B: Proteins Flashcards

Question

Enzymes - Effect of pH

Answer 1

Low pH - **Protonation of NH₂** * changes the shape of the active site --\> prevents formation of ES complex High pH - Denaturisation * **the charge on the R group changes** --\> _responsible for the formation of attractions between the enzyme and substrate_ --\> denaturisation

Answer 2

* Pb, Ag, Hg * **Have a strong affinity for the S-H groups in enzymes**. These groups form the quaternery structure of the enzyme * the shape is altered and the **enzyme can no longer function effectively**

Answer 3

* Shape: long and narrow (**secondary structures**) * Role: structural (**strength and support**) * Solubility: **mostly insoluble** * Sequence: **repetative amino acid sequence** * Stability: **less sensative to changes in pH / temp** * Examples: Collagen, Keratin

Answer 4

* Shape: **rounded / spherical** * Role: functional (**catalyst and transport**) * Solubility: **mostly soluble** * Sequence: **irregular amino acid sequence** * Stability: **more sensative** * Examples: hemoglobin, insulin, catalase

Answer 5

V_max - the point where all the active sites are bound to substrate (enzyme is saturated) * High V_max = fast conversion of substrate per unit tiem

Answer 6

K_m Michaelic Constant - **concentration of substrate when the rate of the reaction has 0.5V_max** * gives indication of how strongly the enzyme is bound to the substrate * **the lower K_m the stronger the bonding between the enzyme and substrate** * K_m does not depend on substrate concentration

Answer 7

* Low temp - molecules of enzyme will have low values of Kinetic E. * High temp - denaturing happens, _enzymes lose its tertiary / quaternary structure_ and will no longer function

Answer 8

* Low pH - protonation of NH₂ --\> NH₃⁺ * High pH - becomes anion --\> ionic bonds in the quaternary structure are disrupted

Answer 9

* React with the S-H bonds preventing the formation of S-S links * can be removed from H₂O supply by host-guest chemistry

Answer 10

* reduces enzyme activity * these substances bind directly and reversibly to the active site without producing products * **they compete with the substrate for the active sites** --\> reduces the numbre of enzyme molecules available to bind with the substrates

Answer 11

* Inhibitor binds reversibly to the enzyme away from the active site at the **allosteric site** --\> **changes the shape of the active site** * The enzyme loses its tertiary structure

Answer 12

* Chromophores * molecules with π electrons ( C=C, C=O, C=-N) * π e^- absorbs energy to excited state; releases energy to return to ground state * wavelength corresponds to wavelength of UV light

Answer 13

A = log₁₀( I₀/I ) = Σlc A = absorbance l = pathlength (1cm) c = concentration moldm^-3 Σ = molar absorptivity constant

Answer 14

1. scan sample to determine the wavelength at which the chromophore absorbs UV 2. make a range of standard solutions of known concentration and measure absorbance at the wavelength above 3. plot the standard curve 4. measure the absorbance of the sample and extrapolate

Answer 15

**weak acid + salt of weak acid** CH₃COOH CH₃COO^-Na⁺ ethanoic acid sodium ethanoate **CH₃COOH \<---\> CH₃COO^- + H⁺** **CH₃COO^-Na⁺ ----\> CH₃COO^- + Na⁺** * adding acid: H⁺ reacts with conjugate base reservoir --\> equilibrium shifts to the left * adding base: OH^- reacts with H⁺ to give H₂O --\> equilibrium shifts to the right, using up acid

Answer 16

**weak base + its salt** NH₃ NH₄Cl ammonia ammonium chloride **NH₃ + H₂O \<---\> NH₄⁺ + OH^-** **NH₄Cl ---\> NH₄⁺ + Cl^-** * adding acid: H⁺reacts with NH_3,moves equilibrium right * adding base: OH^- reacts with NH₄⁺, shifts equilibrium left

Answer 17

**pH = pKa + log([A^-]/[HA])** Ka = [H⁺][A^-] / [HA] [H⁺] = Ka[acid] / [salt]

B: Proteins Flashcards

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