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Week Eight Answers to Formative Exam Explanations
I wrote this as an exercise for my own benefit; however, it can also be used to check why one answer
choice is wrong while another is right after you have taken the formative exam. I’m planning on doing one
of these for each of the remaining formative exams. If you have any questions or if you want to make a
correction, please come talk to me (I’m “that guy” sitting in the front of the room). I try not to bite.
1) The presence of thymidylate in the DNA is one of the major discriminating factors
between DNA and RNA. The other is the presence of the 2’ hydroxyl group.
In accordance with the RNA world hypothesis, we evolved the use of DNA as a stable
and more amenable to proofreading information molecule. When there is an uncatalyzed hydrolysis of amino groups in the bases; guanine is converted to xanthine,
adenine to hypoxanthine, and cytosine to uracil. Proofreading machinery is able to detect
the presence of these bases since they are not present in DNA. If thymine, which doesn’t
have an amine so it cannot be hydrolyzed) wasn’t used (i.e. uracil remained as the DNA
base instead of being methylated) the machinery wouldn’t be able to discriminate what
was the parental and what was the mutant strand. The logic behind the loss of the 2’
hydroxyl group is due to the instability of RNA in alkaline pH. The loss of the 2’ proton
creates a nucleophile that is perfectly positioned to hydrolyze the adjacent 3’
phosphodiester. In fact, this is the mechanism that virtually all RNases use to break apart
RNA.
A – Found in DNA, not RNA
B – Deamination of cytidine makes uracil
C – Synthesis requires methylene-tetrahydrofolate so “antifolates” would inhibit it.
D – Nomenclature battle, thymidine (nucleoside) is not thymidylate (the nucleotide).
2) Gout is the presence of sodium urate crystals (realistically, there are other cations
present in the crystal lattice but this is the composition everyone cites) that occurs at
locations of low oxygen tension (i.e. joints). Anything that leads to an increase in urate
(and if the enzymes converting them are present, anything that increases the amount of
purine bases including adenine, guanine, hypoxanthine, xanthine – including our good
friend 1,3,7-trimethyl xanthine: caffeine).
A – Losing the enzyme responsible for the oxidation of hypoxanthine to xanthine and
then further to uric acid (as a deprotonated molecule it is called urate) will result in
increased levels of hypoxanthine and xanthine, but more importantly, prevent the
accumulation of urate.
B – An overproduction of the phosphoribosyl pyrophosphate will feed-forward into
purine and pyrimidine biosynthesis. Increasing flux through the pathway in a nondividing cell will force a compensating flux out of it (fancy way of saying that if you
build it, you must break it at some point).
C – As we all remember from Dr. White’s exam questions (laugh, wink, laugh), G6PDH
deficiency can present with gout and anemia. The former results since there is reduced
flux of carbon through the oxidative reactions of the pentose phosphate pathway. This
causes a build of glucose-6-phosphate, which will adjust to a new equilibrium through the
trans-ketolase and trans-aldolase reactions to increase ribose-6-phosphate. This is going
to increase the concentrations of PRPP and then the logic is the same as above. Note: this
is not happening in mature erythrocytes since they do not need DNA so are unlikely to
still have the enzymes to recycle the constituents to make it.
D – If the purine salvage pathway is lowered, then more of the nucleotide bases will be
required to be excreted at any given moment since they cannot be recycled.
E – Eating excessive amounts of purines makes you eventually excrete excessive
amounts of purines. Kind of straight forward, but I guess some people may chose it since
it may remind them of how dietary cholesterol causes a reduction of cholesterol
biosynthesis. The difference is that nucleic acids are made in every cell and the liver
doesn’t exactly have a mechanism of controlling nucleic acid levels like it does with
cholesterol.
3)
A – In the purine pathways, formyl-tetrahydrofolate is required twice to form formylamides (i.e. put a carbon unit on the ring), and in the synthesis of thymidylate (a
pyrimidine) methylene-tetrahydrofolate is required to put the methyl group on dUMP
(my mnemonic for keeping the phosphorylation state of this substrate straight deals with
a fecophiliacical sense of humor that I’ll leave you to figure out).
B – Since both nucleotides require ribose, PRPP almost has to be used for the synthesis
of both (the one example of this not being the case is with the salvage pathway where the
pyrimidines are recycled with deoxyribose-1-phosphate).
C – In purine pathway, the bases are built directly on the ribose backbone. In the
pyrimidine pathway, orotate (the fully formed base) is condensed (forming OMP) and
then the pathway is rendered irreversible by decarboxylating to finally form UMP.
D – In reactions that involve amino-group transfers that are not alpha amino acids, the
amino donor is either glutamine (to produce glutamate and ammonia, which can then act
as a nucleophile directly) or aspartate (to produce a succinyl intermediate that can be
beta-eliminated/lyased to form fumarate and leave the nitrogen group where it was). The
use of one reagent over the other depends on the thermodynamics and the time course of
the nitrogen transfer with the aspartate path being essentially irreversible due to
immediate oxidation of fumarate in the TCA cycle. Examples of both techniques are
seen in both pathways.
4) The loss of the hydroxyl group of C2 of the nucleotide diphosphate (NDP) means the
oxidation state of carbon 2 goes from 0 to -2 (Refresher on oxidation states: all you need
to do count the number of electronegative groups and subtract the number of
electropositive groups). This means that a reducing agent is required and the enzymology
shows that the enzyme itself acts as the proximal electron donor by oxidizing two
cysteines. These are reduced after the reaction to form the free thiols again by the
cellular substrate thioredoxin, which is ultimately recreated by NADPH.
The logic behind the regulation of ribonucleotide reductase is straightforward; however,
unfortunately never discussed in class. Under states of high ATP/high energy, the cell
needs to be prepared for the S phase of mitosis. Therefore, ATP acts as the positive
regulator of all dNDP synthesis. If ATP non-discriminately activated all the reduction of
all NDPs, it would eventually remove itself and then turn off the signal too quickly.
Therefore, ATP regulates CDP reduction first; whose product activates UDP reduction.
Coupling of the pyrimidines allows temporal control of their reduction without effecting
the purines, who have other major roles in cell metabolism (protein synthesis and fueling
the Na/K pump) rather than just nucleotide and sugar synthesis. The enzyme is then
activated to reduce GDP by the products of dUDP metabolism (i.e. thymidylate). Finally,
once the concentration of all the other deoxy nucleotides is high enough, ADP can be
reduced. This eventually causes a drop in ATP and a rise in dATP, which turns off the
enzyme. Feedback inhibition of all reduced nucleotides occurs to help set the final
equilibrium concentration of that individual nucleotide.
A – The substrate is NDP. There is little reason for why this must be the case in nature;
however, one could argue that without at least two phosphates, the binding energy to the
enzyme wouldn’t be strong enough with all the flip flopping of the enzymes active site
conformation that occurs in order to select for one base-specific reduction over another
(i.e. it cannot use the hydrogen bonds from the bases to raise the binding energy and
therefore must have other places to bind the substrate)
B – This is the inverse of what is seen. Remember, ATP signals reflect the energy state
of the cell and you should always ask yourself “what would a cell want to be doing if it
has lots of energy” when you see a question/answer like this.
C – There is never a ribose-thymidylate in the cell since thymidylate synthase’s
specificity is for dUMP and there is no way to re-oxidize a deoxy nucleotide base
D – The enzymology of ribonucleotide reductase is fascinating since it is one of a handful
of enzymes that uses an organic radical reaction (i.e. it removes one electron at a time
from the ribose sugar). Due to the ubiquitous nature of DNA, it is safe to assume all
examples of ribonucleotide reductase share an ancestral enzyme (even though some
create the tyrosine radical with an iron-sulfur cluster and others use adenosyl cobalamin).
If someone comes up and demonstrates convergent evolution, then we can still rest safely
saying they are similar since they are doing the same thing and regulated the same way.
E – Though folates are required for nucleotide synthesis, they are not required for the
reduction of the 2’ hydroxyl group of DNA.
5)
A – Both folates and vitamin B12 are water soluble, which classifies them arbitrarily as B
vitamins according to the primitive (i.e. resembling the original) definition. It’s kind of
funny to think that vitamin C is technically a B vitamin according to a nomenclature
purist.
B – Though both are soluble, they are also both examples of ways the body can store
adequate amounts of soluble molecules. Similar to glucose, folates are kept from
passively diffusing out of a cell by the attachment of charged molecules (for glucose that
was a phosphate and for folates it is the poly gamma glutamate tail). Cobalamin is tightly
bound to transcobalamin I in the liver (remember: when a biochemist numbers
something, 99% of the time it is in the order of concentration – the storage form of a
cobalamin binding protein is going to be a higher concentration than the one that
transports it only a short distance from the intestines to the liver so it is the storage form
that is transcobalamin I).
C – The methionine synthase reaction transfers a one carbon group to homocysteine to
make methionine. Though someone could do some paper biochemistry to make methyltetrahydrofolate act as the proximal methyl donor, the evolutionary relationship of
requiring methionine for the translation of all start codons puts the enzymes involved in
making it under extreme selective pressure. Cobalamin is the more “primitive” (see
definition above: I do not make value judgments about organic molecules) and was the
original form of carbon metabolism that hasn’t been replaced through the millions of
years of evolution. In the modern cell, the cobalamin remains permanently bound to the
enzyme and methyl-tetrahydrofolate acts to recreate the methyl-cobalamin when it is
used up during the course of the reaction.
D – Related to the fact that adenosyl cobalamin is used in free radical chemistry (see 4D),
humans have only one enzyme that uses it as a cofactor and it involves a carbon
mutarotation, which is simply an isomerization without a functional group changes.
Methyl malonyl CoA mutase intercalates a methyl group between two carbonyls to form
a methylene. I worded the reaction that way so that people would take a second to think
about how they would try to do that in an organic chemistry lab (“Throw the test tube
away and extract out the poisonous additives from massive amounts of 95% bulk
ethanol?”). Nature answers the question by using a cool radical mechanism than does not
involve any of the folates
E – Methyl cobalamin and methyl tetrahydrofolate are both active cofactors (methyl
refers to CH3 while methylene refers to a bridging CH2).
6)
A – Folate is able to mask most of the anemia symptoms since B12 is only involved in
nucleic acid synthesis by allowing proper recycling of “active” folate. This becomes
redundant if you have an excess of folate. This does not prevent other symptoms that
deal with methyl malonyl CoA metabolism or direct methionine biosynthesis.
B – The spinal cord degeneration (a demyelinating disease) is directly caused by a
cobalamin deficiency and not just a loss of “active” folate.
C – See above for reasoning
D – See above for reasoning
E – Vitamin B12 has a really abstract absorption pathway that requires R proteins (corrin
ring binding proteins), stomach acid, intrinsic factor, and transcobalamin II. Though I’m
sure a kilogram of folate might cause you to vomit out any B12 you are trying to
consume, dietary folate intake is generally independent of cobalamin absorption.
7)
A – The prophase of the first part of meiosis is long in order to allow for proper crossing
over of homologous chromosomes. The sub-stages are leptotene, zygotene, pachytene,
diplotene, and diakenesis. The names represent what is going on since the chromosomes
first condense enough to be viewed as thin lines under the microscope. They then come
together and when the crossing over actually occurs, the lines look fat due to two
DNA/protein complexes in such close proximity. They then start becoming two again
and then finish moving apart. For those who are not masters of the lovely greek
language, then perhaps some analogous wordplay will help us keep them straight (leptin
– leptotene – “thin”)(zygote – zygotene – “come together”)(pachyderm, an elephant –
pachytene – “fat”)(diploid – diplotene – “two”)(diakenesis – cytokenesis – “moving
apart”)
B – It is this mechanism that entails the formation of haploid gametes from a diploid cell
lineage
C – The centromere never technically divides (that would mean the p and the q arms of a
chromosome separate). The synaptonemal complex and the kinetechore do divide in
meiosis I and II, respectively. The synaptonemal complex holds the homologous
chromosomes together while the kinetechore holds the separate centromeres together of
the sister chromatids.
D – If this were true, then we would never halve the genetic content of the cell.
E – In spermatogenesis, this is true. In oogenesis, this is never true. This is also an
example of a “distortion” for those who took a Kaplan course since it is worded in an
extreme way. For instance, even if it said spermatogenesis only, we aren’t always going
to get four perfectly functional daughter cells since sometimes something might go wrong
during cell division and an odd cell will be left non-functional.
8) The acrosome is made out of the Golgi apparatus. The functions of the two organelles
mirror each other. Proteases and glycosyl transferases are present in both in order to
assure proper penetration of the zona pellucida (which is essentially just a tough
extracellular matrix). With this in mind, the Golgi has another function related to its
proteolysis and that is it’s “proof-reading”. The current theory is that the endoplasmic
reticulum and its associated chaperone proteins essentially allow for glycosyl transferases
to mark properly folded proteins by adding N-linked or O-linked sugars. This prevents
access to vulnerable spots in the protein to Golgi apparatus proteases. This shuttle
system assures proper folding and prevents wasting of cellular proteins. The other reason
I add this is to highlight the reason why the ER puts on the majority of the sugars to
proteins, while the Golgi simply modifies them (it makes sense if you have mapped out
that proteins are made first in the cytoplasm, then could be shuttled to the ER, and then to
the Golgi).
9) It is at metaphase that chromosomes are at their thickest and untangled from each
other. Also, it is at this stage that polymerization of the tubulin is halted by colchicine,
though since the mechanism of chromosome motion due to microtubules is thought to
involve both polymerization and depolymerization – its likely all phases (except
interphase) will be slowed down also. Looking at the answer choices, B doesn’t make
sense since it wouldn’t really matter if the prophase was “early” or “late”. Telophase and
anaphase would both show the sister chromatids separated and this is not how people like
to prepare karyotypes (its possible to, but just visualize any of the hundred karyotypes
they gave us in class and you will see that all of them still have the kinetechore’s
together).
10) For those that forgot that quinacrine staining is analogous to G staining, here is some
test taking logic:
A – Telomere staining (aka t stain) is specific for this and if you remember that, you can
rule out this answer choice.
B – Nuclear organizer region is where the ribosomal RNA is being made and ribosomes
are being synthesized nearby. The specific stain for this is a silver stain, which targets
the sugar moieties of the ribosome that are relatively scarce in all of the other proteins in
the nucleus.
C – Heterochromatin is inactive (dark on G stain)
D – Euchromatin is active (light on G stain)
E – Synonymous with D
Looking at C, D & E we are able to reason through that C & D are mutually exclusive.
Since D is essentially saying the same thing as E, that means D and E both cannot be the
right answer and can immediately be eliminated. This does not necessarily mean C must
be true; however, the way most test makers design questions means it is highly likely that
it is
11) If we simply look at the sizes of the chromosomes, and ask “what restriction sites
must this person’s chromosome have to give us these fragments”, the answer is
straightforward. If you have trouble reasoning in hypothetical’s, just assume one of the
sites wasn’t there and then try to figure out what bands it would cause to show up.
A – If the father lacked site C, he wouldn’t have showed a 1200 bp fragment
B – If the father lacked site A, he wouldn’t have showed a 500 bp fragment
C – Since there is no 1000 or 750 bp fragment, we know the father doesn’t have site B.
The mother is heterozygous at site C since we see a 1200, 750 and a combined
(1200+750) site. Soft language (“could be the father”) means we have to see if there is
any fragment that is there that cannot be seen in the mother or the father. There are no
such bands that are seen only in the child so we cannot immediately rule out this man as
the father. If we saw 1500 band in the child only, then we know for certain that the
potential father is not the real father since neither he nor the mother have that fragment.
D – Soft language here means it is tempting to look at it favorably; however,
“inconsistent” refers to there being data to refute a case rather than a lack of positive data
that would have proven a case.
12) Guanine and cytosine have three hydrogen bonds, which are required for the
specificity of the interaction. The stability of the DNA duplex (versus the denatured
DNA) depends on the hydrophobic effect and the removal of the nucleotide bases from
water; however, the difference in stability and density between different sequences of
DNA depends upon what are called pi stacking effects (aka: base stacking). This occurs
because aromatic rings are quadropolar (i.e. not dipolar but not technically nonpolar), and
they have different areas of positive and negative charge that form a very strong
interaction when they are at right angles (as in crystallized benzene) or when slipped
slightly off center (like the rotating step seen in DNA). Since there are three groups
resonating into and out of the aromatic ring in the GC base pairs, favorably changing the
molecular orbital energy overlap, it is more stable than the AT base pair, which has only
two groups resonating into the aromatic rings.
A – That would be a purine, purine base pair and is not seen in Watson/Crick DNA.
B – That is a good base pair (and adenine - thymine is in DNA) but it is not the one
shown here.
C – Adenine does not bind with Cytosine (it would overlap two amino groups and they
would repel due to the partial positive charge pointed at each other from the N-H bond)]
D – Cool
E – See reasoning behind C
13) To intercalate translates into putting something in-between something else and
staying there. For instance, I say “can I shimmy between you two” if I’m trying to get
through a crowd to the other end of a bar, while I say “can I intercalate between you two
[and buy you a drink]” if I’m trying squeeze between two cuties at the bar itself. If you
laughed, awesome; if not, I am sorry.
A – Cationic molecules tend to interact with the phosphates and often represent a
substantial amount of binding energy for protein/nucleic acid interactions (which is why
there are so many Arginine and Lysine residues in nucleoproteins).
B – If there is no hydroxyl group, it cannot hydrogen bond. Some of the lone pair
electrons on the oxygens in the phosphodiesters could be hydrogen bond acceptors, but
that’s hardly enough to warrant a binding interaction since it can happen with just water.
C – Some intercalating agents are highly charged; however, they are not all and the part
that is actually doing the “intercalating” is going into a very hydrophobic environment
and charges are not stable in low dielectric constant environments.
D – While some do interact with the minor groove, many others do not. This is another
example of a distortion answer by saying “exclusively”
E – Boom goes the dynamite
14) If you have trouble playing with nucleotide sequences, I recommend drawing them
out in the proper complement (i.e. 3’ 5’) first and then on another line flipping the
order. The primer for the other strand is going to match the sequence of this strand (i.e.
AGGTC) while the primer for this strand is going to complement the sequence read
starting at the 3’ end (3-CGACT-5). Flipped around, it reads (TCAGC)