MCAT BIOCHEM Chapter 4 Amino Acids

Enzyme Kinetics

- is the study of the rate of formation of products from substrates in the presence of an enzyme

How to find Michaelis Constant (Km) on an enzyme kinetics graph?

- mark the Vmax on the y axis , then divide this distance in half to find Vmax/2 - Km is found by drawing a horizontal line from Vmax /2 to the curve and then vertical line down to the x axis

Ammonium Group

- protonated or acidic form of the amine group - has a pKa between 9-10

Lineweaver-Burk Plot

-graphical representation of enzyme kinetics using the Lineweaver-Burk equation - follows the format y=mx+b - graph is called a double reciprocal plot b/c the y axis is the inverese of the reaction rate (1/v) and the x axis is the inverse of the substrate concentration (1/ [S]) -THINGS TO KNOW: 1. The slope is Km/Vmax 2. The Y intercept is 1/Vmax 3. The X intercept is -1/Km

Two types of covalent bonds between amino acids in proteins

1) Peptide bonds - link AA together in a polypeptide chain 2) Disulfide bridges - bonds between cysteine R-groups

General Amino Acid Structure

All twenty amino acids share the same nitrogen-carbon-carbon backbone but each has a unique side chain 1) Alpha amino group 2)Alpha carboxyl group 3)variable R-group

Question: what effect would a molecule that disrupts hydrogen bonding, such as urea, have on protein structure?

Answer: - The disruption of H bonding disrupts the secondary structure (possibly tertiary, and quaternary) by unfolding alpha helices and B-sheets. - It won't affect primary structure however

1. Covalent modification

- proteins can have several groups covalently bound to them to regulate their activity - e.g addition og a phosphoryl group from an ATP by a protein kinase ( can either activate or inactivate an enzyme)

Cooperativity

- the binding of substrate to one subunit modulates the affinity of other subunits for substrate - cooperative enzymes must have more than one active site - Two types: Positive and Negative

Question: Compound A converts into Compound B ins solution A<=>B. The reaction has the following equilibrium constant: K= [B]eq/[A]eq = 1000. if pure A is dissolved in water at 298 K, will delta G for the reaction A <=> B be bossitive or negative? Is it possible to answer this question without knowing standard delta G?

Answer: - We don't need to calculate standard delta g - based on the Keq there will be 1000 more B than A in solution at equilibrium - If we create a solution with only A , the reaction must move spontaneously toward B

Question: How many unique dipeptides (made from linking two amino acids) can be synthesized using alanine and glycine residues?

Answer: Four (Gly-Gly, Ala-Ala, Gly-Ala, Ala-Gly)

Acidic Amino Acids

Aspartic acid and Glutamic acid - Have carboxylic acid functional groups (pka ~ 4) in their R-group, - 3 functional groups that can act as an acid (2 back bone and the R group) - found in their deprontanated forms in physiological pH (aspartate & glutamate)

Isomerase

Rearranges bonds within a molecule to form an isomer

Oxidoreductase

Runs redox reactions (includes oxidases, reductases, dehydrogenases, and others)

Polar Amino Acids

Serine, Threonine, Tyrosine, Asparagine, Glutamine Cysteine - R-group capable of H-bonds w/ water but does not act as an acid nor base - hydrophilic - The -OH group on Ser, Thr, and Tyr are often modified with a phosphate group via a kinase. This changes the structure and is important fro regulating protein activity - Asparagine and Glutamine are amine derivatives of aspartic and glutamic acid

Question: Which is more oxidized, the sulfur in Cysteine or the sulfur in Cystine?

The sulfur in Cystine is more oxidized

Ligase

forms a chemical bond (e.g. DNA ligase)

Mixed Type Inhibition

occurs when an inhibitor can bind to either the unoccupied enzyme or the enzyme substrate complex - if the enzyme has a greater affinity for the inhibitor in its free from, the enzyme will have lower affinity for the substrate similar to competitive inhibition (Km increases) - inhibitor binds to allosteric site and additional substrate cannot overcome inhibition (Vmax decreases)

How would the lineweaver-burk plot change when a noncompetative inhibitor is added?

- A noncompetitive inhibitor does not affect Km so the x intercept is unchanged - It does however decrease the V max, so the y intercept increases - The slope then decreases

Recognition pocket

- A pocket in the enzyme's structure which attracts certain residues on substrate polypeptides - Usually found near the active site

Parallel B-pleated Sheets

- Beta pleated sheets with adjacent polypeptide strands running in the same direction

Antiparallel B-pleated Sheets

- Beta pleated sheets with polypeptide strands running in opposite directions

Enzymes

- Biological catalyst - increase the rate of reaction by lowering the reaction's activation energy - Do not affect the Delta G between reactants and products - Have a KINETIC role NOT A THERMODYNAMIC one - Classified based on reaction type

Disulfide Bond

- Cysteine is an AA with a reactive Thiol (sulfhydryl, SH) in its side chain. - The thiol of one cysteine can react with the thiol of another cysteine forming a disulfide bond

How do you calculate for the Isoelectric point (pI)?

- For an AA with w functional groups aka glycine just average the pKa's of the two functional groups - Calculating the pI's of more complex molecules is beyond the scope of the MCAT and you are only expected to know how to calculate the pI of a molecule with 2 functional groups

Peptide Bond

- Formed by linking AA together in peptide bonds - Formed between the carboxyl group of the amino acid and the alpha-amino group of another amino acid with the loss of water

Quaternary(4*) Structure: Various Bonds Between Separate Chains

- Highest level of protein structure - Describes interactions between polypeptide subunits - Subunit: A single polypeptide chain that is a part of a large complex containing many subunits (multisubunit complex) - Arrangement of subunits in a multisubunit complex is the quaternary structure - Forces stabilizing quaternary structures are usually the same as 3* (e.g. van der waals, H-bonds, disulfide bonds, and electrostatic interactions)

10 Enzyme Classes

- Hydrolase - Isomerase - Ligase - Lyase - Kinase - Oxidoreductase - Polymerase - Phosphatase - Phosphorylase - Protease

Proteolysis or Proteolytic Cleavage

- Hydrolysis of a protein by another protein - The protein responsible for the cutting is known as a proteolytic enzyme ore protease - Very specific way of cleaving peptide bonds - E.g. The protease trypsin cleaves adjacent to large hydrophobic residues such as phenylalanine (no need to memorize)

Uncompetitive Inhibition

- Inhibitor is only able to bind to the enzyme substrate complex (can't bind before substrate is bound) - bind to allosteric sites - Decreases Vmax by limiting the amount of available enzyme -substrate complex that can be converted into product -By sequestering enzyme bound to substrate, this increases the apparent affinity of the enzyme for the substrate as it cannot readily dissociate (decareasing Km)

Secondary (2*) Structure: Hydrogen Bonds Between Backbone Groups

- Initial folding of a polypeptide chain into shapes stabilized by hydrogen bonds between backbone NH and CO groups - Two most common secondary structure motifs are the Alpha helix and Beta-pleated sheets

Michaelis Constant (Km)

- Is the substrate concentration at which the reaction velocity is half its max - Km is unique for each enzyme and substrate pair and give information on the affinity of the enzyme for its substrate. -Low Km means that not very much substrate is required to get the reaction rate to half the maximum rate; thus the enzyme has HIGH AFFINITY for this particular substrate

Question: CO2 is an allosteric inhibitor of hemoglobin. It dissociates easily when Hb passes through the lungs, where CO2 can be exhaled. Carbon monoxide, on the other hand, binds to the oxygen binding site with an affinity 300 times greater than oxygen; it can be displaced by oxygen but only when there is much more O2 than CO in the environment. Which of the following is correct? I. Carbon Monoxide is an irreversible inhibitor II. CO2 is a reversible inhibitor III. CO2 is a non competitive inhibitor

- Item (I) is false - Item (II) is true - Item (III) is true

Negative Cooperative Binding

- LESS IMPORTANT FOR MCAT - binding if the substrate to one subunit reduces the affinity of the other subunits for substrate

2. Proteolytic cleavage

- Many enzymes are synthesized in inactive forms (zymogens) that are activated by cleavage by a protease

4. Allosteric regulation

- Modification of the active site activity through interactions of molecules with other specific sites on the enzymes (allosteric sites)

Reaction Coupling

- One very favorable reaction it used to drive an unfavorable one - this is possible because free energy changes are additive

Four levels of protein folding that contribute to their final three-dimensional structure

- Primary (1*) Structure: the amino acid sequence - Secondary (2*) Structure: hydrogen bonds between backbone groups - Tertiary (3*) Structure: hydrophobic/hydrophilic interactions - Quaternary(4*) Structure: various bonds between separate chains

Feedback inhibition

- Rather than regulate every enzyme in a pathway , usually there are one or two key enzymes that are regulated, such as the enzyme that catalyzes the first irreversible step in the pathway -There are examples of positive feedback (feed back stimulation ) but negative feedback is by far the most common example of feedback regulation

Primary (1*) Structure: The Amino Acid Sequence

- Simplest level - order of amino acids bonded to each other in the polypeptide chain - Primary structure is the same as sequence - peptide bond determines the 1* structure b/c this is the bond that links the AA

3. Association with other polypeptides

- Some enzymes have catalytic activity in one polypeptide subunit that is regulated by association with a separate regulatory subunit. - Constitutive activity: continuous rapid catalysis if their subunit is removed

Beta-Pleated Sheets

- Stabilized by hydrogen bonding between NH and CO groups int he polypeptide backbone - compared to alpha helices, B-Sheets hydrogen bonding occurs between residues distant from each other in the chain or even on separate polypeptide chains - B-sheet backbone is extended rather than coiled, with side groups directed above and below the plane of the B-sheet - 2 types of Beta sheets: Parallel and Anti-Parallel

Reaction rate (V, for velocity)

- The amount of product formed per unit time, in moles per second (mol/s) - Depends on the concentration of substrate, [S], and enzyme. - If there is only a little substrate then the rate V is directly proportional to the substrate added (e.g. double the amount of substrate then the rate doubles) - Eventually, there is so much substrate that the active sites of the enzymes are occupied much of the time and decreases - Finally there is so much substrate that every active site is continuously occupied and adding more substrate wont increase reaction rate as much. That is the slope of the V vs S curve decreases (the enzyme is saturated, reaction rate =Vmax)

Active Site Model "lock and key hypothesis"

- The substrate and active site are perfectly complementary.

Induced Fit Model

- The substrate and active site differ slightly in structure and that the binding of the substrate induces a conformational change in the enzyme - Gained greater acceptance -Regardless of the model, enzymes accelerate the rate of a given reaction by helping to stabilize the transition state

Noncompetitive Inhibition

- They bind to allosteric sites, NOT the active sites - No matter how much inhibitor you add the substrate will not be displaced from its site of action - Therefore, noncompetitive inhibition DIMINISHES Vmax - Vmax is always calculated at the same enzyme concentration, since adding more enzyme will increase the measured Vmax - It does not affect Km, b/c the substrate can still bind to the active site, but the inhibitor prevents the catalytic activity of the enzyme

Alpha Helix

- all alpha helix have the same well defined dimensions - The alpha helices of proteins are always right handed - There are 3.6 amino acid residues per turn with the alpha carboxyl oxygen of one amino acid residue hydrogen bonded to the alpha amino proton of an amino acid three residues away

Positive Cooperative binding

- binding of the substrate two one subunit increases the affinity of the other subunits for substrate -sigmoidal curve - flat part at the bottom left: at low [S] the enzyme complex has a low affinity for substrate "tense" (adding more substrate increases the rate a little) - Steep part of the curve represent the range of substrate concentrations where adding substrate greatly increases the reaction rate b/c enzyme complex is relaxed - The leveling off at the upper right represents saturation

Competitive inhibition

- molecules that compete with substrate for binding at the active site - resemble the substrate (somewhat) - *can be overcome by adding more substrate* - if the substrate concentration is high enough then the substrate can outcompete the inhibitor - This is why Vmax is not affected, but it takes more substrate - That is why Km increases compared to uninhibited reaction

Four broad categories of amino acids

1)Acidic 2)Basic 3)Non-polar 4)Polar

Different Mechanisms of Enzyme Inhibition

1. Competitive inhibition 2. Non competitive inhibition 3. Uncompetitive inhibition 4. Mixed type inhibition

The activity of key enzymes in metabolic pathways is usually regulated in one or more of the following ways:

1. Covalent modification 2. Proteolytic cleavage 3. Association with other polypeptides 4. Allosteric regulation

The unique side chain of proline causes two problems in polypeptide chains (reasons why proline residues never appear within the alpha helix):

1. The formation of a peptide bond with proline eliminates the only hydrogen atom on the nitrogen atom of proline. This absence of NH bond disrupts the backbone hydrogen bonding in the polypeptide chain 2. The unique structure of proline forces it to kink the polypeptide chain

Zwitterion (dipolar ion)

A molecule with positive and negative charges that balance (net charge is zero)

Question: Since the existence of a dipeptide is less thermodynamically favorable than the existence of isolated amino acids, how are peptide bonds formed and maintained in cells?

Answer, During protein synthesis, stored energy is used to force peptide bonds to form . Once the bond is formed, even though its destruction is thermodynamically favorable, it remains stable because the activation energy for hydrolysis reaction is so high. In other words, hydrolysis is thermodynamically favorable but kinetically slow.

Question: What is the pI of glycine? (pKa of the two functional groups are 9.6 and 2.34)

Answer: (9.6+2.34)/2 = 5.97 ~ 6

Question: Regarding the previous reaction, if pure B is put into solution in the presence of an enzyme that catalyzes the reaction between A and B, which one of the following will be true? A) All the B will be converted into A untill there is 1000 times more A than B B) All of the B will remain as B, Since B is favored at equilibrium C) The enzyme will have no effect, since enzymes act on the transition state and there's no transition state present D. The reaction that produces A will predominate until delta G = 0

Answer: - Choice D is Correct - If only B exists in solution then the bcak reaction producing A will predominate until equilibrium is reached (delta G = 0), regardless of the presence and absence of enzyme. - Note that enzymes do not ac t on the transition state, they act to produce the transition state

Question: If a transition state intermediate possess a transient negative charge, what amino acid residues might be found at the active site to stabilize the transition state?

Answer: - Positive charges would stabilize the negatively charged intermediate. Therefore His, Arg, or Lys. - Alternatively the hydrogen of the NH2 group of glutamine or asparagine could hydrogen bond with the negative charge

Question: The transition state for a reaction possesses a transient negative charge. The active site for an enzyme catalyzing this reaction contains a His residue to stabilize the intermediate. If the His residue at the active site is replaced by glutamate that is negatively charged at pH 7.0 what effect will this have on the reaction, assuming the reactants are present in excess compared to enzyme? A) The repulsion caused by the (-) charge in the glutamate at the altered active site will increase the activation energy and make the reaction proceed more slowly that it would in a solution without enzyme B) Rate of catalysis will be unaffected, but the Equilibrium ratio of products and reactants will change, favoring reactants C) The transition state intermediate will not be stabilized as effectively by the altered enzyme, lowering the rate relative to the rate with catalysis by the normal enzyme D) The rate of catalysis will decrease, and the equilibrium constant will change

Answer: - The statement "assuming that the reactants are present in excess compared to enzyme" adds nothing to the substance of the question LMAO - The substitution would decrease the effectiveness- or destroy all together- the active site of the enzyme . - The transition state then would not be effectively stabilized , and the rate of the reaction, would simply reduce to that f the uncatalyzed reaction - Choice C is Correct

Question: What is the favorable reaction that the cell can use to drive unfavorable reactions?

Answer: ATP hydrolysis - in the lab the standard delta G for hydrolysis of one phosphate group from ATP is -7.3Kcal/mol (favorable reaction) - in the cell the delta G is about -12 Kcal/mol (even more favorable in the cell)

Question: Why are alpha helices favorable structures for hydrophobic transmembrane regions?

Answer: B/c all polar NH and CO groups in the backbone are hydrogen bonded to each other on the inside of the helix , and thus don't interact with the hydrophobic membrane interior - alpha helical regions that span membranes also have hydrophobic R-groups which radiate out from the helix , interacting with the hydrophobic interior of the membrane

Question: Would a protein end up folded normally if you (1) first put it in a reducing environment, (2) then denatured it by adding urea, (3) next removed the reducing agent, allowing disulfied bridges to reform, and (4) finally removed the denaturing agent?

Answer: If you allow disulfide bridges to form while the protein is denatured, it will become locked into an abnormal shape

Question: If a single polypeptide folds once and forms a B-pleated sheet with itself , would this be a parallel or antiparallel B-pleated sheet?

Answer: It would be Antiparallel b/c one participant in the B-pleated sheet would have a C to N direction, while the other would be running N to C

Question: If an enzyme has a reaction rate of 1 µmole./min at a substrate concentration of 50µM and a rate of 10 µmole/min at a substrate concentration of 100 µM, does this indicate the presence of a competitive inhibitor?

Answer: No, the rate increase is greater than linear indicating that it is caused by cooperativity

Question: What is the difference between the situation in vitro (lab) under standard conditions and in vivo (cell) under nonstandard conditions ?

Answer: Q(cell) does not equal K(eq) - This means that the relative concentrations of ATP and ADP+P ar not in equilibrium levels in the cell. - Actually, Q(cell) << K(eq) b/c the cell keeps a high concentration of ATP around

Question: An enzyme may consist of a single polypeptide chain or several polypeptide subunits held together in a ______ structure. (primary, secondary, tertiary or quaternary?)

Answer: Quaternary

Question: What is the difference between the disulfide bridge involved in quaternary structure and one involved in tertiary structure?

Answer: Quaternary disulfide bonds are bonds that form between chains that aren't linked by peptide bonds. Tertiary disulfides are bonds that form between residues in the same polypeptide.

Question: Glycine is the simplest amino acid, with only one hydrogen as its R-group. It's only functional groups are the backbone groups discussed before (amino and carboxyl). What will be the net charge on a glycine molecule at pH 12?

Answer: Since pH 12 represents a very low [H+], both groups will become deprotonated (COO- and NH2), creating a net charge of -1 per glycine

Question: Ribonucleases have eight cysteines that form four disulfide bonds. What effect would a reducing agent have on its tertiary structure?

Answer: The disulfide bridges will be broken and the tertiary.

Question: Assuming pKa of 2 will a carboxylate group be protonated ore deprotonated at pH 1.0?

Answer: The pH is less than the pKa here, so protonation wins over dissociation, and the group will be protonated. (-COOH)

Question: Will the amino group be protonated or deprotonated at pH 1?

Answer: The pH is lower than the pKa for the ammonium group, so the amino group is protonated: NH3+

Question: If a small amount of enzyme in solution is acting at Vmax and the substrate concentration is doubled, what is the new reaction rate?

Answer: The rate will not change and stay at Vmax; adding more substrate wont change the rate because the enzyme is saturated.

Question: the biosynthesis of macromolecules such as DNA and protein are not spontaneous (Delta G > 0) but clearly these reactions do take place. How can this be?

Answer: Thermodynamically unfavorable reactions in the cell can be driven forward by REACTION COUPLING

Question: What if you did the same experiment but in this order: 1, 2, 4, 3

Answer: You would end up with the correct structure.

Question: At pH 6.0 between the pKa's of the ammonium and carboxyl groups, what will be the net charge on a molecule of glycine?

Answer: the carboxyl group will be deprotonated (COO-) with a charge of -1, and the amino group will be protonated (NH3+) with a charge of +1 creating a NET CHARGE of 0 per glycine molecule

Question: In the oligopeptide Phe-Glu-Gly-Ser-Ala, state the number of acid and base functional groups , which residure has a free alpha amino group and which residue has a free alpha carboxyl group.

Answer: - Glutamic acid (acidic) was the only residue in this chain that is acid or base along with the two end functional groups we have 3 acidic/basic functinal groups - Phe has the free alpha amino group - Ala has the free alpha carboxyl group

Question: Which of the following may be considered an example of tertiary protein structure? I. van der Waals interactions between two Phe R-groups located far apart on a polypeptide II. Hydrogen bonds between the backbone amino acid carboxyl groups III. Covalent disulfide bonds between cysteine residues located far apart polypeptide

Answer: - Item I is true: good example of 3* - Item II is false: It describes 2* not 3* - Item II is true: Describes disulfide bonds which is an example of 3*

Question: If during an enzyme catalyzed reaction, an intermediate forms in which the substrate is covalently linked to an enzyme via a serine residue, can this occur at any serine residue or must it occur at a specific serine residue?

Answer: It must occur at a particular serine residue which sticks out into the active site

Question: If the disulfides serve only to lock into place a tertiary protein structure that forms forms first on its own, then what effect would the reducing agent have on correct protein folding?

Answer: The shape should not be disrupted if breaking disulfides is only disturbance. It's just the shape would be less sturdy

Lyase

Breaks chemical bonds by means other than oxidation or hydrolysis (e.g. pyruvate decarboxylase)

Answer to the previous question:

Choice D is CORRECT - Since curve 1 and 3 have the same Vmax , but curve 3 has a reduced rate of product formation , it suggests that curve 3 represents competative inhibition of the enzyme in curve 1 - If 3 was noncompetitive its Vmax would be reduced compared to 1 - In no case will an inhibited enzyme's Vmax be larger than an uninhibited enzyme

Sulfur-Containing Amino Acids

Cysteine and Methionine - Cys contains a thiol (aka sulfhydryl) and is polar - Met contains a thioether and is non-polar

Henderson-Hasselbalch Equation

Describes the relationship between pH, pKa and the position of equilibrium in an acid base reaction *Rules: 1) When pH of the solution is less than the pKa of an acidic group, the acidic group will mostly be in its protonated from 2) when the pH of the solution is greater than the pKa of an acidic group, the acidic group will mosttly be in its deprotonated form

Denaturation

Disruption of a protein's shape without breaking peptide bonds - proteins are denatured by urea (which disrupts hydrogen bonding interactions), by extremes of pH, by extremes of temperature, and by changes in salt concentration

Hydrophobic (Non-polar) Amino Acids

Glycine, Alanine, Valine, Leucine, Isoleucine, Phenylalanine, Tryptophan, Methionine, Proline - either have aliphatic or aromatic side chains. - AA w/ aliphatic side chains: Gly, Ala, Val, Leu, Ile - AA w/ aromatic side chains: Phe, Trp (also Tyr but it's polar) - Tend to associate w/ each other rather than water so they are found in the interior of folded globular proteins - The larger the hydrophobic group the greater the hydrophobic force - Pro's amino group is covalently bound to its non-polar side chain creating a secondary alpha amino group and a ring structure

Hydrolase

Hydrolyzes chemical bonds (includes ATPases, Proteases, etc.)

Protease

Hydrolyzes peptide bonds (e.g. trypsin, chymotrypsin, pepsin, etc.)

Question: The inside of cells is known as a reducing environment because cells possess antioxidants. Where would disulfide bridges be more likely to be found, in extracellular proteins, under oxidizing conditions, or in the interior of cells in a reducing environment?

In the extracellular proteins under oxidizing conditions where they will not be reduced

Basic Amino Acids

Lysine, Arginine, and Histadine - Pka of R groups 10, 12, and 6.5 respectively - Lys and Arg are cationic (protonated) in physiological pH - His' side chain is unique b/c it's pka is close to physiological pH - At pH 7.4 His is either protonated or deprotonated. It can act as an acid too! "His goes both ways" -Amino acids containing -COOH or-NH2 side chains, always anionic (RCOO-) or cationic (RNH3+)at physiological pH

Nine Essential Amino Acids

Lysine, Histidine, Threonine, Valine, Leucine, Isoleucine, Phenylalanine, Tryptophan, Methionine - Cannot be synthesized in the body and must be obtained from the diet

Backbone of the polypeptide

N-C-C-N-C-C pattern

Cystine

Once a cysteine residue becomes disulfide bonded to another cysteine residue, it is called Cystine instead of cysteine

Disulfide Bridges

Play an important role in stabilizing tertiary protein structures

Polymerase

Polymerization (e.g. addition of nucleotides to the leading strand of DNA by DNA Polymerase III)

Phosphatase

Removes phosphate group from a molecule

Two factors Important factors that play a critical role in enzymatic function

Temperature: - as temperature increasess the thermal motion of the peptide and surrounding solution destabilize the structure - If the temperature rises sufficiently the protein denatures and loses its orderly structure pH - several AA possess ionizable -R groups that change charge depending on pH. This can decrease the affinity of a substrate for the active site. - If the pH deviates sufficiently, the protein can denature

Isoelectric point (pI)

The pH at which a molecule is uncharged (zwitterionic)

Substrates

The reactants in an enzyme catalyzed reaction are called substrates

Active Site

The region of an enzyme that is directly involved in catalysis

Kinase

Transfers a phosphate group to a molecule from a high energy carrier, such as ATP (e.g. Phosphofructokinase [PFK])

Phosphorylase

Transfers phosphate group to a molecule from inorganic phosphate (e.g. glycogen phosphorylase)