BIOC 384 Learning Objectives

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BIOC 384 LEARNING OBJECTIVES: BLOCK I

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Page 1: BIOC 384 Learning Objectives

BIOC 384 LEARNING OBJECTIVES: BLOCK I

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LECTURE I

Water, Non-Covalent Interactions, pH

Describe the properties of H2O: its polarity, hydrogen bonding capabilities, solvent properties, and ionization properties.

Property of Water

Description

Polarity Between the hydrogen and oxygen molecules, there is an unequal sharing of electrons, allowing it to become polar because the oxygen

atom has a partial negative charge and the hydrogen molecules have a partial positive charge.

Hydrogen Bonding

Special electrostatic interaction that occurs between a weakly acidic group (donor) and a group with a lone pair of electrons (acceptor).

Water molecules particularly form a strong hydrogen bonding network, allowing it to retain high boiling and melting temperatures as well as a

large heat of vaporization. Solvent

Properties Water contains a strong dipole, making it an optimal solvent for ions and

molecules with charged groups and polar groups. Ionization Properties

Lower attractive force between two oppositely charged ions because of a higher dielectric constant and greater separation of charge from its

polarity.

Explain the features of 4 types of noncovalent interactions (hydrogen bonds, ionic interaction, van der Waals interaction, and hydrophobic interactions), including the effect of solvent polarity and of distance on those interactions. Be able to identify non-covalent interactions on a figure.

Type Description Stabiliza-tion

Energy (kJ/mol)

Length (Å)

Diagram

Hydrogen Bonds

Between a hydrogen atom covalently bonded

to an electronegative atom and a second

electronegative atom

10-30 1.8-3.0

Ionic Interactions

Interactions between ions of opposite charge.

20-30 2.5

Van Der Waals Interactions

Interactions between opposite dipoles

(permanent, temporary or induced dipoles)

1-5 1-2

Hydrophobic

Effect The tendency of

nonpolar groups and molecules to cluster together in aqueous

solution.

5-30 --

Type' Brief'descrip0on'and'example' Stabiliza0on'

Energy'(kJ/mol)'

Length'

(Å)'

Hydrogen'

bonds'

Between'a'hydrogen'atom'covalently'bonded'to'an'

electronega0ve'atom'and'a'second'electronega0ve'

atom'

10+30' 1.8+3.0'

Ionic'

interac0ons'

Interac0on'between'ions'of'opposite'charge' 20+30' 2.5'

van'der'Waals'

interac0ons'

Interac0on'between'opposite'dipoles'(permanent,'

temporary'or'induced''dipoles)'1+5' 1+2'

Hydrophobic'

effect'

The'tendency'of'nonpolar'groups'and'molecules'to'

cluster'together'in'aqueous'solu0on'5+30' ++'

4'Types'of'Noncovalent'(“Weak”)'Interac0ons'

Type' Brief'descrip0on'and'example' Stabiliza0on'

Energy'(kJ/mol)'

Length'

(Å)'

Hydrogen'

bonds'

Between'a'hydrogen'atom'covalently'bonded'to'an'

electronega0ve'atom'and'a'second'electronega0ve'

atom'

10+30' 1.8+3.0'

Ionic'

interac0ons'

Interac0on'between'ions'of'opposite'charge' 20+30' 2.5'

van'der'Waals'

interac0ons'

Interac0on'between'opposite'dipoles'(permanent,'

temporary'or'induced''dipoles)'1+5' 1+2'

Hydrophobic'

effect'

The'tendency'of'nonpolar'groups'and'molecules'to'

cluster'together'in'aqueous'solu0on'5+30' ++'

4'Types'of'Noncovalent'(“Weak”)'Interac0ons'Type' Brief'descrip0on'and'example' Stabiliza0on'

Energy'(kJ/mol)'

Length'

(Å)'

Hydrogen'

bonds'

Between'a'hydrogen'atom'covalently'bonded'to'an'

electronega0ve'atom'and'a'second'electronega0ve'

atom'

10+30' 1.8+3.0'

Ionic'

interac0ons'

Interac0on'between'ions'of'opposite'charge' 20+30' 2.5'

van'der'Waals'

interac0ons'

Interac0on'between'opposite'dipoles'(permanent,'

temporary'or'induced''dipoles)'1+5' 1+2'

Hydrophobic'

effect'

The'tendency'of'nonpolar'groups'and'molecules'to'

cluster'together'in'aqueous'solu0on'5+30' ++'

4'Types'of'Noncovalent'(“Weak”)'Interac0ons'

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Explain the importance of weak interactions for biomolecules. Many biomolecules are amphipathic, and the hydrophobic interactions between lipids and protein determines the structure of membranes. Noncovalent interactions are made and broken, and mediate reactions between proteins and protein as well as proteins and nucleic acids.

Describe the properties of amphipathic molecules. Amphipathic molecules contain polar and nonpolar regions. Polar regions tend to be exposed to water, while nonpolar regions tend to cluster together to minimize interaction and consequent exposure to water molecules.

Relate the state of protonation of a weak acid to its pKa and the pH of its environment. Be able to interpret and sketch a pH titration curve (dominant form when pH < pKa when pH = pKa and when pH > pKa. Lower pH allows for more protonation. The protonation state changes when pH > pKa. When pH = pKa, the two states coexist in equilibria.

Scenario pH < pKa pH > pKa Dominant Form Protonated Un-protonated

Know and use the Henderson-Hasselbach Equation to estimate the percentage of charged molecules in a solution. The Henderson-Hasselbalch equation is as follows: 𝑝𝐻 = 𝑝𝐾! + 𝑙𝑜𝑔

[!!][!"]

. We can determine the pKa by noting two things:

1. pKa = pH at the midpoint of a titration curve 2. The pKa is the pH at which the concentrationof the protonated form equals the

concentration of the deprotonated form ([HA]=[A-]). We can determine the percentage of charged molecules in a solution by noting the ratio of the conjugate base and acid.

Describe the ionization properties of phosphate and what makes it a good buffer in the physiological pH range. It is important to remember the importance of buffers and its ability to maintain small changes in pH against titration. A buffer is simply a mixture of weak acids and conjugate bases. The phosphate buffer is a physiologically important buffer, as it is in the cytoplasm of all cells (𝐻!𝑃𝑂!! = 𝐻! + 𝐻𝑃𝑂!!!), with a pKa of 6.86. It has an effective range of 5.9-7.9. The pH of extracellular fluid and most cytoplasmic compartments is between 6.9-7.4.

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LECTURE II

Thermodynamics

Explain in conceptual terms: change in free energy (ΔG) change in enthalpy (ΔH); change in entropy (ΔS).

Concept Definition Change in Free

Energy (ΔG) The change in free energy of a reaction is the difference of the Gibbs

Free Energy of Products and Gibbs Free Energy of Reactants. Change in

Enthalpy (ΔH) The change in enthalpy is the change of the energy of a system at

constant pressure. Change in

Entropy (ΔS) The change in entropy is the change in the amount of disorder in the system. Thus spontaneous processes have increases in entropy. It is

misinformed to state that the disorder is chaos. Instead, the disorder is the number of ways of arranging the molecules in a system.

Relate ΔG to ΔH and ΔS. We can relate the change in free energy to the change in enthalpy and change in entropy by the following equation: 𝛥𝐺 = 𝛥𝐻 − 𝑇𝛥𝑆, where T standard for the temperature of the system in Kelvin.

Relate the sign of ΔG to the direction in which reaction will go to get to equilibrium. Sign ΔG

Translation Spontaneity Reaction

Direction Negative ΔG < 0 Spontaneous/Favorable/Exergonic Left to Right Positive ΔG > 0 Non-Spontaneous/Non-

Favorable/Endergonic Right to Left,

Reverse of what is written

0 ΔG = 0 Equilibrium No net reaction in either direction

Explain the difference between standard free energy change (ΔG°) and actual or physiological free energy change (ΔG).

Type of Free Energy Change Definition Standard Free Energy Change

(ΔG°) Change in free energy for going from standard

conditions to equilibrium Actual or Physiological Free

Energy Change (ΔG) Gibbs Free energy change under actual conditions are

where the products and reactants are the actual, experimental concentrations.

Calculate the standard free energy change (ΔG°) from the equilibrium constant (keq). We can calculate the standard free energy change from the equilibrium constant by the following formula: ΔG° = −RT ln 𝑘!". At equilibrium, remember that Q = keq.

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Given any two of the following parameters, calculate the third: actual free energy change (ΔG), standard free energy change (ΔG°), actual mass action ratio Q. We can determine the free energy change based on any of these variables from the following formula: 𝛥𝐺 = ΔG° + RT ln𝑄, where 𝑄 = [!"#$%&'(]

[!"#$%#&%'].

LECTURE III

Protein Structure: Amino Acids

Name and use one- and three-letter abbreviations of 20 amino acids. Type of Amino

Acid Amino Acid Chemical

Structure 3-Letter

Abbreviation 1-Letter

Abbreviation Hydrophobic Glycine

Gly G

Alanine

Ala A

Proline

Pro P

Valine

Val V

Leucine

Leu L

Isoleucine

Ile I

Methionine

Met M

Aromatic Phenylalanine

Phe F

Tyrosine

Tyr Y

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Tryptophan

Trp W

Hydrophilic Serine

Ser S

Threonine

Thr T

Cysteine

Cys C

Asparagine

Asn N

Glutamine

Gln Q

Negatively-Charged

Aspartate

Asp D

Glutamate

Glu E

Positively-Charged

Lysine

Lys K

Arginine

Arg R

Histidine

His H

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Define zwitterions. Before discussion of the zwitterion, it is necessary to discuss the biological functions of amino acids and general structure of the amino acid. The amino acid is a building block to proteins, and consequently dictates the folding and stability of proteins. It is involved in the catalysis of reactions and basic component in biological recognition. Amino acids and derivatives also function as hormones and regulatory chemicals, and can be an alternative energy source for cell tissue. They are called amino acids mainly because of their structure. They have an amino and carboxylic acid group attached to a chiral carbon center. In total, there are 20 amino acids in proteins, and each one is unique by their functional or R group. Since they are attached to an optically active (chiral) center, they also have different enantiomers of the molecule (L or D) and thus can be biologically active or inactive. The naturally occurring of the amino acids are of the L stereoisomer configuration. A zwitterion is a neutral molecule with a positive and a negative electrical charge at different locations of that molecule. Since there is a positive and negative electrical charge at different locations, it allows the molecule to have a response in changes of pH (when there are increases or decreases in the number of hydrogen ions/protons).

Know approximate pKas of functional groups, interpretation of their titration curves. We know that biological processes are pH-dependent, and consequently the arrangement of the protons is essential for enzyme activity. Thus we would also know that the amino and carboxyl group of the amino acid will have pKas. The typical pKa of the α-carboxyl group is approximately 2-3, while the pKa of the α-amino group is approximately 8-9. This allows it to have an effect when pH of the environment is altered. When the pH of the environment is altered, these functional groups will be protonated or de-protonated depending. As pH is decreased, the functional groups on the alpha carbon will more likely to be protonated, as the pKa of the α-carboxyl group is between 2-3. Meanwhile, as the pH increases to a more basic environment, the more likely the functional groups on the alpha carbon will be de-protonated. Thus, since this is a

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biochemistry course, this will focus mainly on biological states, where the pH is neutral (pH = 7).

Define pI. The isoelectrical point (or pI) of a molecule is where the molecule carries no net electrical charge. This is mainly the point where the amino acid remains at the zwitterionic form, with the amino group having the positive charge and the carboxyl group having a negative charge. One can calculate the isoelectric point by the following formula: 𝑝𝐼 = !"!!!"!

!, where

pK1 is the equivalence point where the concentrations of the amino acids’ net charges of -1 and 0 are equal, and pK2 of the concentrations of the amino acids’ net charges of 0 and +1 are equivalent.

Know ionization properties of functional groups and what influences them.

Ionization properties of the functional groups are mainly the polar property of certain functional groups that maintain a charged (negatively or positively) or uncharged form according to pH relative to the pKa.

Explain how a pKa is influenced by its surrounding environment (example, ionizable residue in a hydrophobic environment) The pKa is influenced by its surrounding environment the demands of the environment. For example, if it demands protonation or deprotonation, the pKa can be higher or lower.

Determine the net charge of a small peptide. The net charge of an amino can be determined by the sum of the charges of the amino and carboxyl groups as well as the charge of the functional group. For a chain of amino acids (small peptide), it is necessary to account for charges on functional groups as well as determination of the predominant form of the peptide (the majority peptide at that pH). It also may be necessary to utilize the Henderson-Hasselbalch Equation in order to determine the majority charge pH is within 1 pH of the pKa.

Amino"Acid" Data"from"Lehninger,"5th"ed"Table"3:1"—"only"approximate"pKa"is"important;"pKa"varies"inside"protein"

αScarboxyl(group(

Aspartate(

Glutamate(

His;dine( C

HN

C

H

NH

CH

C

HN

C

H

N

CH

COOH COO

7(6(3( 4( 5(2(1(pH(

COOH COO

COOH COO

pKa(=(6.00(

pKa(=(4.25(

pKa(=(3.65(

pKa(=1.8–2.4(

Amino"Acid" Data"from"Lehninger,"5th"ed"Table"3:1"—"only"approximate"pKa"is"important;"pKa"varies"inside"protein"

Cysteine(

αSamino(group(

Tyrosine(

Lysine(

Arginine(

NH3+ NH2

OH O

SH S

C

NH2

NH2

C

NH

NH2

7( 8( 9( 10( 11( 12( 13( 14(pH(

pKa=12.48(

NH3+ NH2

pKa=10.53(

pKa=10.07(

pKa=8.8(–(11.0(

pKa=8.18(

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LECTURE IV

Protein Structure: Peptides

Know what determines the structure and function of proteins. The function of a protein depends on its amino acid sequence. Protein sequence determines its structure and function. Proteins are polymorphic, having sequence variations among human population with little effects on function. Critical regions that are essential to their function are conserved.

Define the primary structure. The primary structure is the order of amino acids in chain (synthesized in the order specified by the sequence of nucleotides in a gene).

Describe the peptide bond: what type of reaction leads to the peptide bond, which atoms are involved. The peptide bond is formed by a condensation reaction between the carboxylic acid and amine groups of two amino acids, which releases H2O. This reaction is catalyzed by the RNA component of ribosomes, which functions as an enzyme. It is mainly a condensation of the carboxylic acid and amino group. Though the bond hydrolysis is thermodynamically favored, the rate is kinetically slow. It requires free energy input (reaction carried out by the ribosome).

Know terms: backbone, carboxyl terminal end (C-term), Amino terminal end (N-term).

Term Definition Backbone The regularly repeating part (everything but the side-chains)

Carboxyl Terminal End (C-Term)

A carboxyl terminal end is a carboxylic acid portion of the amino acid within a peptide. It typically goes from N terminus to C

terminus. (N à C) Amino Terminal

End (N-Term) An amino acid residue is the portion of an amino acid within a

peptide.

Draw a peptide bond – identify atoms that are in the peptide plane. The atoms that are in the peptide plane are:

- Alpha Carbon - Carbon - Oxygen in Carbonyl - Nitrogen - Hydrogens in Peptide Bond - Alpha Carbon in Second Amino Acid

What is unique about the peptide bond that, though it can be mistakably predicted of rotation, there is no

Pep;de(bonds(link(amino(acids(together(

Pep;de(bonds(are(formed(by(a(condensa;on(reac;on(between(the(carboxylic(acid(and(amine(groups(of(two(amino(acids,(which(releases(H2O.(((This(reac;on(is(catalyzed(by(the(RNA"component"of"ribosomes,(which(func;ons(as(an(enzyme.(

Planarity(of(Pep;de(bond(

•  Cα,(C,(O,(N,(H(and(Cα(lie(in(a(plane(•  Rigidity:(No(rota;on(around(the(bond(between(the(

carbonyl(carbon(and(amide(nitrogen(•  Why(?((

hLp://www.biochem.arizona.edu/classes/bioc462/462a/jmol/pep;de/pep;de.html(

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rotation around the bond between the carbonyl carbon and amide nitrogen. Why? Though the peptide bond is typically portrayed as a single bond between the carbonyl carbon and amide nitrogen, the nitrogen still retains a lone pair of electrons, which are adjacent to the carbon-oxygen bond. Thus a resonance structure can be drawn of a double bond linking the carbon and nitrogen, designating a negative charge on the oxygen and a positive charge on the nitrogen. Thus, such resonance can prevent the rotation, because a double bond requires breaking one bond in order to allow such a rotation.

Understand why the trans configuration is favored. The trans (opposite) configuration is favored in approximately 99% of peptides mainly because it is sterically favored. There is less interaction between groups and thus requires less energy.

Identify phi and psi on a drawing of a peptide. Though we cannot rotate around the peptide bond, we can still rotate around the single bonds along the peptide. The angle of rotation around the nitrogen of the amide and alpha carbon in the peptide (N-Cα) is the phi (ϕ) angle, while the angle of rotation between the carbon and the alpha carbon (C-Cα) is the psi (ψ) angle. Each residue contains a particular pair (ϕ, ψ) of angles.

Describe the information in a Ramachandran plot. The combinations of Ramachandran angles cause steric clashes, depending on the size of the side chain functional groups, and thusly vary among sequences. The total possible structures are limited. Dark green represents the allowed combinations of phi/psi à low energy and favorable. The white areas are combinations that are generally not allowed by sterics à high energy and unfavorable. (DARK GREEN = FAVORABLE/LOW ENERGY; WHITE =UNFAVORABLE/HIGH ENERGY)

Pep;de(Bond:((CisSTrans(Isomeriza;on(

•  Cα(carbons(on(same((cis)(or(opposite((trans)(sides(of(“double”(bond((pep;de(bond)(

•  Trans(configura;on(highly(favored((steric)(~(99%(

•  Excep;on:(XSPro,(~10%(of(which(are(cis" Berg(et(al.,(Fig.(2S24((

Cα#

Cα# Cα# Cα#

Why(?(

R(group(

R(group(

A(liLle(terminology(•  Backbone/main:chain:""the(regularly(repea;ng(part((everything(but(the(sideSchains)(Side(chains(are(in(pink((Backbone(HSbonding(groups(are(in(yellow)(

•  Amino"Acid"vs."Amino"Acid"Residue:"

An(amino(acid(residue(is(the(por;on(of(an(amino(acid(within(a(pep;de((example(boxed(in(blue)((

“NSterminus”( “CSterminus”(

•  Naming"pep.des:(Dipep;de((2),(Tripep;des((3),(Oligopep;de((short?),(Polypep;de((long)(

SerSGlySTyrSAlaSLeu(SGYAL(

Residues(are(listed(from(NSterminus(to(CSterminus:(

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Describe and indicate atoms involved in backbone hydrogen bonding. The backbone/main-chain is the regularly repeating portion (everything but the side chains, containing the carboxyl and amino group (O=-CN-H). LECTURE V

Protein Structure: Secondary Structure

Define secondary structure. Secondary structure is the local, regular/recognizable conformations observed for parts of the peptide backbone of a protein (No steric clashes and form good hydrogen bonds). The hydrogen bonding mainly occurs in the backbone itself. In order for it to be stable, there must be no steric clash, the backbone H-bond must be satisfied, and must have stable intermolecular interactions.

List examples of categories of secondary structure that occur in proteins. The categories of secondary structure that occur in proteins are mainly the α-helix and the β-strand. These two are the two most common tyeps of secondary structure. They are abundant mainly because they lie in an area of the Ramachandran diagram where there is no steric clash for adopting these angles (and backbone H-bonds are satisfied).

Describe the α–helix, including what groups serve as hydrogen bond donors and acceptors, chirality of most α–helices in proteins (right- or left-handedness), number of residues per turn, orientation of R groups relative to axis of the helix, the helix dipole (which end is δ+, which end is δ-), packing density of atoms, effect of R groups on the stability. The α-helix is a rod-like structure, having a backbone of regular, repeating rotation residue by residue, meaning that the residue has close to the same (Φ, Ψ). The pitch is approximately 5.4 Å per 360° turn, and spproximately 3.6 Amino acids per turn. They are typically right-handed in proteins, and the R groups radiate outward from the helical cylinder.

The carbonyl group of residue n accepts a hydrogen bond from N-H group of residue n+4, acting as the proton acceptor. The N-H group of residue n donates a hydrogen bond to the oxygen of the C=O group n-4 (acting as a proton donor).

The polarity of the bonds along the helix results in a net helix dipole with the N-terminal end slightly positive and the C-terminal end slightly negative. In proteins, negatively charged amino acids are often found at the amino terminus of the helical segment, and positively charged amino acids are often found at the carboxyl-terminal end. Why? Because it is ideal to place complemtnary charges to maintain attractive forces, rather than incite repulsive forces through placement of identically charged amino acids. The sequence of the amino acid affects the stability of the α-helix, and is often represented by a helical wheel. Each residue in a helix is 100° around the circle from the previous residue. Residues three to four amino acids apart should be on the same side of the helix. Stability is dependent on the distance between certain groups along the helix, because these side chains (R groups) are protruding.

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Describe β-conformation, including which groups serve as hydrogen bond donors and acceptors, and orientation of R groups in a β pleated sheet. The β-strand has the backbone nearly fully extended, with all residues having approximately the same (Φ, Ψ) angles. The distance between adjacent residues is approximately 3.5Å, and thusly more further apartment and more stretched out in comparison to the α helix. The side chains point in opposite directions for adjacent residues in chain. The amino and carboxyl group of the peptide bond point in opposite directions, perpendicular to average direction of extended backbone of chain. In general, there is no predictable relationship in the amino acid sequence for what sections are hydrogen bonded to each other, and the N-H and C=O groups are almost fully hydrogen-bonded. Hydrogen bonds are more or less at right angles to direction of backbone of chain, and are approximately separated by 4.8 Å.

Explain parallel and antiparallel β conformation. Parallel β-conformations are present when the strands are in the same direction. The side-chains line up in register along sheet so there is interaction between strands. However, antiparallel β-conformations are in the opposite direction. Side-chains alternate sides of the sheet along a strand (always pointing away from the backbone).

Identify the most important non-covalent interactions stabilizing the α-helix and β-conformations. The most important noncovalent interaction stabilizing the α-helix and β-conformations is hydrogen bonding along the peptide backbone, not from the functional groups.

Explain what a β-turn is, where β-turns are often found in proteins, and what types of amino acid residues are often found in β-turns. A β-turn (or reverse turns/β hairpins/β bends) are bends in the beta-strand involving glycine, asparagine, or serine (small hydrophilic residues) or proline (with a cyclic functional group with “built-in” elbow/bend in backbone to help start turn). It involves an abrupt change in direction of the polypeptide backbone, often at the surface of the protein. It is stabilized by hydrogen bond across the “step” of the hairpin. A sharp turn in space can cause steric problems with larger amino acid side chains, but is often utilized to connect two antiparallel β strands.

Sequence(affects(α(helix(stability(

15

67

8

9

10

11

12

2

3

4

T

V

V

E A

I

D

R

L

R

D

T

Each residue in a helix is 100 degrees around the circle from the previous residue

Helical(wheel((representa9on(

Residues(3(to(4(apart(in(sequence(are(on(the(same(side(of(helix(

1

2

3

4

5

6

7

TVVEAIDRLRDT

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Identify α-helices and β-strands (or sheets consisting of 2 or more β-strands) on a ribbon depiction of a protein structure. Identify phi/psi combinations indicative of α-helices and β-sheets in a Ramachandran diagram.

Type of Secondary Structure

Ribbon Representation Ramachandran Diagram

α-Helix

β-Sheet

LECTURE VI

Protein Structure: Tertiary/Quaternary Structure

Outline 3 principles guiding folding of water-soluble globular proteins and the generalizations about protein structure resulting from these principles. Relate the principles to real protein structures. The three major principles guiding the folding of water-soluble, globular proteins are:

1. Minimization of solvent-accessible surface area: The polypeptide chain takes the conformation that buries the maximum amount of hydrophobic surface area in the protein interior. This tends to collapse the polypeptide chain into a small volume. Burial of hydrophobic R groups away from water requires at least 2 interacting secondary structural elements, e.g., 2 alpha-helices, or β-α-β loop (uses α helix to connect 2 parallel β strands) or 2 β sheets, etc.

2. Maximization of hydrogen bonding within the protein: Hydrogen bond donors and acceptors must be bonded. On the protein exterior water is available but in the interior they must satisfy each other. The need to form hydrogen bonds can be satisfied by secondary structure formation.

3. Chiral effect: Backbones of L-amino acids tend to twist in a right-handed direction. Both α-helices and β-sheets twist to the right, affecting the connections between secondary structure elements.

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Explain the term amphipathic, with an amphipathic protein α-helix or β-strand as an example. Remember that amphipathic molecules contain both hydrophobic and hydrophilic ends along the protein sequence. It should then be concluded that burying as many hydrophobic groups as possible is the most important driving force in the folding of water-soluble protein, because of the minimization of hydrophobic interactions with water. Secondary structural elements, such as α-helices and β-sheets, are often amphipathic. The R groups on one side are hydrophobic (and are facing interior of the protein), while the R groups on other side is hydrophilic and face the aqueous environment (outside).

Describe how the hydrophobic effect contributes to the folding of globular proteins.

Burying the hydrophobic side chains in amphipathic helices involves burying the hydrophobic R groups towards the interior of the protein on 1 side of the helix while the other side interacts with water. It is also involved in associations with β-strands face-to-face, mediated by side chains and the interactions could be between parts of the same chain or between different chains. Associations of helices and β strands can incur structural motifs, or recognizable patterns or combinations of secondary structural elements. They bury hydrophobic R groups in between “layers” or elements.

Recognize examples (ribbon diagrams) of common folding motifs (frequently encountered combinations of secondary structures) such as coiled coils of α-helices, stacked β-sheets, βαβ elements, β-barrels.

Common Folding Motifs Ribbon Diagram βαβ

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Β-Hairpin

αα

Greek Key Motif: ββββ

Explain the term tertiary structure. Tertiary structure is the 3-dimensional conformation of a whole polypeptide chain in its folded state (includes not only positions of backbone atoms, but of all the sidechain atoms as well.)

Define the terms domain and subunit as they relate to protein structure. Be able to recognize different domains in a ribbon diagram of a single polypeptide chain with 2 or more domains. The domain of a protein is the portion of the protein sequence and structure that can evolve, function, and

exist independently of the rest of the protein chain. It forms a compact three-dimensional structure and can often be stable and folded without assistance. One protein can consist of multiple domains. Domains are structurally independent folding units looking like separate globular proteins but all part of the same polypeptide chain. Larger proteins can consist of 2 or more domains linked together in a single chain. Each domain consists of 2 antiparallel β sheets, with loops between β strands. Domains have a hydrophobic core, where most of the hydrophobic residues are sequestered away from water. The same kinds of bonds and interactions that hold secondary structure elements to form domains also serve to hold complete domains together as larger structures. The shape complementarity becomes an increasingly important factor in higher order structure, allowing many van der Waals interactions. A protein subunit is a protein molecule that assembles with other protein molecules to form a complex. Naturally occurring proteins tend to be multimeric. Subunits can be identical, similar, or even completely different.

Protein(Domains(•  Domains:((structurally(independent(folding(units(looking(

like(separate(globular(proteins(but(all(part(of(same(polypep.de(chain(

•  Larger(proteins:(consist(of(2(or(more(domains(linked(together(in(a(single(chain(

(

Coiled-coil domain (2 helices)

β-sandwich domain (2 sheets)

Have(a(hydrophobic(core,(where(most(of(the(hydrophobic(residues(are(sequestered(away(from(water(

• (3?dimensional(rela.onship(of(the(different(polypep.de(chains((subunits)(in(a(mul.meric(protein(• (Terminology:(each(polypep.de(chain(in(a(mul.chain(protein(=(a(subunit(• (2?subunit(protein(=(a(dimer,(3(subunits(=(trimeric(protein,(4(=(tetrameric,(etc.(

?(homo(dimer(or(trimer(etc.):(iden.cal(subunits((?(hetero(dimer(or(trimer(etc.):(noniden.cal(subunits((

• (Quaternary(structure(is(stabilized(by(the(same(types(of(forces(as(((((((ter.ary(structure:((noncovalent(interac.ons,(or(for(extracellular(proteins(((((((some.mes(disulfide(bonds(

Berg, Tymoczko & Stryer Fig. 3.48 Berg, Tymoczko & Stryer Fig. 3.49

Phage(λ(Cro(protein:(Homodimer(Quaternary(structure(α2#

Hemoglobin:(Heterotetramer(Quaternary(structure α2β2

Quaternary(Structure(

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Describe the structure of fibrous proteins. How are such structures stabilized? Fibrous proteins have many characteristics (supporting its function of structural support for cells and tissues). They contain high proportions of regular secondary structure (often repetitive sequences), such as α-helix or β-pleated sheet. And, as stated before, they have a structural rather than dynamic role because of their rodlike or sheetlike shapes. The structures are stabilized because the helices can wind around each other, forming a supercoil or coiled-coil. This gives it great tensile strength. Left handed supercoils (of right-handed α-helices are found naturally.

Describe in general terms the structure of the polypeptide chain of myoglobin. Myoglobin is a globular protein with a lot of α-helices. It plays a role in oxygen storage, binding O2 in mucle cells. It is a single domain protein, and Heme is buried in hydrophobic core. It is ~70% α-helical and ~30% turns and loops at surface.

Describe the general structure of a globular water-soluble protein, including where in the structure you would expect to find hydrophobic groups and where you would expect to find polar/charged groups. In a globular-water soluble protein, we need to remember the hydrophobic and hydrophilic properties of an amphipathic molecule. On the surface, there are very few hydrophobic residues (and thus dominanted by polar residues), because they minimize their interactions with an aqueous environment. In the interior, however, there are no charged residues, and minimal polar amino acids depending on the function of the protein.

Describe the quaternary structure (of a protein), and be able to describe a protein in terms like “homotetramer”, “heterodimer”, etc. A quaternary structure is the 3-dimensional relationship of the different polypeptide chains (subunits) in a multimeric protein. Each chain in a in a multichain protein is a subunit. Thus, we can name the structure based on the subunits and the number of subunits. For example, a homotetramer consists of 4 identical subunits, while a heterodimer consists of 2 different subunits. For example, the phage λ Cro protein is a homodimer consisting of two α subunits, while hemoglobin is a heterotetramer with two α and two β subunits.

Explain simple rotational symmetry for an oligomeric protein such as a homodimer like the Cro protein or a heterotetramer like hemoglobin. – Be able to use (correctly) the terms “2-fold”, “3-fold”, etc. to refer to simple rotational axes of symmetry and recognize that simple level of symmetry in a protein structure. Individual subunits can be superimposed on other identical subunits (brought into coincidence) by rotation about one or more rotational axes. If the required rotation is 180°, protein as a 2 fold axis of symmetry (as in the Cro repressor protein). If the rotation is 120° for a homotrimer, the protein has a 3-fold symmetry axis. All subunits are related by rotation about a single n-fold rotation axis.

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LECTURE VII

Protein Structure: Dynamics/Structure Determination

Know the three broad categories of intramolecular motions. These categories include a timescale (kinetic component) and an amplitude and directionality (structural component).

Broad Category

Definition Spatial Displacements (Å)

Time Period (seconds)

Atomic Fluctuations

Bond Vibrations N/A 10-15 to 10-11 seconds.

Collective Motions

Groups of covalently linked atoms move as

a unit

0.01 and > 5 10-12 to 10-3

Triggered conformational

changes

Groups of atoms varying in size from

individual side chains to complete subunits move in response to a specific stimuli such

as ligand binding

0.5 and > 10 10-9 to 10-3

Describe where (structurally and surface vs. buried) the most dynamic regions generally occur in proteins. Protein core mobility is dictated by aromatic ring flipping. The most dynamic regions tend to occur on the surface of the protein, where this not much steric clashes. It is much more slower (in terms of flipping rates) as the protein is deeper in the protein. These proteins are tightly packed, thus limiting the amount of space in proximity of the functional group.

Give an example of a large scale conformational change triggered by binding. An example of a large scale conformational change triggered by binding is when the iron-binding protein lactoferrin undergoes a large conformational change on iron binding. The covalent bonds remain intact, thus are not broken in order to attain these conformational changes.

Give an example of how conformational changes can assist protein function. Another example is when hexokinase phosphorylates glucose with ATP, undergoing a large conformational change upon glucose binding. The purpose of this is that it activates the enzyme upon substrate binding. Glucose replaces water in the active site and uses similar weak interactions for binding.

Describe the basic principles behind x-ray crystallography for protein structure determination. Thermal motions are revealed in strucutres determined by x-ray crystallography. The structure is shaded from blue (least mobile) to red (most mobile). We also know that mobility can change in the binding of other molecules. This can display relation to the

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protein’s functions and the role of chaperones in protein folding. X-ray crystallography is a technique that directly images molecules. It is suitable for any size molecule, but requires the protein to be affixed to crystals, which can hinder movement and function. Another method is nuclear magnetic resonance, which measures the environment of atoms inside proteins, and determines which residues are in close proximity. It is limiting, but does not require affixing into crystals. NMR structures are obtained by measuring many interatomic distances and finding the conformation that satisfies them all. A final popular method is cryo electron microscopy, in which the impact of parallel electron beam on the specimen allows viewing of the orientations of the protein. It is low-resolution and is easier to catch molecules in different conformational states. LECTURE IX

Protein Folding

Terminology Term Definition

Denaturation Protein is converted into a random coil without its normal activity. It can be done by altering natural conditions of the protein:

§ Altering Temperature § Changing pH through changing the protein charge

distribution and hydrogen bonding requirements § Detergents to associate with nonpolar residues and

interfere with hydrophobic interactions § Use of ions and small organic molecules that increase

solubility of nonpolar substance in water and disrupting hydrophobic interactions.

Renaturation Protein is converted back to its active, folded conformation. Random Coil Nonspecific polymer conformation where monomer subunits are

disordered while bonded to adjacent units.

Describe the mathematical relationship between the free energy change for a process (ΔG) and the enthalpy change (ΔH) and the change in entropy (ΔS)? We know from thermodynamics that ∆𝐺 = ∆𝐻 − 𝑇∆𝑆, in which ΔG is the change in Gibbs Free Energy, ΔH is change in free energy, T is temperature, and ΔS is change in entropy. We also should remember that in a spontaneous process, the ΔG < 0.

Under “native” (e.g., physiological) conditions, know whether the folded form of a protein is in a higher or lower free energy state than the unfolded state. Protein folding is the process in which a polypeptide chain goes from a linear chain of amino acids with vast number of more or less random conformations in solution to the native, folded tertiary (and for multichain proteins, quaternary) structure. It occurs spontaneously under native conditions. Thus, it is favorable. We should remember that the folding is a spontaneous process. Thus it is a much low Gibbs free energy state in comparison to the Gibbs Free Energy of hydrogen bonding. Note that ΔGfolding is ~-20 to -50 kJ/mol. Small differences in energy are important. A loss of 1

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to 2 hydrogen bonds might shift equilibrium from folded state to unfolded form of the protein. Proteins fold in a defined pathway (or a small number of alternative pathways), and do not search all possible conformations until they arrive at the most stable (lowest free energy) structure. The sequence also specifies the folding pathway. As more interactions form, the structure gets closer and closer to the final folded state.

Describe the effects of a) urea and b) β-mercaptoethanol on protein structure. (They’re different.)

Chemical Structure Effect on Protein Structure Urea

Denaturation of the protein by disrupting the non-covalent

interactions.

β-mercaptoethanol

Reducing Agent by breaking disulfide bridges

Explain the bottom line conclusion of Anfinsen’s experiments that won him the Nobel Prize. From the experiments he found that correct tertiary structure of RNase was reformed, and the right sulfhydryl groups must have been adjacent to each other prior to reoxidation as a result of the backbone refolding correctly, before disulfide bonds formed spontaneously. Thus, he concluded that native tertiary structure is determined entirely by the primary structure, and the native structure is the thermodynamically most stable state for proteins.

Briefly explain amyloid formation. Neurodegenerative disorders promote formation of protein aggregates, amyloid plaques/tangles, and lesions in the brain. Amyloid plaques are mainly protein aggregates that are attribute to neurological pathologies.

Understand how the misfolding and amyloid deposits are related to the function of cellular apparatus for “disposal” of misfolded proteins. Neurodegenerative disorders promote formation of protein aggregates, amyloid plaques/tangles, lesions in the brain. The fibrils resulting from aggregation of different proteins (different diseases) have common structural features. The formation of larger aggregates seems to involve “seeding” with small aggregates, which may be more toxic to the cells than the large fibrils. Anything that increases the concentration of partially folded/misfolded molecules, perhaps with hydrophobic patches exposed on their surfaces that can form inappropriate β-sheet structure, can increase the rate at which aggregates form. For example, the amyloid-β peptide is derived from a larger transmembrane protein called amyloid-β precursor protein. When it is part of the larger protein, the peptide is composed of two α-helical segments spanning the membrane. When the external and internal domains (each

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of which have independent functions) are cleaved off, the remaining and relatively unstable amyloid-β peptide leaves the membrane and loses its α-helical structure. LECTURE X

Evolution/Bioinformatics

Define Term Definition

Homolog Derived from a common ancestor Ortholog Homologs in different species but function is the same Paralog Homologs within the same species, structurally similar but

function is different. Sequence

Alignment Method of arranging sequences of DNA, RNA, or protein to identify

regions of similarity that may be a consequence of functional, structural, or evolutionary relationships between the sequences.

Sequence Identity Match between two amino acids Sequence Template Sequence of one protein with known function that is utilized to

compare the sequence of a protein with unknown function. Convergent Evolution

Two proteins evolve independently to the same or similar structure or same or similar function. The proteins provide common

solutions to biological challenges. Divergent Evolution Proteins from a common ancestor that may acquire different

functions and structural modifications.

Explain how “sliding” sequences and “gaps” are used in sequence alignment. Sliding sequences are important to maximize the number of sequence identities when comparing two sequences. This is important because there may not be enough sequence identities when comparing the 1st amino acid of one protein with the 1st amino acid of another protein. Since insertions and deletions are common through evolution, the introduction of “gaps” in sequence may allow increases in the complexity of the alignment. Think of sequence alignment as like playing Scrabble. In Scrabble, with some degree of simplification, you want to make longer, sophisticated words with the tiles you have. If you have a blank tile, you can make more complex, longer words with much greater ease than without a blank tile. Essentially, gaps are blank tiles, and these gaps help find greater alignment. Ultimately, this would allow for confirmation that the odds of the occurrence due to chance are extremely low, and further confirm divergence.

Know that 25% sequence identity or higher is considered to prove that 2 proteins are homologous and that the correspondence is not a result of chance. Know that less than 15% sequence identity indicates that the relationship between proteins has not been demonstrated.

% Sequence Interpretation > 25 Sequences are homologous

15-25 More analysis needed < 15 Unlikely to indicate similarity

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Know that scoring sequence alignment with substitution matrices is a more sensitive indicator of homology than sequence identity, though both can be used. We need to remember that the scoring scheme involves points for positions occupied by identical residues, where no identity has any points. We also should remember that not all substitutions should be treated equally. Some are typically known as conservative, while others are known as nonconservative. Conservative substitutions involve substituting an amino acid with another amino acid that has the similar properties of the original amino acid. Non-conservative is the opposite, in which the substitution is with another amino acid with dissimilar properties of the original amino acid. The matrix is simple: as the matrix is read downward, the stability goes down with consequence substitution. If conservative substitutions are taken into account, there are more matches, because the properties are retained though the amino acids are substituted. Thus, the substitution matrices are more sensitive for homology than identity.

Two proteins can have the same fold even if all the amino acids are different, those proteins would have a common ancestor. Structure is related to function therefore structures are more evolutionary conserved than sequences. There can be similarities between proteins that have no expectation of relation. Structure only implies that the proteins are descended from a common ancestor, and not so much implying that the proteins descended from a common ancestor.

Form and function are often correlated but consider the fact that proteins with similar shape often do not have parallel functions. Structure similarity suggests that proteins descended from a common ancestor, even though form and function are often correlated. Proteins that have similar structure yet execute completely different functions.

Know that protein structure is more conserved than amino acid sequence, and that this allows homology modeling of protein structure for many related proteins. Conserved positions on a protein are more attended to than the actual amino acid sequence, because conservative substitutions can occur while the function remains intact. Structural and residue sequence homology can be used to identify key residues in either structure or function, as conserved positions across many species tend to have critical roles. Some residues are conserved because they are essential, some are less significant, and others have little specific function.

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LECTURE XII

Protein Laboratory Methods

Define the term assay. An assay is a procedure in molecular biology for testing or measuring the activity of a protein. A quantitative assay may also measure the amount of a substance in a sample. One type of assay is an activity assay, which detects the desired protein in question. The assay needs to take advantage of a property or activity that distinguishes the protein from others in the cell. Ideally, it should be cheap and easy, and the assay should detect the protein of interest from within a complex mixture (as opposed to general detection methods).

Describe the basic steps in protein purification.

The process of protein purification involves finding the source of the protein and then taking advantage of the properties and position of the desired protein to isolate it.

Step 1: Go where the protein is most abundant In order to isolate the protein, we need to extract the protein from the cell. We can fractionate the cell into components, and determine which component is enriched in the protein of interest. We can fractionate the cell via centrifugation, with separation mainly according to weight.

Step 2: Separate the proteins from everything else. We can separate the proteins from the rest of the matter by salt fractionation, to allow the high salt proteins to precipitate out. Changing the salt concentration can allow the protein to diffuse. The dialysis is done to remove the small molecules, while retaining the larger molecules. It will still have a mixture, but it is more organized by smaller and larger molecules, with the larger molecules being retained in the dialysis bag.

Step 3: Take advantage of the protein’s physical and chemical properties. The major technique involved in isolating based on the protein’s physical and chemical properties is chromatography. This involves separating the proteins on a matrix. It detects the presence of the proteins with a UV spectrophotometer. Different proteins

Go where the protein is most abundant

Separate the proteins from everything else

Take advantage of the protein's physical and chemical proteins

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elute at different times. This can be done by column chromatography, ion exchange chromatography, gel filtration chromatography and affinity chromatography.

Relate different techniques used in protein purification to the physical or chemical properties of the protein. Describe the basis for fractionating proteins using various chromatographic techniques (i.e. ion exchange, gel filtration, and affinity chromatography).

Chromatographic Technique Property of Protein Involved Ion Exchange Chromatography Charge

Affinity Chromatography Ligand Gel Filtration Chromatography Size

Ion Exchange We know that proteins are charged, because the amino acids (as monomers of proteins) themselves are charged. By removing charge, we can disrupt the ion interaction by changing the pH. The binding phase is mainly based on charge. Elution with a pH gradient is important because the proteins will elute when the charge is neutralized. Why? The non-covalent interaction being disrupted is ionic interactions. Based on the pI, we can predict how fast the molecule goes down. Positively charged protein bind to the

negatively charged beads, thus become much slower in terms of elution, while negatively charged proteins will be eluted due to repulsive forces.

Gel Filtration There is no binding period in gel filtration chromatography. It is mainly a race to the bottom. Smaller molecules take longer to elute due to their ability to enter the bead (because the beads in the column are small to allow entrance of small molecules). However, larger molecules can pass through the bead

because they cannot interact with the beads. This process can also be utilized to estimate molecular weight.

Affinity The protein has to bind in cells to a ligand, so the ligand can be attached to the column. It must be designed for each individual protein. The protein mixture is added to the column containing a polymer-bound ligand specific to the protein of interest. The unwanted proteins are washed through the column. Then the protein of interest is eluted by the ligand column.Finally, we can elute that with the ligand solution in order to push the remain protein out of the column.

Discuss various methods for detecting and identifying proteins in the lab (UV-light spectra, electrophoresis, mass spectrometry). When is each one used? Why would you choose one over another? We can detect proteins in UV-light spectra mainly because some amino acids can absorb UV light (mainly tryptophan and tyrosine). They can also be detecte by dyes. They are

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hydrophobic and tend to bind to proteins. A color change is associated with more hydrophobic groups. Polyacrylamide gel electrophoresis (PAGE) can allow electricity and a charged media to separate proteins by charge. Charged molecules move in an electric field. Anions migrate to the anode, and the cations migrate to the cathode. In a porous gel, larger molecules are impeded and smaller molecules can pass through more quickly. But the more highly charged molecules can move quickly. Thus, one can add sodium dodecyl sulfate (SDS) in order coat the protein and unfold I, unifying the mass to charge ratio of the proteins. Thus, the separation becomes on the basis of size only. Thus, this can be utilized ot estimate molecular weight. Mass spectrometry involves measuring the mass-to-charge ratio of the charged particles. It can be utilized for determination of the mass, elemental composition, and elucidating chemical structures. It accurate determines the mass of a protein. Often sequencing of peptides can also be done with mass spectrometry.

Relate a protein’s amino acid composition to its pI. The isoelectrical point of the protein is the point where the protein’s net charge is zero. Isoelectric focusing separates proteins based on relative charge. The pH gradient is established in a gel before loading the sample. When the sample is loaded, the voltage is applied. The proteins form bands that can be excised. 2D gels combine SDS-PAGE and isoelectric focusing and can be used to compare all the proteins in two samples. It is often coupled with mass spectrometry to compare the mass to charge ratio. The way to determine the mass of the protein is done by comparison against a database. The table below summarizes the role of pH and pI:

Relationship Overall Charge of Protein Movement pH < pI + Towards - pH > pI - Towards + pH = pI 0 No movement