AngelT
Forum Senior
Topics: 33
Posts: 111
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Could you explain me question number 6?
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| AngelT
Forum Senior
Topics: 33
Posts: 111
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Sorry...I dont know to scan well, so just click twice in the pic to do Zoom..the image it will be higher and clear.
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| AngelT
Forum Senior
Topics: 33
Posts: 111
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Anybody could help me again? 
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| Ivil
Forum Senior
Topics: 8
Posts: 169
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AngelT, in the given L-B graph Km is not changed. So there is no change in affinity. 1/Vmax is decreased. or Vmax is increased. Vmax can increase only if enzyme concentration is increased. So the ans is D. Kinase gene is amplified in the tumor cell leading to increased kinase concentration...
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| study_ing
Forum Fanatic
Topics: 180
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good job ivil 
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| Ivil
Forum Senior
Topics: 8
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Thanks studying.. you r my inspiration..
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Even if you are on the right track, you will get run over if you just sit there...KEEP RUNNING
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| AngelT
Forum Senior
Topics: 33
Posts: 111
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Thanks! Ok..but why the distractor C is wrong?
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| tinabg
Forum Newbie
Topics: 2
Posts: 22
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Inhibitors decrease V max. If this was a non-competitive inhibitor, Vmax would go down, 1/Vmax would go up and the tumor graph would cross the y-axis higher than the normal graph.
In this case the tumor graph crosses the y-axis lower than the normal graph, which means 1/Vmax is decreased, which means Vmax is increased. Vmax is the maximum rate of a reaction at a given enzyme concentration. The only way to increase Vmax is to increase enzyme concentration. The only way to naturally increae enzyme concentration is to induce expression of the gene encoding this enzyme. Did this explain it?
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