3.1.4 Proteins Flashcards

1
Q

What is the general structure of an amino acid?

A

-COOH carboxyl/carboxylic acid group
-R variable side group consists of carbon chain and may include other functional groups e.g. benzene ring or -OH (alcohol)
-NH₂ amine/amino group

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2
Q

Describe how to test for proteins in a sample.

A

Biuret test confirms presence of peptide bond.
1. Add equal volume of sodium hydroxide to sample at room temperature.
2. Add drops of dilute copper (II) sulfate solution. Swirl to mix.
3. Positive result: colour change from blue to purple.

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3
Q

How many naturally occurring amino acids are there and how do they differ from one another?

A

20
They have different R groups

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4
Q

How do dipeptides and polypeptides form?

A
  • Condensation reaction forms peptide bond (-CONH-) and eliminates a molecule of water
  • Dipeptide: 2 amino acids
  • Polypeptide: 3 or more amino acids
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5
Q

How many levels of protein structure are there?

A

4

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6
Q

Define primary structure of a protein.

A
  • Sequence, number and type of amino acids in the polypeptide.
  • Determined by sequence of codons on mRNA.
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7
Q

Define secondary structure of a protein.

A

Hydrogen bonds form between oxygen with a partial negative charge attached to -C=O and hydrogen with a partial positive charge attached to -NH

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8
Q

Describe the 2 types of secondary protein structure.

A

α-helix:
- all N-H bonds on the same side of the protein chain
- spiral shape
- hydrogen bonds parallel to helical axis
β-pleated sheet:
- N-H and C=O groups alternate from one side to the other

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9
Q

Define tertiary structure of a protein. Name the bonds present.

A

3D structure formed by further folding of polypeptide.
Disulfide bridges, ionic bonds and hydrogen bonds

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10
Q

Describe each type of bond in the tertiary structure of proteins.

A
  • Disulfide bridges: strong covalent S-S bonds between molecules of the amino acid cysteine
  • Ionic bonds: relatively strong bonds between charged R groups (pH changes cause these bonds to break)
  • Hydrogen bonds: numerous and easily broken
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11
Q

Define quaternary structure of a protein.

A
  • Functional proteins may consist of more than one polypeptide.
  • Precise 3D structure held together by the same types of bond as tertiary structure.
  • May involve addition of prosthetic groups e.g metal ions or phosphate groups
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12
Q

Describe the structure and function of globular proteins.

A
  • Spherical and compact.
  • Hydrophilic R groups face outwards and hydrophobic R groups face inwards = usually water soluble.
  • Involved in metabolic processes e.g. enzymes and haemoglobin.
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13
Q

Describe the structure and function of fibrous proteins.

A
  • Can form long chains or fibres
  • Insoluble in water
  • Useful for structure and support e.g. collagen in skin.
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14
Q

Outline how chromatography could be used to identify the amino acids in a mixture.

A
  1. Use capillary tube to spot mixture onto pencil origin line and place chromatography paper in solvent.
  2. Allow solvent to run until it almost touches the other end of paper. Amino acids move different distances based on relative attraction to paper and solubility in solvent.
  3. Use revealing agent or UV light to see spots.
  4. Calculate Rf values and match to database.
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15
Q

What are enzymes?

A
  • Biological catalysts for intra and extracellular reactions
  • Specific tertiary structure determines shape of active site, complementary to a specific substrate
  • Formation of enzyme-substrate (ES) complexes lowers activation energy of metabolic reactions.
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16
Q

Explain the induced fit model of enzyme action.

A
  • Shape of active site is not directly complementary to substrate and is flexible
  • Conformational change enables ES complexes to form
  • This puts strain on the substrate bonds, lowering the activation energy
17
Q

How have models of enzyme action changes?

A
  • Initially lock and key model: rigid shape of active site complementary to only 1 substrate
  • Currently induced fit model: also explains why binding at allosteric sites can change shape of active site
18
Q

How could a student identify the activation energy of a metabolic reaction from an energy level diagram?

A

Difference between the free energy of substrate and peak of curve

19
Q

Name 5 factors that affect the rate of enzyme-controlled reactions.

A
  • enzyme concentration
  • substrate concentration
  • concentration of inhibitors
  • pH
  • temperature
20
Q

How does substrate concentration affect rate of reaction?

A

Given that enzyme concentration is fixed, rate of reaction increase proportionally to substrate concentration.
Rate levels off when maximum number of ES complexes form at any given time.

21
Q

How does enzyme concentration affect rate of reaction?

A

Given that the substrate is in excess, rate increases proportionally to enzyme concentration.
Rate levels off when maximum number of ES complexes form at any given time.

22
Q

How does temperature affect rate of reaction?

A

Rate increases as kinetic energy increases and peaks at optimum temperature.
Above optimum, ionic and hydrogen bonds in tertiary structure break = active site no longer complementary to substrate (denaturation).

23
Q

How does pH affect rate of reaction?

A

Enzymes have a narrow optimum pH range.
Outside range, H⁺/OH⁻ ions interact with hydrogen bonds and ionic bonds in tertiary structure = denaturation.

24
Q

Contrast competitive and non-competitive inhibitors.

A

Competitive inhibitors
- Similar in shape to substrate = bind to active site
- Do not stop reaction as ES complex can form when inhibitor is released
- Increasing substrate concentration decreases their effect
Non-competitive inhibitors
- Bind at allosteric binding site
- May permanently stop reaction as triggers active site to change shape
- Increasing substrate concentration has no impact on their effect

25
Outline how to calculate rate of reaction from a graph.
- Calculate gradient of line or gradient of tangent to a point - Initial rate: draw tangent at t=0
26
Outline how to calculate rate of reaction from raw data.
Change in concentration of product or reactant divided by time.
27
Why is it advantageous to calculate initial rate?
Represents maximum rate of reaction before concentration of reactants decreases and 'end product inhibition'.
28
State the formula for pH.
pH = -log₁₀[H⁺]