Proteins - Structure And Function Binding Interactions Flashcards
Definition of a ligand
A ligand is a molecule which binds to a protein at a specific site, often triggering a conformational change that affects the activity of the protein.
Receptor definition
A receptor is the binding partner of a ligand, and is usually involved in signal transduction across membranes
Specific binding definition
Occurs when the binding partner/receptor only interact with specific ligands (biological systems)
Non-specific binding
When the binding partner will interact with a wide range of different ligands, not distinguishing between them
Complex definition
A complex is the combination of a ligand and its binding partner/receptor
Strength of binding definition
Definition of dissociation constant
Definition of equilibrium constant
1) Determined by how much energy is required to separate the binding partner and ligand in the complex.
2)
-A measure of the extent to which a substance is dissociated into ions
-Equal to the product of the concentrations of dissociated ions divided by the concentration of the undissociated substance
3)
-Equilibrium constant expresses the ratio of products to reactants at equilibrium and the extent to which reactants have been converted to products.
-Equal to the product of the product concentrations divided by the concentration of the reactant
Explain the 2 enzyme binding models
1) Lock and Key binding model - the shape of the active site is exactly complementary to the substrate
2) The induced fit binding model - the shape of the active site changes on binding the substrate to give exact complementarity, implying the protein to be flexible and dynamic in shape (more accurate model)
1) Considering the equation A + B <—> C, and using the equilibrium constant, dissociation constant, and your brain, simplify to put in a form of fractional saturation (Y) = a function of [B].
2) By putting the equation in the form Y against a single variable, plot the graph of Y against B
3) State the concentration of B that causes half-saturation
4) Explain mathematically whether a lower or higher dissociation constant would lead to stronger binding
5) Draw Y against log[B]
1) Y = [B] / {Kd + [B]}
2) Hyperbola with asymptote at Y = 1
3) When [B] = Kd = 1/Ke
4) A lower dissociation constant since Y is greater at all concentrations of B
5) sigmoidal curve
Analyse the binding of oxygen to the myoglobin using: Mb + O2 <—> MbO2, Ke, Kd, and Y (fraction of myoglobin saturation). And give the fractional saturation when Kd = [O2]
Plot a graph of fractional myoglobin saturation against Partial Pressure of Oxygen (essentially the concentration but for gases)
1) Y[Mb] = [O2] / Kd + [O2], with Kd = [O2] at half saturation, Y = 0.5
2) hyperbola as usual
What does Kd represent
1) Kd represents the binding affinity of a specific site
2) A higher value of Kd represents more dissociated ions, so lower binding affinity
3) A lower value of Kd represents more undissociated molecules so higher binding affinity
1) How does fractional saturation equation change when there are multiple, q, identical binding sites that do not interact with each other (have the same Kd/binding affinities)
2) draw the hyperbola of Y against [B] and log[B] respectively
1) The equation becomes q[B] / {Kd + [B]}
2) This gives a similar hyperbola with higher values of Y, but a sigmoidal curve for log[B]
Draw diagrams for when there are multiple binding sites, q, that are not identical and do not interact (different Kd and binding affinities)
1) we get a superposition of binding curves with almost 2 sigmoid curves ajdacent to each other
Explain Positive and Negative Cooperativity between different binding sites that interact
1) Positive cooperativity is an allosteric process which occurs when the binding of one ligand increases the affinity of the other binding site to the other ligand (Decreases Kd2, increases affinity and thus increases Y)
2) Negative cooperativity is an allosteric process which occurs when the binding of one ligand decreases the affinity of the other binding site to the other ligand
How does cooperativity and allostery affect the free energy of the products of the 2 step reaction.
Explain the differences between direct and indirect cooperativity
Positive cooperativity lowers the free energy of the product, and therefore favours indirect or direct positive cooperative processes between two ligands which lead to this lower energy
Direct cooperativity usually involves a ligand binding, giving the intermediate a higher affinity to the second ligand
Indirect cooperativity involves a ligand binding, stimulating a separate molecule to stimulate the increase in affinity for the second ligand.
How can positive or negative cooperativity be determined experimentally of a A+B <—> C and C+D <—> E. Given the ligands for this reaction are B and D.
The hill equation can be used to determine this. Y = [L]^n / Kd + [L]^n where n is the minimum number of ligand binding sites. By plotting the Y against [L], if n is larger than 1, then we assume that ligand cooperativity has lead to the increase in affinity of a receptor to the ligand, and has thus lead to higher fractional saturation, Y (still less than or equal to 1)
Compare the structure, location in the body, and the properties of myoglobin and haemoglobin
- Myoglobin is a monomer found in muscle whereas haemoglobin is a tetramer found in blood
- Myoglobin and haemoglobin both have similar high fractional saturations at lung oxygen concentrations but at tissue oxygen concentrations, haemoglobin has a much lower fractional saturation since it unloads oxygen more efficiently to tissues whereas myoglobin is more of an oxygen store
Explain the cooperativity in haemoglobin
- When oxgyen binds to haemoglobin, it pulls Fe2+ ion upwards, leading to a conformational change of the subunit, triggering the other subunits to have a higher affinity to oxygen
- Deoxyhemoglobin exists in the T state but when oxygen binds, positive cooperativity leads to the favouring of the R, relaxed state
Define the three forms of an enzyme and a prosthetic group
- An inactive enzyme is called an apoenzyme
- An active enzyme together with a metal/coenzyme is called a holoenzyme
- An active enzyme together with a metal/coenzyme and substrate is called a complex
Prosthetic group is a non-removable coenzyme
Explain the difference between a cofactor, coenzyme and prosthetic group
Cofactors bind allosterically to an enzyme to modulate the enzyme active site whereas a coenzyme binds to the active site and interacts with substrates .
A prosthetic group is a permanent coenzyme
How do enzymes lower the activation energy of a reaction
Enzymes lowers the free energy of the transition state by forming more favourable interactions with it via hydrogen bonding and electrostatic interactions
Derive the Michaelis Menten equation and explain its assumptions
E + S <—> ES —> E + P
Assumptions:
1. steady state = the rate of formation of the ES is equal to the rate of dissociation of ES (the change in concentration of ES over time = 0
2. Given V0 = k2[ES], we assume that Vmax (maximum rate of enz. Catalysis) = k2[E] since Vmax occurs when [ES] = [E] and there are no free enzymes not undergoing catalysis
Maths:
Make the steady state assumption
Formulate an equation based off the concentration of ES
Substitute a nasty bunch of rate constants to be the michaelis constant
Substitute assumption number 2 into the simplified equation
V0 = Vmax ([S] / Km + [S])
Explain what the michaelis menten equation tells us when we know the concentration of the substrate
- When [S] «_space;Km we can assume that [S] + Km = Km, meaning initial rate is proportional to the [S] (left in the numerator) - this makes sense since the concentration of substrate is limiting the rate of reaction
- When [S]»_space; Km we assume that [S] + Km = [S], and since [S] cancels out the rate of enzyme catalysis does not depend on [S] when in excess, which makes sense because the number of active sites/enzyme activity are the limiting factor
- At constant [S], initial rate, Vo, is linearly dependent on [E] which makes sense because the only thing that can alter rate of reaction when substrate concentrations aren’t changing is the number of active sites catalysing the [S]
How is Km related to Kd
Both Km and Kd represent the dissociation constant for the ES complex.
Kd = [A][B]/[P]
Km shows the ratio between the rates of dissociation and the rate of formation:
Km = k-1 + k2 / k1
High values of Kd and Km both indicate dissociated reactants/products
Explain how the line weaver-Burk plot works
- We invert the Michaelis Menten equation and plot 1/V against [S]+Km / (Vmax*[S])
- This simplifies to 1/V = 1/Vmax + (Km/Vmax)(1/[S]) which is analogous to y = mx + c
- Data for this plot is generated by measuring initial rate at different substrate concentrations
- When 1/[S] = 0, [S] tends towards infinity and V=Vmax or y=1/Vmax
- When 1/V = 0, when rate tends towards infinity, [S] = -Km or x=-1/Km
- Michaelis constant = (k-1 + k2)/ k1
- basically shows the ratio of dissociation of ES / association of ES and is analogous to Kd when we ignore k2 —> k-1/k1
Define turnover number and relate it to Vmax
- The number of substrates that a molecule of enzyme converts to product per unit time when the enzyme is fully saturated
- So the maximum rate of reaction occurs when Vmax = kcat[E]
- So kcat = Vmax/[E]
- Kcat normally range from 1<kcat<10^6 s^-1
How can we measure the efficiency of enzyme catalysed reaction
Kcat / Km = specificity constant
Is equal to k2k1 / k-1 + k2 which shows the ratio between the forwards reaction and dissociation reactions
A high specificity constant would reflect high enzyme efficiency at processing a substrate
Used to compare how well an enzyme metabolises different competing substrates
What is the effect of adding inhibitors on rate of catalysis
Contrast reversible and irreversible inhibitors
- Competitive inhibition has an effect of increasing Km, lowering substrate affinity at low [S] concentrations but effects are countered at high [S]. Vmax is the same.
- However in non-competitive Km stays the same but Vmax falls
- In uncompetitive inhibition the inhibitor strengthens binding to substrate, so Km falls but also Vmax falls since the rate of catalysis is hindered by the strength of the Enzyme-substrate attachment. Inhibition further increases with increasing [S]
- reversible inhibitors bind to enzyme via non-covalent interactions and can dissociate, restoring enzyme activity whereas irreversible inhibitors bind via covalent bonds, causing enzyme to lose activity
Draw graphs showing the different types of inhibition including competitive, non-competitive and uninhibited showing Vmax against [S] including Km labels
Relate Km to the rate of reaction
- Km is the equivalent of Kd ~ k-1/k1 but Km INFLUENCES rate of a reaction whereas Kd is a descriptor of binding dynamics
- When Km is above 1 it means that k-1 is higher than k1 and Km is larger thus rate of reaction is lower
- When km is = [S], rate of reaction is half that of Vmax since the rate of dissociation is equal to the rate of substrate binding (catalysis)
Why do enzymes with interacting binding sites not follow Michaelis Menten kinetics
Which other enzymes don’t follow MM kinetics
- Reaction velocity against substrate conc graphs of allosteric enzymes show sigmoidal curves rather than hyperbolic curves
- Cooperative enzymes have a variable Km since they switch from low affinity T forms to high affinity R forms when substrate conc increases
- Enzymes such as chymotrypsin do not conform to MM kinetics and must be analysed via pre-steady state analysis i.e. measuring within milliseconds of the reaction occurring
What is the purpose of MIchaelis Menten kinetics
Can be used to find Km and Kcat values of an enzyme, which shows the functional properties and roles of a specific enzyme
Give the inhibition Michaelis constants and Vmax for competitive, non-competitive and uncompetitive inhibitors, relating
Competitive: KMapp = Km{1+[I] / KI }
Non competitive: Km is unaffected as active site is not hindered by inhibitor but VMAXapp =
Vmax / {1+[I]/KI}
Uncompetitive: lowers Km and Vmax
Draw 1/V against 1/[S] for each of these remembering that gradient = Km/Vmax
