T1: Biological Molecules - Proteins Flashcards

(35 cards)

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
  • a condensation reaction
  • removing a water molecule
  • between the amine and carboxyl groups (or NH2 and COOH)
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4
Q

What are dipeptides and polypeptides and how is it formed

A
  • dipeptides: 2 amino acids joined together
  • polypeptides: 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, of amino acids joined by peptide bonds in the polypeptide chain
  • 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 secondary protein structures

A
  • folding of polypeptide chains /eg alpha helix / beta pleated sheets
  • due to H bonds between amino acids
  • between the amine group of onne amino acid and the carboxyl of 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, due to interactions between R groups
  • disulfide bonds
  • ionic bonds
  • hydrogen bonds
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10
Q

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

A
  1. Hydrogen bonds – between polar R-groups.
  2. Ionic bonds – between positively and negatively charged R-groups.
  3. Disulfide bridges – strong covalent bonds between sulfur atoms in two cysteine amino acids.
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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 the structure of a protein depends on the amino acids it contains. (5)

A
  1. Structure is determined by (relative) position of amino acid/R group/interactions;
  2. Primary structure is sequence/order of amino acids;
  3. Secondary structure formed by hydrogen bonding (between amino acids);
  4. Tertiary structure formed by interactions (between R groups);
  5. Creates active site in enzymes OR Creates complementary/specific shapes in antibodies/carrier
    proteins/receptor (molecules);
  6. Quaternary structure contains >1 polypeptide chain
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13
Q

Describe how to test for proteins in a sample.

A
  1. Add biuret reagent (sodium hydroxide + copper)
  2. Positive result: colour changes from blue to purple/lilac
    negative result: solution remains blue
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14
Q

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

A
  1. Prepare the chromatography paper by drawing a pencil line near the bottom and placing small spots of the amino acid mixture and known amino acid standards along the line.
  2. Place paper in solvent (e.g., ethanol), ensuring the solvent level is below the spots. Leave it to develop in a covered container.
  3. As the solvent travels, different amino acids move at different rates depending on their solubility
  4. When the solvent has nearly reached the top, remove the paper and mark the solvent front. Dry the paper.
  5. Spray with a locating agent (e.g., ninhydrin) to make the amino acid spots visible.
  6. Compare the positions of the unknown spots with those of the known standards or calculate Rf values and match them to known Rf values.
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15
Q

How do enzymes speed up reactions?

describe graph

A
  • biological catalysts which lower activation energy of reaction.
  • in both intracellular and extracellular reactions.
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16
Q

Explain how the active site of an enzyme causes a high rate of reaction. (3)

A
  1. Lowers activation energy;
  2. Induced fit causes active site (of enzyme) to change shape;
  3. (So) enzyme-substrate complex causes bonds to form/break;
17
Q

Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst. (3)

A
  1. Substrate binds to the active site/enzyme
  2. Active site changes shape (slightly) so it is complementary to substrate
  3. Enzyme-substrate complex forms;
  4. so distorting/breaking bonds in the substrate;
  5. Reduces activation energy
18
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
19
Q

Describe the structure and function of globular proteins

A
  • spherical and compact
  • soluble in water
  • hydrophilic R groups face outwards & hydrophobic R groups face inwards
  • involved in metabolic processes e.g. enzymes and haemoglobin
20
Q

Describe the structure and function of fibrous proteins

A
  • can form long chains or fibres
  • insoluble in water
  • cross-linkages due to hydrogen bonds
  • useful for structure and support e.g. collagen in skin
21
Q

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

A
  • difference between free energy of substrate & peak of curve
22
Q

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

A
  • enzyme concentration
  • substrate concentration
  • concentration of inhibitors
  • pH
  • temperature
23
Q

How does substrate concentration affect rate of reaction?

A
  • As substrate conc. increases, more E-S complexes form.
  • Rate of reaction increases
  • At a certain point, rate of reaction levels off/ plateaus.
  • As all active sites saturated / occupied (at a given time)
24
Q

How does enzyme concentration affect rate of reaction?

A
  • As enzyme conc. increases, rate of reaction increases
  • More enzymes so more available active sites
  • So more enzyme-substrate (E-S) complexes form
  • At a certain point, rate of reaction stops increasing / levels off
  • Substrate conc. = limiting factor (all substrates in use)
25
how does temperature affect rate of reaction?
- As temp. increases up to optimum, rate of reaction increases as more kinetic energy - So more E-S complexes form - As temp. increases above optimum, rate of reaction decreases - Enzymes denature - tertiary structure and active site change shape - As hydrogen / ionic bonds break - So active site no longer complementary - So fewer E-S complexes form
26
How does pH affect rate of reaction?
- As pH increases / decreases above / below an optimum, rate of reaction decreases - Enzymes denature - tertiary structure and active site change shape - As hydrogen / ionic bonds break - So active site no longer complementary - So fewer E-S complexes form
27
A competitive inhibitor decreases the rate of an enzyme-controlled reaction. Explain how. (3)
- Inhibitor is a similar shape to substrate - This binds to the active site - Thus preventing an enzyme substrate complex from Forming
28
Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction. (3)
- 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.
29
RP1 ## Footnote RP1: investigation into the effect of a named variable on rate of enzyme reactions
method - youtube
30
# Common questions 1-4: 1. Describe how temperature can be controlled. 2. Describe how pH can be controlled. 3. Why were the enzyme & substrate solutions left in the water bath for 10 mins before mixing? 4. Describe a control experiment. ## Footnote RP1: investigation into the effect of a named variable on rate of enzyme reactions
1. Use a thermostatically controlled water bath + Monitor using a thermometer at regular intervals and add hot / cold water if temperature fluctuates 2. Use a buffer solution+ Monitor using a pH meter at regular intervals 3. So solutions equilibrate / reach the temperature of the water bath 4. Use denatured enzymes (eg. by boiling) 5. Everything else same as experiment, eg. same conc. / volume of substrate (at start) and enzyme, same type / volume of buffer solution, same temperature
31
# common questions 1. Suggest a safety risk and explain how to reduce this risk. 2. Explain why using a colorimeter to measure colour change is better than comparison to colour standards. 3. Explain a procedure that could be used to stop each reaction.
1. Handling enzymes may cause an allergic reaction. Avoid contact with skin by wearing gloves and eye protection 2. Not subjective + More accurate 3. Boil / add strong acid / alkali -+ denature enzyme. Put in ice - lower kinetic energy so no E-S complexes form. Add high concentration of inhibitor -+ no E-S complexes form
32
Describe how processed data can be presented as a graph
. Independent variable on x axis, rate of reaction on y axis, including units . Linear number sequence on axis, appropriate scale (graph should cover at least half of grid) . Plot coordinates accurately as crosses . Join point to point with straight lines if cannot be certain of intermediate values OR draw a smooth curve but do not extrapolate
33
Explain why the rate of reaction decreases over time throughout each experiment
1. Initial rate is highest as substrate concentration not limiting / many E-S complexes form 2. Reaction slows as substrate used up and often stops as there is no substrate left
34
Describe how the rate of an enzyme-controlled reaction can be measured
1. Measure time taken for reaction to reach a set point, 2. Rate of reaction = 1/ time; example units = s"1 3. Measure concentration / volume / mass / colour of substrate or product at regular intervals throughout reaction 4. Plot on a graph with time on the x axis and whatever is being measured on the y axis 5. Draw a tangent at t = 0 (or any other time for rate at a particular point) 6. Initial rate of reaction = change in y / change in x, example units = cm3s -!
35
state the formula for pH.
pH = −log10 ([H+]) | H+ = hydrogen ion concentration of a solution