Proteins Flashcards

You may prefer our related Brainscape-certified flashcards:
1
Q

What is the general structure of an amino acid?

A
  • COOH carboxyl group
  • R variable side group consists of carbon chain & may include other functional groups e.g. benzene ring or -OH (alcohol)
  • 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

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

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
Differ only by 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 bonds & eliminates molecule of water
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

How many levels of protein structure 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 & 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 form between O (slightly -ve) attached to -C=O & H (slightly +ve) 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 same side of protein chain
- spiral shape
- H-bonds parallel to helical axis
B-pleated sheet
- N-H & C=O groups 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
- disulphide 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
  • Disulphide bridges: strong covalent S-S bonds between molecules of the amino acid cysteine
  • Ionic bonds: relatively strong bonds between charged R groups (p changes cause these bonds to 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 bonds 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 & 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 structure 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 acids in a mixture.

A
  1. Use the capillary tube to spot mixture onto pencil origin line & place chromatography paper in solvent.
  2. Allow solvent to run until it almost touches the other end of the 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 & 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 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 model of enzyme action.

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

How have models of enzyme action changed?

A
  • Initially lock & 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 free energy of substrate & peak of curve.

19
Q

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

A
  • Enzyme concentrations
  • Substrate concentrations
  • Concentration of inhibitors
  • pH
  • Temperature
20
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.

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 & peaks at optimum temperature.
Above optimum, ionic and H-bonds in 3-degree structure break: active site is 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 H-bonds & ionic bonds in 3 degree structure = denaturation.

24
Q

Contrast competitive and non-competitive inhibitors.

A

Competitive:
- similar shape to substrates = bind to active site
- do not stop reaction; ES complex forms when inhibitor is released
- increasing substrate concentration decreases their effect
Non-competitive:
- bind at allosteric binding site
- may permanently stop reaction; triggers active site to change shape
- increasing substrate concentration has no impact on their effect

25
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
26
Q

Outline how to calculate rate of reaction from raw data.

A

Change in concentration of product or reactant / time.

27
Q

Why is it advantageous to calculate initial rate?

A

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