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A LEVEL BIOLOGY > Biological Molecules > Flashcards

Biological Molecules Flashcards

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1
Q

what is the general basic structure of an amino acid

A
  • COOH (carboxyl group)
  • A R variable side group
  • NH₂ amine / amino group
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2
Q

How many naturally occurring amino acids are there and how are they distinguished ?

A
  • 20
  • differ only by side ‘R’ group
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3
Q

Describe how a peptide bond is formed between two amino acids to form a dipeptide (2)

A
  • Peptide bonds are formed from a condensation reaction
  • between the amine and carboxyl groups (or NH2 and COOH)
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4
Q

What is a polypeptide and how is it formed

A
  • condensation reaction forms a peptide bond between 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

Describe the ‘primary’ structure of a protein

A
  • linear sequence , number and type of amino acids in the polypeptide
  • determined by sequence of codons on mRNA
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7
Q

The secondary structure of a polypeptide is produced by bonds between amino acids. Describe how.
[2 marks]

A
  • Hydrogen bonds form between the slightly positive NH group on one amino acid and the slightly negative C=O group on another.
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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 another

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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 bonds
- ionic bonds
- 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 bond: relatively strong bonds between charged R groups (pH changes cause these to break
hydrogen bonds: numerous & easily broken

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11
Q

Define quaternary structure of a protein

A
  • functional proteins which consist of more than one polypeptide
  • may involve the addition of prosthetic groups (moieties)
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12
Q

Describe how to test for proteins in a sample.

A

Biuret Test confirms presence of peptide bond
1. Add equal volumes of sodium hydroxide to sample at room temperatures
2. Add drops of dilute copper (||) sulfate solution . Swirl to mix
3. Positive result: colour changes from blue to purple
negative result: solution remains blue

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13
Q

What are enzymes?

A
  • biological catalysts for intra & extracellular reactions
  • specific tertiary structure determines shape of active site, complementary to a specific sustrate
  • formation of (ES) complexes lowers activation energy of metabolic reactions
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14
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 substrate bonds , lowering activation energy
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15
Q

how have models of enzyme action changed ?

A
  1. Lock and key method : rigid shape of active site complementary to only use 1 substrate
  2. Current induced fit model : explains why binding at allosteric sites can change shape of active site
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16
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 and haemoglobin
17
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
18
Q

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

A
  • Use capillary tube to spot mixture onto pencil origin line & place chromatography paper in solvent.
  • Amino acids move different distances during the mobile phase based on relative attraction to paper & solubility in solvent.
  • Remove paper when the solvent has nearly reached the top and draw a line to show where the solvent has reached, this is the SOLVENT FRONT.
  • The chromatogram is dried in a fumes cupboard and then sprayed with NINHYDRIN SPRAY. When the chromatogram dries, the amino acids will appear as purple spots at different distances up the chromatogram.
  • The unknown amino acid(s) can then be identified by comparing and matching them with the chromatograms of the known standard solutions of different amino acids
  • Calculate Rf values
19
Q

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

A
  • difference between free energy of substrate & peak of curve
20
Q

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

A
  • enzyme concentration
  • substrate concentration
  • concentration of inhibitors
  • pH
  • temperature
21
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 complexed form at any given time
22
Q

How does enzyme concentration affect rate of reaction?

A
  • given that substrate is in excess rate increases proportionally to enzyme concentration
  • rate levels off when maximum number of E complexes form at any given time
23
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 tertiary structure break.
  • active site is no longer complementary to substrate ( denaturation)
24
Q

How does pH affect rate of reaction?

A
  • Below and above the optimum pH of an enzyme, solutions with an excess of H+ ions (acidic solutions) and OH- ions (alkaline solutions) can cause H+ and ionic bonds to break
  • This alters the shape of the active site, which means enzyme-substrate complexes form less easily.
  • denaturation of the enzyme occurs
25
Q

Explain how competitive inhibitor reduce the rate of an enzyme catalysed reaction (3)

A
  • Inhibitor is a similar shape to substrate
  • This binds to the active site
  • Thus preventing an enzyme substrate complex from Forming
26
Q

Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction. (3)

A
  • Attaches to the enzyme at a site other than the active site ( the allosteric site)
  • Changes (shape of) the active site
  • (So active site and substrate) no longer complementary so
    no substrate can bind.
27
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
28
Q

outline how to calculate rate of reaction from raw data

A
  • change in concentration of product or reactant/time
29
Q

why is it advantageous to calculate initial rate?

A
  • represents max rate of reaction before concentration of reactants decreases &’ end product inhibition’
30
Q

state the formula for pH.

A

pH = −log10 ([H+])

H+ = hydrogen ion concentration of a solution

A LEVEL BIOLOGY (9 decks)

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