ing
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please explain your answer. Thanks.
Attached Files:
usmle-cdq.doc (78 KB, 67 downloads)
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| Jinx
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ans is 3
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| robin082006
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Agree, I go with 3
I choose the answer that has the same curve as the control at a limited concentration of enzyme.
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| truedeam59
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I will go for 2 bc as the substrat increases the v max also increases. Not at all sure
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| ing
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Thank you guys.
Edited by ing on 04/02/06 - 08:55 PM
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| ing
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Thank you guys. Here is the answer from kaplan released item.
The correct answer is C.
Halving the amount of enzyme will halve the velocity of the reaction at each higher substrate concentration. Therefore, curve 3 is correct. At very low substrate concentrations, the amount of enzyme is not limiting, and the curve resembles that of the control. This feature is also seen in curve 3.
Curve 1 (choice A) has the same maximum velocity as the control curve, but has a faster velocity at lower substrate concentrations. This pattern can be seen with enzymes with a lower Michaelis-Menten constant (Km).
Curve 2 (choice B) has the same maximal velocity, but has a slower velocity at lower substrate concentrations. This pattern can be seen with competitive inhibition or with enzymes with a higher Km. If the amount of enzyme is halved, the maximal velocity should be only one-half the maximal velocity indicated by the control curve (choice D).
Edited by ing on 04/02/06 - 08:57 PM
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| mjl1717
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good review. Lineweaver Burke is more mathematically palatable
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