Translation timescales: There are roughly 20,000 ribosomes in each E. coli cell, so, if they are working full time, each

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1 Lecture 6: rocesses and Timescales: Rates for the fundamental processes 6.1 Reading Assignment for Lectures 5-6: hillips, Kondev, Theriot (KT), hapter 3 (roblem Set 1 due Friday, Sept. 24) Transcription timescale: We previously estimated the number of protein molecules in E. coli (see Lecture 4, age 4.6) at 3x Notice that this means that the cell must manufacture 300x this number or about 10 9 amino acids over the doubling time of the cell. Assuming that these proteins are not degraded over the cell lifetime, that means that they must be produced at a rate of 3 " = 103 protein molecules/s. To make each protein, you need an mrna. For the typical 300 amino-acid protein, that means about 3x300~10 3 bases/mrna. A typical transcription rate is 40 bases/s, so it takes 1000/40 = 25 s for a single transcription complex to produce one mrna for a typical protein. Translation timescales: There are roughly 20,000 ribosomes in each E. coli cell, so, if they are working full time, each 3 "10 6 proteins one must produce = 150 proteins/ribosome over the 3000 s cell cycle. 20,000 ribosomes It follows that each ribosome requires 3000 = 20 s to produce a typical protein molecule. 150 During this time 300 amino acids are processed, i.e., the incorporation rate is 300 = 15 amino 20 acids/s for each ribosome. omment: The rate of mrna synthesis is one per 25 s, while the rate of protein synthesis is one per 20 s. This makes clear that what is going on ANNT be modeled as: transcription produces 1 mrna followed by translation by ribosome produces 1 protein followed by used mrna discarded. Because the rate of mrna production is (slightly) too low and, in this model, could not keep up with the rate of translation required to replicate the cell. But, actually, the problem with this model is a lot worse than this. Each protein requires 3x300 ~ 10 3 bp for coding. Thus, there could be a maximum of only 5 " # 5 "10 3 distinct proteins coded in the E. coli genome. Actually, fewer, because some of the genome is used for gene control/regulation. But leave this aside for the moment. According to our calculations, each of these coding regions takes about 25 s to transcribe. Suppose that each coding region is read continuously, once every 25 s. That would produce the information for 5x10 3 proteins every 25 s, i.e., 5 "10 3 # = 6 "105 proteins over the doubling time, a factor of 5 less than the 3x10 6 proteins needed for replication.

2 Extra complication: E. coli doesn t need exactly the same number of proteins of each type. learly there is a problem with the model. We can imagine getting around this problem in a variety of ways: translate each gene several times simultaneously use each mrna several times before discarding What does nature do? What actually happens is that each mrna is read successively by a number of ribosomes. (see Fig. below). This shows something about the power of quantitative modeling. 6.2 KT Fig. 3.9 (p. 93) EM Image of simultaneous transcription and translation. Bacterial DNA and its mrna transcript ellular Energy: Mechanical and hemical Equilibrium in the cell Reading Assignment for Lectures 7 9: KT, hapter 5. (we re skipping h. 4) hapter 5 is the first of three chapters devoted to biological applications of thermodynamics and statistical mechanics. This is an area in which your backgrounds are varied. KT does not try to give you a course in these subjects. They take the philosophy that you just jump in (with references to other texts, as required). I am going to try to give you a little more systematic background. E Next: omments on energy and energy units: A. Thermal Energy: Ebarrier x -23 Ethermal=kBTroom=1.38x10-21 x300=4.1x10 J=4.1 pn nm=25 mev (molecular) processes requiring energies of this order of magnitude or less in order to proceed will do so spontaneously; those requiring appreciably more energy will not happen unless additional energy or an appropriate catalyst/enzyme is available.

3 ther units: 6.3 B. Metabolic energy content: 4,200 1 kcal/mole=4.2 kj/mole= 6.02 "10 23 = 7.0 "10#21 J /molecule ~ 1.7 k B T room.. AT equivalents: Adenine is one of the four bases or nucleotides (nucleic acids) which make up DNA: ytosine, Guanine, Thymine, and Adenine. So, the complementary pairs are (G+TA). In RNA the same bases are linked to a sugar to make the nucleosides except that Thymine is replaced by a fifth base Uracil: ytidine, Guanosine, Uridine, and Adenosine. So the complementary base pairs are (G+UA). In the adenosine phosphate family, the sugar has been phosphorylated at the 4 position: The phosphate groups are ionized in solution. ADENINE Nucleotide = Adenylic acid Sugar = pentose (ribose) H 2 4 N 1 Adenosine monophosphate (AM: q=-2) 3 2 H H Adenosine diphosphate (AD: q=-3) ADENINE H 2 4 N H H

4 Adenosine triphosphate (AT: q=-4) 6.4 ADENINE H 2 4 N H H The reactions of phosphorylation and dephosphorylation (hydrolysis) are reversible. The hydrolysis of AT releases energy: Dephosphorylation releases energy, so formation of AT is a way of storing energy. (widely used!!) Thus, AT -4 AD -3 + i -2 + H kcal/mole (number from Alberts, p. 534) Note : E AT =12 kcal/mol=8.3x10-20 J ~ 20 E thermal, which is another useful energy unit, especially when looking at cellular processes. omments: 1. The energy (actually, free energy) release depends on concentrations in the cell. This is a typical number under normal cellular conditions. 2. If reaction needs more energy, then going from the triphosphate to the monophosphate releases even more energy. 3. oupled kinetic traps: (All these processes regulated by enzymes) AT E AT ~20 k B T room AD E useful input material output material E waste

5 4. Another closely related molecule which figures prominently in glycolysis is nicotinamide 6.5 adenine dinulceotide (NAD + ) or in its phosphorylated state NAD +. Note the presence of AD below (circled in red); group circled in blue is nicotinamide ring. nicotine note the ionization site this is the locus of phosphorylation for NAD + This molecule plays an important role in electron transfer, according to the reaction, Loss of electrons = oxidation Gain of electrons = reduction Under normal cellular conditions the high-energy form is the reduced form, which can then (once untrapped) oxidize, thus freeing two electrons (and energy). These, in turn, can be used to reduce an oxidized organic compound (thus increasing its energy). The equivalent energy transfer is 2-3x E AT. Upshot is useful set of energy units and conversions: (see below)

6 Energy units and conversions KEY input k_b= 1.38E-23 J/K definition 1 kt= 4.14E-21 calculated 1 Da= 1.66E-27 kg 1 kcal= 4.19E+03 J 1 kcal/mol= 6.95E-21 J/molecule N_Avogadro 6.02E+23 E_AT 1.20E+01 kcal/mol 6.6 J kt kcal/mol E_AT thermal E_{thermal}=k_{B}T_{room 4.14E E E-02 chemical 1kcal/mol (energy/molecule) 6.95E E E-02 AT E_{AT} 8.34E E E+01 1 = double bond energy 9.94E E E+01 - single covalent bond 5.80E E E+00 hydrogen bond 5-10 kt 2.07E E E-01 trans-gauche isomerisation 2.07E E E-01 VdW (-) 1.24E E E-01