Topic 3 - Enzymes Flashcards

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1
Q

How do enzymes affect the rate of a chemical reaction

A

Catalysts - increase reaction rate, do not alter equilibrium, just alter the rate at which reach equilibrium
They do this by decreasing the activation energy of the reaction - more molecules have the energy for the reaction to occur

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2
Q

What is activation energy

A

Difference between substrate and transition state

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3
Q

How do enzyme-substrate complexes form

A

Binding of enzyme to substrate
Specific religion of enzyme - active site
Binding is stabilised by interaction of amino acid residues surrounding

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4
Q

What are the key features of the active site?

A

Occupies a small part of the enzyme
Formed by amino acids from different parts of the primary sequence
Clefts or crevices - the folds hide active site to prevent water entry to affect reactions
Complementary shape to substrate - lock key or induced
Substrates are bound to enzymes by multiple weak bonds so products can be released

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5
Q

How do you find V0 (initial rate)

A

Tangent to the curve

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6
Q

What does the rate of reaction depend on

A

Substrate concentration

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7
Q

What is the michealis-menten equation

A

V0 = Vmax [S] / Km + [S]

For curve

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8
Q

What is vmax and how do you find it

A

The theoretical maximum rate when all enzyme molecules are saturated with substrate

Extrapolate tangent until level out = vmax

Measured in ..M/min

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9
Q

What is the Km and how do you find it

A

The substrate concentration that gives half the maximal rate of reaction
Low - higher enzyme affinity

Find vmax
1/2 that on V0
Across to curve down to substrate concentration

Measure in mM

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10
Q

1 unit

A

The amount of enzyme that produces 1 micromol of product per min under standard conditions

Rate of enzyme catalysed reaction proportional to concentration of enzyme

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11
Q

What is the lineweaver-burn plot

A

V0 = vmax [S] / Km + [S]

Y = Mc delta T

Eg: y intercept = -1/km
Slope = km/vmax
Intercept = 1/vmax

Straight line

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12
Q

What is competitive inhibition?

A

Binds at active site
Reduce proptiln of enzyme molecules able to bind

Adding enough substrate will overcome effect of inhibitor so no effect on vmax whereas Km increases

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13
Q

What is non competitive inhibitor

A

Bind sat site other than active site
Causes conformational change
Decreases concentration of functional enzyme

Even if add more substarte will not over crime the effect so Km unaffected and vmax decreases

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14
Q

Km is independent of…

A

The concentration of enzyme

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15
Q

If twice as much enzyme used, and halves

A

Vmax double, Km unchanged

Vmax halves, Km unchanged

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16
Q

Less units of enzyme

A

Takes longer for enzyme to break down

17
Q

For individuals on a milk diet, this reaction is rate-limiting for the utilization and metabolism of lactose. Lactose intolerance occurs when the individual is unable to metabolise lactose sufficiently rapidly. Would a suitable response to this condition be to evolve a new enzyme with a lower Km or, alternatively, a higher Vmax or simply to make more of the standard enzyme? Explain your answer.

A

Higher Vmax would do but cannot do that so must change amino acid sequence of the active site to a higher affinity - change gene expression to make more enzymes or mutate

Due to the higher lactose diet - thus not a big impact with Km

18
Q

(a) What is the activity of the enzyme in Units per litre of serum? (1 unit is the amount of enzyme
that produces 1mol of product per minute)

A

0.1ml = 1.t x10-7 ml/min
0.1 ml = 0.15 micromol/min
X10000
1l = 1500 units/litre

19
Q

If states what would activity of enzyme in units per litre of serum be even if ml change

A

Per litre = same as standardised unit

20
Q

Why is ethanol beneficial for methanol poisoning

A

Has lower arm do higher affinity for enzyme so outcompetes meta molecules so ethanol given so not toxic product formed