proteins and enzymes Flashcards

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1
Q

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

A
  1. substrate binds to active site
  2. active site shape changes so it is complementary to substrate
  3. reduces activation energy
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2
Q

suggest what scientists can do to stop enzyme reactions (2)

A
  1. boil solution
  2. denatures the enzyme
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3
Q

how does a competitive inhibitor decrease the rate of an enzyme controlled reaction (3)

A
  1. inhibitor has a similiar shape to substrate
  2. so it binds to active site
  3. prevents enzyme-substrate complex forming
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4
Q

when bread is stale, the starch becomes a competitive inhibitor of amylase in the small intestine, how can eating stale bread reduce weight gain? (3)

A
  1. there is less hydrolysis of starch
  2. into maltose
  3. so less absorption of glucose by the intestine
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5
Q

describe how the structure of a protein depends on the amino acids it contains (5)

A
  1. structure is determined by position of amino acids
  2. primary structure is sequence of amino acids
  3. secondary structure formed by hydrogen bonding
  4. tertiary structure formed by interactions between R groups
  5. this creates active site in enzymes
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6
Q

explain how the active site of an enzymes causes a high rate of reaction (3)

A
  1. reduces activation energy
  2. induced fit causes enzyme active site to change shape
  3. so enzyme-substrate complex causes bonds to form
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7
Q

a dipeptide consists of 2 amino acids joined by a peptide bond.
describe two other ways in which all dipeptides are similar and one way they differ (3)

A
  1. similar: all have amine group at the end
  2. similar: all have carboxyl group at end
  3. difference: have different R groups
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8
Q

describe a biochemical test for a protein (2)

A
  1. add biuret reagent
  2. turns from blue to purple
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9
Q

describe how a non-competitive inhibitor reduces rate of enzyme controlled reaction (3)

A
  1. attaches to the enzyme at allosteric site
  2. changes shape of the active site
  3. active site and substrate no longer complementary so less substrate binds- less enzyme-substrate complexes
    Inhibition cannot be reduced by adding more substrate
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10
Q

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

A
  1. condensation reaction
  2. between amine and carboxyl group
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11
Q

the secondary structure of a polypeptide is produced by bonds between amino acids, describe how. (2)

A
  1. hydrogen bonds
  2. between H in amine group and O in carboxyl group
    forming B-pleated sheets or a- helices
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12
Q

two proteins have the same number and type of amino acids but different tertiary structures, explain why (2)

A
  1. different sequence of amino acids
  2. ionic/hydrogen/disulfide bonds form in different places
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13
Q

how does formation of an enzyme-substrate complex increase rate of reaction? (2)

A
  1. reduces activation energy
  2. due to bending bonds
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14
Q

suggest two control variables when investigating effect of temperature on enzyme reaction (1)

A
  1. enzyme concentration
  2. pH
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15
Q

explain how a change in sequence of DNA bases could result in a non-functional protein (3)

A
  1. changes the amino acid sequence
  2. hydrogen/ ionic bonds form in different places and change tertiary structure
  3. changes active site shape so substrate cannot bind
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16
Q

what is the peptide bond between on 2 amino acids?

A

the C atom of one amino acid and the N atom of another amino acid

17
Q

what is meant by the primary structure of a polypeptide?

A

the sequence of amino acids

18
Q

what is meant by the secondary structure of a polypeptide?

A

formation of hydrogen bonds between positive H in the amine group of one amino acid and negative O in the carboxyl group of another amino acid

forms a-helices and B-pleated sheets

19
Q

what is meant by the tertiary structure of a polypeptide?

A

the 3D folding of the chain due to interactions between R groups on different amino acids

20
Q

what bonds form between R groups within the tertiary structure?

A
  • ionic- between oppositely charged R groups
  • hydrogen- between polar R groups
  • london forces- between non-polar R groups
  • disulfide bridges- between R groups containing sulfur
21
Q

are ionic bonds stronger than disulphide bridges?

A

no, disulfide bridges are stronger and harder to break

ionic bonds are easily broken by changes in pH

22
Q

what determines the 3D structure of a protein

A

the primary structure- sequence of amino acids

23
Q

what is meant by quaternary structure

A

the interaction of two or more folded polypeptides

may contain a prosthetic group e.g haem in haemoglobin

24
Q

what is the biuret reagent made up of

A

sodium hydroxide and dilute copper(II) sulfate

25
Q

describe the differences between a fibrous protein and a globular protein and give an example of each

A

fibrous:
- long chains which run parallel to each other, chains are linked by cross bridges e.g. collagen
- have structural functions
- insoluble

globular:
- spherical shape
- water soluble
e.g. haemoglobin

26
Q

give 3 functions of globular proteins

A

Enzymes- all enzymes are globular proteins as their round shape can be altered to fit their specific active site

Transport proteins- they are soluble so can cross cell membrane. e.g. haemoglobin, which transports oxygen.

Messengers proteins- they are hormones as they are soluble- Regulate the body’s metabolic processes. e.g. insulin

27
Q

give 1 function of fibrous proteins

A

Structural proteins.- they are stable and insoluble- support and protect tissues. e.g. keratin- provides structure to hair and nails
e.g. collagen, a type of connective tissue in the body.

28
Q

why is collagen a strong structural protein

A
  • has hydrogen bonds, lots together are strong
  • Collagen is present as fibres, consist of many collagen fibrils folded around each other.
29
Q

explain why the quaternary structure of collagen makes it a suitable molecule for a tendon

A

has 3 polypeptide chains wound together
forms a strong, rope-like structure that has strength in the direction of the pull of a tendon

30
Q

suggest how the cross linkages between amino acids of polypeptide chains increase the strength and stability of a collagen fibre

A

prevents individual polypeptide chains from sliding past one another

so they gain strength as they act as a single unit

31
Q

the points where 1 collagen molecule ends and another starts is spread throughout the fibre

explain why this arrangement of collagen molecules is necessary for the efficient functioning of a tendon

A

the junctions between adjacent collagen molecules are points of weakness

if they all occurred at the same point, this would be a major weak point and the fibre may break