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It took less time for 25% of cells with cyclin D to be undergoing DNA replication than for 25% of cells without cyclin D.
Use Figure 5 to calculate this time difference as a percentage decrease. Show your working.
(edited 10 months ago)
Reply 1
Original post by saraseax
It took less time for 25% of cells with cyclin D to be undergoing DNA replication than for 25% of cells without cyclin D.
Use Figure 5 to calculate this time difference as a percentage decrease. Show your working.


What are you thinking so far?
If you haven’t managed to figure out the answer yet. It’s 34%.

Why?
- You want to approach it as a percentage change question. And you do this by using the equation ((initial value - final value) / final value) x 100.

You go across for cells with cyclin D at 25% and you will get a value of 11.5 hrs
Go across the graph for cells without cyclin D and you will get a value of 17.5 hrs
Plug this into the equation above, using cells with cyclin D as the initial value.

((11.5 - 17.5) / 17.5) x 100
= -34.28571429
= 34%
Original post by saraseax
It took less time for 25% of cells with cyclin D to be undergoing DNA replication than for 25% of cells without cyclin D.
Use Figure 5 to calculate this time difference as a percentage decrease. Show your working.
Reply 3
Original post by Scienceconsumer
If you haven’t managed to figure out the answer yet. It’s 34%.

Why?
- You want to approach it as a percentage change question. And you do this by using the equation ((initial value - final value) / final value) x 100.

You go across for cells with cyclin D at 25% and you will get a value of 11.5 hrs
Go across the graph for cells without cyclin D and you will get a value of 17.5 hrs
Plug this into the equation above, using cells with cyclin D as the initial value.

((11.5 - 17.5) / 17.5) x 100
= -34.28571429
= 34%

what does it mean by the 25% ? is it just extra information to confuse us or are we actually supposed to use it?

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