Biochem exam 2 Flashcards

1
Q
  • Spatial arrangement
A

Conformation

Native – folded – lowest Free Energy [G]

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2
Q
  • Conformation is stabilized primarily by:
A

tendency to ‘bury’ hydrophobic groups in the interior of the molecule

maximization of hydrogen bonds

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3
Q
  • Structure of the peptide bond:
A

6 atoms: NCC NCC

constrains the protein to certain, allowed conformations

planar

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4
Q
  • Secondary structure
A

α-helix
- right-handed
- stabilized by H-bonds that are aligned roughly parallel with the helical axis
- side chains point out and are perpendicular to the helical axis
- Ala & Leu are strong helix formers
- Pro & Gly are helix breakers. Pro forms a kink

β-sheet
- created by the planarity of the peptide bond and tetrahedral geometry of the α carbon
- held together by the H-bonds between amides and carbonyl groups
- side chains protrude and alternate
- T, V, I, Y, F, and W are found in β-sheet

parallel β-sheet
- H-bonded strands run in the same direction
- results in a bent H-bond which is weaker
- individual strands can be close or distant in the primary structure

anti-parallel β-sheet
- H-bonded strands run in opposite directions
- results in linear H-bonds which is stronger

type 1 β-turns
- find Pro

type 2 β-turns
- find Gly

strands and sometimes helices will have arrows to indicate N and C-terminal of the secondary structure

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5
Q
  • Tertiary and Quaternary structure
A

Major groups: fibrous and globular proteins

Examples of fibrous proteins (know general structural features and properties):

α keratin
- strong
- α-helix
- R-H helix, L-H super-helix (R-H alpha helices wrapped into an LH coiled-coil, stabilized by crosslinked covalent disulfide
- S-S- cross-links
- hair, hooves, nails, horns, outer skin

collagen
- gives tensile strength
- in connective tissues, which are tendons, cartilage, organic matrix of bone, cornea
- strong
- L-H helix and R-H super-helix
- 3 strands
- novel x-links
- each protein folds into an LH helix called alpha chains, not helices
- structure: 3LH helix (which are alpha chains) twisted in a RH manner to give strength
- can find G, A, P HyPro, HyLys

silk fibroin
- anti- parallel β sheets, linear, run in opposite directions, stronger
- fully extended – not stretchable
- no covalent x-links
- rich in A, G
- fully extended - no stretching
- has lots of H-bonding + van der waals but no covalent X links so it is flexible

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6
Q

Myoglobin structure – features: largely α-helical, buried hydrophobic groups, heme

A

globular

binds to Fe, & O2

in muscles

related to hemoglobin but has more affinity for O2

8 alpha helices connected by loops

heme is a porphyrin ring with Fe in the center

prosthetic group (non-protein forming part): heme

proximal histidine is directly attached to the Fe, a distal histidine group hovers near the opposite face

hydrophobic groups are buried

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7
Q
  • Methods of determining 3-D structure:
    X-ray crystallography

NMR spectroscopy

Artificial Intelligence to predict 3-D structure.

A

X-ray crystallography
pros
- no size limits
- high resolution
- well established
cons
- hard for membrane proteins
- have to crystallize
- not dynamic so do not get full protein made. May not represent reality

NMR spectroscopy
pros
- no need to crystallize
- dynamic studies are possible
- interactions with ligands- best for small proteins
cons
- hard for insoluble proteins
- best for small proteins

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8
Q

Motifs and Domains

A

motifs:
- stable arrangement of several secondary structures elements like: alpha helix, beta helix
- usually within a larger protein
- not functional outside of larger protein
- AKA super-secondary structures
- can be found in numerous proteins
- proteins are made of different motifs folded together
- motifs exist in larger proteins and do not retain structure or function when separated from large protein

domains:
- are a region of a larger protein but not in a larger protein and thus can retain structure or function when separated from large protein
- can have 1 or more motifs
- a single protein can have several domains with each one having a certain function

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9
Q
  • Other Globular proteins: know what Motifs and Domains are, and be able to recognize them in a picture of a structure.
A

motif
- within a larger molecule
- beta-alpha-beta

domain
- separated from protein

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10
Q
  • Quaternary Structure:
A

Example = Hemoglobin - 4 polypeptides
2 or more polypeptides associated together

  • these polypeptides are associated via. a side chain side chain interactions - polypeptide backbone

what drives this association
- stability: reduction of surface to volume ratio
- genetic economy + efficiency
- bringing catalytic sites together

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11
Q
  • Know Denaturation, Renaturation and Folding (and the proteins that assist folding); disorder, flexibility
A

dentauration is the loss of structural integrity + function/activity

denaturation is due to: strong acids or bases, organic solvents, detergents reducing agents, [salt], heavy metal ions, temp., mechanical stress

renaturation: break/reform kneading dough

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12
Q
  • Folding defects and disease
A

Not all proteins can fold by themselves
< 100 AA fold autonomously
> 100 AA need assistants from other proteins (ha!) to fold correctly

Some proteins require other molecules – chaperones to promote correct folding.
For example:
- Hsp70 (Heat Shock Protein 70) family protects unfolded proteins from denaturation and aggregation
- Chaperonins promote correct folding
- Isomerases make sure we have the correct stereochemistry

PDI – Protein disulfide isomerase
PPI – Peptide prolyl cis-trans isomerase

normally, misfolded proteins are fixed (remodeled) or degraded (many cellular pathways for this, such as unfolded protein response)

defects in any of the cellular systems (ex, genetic defects) may affect the degree of protein misfolding

alzheimer’s:
The native (correctly folded) β-amyloid is a soluble globular protein which is critical for neuronal growth, survival, and post-injury repair

In Alzheimer’s disease, it is clipped from the cell membrane, fragmented, and then, it misfolds

This misfolding promotes aggregation
Correctly folded helices are lost, and peptides form β strands, β helices, and β sheets  now insoluble

These insoluble plaques collect around the neurons and disrupt their environment and connectivity

Parkinson’s Disease, Lewy-Body Dementia: misfolded α-synuclein forms aggregates  of Lewy Bodies

Huntington’s Disease: genetic mutation that increases CAG repeats (increases #of amino acids) causes misfolding and aggregation  neuronal death

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13
Q
  • Know key terms & concepts: Ligand, Binding Site, flexibility, regulation, enzymes
A

ligand: a molecule reversibly bound

binding site: specific location of protein

flexibility: very important (bretahing_ induced fit

regulation: of binding s important

enzymes: substrate, catalytic (active) site

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14
Q
  • O2-binding proteins: Myoglobin & Hemoglobin: Be familiar with their structures
A

Myoglobin (Mb) is a tertiary structure protein
Myoglobin is a compact globular protein composed of a single polypeptide chain (153 amino acids in length)

Mainly α helices; there are 8. Also some intrinsically disordered regions (flexibility)

Carries and stores oxygen (poorly soluble)for muscles

Contains a heme prosthetic group

Histidines interact with heme and O2

Sterically inhibits oxygen from bindingperpendicularly to the heme plane (specific!)

Shaped/folded to form a “cradle” thatnestles the heme prosthetic group
- Protects iron from oxidation (free radicals  bad!)

In free heme, carbon monoxide (CO) binds 20,000x better than O2

Why? “smagic” (science magic – stuff about HOMOs and LUMOs)

In Mb, heme binds CO only 40X better than O2

The protein structure acts as a gate.

The effect of histidine (His E7), forces ligands to bind at an angle.

Significantly improves O2 vs CO binding
Why not evolve Mb to bind O2 more preferentially and tightly relative to CO?

Myoglobin (Mb)
1 subunit
O2 storage
1 heme group

Hemoglobin (Hb)
4 subunits – 2α and 2β
O2 transport
4 heme groups

(Tense State
No O2 bound
Salt bridges  rigid
more stable in absence of O2, and lower affinity for O2

(Relaxed State
O2 bound
β subunits interact
more stable in presence of O2, higher affinity for O2

Recall that Hb has 4 O2 binding sites

Hb is an allosteric protein: having more than one conformation; binding at one site affects the affinity of another site)

This opens the door for something called cooperativity

With the binding of each O2 molecule, (which changes the conformation of the binding subunity from T to R), the affinity of the whole protein for O2 increases.

Positive Cooperativity: binding of a ligand increases the binding affinity of subsequent ligand

Negative Cooperativity: binding of a ligand decrases the binding affinity of subsequent ligand

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15
Q

+ Know what θ, Ka, and Kd are, and their mathematical relationship

A

Kd = 1/Ka

θ = ([L]/Kd + [L])

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16
Q

+ Know binding affinity of Hb as a function of pO2 - sigmoid binding curve

A

Due to cooperativity, the curve is sigmoidal

low binding affinity for O2

One binding site = hyperbolic
Insensitive to small changes in [O2]
Mb can’t be cooperative. Why? One binding site.

Hb has a sigmoidal curve. Reflects a transition from low-affinity to high-affinity binding. This makes Hb highly sensitive to changes in [O2].

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17
Q

+ Know about the 2 conformations of Hb (R and T) and their different affinities for O2

A

(Tense State
No O2 bound
Salt bridges  rigid
more stable in absence of O2, and lower affinity for O2

(Relaxed State
O2 bound
β subunits interact
more stable in presence of O2, higher affinity for O2

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18
Q

+ Understand Cooperative binding,

A

Positive Cooperativity: binding of a ligand increases the binding affinity of subsequent ligand

Negative Cooperativity: binding of a ligand decrases the binding affinity of subsequent ligand

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19
Q

+ Understand the Bohr Effect: effect of H+ and CO2 on Hb’s affinity for O2

A

Hb can also bind H+, the binding site is different from O2 or CO2 binding site

the [CO2] also influcnes the [H+]: H+ is formed when CO2 recats with Hb or H20

when Hb binds H+, its affinity fro O2 decerases

Hb is the main buffer system in RBCs

the affinity of Hb for O2 is inversely proportional to amount of H+ and CO2 bound

consequences
- peripheral tissues: High [CO2] and [H+]: low affinity
- lungs: low [CO2] and [H+]: high affinity for O2

at lower pH (more H+), there is less affinity for O2
- this is why you hyperventilate when you exercise a lot because your muscles produce lactic acid whch lowers the pH in your muscle causing a decreased affinity for O2 so you need to compensate for that decreased affinity by breathing a lot!

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20
Q

+ Adaptation to high altitude; effect of BPG on the affinity of Hb for O2

A

initially, at lower pO2, in lungs, affinity of Hb for O2 is reduced; so less is delivered to peripheral tissue (30% vs. 38%)

then, within hours, concentration of BPG increases from ~5mM to ~8mM

affinity for O2 in the lungs is reduced slightly, but in peripheral tissue more significantly; so, even though less is bound initially, more of it is released in the peripheral tissue; result: ~37% of bound O2 is delivered to peripheral tissue

so BPG lowers the affinity that Hb has for O2 and this allows Hb to drop off more O2 to the tissues!

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21
Q
  • Cofactors & coenzymes; prosthetic group
A

cofactors help enzymes function

Coenzymes are often vitamins and essential toour diet.

prosthetic group is the non protein forming part of the enzyme

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22
Q
  • Holoenzyme, apoenzyme
A

Holoenzyme: apoenzyme (inactive) +cofactor/coenzyme/metal ion (prosthetic group)

apoenzyme: the protein part of an enzyme

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23
Q
  • Modifying groups: phosphoryl-, glycosyl-
A

Enzymes can be regulated: enzymes can beactivated (phosphorylated) or inactivated (de-)

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24
Q
  • Biomolecules are stable because bio-reactions are slow – they need to be catalyzed
A

A chemical reaction occurs when colliding molecules possess a minimum amount of energy called the activation energy (Ea)
In biochemistry this is called the free energy of activation ∆G‡

Activation energy is the difference between energy levels of the ground state and the transition state (typically higher energy than both ground states)

The rate of a reaction relates to activationenergy.
A higher ∆G‡ corresponds to a slowerreaction.
∆G is the energy difference between substrateand product! NOT CHANGED BY ENZYME

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25
Q
  • Kinetics: Enzymes affect rate (not equilibrium) by lowering activation energy
A

Enzymes LOWER activation energy!

A chemical reaction occurs when colliding molecules possess a minimum amount of energy called the activation energy (Ea)
In biochemistry this is called the free energy of activation ∆G‡

Activation energy is the difference between energy levels of the ground state and the transition state (typically higher energy than both ground states)

The rate of a reaction relates to activationenergy.
A higher ∆G‡ corresponds to a slowerreaction.
∆G is the energy difference between substrateand product! NOT CHANGED BY ENZYME

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26
Q
  • Active Site, Substrate
A

Active Site: location in the enzyme where thereaction occurs

Substrate: substance acted upon (specific!)

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27
Q
  • Slowest step is the rate-limiting step
A

The activation energy “hill”, therefore, is the rate-limiting step!

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28
Q
  • Enzymes typically increase rates by 105-1017-fold
A

yeah

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29
Q
  • B.E. contributes to catalysis by:
A

(1) reducing entropy,

(2) desolvation,

(3) compensating for distortion,

(4) induced fit – change in enzyme conformation

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30
Q
  • Most common types of catalysis:
A

(1) acid-base,

(2) covalent and

(3) metal ion [most enzymes employ more than one)

31
Q
  • Michaelis-Menten:
A

Vo=(Vmax[S])/(Km+[S]) [Know this equation]

32
Q
  • Km where Vo=(1/2)*Vmax
A
  • Km=[S] where Vo=(1/2)*Vmax
33
Q
  • Reversible Inhibition: Competitive, Uncompetitive & Mixed (of which non-competitive [which is rare] is one sub-type)

possible to overcome:
the inhibitor binds to:
inhibitor structure:
inhibitor binds to:
Km:
Vmax:

A

Competitive
possible to overcome: yes
the inhibitor binds to: the active site
inhibitor structure: similar to substrate
inhibitor binds to: enzyme alone
Km: increase
Vmax: unchanged

Uncompetitive
possible to overcome: no
inhibitor binds to: other site
inhibitor structure: different than the substrate
inhibitor binds to ES
Km: decrease
Vmax: decrease

Mixed (of which non-competitive [which is rare]
possible to overcome: no
inhibitor binds to: other site
inhibitor structure: different than the substrate
inhibitor binds to enzyme alone or ES complex
Km: changes (mixed)/unchanged (noncomp)
Vmax: decrease

34
Q
  • Lineweaver-Burk: 1/Vo=(Km/Vmax[S]) + 1/Vmax (don’t need to memorize, but understand graph
A
35
Q

competitive

possible to overcome

inhibitor binds to

inhibitor structure

inhibitor binds to

apparent Km

apparent Vmax

A

yes

active site

similar to substrate

E (enzyme alone)

increases

unchanged

36
Q

uncompetitive

possible to overcome

inhibitor binds to

inhibitor structure

inhibitor binds to

apparent Km

apparent Vmax

A

no

other site

different from substrate

ES complex

decreases

decreases

37
Q

mixed/non-competitive

possible to overcome

inhibitor binds to

inhibitor structure

inhibitor binds to

apparent Km

apparent Vmax

A

no

other site

different from substrate

E or ES complex

changes/unchanged

decreases

38
Q
  • Irreversible inhibitors, Suicide inactivators
A
39
Q
  • Regulatory enzymes: 4 types:
A

1) Allosteric (inhibitory or stimulatory; homotropic & heterotropic modulators; large; sigmoidal curve)
2) Regulated by covalent modification (esp. phosphoryl [kinases/phosphatases])
3) Activated by peptide cleavage (irreversible; “zymogens” or “proenzymes”; examples in class)

40
Q
  • Feedback inhibition
A

if there is enough product, the
product will feedback (or go back) to inhibit the enzyme that helps to make the product

41
Q

α-keratin does not include:

A. Right-hand helices
B. Right-hand super helices
C. Disulfide cross-links
D. Left-hand superhelices

A

B. Right-hand super helices

42
Q

The α chain of collagen is…

A. A left-handed helix
B. A right-handed helix
C. An anti-parallel β-sheet
D. A parallel β-sheet

A

A. A left-handed helix

43
Q

The collagen superhelix consists of _____ helices and is _____-handed.

A. 2, left
B. 2, right
C. 3, left
D. 3, right
E. 4, left
F. 4, right

A

D. 3, right

44
Q

Vitamin C is required for the formation of _____, a component of collagen.

A. Hydroxyproline
B. α-helices
C. Cys-Cys disulfide cross-links
D. hydroxylysine

A

A. Hydroxyproline

45
Q

A stable collection of 2 or more secondary structure elements is a…

A. Domain
B. Motif
C. Polypeptide
D. Prosthetic Group
E. Subunit

A

B. Motif

46
Q

If a protein has quaternary structure, it must have 2 or more…

A. β-sheets
B. Domains
C. Motifs
D. Prosthetic Groups
E. Subunits

A

E. Subunits

47
Q

What fraction of known (parts of) proteins are intrinsically disordered?

A. None: all have stable, well-defined structures
B. 1/3
C. 90-100%

A

B. 1/3

48
Q

How many bonds can the Fe2+ ion at the center of the heme group form?

A. 1
B. 2
C. 3
D. 4
E. 5
F. 6

A

F. 6

49
Q

Which of the following is true of the binding of CO by free heme, compared to the binding of O2 by free heme?

A. It is 20,000x weaker
B. It is 40x weaker
C. It is 40x stronger
D. It is 20,000x stronger

A

D. It is 20,000x stronger

50
Q

Which of the following is true of the binding of CO by heme in Myoglobin compared to the binding of O2 by heme in Myoglobin

A. It is 20,000x weaker
B. It is 40x weaker
C. It is 40x stronger
D. It is 20,000x stronger

A

C. It is 40x stronger

51
Q

An antibody binds an antigen with a Kd of 5x10-8 M. At what antigen concentration will half of the antibody binding sites be occupied?

A. 1x10-8 M
B. 2.5x10-8 M
C. 5x10-8 M
D. 1x10-7 M

A

C. 5x10-8 M

52
Q

An antibody binds an antigen with a Kd of 5x10-8 M. At what antigen concentration will Ө = 0.8?

A. 1x10-8 M
B. 2x10-8 M
C. 4x10-8 M
D. 5x10-8 M
E. 2x10-7 M

A

E. 2x10-7 M

53
Q

At an antigen concentration of 2.55x10-8 M, what fraction of the binding sites will be occupied?

A. 10%
B. 25%
C. 33%
D. 50%
E. 75%

A

C. 33%

54
Q

At an antigen concentration of 1.5x10-7 M, what fraction of the binding sites will be occupied?

A. 20%
B. 25%
C. 33%
D. 50%
E. 75%

A

E. 75%

55
Q

When [CO2] increases, the affinity of Hb for O2…
A. Decreases
B. Stays the same
C. Increases

A

A. Decreases

56
Q

When the pH increases, the affinity of Hb for O2…

A. Decreases
B. Stays the same
C. Increases

A

C. Increases

57
Q

When the [BPG] increases, the affinity of Hb for

A. Decreases
B. Stays the same
C. Increases

A

A. Decreases

58
Q

A study of the binding of Hormone Z by Receptor Protein 1022, yielded the following data. What is the Kd for the binding of Z to P-1022?

A. 0.5 x 10-9
B. 1.0 x 10-9
C. 4.0 x 10-9
D. 10 x 10-9

A

C. 4.0 x 10-9

59
Q

Enzymes increase reaction rates by…

A. Shifting equilibrium towards substrate(s)
B. Shiftingequilibriumtoward product(s)
C. Decreasing ∆G
D. Increasing ∆G
E. Decreasing∆G‡
F. Increasing∆G‡

A

E. Decreasing∆G‡

60
Q

the binding site of an enzyme is most complementary to the…

A. Substrate
B. Transition State
C. Product
D. Equilibrium State

A

B. Transition State

61
Q

Which is ∆G‡cat?

A. 1 B. 2 C. 3

A

C. 3

62
Q

Estimate the Vmax of this enzyme.

A. 2.5 x 10-6
B. 1.0 x 10-5
C. 1.0 x 10-2
D. 28
E. 70
F. 141

A

F. 141

63
Q

Estimate the Km of this enzyme.

A. 2.5 x 10-6
B. 1.0 x 10-5
C. 1.0 x 10-2
D. 28
E. 70 F. 141

A

B. 1.0 x 10-5

64
Q

Working with the M-M Equation: At which [substrate] would the enzyme operate at 1⁄4 of the max. rate?

A. 0.0013 M
B. 0.0017 M
C. 0.0025 M
D. 0.0050 M
E. 0.010 M

A

B. 0.0017 M

65
Q

Working with the M-M Equation: What will the rate be at [S] = 1⁄2Km?

A. 1/4 Vmax
B. 1/3 Vmax
C. 1/2 Vmax
D. 2 Vmax
E. 4 Vmax

A

B. 1/3 Vmax

66
Q

What is the mechanism of action for penicillin?

A. Competitive inhibition of trypsin
B. Inhibition of β-lactamase
C. Inhibition of bacterial transpeptidase
D. Inhibition of clavulanic acid

A

C. Inhibition of bacterial transpeptidase

67
Q

A new drug candidate is found to decrease both the Vmax and the Km of the enzyme MCPHSase. What type of inhibitor is it most likely to be?

A. Competitive
B. Uncompetitive
C. Mixed
D. Non-competitive
E. Irreversible

A

B. Uncompetitive

68
Q

When does an uncompetitive inhibitor bind to the enzyme? Before or after the substrate is bound?

A. Before
B. After
C. Before or After

A

B. after

69
Q

Another drug candidate is found to decrease the Vmax of the enzyme with no effect on the Km. What type of inhibitor is this molecule?

A. Competitive
B. Uncompetitive
C. Mixed
D. Non-competitive
E. Irreversible

A

D. Non-competitive

70
Q

Another drug candidate increases the Km of the enzyme with no effect on the Vmax. What type of inhibitor is this molecule?

A. Competitive
B. Uncompetitive
C. Mixed
D. Non-competitive
E. Irreversible

A

A. Competitive

71
Q

What type of inhibitor was used in this series of experiments?

A. Uncompetitive
B. Competitive
C. Mixed
D. Non-competitive
E. Irreversible

A

B. Competitive

72
Q

What type of inhibitor is being used?

A. Competitive
B. Uncompetitive
C. Non-competitive
D. Other Mixed
E. Irreversible

A

C. Non-competitive

73
Q

Enzymes that catalyze the addition of phosphoryl groups to proteins are called…

A. Enteropeptidases
B. Kinases
C. Phosphatases
D. Enolases

A

B. Kinases

74
Q

The inactivated precursor of angiotensin is activated through…

A. An allosteric modulator
B. Phosphorylation
C. Dephosphorylation
D. Proteolytic cleavage

A

D. Proteolytic cleavage