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AS Biology > Cheater 3: Enzymes > Flashcards

Cheater 3: Enzymes Flashcards

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1
Q

Enzymes are

A

Biological catalysts that speed up rate of metabolic reactions

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2
Q

What type of proteins are enzymes

A

Tertiary

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3
Q

Surgically enzymes are _ proteins nature being _ because of _

A

Globular
Soluble
Hydrophilic R groups outside of 3d coil shape

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4
Q

2 types of enzymes, description and examples

A

Intracellular- secreted within cells, eg rubisco atp synthase
Extracellular- outside cell, eg pancreatic lipase, pepsin

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5
Q

What is lock and key hypothesis

A

Both enzyme and substrate are rigid/ don’t change shape

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6
Q

Definition of active site

A

Region, usually depression/cleft in the surface of an enzyme, to which substrate can bind to

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7
Q

Shapes of active site substrate are

A

Complementary, not same!

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8
Q

Explain the lock and key model

A

Random movement of enzyme and substrate brings the substrate into active site. A enzyme-substrate complex-bond-is temporarily formed. R groups of amino acids in active site interact with substrate. This breaks substrate apart. Enzyme-product complex briefly formed before product molecules leave.

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9
Q

What happens to an enzyme after changing substrate to products

A

It is unchanged, ready bind with another substrate

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10
Q

Which theory has replaced lock and key hypothesis

A

Induced fit hypothesis

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11
Q

What is the induced fit hypothesis

A

Both active site-most time and substrate-some time change shape

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12
Q

What is activation energy

A

Extra energy needed by the substrate to be converted into products

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13
Q

2 ways in which activation energy can be provided

A

Heating

Enzymes

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14
Q

How can heating increase rate

A

Increase energy of reactants/substrate

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15
Q

How do enzymes provide activation energy

A

Lower it

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16
Q

Reason enzymes lower Ea and how

A

Temperature can’t always be raised to give activation energy. And hence enzymes decrease it of the reaction which they catalyse.

They hold the substrate in such a way that molecules can react more easily to convert to products

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17
Q

Which enzyme is involved in basic H2O2 breakdown

A

Catalase

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18
Q

2 ways in which rate of reaction can be calculated

A

1) amount of substrate disappeared
2) amount of product formed
PER UNIT TIME

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19
Q

Rate and time are _ proportional

A

Indirectly

R α 1/t

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20
Q

H2O2 is toxic. State it’s breakdown equation

A

H2O2–>H2O+O2

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21
Q

State the trend in the breakdown of H2O2 and reason for the same

A
  • •• reaction begins very fast. As soon as enzyme and substrate mixed bubbles of O2 released quickly and large volume of it collected in first mins ••• because when they’re first mixed there are a large no of substrate molecules. Virtually every enzyme has substrate in active site •••
  • •• rate of O2 released gradually slows and eventually stops ••• because substrate is being used up and enzyme is waiting for substrate to hit active site as fewer substrate is there
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22
Q

Rate of reaction depends on

A

How many enzymes enzymes can covert the substrate into product and release it

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23
Q

Which part of curve is best to calculate rate

A

Initial

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24
Q

Briefly explain 3 methods to calculate rate of reaction

A

1) gradient of straight line
2) gradient of tangent
m=Δy/Δx
3) Check volume produced in first 30s and multiply by 2 to get for 60s ie a min

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25
Q

Which factors affect rate of enzyme catalysed reaction

A

Enzyme concentration
Substrate concentration
Temperature
pH

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26
Q

When investigating effect enzyme concentration what should be kept constant

A

Substrate concentration

Volume of setup, done by adjusting water

27
Q

Why do all graphs meet though at different times for diff enzyme conc

A

Substrate amount is same so product made will also be same

28
Q

Why is it best to use in it all rate to compare enzyme concentrations

A

Once reaction is under way with different zym concentrations amount of substrate in each reactions varies bcos it’s converted at different rates.
Only in the beginning can we be sure that rate differences are caused ONLY by zym conc

29
Q

Reaction rate is _ proportional to enzyme conc

A

Directly

30
Q

Why is rate α zym conc

And how does initial rate increase with zym conc and under what condition

A

More the enzyme, more active sites available for substrate.

And as long as there is plenty of substrate initial rate increases linearly with enzyme conc

31
Q

What breaks down starch to maltose

A

Amylase

32
Q

Why is it difficult to identify rate of the course of reaction of starch breakdown

A

Both starch and maltose are colorless

33
Q

How can we measure rate of starch breakdown

A

Rate at which starch disappears by taking samples at known times and adding iodine. Using a colorimeter to measure intensity of blue-black color as a measure of amount of starch left
Plot graph for starch left against time to calculate initial rate

34
Q

For graphs of enzyme conc while measuring O2 formation of starch disappearance, which graphs will be at the top and bottom

A

Formation : highest conc will be top with highest rate of production
Disappearance : highest conc will be inside with highest rate and hence reaching minimum sooner

35
Q

How are enzyme conc vs rate graphs (formation of disappearance)

A

Linear

36
Q

Effect of substrate conc is normally measured with reference to

A

Product formation

37
Q

State the explanation for the graph of substrate vs initial rate

A

As substrate conc increases initial rate also increases. However, if substrate is increased further, it would not be as though there is mores substrate and hence more often bind to active site. Instead, as enzyme conc is constant, there is a point where every active site is occupied and cannot work faster as substrates are queuing up for vacant active sites

38
Q

What does Vmax indicate and when is it achieved

A

Maximum rate of enzyme catalysed reaction

When all active sites are bound to substrate. Enzyme is saturated with substrate

39
Q

What is Km

A

Michaelis Menten constant

Substrate concentration at which half of maximumrate of reaction, 1/2Vmax, is achieved

40
Q

What 2 things does Km tell us

A
  • half the active sites of given conc of enzyme are saturated with substrate
  • gives info about affinity os enzyme for substrate ie Km α 1/affinity of e-s
41
Q

How to calculate Vmax?

A

Since Vmax is achieved at infinite substrate conc it is impossible to accurately measure Vmax from graph.
Instead of plotting [s] on X axis and [v] on y axis, we plot 1[s] and 1/[v] and thus 1\infinity=0.

DOUBLE RECIPROCAL PLOT
Is straight line, easier to understand
1/Vmax is y intercept (where [s]=0 ie infinity). And from there calculate Vamx

42
Q

What is turnover rate

A

Rate at which substrate molecules are converted to product per unit time.

43
Q

What is carbonic hydrase and where is it found

A

H2CO3
Fastest enzyme to remove 600,000 CO2 molecules from repairing tissue
In RBC

44
Q

Significance of Vmax and Km values (6)

A
  • make computer models of biochemical pathways. Behavior of cells to predict reaction pathway and how enzymes will interact. Consequences of changing conditions (temp, ph, inhibitor) can be built in model
  • presense of enzyme for diff substrate compared quantitatively
  • understand what affects enzyme efficiency and design better catalysts with genetic engineering
  • performance of a commercially important enzyme from diff organisms compared
  • calculation can be applied to other fields (eg biochemistry, antibody-antigen engineering)
  • knowing Km means proportion of active site occupied by substrate can be calculated for any substrate conc

(for scientists^)

45
Q

What is optimum temperature

A

Temp at which enzyme catalyses a reaction the maximum rate

46
Q

How is rate at high and low temp

A

H- enzyme and substrate molecules move faster and collide more. Frequently so substrate is converted to products more often

L- reaction is very slow bcos substrate does not often collide with active site hence binding is rare

47
Q

What is the optimum temp

A

Rate highest at 40°C

48
Q

What happens after optimum temp

A

Structure of enzyme vibrates so energetically that hydrogen bonds holding it in precise shape begin to break

49
Q

Enzyme losing shape and activity is called _ and is _

A

Denaturation

Irreversible

50
Q

State the chain of bacteria which thrive at higher optimum temps

A

Thermus Aquaticus (bacteria) —> Taq polymerase (enzyme) 75-80°C —> PCR-polymerase chain reaction (gene/DNA amplification)

51
Q

Human enzymes optimum temp is _. Our body is kept at _ to ensure _ reactions occur at _ to their _ rate. Slight rise in temp is dangerous because enzymes would _

A 
40°C
37°C
Enzyme catalysed 
Close to
Maximum
Denature
52
Q

What happens to amino acid when pH decreases/increases and is neutral

A

Decrease ie acidic
Amine NH2 ionised to NH3+

Increase ie alkaline
Carboxylic acid ionised ti COO-

Neutral ie 7
Net charge=0

53
Q

What in general explains why enzymes change with pH changes

A

They are proteins so the ionic bonds of their amino acids are disturbed causing overall charge

54
Q

Optimum ph

A

7 max rate

55
Q

what is pH

A

Measure os concentration of Hydorgen ions in a solution

56
Q

How exactly does pH affect enzymes

A

Interacts with R groups of amino acids by affecting ionisation (+/-) of groups which in turn affects 3d arrangement of enzyme. Shape of active site changes and therefore reduce chance of substrate fitting

57
Q

pH changes are _ just like temp changes

A

Irreversible

58
Q

Function of a buffer

A

Has particular pH

Maintains it’s pH even when reactions cause pH changes

59
Q

Method to prepare immobilized enzymes

A

Enzyme mixed with sodium alginate solution
Little droplets of this mixture then added ti solution of calcium chloride
Na and CaCl2 react ti form jelly which turns each droplet into bead with enzyme which is immobilized

60
Q

What is competitive inhibition

A
  • Inhibitor has similar shape to substrate and fits into active site.
  • Is reversible and can be achieved by increasing conc of substrate
  • Increase in Km, but Vmax unchanged
61
Q

Graphs of competitive inhibitor

Recall graphs

A

As substrate conc increases competitive inhibitor and no inhibitor rates increase. But substrate rate is more

In reciprocal, 1/v value is same so they intersect same place at Y axis
However 1/km is less for no inhibitor

62
Q

What is non competitive inhibition

A
  • Shape of inhibitor not similar to substrate
  • Inhibitor binds to some other part of enzyme, allosteric site, rather than active site
  • While inhibitor is bound, it disrupts normal arrangement of hydorgen bonds and hydrophobic interaction holding enzyme in 3d shape. Distortion ripples across molecules to active site making it unsuitable to substrate
  • Reversible
  • Decreases Vmax, Km unchanged
63
Q

How are graphs of non competitive inhibitor and non inhibitor ie substrate

A

As [S] increases both V increase but non competitors is less than substrate

In reciprocal plot
Substrate has higher 1/Vmax, but intersect X axis at 1/Km

64
Q

What is feedback mechanism?

A

Eg
S1>e1>P1>e2>P2>e3>P3 to S1 is e1

As levels of P3 rise e1 inhibition increases. So less P1 is made and hence less P2, P3.
As P3 level falls, function of e1 increases so Ps are produced again ti restart cycle.
This end product inhibition finely controls levels of P3 between narrow upper and lower limits.

AS Biology (11 decks)

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