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Mazyrski [523]
3 years ago
10

BIOCHEMISTRY HELP! You and a lab partner are performing on several molecular analogs to determine Km. One is a substrate analog,

one is a transition state analog, and the last is a product analog. The data you obtain shows that the three Km's are 0.0013 M, 0.025 M, and 0.0045 M (in no particular order). Given what you know about the induced-fit model of catalysis, which Km would you hypothesize to belong to which analog, and why?
Chemistry
1 answer:
Alina [70]3 years ago
5 0

Answer:

Michaelis constant is known as km which is the substrate concentration that encourages the compound to work at half maximum velocity represented by Vmax/2. Michaelis constant is inversely related to the substrate and the affinity of the enzyme.  

Induced fit model: The premise of the purported induced fit hypothesis, which expresses that the attachment or association of a substrate or some other atom to an enzyme causes an adjustment to the enzyme in order to fit or restrain its activity.  

In substrate, analog Km or Michaelis constant will be high as the substrate will stay because of analogs inhibit activity.

In the transitional state, analog Km will be in the middle of the substrate and product analogs. Progress state analogs are synthetic mixes with a structure catalyzed reaction that looks like the progressing condition of a substrate atom in a compound enzyme.  

In item simple thus Km is the least.  

0.0013 M = product ananlog,

0.025 M=Transition state, and

0.0045 M = Substrate analog

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The equation for the protonation of the base pyridine is the following:

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C₅H₅N  +  H₂O  ⇄  C₅H₅NH⁺  +  OH⁻

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After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

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                = x . x / 0.64-x

1.7 × 10⁻⁹ (0.64-x) - x² = 0

Solving the above quadratic equation for x, we have :

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Now, We can calculate the pH, after taking the positive value, x₂, (concentrations cannot be negative) and entering into above equation :

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<em>pH = </em>14 + log[OH⁻]

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