1.4-Proteins + 1.5-Enzyme Action Flashcards

1
Q

What is the general structure of of an amino acid?

A

COOH - carboxylic acid group.
R group - consists of carbon chains and other functional groups.
NH2 - Amino Group

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Describe how to test for proteins in a sample.

A

The Biuret test confirms presence of peptide bonds.
1. Add an equal volume of sodium hydroxide to sample at room temperature.
2. Add drops of silute copper (II) sulfate solution. Swirl to mix.
3. Positive result: colour changes from blue to purple.
4. Negative result: solution remains blue.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

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

A

20
They only differ by side ‘R’ group.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

How do dipeptides and polypeptides form?

A
  • Condensation reaction forms peptide binds and eliminates molecule of water.
  • Dipeptide: 2 amino acids
  • Polypeptides: 3 or more amino acids.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

How many levels of protein structures are there?

A

4

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Define ‘primary structure’ of a protein.

A
  • Sequence, number ad type of amino acids in the polypeptide.
  • Determined by sequence of codons on mRNA.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Define ‘secondary structure’ of a protein.

A

Hydrogen bonds from between O δ- (slightly negative) attached to -C=O & & H δ+ (slightly positive) attached to -NH.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Describe the 2 types of secondary protein structure.

A

a-helix:
- all N-H bonds on the same side of protein chain.
- spiral shape.
- H-bonds parallel to helical axis
B-pleated sheet:
- N-H & C=O alternate from one side to the other.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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
- Hydrogen bonds

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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 break)
  • Hydrogen bonds: numerous & easily broken.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Describe the structure and function of globular proteins.

A
  • Spherical and compact
  • Hydrophilic R groups face outwards & hydrophobic R groups face inwards = usually water soluble.
  • Involved in metabolic processes e.g. enzymes & haemoglobin.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Describe the stucture and function of fibrous proteins.

A
  • Can form long chains of fibres
  • Insoluble in water
  • Useful for structure and support e.g. collagen in skin
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Outline how chromatography could be used to identify the amino acides in a mixure.

A
  1. Use capillary tube to spot mixture onto pencil origin line & place chromatography paper in solvent.
  2. Allow solvent to run until it almost touches other end of paper. Amino acids move different distances based on relative attraction to paper & solubility in solvent.
  3. Use revealing agent or UV light to see spots.
  4. Calculate the Rf values and match to database.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

What are enzymes?

A
  • Biological catalysts for intra & 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.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Explain the induced fit modle 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 substrate bonds, lowering the activation energy.
17
Q

How have models of enzyme action changed?

A
  • Initially lock and key model: rigid shape of active site complementary to only 1 substrate.
  • Currently induces fit model: also explains why binding at allosteric sites can chage 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 free energy of substrate & 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
19
Q

How does substrate concentration affect rate of reaction?

A
  • Given that enzyme concentration is fixed, rate increases proportionally to substrate concentration.
  • Rate levels off when maximum number of ES complexes form at any given time.
20
Q

How does enzyme concentration affect rate of reaction?

A
  • Given that substrate is increases proportionally to enzyme contentration.
  • Rate levels off when maximum number of ES complexes form at any given time.
21
Q

How does temperature affect rate of reaction?

A
  • Rate increases as kinetic energy increases & peaks at optimum temperature.
  • Above optimum, ionic & H-bonds in 3° structure break = active site no longer complementary to substrate (denature).
22
Q

How does pH affect rate of reaction?

A
  • Enzymes have a narrow optimum pH range.
  • Outside range, H+/OH- ions interact with H-bonds and ionic bonds in 3° structure = denature.
23
Q

Contrast competitive and now competitive inhibitors.

A

Competitive Inhibitors:
- Similar shape to substrate = bind to active site.
- Do not stop reaction; ES complex forms when inhibitor is released.
- Increasing substrate concentration decreases their effect.
Non-competitive Inhibitors:
- Bind at allosteric binding site.
- May permanently stop reaction; triggers active site to change shape.
- Increasing substrate concentration has no impact on their effect.

24
Q

Outline how to calculate rate of reaction from a graph.

A
  • Calculate gradient of line or gradient of tangent to a point.
  • Initial rate: draw tangent at t=0.
25
Q

Outline how to calculate rate of reaction from raw data.

A

Change in concentration of product or reactant / time.

26
Q

Why is it an advantage to calculate initial rate?

A

Represents maximum rate of reaction before concentration of reactants decreases & ‘end product inhibition’.

27
Q

Formula of pH

A

pH = -log₁₀ [H⁺].