How does the total enzyme concentration affect turnover number and Vmax? Enzymes

Question

How does the total enzyme concentration affect turnover number and Vmax? Enzymes with a kcat/Km ratio of about 10^8 M-1s-1 are considered to show optimal catalytic efficiency. Fumarase, which catalyzes the reversible-dehydration reaction fumarate + H2O leftrightarrow malate has a ratio of turnover number to the Michaelis-Menten constant, (kcat/Km) of 1.6 times 10^8 for the substrate fumarate and 3.6 times 10^7 for the substrate malate. Because the turnover number for both substrates is nearly identical, what factors might be involved that explain the different ratio for the two substrates? The benefit of measuring the initial rate of a reaction V_0 is that at the beginning of a reaction: A) [ES] can be measured accurately. B) changes in [S] are negligible, so [S] can be treated as a constant. C) changes in K_m are negligible, so K_m can be treated as a constant. D)_ V_0 = V_max. E) varying [S] has no effect on V_0.

Explanation / Answer

Answer (1)

In enzymology, turnover number (also termed kcat) is defined as the maximum number of molecules of substrate that an enzyme can convert to product per catalytic site per unit time and can be calculated as follows: kcat = Vmax/[E]T (see Michaelis-Menten kinetics). For example, carbonic anhydrase has a turnover number of 400,000 to 600,000 s-1, which means that each carbonic anhydrase molecule can produce up to 600,000 molecules of product (CO2) per second.

Mathematically it means turnover number is inversely proportional(the bigger one gets the smaller the other one is) to enzyme concentration and directly proportional(both get big, or get small at the same time) to Vmax.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.