Enzymes

Question 1

Enzyme Active Site

Answer 1

• Enzymes have specific active site complementary in shape and charge to substrate
• Effective collisions between enzyme and substrate form a temporary enzyme substrate complex (frequency of this affects rate of reaction)
• Contact/binding residues determine enzyme-substrate specificity and bind reversibly with substrate and position them in correct orientation
• Catalytic residues catalyse conversion of substrate to product
• Structural residues maintain overall 3D conformation of protein
• Lock and Key Hypothesis: Enzyme and substrate fit exactly in specific conformation and charge
• Induced Fit Hypothesis: Substrate binding induces conformation change of active site to be more precise fit for substrate

Question 2

How Ea lowered on molecular basis

Answer 2

SPOMA
1. Strains bonds to be broken
2. Proximity in aligning reactants next to each other
3. Orientation exposes reactants to chemical attack
4. Microenvironment favourable for reactants to react
5. Acid-base catalysis

Question 3

Explaining Trend for Enzyme-Catalysed Reactions

Answer 3

1. At t0, concentration of substances is highest with highest frequency of effective collisions
2. Rate of formation of enzyme-substrate complexes is highest
3. As time progresses, less substrate remains = less effective collisions = lower rate of reaction until graph plateaus with no reaction

Question 4

Explain how rate of enzyme reaction varies with one manipulated variable

Answer 4

1. Carry out at least 5 experiments with change in only one independent variable
2. Measure progress of dependent variable over fixed time intervals
3. Find initial rate by finding gradient of tangent at t0

Question 5

Explain how temperature affects rate of reaction

Answer 5

1. At low temps increase in temp leads to increase in KE of enzyme and substrate, increasing frequency of effective collisions and rate of formation of enzyme-substrate complexes and number of molecules with sufficient energy to overcome Ea barrier, increasing reaction rate
2. At optimum temperature, rate of reaction is maximum
3. Beyond optimum temp, intramolecular vibrations increase and break H bonds, ionic bonds and hydrophobic interactions that stabilise tertiary 3D conformation, causing enzyme to denature.
4. Denaturation causes loss of specific functional 3D conformation of active site, such that it’s no longer complementary in shape and charge to substrate

Question 6

Explain how pH affects enzyme reaction rate

Answer 6

1. Each enzyme’s optimal pH is when rate of reaction is maximum, any deviation from its peak decreases the rate of reaction
2. Excess [H+] or [OH-] disrupts ionisation of charged R groups
3. This causes disruption of ionic and hydrogen bonds causing loss of specific functional 3D conformation of active site, such that it’s no longer complementary in shape and charge to substrate, causing denaturation
4. pH may also change the specific charge of the R groups of catalytic residues so catalytic activity may be lost
5. pH may also change specific charge of R groups of contact/binding residues such that substrate cannot bind to enzyme and no enzyme-substrate complex can form

Question 7

Explain how enzyme concentration affects reaction rate

Answer 7

1. At low [E], [E] is limiting. Thus as [E] increases, frequency of effective collisions increases, rate of ES complex formation and reaction increases proportionally
2. At high [E], [E] no longer limiting factor, [S] limiting instead

Question 8

Explain how substrate concentration affects rate of reaction

Answer 8

1. At low [S], [S] is limiting as enzyme active sites are readily available, Thus as [S] increases, frequency of effective collisions increases, rate of ES complex formation and reaction increases proportionally
2. At high [S], [S] no longer limiting factor as all available active sites saturated, [E] limiting instead

Question 9

Michaelis-Menten Curve

Answer 9

• Km is the X variable, the concentration of substrate required to make reaction attain half its max rate or 1/2 Vmax (Y variable)
• Km is always the same for a particular enzyme
• Low Km = high affinity of enzyme to substrate, low [S] needed to attain 1/2 Vmax

Question 10

Competitive Inhibition

Answer 10

1. Similar shape and conformation to substrate, complementary to active site
2. They compete for active site, binding temporarily and reversibly, thus decreasing availability of enzyme active site for substrate binding
3. Inhibition can be overcome by high [S]

Question 11

Non-Competitive Inhibition

Answer 11

1. Bears no structural similarity to active site, binds instead to site other than active site, to alter conformation of active site such that substrate can no longer bind to active site
2. This decreases enzyme availability by forming inactive enzyme-inhibitor complexes, decreasing rate of reaction
3. Cannot be overcome by high [S]

Question 12

Allosteric Regulation

Answer 12

1. Binding of allosteric activator at allosteric site stabilises enzyme into functionally active conformation such that active site has a higher affinity for substrate
2. Binding of allosteric inhibitor at allosteric site stabilises enzyme into functionally inactive conformation such that active site has a lower affinity for substrate. However, this can be overcome by high [S]
3. Binding of substrate exhibits cooperativity with S-shaped sigmoid curve

Common Example is feedback/end-product inhibition, the binding of end product of reaction to original enzyme at allosteric site to prevent cell from wasting resources (eg isoleucine)