BCHM 312: Exam 3 Content - Enzymes and Carbs

2 types of ligand and enzyme interactions

1. molecular structure of ligand is unchanged after binding with enzyme

2. molecular structure of ligand is changed after binding with enzyme

How do enzymes act as a catalyst?

Increases reaction rate by decreasing activation energy of reaction

Enzyme Selectivity

3D structure of active site is specific for substrate geometry

Do enzymes affect kinetics or thermodynamics?

They only affect kinetics due to only affecting reaction rates, not the equilibrium concentration

What occurs at transition state?

Bonds are broken or made; least stable chemical species, so it decays into substrate or product

Carbonic Anhydrase Chemical Mechanism

V0 (initial velocity)

• as [substrate] increases, initial velocity increases (1st order kinetics; linear increase)

• at low [s], indicates 1st order kinetics

• at high [s] (aka approaches Vmax), indicates 0th order kinetics due to unchanging velocity

Saturation Effect

all enzymes are bound to a substrate, hence being saturated; increasing [s] has no more effect

Equilibrium Assumptions

1. The rate of the overall reaction depends on step 2 (aka the rate-limiting step; k2 < k1)

2. [Product] is small, so conversion back to ES is unlikely

3. Steady-State Assumption

• [E] decreases, [ES] formation and breakdown rate is equal

Km

[substrate] when reaction velocity is half of its Vmax

Michaelis-Menten Eqtn

V0 = ((Vmax)([S])/(Km+[S])

Smaller Km value

higher enzyme-substrate affinity

Vmax eqtn

Vmax = K₂[E_t]

Kcat

• denotes efficiency of enzyme (how many substrates are multiplied into products)

• k2 = kcat

Catalytic Coefficient of Enzyme

Kcat / Km

Lineweaver-Burk Plot eqtn

(1/V₀) = (K_m/V_max)(1/[S]) + (1/V_max)

y = m x + b

Calculating Km and Vmax from LB Plots

1. Set eqtn of line equal to 0

2. solve for x

3. km = 1/x

4. to find Vmax, divide Km from slope of line

Types of Enzyme Inhibition

1. Competitive

2. Uncompetitive

3. Noncompetitive

4. Mixed

Reversible Inhibition

• can bind and unbind from enzyme active site

• able to bind/unbind due to IMFs and steric complementarity to active site

• affects

  • Km

  • Vmax

Competitive Inhibition

• type of reversible inhibitor

• competes with substrate for active site directly

• effects

  •  km increased by α

    • α = 1+ ([I]/k_I)

  • V_max unaffected due to direct competition and can be changed when [S] > [I]

• LB plots: lines change in slope, same y-int

For a competitive inhibitor, what would happen to the LB plot eqtn if [I] increased by 3?

Since [I] increases, α increases, Km increases, which increases the slope of the LB plot. Y-int (aka vmax) is unchanged

Uncompetitive Inhibition

• inhibitors bind to ES complex, not the active site

• allosteric

• effects

  • Km dec by α’ (Km / α)

  • Vmax dec by α’ (Vmax / α)

    • α’ = 1 + ([I] / (1/k_I))

• creates ESI complex, preventing products from forming

• Vmax decreases because total enzymes are being removed

• Km decreases due to now Vmax reduced by α, less substrate needed to reach half max velocity

• LB plots: lines change in left or right shift (parallel lines)

Mixed Inhibition

• inhibitors bind to a site that is not the active site

• also binds to either free enzymes and ES complexes

• effects

  • Vmax dec by α’

    • Vmax / α’

  • Km inc or dec by α’

    • (α)(Km) / (α’)

• LB plot

  • y-int: 1/vmax

  • x-int: (-1/km)

  • intersect in the second quadrant of LB plot

Noncompetitive Inhibition

• do not bind to the enzyme active site

• effect

  • Vmax dec by α’

  • Km unaffected because α = α’

• LB plot

  • y-int: 1/Vmax

  • x-int: same

  • intersects at x-int, but different y-int

Carbohydrates

• units: monosaccharides

• carb digestion is central pathway to make ATP in cells

• Glycolysis: glucose used to make ATP

• Glucogenesis: ATP used to make glucose

Monosaccharides

• contains

  • aldehyde or ketone

    • aldehyde = aldose

    • ketone = ketose

  • 2+ hydroxyl groups

Number of Carbons in Monosaccharides

• Naming

  • tetrose = 4

  • pentrose = 5

  • hexose = 6

• Numbering

  • Aldose

    • C1 is carbonyl carbon

  • Ketose

    • C1 is C on shorter side of ketose

Chiral Isomers

• enantiomers (nonsuperimposable)

• chiral center is found on most distant carbon from carbonyl carbon

• D isomers

  • hydroxyl group is on right side

• L isomers

  • hydroxyl group is on left side

If monosaccharide has 5+ carbon backbone

occur as cyclic structure

• covalent bond between oxygen of hydroxyl and carbonyl carbon

• cyclization creates new chiral center, creating anomers

• differentiates in anomeric carbon (clockwise to O in ring)

• α = Long OH and anomeric OH faces opposite way

• β = Long OH and anomeric OH faces same way

Ring nomenclature

5 membered ring: furanoses

6 membered ring: pyranoses

Monosaccharide Nomenclature Steps

1. Give anomeric configuration (α or β)

2. Give stereoisomer (L or D)

3. Give initial prefix name 

   1. Give name based on ring at end

From Fischer to Ring

1. O in top right corner

2. 1st C in top of Fischer is anomeric carbon (squiggle for no preference in OH configuration)

3. Clockwise for next carbon & place hydroxyls

• Left = Above plane

• Right = Below plane

1. Place terminal CH2OH

Disaccharide Naming Steps

1. Give alpha or beta for first monosaccharide, then stereoisomer.

2. Give original prefix and suffix by number of carbons in ring (furanosyl or pyranosyl)

3.  In parentheses, anomer C# to anomer C# from left to right. 

4. Anomer then stereoisomer.

5. Prefix and suffix by carbons in ring (furanoside or pyranoside).