- What organ secretes chymotrypsin and what is it secreted as?
Serine protease → contains a serine residue
Pancreas Pro-eznyme chymotrypsinogen
An aromatic side chain (or bulky hydrophobic)
e.g.'s phenylalanine, tyrosine or tryptophan
- What is GPNA?
Carboxyl side of the peptide bond (between CO and NH)
Artificial enzyme substrate known as N-Glutaryl-L-phenylalanine p-nitroanilide
N-glutaryl-L-phenyl alanine and the bright yellow product, p-nitroaniline
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- How is the Km biochemically useful?
The concentration of substrate at which a particular enzyme works at half its maximal velocity
Allows us to compare the strength of Enzyme-substrate complexes
- What does a high Km indicate?
Tight binding of a substrate to an enzyme
Indicative of weak binding
Vmax tells us the maximum velocity at which chymotrypsin can cleave peptide bonds
If we know the enzyme concentration used to derive Vmax, then by dividing Vmax by the enzyme concentration we can obtain the number of peptide bonds that chymotrypsin can cleave in a second This is the turnover number (also known as Kcat).
E-S complexes being formed and consumed at the same rate
Reaction velocity remains constant
x axis - 1/[S] (Substrate concentration)
y axis - 1/V (velocity)
- How do you find the Vmax from the Lineweaver-Burk plot?
Extrapolating the straight line graph until it crosses the x-axis gives a 1/[S] value of –1/KM
As [S] increases, 1/[S] tends towards zero, so that 1/Vmax is determined as the value of 1/V0 when 1/[S] = 0 i.e. the intercept on the y-axis
Indole → competitive inhibitor as it binds within the active site of chymotrypsin
Indole therefore increases Km In contrast, the maximal velocity (Vmax) will be unaffected
- Why is the Km increased during competitive inhibition?
Because increasing amounts of substrate can swamp the inhibitor (present in fixed concentration)
Allowing the enzyme to effectively not see the inhibitor at high substrate concentrations
It takes more substrate to get the competitively inhibited reaction to Vmax/2
Km is a measure of the affinity of the enzyme for its substrate and this can only be measured by active enzyme
The fixed amount of inactive enzyme in non-competitive inhibition does not affect the Km and the Km, therefore is unchanged