biochem exam 1

Henderson-Hasselbalch equation

*pH = pKa + log{{A-]/[HA]}* (hint: AHA, not HAA) Example: Acetylsalicylic Acid- only neutral form penetrates cells (pKa=3) -Stomach (pH=2) 2 = 3 + logAHA -1 = logAHA 1 = logHAA 10=[HA]/[A] 10x[A]= HA (more HA than A) 10/11 is protonated (HA) & 1/11 is unprotonated (A) protonated = neutral form. so majority dissociates and penetrates cells in stomach (can cause pain) -Intestine (pH=7) 7 = 3 + logAHA 4 = log AHA 1000 = [A]/[HA] 1000x[HA] = [A] (more A than HA) 10000/10001 (99.99%) is unprotonated (A) & 1/10001 is protonated (HA) almost none is protonated (neutral form) so it doesn't penetrate and hurt cells Enteric coating- protects drug from dissolving in stomach, dissolves in intestine instead. -can delay drug's reaction -drug targeting

Arg-Gly-Lys-Phe pKas: 2.18, 9.6, 10.53, 12.48 (L4)

+NH3-Arg(+NH3)-Gly-Lys(+NH3)-Phe-COOH Charge: +3 +3 (2.18) +2 (9.6) +1 (10.53) 0 (12.48) -1 PI = (10.53 + 12.48) / 2 = 11.51

Lys-Glu-Gly pKas: 8.97, 10.42, 3.67, 2.12 (L4)

+NH3-Lys(+NH3)-Glu(OH)-Gly-COOH Charge: +2 +2 (2.12) +1 (3.67) 0 (8.97) -1 (10.42) -2 PI = (3.67 + 8.97) / 2 = 6.32

Proton Hopping

- *no free H+*, always forms H3O+ - autoprotolysis: proton transfer b/w two identical water molecules -protons hop in natural chain of water molecules until water accepts H+ and becomes H3O+ -results in fast acid/base rxns

Electrostatic Interactions

-Attractive and Repulsive forces -takes more energy to keep charges of same sign together (likes repel) as opposed to opposite charges (opposites attract) -opposite signs, energy released spontaneoulsy, bond is stable -tend to be on surface -water is released

Salt Bridge

-H bonding plus electrostatics -strongest -occurs b/w opposite charges in close proximity

Hydrophobic Interaction

-H2O released from apolar groups which increases randomness (S>0) -NOT true bonds -strength determined by # of water molecules released -facilitates protein folding-->decrease entropy (but overall S>0) -side chains can swivel to facilitate protein folding -stabilize proteins and membrane bilayer Gibbs equation: G=H-TS -water binds to apolar groups -H for water binding < 0 -H for water release > 0 -G>0 is unfavorable -G<0 is favorable Temperature -at 5 degrees, G can be > 0 -at 37 (body temp), G can be < 0 -all depends on TS (not just S)

Hydrogen Bonds

-basically an acid-base interaction R-N-H-----O=C-R -b/w H and O, or H and N -creates network of H-bonds used for structure - alpha helix or beta sheets create scaffold that dehydrates interior by positioning side chains to maximize hydrophobic interactions -essential for: enzyme substrate interactions, receptor-ligand interactions, RNA & DNA base-pairing, fast kinetics (easy to break/make) -NOT a major driving force in protein folding (2 bonds broken/2 bonds made)

Water Cycle

-continuous exchange of water within hydrosphere -exchange b/w atm., soil water, surface water, groundwater, plants, animals - evaporation - transpiration - precipitation - run-off

Enzyme

-increase reactivity -specific and controllable -w enzyme: 1 sec. w/o enzyme: 1 billion secs (30 years) -Example: Glucose Phosphorylation HK specifically phosphorylates at position-6 HK regulated by glucose-6-P

Water

-large dipole moment -so polar, interactions so strong, high heat capacity (thermal mass- resistant to temp. change) -high B.P. relative to m.w. -high viscosity relative to m.w. -strong interactions w ionic molecules -special interactions w nonpolar molecules accessible M.P -occurs in temp. range on Earth -H2O freezes, expands, breaks plants, allows C atoms to re-enter carbon cycle low B.P. relative to Earth's temp. -evaporative heat loss -keeps atm. moist -moves big H2O quantities as clouds, precip., run-off -erosion remodels Earth's landscape, inc. SA, exposing mineral nutrients

Buffers

-obey Le Chatelier's Principle -most effective when pH=pKa -effective buffer zone: +/- 1 from pH -important: *bicarbonate* regulates blood pH catalyzed by carbonic anhydrase high blood pH- breathe into bag low blood pH- hyperventilate

Van der Waals Interactions

-short-lived bonding interaction b/w oscillating dipoles of non-bonding e- -depends on distance b/w molecules: operate in short distances -weak bond -in protein's dehydrated interior, numerous interactions contribute to protein stability

H-Bond

-stable interaction occurring when 2 electroneg. atoms share one H-atom -always being made and broken -flickers -gives liquid water so much stability that it must be heated to 100 degrees to break all H-bonds -linear bonds are stronger than bent -liquid has 3.4 H-bonds on average -ice has 4

Pi Stacking Interactions

-stacking aromatic rings -aromatic rings have delocalized pi electrons above and below plane of ring -interact w quarternary ammonium ions to form stable bonds -important for substrate interactions and receptor-agonist interactions -increase amount of protonated side chains, increasing acidity, lower pKa -nicotine binds and blocks receptor

H2O deactivates nucleophilic rxns

-water crowds e-rich and e-deficient reactants, so NR -requires enzyme catalyst w/ reactants -increase concentration, reactivity (orient substrate), and speed -stabilize T.S. (lower EA) on/off switches -NR w/o enzyme -reduces toxic side rxns -effective regulation

Ramachandran plot (L5) *look at blank plot & name all positions

1. Antiparallel B-sheet 2. Parallel B-sheet 3. Collagen Triple Helix 4. Right-twisted B sheet 5. Left-handed alpha-helix 6. Right-handed alpha-helix

How to solve pI questions (L4)

1. look out for these 7 amino acids *will have 3 pKa values* neg charged: Asp (OH) and Glu (OH) pos charged: Lys (+NH3), Arg(+NH3), His(+NH3) other: Cys(SH) and Tyr(OH) 2. Cross out pKas you won't use on left end: need one (8-10), cross out one (1.5-3), leave others middle: cross out two (1.5-3 and 8-10) right end: need one (1.5-3), cross out one (8-10), leave others 3. Protonate all of them NH2 --> +NH3 O- --> OH COO- --> COOH 4. See what charge you start with 5. Take off H+ by lowest pKas 6. Stop when overall charge is -1 7. Use pKa before and after charge 0 for Eq pI = (pK1 + pK2) / 2

Determine the ratio of protonated and unprotonated A. pH = 4 & pKa = 6 B. pH = 7 & pKa = 4 (L1)

A. Use H-H eq. pH = pKa + log{[A]/[HA]} 4 = 6 + log{[A]/[HA]} -2 = log{[A]/[HA]} .01 = [A]/[HA] A:HA .01:1 protonated: 1/(1+.01) = 99% unprotonated: .01/(1+.01) = 1% B. Use H-H eq. pH = pKa + log{[A]/[HA]} 7 = 4 + log{[A]/[HA]} 3 = log{[A]/[HA]} 1000 = [A]/[HA] A:HA 1000:1 protonated: 1/(1+1000) = .1% unprotonated: 1000/(1+1000) = 99.9%

What is Mad Cow's disease? A) A protein misfolding disease in which a misfolded protein can interact with a normal protein of the same type and cause the formation of amyloid fibrils B) A disease caused by a virus that affects the brain of cows C) A protein misfolding disease in which misfolded proteins bind to the active site of other enzymes and cause the release of toxins that affect the brain of cows D) A disease caused by a bacteria that affects the brain of cows and that can be passed to humans through the consumption of cow milk E) None of the above are correct (L6)

Answer: A

Which of the following diseases are related to protein misfolding? A) Alzheimer's B) Schizophrenia C) Flu D) Cough E) All of the above are true (L6)

Answer: A

Which of the following is NOT true? A) Structural similarity between two proteins requires extensive sequence identity B) Protein folding patterns are conserved during evolution C) Muscles use myoglobin to store oxygen while red blood cells use hemoglobin to transport oxygen D) Hb and Mb are both 8-helix polypeptides E) All of the following are true (L7-8)

Answer: A Explanation: Although hemoglobin and myoglobin have very similar structures, they only share 27 identical residues. This means that primary structure alone is not just the only factor in protein structure.

Which statement is FALSE regarding enzyme function? A) The active site is responsible for defining the equilibrium constant. B) They increase reaction rates without altering equilibrium. C) They exhibit desolvation, facilitating nucleophilic attacks. D) They greatly improve in-line displacement of leaving group. E) The active site tends to be stereospecific. (L9-10)

Answer: A Explanation: Answer choice "A" is incorrect because the equilibrium constant is defined by the relative concentrations of substrates and products present at equilibrium in a biochemical reaction.

The white region of a Ramachandran plot indicates A) Phi and psi angle combinations that are not typically seen in nature B) Antiparallel β-Pleated sheets C) The gly - pro - __ - gly - pro - ___ pattern found in tropocollagen D) The most common phi and psi angle combinations E) Which amino acids are present in a polypeptide and their location in the sequence (L5)

Answer: A Explanation: Blue is seen; White is not seen.

Name the following amino acid and choose its correct characteristic at a pH 7.0. *chain of AA pictured: -CH2-CH2-CH2-NH2 A. K; basic amino acid B. K; acidic amino acid C. L; basic amino acid D. L; acidic amino acid (L2)

Answer: A Explanation: Picture shows Lysine. Have to know 1-letter code is K (L is for leucine). Lys is a positively charged amino acid so basic (acid would be negatively charged ones).

Select the FALSE statement below regarding both myoglobin and hemoglobin A) Myoglobin contains one heme group while hemoglobin contains four B) Myoglobin is primarily found in the liver and promotes oxygen uptake withinhepatocytes C) Myoglobin is only present as a tertiary structure D) Hemoglobin is comprised of four subunits E) More than one statement is false (L7-8)

Answer: B

The omega bond of the peptide does not rotate because: A) It is a shorter bond than both psi and phi B) It has double bond character (40%) due to resonance C) It is not coplanar unlike the other bonds in the peptide D) It is part of the backbone and therefore does not participate in any rotation E) All of the above are correct (L5)

Answer: B

You are a healthcare professional at Shands and receive a case of a patient being exposed to high levels of carbon monoxide while working at the construction site. How does carbon monoxide poisoning affect circulation? A) Carbon monoxide causes oxygen binding to become parabolic as the patient gets exposed to the toxic gas B) Carbon monoxide and oxygen compete for the same heme groups on hemoglobin C) Carbon monoxide binds 20000X more tightly than oxygen onto hemoglobin D) Carbon monoxide present at low levels causes surrounding oxygen molecules to lose affinity for hemoglobin E) Two of the above statements are true (L7-8)

Answer: B

Calculate [P·L]/[Ptot], if [L] = 3Kd A) 0.85 B) 0.75 C) 0.38 D) 0.67 E) 0.94 (L7-8)

Answer: B Explanation: Sat. Eq. = 1/(1+Kd/[L]) 1/1.333 = .75

Which of the following is an INCORRECT pairing of enzyme activity and how they promote catalysis? A) Metal ion catalysis exploits Lewis acid and redox properties. B) Orientation involves binding a substrate in ways that stabilizes chemical and conformational intermediates of catalysis. C) Acid-base catalysis involved the donating/abstracting of protons between enzyme and substrate. D) Conformational catalysis creates a dynamic environment that facilitates chemical changes during catalysis. E) Transition-state stabilization is a combination of enzyme properties that lower the activation energy for each step in a catalytic mechanisms. (L9-10)

Answer: B Explanation: All of the listed statements are true pairings except for the choice regarding orientation. Orientation is the act of arranging the active-site for optimal activity. It is in fact induced-fit that stabilizes the intermediates of catalysis.

Which statement about Km is TRUE? A) Km and Kd can never be equal. B) Km is the concentration of substrates at half the Vm C) Km can be used to calculate affinity. D) Km is a thermodynamic parameter. E) Km cannot be calculated at steady-state enzyme activity. (L11)

Answer: B Explanation: By definition, this is what Km is. Only Kd can be used to determine the affinity directly. Km is not a thermodynamic parameter but a kinetic parameter. Finally, all Michaelis-Menten related equations must only be calculated in under the steady-state assumption, including calculations of Km.

Secondary structures are essential to the proper structure and function of proteins. Which of the following statements is false regarding secondary structures? A) In an alpha-helix, there are 3.6 residues per turn B) Twists are short sequences that connect alpha-helices and/or beta-sheets and allow proteins to be compact and stable C) Beta-pleated sheets can be either parallel or anti-parallel D) Anti-parallel beta-sheets are more stable than parallel beta-sheets due to their 180 degrees hydrogen bonds E) The allowable torsion angles between residues in each secondary structure can be identified in specific blue areas of a Ramachandran Plot (L5)

Answer: B Explanation: They are called TURNS, not twists. Everything else is true.

The statement "[P] = 0 throughout, measurements such that: 'EX-to-E+P' is effectively unidirectional," _________________. A) Is the primary conclusion for the Lineweaver-Burk equation. B) Is an assumption of the Michaelis-Menten equation. C) Has been proven incorrect since it was first "discovered." D) Applies only to irreversible inhibitors. E) Is a steady-state assumption. (L11)

Answer: B Explanation: This statement is an assumption of the Michaelis-Menten equation. It is one of four assumptions made in order to accurately predict the enzyme kinetics. It is not the steady state assumption, which states that the concentrations of the intermediates of a reaction remain the same even when the concentrations of starting materials and products are changing.

What type of enzyme would catalyze this rxn? *ribulose-->xylulose A) Hydrolase B) Isomerase C) Ligase D) Oxidoreductase E) Ergase (L9-10)

Answer: B Explanation: catalyze intramolecular rearrangements

Proline differs from other amino acids in that A) It can only be formed by post translational modifications to an amino acid sequence B) It most commonly exists in the cis configuration C) It has a side chain that is both non-polar and hydrophobic D) It is made as L-proline E) It is capable of absorbing significant amounts of UV light at a wavelength of 280 nm (L5)

Answer: B Explanation: Because of its cyclic side chain, proline is able to exist in the cis configuration. This is its most common form. The other amino acids are only capable of being trans.

Which of the following is false concerning protein folding? A) A protein's secondary structure begins to form before the polypeptide chain has been released from the ribosome. B) Hydrophobic interactions and hydrogen bonding are thermodynamic driving forces in the folding of proteins. C) Phi and psi angles can be used to accurately predict the secondary structure of a protein. D) Hydrogen bonding in alpha helices and beta sheets allows for the interior of the protein to be dehydrated. E) None of the above. (L5)

Answer: B Explanation: When hydrogen bonding occurs in protein folding, the number of hydrogen bonds being formed equals the number of hydrogen bonds being broken. Because of this, the energy change is 0. But hydrophobic interactions are driving forces

When climbing a mountain what molecule will be released due to low blood pH at higher altitudes? A) Cortisol B) Epinephrine C) 2,3 - Bisphosphoglycerate D) Glucagon E) Insulin (L7-8)

Answer: C

Which of the following answer choices is FALSE? A) Most proteins are spherical B) Proteins are long polypeptides that range in molecular weight C) The folding of a protein is a time-ordered process. D) Proteins can function as antibodies. E) Some proteins are structureless but can gain a 3D structure after binding to another well-folded protein (L5)

Answer: C

Which of the following statements is TRUE regarding covalent modification: A) Serine, Threonine, and Tryptophan may be phosphorylated B) Phosphatases catalyze addition of phosphate groups to enzymes C) Proteases catalyze irreversible cleavage of polypeptides D) Protein Kinases do not use ATP E) More than one of the above is correct (L12)

Answer: C Explanation: The slide states "Some modification reactions are irreversible (proteases)."

Which aromatic structured amino acid absorbs the least amount of UV light? A. Proline B. Histidine C. Phenylalanine D. Tyrosine E. Tryptophan (L2)

Answer: C Explanation: The three aromatic amino acids are: phenylalanine, tyrosine, and tryptophan (crosses out A and B). we know Trp absorbs the most (bc biggest?), so Phe has to absorb the least (bc it is the smallest?). IF it didn't say aromatic, A would be the answer

Which statement is FALSE regarding enzymatic regulation: A) Phosphorylation is an example of reversible covalent modification B) The initial enzyme (E1) in a metabolic pathway is usually subject to feedback inhibition C) Activators induce a conformation change in the enzyme that decreases Kcat D) Multi-subunit enzymes often show cooperativity E) The initial enzyme (E1) catalyzes the committed step in a pathway (L12)

Answer: C Explanation: The slide states that "Activators induce a conformational change that increases substrate binding and/or Increases Kcat."

During the 2018 college football Peach Bowl, our very own Feleipe Franks (Gator football player) sprinted down the field for a touchdown. During this short bout of physical activity, his breathing increased dramatically. This respiratory compensation is noted as a powerful acid-base regulator. Which is true regarding the statements below? A) Franks had a decrease in H+ ion concentration in his peripheral tissues B) Franks had an increase in H+ ion concentration in is peripheral tissues due to carbonic acid binding to hemoglobin's N-terminal groups. C) Franks' increased breathing rate implies that acidosis occurred. D) The Bohr Effect explains why Franks' hemoglobin-O2 affinity increased during his run. E) More than one of the above is correct. (L7-8)

Answer: C How I think through it: when we exercise, we make lactic acid, more acidic, more H ions, lower pH muscles are using more O2 hemoglobin affinity for O2 would decrease so they can drop it off better at the tissues Explanation: The important piece of information to take away from the football scenario is that physical activity occurred, and pulmonary ventilation has increased. If physical activity occurred, then Franks' metabolic activity has increased. Lecture indicates that a high metabolic activity lowers peripheral tissue pH, thus increasing H+ ion concentration (eliminates "A"). It also indicates that it is bicarbonate (HCO3-) that binds to Hb's n-terminal groups, not carbonic acid (eliminates "B"). Moreover, the Bohr Effect describes how protons antagonize O2 binding, thus decreasing Hb-O2 affinity. This makes sense because you want the peripheral tissue to receive O2, which means that O2 needs to be unloaded form Hb (eliminates "D"). Answer choice C is correct because acidosis is when the blood has excess H+ ions. If there is an excess amount of H+ ions, this will cause a shift in favor towards CO2, which needs to be exhaled out of the body (one reason why Franks' breathing rate increased).

In order for proteins to fold, they must follow the laws of thermodynamics. Which of the following drive protein folding? A) The local entropy of protein folding B) The enthalpy of hydrogen bonding C) The entropy of the hydrophobic effect D) The enthalpy of the hydrophobic effect E) More than one of the above drive protein folding (L6)

Answer: C Short explanation: protein folding decreases S, but water releasing from hydrophobic interactions increases S. so overall, S is positive Long explanation: Remember that ΔG= ∑ΔH- TΔS and that a negative ΔG is favorable. You therefore want negative enthalpy and positive entropy. ( - = - +) Protein folding leads to reduced local entropy (-ΔS), which is a positive -TΔS and therefore unfavorable. The net enthalpy of hydrogen bonding equals zero, so it does not have a thermodynamic effect on protein folding. (+ = 0 +) The release of water attending burying hydrophobic residues leads to an increase in entropy (+ΔS) and therefore a -TΔS unit, which is favorable and this makes C correct. The enthalpy of the hydrophobic effect, however, is unfavorable. (- = + +) This brings an important fact that it only matters that the overall ΔG is negative for protein folding to occur, and not that every component that dictates the thermodynamics of the system is favorable.

What order of events for the GroEL-GroES complex mechanism in protein folding is correct? I. Complex recognizes a target protein by its exposed hydrophobic region II. ATP hydrolysis within the interior chamber occurs activating a conformational change in the target protein III. Polypeptide is inserted in through the GroES cap IV. Complex flips allowing the native protein out V. Protein is forced into a compact state A) I, III, II, IV, V B) I, II, III, IV, V C) I, III, II, V, IV D) III, I, V, II, IV E) III, I, II, V, IV (L6)

Answer: C) I, III, II, V, IV

Hemoglobin is an example of which type of protein structure? A) Primary B) Secondary C) Tertiary D) Quaternary E) Hemoglobin is not a protein (L7-8)

Answer: D

In order for drugs to effectively carry out its desired function, the drug must be sterically complementary to the active site of the enzyme. Structures of these drugs are similar to the structure of the_______. A) Active enzyme B) Substrate C) ES complex D) Transition state E) Receptor tyrosine kinase (L9-10)

Answer: D

Which of the following is TRUE? A) Chaperone proteins (foldases) help proteins fold spontaneously B) Protein folding takes place after proteins are released from the ribosome/mRNAcomplex C) Water release due to hydrophobic interactions between amino acid residues in apeptide causes a decrease in entropy D) Proteins can denature upon exposure to organic solvents, urea, detergents, andheat E) Protein folding is an all-or-nothing two step process (L6)

Answer: D

What fraction of lysine has its R group protonated at pH= 9.8. Assume lysine has pKa values of 2.16, 9.06, and 10.78. A. 30% B. 65% C. 86% D. 91% (L1)

Answer: D 2.16 would correspond to the C-terminus (b/w 1.5-3) and 9.06 corresponds w N-terminus (b/w 8-10). so for this problem, we use 10.78 as pKa as it refers to the R group the question is asking about. Use H-H eq. pH = pKa + log{[A]/[HA]} 9.8 = 10.78 + log{[A]/[HA]} -.98 = log{[A]/[HA]} .105 = [A]/[HA] A:HA .105:1 fraction that is protonated: 1/(1+.105)= .905 = 91%

Why does an uncatalyzed ester hydrolysis reaction take longer than a enzyme-catalyzed ester hydrolysis reaction? A) An enzyme-catalyzed ester hydrolysis requires the enzyme to always return to its original form after each round of catalysis. B) Enzymes are grouped into six classes. C) The uncatalyzed ester hydrolysis is due to ribosome catalysis rather than enzyme catalysis. D) Hydrolyzing an ester without an enzyme requires overcoming a high energy barrier due to an unstable oxy-anion intermediate. E) More than one of the above is correct. (L9-10)

Answer: D Explanation: Answer choices "A" and "B" are correct statements, but they do not answer the question of why uncatalyzed ester hydrolysis reactions take longer to proceed. "C" is incorrect because enzyme catalysis via RNA molecules are rare. An example of RNA molecule-enzyme catalyze reactions are when ribosomes catalyze peptide bond formation.

The reaction shown is catalyzed by what type of enzyme? *picture is rxn: glucose-6-phosphate --> fructose-6-phosphate A) Oxidoreductases B) Transferases C) Ligases D) Isomerases E) Ergases (L9-10)

Answer: D Explanation: Isomerases catalyze intramolecular rearrangements. Glucose-6-P ⇌ Fructose-6-P was listed as an example of an Isomerase.

A given polypeptide sequence: Val-Ala-Arg-Ile-Pro-Asn-His-Leu-Lys-Trp-Cys-Met-Ser-Thr-Gly-Glu What is the fragment of the polypeptide that will be released from the active site of Chymotrypsin after Phase 1 only? A) Val-Ala-Arg B) Val-Ala-Arg-Ile-Pro-Asn-His-Leu-Lys-Trp C) Ile-Pro-Asn-His-Leu-Lys-Trp-Cys-Met-Ser-Thr-Gly-Glu D) Cys-Met-Ser-Thr-Gly-Glu E) Val-Ala-Arg-Ile-Pro-Asn-His-Leu-Lys-Trp-Cys-Met-Ser-Thr-Gly-Glu (L9-10)

Answer: D Explanation: Remember that in the first phase of the chymotrypsin mechanism, the C terminal fragment is the first product. The N terminal side still has enzyme attached to it at this point.

Where is hemoglobin most likely to be found in its R state? A) The capillaries B) Cerebrospinal fluid C) Lymph D) The lungs E) The brain (L7-8)

Answer: D Explanation: Since the partial pressure of oxygen is high in the lungs, more hemoglobin is likely to be saturated. The capillaries are where hemoglobin lose their binding affinity to oxygen due to low partial pressure in that area. While the brain uses oxygen extensively, hemoglobin will also lose its affinity for oxygen since the brain will be needing it for cellular respiration (outside the brain cells, partial pressure of oxygen is low). Lymph and cerebral spinal fluid do not contain red blood cells.

An individual is found to have a deletion of BOTH amino acid residues His57 and Asp102 in the active site of chymotrypsin. How would this change chymotrypsin's ability to cleave peptides? A) The reaction would have a slightly higher ΔG, but still would occur since Ser102 is the primary catalytic residue and is intact. B) The hydrophobic pocket would not bind the substrate's aromatic residue. C) The reaction would likely proceed past the first tetrahedral intermediate, but then stop immediately after. D) The substrate would bind, but could not be attacked by Ser195 E) The acylated intermediate forms would become more stable than the product. (L9-10)

Answer: D Explanation: The Asp and His act to hold protons for the Ser as Ser performs much of the chemical reactions with the substrate. Without these two residue, the substrate would still bind the hydrophobic pocket, as seen in step 1, but Ser itself could not attack the substrate so no tetrahedral intermediate would be formed.

Felipe Franks went on a vacation in Denver, CO. While hiking up a mountain, Franks' noticed he was more tired than normal. After a few days of hiking, Franks' no longer felt fatigued while hiking. What is the cause for this acclimatization? A) 2,3-Bisphophoglycerate binds to oxy-Hb, stabilizing the T-state, thus favoring O2 dissociation. B) 2,3-Bisphophoglycerate binds to oxy-Hb, stabilizing the R-state, thus favoring O2 dissociation. C) 2,3-Bisphophoglycerate binds to deoxy-Hb, stabilizing the R-state, thus favoring O2 dissociation. D) 2,3-Bisphophoglycerate binds to deoxy-Hb, stabilizing the T-state, thus favoring O2 dissociation. E) None of the above (L7-8)

Answer: D Explanation: The slide states "BPG binds to deoxy-Hb, stabilizes the T-state & favors O2 dissociation"

Raising [S] in the presence of an enzyme under Michaelis-Menten assumptions __________. A) Cannot fully reverse noncompetitive inhibition. B) Can fully reverse competitive inhibition. C) Will get v to approach Vm D) All of the above are correct. E) Only a and b are correct. (L11)

Answer: D Explanation: By increasing [S] with noncompetitive inhibitors, the inhibitor will still bind to the enzyme regardless of substrate concentration, therefore the inhibition cannot be fully compensated for. Conversely, with competitive inhibitors, increasing [S] there will be enough substrates to outcompete the inhibitor binding directly to the enzyme, therefore reversing inhibition of competitive binding inhibitors. Finally, by increasing the [S], there will be an increase rate of reaction, but this is limited by the enzyme's Vm

You are a physician at UF Health and a patient comes in with what you suspect to be Mad Cow Disease. What should you look for to confirm your theory? A) The formation of soluble aggregates within or outside cells called amyloids B) Normal PrPSc interacting with misfolded PrPC that then cause the PrPSc to misfold C) The aggregation of misfolded prion proteins into amyloid fibrils in the Peripheral Nervous System D) The conversion of α-helices to β-sheets E) None of the above can confirm your assumption (L6)

Answer: D Explanation: Prions cause the conversion of *PrPC (normal)* with α-helices to become PrPSc (misfolded) prions with β-sheets. The aberrant prions promote further normal prion misfolding. This leads to the formation of *insoluble* amyloid fibrils in the *Central* Nervous System

The C-N _______ bond is shorter than the C-N _______ bond due to _______ . A) Peptide; amide; its ability to form hydrogen bonds with other molecules B) Peptide; amide; post translational modifications to the N-terminus of the protein C) Amide; peptide; its ability to form hydrogen bonds with other molecules D) Peptide; amide; its double bond character E) Amide; peptide; its double bond character (L5)

Answer: D Explanation: The peptide bond has a resonance structure in which a double bond is present between the C and N. Double bonds are shorter than single bonds.

Which substance would not denature a protein? A) Organic solvents B) Urea C) Detergents D) Heat E) All of the above would denature proteins (L6)

Answer: E

All of the following are true in regards to hemoglobin and myoglobin EXCEPT: A) Hemoglobin and myoglobin are heme proteins that bind a ligand, molecular oxygen. B) Myoglobin has a similar tertiary structure to the monomer subunit of hemoglobin. C) Myoglobin and hemoglobin subunits do not share much primary structure. D) Hemoglobin functions to transport oxygen from the lungs to the tissues, while myoglobin functions to store it in the tissue. E) None of the above (L7-8)

Answer: E Explanation: All are true

When climbing up a tall mountain from ground level, what are some of the physiological changes that might occur in one's body that differ from the top of the mountain compared to the bottom? A) Blood pH increases B) 2,3-bisphosphoglyercate is inhibited C) Respiratory rate decreases D) The R state of hemoglobin is favored E) None of the above (L7-8)

Answer: E Explanation: At high altitudes, the partial pressure of O2 will have decreased. This will lower the pH of the blood. Since pH is lowered, respiratory rate should increase to exhale more CO2 and shift the carbonase reaction towards the reactants, making the buffer solution less acidic in the blood. At low pH, 2,3-bisphosphoglycerate is released and plays a role in stabilizing the T state so oxygen can be released more efficiently in the blood where O2 is needed.

Which of the following is false regarding protein folding? A) The first step in one model is nucleation, where local secondary structures are formed B) Folding is a massive parallel search of conformation space C) The second model states that folding is initiated by the rapid collapse into a molten globule state D) Following the formation of a molten globule, local bonding interactions occur rapidly and cooperatively E) The scientific community agrees that folding can only follow one idea, but they have not determined which is accurate yet (L6)

Answer: E Explanation: Folding is most likely a mix of both ideas

Which of the following is not a type of secondary structure? A) Collagen triple helix B) β-Pleated sheets C) α-Helix D) Turns E) All of the above are types of secondary structures (L5)

Answer: E Explanation: Secondary structures include: helices, beta sheets, and turns. Collagen triple helix is a type of helix, so it does count.

Which group of enzymes would be responsible for catalyzing the following chemical reaction? Histine--> Urocanate + ammonia *Urocate- same but w/o ammonia and now a double bond A) Transferases B) Ergases C) Hydrolases D) Isomerases E) None of the above (L9-10)

Answer: E Explanation: The molecule represented in the figure is histidine and the chemical process shows it losing an amino group. The following reaction is carried out by a lyase, more specifically a ​histidine ammonia-lyase. Lyases are involved in the addition or removal of functional groups. The enzyme wouldn't be a hydrolase because water is not present during the reaction.

On Mars, an alien life-form discovers that peptides are assembled the same way as on Earth. However, when naming the torsion angles with Greek letters, it uses the same letter for the peptide bond but switches the other two. In this scenario, which of the following statements is true? A) The peptide bond refers to the bond between the carbonyl group and the alpha carbon B) Phi refers to the bond between the alpha carbon and the amino group C) Pi refers to the bond between the carbonyl group and the alpha carbon D) Psi is the peptide bond E) Omega is the bond between the carbonyl group and the amino group (L5)

Answer: E Explanation: The omega bond is the peptide bond, which is the bond between the carbonyl group of one amino acid and the amino group of another. On Mars, phi and psi are switched, so ​in this scenario​ phi refers to the bond between the carbonyl group and the alpha carbon, while psi refers to the bond between the alpha carbon and the amino group. The pi bond angle does not exist.

Chymotrypsin Q: What makes His57 a better base? (L9-10)

Asp

Chymotrypsin Q: Which AA are involved in catalytic triad? (L9-10)

Asp, His and Ser

Chymotrypsin Q: Where does chymotrypsin cleave? (L9-10)

C-terminal of aromatic groups

Noncovalent Interactions

Essential to life: -the strengths of noncovalent bonding interactions (not bonds) are "tunable" -made/broken readily -weak, but numerous enough to make them effective -weaker than a C-C single bond -nature allows proteins to fold/unfold correctly -determine: protein structure and function, enzyme specificity, DNA structure and function, membrane formation and stability, and nearly every other vital process in living systems

Chymotrypsin Q: Which AA acts as acid/base in Phase 1 and 2? (L9-10)

His

Hydrolysis

Ionic (uncatalyzed) - NaCl + 8H2O --> Na+(H2O)4 + Cl-(H2O)4 Covalent (catalyzed) 1. Carbonic Anhydrase 2. Glucose-6-Phosphate (opposite of glucose phosphorylation)

Chymotrypsin Q: What are the products formed in Phase 1 and 2? (L9-10)

Phase 1: NH2-R (C-terminal pdt) Phase 2: serial hydroxyl (N-terminal pdt)

Chymotrypsin Q: What acts as the nuc in Phase 1 and 2? (L9-10)

Phase 1: Ser Phase 2: OH

Acetylsalicylic acid- only neutral form of aspirin (pKA = 3.0) penetrates cells. Calculate percentage of neutral form in the stomach (pH 2.0) and intestine (pH 7). (L1)

Stomach: Use H-H eq. pH = pKa + log{[A]/[HA]} 2 = 3 + log{[A]/[HA]} -1 = log{[A]/[HA]} 1 = log{[HA]/[A]} 10 = [HA]/[A] HA:A 10:1 protonated (neutral form): 10/11 = 90.9% Intestine: Use H-H eq. pH = pKa + log{[A]/[HA]} 7 = 3 + log{[A]/[HA]} 4 = log{[A]/[HA]} 10000 = [A]/[HA] A:HA 10000:1 protonated (neutral form): 1/10001 = .009%

Aspirin has a pKa of 3.4. What is the ratio of A- to HA in: A) the blood (pH= 7.4) B) the stomach (pH= 1.4) (L1)

Use H-H eq. A) Blood pH = pKa + log{[A]/[HA]} 7.4 = 3.4 + log{[A]/[HA]} 4 = log{[A]/[HA]} 10000 = [A]/[HA] A:HA 10000:1 B) Stomach pH = pKa + log{[A]/[HA]} 1.4 = 3.4 + log{[A]/[HA]} -2 = log{[A]/[HA]} 2 = log{[HA]/[A]} 100 = [HA]/[A] HA:A 100:1 A:HA .01:1

The pH of the given solution of lactic acid and lactate is 4.30. Calculate the pKa of lactic acid, when the concentration of lactic acid and lactate are .020M and .073M respetively. (L1)

Use H-H eq. pH = pKa + log{[A]/[HA]} 4.3 = pKa + log[(.073/.020)] 4.3 = pKa + .56 pKa = 3.74

pH formulas to know

pH= -log [H3O+] Kw= [H3O=][OH-]= 10^-14M^2 *pH does not measure proton concentration [H+], measures proton activity* water has pH ~7, not 7.00 exactly

Calculate the fraction of histidine that has an imidazole side chain protonated at pH 7.3. The pKa values for hisatine are pK1= 1.82, pK2= 6.0, pK3= 9.17. (L1)

pK1 would correspond to the C-terminus (b/w 1.5-3) and pK3 corresponds w N-terminus (b/w 8-10). so for this problem, we use pK2 as it refers to the side chain the question is asking about. use H-H eq. pH = pKa + log{[A]/[HA]} 7.3 = 6 + log{[A]/[HA]} 1.3 = log{[A]/[HA]} 20 = [A]/[HA] A:HA 20:1 fraction that is protonated: 1/(20+1)=.048= 4.8%

What is this peptide's isoelectronic point? Gly-Tyr-Asp-His-Ser-Arg-Glu pK Gly: 2.2, 9.6 pK Tyr: 2.1, 9.3, 10.7 pK Asp: 2.3, 3.65, 9.7 pK His: 2.1, 6.0, 9.5 pK Ser: 2.2, 9.8 pK Arg: 2.4, 9.6, 12.5 pK Glu: 2.1, 4.25, 9.7 (L4)

pKas we can use: Gly- 9.6 (N-terminus) Tyr- 9.3 (OH group) Asp- 3.65 (OH group) His- 6.0 (NH3+ group) Ser- NA Arg- 12.5 (NH3+ group) Glu- 2.1 (C-terminus) and 4.25 (OH group) Charge:+3 +NH3-Gly-Tyr(OH)-Asp(OH)-His(+NH3)-Ser-Arg(+NH3)-Glu(OH)-COOH pKa= 2.1 Charge: +2 +NH3-Gly-Tyr(OH)-Asp(OH)-His(+NH3)-Ser-Arg(+NH3)-Glu(OH)-COO- pKa= 3.65 Charge: +1 +NH3-Gly-Tyr(OH)-Asp(O-)-His(+NH3)-Ser-Arg(+NH3)-Glu(OH)-COO- pKa= 4.25 Charge: 0 +NH3-Gly-Tyr(OH)-Asp(O-)-His(+NH3)-Ser-Arg(+NH3)-Glu(O-)-COO- pKa= 6.0 Charge: -1 +NH3-Gly-Tyr(OH)-Asp(O-)-His(NH2)-Ser-Arg(+NH3)-Glu(O-)-COO- *didn't use rest of pKas bc I already have the two pKas around 0 that I need* pI = (pK1 + pK2) / 2 pI = (4.25 + 6.0) / 2 = 5.125

What is the isoelectronic point of the following residues? Leu-Lys-His-Ala pKa of ionizable groups: Leu: 2.34, 9.72 Lys: 2.18, 8.95, 10.53 His: 2.47, 6.0, 8.79 Ala: 2.11, 9.63 (L4)

pKas we can use: Leu- 9.72 (N-terminus) Lys- 10.53 His- 6.0 Ala- 2.11 (C-terminus) Charge: +3 +NH3-Leu-Lys(+NH3)-His(+NH3)-Ala-COOH pKa= 2.11 Charge: +2 +NH3-Leu-Lys(+NH3)-His(+NH3)-Ala-COO- pKa= 6.0 Charge: +1 +NH3-Leu-Lys(+NH3)-His(NH2)-Ala-COO- pKa= 9.72 Charge: 0 NH2-Leu-Lys(+NH3)-His(NH2)-Ala-COO- pKa= 10.53 Charge: -1 NH2-Leu-Lys(NH2)-His(NH2)-Ala-COO- pI = (pK1 + pK2) / 2 pI = (9.72 + 10.53) / 2 = 10.13

What is the isoelectronic point of the following residues? (pKas in the increasing order: 2.3, 9.6, 10.5, 12.5). Arg-Val-Lys (L4)

pKas we'll use: 2.3 = C-terminus 9.6 = N-terminus 10.5 = Lys's R group 12.5 = Arg's R group Charge: +3 +NH3-Arg(+NH3)-Val-Lys(+NH3)-COOH pKa= 2.3 Charge: +2 +NH3-Arg(+NH3)-Val-Lys(+NH3)-COO- pKa= 9.6 Charge: +1 NH2-Arg(+NH3)-Val-Lys(+NH3)-COO- pKa= 10.5 Charge: 0 NH2-Arg(+NH3)-Val-Lys(NH2)-COO- pKa= 12.5 Charge: -1 NH2-Arg(NH2)-Val-Lys(NH2)-COO- pI = (pK1 + pK2) / 2 pI = (10.5 + 12.5) / 2 = 11.5

Glu-Arg-Ala-Lys-Arg pKas: Glu: 2.1, 9.5, 4.1 Arg: 1.8, 9.0, 12.5 Ala: 2.4, 9.9 Lys: 2.2, 9.1, 10.5 (L4)

pkas we'll use: Glu: 9.5(N-terminus), 4.1 (OH) Arg: 1.8 (C-terminus), 12.5 (+NH3) Ala: NA Lys: 10.5 (+NH3) Charge: +4 +NH3-Glu(OH)-Arg(+NH3)-Ala-Lys(+NH3)-Arg(+NH3)-COOH pKa= 1.8 Charge: +3 +NH3-Glu(OH)-Arg(+NH3)-Ala-Lys(+NH3)-Arg(+NH3)-COO- pKa= 4.1 Charge: +2 +NH3-Glu(O-)-Arg(+NH3)-Ala-Lys(+NH3)-Arg(+NH3)-COO- pKa= 9.5 Charge: 1 NH2-Glu(O-)-Arg(+NH3)-Ala-Lys(+NH3)-Arg(+NH3)-COO- pKa= 10.5 Charge: 0 NH2-Glu(O-)-Arg(+NH3)-Ala-Lys(NH2)-Arg(+NH3)-COO- pKa= 12.5 Charge: -2 NH2-Glu(O-)-Arg(NH2)-Ala-Lys(NH2)-Arg(NH2)-COO- pI = (pK1 + pK2) / 2 pI = (10.5 + 12.5) / 2 = 11.5

His-Gly-Val-Cys-Met pkas: His: 1.8, 6.0, 9.3 Gly: 2.4, 9.8 Val: 2.3, 9.7 Cys: 1.9, 10.7, 8.4 Met: 2.1, 9.3 (L4)

pkas we'll use: His: 6.0 (+NH3), 9.3 (N-terminus) Gly: NA Val: NA Cys: 8.4 (weird exception-SH) Met: 2.1 (C-terminus) Charge: +2 +NH3-His(+NH3)-Gly-Val-Cys(SH)-Met-COOH pKa= 2.1 Charge: +1 +NH3-His(+NH3)-Gly-Val-Cys(SH)-Met-COO- pKa= 6.0 Charge: 0 +NH3-His(NH2)-Gly-Val-Cys(SH)-Met-COO- pKa= 8.4 Charge: -1 +NH3-His(NH2)-Gly-Val-Cys(S-)-Met-COO- *didn't use last pKa bc I already have the two pKas around 0 that I need* pI = (pK1 + pK2) / 2 pI = (6.0 + 8.4) / 2 = 7.2