Exam II

Question 1

what stabilizes the structures in the secondary structures

Answer 1

hydrogen bonds

Question 2

what is the configuration of side chains in alpha helix?

Answer 2

facing away from the center

Question 3

what configuration do most peptide chains take (cis or trans)?

Answer 3

trans, exception is proline, which sometimes forms cis (10%) d/t less steric hindrance.

Question 4

how is the backbone of a polypeptide chain described

Answer 4

by torsion angles, aka dihedral angles

Question 5

the backbone of a polypeptide chain is described as rigid? What causes this?

Answer 5

The peptide group has a rigid, planar structure because of resonance interactions that give the peptide bond 40% double bond character

Question 6

Both the α helix and the β sheet are called regular secondary structures because ?

Answer 6

they are composed of sequences of residues with repeating φ and ψ values.

Question 7

T/F: the alpha helix is always right handed

Answer 7

true because it’s the only conformation of helix that is favorable according to the ramachandran diagram

Question 8

where are the R groups pointing in alpha helix?

Answer 8

outward and downward

Question 9

what is the central difference between beta sheets and alpha helices?

Answer 9

in beta sheets hydrogen bonding is between neighboring polypeptide chains rather than within one polypeptide

Question 10

where are the side chains pointing in beta sheets?

Answer 10

perpendicular to the sheet and alternating sides, such that 2 R groups will be on the same side every 7 angstroms

Question 11

what accounts for the pleated appearance of the beta sheet?

Answer 11

the conformations of deal hydrogen bonding make the structure deviate from the phi and psi angles of 180° of its fully extended form

Question 12

Parallel β sheets containing fewer than five strands are rare. Why is that?

Answer 12

This observation suggests that parallel β sheets are less stable than antiparallel β sheets, possibly because the hydrogen bonds of parallel sheets are distorted compared to those of the antiparallel sheets

Question 13

The geometry of a particular β sheet is said to be a compromise between

Answer 13

optimizing the conformational energies of its polypeptide chains in terms of R group interactions and preserving its hydrogen bonding.

^{because the interactions between chiral R groups causes a twist in the sheet that distorts and weakens the β sheet’s interchain hydrogen bonds.}

Question 14

T/F: the link between tandem parallel strands must be a crossover connection that is out of the plane of the β sheet

Answer 14

true

Question 15

what is a β bend

Answer 15

• Polypeptide segments with regular secondary structure such as α helices or the strands of β sheets are often joined by stretches of polypeptide that abruptly change direction.
• Such reverse turns or β bends (so named because they often connect successive strands of antiparallel β sheets) almost always occur at protein surfaces.
• They usually involve four successive amino acid residues arranged in one of two ways, Type I and Type II, that differ by a 180° flip of the peptide unit linking residues 2 and 3
• In Type II turns, the oxygen atom of residue 2 crowds the Cβ atom of residue 3, which is therefore usually Gly.
• Residue 2 of either type of turn is often Pro since it can assume the required conformation.

Question 16

Keratins have been classified as either α keratins, which occur in 1 , or β keratins, which occur in 2

Answer 16

1 mammals

2 birds and reptiles

Question 17

What 2 factors are principally responsible for the structure of a coiled coil?

Answer 17

1. Its primary structure: the central segment of the polypeptide units in a coiled coil each have a 7-residue pseudorepeat of a-b-c-d-e-f-g, where a and d are non-polar, which means that each helix has a hydrophobic strip. These hydrophobic strips associate with one another.
2. Because the 3.5-residue repeat in α keratin is slightly smaller than the 3.6 residues per turn of a standard α helix, the two keratin helices are inclined about 18° relative to one another, resulting in the coiled coil arrangement

Question 18

in a coiled coil, each alpha helix is _____1____handed, and the coiled coil itself is _______2____handed

Answer 18

1. right
2. left

Question 19

E and G are involved made up of oppositely charged amino acids and are involved in ionic interaction at the right pH. What does this do?

Answer 19

The ionic interactions cause the polarity to be cancelled out which leads to stability, because there will be no polar groups available for interactions outside of the coiled coil.

these ionic interactions are aka “salt bridges”

Question 20

What’s the hydrophobic pocket and how does that relate to the “knobs in hole” model?

(on coiled coil)

Answer 20

The hydrophobic pocket is the region created by the hydrophobic side chains on “a” and “d” where the 2 helices come together. This pocket is the place of insertion for the “d” amino acid, leading to the knobs in hole model, where “d” is the knob and the hydrophobic pocket is the hole.

Question 21

Describe each level of structure of a mammalian hair

Answer 21

• The N- and C-terminal domains of each polypeptide of an alpha helix facilitate the assembly of coiled coils (dimers) into protofilaments (using ionic interaction?), each protofilament containing several coiled coils, arranged head to toe and in parallel lines. 2 protofilaments constitute a protofibril.
• Four protofibrils constitute a microfibril, which associates with other microfibrils to form a macrofibril.
• A single mammalian hair consists of layers of dead cells, each of which is packed with parallel macrofibrils.

Question 22

quarternary structure involves which types of interactions?

Answer 22

all the same interactions as were present in tertiary, disulfide bonds, Van der Walls, and ionic bonds, but involves interactions between multiple polypeptide chains

Question 23

what is the simplest possible type of protein in quarternary structure?

Answer 23

dimer

Question 24

what are the 2 types of proteins in the human body

Answer 24

fibrous

• alpha karatin
• collagen

globular

• transmembrane protein
• hemoglobin
• myoglobin

Question 25

2,3 BPG can only bind to deoxy hemoglobin because ?

Answer 25

when oxygen bonds to hemoglobin, it creates a conformational change theat narrows in the cavity that the 2,3 BPG would take up

Question 26

what is the effect of 2,3 BPG

Answer 26

It makes oxygen unlikely to bind to hemoglobin which pushes the curve to the right towards dissociation, so that enough oxygen is being removed from Hgb to be delivered to tissues

Question 27

what's the mechanism of 2,3 BPG

Answer 27

2,3 BPG is a negatively charged molecule formed during cell glycolisis (produced by exercising cells) so it is attracted to the positively charged AAs that exist in between the beta subunits of Hgb. As long as there is enough space in that cavity between the subunits, which only occurs in T state (deoxy), it will bind and block binding of oxygen

Question 28

salt bridges between AAs stabilize T o R state?

Answer 28

T state

Question 29

What is the Bohr effect?

Answer 29

\*double check: It is the total phenomenon of how Hydrogen and CO2 together enhance oxygen transport by modifying hemoglobin's affinity for oxygen: Broken down into 2 components: 1. **role of H⁺** * ^​​hydrogen acts to stabilize the T state by binding to the C terminal histadine of beta subunit chain and N terminal residues of alpha subunit chain, thereby rendering them positively charged and able to participate in ionic interactions with the surrounding amino acids. These ionic interactions stabilize the T state, which lowers its affinity for oxygen - promotes unloading of oxygen 2. **role of CO²** * **^​​**CO2 reacts with the N terminal ends of subunits to form cabamates, which are molecules with R-NH-COO^- end groups. This releases a hydrogen, which further contributes to the Bohr effect * CO2 also participates in the blood buffer reaction, in which CO2 joins with water to form HCO3^- and H+, which also contributes to the Bohr effect.

Question 30

What are the steps in the conformational change from T to R

Answer 30

1. In the T state, the heme group is out of plane (Fe II) is out of heme plane because of doming of phorphyrin group toward HisF8 and because the Fe--N_porphyrin bonds are too long. When oxygen binds it causes **shortening of the Fe--N_porphyrin bonds, which allows the Fe II to fall in plane.** 2. The Fe II drags the His F8 along with it (bc they are covalently linked) and in order to avoid steric class, **His F8 tilts** 3. knob and groove interaction: changes in tertiary structure are coupled with shift in arrangement of the 4 subunits such that **the contact between His 97 of beta chain and the other AAs is moved from a His 97-Thr 41 contact to a His 97-Thr 38 contact.** This shift is larger than the other shifts that take place because His 97 (a knob) has to dodge the knob on the alpha subinit to avoid steric clash. 4. The C terminal residues of each subunit are capable of participating in ionic interactions in the right conditions, which stabilizes the T state. The conformational shift that occurs with T→R of all the hemoglobin subunits **tears apart these interactions.** (catalyzed by formation of Fe-O2 bonds)

Question 31

K (in hemoglobin and myoglobin) is defined as

Answer 31

p50: the oxygen pressure at which myoglobin is 50% saturated (ie when Y_O2 is 0.5)

Question 32

T/F the greater the p50, the more efficiently hemoglobin or myoglobin binds oxygen

Answer 32

false. the lower the p50, the more hemoglobin binds oxygen since it means it requires a low concentration of oxygen to be half saturated

Question 33

what is the major determinant of native protein structure

Answer 33

hydrophobic effect after that in order the causes are ionic interactions then volume of side chain

Question 34

T/F: native proteins are not very stable under physiological conditions

Answer 34

True. It's structure is maintained through a delicate balance of powerful countervailing forces

Question 35

Why are glycine and proline often found within a β turn?

Answer 35

A β turn results in a tight 180° reversal in the direction of the polypeptide chain. Glycine is the smallest and thus most flexible amino acid, and proline can readily assume the cis configuration, which facilitates a tight turn

Question 36

T/F: native proteins are very stable under physiological conditions.

Answer 36

False: Thermodynamic measurements indicate that native proteins are only marginally stable under physiological conditions.

Question 37

\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_, is the major determinant of native protein structure.

Answer 37

_The hydrophobic effect_, which causes nonpolar substances to minimize their contacts with water, is the major determinant of native protein structure.

Question 38

what does it look like when the phi and psi angles are both 180 degrees?

Answer 38

correction to diagram: Φ (phi) is actually -180° because C=O group is pointed down for that AA in that rotation (up is positive 180°)

Question 39

what defines a "sterically forbidden" angle

Answer 39

phi and psi angles that would bring atoms closer than the corresponding van der Waals distance

Question 40

what does the ramachandran diagram look like?

Answer 40

y axis is psi angles (mnemonic, psi looks like a y) and x axis is phi angles. most of the diagram is white meaning no angles allowed. only 3 small regions of diagram are physically accessible to most residues.

Question 41

the 2 exceptions to the typical premissible angle rule are:

Answer 41

proline and glycine proline causes restriction of typically allowed angles on Ramachandran diagram glycine expands the limits of typically allowed angles on Ramachandran diagram

Question 42

the alpha helix and the beta sheet are called regular secondary structures because?

Answer 42

they are composed of sequences of residues with repeating Φ and Ψ values

Question 43

In a Ramachandran diagram, a larger area represents sterically allowed torsion angles of phi and psi that are allowed in _____ rather than in ______ because there is greater opportunity for separation of amino acid side chains. ## Footnote A) secondary structure…tertiary structure B) α helix…β sheet C) β sheet…α helix D) tertiary structure…secondary structure E) none of the above

Answer 43

C) β sheet…α helix

Question 44

The α helix is \_\_\_\_1\_\_handed. The α helix, which ideally has ϕ = ? and ψ = ? has \_\_\_\_?\_\_\_\_ residues per turn and a pitch (the distance the helix rises along its axis per turn) of\_\_\_\_?\_\_ The α helices of proteins have an average length of \_\_\_\_?\_\_\_ residues, which corresponds to more than three helical turns, and a length of \_\_\_\_\_

Answer 44

1: right ϕ = −57° ψ = −47° 3.6 residues per turn pitch of 5.4 Å average length of ∼12 residues 12 residues= length of ∼18 Å

Question 45

An alpha helix will fold in on itself if its R groups are all/mostly __________ and is in cytosol – forms plaques

Answer 45

non-polar

Question 46

What makes the backbone of an alpha helix or beta sheet solid and hard to break apart?

Answer 46

Hydrogen bonding cancels out polarity, because it is using it up, which makes the core structure very solid and hard to break apart

Question 47

T/F: The core of the helix has atoms that are in van der Waals contact.

Answer 47

True

Question 48

When one AA changes (at position n), what happens to the (n+4)th from an evolutionary standpoint?

Answer 48

AA at the n+4 position changes through conservative substitution

Question 49

why are amyloids so strong?

Answer 49

you have groups stacked on top of each other, (esp aromatic stacking) similar to DNA, which creates a similar effect to cell well structure (phosphate on exterior, hydrophobic domain on interior)

Question 50

Keratin is a mechanically durable and relatively unreactive protein that occurs in all higher vertebrates. It is the principal component of their outer epidermal (skin) layer and its related appendages, such as hair, horn, nails, and feathers. Keratins have been classified as either α keratins, which occur in \_\_\_\_1\_\_\_, or β keratins, which occur in \_\_\_\_2\_\_\_\_

Answer 50

1. mammals 2. birds and reptiles.

Question 51

Beta sheets are said to be the product of a compromise between optimizing the conformational energies of its polypeptide chains and preserving its hydrogen bonding. Why?

Answer 51

Beta sheets are said to be undergo a slight twist because of **interactions between chiral L-amino acid residues** in the extended polypeptide chains. The twist distorts and weakens the β sheet's **interchain hydrogen bonds.** The geometry of a particular β sheet is thus a compromise between optimizing the conformational energies of its polypeptide chains and preserving its hydrogen bonding.

Question 52

In the primary sequence of a polypeptide chain, if every 4th AA is hydrophobic, it is reasonable to assume that it is involved in a \_\_\_\_\_\_\_\_\_

Answer 52

coiled coil

Question 53

ionic interactions result in a canceling out of polarity, this has the net effect of

Answer 53

increased stability

Question 54

T/F: Collagen, which occurs in all multicellular animals, is the most abundant vertebrate protein.

Answer 54

True

Question 55

which 3 non-standard AAs require vitamin C to be formed?

Answer 55

Question 56

what factor is responsible for the fact that collagen is not made up of alpha helices?

Answer 56

Collagen's Pro residues prevent it from forming an α helix (Pro residues cannot assume the α-helical backbone conformation and lack the backbone N—H groups that form the intrahelical hydrogen bonds)

Question 57

The peptide groups of collagen are oriented such that the N—H of each Gly makes a strong hydrogen bond with the carbonyl oxygen of \_\_\_\_\_\_residue on a neighboring chain

Answer 57

X (proline)

Question 58

what factor is responisble for the great tensile strength of collagen

Answer 58

The twist in the helix cannot be pulled out under tension because its component polypeptide chains are twisted in the opposite direction

Question 59

Collagen is synthesized as a \_\_\_\_\_\_\_\_\_

Answer 59

pre-pro-protein, which means it's not ready for use until it’s modified. E.g. Lysyl oxidation

Question 60

what does lysyl oxidase do?

Answer 60

converts Lys residues into allysine, which is able to react with His, Lys, and Hyp to form 4 way covalent cross-linkages between collagen strands

Question 61

Which of these characteristics does not describe the β sheet? Option A: Amino acid side chains are located both above and below the sheet. Option B: The sheets contain as few as two and as many as 22 polypeptide chains. Option C: β sheets have a pleated edge-on appearance. Option D: Parallel β sheets containing fewer than five chains are the most common. Option E: They can exist in either parallel or antiparallel configurations.

Answer 61

Option D: Parallel β sheets containing fewer than five chains are the most common.

Question 62

Hydrogen bonds and maximum separation of amino acid side chains make the \_\_\_\_\_very stable and energetically \_\_\_\_\_\_\_\_\_\_\_\_\_\_. Option A: α helix, unfavorable Option B: α helix, favorable Option C: β sheet, unfavorable Option D: α helix and β sheet, favorable Option E: β sheet, favorable

Answer 62

Option D: α helix and β sheet, favorable

Question 63

Which pairs of amino acid side chains could interact in the interior of a protein only via van der Waals interactions? Option A: Val, Leu. Option B: Ser, Ser. Option C: Gln, His. Option D: Arg, Thr.

Answer 63

Option A: Val, Leu.

Question 64

Which statement about disulfide bonds is false? Option A: Disulfide bonds can occur within a polypeptide chain. Option B: Disulfide bonds can occur between two different polypeptide chains. Option C: Disulfide bonds can occur between either Cys or Met residues. Option D: Disulfide bonds help stabilize extracellular proteins. Option E: Disulfide bonds are rare in intracellular proteins because the cytoplasm is a reducing environment. Option F: Disulfide bonds are not essential for stabilizing a folded protein. Option G: Can be reductively cleaved by 2-mercaptoethanol.

Answer 64

Option C: Disulfide bonds can occur between either Cys or Met residues.

Question 65

what is a molten globule?

Answer 65

A collapsed state of a protein in which nonpolar groups are largely buried, but they are not well packed.

Question 66

what are molecular chaperones?

Answer 66

Essential proteins that bind to unfolded and partially folded polypeptide chains to prevent the improper association of exposed hydrophobic segments that might lead to non-native folding as well as polypeptide aggregation and precipitation. All of these molecular chaperones operate by binding to an unfolded or aggregated polypeptide's solvent-exposed hydrophobic surface and subsequently releasing it, often repeatedly, in a manner that facilitates its proper folding. \*a chaperonin is a larger multisubunit complex that performs a similar function but acts on smaller proteins that enter into its structure rather than being surrounded by it

Question 67

T/F: Heat shock proteins (Hsp) : Catalyze the formation of native disulfide bonds.

Answer 67

FALSE

Question 68

In the experiment of Christian Anfinsen, which condition permitted Ribonuclease A to renature? Option A: The presence of O2 at pH 8.0 during renaturation. Option B: The use of 8 M urea to denature the protein. Option C: The presence of 2-mercaptoethanol during denaturation. Option D: Exposure to high temperatures during renaturation.

Answer 68

Option A: The presence of O2 at pH 8.0 during renaturation.

Question 69

Which of the following is not a disease that is caused by protein misfolding? Option A: Creutzfeldt-Jakob disease. Option B: Down's syndrome. Option C: Bovine spongiform encephalopathy. Option D: Alzheimer's disease.

Answer 69

Option B: Down's syndrome.

Question 70

Noncovalent forces that stabilize protein structure include

Answer 70

hydrogen bonding ionic interactions/salt bridges van der Waals forces hydrophobic effect electrostatic interactions with metal ions

Question 71

Which of the following is **not** a requirement for the structural determination of a protein using two-dimensional (2D) NMR spectroscopic techniques such as NOESY? Option A: A low molecular mass no greater than ~40 kD. Option B: The protein's primary amino acid sequence. Option C: Known geometric constraints such as covalent bond distances and angles. Option D: The ability of the protein to crystallize.

Answer 71

The ability of the protein to crystallize.

Question 72

Which of the following best describes the cause of Creutzfeld-Jakob Disease (a disease which human can develop with symptoms similar to those of mad cow disease)? Option A: Aggregation of a misfolded protein. Option B: Aggregation of random coil regions on a protein. Option C: Ingestion of ammonium salts. Option D: The serious side effects of experimental treatment with Quinacrine. Option E: All are potential causes Creutzfeld-Jakob disease.

Answer 72

Option A: Aggregation of a misfolded protein.

Question 73

Which statement is **not** a consequence of the hydrophobic effect? Option A: Nonpolar side chains are buried in the interior of a protein. Option B: The entropy of water increases when proteins fold. Option C: Metal ions can function to stabilize folded proteins. Option D: The folding of proteins is generally a favorable process.

Answer 73

Option C: Metal ions can function to stabilize folded proteins.

Question 74

T/F: Lysyl oxidase, the enzyme that converts Lys residues to those of the aldehyde allysine, is the only enzyme implicated in this cross-linking process.

Answer 74

true

Question 75

Cross-linking of collagen involves Lys residues being converted to the allysine, where the lysine side chain amino group is converted to an \_\_\_1\_\_\_\_. Cross-linking of allysine residues in collagen occurs via an \_\_\_\_\_2\_\_\_\_\_\_ reaction between two allysine side chains.

Answer 75

1. aldehyde 2. aldol condensation

Question 76

A chaperonin Option A: helps fold some proteins in their lowest energy state. Option B: is required for all proteins to fold properly. Option C: mediates the unfolding of proteins. Option D: is required for protein denaturation. Option E: counteracts the laws of thermodynamics.

Answer 76

helps fold some proteins in their lowest energy state.

Question 77

In general molecular chaperone proteins function by Option A: mediating disulfide bond formation. Option B: synthesizing new proteins when one is misfolded. Option C: preventing premature folding by binding hydrophobic regions of the protein. Option D: enhancing salt bridge formation. Option E: none of the above.

Answer 77

Option C: preventing premature folding by binding hydrophobic regions of the protein.

Question 78

Which of the following occurs first when folding a disordered polypeptide chain into a stable protein formation? Option A: formation of a low energy state Option B: association of ordered subunits Option C: aggregation of hydrophobic regions in the protein Option D: tertiary structure refinement Option E: formation of a low entropy state

Answer 78

Option C: aggregation of hydrophobic regions in the protein

Question 79

Which of the following amino acids combinations have side chains with groups that have the greatest ability to stabilize the tertiary structure of a protein? A) Lys and Arg B) Cys and Glu C) Glu and Lys D) Gln and Glu E) Pro and Asp

Answer 79

C) Glu and Lys because they form an ion pair/salt bridge

Question 80

Does the structure of a protein in a crystal accurately reflect the structure of the protein in solution, where globular proteins normally function?

Answer 80

yes - crystalline proteins assume very nearly the same structures that they have in solution because a protein molecule in a crystal is essentially in solution because it is bathed by solvent of crystallization over all of its surface except for the few, generally small patches that contact neighboring protein molecules.

Question 81

Groupings of secondary structural elements are called

Answer 81

supersecondary structures or motifs

Question 82

The \_\_\_\_\_\_\_\_is the fundamental unit of protein evolution

Answer 82

domain

Question 83

protein structures can be grouped into families by examining ?

Answer 83

the overall paths followed by their polypeptide chains.

Question 84

The most common protein domains are **evolutionary sinks,** which are:

Answer 84

Domains that arose and persisted because of their ability (1) to form stable folding patterns; (2) to tolerate amino acid deletions, substitutions, and insertions (3) to support essential biological functions.

Question 85

T/F: Structure Is Conserved More Than Sequence

Answer 85

TRUE approximately 200 different folding patterns account for about half of all known protein structures.

Question 86

T/F: Polypeptides with similar sequences tend to adopt similar backbone conformations.

Answer 86

TRUE This is certainly true for evolutionarily related proteins that carry out similar functions. e.g. Cytochrome C: the parts that are conserved are the ones that hold the heme so that the function can be carried out.

Question 87

Most/few? proteins, consist of more than one polypeptide chain.

Answer 87

Most, particularly those with molecular masses \>100 kD,

Question 88

T/F: All residues in a typical globular protein are found in alpha-helices, beta-sheets, or reverse turns.

Answer 88

FALSE: that answer doesn't acount for combined structures

Question 89

Which of the following is not a requirement for the structural determination of a protein using two-dimensional (2D) NMR spectroscopic techniques such as NOESY? Option A: The ability of the protein to crystallize. Option B: A low molecular mass no greater than ~40 kD. Option C: The protein's primary amino acid sequence. Option D: Known geometric constraints such as covalent bond distances and angles.

Answer 89

Option A: The ability of the protein to crystallize.

Question 90

Structure Comparisons Reveal Evolutionary Relationships: Most proteins have a related sequence to other proteins, since evolution tends to conserve the sequences of proteins rather than their structures.

Answer 90

FALSE: structures conserved more than sequences

Question 91

why do we even need myoglobin and hemoglobin?

Answer 91

Even though oxygen can pass instantly through cells, O2 can only go 1mm, so in order for it to get it to the inner part of cell, you need something else Another problem that must be overcome: Oxygen is poorly soluble, so you need something to capture it and keep it present In blood so that it can be carried in non-dissolved form.

Question 92

what is the oxidation state necessary for heme to bind oxygen?

Answer 92

Fe (II) +

Question 93

T/F: ## Footnote In addition to O2, certain other small molecules such as CO, NO, and H2S can bind to heme groups in proteins.

Answer 93

True in fact, These other compounds bind with much higher affinity than O2, which accounts for their toxicity. CO, for example, has 200-fold greater affinity for hemoglobin than does O2.

Question 94

Myoglobin increases the \_\_\_\_\_\_\_\_\_\_of O2 in muscle cells, acting as a kind of molecular bucket brigade to boost the O2 diffusion rate.

Answer 94

effective solubility

Question 95

the K_eq of myoglobin can be defined as?

Answer 95

p50 that is, the oxygen pressure at which myoglobin is 50% saturated.

Question 96

what is normal p50 for myoglobin?

Answer 96

pO2 of 2.8 torr

Question 97

Why is Myoglobin, a useful model for other binding proteins?

Answer 97

Even proteins with multiple binding sites for the same small molecule, or ligand, may generate hyperbolic binding curves like myoglobin's. A hyperbolic binding curve occurs when ligands interact independently with their binding sites.

Question 98

T/F: the α and β subunits of Hgb do not need to be structurally related

Answer 98

false, they do, that is a requirement for their substrate-binding activity structurally related refers to their tertiary form

Question 99

How come the "knob and groove" shift is so extensive compared to other positional shifts during T to R transition?

Answer 99

An intermediate position would be severely strained because it would bring His 97 and Thr 41 too close together (i.e., knobs on knobs).

Question 100

explain the small slope portion of the Hgb curve

Answer 100

The initial slope of the oxygen-binding curve is low, as hemoglobin subunits independently compete for the first O2

Question 101

mathematically speaking? why wouldn't a hyperbolic curve be sufficient to sustain life?

Answer 101

the difference between O2 sat at venous end and O2 sat at arterial end is not sufficient. Hyperbolic would lead to 80% saturation at O2 of 100 (arterial). This equates to a displacement amount of 30% (assuming saturation at pO2= 30 is Y=0.5)

Question 102

How can a hgb subunit be in the R state even when no O2 is bound to that particular subunit?

Answer 102

mechanical transmission from at least one subunit that is bound to O2

Question 103

T/F unliganded subunits in R state have higher affinity for oxygen

Answer 103

True (was caused by a different subunit being liganded) this phenomenon accounts for the sigmoidal curve

Question 104

Troponin is an \_\_\_\_\_\_\_\_motif and the Calcium binds in the what part of the structure?

Answer 104

alpha alpha motif in the loop region between the 2 alpha alpha components can bind calcium

Question 105

Which one of these characteristics is not true for the α helix? A) There are 3.6 amino acids per turn. B) There is a requirement for glycine every third amino acid residue. C) A hydrogen bond forms between the carbonyl oxygen of the nth amino acid residue and the - NH group of the (n + 4)th amino acid residue. D) Proline is typically not found in the α helix. E) It is right-handed.

Answer 105

B) There is a requirement for glycine every third amino acid residue. this is only a requirement for collagen (Gly-Pro-Hyp repeating units)

Question 106

16. Which statement below does not describe fibrous proteins? A) Domains have a globular fold. B) These proteins usually contain only one type of secondary structure. C) These proteins usually exhibit structural or protective characteristics. D) These proteins have usually elongated hydrophilic surfaces. E) These proteins are usually insoluble in water.

Answer 106

A) Domains have a globular fold.

Question 107

17. Which of the following changes would not alter the functional characteristics of α keratin? A) Increasing the number of residues per turn to 4.1 while maintaining the same amino acid sequence. B) Substitution of a hydrophilic amino acid for a hydrophobic amino acid at position a and d of the 7-residue pseudorepeat. C) Decreasing the number of cysteine amino acids within each protofilament. D) Changing the environment surrounding the protein to one that is more reductive. E) All of the above would alter the functional characteristics of keratin.

Answer 107

E) All of the above would alter the functional characteristics of keratin.

Question 108

Which of the following statements is true regarding collagen? A) The inability to hydroxylate proline results in the inability to synthesize collagen. B) The α helical structure is ideal for intertwining 3 filaments. C) Hydrogen bonds between the ─OH groups of Hyp residues stabilize the helix. D) The requirement for glycine every 3rd amino acid is essential for the triplet helix formation. E) On average, there is one proline for every hydroxyproline.

Answer 108

D) The requirement for glycine every 3rd amino acid is essential for the triplet helix formation.

Question 109

19. Which of the following gives the best example of a nonrepetitive structure in a protein? A) a random sequence of 12 amino acids with high Pα values forming an α helix B) an amino acid sequence with the following pattern "…a-b-c-d-e-a-b-c-d-a-b-c-d…" C) a 13 residue α helix with a Gln at position n+12 which hydrogen bonds to a residue at position n+10 D) All of the above statements describe nonrepetitive protein structures. E) None of the above describe nonrepetitive protein structures.

Answer 109

C) a 13 residue α helix with a Gln at position n+12 which hydrogen bonds to a residue at position n+10

Question 110

20. Which of the following is true regarding crystalline proteins? A) Many crystallized enzyme proteins remain catalytically active. B) The diffractive pattern observed during X-ray exposure to the crystal can be used to calculate the electron density map of the crystalline protein. C) The larger region indicating electron density with in the electron density map, the more accurate the structure determination. D) A and B are true. E) A, B, and C are true.

Answer 110

D) A and B are true.

Question 111

21. In the absence of ascorbic acid, prolyl oxidase is unable to oxidize proline residues in collagen to hydroxyproline, resulting in: A) lathyrism B) prion diseases C) amyloid formation D) scurvy E) allysine

Answer 111

D) scurvy

Question 112

23. Which of the following amino acids combinations have side chains with groups that have the greatest ability to stabilize the tertiary structure of a protein? A) Lys and Arg B) Cys and Glu C) Glu and Lys D) Gln and Glu E) Pro and Asp

Answer 112

C) Glu, Lys

Question 113

24. The low pH found in the gut can enhance the digestibility of dietary protein by causing \_\_\_. A) amide hydrolysis B) protein denaturation C) disulfide reduction D) prion formation E) cysteine oxidation

Answer 113

B) protein denaturation

Question 114

26. Imagine that a researcher treated a protein with a high concentration of a chaotropic agent. Which of the following is the most likely result of the treatment? I. Nonpolar portions of the protein become more soluble. II. The protein begins to denature , III. The protein stability increases due to hydrophobic collapse, A) I, II, III B) I, II C) II, III D) I, III E) II

Answer 114

B) I, II

Question 115

29. Conventional one dimensional NMR spectroscopy is not generally an effective tool for determination of protein structure because… I. Proteins (including small proteins) have a high number of hydrogen atoms. II. NMR requires a high quality protein crystal. III. The NMR spectra exhibit high peak overlap. A) I and II B) II and III C) I and III D) I, II, and III E) III only

Answer 115

C) I and III

Question 116

30. When comparing similarities among multiple protein structures, which of the following is false? A) Proteins with the same function from a different species are likely to have similar motifs. B) Proteins with the same function from different species are likely to be more similar in sequence than in structure. C) An effective protein motif isl likely be observed in multiple proteins. D) Proteins with the same motifs are likely to perform similar functions. E) None of the above statements are false.

Answer 116

B) Proteins with the same function from different species are likely to be more similar in sequence than in structure.

Question 117

31. The structure and sequence of a protein of unknown function was examined. Which of the following provides the best prediction of the protein's function? A) the observation of several disordered α helical domains. B) the observation of multiple protein subunits. C) the observation of motif known as the Rossmann fold. D) the observation of a large number of random coil regions. E) All of the above offer excellent prediction of the protein's function.

Answer 117

C) the observation of motif known as the Rossmann fold.

Question 118

32. The Protein Data Bank (PDB) is a database provides structural information about proteins that may be useful for which of the following? I. A researcher studying the changes in protein fold associated with prions. II. A researcher classifying structural elements by function. III A researcher designing a compound to bind tightly to a particular region in the protein. A) I only B) II only C) III only D) I, II E) I, II, III

Answer 118

E) I, II, III

Question 119

33. Noncovalent forces that stabilize protein structure include all of the following except \_\_\_\_\_\_. A) the hydrophobic effect B) salt bridges C) electrostatic interactions with metal ions D) hydrogen bonding E) disulfide bridges

Answer 119

E) disulfide bridges

Question 120

34. The classic experiment demonstrating that reduced and denatured RNase A could refold into the native form demonstrates that \_\_\_\_\_\_\_. A) 1° structure can determine 3° structure B) denaturation does not disrupt protein 2° structure C) disulfide bonds do not stabilize folded proteins D) All of the above. E) None of the above

Answer 120

A) 1° structure can determine 3° structure

Question 121

35. The first step in the folding of disordered polypeptides into ordered functional protein is the formation of \_\_\_\_\_\_. A) 1o structure B) 2° structure C) 3° structure D) 4° structure E) hydrogen bonds

Answer 121

B) 2° structure

Question 122

36. Evolutionary processes have A) increased the stability of 4° structures. B) decreased the number of subunits. C) increased similarity amount 1° structures. D) enhanced efficient folding pathways. E) all of the above

Answer 122

D) enhanced efficient folding pathways.

Question 123

39. Which of the following would be most stable based on the information you have learned about protein structure? A) a loop region with 8 amino acids B) a β sheet region made up of amino acids Val, Ile, Phe C) an α helix made up of Cys, Pro, and Phe D) a β hairpin with 12 amino acids E) All have equal stability.

Answer 123

B) a β sheet region made up of amino acids Val, Ile, Phe

Question 124

42. A chaperonin A) helps fold some proteins in their lowest energy state. B) is required for all proteins to fold properly. C) mediates the unfolding of proteins. D) is required for protein denaturation. E) counteracts the laws of thermodynamics.

Answer 124

A) helps fold some proteins in their lowest energy state.

Question 125

43. A helix has hydrogen bonds between the carbonyl group from residue “n” and the amino group of residue “n+6,” which of the following is TRUE? A) It has 3.6 residues per turn. B) It is a random coil, not a helix. C) It is an α helix. D) It has more residues per turn than an α helix. E) It has fewer residues per turn than an α helix.

Answer 125

it has more residues per turn than an alpha helix

Question 126

44. Which of the following contribute to the minimization of energy that occurs with protein folding? A) orientating amino acid groups to maximize hydrogen bonding B) folding hydrophobic groups towards the exterior of the protein C) burying polar groups towards the interior of the protein D) extensive cavity formation E) all of the above

Answer 126

A) orientating amino acid groups to maximize hydrogen bonding

Question 127

49. Based on what you know about fibrous protein structure and sequence, what type of fibrous protein is this sequence most likely to from (You can assume that the protein is longer than what is shown and is repeating as shown, also note the polarity of each amino acid.)? Val – Cys – Lys – Val - Cys – Ala – Cys - Val – Cys – Lys – Val - Cys – Ala – Cys A) α keratin B) β keratin C) collagen D) pleated collagen E) This sequence cannot be from any of the structural proteins.

Answer 127

A) α keratin

Question 128

50. Which of the following structural proteins has the greatest elasticity? A) αkeratin B) β ketatin C) collagen D) pleated collagen E) A and B are equal

Answer 128

D) pleated collagen

Question 129

53. When considering fibrous proteins, which of the following statements is TRUE? A) Noncovalent interactions contribute to the strength of all of these proteins. B) All of them consist of α helix structure. C) All of them require vitamin C. D) Decrease in amounts of any of them cause scurvy. E) All of these are true of fibrous proteins.

Answer 129

A) Noncovalent interactions contribute to the strength of all of these proteins.

Question 130

54. In a Ramachandran diagram the region representing the angles of φ and ψ that correspond to those commonly made by an amino acid that favors a left-handed α helix are different from those angles commonly made by an amino acid that favors right-handed α helix formation. Which of the following statements provides a plausible explanation for this difference? A) Groups which would normally undergo high steric hindrance in the right-handed arrangement are separated maximally in the left-handed arrangement. B) Left-handed helices have smaller pitch than right-handed helices. C) The peptide backbone can coil tighter in the left-handed helices than in the right-handed helices. D) Left-handed helices exhibit cyclic symmetry, while righthanded helices are asymmetric. E) All of the above are plausible explanations.

Answer 130

a

Question 131

55. Which of the characteristics of collagen structure listed below contrubute to the tensi le strength of collagen? I. Collagen is made up of a triplet helix where 3 left-handed helices twist together in a right handed sense. II. Collagen includes at repeating sequence of amino acids with glycine every 3 amino acids in a helix with about 3 amino acids per turn. III. The three left-handed helices are staggered to allow close packing between glycine residues and rigidity from the bulky and inflexible proline/hydroxyproline. A) I B) I and II C) I, II, and III D) II and III E) I and III

Answer 131

C) I, II, and III

Question 132

58. Protein dynamics is a field of study that examines the movements with in a protein. Which type of protein structure determination would be most useful to study this type of change? A) X-ray crystallography B) Nuclear Magnetic Resonance (NMR) C) X-ray absorption spectroscopy D) A and B E) B and C

Answer 132

B) Nuclear Magnetic Resonance (NMR)

Question 133

59. What observation about protein refolding or renaturation helped to solidify the connection between primary amino acid sequence and 3-D structure? A) Spontaneous refolding of proteins into their native state under physiologic conditions. B) Assisted refolding of proteins into their native state under laboratory conditions. C) Identification of thermostable proteins than maintain their native state in adverse temperatures. D) A and B E) B and C

Answer 133

A) Spontaneous refolding of proteins into their native state under physiologic conditions.

Question 134

60. Molecular chaperones bind to unfolded or partially folded polypeptide chains in order to accomplish which of the following? A) ensure that improper aggregation of hydrophobic segments does not occur B) engulf the protein in order to ensure that the protein is not damaged by heat denaturation C) facilitate native folding by exposing hydrophobic segments of the protein as it is synthesized D) facilitate aggregation of multiple subunits of a protein during synthesis E) All of the above are accomplished by molecular chaperones.

Answer 134

A) ensure that improper aggregation of hydrophobic segments does not occur

Question 135

57. When solving a protein structure using X-ray crystallography, the crystallographer generates a 3-D grid called an electron density map based on the observed diffraction pattern. The higher the resolution, the more detailed the electron density map and therefore the easier it is to identify what atoms (and therefore what amino acids) are in a given position. Based on the three choices below, in which of the following groups could the two of amino acids be the easiest to differentiate regardless of resolution? I. Leucine vs. Isoleucine II. Phenylalanine vs. Alanine III. Glutamate vs. Glutamic acid A) Those in both groups I and II could be differentiated B) Those in both groups I and III could be differentiated C) Only those in group II could be differentiated D) Only those in group III could be differentiated E) Only those in groupI could be differentiated

Answer 135

II. Phenylalanine vs. Alanine because they are the most different from eachother

Question 136

21. Myoglobin’s secondary structure is primarily composed of \_\_\_\_\_\_\_\_\_\_\_\_\_\_. A) parallel β-sheets B) antiparallel β -sheets C) α-helices D) α-bends E) β-helices

Answer 136

C) α-helices

Question 137

16. Hemoglobin's p50 value is about ______ as great as myoglobin's p50 value. A) one-tenth B) half C) twice D) ten times E) twenty times

Answer 137

D

Question 138

26. In sickle-cell anemia, the negatively charged glutamic acid residue is replaced by the neutral amino acid \_\_\_\_\_\_\_\_\_\_\_\_. A) tyrosine B) lysine C) valine D) adenosine E) glycine

Answer 138

C

Question 139

27. Which of the following increases the affinity of hemoglobin for oxygen: A) an increase in BPG concentration. B) the formation of N-terminal carbamates. C) increasing pH. D) all of the above E) none of the above

Answer 139

C

Question 140

Where are irregular secondary structures (loops) generally found in soluble globular proteins and why? - On the surface so that they can interact with the solvent. - On the surface because they are less compact. - In the core of the protein so that they can interact with hydrophobic groups. - In the core of the protein because they connect β-strands and -helices.

Answer 140

On the surface so that they can interact with the solvent.

Question 141

a zymogen is

Answer 141

a protein that is not yet active. prosegment needs to be cleaved

Question 142

how do karatin and collagen compare?

Answer 142

Question 143

do parallel sheets in secondary structure have dipoles?

Answer 143

no because everything is H bonded, which cancels out the polarity

Question 144

Treadmilling refers to myosin heads walking along actin microfilaments. actin and myosin filaments sliding along each other. actin monomers moving through a microfilament from the (+) end to the (-) end. synthesis and degradation of actin monomers. the interactions between actin and tropomyosin

Answer 144

actin monomers moving through a microfilament from the (+) end to the (-) end.

Question 145

Which statement about antigen-binding sites in antibodies is false? An antigen-binding site on an IgG is formed by the amino-terminal ~110 amino acids of a light chain and the amino terminal ~110 amino acids of a heavy chain. An antigen-binding site on an IgG is formed by the variable region of a light chain and the variable region of a heavy chain. The antigen-binding site is composed of two Ig folds. Antigen-binding specificity is determined by the sequences of the hypervariable sequences in both the light chain and the heavy chain. Antigen binding specificity is determined exclusively by the sequences in the carboxy-terminal ~110 amino acids in the light chain and the heavy chain

Answer 145

Antigen binding specificity is determined exclusively by the sequences in the carboxy-terminal ~110 amino acids in the light chain and the heavy chain

Question 146

Where are disulfide bonds found in immunoglobulins? They are found between the beta sheets in many of the domains. They link the light chains to the heavy chains. They link the heavy chains to each other. They are found in all of the places listed above.

Answer 146

They are found in all of the places listed above.

Question 147

What BEST distinguishes irregular secondary structure from regular secondary structure? - Irregular secondary structure is disordered (random) whereas regular secondary structure is ordered. - In a given protein the conformation of irregular secondary structure is unknown whereas the conformation of regular secondary structure is known. - Unlike regular secondary structure, successive residues in irregular secondary structure do not have the same backbone configuration. - Unlike regular secondary structure, the backbone of irregular secondary structure does not form any hydrogen bonds.

Answer 147

Unlike regular secondary structure, successive residues in irregular secondary structure do not have the same backbone configuration.

Question 148

In protein X-ray crystallography, X-ray radiation is utilized because \_\_\_\_\_\_\_. - X-ray lenses are very precise - X-ray diffraction patterns are most readily interpreted - the colors of protein crystals block visible light - the wavelength of X-ray radiation is close to the observed covalent bond lengths - X-rays are effectively diffracted by the atomic nuclei

Answer 148

the wavelength of X-ray radiation is close to the observed covalent bond lengths

Question 149

Which of the following is NOT a role of histidine in hemoglobin? - The distal histidine occupies the 6th coordination position of Fe2+. - Histidine residues become protonated as part of the Bohr effect. - The proximal histidine occupies the 5th coordination position of Fe2+. - Protonated histidine residues aid in BPG binding

Answer 149

The distal histidine occupies the 6th coordination position of Fe2+

Question 150

Why is hemoglobin's affinity for oxygen sensitive to small changes in pH (the Bohr effect)? - Histidine side chains in the central cavity of hemoglobin are charged at lower pH, decreasing BPG binding. - The affinity of the proximal histidine for the heme Fe2+ ion is pH-dependent. - The distal histidine becomes charged at lower pH, forcing the oxygen out of its binding pocket. - Histidine side chains in hemoglobin become charged at lower pH forming salt bridges that stabilize the T state.

Answer 150

Histidine side chains in hemoglobin become charged at lower pH forming salt bridges that stabilize the T state.

Question 151

what's the F helix?

Answer 151

the helix on hemoglobin and myoglobin to which His F8 is attached that moves during transition from T to R state

Question 152

If hemoglobin is in the T state, what happens at one -subunit when the other -subunit binds oxygen? - It maintains the T state until its β-subunit binds oxygen. - Its N-terminal becomes protonated. - Its F-helix moves. - It decreases its affinity for oxygen.

Answer 152

Its F-helix moves

Question 153

One of the adaptations to high altitude is an increase in the concentration of BPG in red blood cells. What effect does this have on the oxygen binding curve of hemoglobin and why? - The curve is shifted to the left because hemoglobin binds oxygen more tightly. - The curve is shifted to the left because hemoglobin has a lower Kd (dissociation constant). - The curve is shifted to the right, because hemoglobin has a lower affinity for oxygen. - The curve is shifted to the right, because hemoglobin has tighter oxygen binding.

Answer 153

The curve is shifted to the right, because hemoglobin has a lower affinity for oxygen.

Question 154

Which of the following statements about hemoglobin is TRUE? - In the Bohr effect the binding of oxygen to hemoglobin is increased by the presence of H+ ions and CO2. - Hemoglobin differs from myoglobin because it contains more β-pleated sheet structure. - The affinity of fetal hemoglobin for -oxygen is higher than that of maternal hemoglobin. - Variations in the primary structure of hemoglobin always result in genetic diseases.

Answer 154

The affinity of fetal hemoglobin for -oxygen is higher than that of maternal hemoglobin

Question 155

12. Which of the following is NOT a role for a histidine residue in hemoglobin? A. The proximal histidine occupies the 5th coordination position of Fe²⁺. B. Histidine residues become protonated as part of the Bohr effect. C. Protonated histidine residues aid in BPG binding. D. The distal histidine occupies the 6th coordination position of Fe²⁺.

Answer 155

D. The distal histidine occupies the 6th coordination position of Fe²⁺.

Question 156

15. A newly‐identified protein shows cooperative binding of a ligand. Which of the following statements about that protein is FALSE? A. A graph of percent saturation versus ligand concentration will have a sigmoidal shape. B. Ligand binding affects the tertiary structure of the protein. C. The protein has two ligand binding sites. D. The protein exists in two conformational states with different affinities for the ligand.

Answer 156

C. The protein has two ligand binding sites. ^this is false because cooperative binding relies on allosteric model. In an allosteric model you need at least 3 (1 for the allosteric inhibitor, 1 for the substrate binding site, and 1 for the other substrate binding site that is to be bound cooperatively)

Question 157

Which of the following is a role of histidine in myoglobin? - A histidine residue occupies the 6th coordination position of Fe2+. - A histidine residue forms a hydrogen bond with oxygen. - Histidine residues become protonated as part of the Bohr effect. - Protonated histidine residues aid in -BPG binding. - All of the above.

Answer 157

A histidine residue forms a hydrogen bond with oxygen. \*myoglobin not hemoglobin

Question 158

34. Which of the following is _not true_ for the symmetry model of allosterism: A) the protein is an oligomer of symmetrically (or pseudosymmetrically) related subunits. B) the oligomer can exist in two conformational states which are in equilibrium. C) the ligand can bind to a subunit in either conformation. D) the molecular symmetry of the protein is conserved during the conformational change. E) none of the above.

Answer 158

E) none of the above.

Question 159

definition of allosteric interactions

Answer 159

allosteric effects are those in which the binding of a ligand at one site affects the binding of another ligand at another site in a way that requires interactions among the subunits of oligoemeric proteins

Question 160

how do the curves for Hb +BPG and the Hb plus CO2 differ?

Answer 160

BPG has a stronger effect of stabilizing the T state than CO2 alone, so it's curve is shifted more towards the right. But together the curve is shifted farthest to the right and is identical to the curve for whole blood (meaning the hypothetical matches actual)

Question 161

22. Hemoglobin's affinity for oxygen is sensitive to small changes in pH (the Bohr effect). Which of the following amino acid(s) is/are primarily responsible for this? A. Lysine residues in the protein's central cavity. B. Histidine residues in the protein's central cavity. C. The distal histidine. D. A and B (While Lys residues and the subunit N-termini also contribute to BPG binding and stabilization of the T state, His residues (pKa ~6) are most affected by small changes in pH.) E. A, B, and C

Answer 161

B. Histidine residues in the protein's central cavity.

Question 162

18. If curve 3 represents the binding of oxygen to normal hemoglobin, which curve represents the binding of oxygen to hemoglobin when a His that interacts with BPG is mutated to a Gly?

Answer 162

curve 2

Question 163

1. A protein reversibly binds two different ligands, Y and Z. If the K_d for ligand Y binding is twice the K_d for ligand Z binding, then: A. The binding affinity for Y is higher than that for Z. B. The binding affinity for Y is lower than that for Z. C. The binding affinity for Y and Z is the same. D. No statement can be made regarding their relative affinities.

Answer 163

_B. The binding affinity for Y is lower than that for Z._

Question 164

Which of the following statements correctly describes the interaction between an allosteric protein and an allosteric effector? - The effector binds reversibly at a specific site on one subunit of the protein, causing a global change in conformation. - The effector binds covalently at a specific site on the protein, causing a global change in shape. - The effector activates the protein by causing it to switch from its T (low affinity) to R (high affinity) form. - The effector binds non-specifically to one subunit and through induced fit initiates cooperativity between the subunits.

Answer 164

The effector binds reversibly at a specific site on one subunit of the protein, causing a global change in conformation.

Question 165

If a Lys residue that interacts with 2,3-bisphosphoglycerate (BPG) in the central cavity of hemoglobin is changed to a Ser residue, how would this affect hemoglobin behaviour? - Oxygen binding would be less sensitive to pH. - Oxygen binding would be more sensitive to pH. - The T state would be less stable. - The T state would be more stable. - Both A and C would occur.

Answer 165

-The T state would be less stable. Lys not His.. if it were histidine, A would also be true

Question 166

What is the fractional saturation of myoglobin at pO2 = 2.8 torr, if p50 = 2.8 torr? 0. 50 0. 28 1. 00 2. 80

Answer 166

0.50

Question 167

what equation do you use to solve this problem: What is the fractional saturation of myoglobin at pO2 = 7.2 torr, if P50 = 2.8 torr?

Answer 167

use Y_O2=pO2/K + pO2 where K is the value of p50 in torr.

Question 168

Which of the following statements does not apply to the K value in the equation for the oxygen binding curve of myoglobin? 1. -It is defined as that oxygen partial pressure at which half of the oxygen binding sites are occupied. 2. -It is the value of pO2 at which Y = 0.5. 3. -It is a measure of the affinity of myoglobin for oxygen. 4. -If Y \> K, then myoglobin is less than 50% saturated with oxygen. 5. -It is numerically equal to p50.

Answer 168

If Y \> K, then myoglobin is less than 50% saturated with oxygen.

Question 169

Myoglobin and a single chain of hemoglobin have similar ______ structures. primary secondary tertiary quaternary none of the above

Answer 169

tertiary

Question 170

Which of the statements about muscle contraction is correct? During muscle contraction the sarcomere becomes shorter. During muscle contraction the I band becomes shorter. During muscle contraction the H zone becomes shorter. During muscle contraction the distance between the Z disk and the M disk becomes shorter. All of the answers above are correct.

Answer 170

All of the answers above are correct.

Question 171

Which of the statements about muscle contraction is not true? During muscle contraction the sarcomere becomes shorter. During muscle contraction the I band becomes shorter. During muscle contraction the H zone becomes shorter. During muscle contraction the A band becomes shorter. During muscle contraction the distance between the Z disk and the M disk becomes shorter.

Answer 171

During muscle contraction the A band becomes shorter.

Question 172

features of myosin structure

Answer 172

Myosin is a heterohexamer. Myosin contains two globular heads at the N terminus Myosin contains six different polypeptides: 4 light chains and 2 heavy chains Myosin's coiled coil has an *abcdefg* repeating unit like Karatin It undergoes conformational changes as it hydrolyzes ATP

Question 173

During muscle contraction myosin heads walk along thick filaments toward the M disk. walk along thick filaments toward the Z disk. walk along thin filaments toward the M disk. walk along the thin filaments toward the H zone. walk along the thin filaments toward the Z disk.

Answer 173

walk along the thin filaments toward the Z disk.

Question 174

Select the statement that best matches the following: Fab fragments Antibodies produced by hybridoma cells. Diverse group of proteins also known as antibodies. Fragments of proteolyzed IgG that have antigen-binding sites. Cells that mediate cellular immunity. Cells that mediate humoral immunity.

Answer 174

Fragments of proteolyzed IgG that have antigen-binding sites.

Question 175

Which of the following statements is not true about IgG? 1. The antigen-binding site is located at the tip of each Fab fragment in a crevice between its VL and VH domains 2. Has a Y-shaped general structure. 3. The heavy chain of an IgG molecule is composed of four similar domains made of beta sheets. 4. Has two antigen binding sites that can recognize and bind different antigens. 5. The antigen binding sites have an immunoglobulin fold motif (a sandwich composed of three- and four-stranded antiparallel sheets that are linked by a disulfide bond).

Answer 175

Has two antigen binding sites that can recognize and bind different antigens.

Question 176

The secondary structure of myoglobin consists of \_\_\_\_#\_\_\_ α-helices and the connecting loops.

Answer 176

8

Question 177

Which of the following represents the true protomer of hemoglobin? A. alpha B. Alpha₂ Beta₂ C. Alpha Beta D. Beta

Answer 177

C. Alpha Beta (but mostly Alpha)

Question 178

An intial response in adapting to a higher altitude is increased BPG synthesis in erythrocytes. This increase in BPG causes the O2-binding curve of hemoglobin to shift from its sea-level position to a _______ affinity position facilitating better O2 absorption and transport at the elavated altitude.

Answer 178

lower

Question 179

His F8 is the proximal (attached to Fe II) or distal (attached to O2)histidine?

Answer 179

proximal -- attached to Fe II

Question 180

Muscle fibers are composed of parallel bundles of

Answer 180

myofibrils

Question 181

T/F In striated muscle, the longitudinal axis of the fiber is ## Footnote perpendicular to the striations.

Answer 181

true

Question 182

T/F: The ability of an antibody to recognize antigens resides in three loops in the variable domain of the immunoglobulin fold. The basic immunoglobulin structure must accommodate an enormous variety of antigens. Most of the amino acid variation among antibodies is concentrated in these three short segments, called hypervariable sequences that line an immunoglobulin’s antigen-binding site, so that their amino acids determine its binding specificity.

Answer 182

All true

Question 183

T/F β Loops typically occur on the protein surface and may have important roles in biological recognition processes.

Answer 183

true

Question 184

In a 12-residue α-helix, how many backbone hydrogen bonds will the 5th amino acid make and how many backbone hydrogen bonds will the 10th amino acid make? Option A: 2 and 1. Option B: 3 and 2. Option C: 0 and 0. Option D: 1 and 1.

Answer 184

Option A: 2 and 1.

Question 185

Which of the following pairings of a supersecondary structure / motif and a statement describing it is INCORRECT: Option A: β-hairpin motif: two anti-parallel β-strands connected by a tight turn. Option B: immunoglobulin fold: two βαβαβ units that combine to form a dinucleotide-binding site. Option C: β-barrels: β-sheets that are rolled-up to form a continuous, circular sheet, with the first strand adjacent to the last strand Option D: αα motif: two successive antiparallel α-helices packed against each other with their axes slightly inclined. Option E: βαβ motif: an α-helix connecting two parallel strands of a β-sheet.

Answer 185

Option B: immunoglobulin fold: two βαβαβ units that combine to form a dinucleotide-binding site.

Question 186

Question NumberQ 10: Which of the following statements about domains found in multi-domain proteins is false? Option A: They often have specialized functions. Option B: They resemble small, globular proteins. Option C: They contain 20-50 amino acids. Option D: They are often structurally independent.

Answer 186

Option C: They contain 20-50 amino acids.

Question 187

Alpha helices are often flanked by residues such as Asn and Gln, whose side chains can fold back to form hydrogen bonds with one of the four terminal residues of the helix, a phenomenon known as _________ .

Answer 187

helix capping

Question 188

\_\_\_\_\_\_\_\_\_\_\_\_\_residues occur mostly on the surfaces of proteins

Answer 188

charged polar

Question 189

a\_\_\_\_\_\_\_\_\_\_ is one of two or more identical units of an oligomeric protein

Answer 189

protomer

Question 190

The conformational flexibility of proteins, leading to displacements of atoms up to ˜2 Angstroms, is known as

Answer 190

breathing

Question 191

T/F Incredible as it may seem, thermodynamic measurements indicate that native proteins not very stable under physiological conditions.

Answer 191

true

Question 192

The individual helices in the collagen triple-helix are connected by\_\_\_\_\_\_\_\_\_\_

Answer 192

interchain hydrogen bonds

Question 193

•polypeptide subunits associate with a specific geometry, and the spatial arrangement of these subunits is known as a protein's \_\_\_\_\_\_\_\_\_?

Answer 193

quaternary structure.

Question 194

protomers

Answer 194

Proteins with more than one subunit are called oligomers, and their identical units are called protomers.

Question 195

T/F: A protomer may consist of one polypeptide chain or several unlike polypeptide chains.

Answer 195

True what matters is that the protomer is identical to other protomers

Question 196

What type of symmetry can proteins in quaternary strucutre have?

Answer 196

Proteins can have only rotational symmetry. In the simplest type of rotational symmetry, cyclic symmetry, protomers are related by a single axis of rotation \*they cannot have inversion or mirror symmetry

Question 197

Hydropathy= what? The greater a side chain's hydropathy, the \_\_\_\_\_\_\_\_\_likely it is to occupy the interior of a protein.

Answer 197

Difference between hydrophobic and hydrophilic effect on a region. A large positive hydropathic index indicates a hydrophobic region of the polypeptide, whereas a large negative value indicates a hydrophilic region. "more"

Question 198

How does Hydrogen bonding affect tertiary structure?

Answer 198

it fine-tunes tertiary structure by “selecting” the unique native structure of a protein from among a relatively small number of hydrophobically stabilized conformations.

Question 199

Which stabilizing factor is rare in intracellular proteins because the cytoplasm is a reducing environment.

Answer 199

disulfide bridges

Question 200

The low conformational stabilities of native proteins make them easily susceptible to denaturation

Answer 200

True

Question 201

how do chaotropic agents work and what are the agents?

Answer 201

* Chaotropic agents are ions or small organic molecules that increase the solubility of nonpolar substances in water. * Their effectiveness as denaturants stems from their ability to disrupt hydrophobic interactions guanidium ion and urea

Question 202

How does temperature denature a protein?

Answer 202

Causes a protein's conformationally sensitive properties to change abruptly Most proteins have melting temperatures that are well below 100°C.

Question 203

steps and conclusion in Anfinsen experiment

Answer 203

1. add 8 M urea to denature the protein by increasing solubility of nonpolar regions, allowing the protein to expand and add mercaptoethanol to break the disulfide bonds 2. remove the denaturants simultaneously + oxidizing environment= **protein renatures into native active form** 3. Then, he wanted to know whether it was the disulfide bond or the hydrophobic interaction that was more important. So start step 1 again. 1. add denaturants 2. remove ONLY mercaptoethanol first ⇒random disulfide bonds form 3. remove urea next ⇒ scrambled protein 4. add a SMALL amount of mercaptoethanol so that the disulfide bonds are not reduced but merely exchanged⇒**native protein forms** **Conclusion:** upon removal of urea in the second part of the experiment, the hydrophobic regions made appropriate interactions/folding on their own without the correct disulfide bonds. Therefore the hydrophobic effect NOT the disulfide bonds was the driving force in determining the shape and stability of the protein.

Question 204

hydrophobic collapse

Answer 204

step during protein folding in which hydrophobic portions condense into cores, yelding the molten globule

Question 205

molten globule

Answer 205

a species that has much of the secondary structure of the native protein but little of its tertiary structure.

Question 206

how many subunits in GROEL

Answer 206

7 on each ring and they are mechanically linked

Question 207

within the GroEL/GroES structure, It requires an average of 24 folding cycles for a protein to attain its native state Total Cost = \_\_\_\_\_\_ATPs Cost per cycle: \_\_\_\_\_\_\_

Answer 207

168 7 per cycle

Question 208

What is the equivalent structure for GroEL/GroES in Eukaryotic cells?

Answer 208

* Eukaryotic cells contain the chaperonin TRiC, with double rings of eight nonidentical subunits, each of which resembles a GroEL subunit. * However, the TRiC proteins contain an additional segment that acts as a built-in lid, so the complex encloses a polypeptide chain and mediates protein folding without the assistance of a GroES-like cap.