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MCAT Psychology, Potomac Watch
-The insula of the brain is implicated in conscious urges and emotions. Therefore smokers with brain damage involving the insula would be more likely to quit smoking more easily without relapse than smokers without brain damage to the insula -Avolition, Avolition amounts to the overall lack of drive to perform activities and pursue objectives. For example, people with this symptom may not have the will to run errands or perform a task, even when those tasks will bring obvious advantages to their lives (e.g., cooking). It is thus a negative symptom -An allosteric enzyme has a site other than the one for the substrate at which a molecule (not the substrate) that directs the function of the enzyme can bind
- oxygen in H2O is δ- and thus will be attracted to a δ+ atom. The carbon double bonded to the nitrogen must be δ+ -First, exactly why is DNA negatively charged? The reason why DNA is negatively charged is the phosphate group that makes up every nucleotide (pentose + nitrogenous base + phosphate). When forming part of the phosphodiester bond, the phosphate group (circled in blue below) retains 1 of 2 negatively charges (the other being lost to form the other ester bond to a new pentose, which is the reason for the name "phospho-di-ester")
Brain Functioning Heart Physics Senses, Eyes, Ears, Tongue, Skin Sleep Stages, Radiation Waves Lengths
Chemistry Nomenclature (Zcis, Etrans) (R Clockwise, S Counterclockwise) DNA Structure, Amino Acids, DNA Nitrogen Bases (AG, TUC) ATP Cycles, Phosphagen Glycolytic Oxidative Colligative Properties
gamma > beta > alpha
Homophily, Relationships from Shared Characteristics
Avolition amounts to the overall lack of drive to perform activities and pursue objectives. For example, people with this symptom may not have the will to run errands or perform a task, even when those tasks will bring obvious advantages to their lives (e.g., cooking). It is thus a negative symptom
Apoptosis is the process of programmed cell death that can occur in multicellular organisms
The Stressor, Cognitive and behavioral approaches to the treatment of anxiety disorders usually refer to the stimulus that brings anxiety episodes
The zygote, undergoes a split for monozygotic twins, whereas dizygotic twins result from two ova
Game Theory, The idea that strictly rational and selfish decisions can be prejudicial for all the parties involved was offered by which cooperation theory?
Sex linked, means that the gene in question is present on the X chromosome
Adrenaline
Which of the following hormones found in the human menstrual cycle are produced in the ovary? 2 And 4 -You must be familiar with the graph of the menstrual cycle in order to know which curve refers to which hormone. There are four hormones involved: luteinizing hormone (LH), follicle- stimulating hormone (FSH), estrogen, and progesterone. If the menstrual cycle is understood well, you should immediately know that both estrogen and progesterone are secreted by the corpus luteum, which comes from the ovary. Alternatively, if the pituitary hormones are known, it is easy to eliminate LH and FSH because they are secreted by the anterior pituitary
seminiferous tubules → epididymis → vas deferens
Ivas deferens → ejaculatory duct → urethra
- Not, gametogenesis → seminal vesicles → seminiferous tubules
The process is summarized in the following diagram which also includes the presence of ions that are typically released when binding occurs
-Affinity does not solely depend on electrostatics but may include, depending on the situation and/or to one degree or another, H-bonding, Van der Waal’s interactions (relatively very, very weak), and the shape of the molecules (e.g. a pocketed/cup-shaped/invaginated site matching a relatively large or protuberant ligand or substrate, leading to the imagery of ‘lock and key’ similar to the image above, but more realistically described as ‘induced fit’). In this particular question, we only have information regarding electrostatics: the attraction of negatively charged to positively charged molecules
nitrogen is an integral part of the subunits (nucleotides: BCM 1.5, BIO 1.2.2) that polymerize to form DNA. After part 1 of the experiment, essentially all bacterial DNA will contain the isotope of nitrogen 15N. Thus, the original molecules contain only 15N. By semi-conservative replication in the presence of only 14N (parts 2 and 3 in the experiment), the first new molecules will each contain one strand of 15N (= the parent or template) and one strand of 14N (= the new or daughter strand). In the second replication, the 15N/14N molecule created in the first replication will be separated and used as a template. One strand of 15N will be used as a template for a new 15N/14N molecule, and one strand of 14N will be used as a template for a new molecule that will contain only 14N. This will produce equal amounts of the 15N/14N and the 14N molecules in the second generation.
On the Surface: The information provided in the first paragraph applies to all subsequent generations: "The method of DNA replication proposed by James Watson and Francis Crick is known as semi-conservative replication since each new double helix retains one strand of the original DNA double helix". That same formula is applied twice to make 2 generations.
So if you have been only growing in N15, you begin with 100% N15. The first generation in N produces 100% combinations of N15/N14. Now the 2nd generation, you get 50% N15/N14 and 50% N14/N14.
In color, here is N15 in blue and N14 in red (note that the question is asking about the 2nd generation, F2, but just as an academic exercise, we have included the 3rd generation, F3):
If conservative replication had occurred , the initial strand, which contained only 15N (see previous question), would not separate to form templates for two new strands (conservative = to resist or oppose change). The replication would produce no DNA molecules containing a mix of the old strand (15N) and the new strand (14N) [see Figure 1]. After one generation there would be the original old strand with 15N and an equal number of new daughter strands with 14N
Dispersive replication takes the original molecule and produces two new molecules that are both mixtures of the old and new strands (see Figure 1). Therefore, the centrifuge of a mixture that had replicated in this manner would show only a large band of 15N/14N molecules
Which of the following statements could be held LEAST accountable for DNA maintaining its helical structure? -Unwinding the helix would separate the base pairs enough for water molecules to enter between the bases, making the structure unstable -(Not The helix is stabilized by hydrogen bonds between bases, The sugar phosphate backbone is held in place by hydrophilic interactions with the solvent, C–G pairs have three hydrogen bonds between them, but A–T pairs only have two)
RNA Polymerase , Which of the following enzymes is most important in RNA synthesis during transcription? -(Not DNA polymerase, RNA replicase, Reverse transcriptase) -You are asked for an enzyme that will help synthesize a chain of RNA. You can then narrow it down to a choice that includes the word RNA. The two choices are either replicase or polymerase. In producing a strand of RNA, nothing is being replicated. A chain (polymer) of RNA is being made that is complementary, but not identical to, the strand of DNA. It is easy to conclude that RNA polymerase must be involved
(1) must be the gene with introns/exons (exons: coding regions; introns: noncoding regions), which according to the central dogma of biology is transcribed in the nucleus to (2) mRNA (BIO 3.0; of course, in RNA, thymine is replaced with uracil). RNA is modified leaving out the introns (4) and leaving the exons (3), creating functional mRNA, which is translated in the cytosol on a ribosome (which can be free or associated with rER) into (5) a primary structure/pre-protein that undergoes folding/modifications, producing the final protein p product (6). It takes three nucleotide residues to code for one amino acid (i.e., the "triplet" code), and thus their numbers are not "twice" but rather three times.
There is also Kd, the dissociation constant for the formation of the enzyme–substrate complex, and logically, lower Kd means higher affinity (in fact, Kd is the inverse of the association constant Ka, which is unrelated to the acid dissociation constant Ka).
Conditions exist in which Km is equal to Kd (common for the MCAT); thus by inference, a low Km means a high affinity and therefore, for a given substrate concentration, a high velocity.
For competitive inhibition there is one more important constant, the Ki, or inhibition constant, which is the dissociation constant for the enzyme–inhibitor complex. Thus Ki can be defined for the MCAT as the concentration of inhibitor that is required to decrease the maximal rate of the reaction to half of the uninhibited value. In practical terms, the lower the Ki, the lower the concentration of inhibitor needed to lower the rate
the IR absorption peaks. The absolute minimum to memorize are the bands for an alcohol (OH, 3200 - 3650) and that for the carbonyl group (C=O, 1630 - 1780) because these are the two most encountered functional groups in MCAT organic chemistry.
Bottle II has a peak at 1710 (carbonyl) 3333 - 3500 (hydroxyl) = carboxylic acid (i.e. benzoic acid). Bottle IV has a peak at 3333 so it must be the alcohol. Without doing anything else, there is only one possible answer, D.
{For fun, draw the structures of the four compounds; allyl,ORG 4.2, add -OH to make it an alcohol; benzoic acid - ORG 8.1; the four carbon ketone 2-butanone (= methyl ethyl ketone) and the four carbon aldehyde butyraldehyde (= butanal) - ORG 7.1}
proton NMR. Not including the reference peak at zero, there is only one peak! This means two things :
(i) each H must be living in an environment which is identical to any other H in the entire molecule (only answer choice B. is possible); and
(ii) since there is only one peak, to calculate the number of H's n on an adjacent carbon: n + 1 =
- Thus n is zero indicating that each H is attached to a carbon whose neighboring carbon has no attached H's (again, only option B. is possible)
cyclohexanol which is a cyclic alcohol with six carbons. Now we look at Figure 1 to see where the four possible answers are. Note that in order for I or II to be possible, there would have had to be a reaction with the 5% sodium bicarbonate. However, Table 1 shows that a sodium bicarbonate test is positive with carboxylic acids (which is not cyclohexanol!).
Now let's compare III and IV in Figure 1. The only difference between the two is their response to the Lucas test. From Table 1 the Lucas test is positive with alcohols with five or less carbons (which is not cyclohexanol!)
Ketone, A water soluble unknown is unreactive in the presence of sodium bicarbonate, gives a positive 2,4-DNP test and negative Fehling's and Iodoform tests. In which of the following classes should this compound be classified? -(Not Aldehyde, Carboxylic acid, Alcohol) -only one test is positive - the 2,4-DNP test. By looking at Table 1, we see that aldehydes and ketones are the only two possibilities. But how can we differentiate the two? Easy! Also in Table 1, we see that Fehling's is positive with aldehydes. Since Fehling's is negative in our problem, then the answer must be ketones
Warfarin is a ‘blood thinner’ (= anticoagulant) used in the treatment of disorders that require the dissolution of a blood clot or for the prevention of clotting as suggested by the passage. A patient with deep vein thrombosis will be receiving warfarin to function as a blood thinner. The data presented in Tables 1 and 2 reveal that Cu-PTSM and Cu-ATSM are both readily displaced from HSA by warfarin, as indicated by increasing amounts of free Cu-PTSM or Cu-ATSM in the presence of increasing molar ratios of warfarin to HSA. Conversely, Table 2 shows that increasing molar ratios of warfarin to HSA did not displace Cu-ETS significantly; therefore, Cu- ETS would be the optimal choice in these patients
Which of the following statements is NOT true concerning a graph of reaction rate v vs. substrate concentration [S] for an enzyme that follows Michaelis-Menten kinetics? At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km -(Yes, Km is the [S] at which v = ½ Vmax, As [S] increases, the initial velocity of reaction, v, also increases, The y-axis is a rate term with units such as μM/min)
dideoxynucleoside triphosphates are able to be used in the Sanger method because they lack a: hydroxyl group on C3 of their ribose component -(Yes, hydroxyl group on their phosphoric acid component, hydroxyl group on C1 of their ribose component, hydroxyl group on C5 of their ribose component)
-the method by which DNA is linked together. You should know that the chain is made by additions to the hydroxyl group of the third carbon in the sugar. Thus if the nucleotide will not accept a bond (= phosphodiester; BCM 1.5, 1.6; BIO 1.2.2), then the hydroxyl group at C3 must be missing.
inhibitory ability. The highest to lowest concentration listed above is 7 μM > 10 nM > 8 nM > 100 pM. Therefore, 100 pM is the lowest concentration listed, so Compound 5 is the most potent inhibitor.
Disambiguation: The association constant Ka is unrelated to the acid equilibrium constant Ka. Context during the exam will permit you to clearly identify which Ka is being assessed. Even if you have never heard of Ki (which of course would no longer be the case, especially because there is a good chance that it will come up on the real exam), you could have reasoned the answer. The question asked what is “more potent” than 7 nM and 3 answer choices have higher values and only one answer choice has a lower value (‘one of these things are not like the other’), and so only one answer choice is possible.
The Bottom Line: When there is tight binding, Ki, KM, and KD (the dissociation constant = 1/(association constant) = 1/Ka), are - all 3 - relatively small values. Of course, Ka, would be relatively large because it is the inverse of KD, and more importantly, it is the association constant
Tertiary, It has been found that proinsulin, the precursor molecule to insulin, contains a portion that is stabilized by disulfide bonds after folding. Such disulfide bonds pertain to what level of protein structure? -(Not, Primary structure, Secondary structure, Tertiary structure, Quaternary structure)
-Disulfide bonds are sulfur-sulfur covalent bonds that help to stabilize the tertiary structure of proteins. Disulfide bonds are formed between the side chains of cysteine by oxidation of two - SH (thiol) groups to form a disulfide bond (S-S) = disulfide bridge.
Primary structures involve the amino acid sequence with the amino acids held together typically by amide bonds. Note that it is possible to have cys-cys residues involved in disulfide bonds in the primary structure but the question stem specifically mentions "folding" which refers to a higher order structure.
Secondary structures involve the conformation of the polypeptide backbone through hydrogen bonds. Tertiary structures pertain to the 3 dimensional folding of one polypeptide chain involving both covalent (sulfur bridges; answer choice C.) and non-covalent bonds, whereas quaternary structures involve the 3 dimensional folding of more than one polypeptide
Chemistry Nomenclature
-The structure of glucose shows six carbons (this should be consistent with your knowledge of the molecule!). Only answer choices A and C have six carbons. The structure given never has two hydroxy groups attached to the same carbon -Hydrolysis breaks a molecule in 2 (by the action of water), while condensation means 2 molecules become one. ‘Acetyl’ refers to a methyl group bonded to carbonyl (C=O) and thus it is a 2-carbon unit. Adding three 2-carbon units together makes a 6-carbon unit.
Going Deeper: A coenzyme in a nonprotein molecule that cannot catalyze a reaction on its own but can bind with an enzyme (apoenzyme) to make the enzyme active (holoenzyme).
We have already discussed the fact that ‘acetyl’ is a 2-carbon unit. In fact, Ac-CoA can be considered as a courier in the body for the transport of 2 carbons for numerous biosynthetic reactions (oxidation to produce energy and carbon dioxide; building fatty acids from carbohydrates, not surprisingly most fatty acids in humans have C chains divisible by 2; etc.
-The key is to assess each carbon to see if it is bonded to 4 different substituents (= chiral). Ignore any carbon that is double bonded (because it is attached to the same thing twice) and ignore any carbon that is only showing 2 bonds (because it must be attached to H twice). All the other carbons in lovastatin, except those in methyl groups, are chiral.
Remember, chirality is not about being bonded to 4 different atoms, it requires you to evaluate any difference in the 4 substituents. You can look at the explanation to the configuration question to see a diagram with the labeling of the 8 chiral centers. A common question students ask is: if you are comparing 2 sides, when do you stop? You never stop! You continue until you find any point of difference. If there is perfect symmetry then there is no difference
-First a note regarding nomenclature: The shortest peptides are dipeptides, consisting of 2 amino acids joined by a single peptide bond, followed by tripeptides (3 amino acids, 2 peptide bonds), tetrapeptides (4 amino acids, 3 peptide bonds), pentapeptides (5 amino acids, 4 peptide bonds), etc. A peptide (= amide) bond is a covalent bond formed when the carboxyl group of one amino acid reacts with the amino group of another. In other words, focus on the number of times that you see C=O connected to N, as long as that bond is WITHIN the molecule (i.e. between amino acids). Since there are 3 such instances in the molecular structure provided, given the information provided above regarding 3 peptide bonds, the molecule must be a tetrapeptide. Going from left to right in the structure in the image, here are the 4 amino acids: Tyr-Pro-Phe-Pro-NH2.
Going Deeper: The rules above apply to linear oligo/polypeptides. However, cyclic oligo/polypeptides will always have 1 additional bond where the amino end and carboxyl end bond to complete the ring.
Beyond Required Assumed Knowledge (but the following includes some commonly explored ideas in MCAT Biochemistry passages): The tetrapeptide in the image is ‘morphiceptin’. It is an
iii. If the difference between the two groups is due to the number of otherwise identical atoms, the higher priority is assigned to the group with the greater number of atoms of higher atomic number or mass.
iv. To assign priority of double or triple bonded groups, multiple-bonded atoms are considered as equivalent number of single bonded atoms:
- The chiral carbon is then evaluated from the side opposite the atom or group with the lowest priority. If the order of the other three atoms or groups in decreasing priorities is clockwise, the arrangement is designated R; if counterclockwise, the arrangement is designated S.
Figure 1A. A part of the hexahydro-naphthalene ring can be rewritten as that shown on the right according to the rules. Please note that this is just to enable visualization of the molecule in order to assign E and Z configurations.
Assigning configuration to the second double bond from the left is fairly simple (just as it is explained above). It is the first double bond from the left that is tricky. To assign its configuration, it will be useful to draw out that part of the molecule with the rules clearly applied. For example, when the rules are applied, the molecule can be written as shown on the right in Figure 1A.
If you pay attention to the second carbon of the C=C of the first double bond from the left, it is attached to C=C (pink), which should be read as CC. Now you see that C2 of the double bond is attached to 2 carbons (orange and pink). The orange carbon is attached to two other carbons and a hydrogen. The pink carbon is read (according to the rules) as if it is attached to two other carbons and a hydrogen. So, the orange and pink carbons are similar in terms of priority up to that point.
Then, when you look at the next atom, the orange carbon is attached to a carbon that is attached to an oxygen while the pink carbon is attached to a carbon that is attached to another carbon only. Between these two, oxygen wins the higher priority due to its higher atomic number. Hence, the orange side of the molecule gets priority and it is counted as Z
one fewer S configuration chiral center , In the segments of lovastatin that are outside of the hexahydro-naphthalene moiety, as compared to R configuration chiral centers, lovastatin has:
On the Surface: Consider reviewing the priority rules in the explanation to the previous question. Notice that outside of the bicyclic hexahydro-naphthalene moiety, there is 2 R and 1 S configuration thus “one fewer S configuration”.
Going Deeper: Let’s discuss the evaluation of some of the different chiral carbons. There are 4 chiral centers where the H, which is the lowest priority group is already pointing away from the viewer so you can begin by assigning the priority for these 4 chiral carbons because assigning the configuration will be simple. Like the spokes of a wheel, if you see 1-2-3 and it would be as if you were turning Right, then the configuration is R, and the opposite for S.
If, however, hydrogen is pointing towards you, then assign R/S and then whatever result you get, the correct answer will be the opposite. To avoid confusion, one of the assignments of priority was placed in brackets for one of the (S) configurations. Of the 4 chiral carbons with H pointing away, 3 are (S) and 1 is (R).
The other 4 chiral carbons have the H is pointing towards the viewer. This means that whatever configuration that it appears to be, it is actually the opposite.
Argininosuccinase , According to the information provided, a conclusion that can be made with certainty is that neither mutant strain P nor Q have the defective enzyme:
You should recognize that enzymes typically end with – ase and so you can see that the figure provided indicates the 3 enzymes that are catalysts for the three reactions as shown. Now let’s reinterpret the question: if neither P nor Q have a defective enzyme X, then both P and Q must have a functioning enzyme X (the other interpretation is that they have no enzyme but that is not consistent with the data in the table provided). In other words, "Which of the answer choices corresponds to an enzyme that is fully functional in strains P and Q?"
-when considering linear polypeptides that there must be an amino end and a carboxyl end, thus RRPP is a different molecule than PPRR (note the IUPAC official abbreviations: arginine is abbreviated as arg or R, proline is abbreviated as pro or P). Thus there are 6 possibilities: RRPP, RPPR, RPRP, PPRR, PRPR, PRRP.
Side Notes: If you have a statistics background, you can do “4 choose 2” = 4!/2!2! = (4)(3)(2)(1)/4 = 24/4 = 6.
The ability to form multiple H-bonds make arginine ideal for binding negatively charged groups. For this reason, arginine prefers to be on the outside of proteins, where it can interact with the polar environment.
A tetrapeptide is an oligopeptide since it only consists of four amino acids joined by peptide bonds. Tetrapeptides can be pharmacologically active including affinity to receptors involved in protein-protein signaling. Both linear and cyclic tetrapeptides can be found in nature
On the Surface: Molecules have free rotation around sigma (single) bonds (= 'conformation'). The image in the question stem is identical to the molecule labelled 'squalene' in Figure 1 viewed as a different stereoisomer (more specifically, a different conformer but certainly the same molecule).
In the absence of a better analogy (!!), you can imagine squalene in Figure 1 as a snake with its body straightened/elongated while squalene in the question stem is the identical snake but sidewinding. You can verify that they are identical by counting carbons and noting the position of double bonds and branches.
Going Deeper: The molecule in the question stem (and in Figure 1, squalene) has 30 carbons, so it is a triterpene (according to the rules given in the passage, Table 1; notice that C10 is "mono," C20 is "di," C30 is "tri," etc.). It is certainly a precursor for steroid biosynthesis, but the image is linear, and steroids are characterized by having four cyclic rings (in the four-ring structure, three are composed of six carbons—rings A, B, and C—followed by one with five carbons—ring D). However, looking at the conformer of squalene in the question stem, one can easily imagine how squalene can be a biochemical precursor to steroid-type compounds including cholesterol, steroid hormones (testosterone, estradiol, etc.), and vitamin D.
The Z configuration (“zame zide”) is where the highest-priority groups are on the same side of the double bond (cis) and one can quickly notice that all the double bonds are E ("epposite" side highest priority, or trans; 6E, 10E, 14E, 18E).
Carefully compare the molecule provided in the question stem to the representation of squalene in the passage (same molecular formula, just a different representation of the molecule).
Note that squalene is NOT conjugated. To be conjugated, there would be one single bond between double bonds, and due to resonance, the structure would have increased stability
Answer choice A could have reasonably occurred because it must be causing a reduction reaction (i.e. hydrogens must have been gained by a substrate), and since the number of double bonds is reduced from squalene to lanosterol in Figure 1, we expect that there was a donor for those hydrogens (note that on the MCAT, NADH/NADPH/FADH2 are all common reducing agents, which means that they can all be oxidized in the course of a reaction as they reduce a substrate).
Half reaction (simplified): H2C=CH2 + 2e- + 2H -> H3C-CH The reduction of the double bond by the addition of hydrogen is also called hydrogenation (e.g. the hydrogenation of unsaturated fats including trans fats). {Classic MCAT: What is the change in hybridization of carbon in the above reaction? Consider your response then see below. And please don't think that this is too basic for the MCAT: the AAMC brings the full range of straightforward and challenging MCQs.}
A condensation reaction, aka dehydration synthesis, is when two molecules or moieties (functional groups) combine to form a larger molecule, together with the loss of a small molecule. This is illustrated in the first step in Figure 1 (note the elimination of phosphate groups).
An oxidation reaction in biochemistry and organic chemistry is typically when the oxygen content increases (e.g., Figure 1 in the last step) or the hydrogen content decreases (note that the other important definition for both biochemistry and general chemistry is that oxidation is the loss of electrons).
Going Deeper: Yes, it is more complicated than as described; hence, the “reasonably” in the question stem. In particular, the issue of reduction is more complex; although, a reducing agent is used more than once in the series of reactions, many of the double bonds are lost in creating the ring structure (like a Diels–Alder reaction, which is beyond presumed knowledge).
For the mini-bonus question (!!): The hybridization of carbon changed from sp2 - with its double bond - to fully saturated (4 single bonds) sp3 (i.e. where carbon was in the center of a trigonal planar arrangement and then converted to being in the center of a tetrahedron)
-First we need to determine which reaction involves water as a reactant. A quick look at Figure 1 reveals H2O on the third line. Of course, oxygen in H2O is δ- and thus will be attracted to a δ+ atom. The carbon double bonded to the nitrogen must be δ+. Furthermore, since nitrogen is δ- and the hydrogens in H2O are δ- then their mating is also inevitable. In summary, the hydrogens in H2O help form the amino group -NH2 while the oxygen in H2O, which we are tracing, helps form the second carbonyl group, C=O, in the illustrated product (line 3, Figure 1). The forward and reverse reactions are represented below including the tracing of oxygen in the reaction; note that when water breaks (i.e. lyses) 1 molecule creating 2 molecules, this is called "water lysis" or more formally, hydrolysis (BCM 1.2.1).
Now if you were able to get the previous question correct then this question is just a follow-up. The amino acid contains a δ- nitrogen in the amino group which is a nucleophile. The nucleophile is attracted to the δ+ carbon of the carbonyl group in benzoyl chloride. The weakest bond breaks (the pi bond within the double bond, ORG 1.3.1), and the electron pair lands on the electronegative oxygen. Now carbon is surrounded by three electronegative atoms making carbon verrrrry δ+! This is an unstable situation!
Thus the negatively charged free electrons on oxygen quickly mate with the verrrrry δ+ carbon. However, since carbon can only be bonded four times, one of its substituents has to go!Since chloride is an excellent leaving group (see the last paragraphs of ORG 6.2.4), we are left with an amide (ORG 9.3).Draw the mechanism on your scratch paper.
Why was the medium "dilute aqueous sodium hydroxide"?Once again we turn our attention back to the previous question.Base treatment of an amino acid increases the rate of a nucleophilic reaction of the free amino group.
Would the "dilute aqueous sodium hydroxide" create other products in the reaction?Because it is dilute we would expect very little contamination or by-products. But just for fun, let's treat both our reactants with concentrated aqueous sodium hydroxide.
Now NaOH in water (= aqueous) means Na+ and OH- (CHM 5.2), which simply means the nucleophile OH- since Na+ is a spectator ion (ORG 1.6).Note that both compounds have a carbonyl group which are internationally popular in attracting nucleophiles!Treat your preceding reactants with OH-.The nucleophile is quickly attracted to the δ+carbon of the carbonyl group (i.e. ORG 8.1).The weakest bond breaks (the p bond within the double bond, ORG 1.3.1), and the electron pair lands on the electronegative oxygen.Now carbon is surrounded by three electronegative atoms making carbon verrrrry δ+ !This is an unstable situation! {does this sound familiar?!}
Thus the negatively charged free electrons on oxygen quickly mate with the verrrrry δ+ carbon.However, since carbon can only be bonded four times, one of its substituents has to go!In the case of the amino acid, the hydroxyl group leaves simply re-establishing the carboxylic acid.In the case of the acid chloride, chloride is an excellent leaving group (see the last paragraphs of ORG 6.2.4), thereby creating a carboxylic acid
Which of the following sites would be ideal for tritium (3H) labeling? -Radiolabeled ligands represent a sensitive method for probing receptor binding. The interaction between norepinephrine and the adrenergic receptors is to be examined. -On the Surface: If you put the radiolabel 3H somewhere that can be exchanged (removed and replaced by normal hydrogen, H) then you won’t be following the norepinephrine any more (instead you will be following the molecule that accepted the tritium). Of course, in biological systems, the solvent is water. Water can protonate and it can deprotonate. BUT, water cannot remove a proton from the very strong C-H bond. Thus answer choice B is correct.
Going Deeper: The lone pair of electrons on N and O make it possible to accept a proton and once protonated, a proton can be released which means that if those atoms are tritiated, they may lose the radiolabel. Carbon has no lone pair and the C-H bond is very strong which makes labeling of the carbon a very reliable way to follow the activity of the ligand under examination
the thermodynamic product (the one produced with the greater % under thermodynamic control, ‘thermodynamic mixture’) is the more stable thermodynamic product! But, sometimes, the real MCAT will be that straightforward! But sometimes not (!!) so let’s get some background
Side note: Choice III is not a direct response to the question regarding “relative thermodynamic stabilities” and the statement in choice III is factually unknowable based on the information provided and, in fact, it is often incorrect as a general statement.
Going Deeper: Keep in mind that ‘thermo-‘ relates to heat, and ‘kinetic’ relates to movement (which in this case relates to speed).
A chemical reaction is usually influenced by two factors:
the relative stability of the products (i.e. related to heat/temperature, thermodynamic factors).
the rate of product formation (i.e. related to speed, kinetic factors).
A reaction that is under thermodynamic control is typically done in higher temperatures relative to that under kinetic control. As a result, the temperature provides enough energy for the reaction to be reversible and favor the more stable product.
At low temperature, the reaction is likely to be under kinetic control (rate, irreversible conditions) and the major product is formed as a consequence of the fastest reaction. At a relatively high temperature, the reaction is likely under thermodynamic control (equilibrium, reversible conditions) and the major product is the more stable product (side note: notice that the image of the reaction in the question step shows double-sided arrows between indicating equilibrium conditions)
The reduced species of the electrochemical equilibrium with the most negative Eo value is the strongest reducing agent (CHM 10.1). Memory aside (!), it is of value to note that a reducing agent reduces the other substance, thus a reducing agent is oxidized. Note that only answer choices B. and C. are oxidized (= lose electrons). When you write the two relevant equations as oxidations, instead of reductions like the table provided, you will note that only answer choice B. has a positive Eo value indicating the spontaneous nature of the reaction. The table provided
