AQA A Level Biology Recombinant DNA Tech

https://pmt.physicsandmathstutor.com/download/Biology/A-level/Topic-Qs/AQA/8-Control-of-Gene-Expression/Set-C/8.4%20Recombinant%20DNA%20QP.pdf

For question 3d how did they work out the ratio please?It's 1.4:1

Reply 1

bububbles

Original post by anon25x

Use the graph to find C. You can assume that the maximum fluorescence occurs when there is maximum DNA concentration.
The maximum fluorescence on the graph is approximately 0.48, meaning that 50% of this is 0.24.

Follow the X axis where Y=0.24 (approx.) to until you intercept curves A and B, to find the number of cycles required (C)
For A you find that C = 19 (approx.)
For B you find that C = 27 (approx)

RNA in a sample = 1/C
RNA in sample A = 1/19 = 0.05263...
RNA in sample B = 1/27 = 0.03704....

to find the ratio:
0.05263 / 0.3704 = 1.42...
Ratio of RNA in A:B = 1.4:1

Hope that helps :smile: