HW_7_final_2011

# Make sure to state the assumptions you use in

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total number of polymerases in the cell? Make sure to state the assumptions you use in answering this question since in the end, we are making estimates and more important than the numbers is the rationale behind the analysis. (b) Translation by the numbers. We earlier considered the translation process and noted that the ribosome can synthesize new proteins at a rate of roughly 20 aa/s. Following logic similar to that we used for transcription, use the fact that the footprint of the ribosome on the mRNA is roughly 35 nucleotides wide and determine the maximal rate at which new proteins could be synthesized? Given the lifetime of the mRNA molecule itself (see fig. 1), how many proteins can be produced off of this transcript during its lifetime? If you use this as the mean number of proteins per gene, how many total proteins would you estimate in E. coli ? This is a serious overestimate. 2

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Average number of molecules per cell 10 -1 10 0 10 1 10 2 10 3 10 4 10 5 50 100 150 200 250 Frequency (number of protein species) Lu et al. (2007) Taniguchi et al. (2010) Figure 2: Different measurements of the mean number of proteins per E. coli cell. The measurements are from fluorescence (green) and mass spec. (blue). Though the two measurements don’t agree, we see that a rough rule of thumb is that the average gene has a protein copy number of roughly 1000. (Note to class: the weird right to left numbers on the x-axis has to do with the fact that normally this graph is shown on its side.) In fact, we should probably think of roughly half of the genes as being ac- tively synthesized with each such gene coming in at a mean protein copy number of roughly 1000 proteins (see fig. 2). What does this imply about the protein synthesis rate per gene and what can you say on the basis of this numbers about the total number of ribosomes in the cell? 2. Measuring gene expression using enzymatic reactions. As discussed in class, a widespread strategy for quantifying levels of gene expression relies on the use of enzymes as reporters. β -galactosidase (also known as LacZ and we will use those names interchangeably), the enzyme encoded by the lacZ gene, is one popular choice for such assays. In the cell, this enzyme breaks down lactose, but it can also be used with alterna- tive substrates such as ONPG. When ONPG is cleaved by β -galactosidase 3
one of the resulting products is the molecule ONP which turns the solution yellow. To measure the amount of β -galactosidase in a population of cells, the cells are lysed and then substrate is added. By subsequently monitor- ing the increase in “yellowness” (given by the optical density at 420 nm, OD 420 ) of the reaction as a function of time the amount of enzyme per cell can be calculated by exploiting the fact that this rate is a linear function of the concentration of β -galactosidase itself. In this problem, we flesh out those ideas captured in words with a mathematical description which also shows us the set of conditions under which the measurement method is valid.

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