5Biological molecules test w solutions

ORGANIC CHEMISTRY TOPICAL: Biological Molecules Test Time: 23 Minutes* Number of Questions: 18 * The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 Nb 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac † 227.0 104 Unq (261) 105 Unp (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 † 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Biological Molecules Test Passage l (Questions 1–6) Sucrose, a disaccharide commonly employed in the food industry, easily undergoes acid-catalyzed hydrolysis to form D-glucose and D-fructose: CH2 OH O HO HO H CH2 OH O H+ HO OH O CH 2OH HO Sucrose CH2 OH O CH2 OH O HO HO OH + HO OH CH 2OH OH Pure sucrose, distilled water, and concentrated hydrochloric acid were heated to reflux for approximately hours (The optical rotation of sucrose was measured using a polarimeter prior to heating.) The reaction mixture was then cooled to room temperature, transferred to a volumetric flask, and diluted with distilled water The optical rotation of the product mixture was then determined In addition, a small sample of both sucrose and the product mixture were reacted with Benedict’s reagent (an aqueous solution of Na2CO3, CuSO4, and sodium citrate) Benedict’s reagent is used to test for the presence of aliphatic aldehydes and is often employed to test for diabetes; a positive result is indicated by the formation of the brick-red precipitate Cu2O The results of the experiment are as follows: HO D– Glucose Table Specific rotation [α]D D–Fructose (I) Figure Under acidic conditions, D-glucose undergoes mutarotation giving an equilibrium mixture of α - and βanomers, whereas D-fructose exists as four different isomers (isomer IV being the predominant form): OH HOCH2 O O OH CH 2OH HO OH CH 2OH OH Solution before hydrolysis Solution after hydrolysis +66/5° Negative –19.9° Positive —The specific rotation for D-(+)-glucose is +52.7°, while that of D-(–)-fructose is –92.4° In sucrose, shown below, which bond is an αglucosidic linkage? OH HO II O OH OH I IV I CH 2OH HOCH2 O OH CH2OH O OH HO HO CH2OH HO Benedict’s Test HO H CH OH O HO OH O CH2 OH OH HO V III IV III II Figure A student carried out the hydrolysis of sucrose using the following procedure: A B C D I II III IV GO ON TO THE NEXT PAGE KAPLAN MCAT Sucrose does NOT produce a positive test when treated with Benedict’s solution because: A the carbonyl functionalities of both subunits are involved in the formation of a glycosidic linkage B it undergoes mutarotation C the molecule is not oriented correctly for a reaction to occur D disaccharides not possess free hemiacetal or hemiketal groups Which of the following is another name for sucrose? I II III IV A B C D α-D-Glucopyranosyl β-D-fructofuranoside α-D-Glucofuranosyl β-D-fructopyranoside β-D-Fructofuranosyl α-D-glucopyranoside α-D-Fructofuranosyl β-D- glucopyranoside I only I and III only II and III only II and IV only In the experiment, what would happen if only part of the reaction mixture was transferred to the volumetric flask? A The observed rotation would decrease specific rotation would remain the same B The observed rotation and the specific would decrease C The observed rotation and the specific would increase D The observed rotation would increase specific rotation would remain the same and the rotation rotation and the Which of the isomers shown in Figure are αanomers? A B C D II and III only II and IV only III and IV only III and V only The experiment described in the passage was repeated using maltose instead of sucrose The key difference in the results was that the Benedict’s test before hydrolysis was positive This proves that: A B C D maltose contains a more reactive glycosidic bond sucrose is a reducing sugar sucrose will hydrolyze more readily than maltose maltose contains a free hemiacetal group GO ON TO THE NEXT PAGE as developed by Biological Molecules Test Passage II (Questions 7–13) Aerobic organisms are defined as those that require oxygen to generate energy Involved in this energy generating process are cytochromes: mitochondrial ironcontaining proteins that facilitate the transfer of electrons One cytochrome that has been extensively studied is cytochrome c This molecule consists of a polypeptide chain approximately 100 amino acids long It folds in such a way as to allow the covalent attachment of an ironcontaining heme group When the original cytochrome c protein was treated using the Sanger method, the 2,4-dinitrophenyl derivative of glycine was observed In addition, the cytochrome c portion shown in Figure was treated with a proteolytic enzyme specific for the N-terminal end of amino acids that possess an amino side chain Five fragments were produced in equal amounts, which were separated purified and analyzed by electrophoresis The results are shown below Selective hydrolysis of the protein portion of cytochrome c produces the polypeptide chain whose amino acid sequence is shown below: Gly-Asp-Val-Glu-Lys-Gly-Lys-Lys-Ile-Phe-Ile-Met-LysCys-Ser A chemist attempted to analyze the protein portion of cytochrome c by employing the Sanger method This technique is used to identify the N-terminal amino acid residue in a polypeptide and involves the use of 2,4dinitrofluorobenzene (DNFB) In the presence of a mild base, a nucleophilic aromatic substitution reaction occurs yielding a labeled polypeptide Subsequent hydrolysis then gives a mixture of amino acids, and the N-labeled amino acid can now be separated and identified: O 2N HNCHCO F + H2NCHCO R etc Isoelectric point (pI) 2.92 9.62 9.74 9.74 9.74 The following data was also obtained to assist the chemist in identifying the fragments Table pK R Molecular COOH pK a NH3+ group weight Glycine (Gly) 2.34 9.63 — 75 H Valine (Val) 2.32 9.62 — 117 CH(CH3)2 HCO – Glutamic 2.19 9.67 4.25 147 (CH2)2COOH (–HF) acid (Glu) 2.10 9.8 3.9 133 CH2COOH Lysine (Lys) 2.18 8.95 10.53 146 (CH2)4NH2 Isoleucine (Ile) 2.36 9.68 — 131 CH(CH3)C2H Phenylalanine 1.83 9.13 — 165 CH Ph 2.28 9.21 — 149 (CH2)2SCH Cysteine (Cys) 1.96 10.8 8.3 121 CH 2SH Serine (Ser) 2.21 9.15 — 105 CH2OH R' Name Aspartic NO2 pK a R group acid (Asp) 2,4-Dintrofluorobenzene O 2N Table Molecular weight (Daltons) 418 336 203 146 650 Fragment number Polypeptide NHCHCO NHCHCO R R' etc H 3O + (Phe) Methionine (Met) NO2 Labeled polypeptide + NHCHCOOH + H 3NCHCO2 – O 2N R R' NO2 Labeled amino acid Unlabeled amino acid Reaction GO ON TO THE NEXT PAGE KAPLAN MCAT If the fragments shown in Table are subjected to electrophoresis at a pH of 4, in which direction will fragment move? A B C D Toward the cathode Toward the anode It will not move The direction of migration cannot be determined without more information Which of the following amino acids would possess a negatively charged R group at pH 6–7? A B C D Aspartic acid and glutamic acid Lysine Cysteine and lysine Serine Which of the amino acids listed below has the most basic functional group on its side chain? Which of the following fragments are unlikely to be formed when the polypeptide in Figure l is treated with the side-chain-specific proteolytic enzyme? A B C D Lys-Gly Gly-Asp-Val-Glu Gly-Asp-Val Lys-Cys-Ser A B C D Aspartic acid Lysine Phenylalanine Glycine What is the isoelectric point of aspartic acid? A B C D 9.80 6.80 5.95 3.00 From the data in Table and 2, Fragments and are respectively: A B C D Gly-Lys-Lys-Lys and Phe-Ile-Met Gly-Asp-Val-Glu and Lys-Cys-Ser Lys-Gly and Lys-Cys-Ser Val-Glu and Lys-Ile-Phe-Ile-Met 1 The reaction of DNFB with the polypeptide is carried out under basic conditions rather than acidic conditions because: A it ensures that the terminal carboxyl functionality on the polypeptide is protonated and is free to react B it ensures that the terminal amine functionality on the polypeptide is protonated and is free to react C it ensures that the terminal amine functionality on the polypeptide is deprotonated and is free to react D it ensures that the terminal carboxyl functionality on the polypeptide is deprotonated and is free to react GO ON TO THE NEXT PAGE as developed by Biological Molecules Test 1 All of the bonds listed below are apparent in the secondary and tertiary structure of a protein EXCEPT: Questions 14 through 18 are NOT based on a descriptive passage A B C D Which would be the best representation of a cross linkage in the following polypeptide: Alal-Thr-Tyr-Pro-His-Gly-Ser-Phe-Ala-Cys-Met-LysGin-Asp-Asp-Cys-Arg-Glu-Glu19 A C 1 hydrogen bonding hydrophobic interactions electrostatic interactions peptide bonds Which of the following pairs of side chain –R groups would tend to associate with each other? A N CH2 and CH2OH (CH 2) COOH and CH2 N O B C=O 14 B 19 19 D 1 C CH3 CH 10 NH S 14 D S O=C CH3 CH3 and CH CH2CH3 CH3 and CH2COOH CH3 CH 19 19 The localized bending and folding of a polypeptide backbone in a protein molecule is usually referred to as its: A B C D primary structure secondary structure tertiary structure quaternary structure Which of the following bonds is a peptide linkage? A C O C O H O H NH ••• O C N C O B D C KAPLAN END OF TEST MCAT ANSWER KEY: C A A B D 8 10 B A C D B 11 12 13 14 15 C A B D B 16 17 18 D D C as developed by Biological Molecules Test BIOLOGICAL MOLECULES TEST EXPLANATIONS Passage I (Questions 1–6) The correct answer to question is choice C This question requires you to have some outside knowledge of the types of linkages found in carbohydrates, specifically where the alpha-glucosidic linkage is in sucrose You should be able to see that in sucrose, glucose and fructose are present as glycosides Remember that a glycoside is a carbohydrate in its acetal or ketal form, in which the anomeric hydroxyl group is replaced with an alkoxy group This alkoxy group could be an oxygen attached to another monosaccharide or a simple alkyl group So, in the sucrose molecule you have a glucoside sub-unit (a glucose molecule as its corresponding acetal) and a fructoside (a fructose molecule as corresponding ketal) Therefore, a glucoside bond would be one in which the glucose sub-unit is attached to an alkoxy group You are also told that the linkage is alpha and again, if you visualize the glucose portion on its own, you should know that the alpha corresponds to the downwards or axial orientation of the hydroxyl group on the anomeric carbon In sucrose, this hydroxyl group is replaced by an oxygen linked to fructose confirming glucose is in the form of an acetal In summary then, you are looking for a linkage between glucose and its acetal group which is oriented axially or downwards on the page The only choice that fits this criteria is III, making choice C the correct response Now for the wrong answers Choice A is incorrect because even though this bond is present in the glucoside portion of the molecule, it is not a glucosidic linkage This bond is simply a carbon-oxygen bond within the glucose sub-unit Choices B and D are wrong due to the fact that they are part of the fructose sub-unit Choice D is a carbon-carbon bond while choice B is a beta-fructoside linkage This linkage is opposite to the alpha-glucoside linkage in that this time it's the glucose attached to the oxygen that is the ketal moiety, and this group is oriented equatorially which is the beta position So again, the correct answer is choice C The correct answer to question is choice A This is another question in which you really have to know your carbohydrates This particular question has a more experimental bias because you have to know about Benedict's reagent and what it tests for You should know that Benedict's is a common reagent used to test for reducing sugars; in other words, those sugars that contain hemiacetal or hemiketal groups Benedict's reagent is a blue alkaline solution containing a cupric tartrate complex and when it reacts with the reducing sugar, a brick-red copper oxide precipitate is formed characteristic of a positive test This reaction only occurs when hemiacetals and hemiketals exist in equilibrium with their open chain forms These open chains possess free carbonyl groups which can undergo oxidation and so reduce Benedict's Anyway, back to the question Sucrose doesn't give a positive Benedict's test since it doesn't possess any free carbonyl groups Instead, these groups are tied up in forming the glycosidic bond between the two monosaccharide sub-units, so choice A is the correct answer Choice B is incorrect as sucrose will not undergo mutarotation As I said before, sucrose doesn't possess any free hemiacetal or hemiketal groups and the fact that it doesn't undergo mutarotation is a consequence of this Choice C is wrong as the orientation of the molecule has nothing to with the fact that it will not reduce Benedict's Choice D is wrong because some disaccharides DO possess free hemiacetal and hemiketal groups For example, maltose consists of two glucose residues, one of which is present in the hemiacetal form Again then, choice B is the correct answer For question 3, the correct answer is choice A Yet again, this question has an experimental bias to it, so you have to draw upon outside knowledge You also need to know the equation commonly used in polarimetry namely that the specific rotation is equal to the observed rotation divided by the concentration of the solution multiplied by the length of the tube Partial transfer of the reaction mixture into the volumetric flask would result in a lower concentration of products This would mean that the observed optical rotation would decrease eliminating choices C and D Since both the concentration of the product and the optical rotation decrease, the specific rotation will remain the same This is logical if both the numerator and the denominator decrease, but looking at it from a more qualitative point of view, you should remember that for an optically active substance at a given temperature, the specific rotation only has one value So, choice B can be eliminated and again, choice A is the correct answer The correct answer to question is choice B Alpha anomers are defined as those in which the hydroxyl group on the anomeric carbon in the ring is oriented axially Let's take a look at each individual anomer and figure out whether they are alpha or beta in their configuration For structures II and III, the anomeric carbon has a -CH2OH and a hydroxyl substituent attached to it In isomer II, the hydroxyl substituent is oriented axially to the ring Therefore, this can be classed as an alpha-anomer, and so choices C and D can immediately be eliminated In isomer III, the hydroxyl substituent is oriented equatorial to the ring and so this is a beta anomer; choice A is out too This leaves only choice B as the logical choice, but let's look at isomers IV and V anyway In these, the anomeric carbon is again the one with the CH2OH and the OH group The key difference here though is that the ring is 6-membered, not 5-membered In isomer IV, the hydroxyl substituent is oriented axially, so it is an alpha anomer, whereas in isomer V, it KAPLAN MCAT is oriented upwards or equatorially, making it a beta-anomer, so the alpha anomers are II and IV making choice B the correct response The correct answer here is choice D As I mentioned at the end of the explanation to question 2, maltose is a disaccharide, but it differs from sucrose in that it is a reducing sugar Maltose is made up of two glucose sub-units, but only one of the units is a glucoside The other sub-unit is present in the hemiacetal form, hence it can mutarotate Mutarotation is characterized by an equilibrium between alpha and beta anomers; the conversion between these two proceeds through an open chain form which possesses a free carbonyl group that can reduce Benedict's reagent giving the characteristic brick-red precipitate Therefore, choice D is the correct answer as maltose contains one free hemiacetal group Choice A is incorrect since Benedict's test has nothing to with the glycosidic bond, so you should be able to eliminate this answer choice right away Choice B is also incorrect because as I said before, sucrose is not a reducing sugar, due to the lack of hemiacetal and hemiketal groups Choice C is also wrong Although it is stating the opposite to choice A, it is still an incorrect statement because it is not relevant to the question anyway Again, choice D is the correct answer For question 6, the correct answer is choice B A common way to name cyclic monosaccharides is according to the ring size, and if you recall heterocyclic chemistry, you will know that pyran corresponds to a 6-membered ring, while furan corresponds to a 5-membered ring It's logical then that a five-membered fructose ring would be a fructofuranose, while a 6-membered glucose ring would be a glucopyranose, but as I said before, the sub-units in sucrose are glycosides, so the acetal forms of glucose and fructose would be glucopyranoside and fructofuranoside Straight away then, you can eliminate choice II, since this states that glucose will be a furanoside, not a pyranoside and that fructose will be a pyranoside, not a furanoside so choices C and D are wrong You should also be aware that the prefix for a glycoside is -osyl, so that fructofuranoside is fructofuranosyl and so on Sucrose can therefore be named as a fructofuranosyl-glucopyranoside or a glucopyranosyl-fructofuranoside; statements I, III and IV are possible answer choices The last thing to look at is the alpha and beta designations The glucoside has an alpha linkage, while the fructoside has a beta linkage If you can't remember this, rewind the tape to the explanations to questions or where I discuss these anomers in more detail You should be able to see that choice IV is wrong, as this describes the fructofuranoside as alpha, while the glucopyranoside is beta This confirms answer choice D as being wrong, but you could have discarded this on the strength of what we discussed earlier As sucrose can be named as a fructofuranoside or a glucopyranoside, statements I and III are equally viable The ring size and the anomeric designations are correct, and so choice A can be eliminated, leaving B as the correct answer Passage II (Questions 7–13) The correct answer here is choice A From Table 1, you can see that the isoelectric point of fragment is 9.74 The isoelectric point is defined as the point where the negative and positive charges in the amino acid cancel out, so the molecule has a zero net charge and will not move when placed in an electric field At a pH below 9.74, the molecule will be protonated and hence positively charged, whereas at a pH higher than 9.74, the molecule will be deprotonated and so negatively charged Since fragment is in a buffer at pH (which is below its isoelectric point of 9.74) it will be positively charged and so will migrate to the negatively charged electrode; namely the cathode This makes choice A the correct response Choice B is incorrect since the buffer would have to be at a pH above 9.74 in order for fragment to migrate toward the anode Choice C is incorrect because as I said, an amino acid that will not move when placed in an electric field is one which is at its isoelectric point The only point which fragment would be at its isoelectric point would be if the pH of the buffer was 9.74 Obviously from the question stem it isn't, so choice C can also be discarded Finally, choice D is wrong since you can tell just from the fragment's isoelectric point and the pH of the environment which way the molecule will move Again, choice A is the correct response The correct answer to question is choice C In order to answer this question, you need to look to the discussion of the proteolytic enzyme just before Table in the passage Here it states that the enzyme is specific for the Nterminal end of an amino acid (conventionally written on the left) which itself possesses an amino or basic side chain Out of all of the amino acids listed, the only one that possesses an amino side chain is lysine Therefore, we would expect the proteolytic enzyme to cleave the polypeptide at the N-terminal residue of lysine; on the left hand side Looking at the answer choices, you can see that choice A is incorrect, since the polypeptide in Figure could be cleaved between the glutamic acid and the lysine residue and the glycine and lysine residue to form the fragment shown The fragment in choice B also forms due to cleavage of the glutamic acid-lysine linkage alone (but obviously it's an end fragment since it constitutes the first four amino acids in the polypeptide, so no other linkages have to be cleaved) As in choice B, the sequence in choice D is the end fragment of the polypeptide; more specifically the last three amino acids on the right hand side of the molecule This can also be formed by cleavage of the methionine-lysine linkage, so D is also out This leaves choice C as the correct response In order to form this fragment, the enzyme would have to cleave the polypeptide at the N-terminal end of glutamic acid This amino acid possesses a carboxyl or acidic side chain but you are told that the 10 as developed by Biological Molecules Test proteolytic enzyme is specific for amino acids with amino side chains, so cleavage will not occur here Again the correct answer is choice C The correct answer here is choice D In order to answer this question, you need to know how to calculate the isoelectric point of an amino acid from the pKa values provided In a simple amino acid where there is only one amino group and one carboxyl group, it is relatively easy to calculate the isoelectric point; it is simply the average of the two pKa values for the amino acid; the pKa of the carboxyl group and the pKa of the protonated amino group However, the situation starts to get a little more complex if the amino acid possesses an acidic or basic side chain Aspartic acid indeed has an acidic side chain, so the calculation has to be adjusted to account for this To calculate the isoelectric point, you have to look at which forms of the amino acid are involved in formation of the zwitterion Remember that a zwitterion contains the carboxyl group in its deprotonated form and the amino group in its protonated form; these two charges cancel each other out to give a neutral molecule Well, at low pH, the carboxyl group has to be deprotonated in order to form the zwitterion, so you certainly use the acid dissociation constant of this group Upon increasing the pH, the zwitterion is formed The only other proton that can now dissociate is that of the acidic side chain to produce the monoanionic form of the amino acid; therefore the acid dissociation constant of the chain also has to be used Therefore, the isoelectric point of aspartic acid is equal to the acid dissociation constant of carboxyl group (2.1) plus the acid dissociation constant of the acidic side chain (3.9) divided by two This gives a value of which corresponds to choice D Choice A is incorrect as this value simply corresponds to the acid dissociation constant of the NH3+ group Choices B and C are wrong due to incorrect combinations of acid dissociation constants Choice B is the average of the acid dissociation constants for the amino group and the carboxyl group whereas choice C is the average of the acid dissociation constants for the amino group and the acidic side chain The pKa of the amino group is only used in the calculation if the amino acid is neutral or basic; since aspartic acid is neither one of these, choices B and C can be discarded Again, choice D is the correct response 10 The correct answer to question 10 is choice B The easiest way to work out the nature of fragments and is to use the molecular weight data given in Table Looking at choice B you can see that the molecular weight of the first fragment is equal to 418 Simply add the molecular weights of the individual amino acids (75, 133, 117, and 147) and subtract 18 for each peptide bond that is formed (as peptide bonds are formed between amino acids by a condensation reaction resulting in the loss of water) This gives you a grand total of 472 minus 54 for the peptide linkages to give a value of 418 If you apply the same method to the second fragment in choice B you get 372 minus 36 for the peptide linkages to give a total of 336 All of the other answer choices possess molecular weights that not correspond to fragments and in Table Again the correct answer is choice B 11 For question 11, the correct answer is choice C You are told that the reaction of 2,4dinitrofluorobenzene with the polypeptide proceeds through a nucleophilic aromatic substitution mechanism You may not be too familiar with this process; electrophilic aromatic substitution is probably the reaction you have encountered Let's look at these two different processes Since aromatic rings are electron rich, they are susceptible to attack by species that love electrons; electrophiles The aromatic ring consists of pi electrons, arranged in a conjugated system However, the electrons delocalize throughout the ring and so the ability of the ring to behave as isolated double bonds is lost This means that substitution is more likely to occur over addition The substitution versus addition property of aromatic rings is something that cannot change, but as seen in Reaction 1, the nature of the attacking species is Here the species attacking the ring is a nucleophile; something that loves positive charges But how can this occur when we know that the aromatic ring is electron rich? Well, you can see that not only are there two highly electron withdrawing nitro groups attached to the ring in DNFB, but there is also a fluorine attached This means that the carbon which is attached to the fluorine has a lot of electron density pulled away from it; so much in fact that it becomes positively polarized In other words, this carbon is susceptible to nucleophilic attack The perfect nucleophile is the nitrogen with the lone pair of electrons at the N–terminal end of the polypeptide, which attacks the aromatic ring and substitutes the fluorine So where does the base come in to play? Since the attacking nucleophile is the nitrogen with the lone pair, the base ensures that the terminal amine functionality remains deprotonated (in other words as –NH2), so the lone pair can attack the aromatic ring: choice C is the correct response If the reaction was carried out under acidic conditions, the amine functionality would be protonated to form –NH3+ There is no lone pair on the nitrogen here; it has been used to form an additional nitrogen-hydrogen bond, so the nitrogen isn't nucleophilic anymore Therefore, protonation of the amine functionality is out, making choice B incorrect Choices A and D are incorrect as the terminal carboxyl functionality does not react with DNFB Again, it is the nitrogen in the terminal amine that reacts Again, choice C is the correct response 12 The correct answer here is choice A At biological pH to 7, R groups which contain a carboxyl group would ionize to a carboxylate anion (if the pKa for this group is below 6) From Table 2, you can see that both aspartic and glutamic acid contain carboxyl side chains and their pKas are below 6; at biological pH, these groups will ionize to form a negatively charged R group Choice A is the correct response Choice B is incorrect because lysine possesses a basic side chain, and since the pKa of this group is 10.53, it would be protonated and hence positively charged at pH to Lysine is also included in choice C making it an incorrect response KAPLAN 11 MCAT Finally, choice D is incorrect since the R group for serine doesn't even possess a pKa, so you cannot tell whether it will even dissociate at pH to Again, choice A is the correct response 13 The correct answer to question 13 is choice B As I discussed in question 12, lysine has a basic side chain which has a pretty high pKa value In other words, the protonated form of this side chain is a weak acid This makes the free –NH2 a strong base and choice B the correct response Basic R groups are recognized by the presence of the amino group in the R group structure Of all of the amino acids in the passage, only lysine possesses this basic side chain Glycine has a side chain consisting of hydrogen which is definitely not as basic, so choice D is wrong Phenylalanine possesses a side chain which isn't even polar, so C is also wrong Finally, choice A is wrong as aspartic acid contains an acidic side chain; obviously a much weaker base than the amino side chain in lysine Again, choice B is the correct response Discrete Questions 14 The correct answer here is choice D This bond is characteristic of a disulfide linkage which contributes to the primary structure of a protein Disulfide linkages often arise due to the covalent interaction between two cysteine residues; this is the case here, where the cysteine residues numbered 10 and 16 cross link Ester linkages such as that shown in choice A not arise between polypeptide chains, so this can be discarded In addition, peptide linkages like that shown in choice B also not arises between chains These linkages arise within peptide chains to link amino acids; but not between peptide chains to link amino acids Therefore, choice B is incorrect as well Finally, choice C is wrong as a 'dinitride' linkage does not occur at any level of protein structure Again, choice D is the correct response 15 The correct answer to question 15 is choice B A polypeptide chain can undergo short range bending and folding to form sheets or helices These structures arise as the peptide bonds can assume partial double bond character and so adopt different conformations The arrangement of groups around the relatively rigid amide bond can cause R groups to alternate from side to side and hence interact with each other In addition, the carbonyl oxygen in one region of the polypeptide chain could become hydrogen bonded to the amide hydrogen in another region of the polypeptide chain This interaction often results in the formation of a beta-pleated sheet or an alpha helix Localized bending and folding of a polypeptide does not constitute a protein’s primary structure The primary structure of a protein is the amino acid sequence; individual amino acids are linked to each other usually through peptide linkages Therefore, choice A is incorrect Choice C is incorrect since the tertiary structure of a protein is the 3–D shape that arises by further folding of the polypeptide chain Usually these non random folds are superimposed on an alpha helix and serve to give the protein a particular function Finally, choice D is wrong since quaternary structure is the spatial arrangement between two or more associated proteins Again, choice B is the correct response 16 For question 16, the correct answer is choice D The –CO-–NH– linkage that arises between individual amino acids are known as amide linkages or more commonly, peptide linkages This linkage forms when the carboxyl group of one amino acid reacts with the amino group of another This process is characteristic of a condensation reaction and so results in the loss of water as well as peptide bond formation These bonds can link an amino acid sequence to form a dipeptide, tripeptide or a polypeptide Choice A is incorrect because this is an ester bond Choice C is also wrong because this is an ether bond Finally, choice B is incorrect because this bond is a hydrogen bond While this bond may be important in determining the secondary and tertiary structures of proteins, it does not hold amino acids together in the chain Again the correct answer is choice D 17 The correct answer to question 17 is choice D As I discussed earlier, peptide bonds link the primary sequence of amino acids in a protein These bonds form between individual amino acids that can then go on to form a peptide chain Peptide bonds are never observed when a polypeptide bends or folds to form a secondary structure In addition, when the polypeptide folds into a three dimensional shape (which is the tertiary structure of a protein) peptide bonds not form Choice A is incorrect because hydrogen bonds are involved in both the secondary and tertiary structure of a protein In the secondary structure, the polypeptide chain can fold so as to allow the carbonyl oxygen and the amine hydrogen to lie in close proximity to each other As a result, hydrogen bonding can occur to form sheets, helices or turns Likewise, hydrogen bonding may serve to stabilize the tertiary structure of a protein Hydrophobic interactions can also play an important role in the tertiary structure of a protein For example, in an aqueous environment, the hydrophobic side chains of the amino acids may interact so as to arrange themselves towards the inside of the protein Therefore, choice B is also incorrect Finally, choice C is wrong since interactions between charged groups (often called electrostatic interactions) can arise, especially in the tertiary structure of a protein Again the correct answer is choice D 18 The correct answer here is choice C The association rule in proteins is similar to the solubility rule you learned in general chemistry Groups of similar polarity will tend to group together and influence the secondary and tertiary structure of a protein Choice C is correct since both groups are hydrocarbons and non polar in nature If these were side 12 as developed by Biological Molecules Test chains in two different amino acids, we may expect them to interact on the interior of the protein, thus contributing to its tertiary structure Choice A is incorrect since the first side chain is hydrophobic and the second is polar Remember we can recognize polar groups by the presence of electronegative atoms such as oxygen and nitrogen Anyway, this polar–non polar interaction will not occur making choice A incorrect In choice B, the first side chain is polar whereas the second is non-polar; these two also will not interact Choice D is the same as choice A in that the first side chain is hydrophobic and apolar whereas the second is polar These side chains will not interact and again, choice C is the correct response KAPLAN 13 ... B D C KAPLAN END OF TEST MCAT ANSWER KEY: C A A B D 8 10 B A C D B 11 12 13 14 15 C A B D B 16 17 18 D D C as developed by Biological Molecules Test BIOLOGICAL MOLECULES TEST EXPLANATIONS Passage... Na2CO3, CuSO4, and sodium citrate) Benedict’s reagent is used to test for the presence of aliphatic aldehydes and is often employed to test for diabetes; a positive result is indicated by the formation... maltose contains a free hemiacetal group GO ON TO THE NEXT PAGE as developed by Biological Molecules Test Passage II (Questions 7–13) Aerobic organisms are defined as those that require oxygen