NOTE: THIS HANDOUT IS NON-EXCLUSIVE AND MAY CONTAIN SOME INFORMATION THAT IS NOT
REQUIRED FOR YOU TO KNOW, AND IT MAY BE MISSING INFORMATION THAT IS REQUIRED.
UNFORTUNATLY, I CAN NOT PREDICT OR DETERMINE EXACTLY WHAT IS GOING TO BE ON THE EXAM SO
THIS HANDOUT IS JUST SUPPLMENTARY. READING THE CHAPTERS IS THE BEST TOOL! THANK YOU
Week 7
Enzyme: increases rate of chemical reaction, decreases activation energy
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How? Binding to the transition state of the substrate (L. Pauling 1946)
Chemical reactions -> proceed only in the direction of a loss of free energy.
E.g. Glucose when combined with oxygen will spontaneously burn into H20 and CO2. But why does the
sugar in the cabinet catch on fire. Because the activation energy barrier is too high and energy needs to
be added to the substrates before they can get over the barrier.
When stuff burns -> the molecules are dispersed which increases the entropy of the molecule which is
energetically favored and is a loss of free energy. I.e. the products free energy state is lower than the
ground state of the substrate.
Enzymes -> Put energy into the substrate to allow them to overcome the activation energy barrier.
- Do this over and over again and very fast and efficient.
- Enzymes are catalyst (substances that lower the ∆G≠ “activation energy”)
- Ground state = the contribution to the free energy of the system by an average molecule (S or P) under
a given set of conditions.
- Transition state = fleeting molecular moment. Not the same as an intermediate.
-Catalyst affects the rate in both directions. Just reach equilibrium faster by increasing the rate.
- E+S -> E:S -> E:P -> E+S
-ES and EP are intermediates.
Active site -> pocket that is lined by specific amino acids
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Interaction between a.a. + substrates that allows the RXN to go forward.
How? -> Selectively stabilize the transition state (fleeting molecular moment)
Enzymes have a higher affinity for the transition state than they substrate or product.
In the active site they, align various reacting groups, stabilize transient unstable charges and,
rearrange bonds
This lowers the activation energy barrier.
The energy cost of activation is made up for by favorable interactions during the transition state,
which pays the energy required for the activation barrier.
This is known as the binding energy, which is released during the Enzyme substrate interaction
in the transition state.
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The binding energy is a combination of forces, the same forces that stabilize protein structure
-> hydrogen bonding, ionic interactions, and hydrophobic interactions.
This binding confers both specificity and catalysis.
These weak interactions are optimized during the transition state of the reaction. Enzymes are
not complementary to the substrates but to the transition state substrates pass during
conversion to products.
Important: Binding energy is the major source of free energy used by enzymes to lower the activation
energy of reactions.
JUST A SIDE NOTE
I.
Evidence for transition state complementarily.
Transition state analogs ->bind it an enzyme more tightly than substrate.
-Can be used to design pharmacological agents such as enzyme inhibitors.
Catalytic Antibodies -> antibodies that can catalyze and enzymatic reaction
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Usually antibodies do not catalyze reactions, only bind things.
However catalytic antibodies are produced when animals are immunized with hapten molecules
that are specially designed to elicit antibodies that have binding pockets capable of catalyzing
chemical reactions.
Using haptens that are transition state analogs -> the antibodies that are created bind an
epitope which is the transition state of a chemical reaction. Therefore they are able to catalyze
enzymatic reactions.
ENZYMES CONT…
I.
Serine Proteases ____________(are hydrolases) -> A class of enzymes that use _______
hydrolysis to break ___________peptide bonds to create two smaller peptides. These include a
______serine in the active site.
II.
4 essential and conserved features of serine proteases:
1.) Catalytic Triad_________ -> This includes 3 key residues – Asp, His, Ser,
a. The three residues are very close to one another in the active site, but actually far apart
in the 1° sequence.
a) Ser___ -> Most important in the reaction. Acts as the nucleophile and forms a
temporary covalent bond with the substrate.
b) His___ - Acts as a strong base first and then later as an acid. Removes the proton from
Ser, turning it into a strong nucleophile.
c) Asp___ – Helps stabilize the positive charge on His by being a temporary proton
acceptor, and turns His into a strong base.
2.) Oxyanion Hole________ -> 2 hydrogen bonds form that come from main chain N-H form with
the negative oxygen on the tetrahedral intermediate. Helps stabilize the transition state.
3.) Non-Specific substrate binding ___________-> Hydrogen binding a long the main chain of the
substrate -> helps to stabilize the transition state.
4.) Specificity pocket _____________-> Complementary to whatever side chains the enzyme
cleaves after.
Note: These general features are conserved among many species.
Subtilisin_________ (bacterial protein) – protein with completely different structure, different a.a.
sequence, but the active site has the some configuration, and same mechanism, an example of
convergent evolution.
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4  helices surrounding 5 parallel -strands -> Alpha/Beta Barrel.
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Active site: C-end of the central -strands
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III.
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Catalytic triad: S,H, D
Different types of serine proteases
Trypsin_______ _-> Cleave after larger positively charged side chains. (K and R) -> Specificity
pocket includes – Asp (small negatively charged)
Elastase _________-> Cleaves after small uncharged side chains. (G, A, V) -> Specificity pocket
includes large hydrophobic residues (Thr + Val)
Chymotrypsin ________-> Cleaves after large aromatic side chains. (F, W, Y) –Specificity pocket
includes Hydrophobic glycine and also a serine.
More detail on chymotrypsin.
Chymotrypsin – cleaves on the C-Terminal after aromatic residues. General mechanism can be broken
down into two phases.
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Acylation phase__________ -> In this phase the peptide bond is ________ broken, and an ester
linkage is formed between the substrate and the enzyme. Known as an covalent _________
acyl-enzyme intermediate.
Deacylation phase ___________-> In this phase _______ water enters and the ester linkage is
hydrolyzed and the non-acyl enzyme is regenerated.
-Important residues to remember -> Asp 102, His 57, Ser 195. ( don’t think you need to know
numbers, not 100% sure.)
Structure of chymotrpysin -> Two domain protein -> Right at the interface where the two domains
meet is where the active site is located in the 3-D structure. All the key residues are very far apart in
the 1° but come together very close in the active site.
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2 domains
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Each domain: antiparalled  -barrel, six  -strands
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IV.
Active Site: 2 loop regions from each domain
CHYMOTRYPSIN Mechanism -> FOR FIGURE SEE ATTACHMENT AT THE BOTTOM
NOTE: This mechanism and the figure attached at the bottom is a little more detailed than the
book. However I feel it’s a little more helpful to try to go through the full mechanism a couple
times. You can try it once or twice; you don’t have to memorize everything but DO know the
level of detail the book presents.
1. Upon binding of the target protein, the carboxylic group (-COOH) on Asp102 forms a low-barrier
hydrogen bond with His57, increasing the pKa of its imidazole nitrogen from 7 to about 12. This allows
His57 to act as a powerful general base, and deprotonate Ser195.
2. The deprotonated Ser195 serves as a nucleophile, attacking the carbonyl carbon on the C-terminal
side of the residue and forcing the carbonyl oxygen to accept an electron, and transforming the sp2
carbon into a tetrahedral intermediate. This intermediate is stabilized by an oxanion hole, which also
involves Ser195.
3. Collapse of this intermediate back to a carbonyl causes His57 to donate its proton to the nitrogen
attached to the alpha carbon of the leaving peptide. The nitrogen and the attached peptide fragment (cterminal to the F W or Y residue) leave by diffusion.
4. A water molecule then donates a proton to His57 and the remaining OH- attacks the carbonyl carbon,
forming another tetrahedral intermediate. The OH is a poorer leaving group than the C-terminal
fragment, so, when the tetrahedral intermediate collapses again, Ser195 leaves and regains a proton
from His57.
5. The cleaved peptide, now with a carboxyl end, leaves by diffusion.
V.
Some ways to study enzyme mechanisms:
Site Directed Mutagenesis___________ -> Change or delete a ___________particular amino acid and
see how it effects the enzyme or mechanism
Eg. Chymotrypsin -> eliminate serine, reduce catalytic activity by 2 million times.
VI.
Eliminate his/asp – reduce catalytic activity by 300,000 times.
Asp only – 30,000 times.
Discover the ser is the most important.
Enzyme Regulation:
1.) Allosteric control (other space)_____________: Mechanism where enzyme activity is__________
either increase (+) or decreased (-) by reversible non-covalent binding of some molecule at another site
distinct from the active site. Usually multi-subunit and has a regulatory_______ subunit and
_________catalytic subunit.
-The Allosteric modulator or effector can be a small metabolite or co-factor.
-This binding causes _______ a Conf. change in the active site of the enzyme
-This type of regulator is often found in metabolic pathways -> feedback inhibition.
THE GREY STUFF IS NOT IMPORTANT TO KNOW I DON’T THINK
2.) Reversible covalent modification__________: Many types (ADP-Ribosylation, Methylation,etc.), vast
majority is ____________phosphorylation.
->Kinases ________: Use ATP to add a terminal phosphate onto a Tyr, Ser, Thr or rarely His.
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Can be __________active or non-active when phosphorylated.
Many enzymes are regulated by multiple phosphorylation -> fine tuning________ of control.
(not always binary on/off).
Can be ________hierarchical.
Kinases recognize___________ consensus sequences -> e.g. PKA – X-R-(R/K)-X-(S/T)-B-
->Phosphatase_________: remove phosphates. (less specific)
IMMUNE SYSTEM
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Immune system recognizes, destroys and remembers foreign parasites while ignoring the host’s
cells and proteins
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Recognizes foreign molecules = Antigen
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B cells release Immunoglobulins to circulation
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T cells have T cell Receptor that recognizes when presented on MHC class 1 protein
IMMUNE SYSTEM
Important functions -> Distinguish between self and non-self
1.) Specific recognition of foreign particles
2.) Ability to destroy foreign microbes
3.) Memory -> allow more of a rapid + stronger response the 2nd time.
Two main branches
1.) _________Humoral (fluid) -> bacteria and viruses in blood (antigens) + antibodies
2.) ________Cellular -> cells infected with viruses, T-cell receptors interact with MHC molecules that present
viral peptides to the T-Cell.
Definitions:
_______antigen- any molecule of pathogen that can elicit an immune response.
_____________antigenic determinant(epitope)- specific site on antigen that specifically recognized by the immune
molecule. (B cell receptor)
________antibody___(fill) -> proteins present in the bodily fluids of vertebrates used by the immune system to
identify and neutralize foreign objects, such as bacteria and viruses.
How does body make anybodies? Antigens bind to a specific receptor on the B-Cell surface, triggers a conal
response and the secretion of soluble antibodies (soluble version of the membrane surface receptor).
Structure of a classic antibody -> The Immunoglobulin G (IgG).
Structural features-> 4 poly peptide chains,2 heavy, 2 light,
Heavy has 4 domains -> 3 constant,1 variable.
Light has 2 domains-> 1 variable, 1 constant
Hinge region -> digest with papain -> cut molecule into 2 regions -> Fc= crystallizes easily, and Fab -> antigen
binding fragment
All domains have a immunglobulin fold motif -> similar structure = beta sandwhich linked by adisulfide bond. ->
variable region has 3 loops that connect the beta stands -. Known as CDRs (complementary determining regions)
or hypervariable regions.
These regions determine the specificty of the molecule that the antibody binds. -> association of the hypervariable
loops of the heavy and light domains. -> what ammino acid resiudes are on the loops.
Binding can be high affinity of low affinity -> different types of 3-D shapes formed by the loops. Pockets, grooves,
flat surface.
Specifity of an IG determined by
1.) a.a. sequence of the hypervariable regions
2.) a.a. size of the hypervariable regions.
Induced Fit -> enzymes are flexible -> antigen binding causes the shape of the antibody to change inducing a
better fit.
Cell mediated response -> Cell infected by a virus presents viral peptides onto its MHC protein -> The T Cell
receptor on the T-Cell reads those peptides.
MHC = M_____ H_____ C______ Major Histocompatibility Complex -> found on virtually all our cells.
(polymorphic)
Two main MHC classes.
Class I MHC proteins mainly
present foreign peptides to cytotoxic T cells
class II MHC proteins mainly
present foreign peptides to helper and regulatory T cells
MHC Class I -> Structure = 2 poly peptide chains -> 4 extracellular domains.
Heavy chain -> 3 domains – α1, α2, α3
Light chain -> 1 domain -> Beta2
- Each of the two domains closest to the plasma membrane (a3 and b2)
resembles a typical Ig domain while the two
domains farthest from the membrane (a1 and a2) are very similar to
each other (alpha beta domains) and together form a peptide-binding groove at the top of
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All TCRs recognize a helices of MHC
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Foreign peptide lays in cradle between helices
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High specificity binding of TCR to foreign peptide within MHC cradle
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Recognition motif is 2 a helices + 4 antiparallel b strands
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Recognition motif connected to TM domain by b sandwich
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MHC class 1 is monomeric
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MHC class 2 is dimeric
MEMBRANE
A hydropathy plot of the amino acid sequence of an erythrocyte membrane protein begins with a region
of high negative hydrophathy index, followed by three regions of high positive hydropathy index, and
ends with a region of high negative hydropathy index. (Each of the regions of high positive and high
negative hydropathy index spans more than 20 residues.) What can you predict about the topology of
this membrane protein?
I.
A.
It has three transmembrane domains.
B.
At least one domain of the protein faces inside the cell.
C.
It is a peripheral membrane protein.
D.
It has two domains that span the membrane.
BIOLOGICAL MEMBRANES
a. BASIC PHOSPHOLIPID STRUCTURE
II.
a. Made up of proteins and lipids -> protein/lipid composition can vary among cell types.
b. Membranes are very THIN, around 4.5 nanometers, or 45 Angstroms
c. Membrane proteins are harder to crystallize and need to be removed and crystallized
with detergents, which mimic the hydrophobic environment of the lipid bilayer.
d. The lipid bilayer is held together by mostly non-covalent interactions, primarily ->
Hydrophobic effect.
III.
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e. Membrane Lipids are _________amphiphatic molecules -> have both a polar and a nonpolar side
i. Soap and detergents are amphipathic like membranes, eg. Sodium dodecyl
sulfate (SDS).
f. 3 Main Types of membrane lipids: 1.) Phospholipids 2.)Glycolipids 3.) Cholesterol
i. 2 types of phospholipids: 1.) Glycerophospholipids 2.) Sphingophospholipids
 The four main specific phospholipids in membranes:
g. Phosphatidylethanolamine, Phosphatidylserine (-), Phosphatidylcholine, Sphingomyelin
Detergent vs. phospholipid
The detergent will form a micelle because the single fatty acid tail -> allows for tighter packing
into the core
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IV.
VII.
The phospholipid will form the classic bilayer.
A liposome is an artificial lipid bilayer
Cholesterol -> Planar, 4 steroid ring with a polar hydroxyl –OH head group. Ampipathic.
a. Abundant -> up to 50% of membrane. (vary among cell types)
b. Both increases and decreases the fluidity of membranes -> allows for tighter packing of
the phospholipids, and at the same time prevents crystallization of the phospholipids at
low temperatures. -> NOTE ACCORDING TO MONTALS SLIDE HE ONLY MENTIONS THAT
CHOLESTEROL MAKES MEMBRANES MORE RIGID.
Asymmetry, Movement:
a. Membrane is very asymmetric: eg. PS is only found on the inner leaflet -> important for
cell signaling. (inner is negative). Glycolipids are only found on the outside.
b. Movement: 3 types:
1.) uncatalyzed trans bilayer -> rare and slow
2.)Rotation and flextion
3.) uncatalyzed lateral diffusion -> lots and fast , single phospholipid goes all the ways
around a cell in 1 sec.
Fluidity: Affected by temperature and lipid composition
a. Phase Transition Temperature -> temp. where lipids go from fluid to the para-crystaline
state.
b. Composition of lipids -> degree of saturation of the hydrophobic tail
i. More saturation = higher phase transition temperature
ii. Less saturation = more double bond order -> lower phase trans. Temp
iii. Hydrophobic tail length -> longer tail = higher phase trans. Temp.
iv. Why -> with more saturation, lipids can pack more tightly. Double bond introduces
kinks.
v. Bacterial can change lipid comp with varying environmental temperatures.
c. Cholesterol -> Modulator, prevents crystallization at temp below phase trans. Increases
rigidness at temp above phase trans.
i. How? Inhibit the close packing of the hydrophobic tails.
d. Temperature ->
i. Low temperature membranes become less fluid.
ii. High temp membranes become more fluid.
Question: would you expect a polar bear to have more/less unsaturated fatty acids that a cold water
fish? Why?
e. Integral Membrane Proteins -> Amphiphatic , transmembrane proteins
i. Firmly associated with the membrane via lots of hydrophobic interactions.
ii. Can be removed via detergents.
iii. Can be single span or multi-span.
iv. Allow the cell to communicate with the outside to the inside of the cell.
A. Integral Membrane proteins -> are usually beta barrels or alpha helices. Why? Because main
chain is hydrophilic and the you need to have hydrophobic interactions with the interior of the
membrane and maximize intrachain hydrogen bonding.
a. Integral membrane proteins are harder to crystallize.
b. Hydropathy Index: Experimentally determine the polarity of amino acid residues. ->
move from a non-polar solvent to a polar one. Determine Free E change.
i. Negative -> hydrophilic
ii. Positive -> hydrophobic
c. Hydropathy plot -> used to predict regions that might cross the membrane.
i. Plot hydrophathy index ( y axis) vs. Residue number (x-axis).
ii. Average length of transmembrane alpha helix = 20 residues.
Can extract/ isolate integral membrane proteins and study in artificial membranes
BACTERIORHODOPSIN
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Found in the membrane of bacterium Halobacterium Halobium -> 248 a.a. and binds retinal, the
same photosynthetic pigment we use in our eyes to capture light.
Bacteriorhodopsin uses light energy to pump protons from the cytosol to the extracellular
space.
7-Transmembrane helices span the hydrophobic membrane -> The helices are tilted about 20° in
respect to the membrane -> characteristic of a helical bundle.
When retinal absorbes a photon it undergoes a trans to cis isomerization at bond position 11
and protons are subsequently pumped across the membrane.
Retinal is bound in a pocket of bacteriorhodopsin, and the pigment forms a Schiff base with a
lysine residue -> LYS 216. i.e. it is covalently linked to the nitrogen atom of a lysine side chain
that is protonated and therefore has a positive charge.
The lysine is positioned in a channel near the cytosolic side that is lined with hydrophobic
residues with the expectation of a asp 96 which is important for pumping protons -> the
extracellular side of the channel is wider and is lined with hydrophilic amino acids including asp
85
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Tense (T) – binds trans-retinal
Relaxed (R) – binds cis-retinal
PROTON RELAY MECHANISM
Short version
R state: H+ transferred from Asp 96 to Schiff base and H+ from Asp 85 is transferred to Extracellular
Space  H+ transferred from Cytoplasm to Asp 96
Long Version
Tense State binds trans-retinal -> photon hits and retinal is coverted into cis, this causes to the Schiff
base to change its position relative to the Asp85, whicn induces transfer of the Schiff base proton to the
Asp86-> once this transfer has occurred to protein undergoes a T State to R state conformational change
-> Asp 85 then delievers the extra proton to the EC space, and Asp 96 reprotonates the Schiff base,
which subsequently reverts to the trans state and the protein goes back to the T state ready for another
cycle.
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Why is retinol purple in the membrane but orange when outside the membrane?
o Because the environment is different when its in the membrane.
PORIN Protein BETA BARREL
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Generally found in bacteria
Homotrimeric protein.
Each monomer is made of a 16 stranded Up/Down β Barrel.
Subunits interact by polar loop and hydrophobic side chain interactions.
Periplasmic End: short and smooth
Cytoplasmic End: funnel shaped, made of long loops with hydrophilic residues.
Eyelet = loop connecting β5 and β6 extending into the central open cavity of the barrel.
Cavity is lined with charged residues, arranged in an electrochemical gradient.
Different from RBP, because this time the inner barrel is hydrophilic to allow polar molecules to
pass through the membrane, and the outside is hydrophobic to interact with the hydrophobic
membrane.
Notice the alternating hydrophobic hydrophillic residues, similar to RBP but just reversed.
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Aromatic Belt – mainly Phe and Tyr
has polar and hydrophobic character to interactive with the hydrophilic space of the membrane
extracellular solution and hydrophobic internal space of the membrane.
Helps stabilize protein in membrane.
Bacterial. K+ potassium channel (KCSA)
i. 4 identical transmembrane units, which form a central pore -> homotetrameric
ii. Functional Channel requires four subunits.
iii. Each monomer is made of 2 transmembrane α-helices, 1 pore helix and a
cytosolic tail.
iv. Selectivity Filter – formed from the backbone carbonyl Oxygens from the loops
connecting pore helix to the inner transmembrane helix. The structure is
stabilized by the packing of residues from the loop into the pore helix.
v. Negatively charged residues concentrate on the cytsolic side to attact potassium
and repel anions. Selectivity filter backbone carbonyls are precisely aligned so
they replace favorable interaction with dehydrated potassium. Sodium is too
small, carbonyl interactions not favorable enough.
 K (OH2)8+ = hydration shell (8 H2O needed to stabilize K+ ion)
o Within the K+ Ion Channel only:
Resolvation Energy ≥ Desolvation Energy
 Na (OH2)6+ = hydration shell (6 H2O needed to stabilize Na+ ion)
o Within the K+ Ion Channel only:
Resolvation Energy <<< Desolvation Energy
 As a K+ ion enters the selectivity filter, repulsive force from the next K+
ion entering the filter will push the first ion through the protein out of
the cell.
 Some distances: selectivity filter is 3A wide, and 12A long. The protein is
about 34A across the membrane, and the bigger opening is 10A wide.
K channel Images.