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For some reason I'm having a lot of trouble setting up this problem. Any help would be greatly appreciated!
Louis Pasteur discovered that by depriving a culture of yeast cells access to oxygen, their growth rate decreased by a factor of 6, while their glucose utilization increased. Under aerobic conditions, yeast can synthesize 36 NTPs per molecule of glucose oxidized. the rate of glucose consumption under anaerobic conditions would increase by approximately what factor?
A) 2
B) 3
C) 4
D) 6
Answer:
B is correct.
The faster the growth rate of an organism, the shorter its doubling time. the guiding principle in problems such as these is that growth rate is directly proportional to the amount of atp available to each cell per unit of time. in the presence of plenty of O2, eukaryotic organisms (such as yeast) can make eithr 36 or 38 atps. for the purpose of this question, we were to assume 36 ntps per molecule of glucose oxidized (under aerobic conditions). when oxygen is removed, the yeast cells switch over to anaerobic metabolism, and the net yield of ntp per molecule of glucose falls to just 2. this is called anaerobic fermentation.
Depriving yeast cultures of O2 causes each cell to produce 18 times less NTP per molecule of glucose consumed (from 36/2=18) pasteur found that yeast cells grew 6 times less rapidly under these conditions implying that they were forming ntps at a rate of only 6 times less rapidly than before. this means they must be consuming glucose more rapidly than before. in particular they must be consuming glucose at a rate approx 3 times faster (from 18/x=6).
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what i'm confused about is how they set up 18/x=6. I dont understand why they set it up this way when both of the values are related to how many times less ntp or how much less the rate is compared to aerobic. If anyone could help clarify this that would be great! Thank you!
Louis Pasteur discovered that by depriving a culture of yeast cells access to oxygen, their growth rate decreased by a factor of 6, while their glucose utilization increased. Under aerobic conditions, yeast can synthesize 36 NTPs per molecule of glucose oxidized. the rate of glucose consumption under anaerobic conditions would increase by approximately what factor?
A) 2
B) 3
C) 4
D) 6
Answer:
B is correct.
The faster the growth rate of an organism, the shorter its doubling time. the guiding principle in problems such as these is that growth rate is directly proportional to the amount of atp available to each cell per unit of time. in the presence of plenty of O2, eukaryotic organisms (such as yeast) can make eithr 36 or 38 atps. for the purpose of this question, we were to assume 36 ntps per molecule of glucose oxidized (under aerobic conditions). when oxygen is removed, the yeast cells switch over to anaerobic metabolism, and the net yield of ntp per molecule of glucose falls to just 2. this is called anaerobic fermentation.
Depriving yeast cultures of O2 causes each cell to produce 18 times less NTP per molecule of glucose consumed (from 36/2=18) pasteur found that yeast cells grew 6 times less rapidly under these conditions implying that they were forming ntps at a rate of only 6 times less rapidly than before. this means they must be consuming glucose more rapidly than before. in particular they must be consuming glucose at a rate approx 3 times faster (from 18/x=6).
----
what i'm confused about is how they set up 18/x=6. I dont understand why they set it up this way when both of the values are related to how many times less ntp or how much less the rate is compared to aerobic. If anyone could help clarify this that would be great! Thank you!