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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)