Proteins

Question 1

Briefly explain the central dogma

Answer 1

Central dogma is the theory that states the one way flow of genetic information, from DNA to RNA to protein.
The steps involved includes transcription of DNA to mRNA, and then translation of mRNA to amino acid sequence, which is then folded to form protein structures

Question 2

Structure of amino acids

Answer 2

• Amino acids contain a tetrahedral carbon atom
• Amino acids joined via peptide bonds
• There are 20 common amino acids, some of which rarely or do not occur in proteins (at lesser frequency)

Question 3

Common structure of amino acids (have 3 things)

Answer 3

α-carbon of the amino acid is bound to:
1. Amino group
2. Carboxyl group
3. Side chain (R group)
(4. H atom)

α-amino acids are distinguished by the R group, except for proline and its derivatives

Question 4

What is the acid strength indicator?

Answer 4

pKa is the number that describes the acidity of a particular molecule and measures the strength of an acid.

The lower the pKa, the stronger the acid and the greater its ability to donate its protons, describing the acidity.

Question 5

What happens when pH is higher or lower than the pKa?

Answer 5

• When pH of system is more than pKa:
  The compound is more acidic than the system and is deprotonated
• When the pH of the system is less than pKa:
  The system is more acidic than the compound and the compound is protonated

Question 6

What is the pKa of the carboxyl group and the amino group? What happens at the physiological pH?

Answer 6

The pKa of the carboxyl group is about 2

The pKa of the amino group is about 10

At the physiological pH, both the carboxylic group and the amino group will be ionized to yield the zwitterion form, where the amino acid possesses positive and negative charges

Question 7

Where are hydrophobic amino acids found?

Answer 7

In the core of a protein, shielded away from water

Question 8

Common traits of non-polar amino acids?

Answer 8

Aliphatic R groups

R groups with no charges, mostly H or CH

Question 9

Common traits of amino acids with polar, uncharged R group?

Answer 9

R group of amino acid contains an electronegative atom, which results in polarity of the R group (e.g. F, O, N, S)

The -OH or -SH groups on the R group act as nucleophiles and often play key roles in enzyme activity

The side chains of polar amino acids can also engage in hydrogen bonding

Question 10

Common traits of basic amino acid?

Answer 10

Positively charged R groups, side chains are almost always positively charged under physiological conditions

*Resonance stabilization of the protonated side chain of guanidino group of arginine makes it the most basic amino acid

Question 11

Common traits of acidic amino acid?

Answer 11

Negatively charged R groups
Typically carries negative charges at pH 7.4

*Important in electrostatic interactions in metal ion binding (of Mg2+ and Ca2+)

Question 12

Common traits of hydrophobic amino acids?

Answer 12

Aromatic R groups

*Phenylamine is the most hydrophobic amino acid

Question 13

Describe the disulfide bond formation of 2 cysteines

Answer 13

Disulfide bond formation is a reversible reaction via oxidation of 2 molecules of cysteine (2H+ + 2e- released)

Disulfide bonds between cysteine molecules stabilizes the structures of many proteins

Question 14

Spectroscopic properties of amino acids

Answer 14

Aromatic amino acids residues can absorb UV light due to delocalized pi-electrons

Absorbance at 280nm is good diagnostic device for proteins, and absorption can be used to measure protein concentration by beer-lambert law, A=ƐCL
where A = absorbance
Ɛ = molar absorption coefficient
C = concentration of substance
L = light path length

Question 15

Structure of polypeptides

Answer 15

Proteins are polymers of amino acids joined head-to-tail through the formation of covalent peptide bonds

The peptide bond formation results in the release of water

The peptide backbone of a protein consists of the repeated sequence of -N-Cα-C₀-
N = amide nitrogen of amino acid
Cα = alpha carbon
C₀ = carbonyl carbon of amino acid

Question 16

How is peptide bonds formed?

Answer 16

peptide bonds are formed between the carboxyl group of an amino acid and the amino group of another amino acid, with the release of 1 H2O (1 H from the amino group, 1 OH from the carboxyl group)

Question 17

Structure of peptide bonds (have 5)

Answer 17

• Usually found in trans configuration
• Has partial (40%) double bond character
• 0.133 nm (1.33 Å) bond length => shorter than a single bond but longer than a double bond
• Due to the double bond character, the 6 atoms of the peptide bond group are always planar
• NH group is partially positive while the O of carboxyl group is partially negative

Question 18

Why is the trans configuration of peptide bonds strongly favoured?

Answer 18

Cis configuration has steric hindrance, due to the R groups on adjacent α-carbons can sterically interfere

Question 19

Are there instances where the cis configuration is preferred?

Answer 19

In the sequence X-Pro, where X is another amino acid and has peptide bonds with proline

Question 20

Briefly describe the structural hierarchy of proteins (1°, 2°, 3°, 4°)

Answer 20

1°: Sequence of amino acid in the polypeptide chain - peptide bonds between amino acid residues

2°: Basic pattern of hydrogen bonding to form α-helices and β-sheets - hydrogen bonds between the O and NH of the peptide bonds

3°: 3-dimensional structure as defined by the atomic coordinates, due to protein folding - non-covalent bonds (hydrogen bond, disulfide bonds, hydrophobic interactions, ionic bonds) between R groups of amino acid residues in the same chain

4°: 3D structure defined by multiple polymer chains interacting with one another to form subunits/multi-protein assembly - non-covalent bonds between R groups of amino acid residues of different chains

Question 21

What are the elements of 2° structure in proteins and how are they formed?

Answer 21

Amide or peptide bond planes are joined by tetrahedral bonds of the α-carbons

The rotation parameters are φ and ψ

Angle about the alpha carbon and nitrogen bond is denoted as φ (phi)
Angle about the alpha carbon and carbonyl carbon bond is denoted as ψ (psi)
The entire path of the peptide backbone is known if all φ and ψ angles are specified

Question 22

Why are some values of φ and ψ not allowed?

Answer 22

Many possible conformations about an α-carbon between peptide planes are forbidden due to steric hindrance

φ is the torsion angle C1-N-Cα-C2, which is the angle between the planes formed by C1-N-Cα and N-Cα-C2

ψ is the torsion angle N1-Cα-C-N2, which is the angle between the planes formed by N1-Cα-C and Cα-C-N2

Question 23

Classes of 2° structures

Answer 23

2° structures are local structures which are stabilized by systemic hydrogen bonds between peptide bonds
Classes include:
- Alpha helix
- Other types of helix
- Beta sheets (composed of beta strands)
- Tight turns (beta bends or beta turns)
- Beta bulges

Question 24

Parameters of an α-helix (3.6₁₃ helix) (Have 6)

Answer 24

• A loop of 13 atoms is formed
• There are 3.6 amino acid residues per turn of the helix
• Hydrogen bonds tend to be linear, nearly ideal, hence giving intrinsic stability (almost parallel to the helix axis)x
• Carbonyl oxygen is bonded to the amide proton 4 residues away in the direction of the C terminus
• The 3₁₀ helix contrast has exactly 3 residues per turn and a 10 atom hydrogen-bonded loop
• Torsion angles: around φ: -60° and around ψ: -45°

Question 25

What is a Pitch (p) and how is it calculated?

Answer 25

Pitch (p) is the distance along the helix axis that is translated to 1 turn of the helix pitch = residues per turn x rise per residue

Question 26

What does the degree of coiling of polypeptide chain affect?

Answer 26

The degree of coiling of polypeptide chain affects the type of helix formed, depending on the hydrogen periodicity 3₁₀, α, and π helices have 3.0, 3,6, and 4.1 residues per tun respectively, with 3₁₀ being the most tightly coiled

Question 27

What is the structure of π helices and where is it believed to be adapted from?

Answer 27

The π helix has a large number of residues per turn, found in about 15% of known protein structures π helix are believed to be an evolutionary adaptation derived from the insertion of a single amino acid into an α-helix

Question 28

Direction of strands β-sheets

Answer 28

β-strands in a β-sheet may be parallel or antiparallel

Question 29

Parameters for β-sheets

Answer 29

- A sheet is composed of 2 or more strands - Each residue in the strand is rotated 180° with respect to the preceding one - Side chains alternate between being below vs above the plane

Question 30

How are the arrangements different between the parallel and antiparallel β-strands in β-sheets?

Answer 30

Antiparallel: - N-terminus to C-terminus orientations of the β-strands are in opposite directions - The hydrogen bonds between antiparallel strands are linear Parallel: - N-terminus to C-terminus orientations of the β-strands are in the same direction - The hydrogen bonds between parallel strands are not linear

Question 31

Function of β-turns (aka β-bends or tight turns) (proline and glycine are prevalent in β-turns)

Answer 31

- Allow the peptide chains to reverse direction - 4 residues are required to form a β-turn - Oxygen of carbonyl group of on of the residue is H-bonded to the amide proton of a residue 3 residues away - There are 2 principle forms of the β-turn: Type I (more common) and type II In type 1, all residues lie in the same plane and the hydrogen bonding is closer while in type 2, the 3rd position is always a glycine and the hydrogen bond is more loose

Question 32

What is a Ramachandran diagram?

Answer 32

Ramachandran diagrams show the sterically reasonable values of the angles of φ and ψ, with blue regions indicating the particularly favourable values for these angles *The blue regions do not include the angles allowed for glycine, which entails larger region of freedom due to the flexibility from not having a side chain (less steric hindrance)

Question 33

Structure and function of fibrous proteins - the structural material of cells

Answer 33

Structure: - Have a filamentous or elongated form - Formed from protein subunits that are capable of polymerizing (quaternary interactions) - Amino acid sequences of the proteins favour a particular kind of secondary structure, which confers on each protein a particular set of appropriate mechanical properties Function: - Plays structural roles in animal cells and tissues, which includes the major proteins of skin and connective tissues, and animal fibers like hair and silk

Question 34

α-keratin is a fibrous protein found in hair, fingernails, claws, horns, and beaks. Describe the structure of α-keratin

Answer 34

- Polypeptide of α-keratin consist of 300+ amino acids, 500nm long, α-helical rod segments capped with globular N- and C- terminal domains - The sequence promotes the association of helices to form coiled coils - Very unusually high content of cysteine residues allow for extensive cross linking between chains, which gives extra strength to the fiber

Question 35

Structure of collagen - a triple helix structure (have 8)

Answer 35

1. Super high glycine and proline content (about 55%) 2. Modified amino acids present: 4-hydroxyproline (4hyP) 3. Proline disrupts α-helices and β-sheets 4. Sequence ideally suited for the triple helix, where 3 helical strands, with 3.3 residues per turn, intertwined with each other 5. Much more extended than α-helix, with a rise per residues of 2.9Å (very elongated) 6. Long strands of predominantly Gly-Pro-4hyP 7. Glycine residue is commonly found on the 3rd residue due to its small side chain fitting into the crowded center of the triple helix 8. Pro and 4hyP 'fix' the geometry to stabilize the triple helix

Question 36

Fibroin is a β-sheet protein found in silk produced by insects and spiders. The structure of fibroin is as shown: 1. Alternating sequence of almost half ___ residues (mostly ___-___ repeats, sometimes ___-___) 2. Since residues of a β-sheet extend ___ above and below the plane of the sheet, all ___ will be one one-side while all ___/___ will be on the ___ side, allowing for strong Van der Waals interactions from ___ and ___ ___ of small side chains - Allows for ___ ___

Answer 36

1. Alternating sequence of almost half glycine residues (mostly gly-ala repeats, sometimes gly-ser) 2. Since residues of a β-sheet extend alternately above and below the plane of the sheet, all glycine will be one one-side while all alanine/serine will be on the other side, allowing for strong Van der Waals interactions from interdigitation and close packing of small side chains - Allows for tight packing

Question 37

Domains are 2 or more recognizable and distinct structures of their own ___ ___. Domains usually consist of a single ___ portion of the polypeptide or an entire chain, in cases where ___ ___ assemble together

Answer 37

Domains are 2 or more recognizable and distinct structures of their own functional group. Domains usually consist of a single continuous portion of the polypeptide or an entire chain, in cases where multiple polypeptides assemble together

Question 38

How do polypeptides fold into globular structures? Anfinsen's experiment proved that ___ ___ sequence determines protein structure by finding out that ribonuclease can be ___ by treatment with urea and β-mercaptoethanol (HOCH2CH2SH) to ___ the disulfide bonds and can be ___ under appropriate ___ conditions.

Answer 38

Anfinsen's experiment proved that amino acid sequence determines protein structure by finding out that ribonuclease can be unfolded by treatment with urea and β-mercaptoethanol (HOCH2CH2SH) to reduce the disulfide bonds and can be restored under appropriate renaturing conditions.

Question 39

What is the folding and stability of a protein structure depend on?

Answer 39

Folding and stability is a delicate balance of weak non-covalant forces that weigh out to favour the folded state - Stability depends on Gibb's Free Energy, ΔG=ΔH-TΔS, where S describes the microstates associated with the protein and solvent (entropy) and H is the internal energy components of protein solvent -> ΔG must be < 0 for protein structure to be stable *ΔG of folded and unfolded = (ΔH of protein + ΔH of solvent) - T(ΔS of protein + ΔS of solvent)

Question 40

Name the 3 critical generalizations of the stability of a protein structure.

Answer 40

1. Protein conformation entropy (ΔS<0): - The contribution of protein entropy favours unfolded state 2. Solvent entropy (ΔS>0) - Higher entropy of solvent favours the folded state by contributing to the burying of hydrophobic groups 3. Protein enthalpy (ΔH<0) - Stable structure has lower energy from intramolecular interactions in the folded state

Question 41

How does protein conformation entropy contribution favour the unfolded state?

Answer 41

ΔS<0 coincides with protein folding going from very disordered state of many microstates to well-defined, few microstates state. There must be a massive investment of energy to compensate for the highly unfavourable change in microstates

Question 42

How does hydrophobic effect affect the protein folding?

Answer 42

When hydrophobic groups associate with each other, ΔS>0 When hydrophobic groups are sequestered away from water, H2O does not form interactions with the protein and is not in an orderly arrangement around the protein, but rather, the H2O is in an unorderly arrangement in the solvent, increasing entropy. *Hydrophobic molecule forces H2O in surrounding to be highly ordered, thus the lack of hydrophobic group ensures water is not arranged orderly

Question 43

How does the solvent entropy contribute to the stability of folded state of protein?

Answer 43

Solvent entropy contributes by burying hydrophobic groups in the folded state - Hydrophobic amino acids will always be shielded from the solvent and form the core of the protein while the hydrophilic group form the surface of the protein. Thus, presence of solvent ensures that the hydrophobic and hydrophilic groups stay in their conformation, contributing to the stability of the folded state

Question 44

How does intramolecular interactions of protein contribute to the stability of the folded state of protein?

Answer 44

Protein enthalpy is favourable as it is going from an open unfolded state of little intramolecular interactions, to a defined folded state of more favourable non-covalent bonding interactions

Question 45

What process disrupts the weak forces that preserves protein structure and function?

Answer 45

Denaturation by external stress like heat or chemical treatment, resulting in a loss of structure and function of a folded macromolecule *melting point: temperature at which 505 of the protein is folded & 50% is unfolded

Question 46

What are the 2 denaturants used for protein denaturation and what was used to monitor either of the denaturation?

Answer 46

Guanidinium chloride or urea are denaturants used in high concentrations Circular dichroism (CD) or Trp fluorescence is used to monitor denaturation

Question 47

Protein misfolding must be ___ as it represents a ___ of energy and poses a danger to the cell by virtue of the potential to form ___ ___. Some protein folding problems underlie a class of disease referred to as ___-___

Answer 47

Protein misfolding must be avoided as it represents a waste of energy and poses a danger to the cell by virtue of the potential to form large aggregates. Some protein folding problems underlie a class of disease referred to as amyloid-based

Question 48

In Alzheimer's disease, protein misfolding led to the manifestation of the disease. Describe the process and effect of the misfolding.

Answer 48

The molten globule of amyloid polypeptides are partially unfolded, as the α-helix becomes β-strand. Hydrophobic residues that would normally be buried in the native structure become nucleation sites for polymerization after cleavage and misfolding This then results in the protein becoming aggregates to produce fibrils, resulting in Alzheimer's

Question 49

___ and ___ are heme- proteins whose roles are related to their ability to ___ to O2 and carry out functions for oxygen ___ and ___ Mb has a ___ subunit while Hb has ___ subunits of ___ α subunits and ___ β subunits

Answer 49

Myoglobin and hemoglobin are heme- proteins whose roles are related to their ability to bind to O2 and carry out functions for oxygen storage and transport Mb has a single subunit while Hb has 4 subunits of 2 α subunits and 2 β subunits

Question 50

Why are Hb subunits structurally similar to Mb?

Answer 50

Mb and Hb evolved from the same ancestral gene via gene duplication, and then they diverge, assuming specialized roles in O2 storage vs transport

Question 51

What yields a heme prosthetic group from protoporphyrin IX?

Answer 51

Fe2+. The binding of Fe2+ to the protoporphyrin IX yields heme, which is necessary for a binding, catalytic, or structural function

Question 52

Fe2+ can form bonds having ___ character with ligands that are also ___ , in particular aromatic ring nitrogen groups, including those of ___ and the ___ systems. Fe2+ assumes an ___ coordination geometry, with 6 ligands and 4 porphyrin N groups (a proximal His ligand and a sixth location for binding O2)

Answer 52

Fe2+ can form bonds having covalent character with ligands that are also polarizable, in particular aromatic ring nitrogen groups, including those of histidine and the porphyrin systems. Fe2+ assumes an octahedral coordination geometry, with 6 ligands and 4 porphyrin N groups (a proximal His ligand and a sixth location for binding O2) *N=C delocalizes e- towards N, allowing the sharing of e- with Fe and forming Fe-N

Question 53

How does the binding of CO and O2 to Fe differ?

Answer 53

Carbon monoxide binds to free heme in linear conformation while O2 binds in a bent (120°) configuration O2 binds to Fe2+ in the same ideal fashion in both Mb and Hb, which also allows O2 to form weak interaction with the distal histidine, stabilizing the O2 binding

Question 54

Describe the O2 transport vs storage problem

Answer 54

Mb has a very high affinity for O2 at all O2 partial pressure, making it ideal for an O2 storage protein Hb must bind to O2 in lung and release O2 in the tissues. Hb becomes saturated with O2 in the lungs where Po2 is about 100mmHg, and then release O2 from Hb when Po2 is about 30 mmHg in the tissues Binding of O2 must be cooperative, where the binding of O2 to one subunit increases the affinity of O2 for all subunits by conformational changes

Question 55

Why Mb cannot be used for O2 transport?

Answer 55

Mb binds to O2 so strongly that only little would be released in tissue despite the fall in pressure. The O2 affinity at low partial pressure is too high

Question 56

Cooperativity is an allosteric effect, briefly describe the main mechanism of allosteric effect

Answer 56

Allostery corresponds to the altering of conformation and/or dynamics of a macromolecule at a distant location Binding of an effector molecule at a distant location leads to structural changes that promote substrate binding and catalytic activity

Question 57

O2 binding to Mb vs Hb, comparing the curves

Answer 57

Cooperative vs non-cooperative O2 binding curve differs by the shape! Hb-O2 binding curve has a sigmoidal shape where there is a weak binding area (near the bottom), a transition state from weak to strong binding state (at the middle), and a strong binding state (at the top), as Po2 increases Mb-O2 binding curve is a linear up curve that is exponential to the max O2 yield

Question 58

Describe the T-R transition in Hb

Answer 58

Tense state: low affinity for O2, favoured by lack of bound O2 Relaxed state: high affinity for O2, favoured by O2 binding In deoxy state, heme has a dome shape. The binding of O2 ligand pulls the iron into the heme plane, flattening the heme and causing strain. Shift in orientation of His F8 relieves the strain, pushing Val FG5 to the right

Question 59

Explain Bohr effect in tissue, where Hb-O2 binding is pH dependent

Answer 59

As O2 is consumed in tissues, Co2 is produced and must be transported back to the lungs. Carbonic anhydrase converts Co2 to carbonic acid, lowering the pH of the tissue which reduces the O2-binding affinity of Hb, enhancing Hb ability to release O2. (Lower pH promotes release of O2 by protonating Hb, which stabilizes the deoxy (T) state) Uptake of protons in T state of Hb helps drive the reaction of turning Co2 to carbonate, which can move freely in the bloodstream to be excreted by lungs

Question 60

Explain Bohr effect in lungs

Answer 60

The high o2 concentration in the lungs promotes the oxy (R) state of Hb, whereby protons are shed and the carbonates form carbon dioxide, an insoluble gas expelled from the lungs

Question 61

What are the 3 methods used to determine the 3-D structure of proteins?

Answer 61

1. X-ray crystallography (X-ray) - Dehydrate protein (purified protein) is concentrated an ideal conditions sought for crystallization - Crystal structure is determined by X-ray diffraction - diffraction pattern is translated to 3D distribution of electron density 2. Nuclear Magnetic Resonance Spectroscopy (NMR) - Macromolecule structure can be determined in solution with multidimension NMR analysis of stable isotope-labelled samples 3. Cryo electron Microscopy (cryo-EM) - Electron microscope used to see individual amino acids in protein