AQA AS Biology paper 2 2018

Hi - I just had a go at those questions so I'll do my best to explain them!
*****ignore the dashes in the equations, they are there so that all of the words are in the right place*****

5.2
This question is based on the fact that the number of cells in a certain stage of the cell cycle is proportional to the length of that stage of the cell cycle.
Applied to this question, it means that the number of cells in anaphase will be proportional to the length of anaphase.
This means you can use the following formula to work out the length of anaphase:

Number of cells in anaphase in Figure 2 = Length of anaphase
Total number of cells in Figure 2-------------Length of whole cell cycle

This formula is definitely worth learning if you haven't already
(they are fractions by the way!)

In Figure 2 there are 3 cells in Anaphase, all found around the centre of the picture (two are touching right in the middle and the third is a little bit to the bottom right of these two cells). We can tell that these are in anaphase because their chromatids have been separated and pulled to opposite ends of the cell.
The question says that there are 32 total cells in Figure 2
It also says that the whole cycle lasts 1080 minutes

So now we have everything we need to work out the answer, by setting out the equation like this:

3 (cells in anaphase) = length of anaphase
32 (total cells)----------------1080 (total length)

3 divided by 32 is 0.094, so if we multiply this by 1080 then we get the length of anaphase, which is 101.25 minutes

We then have to work out the percentage difference between this answer and the scientists' answer (which is 105 minutes):

The difference between 101.25 and 105 is 3.75.
We can then work this out as a percentage of 101.25 by dividing 3.75 by 101.25 and then multiplying this value by 100
The answer to this is 3.7%, which gets both the marks.

7.3
This is about the rate of an enzymatic reaction.

In order to work out rate, we need to find the formation of product or loss of substrate in a given time.
In this instance it is the loss of substrate because there is a decrease in the mass of cellulose.
We are told that all 3 solutions initially contain 15g of cellulose, and we can see from the table that when both the enzymes are present there is 9.2g of cellulose left after 48 hours (we are looking at the third column as this is where both enzymes are present).
Therefore, we can work out the change in mass, which is 5.8g.

We are given the time (48 hours), but we need to convert this into minutes by multiplying 48 by 60, giving 2880 minutes

We can then divide 5.8g by 2880 minutes to find the rate of the reaction in g/min: 0.00201 g/min
We then need to convert this into standard form. This is quite straightforward, but if you are unsure on it then an easy rule for numbers which are less than one is that the power used is equal to the number of 0s before the first digit.
So in this instance there are 3 zeros before the first digit (2). Therefore, in standard form it is 2 x 10^-3 (notice the power of -3)

Therefore, the final answer is 2x10^-3.

I really hope this helped - if you want me to clarify anything in this reply then just ask and I am happy to do it on pm if you prefer