Chapter 3: Problem 5
Cell cycle and number of ribosomes Read the paper "Ribosome content and the rate of growth of Salmonella typhimurium" by R.E. Ecker and M. Schaechter (Biochim. Biophys. Acta 76,275,1963 ). (a) The authors show that their data can be fit by a simple assumption that the rate of production of soluble protein, \(P\) is proportional to the number of ribosomes. But this leaves out the need for the ribosomes to also make ribosomal protein, in an amount, \(R\). The simplest assumptions are that the total protein production rate, \(\mathrm{d} A / \mathrm{d} t,\) is proportional to the number of ribosomes, and the total amount of non-ribosomal protein (also knows as soluble protein) needed per cell is independent of growth rate. Given these assumptions, how would the ratio \(R / A\), with \(A\) the total protein content, depend on the growth rate? Does this give a maximum growth rate? If so, what is it? (b) With a 3000 s division time for \(E .\) coli, about \(25 \%\) of its protein is ribosomal. Note that the microbe Salmonella typhimurium considered here is very similar to \(E .\) coli. Using these numbers and your results from above, what fraction of the protein would be ribosomal for the highest growth rate studied in the paper? How does this compare to their measured ratio of ribosomes to soluble protein, \(R / P\) at these growth rates? How does the predicted \(R / P\) at high growth rates change if you now assume that \(\mathrm{d} P / \mathrm{d} t\) is proportional to \(R\) as they did in the paper? (c) Another factor that needs to be taken into account is the decay of proteins. If all proteins decayed at the same rate, \(\gamma\) how would this modify your results from (a)? How does the predicted functional from of R/A versus growth rate change? Explain why the data rule out \(\gamma\) being too large and hence infer a rough lower bound for the lifetime, \(1 / \gamma,\) of "average" proteins.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.