Hi Joshua,

I am also not from a math background, but I have done some cell culturing that kind of resembles the steps in this question. But before anything else, I do think that this would be a “low-yield” question as it requires us to do a lot of math. So, definitely something to flag and be done after answering the rest of the bio/biochem section.

1. Identify how many E.coli cells are in the 0.01 L solution (don’t bother the 2 L, it’s a distractor). The passage says that there are 10^9 cells/mL, so convert 0.01 L to mL which gives 10 mL. Then, we multiply 10 mL with 10^9 cells/mL to get 10^10 cells. So, in 0.01 L solution there are 10^10 cells

2. These 10^10 cells still have water in them. According to the passage, the wet mass of 1 bacterium cell is around 1 pg (equivalent to 10^-12 g), so technically we have 10^10 pg wet cell mass.

3. The author dried all the bacterial cells, removing the water and giving us a dry cell mass. According to the passage, dry cell mass is 20% of wet cell mass. So, we multiply 20% or 0.2 to 10^10 pg. This would give us a dry cell mass of 2 x 10^9 pg.

4. The question stem asked for “dry mass of proteins.” The passage says that half of the dry cell mass is carbon while the other half is protein. So, we divide 2 x 10^9 pg by half which gives us 1 X 10^9 pg proteins. With this, we could eliminate option A because this is the dry cell mass with BOTH carbon and protein.

5. We convert 1 x 10^9 pg dry cell protein to grams by dividing it with 10^12 (recall that 1 g = 10^12 pg). This would give us 1 x 10^-3 or 0.001 g (option D)

Hope these help!