a Exam02 SP1O KEY (1) [Compatibility Mode] a Search in Document Home Insert Desi

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a Exam02 SP1O KEY (1) [Compatibility Mode] a Search in Document Home Insert Design Layout References Mailings Review View + Share A Times New Ro? 11 A? A??. · Heading 1 Pane 25. The following table represents data collected from a study on enzyme X. From this data determine what the Vx and the Km is for this particular enzyme. Then answer the following Substrate concentration (mM) (umol/min) 217 325 433 488 647 650 651 10 20 40 60 1,000 2.000 3,000 Vmai: Kn: IfI were to compare this enzyme to another enzyme (call it Y) that obtained the same Vm using ET] and had a Ka double the size of this onc which would you expect to be more efficient, as measured by the koaK ratio, Y or X? Please explain your (a) Page 9 of 9 1845 Words k English (US ?Focus- 139%

Explanation / Answer

according to the table after 1000 mM the Vo is constant. so the Vmax is 647. and Km is 20 mM.

now

Vmax = Kcat*[Et]. for enzyme Y Kcat is double compared to X.

and its Km is also double as compared to X

the ratio of Kcat/Km is equal for both the enzyme so both are equally efficient.

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