Each amino acid must be attached to a tRNA by an aminoacyl tRNA synthase in the cytosol. This process requires 2 high-energy phosphate bonds per amino acid, or 900 high-energy bonds total.Translation elongation requires 2 high-energy phosphate bonds per amino acid incorporated. One is used in bringing the activated tRNA to the A site, and one is used in translocation. So elongation requires 900 high-energy bonds total.
Translation initiation requires 1 high-energy phosphate bond per polypeptide.
TOTAL = 2(450) + 2(450) + 1 = 1801 high-energy phosphate bonds per 450-amino-acid monomer
NOTE: For the exam, you will need to be able to describe where in the process of translation these high-energy bonds are used, not simply to remember that there are 4 per amino acid.
CORRECTION: as pointed out to me by alert Bio200 students, the first amino acid of the protein is positioned in the P site of the ribosome by the process of initiation. So incorporation of that first amino acid (Met) requires 1 high-energy bond for the initiation factor and 2 for its aminoacyl tRNA synthase, for a total of 3 (not 5). The other amino acids require 4 each. So the total should be (1*3) + (449*4) = 1799
2. (2 pts) A student lab group has observed flagellar regeneration in Chlamydomonas after a pH shock and determined the rate of growth to be 0.1 µm per minute per flagellum. Review your calculations from lab (pp. 26-28 of your lab manual). Assume that all the necessary mRNA is already present (i.e. no transcription is required), but that all tubulin protein in these flagella must be synthesized from amino acids. How many high-energy phosphate bonds would have to be consumed per minute by a Chlamydomonas cell in order to produce enough new alpha and beta tubulin protein to sustain this rate of flagellar growth?
0.1µm/min * 1000 nm/µm * 1 dimer/8 nm of protofilament * 233 protofilaments/ flagellum * 2 flagella/cell = 5825 dimers/min5825 dimers/min * 900 amino acids/dimer * 4 high-E phosphate bonds/amino acid = 20,970,000 high-E bonds per min for elongation
1 high-E phosphate bond per monomer for initiation: 5825 dimers/min * 2 monomers/dimer = 11,650 high-E bonds for initiation
(notice that the number of high-E bonds for initiation doesn't affect the final answer, within a reasonable margin of error)
TOTAL = about 21,000,000 high-E bonds per minute
CORRECTION: The first amino acid in each monomer requires 3 high-E phosphate bonds, not 5. So the total should be 20,970,000 minus 11,650, not plus. This is still about 21,000,000 high-E phosphate bonds per minute.
3. (2 pts) Assume that each tubulin mRNA (either alpha or beta) can accomodate about 17 ribosomes at one time (spaced about 80 nt apart from each other along the 1350 nt coding region of the mRNA). Assume also that it takes about 1 minute for a ribosome to synthesize a full-length tubulin monomer, either alpha or beta (a fairly typical rate of protein synthesis), so that a given mRNA can produce about 17 tubulin monomers per minute. How many molecules of tubulin mRNA (both alpha and beta) would have to be simultaneously translated at this rate in order to support the rate of growth described in #2?
5825 dimers/min (see above) * 2 monomers/dimer * 1 mRNA/17 monomers/min = 685 molecules of mRNA simultaneously translated, each at a rate of 17 monomers per minute
4. (no points, no answer required) Think about your answers to #1-#3. Isn't that amazing?? And this is just one example of a typical cellular process. The amount of frantic activity in which cells must engage in order to stay alive from minute to minute is truly mind-boggling, I think.
5. (4 pts) If you were going to do research related to cystic fibrosis, what experimental question would you want to address? Describe the question and explain its importance. (This can be a question discussed in class or one of your own choosing. There is no "right answer." Questions that are relevant to CF and that are clearly described and explained will receive full credit.)
Maintained by: bkbaxter@lclark.edu
Updated: 4 Apr 01