e. N/A; all of the above are principle uses of proteins

Transcription

1 Chemistry 131 Exam 3 Practice Proteins, Enzymes, and Carbohydrates Spring 2018 Name KEY The following exam contains 44 questions, valued at 2.6 points/question 1. Which of the following is a protein? a. Amylase b. Cellulase c. Phosphofructokinase d. All of the above e. None of the above While not all enzymes have a name that ends in ase, if you see such a suffix you can rest assured an enzyme is being referred to. Enzymes of course, are one of the principal classes/uses of proteins 2. Which of the following is not a principal use of proteins? a. Structure b. Movement c. Nutrient storage d. Regulation of gene expression e. N/A; all of the above are principle uses of proteins opefully by now you are beginning to appreciate that the information which directs the type of cell and how it will function under a given set of conditions (as well as the lifespan of the cell) is dictated by the sequences of bases in the DNA. ow that function is actually carried out is under the control of the expressed proteins. As a result, they do just about everything a cell needs to have done 3. In terms of variety in the fully developed organism, which of the following protein classes has the most members? a. Motor proteins b. Nucleic acid binding proteins c. Signal transduction proteins d. Enzymes e. All have approximately the same number of members The other major classes are enzymes (controlling chemical reactions in the body) and signal transduction proteins (allowing cells to communicate/orchestrate complex function)

2 4. Which of the following is not a major classification of proteins/peptides? a. ormone b. Transport c. Contractile d. Immunoglobulin e. N/A; all are major classifications of proteins Again, proteins carrying out a wide variety of necessary tasks 5. What is the abbreviation for the pentapeptide tryptophan-tyrosine-threonineglutamine-glutamic acid? a. Try-Tyr-Thr-Glm-Glu b. Trp-Tyr-Thr-Gln-Glu c. Trp-Tyr-Thr-Glm-Glu d. Try-Tyr-Thr-Gln-Glt e. Trp-Trs-Thn-Glm-Glt 6. At near neutral, physiologic p, which of the following amino acids has a net positive charge? a. Asparagine b. Glutamine c. Lysine d. Proline e. (a) & (b) Asparagine unlike arginine does not possess a basic amino acid side chain (amides are polar but not basic). As a primary amine, lysine is basic. At physiologic p (usually taken as blood p = 7.4) nitrogen bases (aniline-like bases excluded) exist in their conjugate acid forms and as such bear a positive charge. The negatively charged carboxylate portion of the amino acid can only cancel out 1 of the 2 positively charged nitrogens in lysine so it is net positive. 7. At near neutral, physiologic p, which of the following amino acids has a net negative charge? a. Aspartic acid b. Asparagine c. Alanine d. (a) & (b) Really an extension of the logic in #6, with the aspartic acid molecule bearing 2 carboxylate groups and only one ammonium at p 7.4

3 8. Which of the following amino acid side chains is polar neutral? a. Cysteine b. Glutamine c. Threonine d. Tyrosine A little tricky since one might think cysteine is non-polar given the low electronegativity difference between C and S, which does hold true for methionine. owever, the S- bond is weakly acidic, which is enough to help polarize the thiol 9. Which of the following amino acids contains a benzene ring, either alone or part of a greater aromatic structure? a. Phenylalanine b. Tyrosine c. Tryptophan d. (a) & (b) Ah yes, biochemistry as applied organic chemistry. The 4 th and final aromatic amino acid is histidine, which makes it harder to keep the hydrogen it captures as a base it tends to lose or pick up hydrogen around physiologic p, which is why it is so common in enzyme active sites, aiding in catalysis by acid-base mechanisms 10. Which amino acid cannot be assigned a D or L designation? a. Leucine b. Isoleucine c. Proline d. Phenylalanine e. Glycine In order to exist as a pair of handed isomers (non-superimposable mirror images) the carbon in question must be bonded to 4 different groups. In glycine, since the side chain is hydrogen, there are 2 hydrogens on the alpha carbon, and the possibility does not exist. For the others, know we specifically use L amino acids (and D sugars). You should know how this is determined

4 11. Please draw 2 different amino acids of your choosing, name them, and show them reacting to form a dipeptide. 2O Serine Cysteine If you have to carry out such a process on the exam (and in all likelihood you will) I hope you will have the good taste to not use these amino acids (even though they are pretty bitchin as far as amino acids go and a couple of my favorites). For the level of this class you do not need to indicate reaction conditions, and you may dispense with the need to form an acid derivative first Bonus (2.5 EC): Post-translational modifications aside, why is the amino terminus the first amino acid in a protein? The amino acid called for by the 3 base sequence (codon) on the messenger ribonucleic acid (mrna) is carried to the ribosome/protein factory by a specific transfer RNA (trna) via an ester linkage with the last ribose sugar, leaving the amino portion free. The amino portion of the 2 nd amino acid carried in (also via ester formation with its trna) then attacks the ester, forming an amide and liberating the alcohol portion (which in this case happens to be the entire trna). The 2 nd trna now has a string of 2 amino acids attached and the process repeats itself, with the amino group on the first amino acid remaining free (see lecture 14, page 9) 12. Which of the following is a major protein secondary structure? a. -helix b. -helix c. -helix d. -helix e. (a) & (b) Definitely need to know the 2 major structural motifs of proteins

5 13. Which of the following intramolecular forces is the principal determinant of the secondary structure of proteins? a. ydrophobic interactions b. ydrogen bonding c. Disulfide linkages d. Salt bridges e. Caging by water 14. Which of the following intramolecular forces does not determine the tertiary structure of proteins? a. ydrophobic interactions b. ydrogen bonding c. Disulfide linkages d. Caging by water e. N/A; All of the above forces determine the tertiary structure It will be important to your fortunes to know the forces that aid in stabilizing the tertiary structure of proteins as well as which forces are imparted by the various denaturing agents 15. Which of the following amino acids greatly stabilizes tough proteins like those found in hooves, claws, and horn by forming covalent bonds? a. Methionine b. Cysteine c. Glutamic acid d. Aspartic acid e. Arginine Recall perms work by breaking the covalent disulfide linkages by reducing them to free thiols, followed by shaping then reoxidation for those rams than always wanted to be unicorns (twonicorns?) 16. What is the main mechanism by which detergents alter the tertiary structure of proteins? a. Disruption of hydrophobic pockets b. Interference with hydrogen bonding c. Breaking of disulfide linkages d. Allowing the protein to fold in a lower energy state

6 This question isn t too hard if you focus on what detergents do generally they act as molecular diplomats allowing non-polar dirt and oils (hydrophobic substances) to mix with highly polar water. They can do this because they are amphipathic -they have both polar and non-polar regions - which make them amphiphilic lovers of both polar and non-polar substances. In short, they help bring compounds together that want nothing to do with each other, since they interact well with both! In the case of proteins, allowing the hydrophobic portion which is typically trying to hide in the interior away from the water to interact well with water totally changes the 3-dimensional shape of the protein 17. What is the main mechanism by which oxidizing agents alter the tertiary structure of proteins? a. Disruption of hydrophobic pockets b. Interference with hydrogen bonding c. Breaking of disulfide linkages d. Allowing the protein to fold in a lower energy state Refer to question #15. Yes, one could argue that the disulfide linkage is already oxidized (you should be able to rationalize this by comparison to the free thiol form of cysteine). owever, some of the cysteine thiol side chains are not linked together and adding an oxidizing agent increases the number that are. One could then argue this should stabilize the existing protein structure, which leads to a very important point if you want to understand proteins proteins move! (see YouTube - protein conformational dynamics). Additional disulfide linkages change the ability of proteins to move and thus their overall tertiary structure in real time 18. What is the main mechanism by which heat alters tertiary protein structure? a. Disruption of hydrophobic pockets b. Interference with hydrogen bonding c. Breaking of disulfide linkages d. Allowing the protein to fold in a lower energy state Remember that proteins are made in the ribosome, which extrudes the first portion of the protein into the endoplasmic reticulum as it is being made in such a specific environment, the protein may not fold in the lowest energy shape heat allows a lower energy state to be realized, such as in the heating of egg white classic!

7 19. Which of the following is not a primary function carried out by enzymes? a. Catalyze chemical reactions b. Control the amount of substances produced c. Make specific products from specific substrates d. Directly provide structure for the organism e. Allow energetically unfavorable reactions to occur by coupling them to favorable ones O.K. not a great question, but hopefully you recognize enzymes are proteins that catalyze reactions and are not structural proteins like collagen and elastin. Enzymes do indirectly provide structure, since as proteins, collagen and elastin require the same enzymes all synthesized proteins require (such as RNA Polymerase II to synthesize the mrna that codes for them, and the aminoacyl trna synthetases that load the amino acids onto transfer RNAs) as well as specialty enzymes that gives these remarkable structural proteins their final form (such as the collagenase that converts procollagen to collagen 20. Which of the following is not one of the major subclass of enzymes? a. Lipase b. Dehydrogenase c. Decarboxylase d. Wisase e. N/A; all are major classes of enzymes i-larious! 21. Which of the following is likely to be a critical residue in a protease? a. Leucine b. Isoleucine c. Methionine d. Tryptophan e. Serine A large number of protease (protein degrading) enzymes use serine to attack the peptide/amide bond so hydrolysis can ultimately occur (see the chymotrypsin figure on page 3 of the lecture 16 notes) 22. What functional groups are generated when a lipase acts on a triglyceride substrate? a. Carboxylic acid and alcohol b. Carboxylic acid and amine c. Alcohol and amine d. Aldehyde and alcohol e. emiacetal and alcohol

8 Probably a good idea to know what you are doing (or have done) in the lab. Lipases hydrolyze fats into 3 free fatty acids and glycerol (1,2,3-propanetriol). Really nothing more than a fancy ester hydrolysis! 23. Which of the following amino acids is most likely to be found in the active site of an enzyme? a. Phe b. Asn c. Gln d. is e. Met Since histidine is equally happy to have an extra hydrogen as not at p = 7, it is very good at aiding in acid-base catalysis in the enzyme active site. It is also good at coordinating metals that are needed at times in the enzyme active site in order to confer specific reactivity 24. Please draw the curve for reaction velocity vs. substrate concentration in an enzyme catalyzed reaction. Indicate where Km is on the S axis and where enzyme is completely populated with substrate. You should have marked out the graph where the curve flattens out for complete enzyme population and placed Km under the vertical dashed line on the x-axis 25. Briefly explain how you could distinguish competitive from non-competitive inhibition The influence of an inhibitor can be removed by adding more substrate in competitive inhibition. In non-competitive inhibition, adding more substrate does not restore enzyme activity, since the inhibitor and substrate do not bind the same site

9 26. In terms of Vmax and KM, what is the effect of adding a competitive inhibitor? a. Vmax decreases, KM decreases b. Vmax decreases, KM unchanged c. Vmax decreases, KM increases d. V max unchanged, K M increases e. Vmax unchanged, KM decreases Since more substrate is necessary to compete away the inhibitor, and since K m tells us the substrate conc. necessary to make the enzyme run at half maximal velocity, the K m must rise. Ultimately V max can be restored so it does not change 27. What is the term used to describe the product of a series of biochemical reactions inhibiting a key enzyme in the process? a. Feedback control b. Feedbag control c. Irreversible inhibition d. Allosteric control e. Proenzyme paralysis The inhibition of phosphofructokinase the critical control point enzyme in the breakdown of glucose for energy by ATP, the final derived high energy currency of the overall process is a great example of this 28. Which of the following is not one of the uses of carbohydrates in the body a. Structure b. As energy source c. Component of cell surface/cellular recognition d. Information storage and transfer e. N/A; all are uses of carbohydrates Another less than terrific question certainly structural carbohydrates exist, and you should be well aware that cellulose is the principal structural carbohydrate in biochemistry as a whole. owever, in the body is a different question and one has to consider acidic polysaccharides to find structural examples such as hyaluronic acid, which is a critical component of cartilage and has numerous other roles as a component of the extracellular matrix 29. Please draw the aworth projection of glucose in the form.

10 30. Please draw the Fischer projection of glucose. O CO C Please draw the aworth projection of maltose. The beta form is shown of course this pertains to the hemiacetal on the right glucose (note the up, down, up, down structuring indicative of glucose) since in order to be maltose, the glycoside bond must be in the alpha orientation Bonus (2.5 EC): Please draw the structure for the -glucuronide formed between glucuronic acid and the intravenous anesthetic agent propofol 1 Propofol Notice the beta (or ) in -glucuronide is a glycosidic linkage between the glucuronic acid and non-polar compound being removed 1 Propofol forms a glycosidic linkage with glucuronic acid at C1. If you can draw glucose, you can draw glucuronic acid

11 32. Which of the following is not a monosaccharide? O CO C 2 O O C 2 C 2 C 2 O C 2 I II III IV a. I b. II c. III d. IV e. N/A; all shown are monosaccharides No aldehyde or ketone? Not a [transportable] monosaccharide. The polyol structure on the far right is sorbitol. 33. Which of the following is a ketohexose? O CO C 2 O O C 2 C 2 C 2 O C 2 I II III IV a. I b. II c. III d. IV e. None of the above are ketohexoses Only one of the above is a ketone with 6C (fructose). The carbohydrates portion of the exam is likely the easiest 34. Which of the following is not a disaccharide? a. Fructose b. Maltose c. Sucrose d. Lactose e. N/A; all are disaccharides Simply must know the major players in terms of the mono-, di-, and polysaccharides we have discussed (refer to the exam review for what you must know about them).

12 35. Which of the following is not a reducing sugar? a. Glucose b. Galactose c. Fructose d. Ribose e. N/A; all are reducing sugars The question is slightly tricky since fructose is a ketohexose that must isomerize to the aldohexose glucose in order to act as a reducing sugar. Really though, all of the above are monosaccharides and all of the monosaccharides we have studied are reducing sugars, since there is nothing to lock up the hemiacetal or hemiketal form. 36. Which of the following is a reducing sugar? a. Sucrose b. Maltose c. Amylose d. Amylopectin e. N/A; all are reducing sugars The plant starch components amylose and amylopectin are too large to be reducing sugars (only one terminal hemiacetal amongst a couple thousand linked glucose units) and the disaccharide sucrose ties up both hemiacetal and hemiketal by linking them together 37. In terms of organic functional groups, formation of a glycosidic linkage is equivalent to formation of a(n) a. Acetal b. Amide c. Ester d. Ether e. emiacetal Bear in mind the formation of the cyclic structure coincides with hemiacetal (or hemiketal) formation, and the formation of linked species corresponds to full acetal formation. See question #38 as well 38. What functional groups are generated when a glucosidase (e.g. amylase) hydrolyzes a substrate? a. Carboxylic acid and alcohol b. Carboxylic acid and amine c. Alcohol and amine d. Aldehyde and alcohol e. emiacetal and alcohol

13 39. Which of the following is not composed entirely of glucose units? a. Sucrose b. Maltose c. Amylose d. Amylopectin e. N/A; all are composed of glucose Sucrose contains fructose, lactose contains galactose, and maltose is pure glucose goodness 40. Which of the following contains a -glycosidic linkage? a. Sucrose b. Lactose c. Cellulose d. yaluronic acid ighly tricky (unless you solved the problem by deduction) since while it is true that in terms of the glucose component sucrose does feature an (1,2) glycosidic bond. owever, where the fructose is concerned the bond is in the orientation, the anomeric (look it up) C in each case being involved in linking the 2 monosaccharides together 41. What are the 2 components of starch? a. Glycogen and amylopectin b. Glycogen and amylose c. Cellulose and amylopectin d. Cellulose and amylose e. Amylopectin and amylose 42. Which of the following polysaccharides are branched? a. Glycogen and amylopectin b. Glycogen and amylose c. Cellulose and amylopectin d. Cellulose and amylose e. Amylopectin and amylose

14 43. Which of the following polysaccharides form both -1,4 and -1,6-glycosidic linkages? a. Glycogen and amylopectin b. Glycogen and amylose c. Cellulose and amylopectin d. Cellulose and amylose e. Amylopectin and amylose Really just an extension of #42 note that for glucose, the aldehyde C is position 1, position 5 attacks it to cyclize the molecule, and position 6 is not part of the ring as such it makes for a convenient (from an access/bulk/steric constraint standpoint) site to branch the polymer 44. Which of the following polysaccharides forms difficult to digest fibers? a. Amylose b. Amylopectin c. Glycogen d. Cellulose Pretty amazing the difference in structure and resulting function just by changing the glycosidic linkage from to, since all of the above are made completely from glucose units