. The SP, and the sequence required for its removal (recognized by signal peptidase)
should be near the amino end of the normal secretory protein, so they should also be on the amino end of the
recombinant protein. Therefore the recombinant protein should follow the same path as the secretory protein.
The recombinant protein will have an NLS, but it will not be functional (see below).
Histones are made in the cytoplasm and enter the nucleus, so a gene for a histone should code for an
NLS. Therefore the recombinant gene should code for an NLS. However the NLS cannot be read. The
recombinant protein will have an SP and enter the ER as it is made. The peptide section with the NLS will be
inserted into the ER as soon it is made; it will never be outside in the cytoplasm. The proteins that bind to the
NLS (importin, etc.) and shuttle the protein to the nucleus are all cytoplasmic. They can’t reach a protein inside
the ER, so the NLS will not have any effect.
Question 6 – Questions & Answers:
For this page, circle ALL correct answers for each question.
6. Pax6 is a transcription factor (TF) which drives crystallin expression in the lens and somatostatin expression in the
pancreas. A. What is a reasonable explanation as to how Pax6 binds the DNA of both of these genes to activate
transcription?
(Pax6 binds to similar DNA sequences in both genes) (Pax6 binds to totally different DNA sequences in both genes)
(Pax6 works with a different combination of TFs in lens & pancreas)
(Pax6 works with RNA Polymerase II in lens & RNA polymerase I in pancreas). Explain your choice(s)
Explanation: Pax6, a TF, would recognize a similar DNA sequence in both genes, which is conferred by the amino
acids in Pax6’s DNA binding domain. However, PAX6 works with a different set of TFs in the different cell types.
For example, activation of crystallin in the lens requires PAX6 and TFs “A” & “B” and activation of
somatostatin in the pancreas requires PAX6 and TFs “C” & “D”.
B-1. Pax6 could bind the DNA in which of the following regions relative to the transcription initiation site of crystallin?
(-2000) (introns) (-25000) (+15000) (+1500) (none of these) (any of these).
B-2. Pax6, Sox2, and L-Maf all need to bind the enhancer to drive expression of crystallin in the lens. When activating
crystallin in the lens, Sox2 and L-Maf are (trans-factors) (cis-elements) (proteins) (RNA) (DNA) (TFs) (receptors)
(ligands) (nuclear) (cytoplasmic).
B-3. The binding sites that Pax6, Sox2 & L-Maf bind to activate crystallin transcription in the lens are (Protein) (RNA
(DNA) (nuclear) (cytoplasmic) (a cis-element) (a trans-factor).
Diagram and label this crystallin regulatory region. Include the cis-element(s) and trans-factor(s) upstream of the
transcription initiation site.
Explanation & Diagram for B are on next page.
Diagram for Problem 6-B:
TFs:
(or trans-factors)
Enhancer
(or cis-element)
+1
(or transcription initiation site)
Explanation: You had to draw a diagram and label the following to get full credit:
1. Enhancer or cis element
2. TFs or trans-factors – labeling Pax6, Sox2 & L-Maf wasn’t enough
3. +1 or transcription initiation site
C. After Pax6, Sox2 & L-Maf bind the crystallin regulatory regions they recruit a histone acetlytransferase which
i. Switches the chromatin configuration from (euchromatin to heterochromatin) (heterochromatin to euchromatin)
(loosest euchromatin to loose euchromatin) (loose euchromatin to loosest euchromatin) (none of the above),
ii. AND (increases transcription) (decreases transcription) (pauses transcription elongation) (modifies the RNA).
Explanation: The enhancer, when Pax6, Sox2 & L-Maf bind, switches from loose to loosest euchromatin,
increasing transcription. Regions of the DNA that are sometimes transcribed are always euchromatin, never
heterochromatin.
D. The region where Pax6, Sox2 & L-Maf are bound, if treated with the appropriate DNase, would
(represent a hypersensitive site) (be digested more than naked DNA) (remain intact)
(be less digested than the crystallin coding region) (be more digested than the crystallin coding region). Explain
briefly
Explanation: The enhancer while bound by Pax6, Sox2 & L-Maf has a paucity of nucleosomes and so is relatively
exposed to DNase. Treatment of this enhancer with DNase, results in a hyper sensitive site. In contrast, the
transcribed region of the crystallin gene (looser) has less nucleosomes than when this gene isn’t actively
transcribed (loose) but has more nuclesomes than the enhancer (loosest).
E. For somatic nuclear transfer to succeed, which of the following needs to be reset if the somatic nucleus is taken
from a heart muscle cell? (transcription levels of Oct4) (histone methylation) (DNA methylation)
(which TFs are bound to the DNA) (transcription levels of Nkx2.5 – the master regulatory gene for heart)
(all of the above) (none of the above). Explain briefly.
Explanation: In order for somatic nuclear transfer to work, the heart cell’s DNA needs to reverse the
differentiation process, so that the DNA expresses the correct genes for totipotent and can progress through the
ensuing stages of potency (pluripotency, multipotency, etc.) to form the entire organism. For this to happen all of
the above choices must be reset or changed. Reseting of the potency is performed by the various components
present in the egg cytoplasm.
For a pdf of the last page of the exam, including the structure of the DMT gene, go to
http://www.columbia.edu/cu/biology/courses/c2006/exams11/exam2-lastpage.pdf