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BIOC 2580 > Lecture Slides III > Flashcards

Lecture Slides III Flashcards

1
Q

What does chymotrypsin do?

A
  • Catalyzes hydrolysis of a substrate peptide at C-terminus of Phe, Tyr or Trp until Pro
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2
Q

How does hydrolysis by chymotrypsin work?

A
  • Binds weakly to peptide chain upstream of target AA
  • Targeted AA (Phe, Tyr or Trp) fits into binding pocket so substrate binds more tightly
  • If substrate binding is good fit, targeted peptide bond lines up with catalytic components
  • Large, nonpolar binding pocket
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3
Q

How does peptide hydrolysis by H2O (without catalyst) work?

A
  • H2O acts as nucleophile (lone pair donates to electron deficient C)
  • C maintains 8 valence e- by allowing upper O to take back bond
  • Leads to oxyanion transition state
  • Transition group may break down (with N as leaving group)
  • If N takes back excess electrons, peptide bond breaks
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4
Q

Is neutral O a good nucleophile? Why/why not?

A
  • No
  • Tends to hold onto its own electrons
  • Makes unfavourable O+ transition state
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5
Q

What formation is the C in the oxyanion transition state?

A
  • sp3 or tetrahedral
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6
Q

How does the transition state break down in peptide hydrolysis by H2O?

A
  • Oxyanion O returns bond to C

- Carboxylate C must give up bond to maintain 8 valence electrons

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7
Q

What is another way that the transition state can be broken down with peptide hydrolysis by H2O?

A
  • Carboxylate C may give up excess bonding electrons to original nucleophilic O
  • C-O bond breaks, and H2O is a good leaving group
  • Reactants are back to starting point (no net reaction)
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8
Q

How does chymotrypsin make peptide hydrolysis easier?

A
  • Reaction is broken down into 2 easy steps instead of one difficult one
  • Nucleophilic group in enzyme attacks peptide C=O to split off C-terminal, but N-terminal half is covalently bonded to enzyme group (acyl-enzyme intermediate)
  • Brings H2O to release N-terminal half, restores enzyme group to its original state
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9
Q

What is the catalytic triad?

A
  • Better nucleophile used by chymotrypsin
  • 3 AAs line up side by side and cooperate for maximum effectiveness
  • Asp 102, His 57, Ser 195
  • Combined effect makes Ser 195 into better nucleophile
  • Transition state is stabilized by oxyanion hole
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10
Q

What does Asp 102 attract?

A
  • It has a negative charge, and favours a positive charged partner
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11
Q

Describe His 57

A

Positive if it could capture H+

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12
Q

Describe Ser 195

A

Could give up H+ if it shares lone pair with suitable atom

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13
Q

Describe catalytic reaction step 1 of chymotrypsin

A
  • His 57 removes H+ to help Ser
  • -ve charge of Asp helps His act as a base
  • Oxyanion hole helps pull O- into transition state by H-bonding to backbone NH groups of Ser and Gly 193
  • H-bonds are aimed at location matching a tetrahedral C
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14
Q

Describe Step 1: Nucleophilic Attack (3 steps)

A
  • His 57 acts as general base, removing H+ from Ser 195
  • Ser 195 becomes a better nucleophile and attacks peptide C=O
  • Negative charge on Asp 102 delocalizes positive charge on His 57
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15
Q

Describe the formation of the First Transition State (with chymotrypsin)

A
  • Oxyanion hole pulls O- into transition state
  • Complementary to transition state
  • Favours tetrahedral carboxyanion configuration
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16
Q

Describe the breakdown of the 1st transition state/formation of acyl-enzyme intermediate

A
  • NH group of substrate acts as the leaving group
  • His 57 acts as general acid, donating H+ to leaving group
  • C-terminal peptide leaves
  • N-terminal peptide remains covalently bound (ACYL-ENZYME INTERMEDIATE)
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17
Q

Describe Step 2: Nucleophilic Attack (chymotrypsin)

A
  • Water enters catalytic site
  • His 57 acts as general base, removing H+ from water
  • Water becomes better nucleophile (by becoming OH-) and attacks acyl-enzyme C=O
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18
Q

Describe the Formation of the Second Transition State (chymotrypsin)

A
  • Oxyanion hole stabilizes transition state configuration
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19
Q

Describe the breakdown of the transition state/formation of products

A
  • His 57 acts as a general acid, donating H+ to Ser 195
  • Breaks acyl-enzyme bond
  • N-terminal peptide leaves
  • Catalytic triad is regenerated
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20
Q

How was the chymotrypsin catalytic cycle determined?

A
  • Structural studies and mutation studies that replaced selected amino acids (not as fast)
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21
Q

What other enzymes have an identical catalytic mechanism to chymotrypsin?

A
  • Trypsin

- Elastase

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22
Q

Enzymes speed up reaction rate in proportion to _____.

A
  • Amount of enzyme present
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23
Q

What is enzyme assay?

A
  • Process of measuring enzyme-catalyzed reaction rate
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24
Q

What is enzyme kinetics?

A
  • Mathematical analysis of how rate varies as a function of substrate concentration
  • Kinetics can be used to test reaction mechanisms
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25
Q

Why is a better way of analyzing trypsin products necessary? What is this other method?

A
  • Direct analysis is too time consuming
  • Artificial substrates provide better way
  • Artificial substrate is molecular ‘lookalike’
  • Reaction product is distinctly coloured - easy to measure
  • Trypsin recognizes Lys, but peptide after Lys is less critical
  • Accepts any primary amino group in position after Lys/Arg, won’t accept Pro (which has secondary amino group)
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26
Q

Do some natural substances show UV absorbance change after conversion to product?

A
  • Yes
  • Lactate dehydrogenase uses 2 H* atoms (H: and H+) from reduced form of NADH and transfers them to pyruvate to form lactate
  • Uses up NADH (which absorbs UV light at 340 nm whereas NAD+ does not absorb)
  • Overall absorbance decreases as reaction proceeds
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27
Q

What are chromophores and where are they found?

A
  • They are parts of molecules with conjugated double bonds (e.g. N-C=C-C=O or aromatic rings)
  • Coloured or UV-absorbing molecules contain chromophores
  • Coloured absorb b/n 400-700nm
  • Natural biochemical chromophores frequently absorb in UV range (200-400nm)
  • Larger chromophores absorb at longer wavelength
28
Q

How is absorbance measured?

A
  • With spectrophotometer
  • Beam has intensity Io, after light is absorbed intensity is I
  • A = log10(Io/I)
29
Q

What is the Beer-Lambert Law?

A

A = ECl

30
Q

What is the rate of reaction?

A
  • Change in [substrate] or [product] per unit time
  • conc substrate used/time
  • conc product formed/time
31
Q

What is enzyme activity?

A
  • Moles substrate converted per unit time

- rate x volume

32
Q

What does enzyme activity represent?

A

Quantity of enzyme present

33
Q

What is 1 enzyme unit?

A

1 micromol/min

34
Q

What is 1 katal?

A

1 mol/s

35
Q

What is specific activity?

A
  • activity per unit mass of enzyme (activity per unit mass of total protein)
  • Moles substrate converted per unit time per unit mass of enzyme
36
Q

What are the units for specific activity?

A

micromol min^-1 mg^-1

37
Q

What does specific activity measure when 2 different pure enzymes are compared?

A
  • Enzyme efficiency
38
Q

What does specific activity measure when pure and impure samples of the same enzyme are compared?

A
  • Enzyme purity
39
Q

What is molar activity?

A
  • Activity per mole of enzyme

= specific activity x molar mass of enzyme

40
Q

What is molar activity equal to?

A
  • Turnover number

- Number of catalytic reaction cycles per molecule of enzyme per second

41
Q

What is a progress curve?

A

Plot substrate or product concentration over period of time

  • Initial rate Vo is taken from slope of curve at t = 0
  • Measure several rates at different initial [S]
42
Q

What is the zero-order rate law?

A

Rate = k[S]^0

43
Q

What is the first-order rate law?

A

Rate = k[S]

44
Q

What is the second-order rate law?

A

Rate = k[S]^2

45
Q

Does an enzyme reaction follow a simple rate law?

A

No

46
Q

What are the two steps of an enzyme reaction?

A

Binding and Catalysis

E + S ES E + P

47
Q

What does the initial rate equal in an enzyme reaction?

A

Vo = k2[ES]

48
Q

Derive the Michaelis-Menten equation

A 
ES is formed at rate = k1[E][S]
ES breaks down at rate = k2[ES] +k-1[ES]
At steady state, (k2+k-1)[ES] = k1[E][S]
Let (k2+k-1)/k1 = Km
Km[ES] = [E][S]
Km[ES] = ([E]total-[ES])[S]
[ES](Km+[S]) = [E]total[S]
[ES]= ([E]total[S])/(Km+[S])
Since Vo=k2[ES]...
Vo = (k2[E]total[S])/(Km+[S])
When 100% of enzyme is occupied by substrate...
Vo=(Vmax[S])/(Km+[S])
49
Q

What are the 2 constants of every enzyme?

A

Vmax and Km (Michaelis constant)

50
Q

What is Vmax and what does it tell you about an enzyme?

A
  • Upper limit for rate
  • A pseudo-constant (only constant if amount of enzyme is fixed)
  • Vmax = k2[E]total (k2, rate constant for catalytic step, is the true constant and is turnover number)
51
Q

What is Km and what does it tell you about an enzyme?

A
  • [S] at which rate Vo is equal to 50% of Vmax
  • A low Km indicates that enzyme binds and utilizes substrate well
  • A high Km indicates that enzyme binds and utilizes substrate poorly
52
Q

What do Km and Vmax tell you about an enzyme in general?

A
  • Properties
  • Vmax indicates catalytic rate when 100% of enzyme is occupied by substrate
  • Higher = faster reaction
  • Km indicates how well substrate fits catalytic site
  • Higher = poorer recognition
  • Different Km values for each substrate
53
Q

What kind of curve is the Michaelis-Menten equation?

A
  • Hyperbolic curve (reaches asymptote), approaches Vmax gradually
54
Q

Why does real experimental data often show scatter?

A
  • Measurement errors
55
Q

What converts M-M equation into a straight line form?

A
  • Linear transformations

- Lineweaver-Burk method

56
Q

What is the Lineweaver-Burk or double reciprocal plot?

A
  • Take reciprocals of M-M equation
  • 1/Vo = (Km/Vmax)(1/[S]) + 1/Vmax
  • y-int = 1/Vmax
  • x-int = -1/Km
57
Q

What are inactivators? How do they affect catalysis?

A
  • React with enzymes irreversibly
  • Inactivation results from covalent chemical reaction b/n inactivator and enzyme
  • Rxn destroys catalytic activity
  • Simple stoichiometric relationship b/n inactivator and enzyme
  • Many are highly toxic (ex. nerve gases inactivate acetylcholinesterase)
58
Q

What are inhibitors? How do they affect enzyme catalysis?

A
  • Bind to enzymes reversibly
  • Inhibiter decreases enzyme activity w/out destroying catalytic function of enzyme
  • Activity restored if inhibitor is reduced
  • Non-covalent binding
  • Degree of inhibition is governed by binding equilibrium
59
Q

What is competitive inhibition?

A
  • Affects ability to bind substrate
  • Binds to unoccupied enzyme
  • Formation of EI complex makes E less available
  • I and S compete for E (high [S] can overcome competitive inhibitor)
  • S and I share bonding site (similar chemical structure)
60
Q

What is non-competitive inhibition?

A
  • Affects catalytic rate
  • Binds to both E and ES
  • Less ES to undergo catalysis
  • Substrate can bind to EI without yielding product
  • Binding site is different than substrate
  • May disorganize catalytic component
61
Q

What regulates enzyme activity?

A
  • Inhibitors
  • Drugs (ex. acetylsalicylic acid/aspirin inhibits cyclo-oxygenase enzymes that make prostaglandins)
  • Ex. stratins (lipitor) inhibit key liver enzyme involved in cholesterol biosynthesis
62
Q

How does competitive inhibition affect enzyme behaviour?

A
  • No effect on Vmax
  • Km is increased
  • Inhibition factor = 1 + [I]/Ki
  • Ki is [I] that causes Km to double
63
Q

How is competitive inhibition seen on a Lineweaver-Burk plot?

A 
- Different lines = different [I]
As [I] increases...
- Same y-int
- x-int gets smaller
- Slope increases
64
Q

What is mixed inhibition?

A
  • If EI and EIS steps have different Ki
65
Q

How does non-competitive inhibition affect enzyme behaviour?

A 
- Km is unchanged
As [I] increases...
- Vmax decreases
- V'max = Vmax/(1+[I]/Ki)
- Ki is [I] that causes Vmax to halve
66
Q

How is non-competitive inhibition seen on a Lineweaver-Burk plot?

A 
As [I] increases...
- Same x-int
- y-int gets bigger
- Slope increases
MIXED INHIBITION
- Lines meet above x-axis

BIOC 2580 (5 decks)

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