Chapter 2 Biological Structures: Rulers at Many Dierent Scales 2.1 Revisiting the E. coli mass We made a simple estimate of the mass of an E. coli cell by assuming that such cells have the same density of water. However, a more reasonable estimate is that the density of the macromolecules of the cell is 1.3 times that of water (BNID 101502, 104272). As a result, the estimate of the mass of an E. coli cell is o by a bit. Using that two-thirds of the mass is water and that the remaining one-third is macromolecular, compute the percentage error made by treating the macromolecular density as the same as that of water. Our approximate estimate gives a mass of E. coli of approx ME.coli = water VE.coli . (2.25) The better estimate is built by noting ME.coli = 2 1 water VE.coli + 1.3water VE.coli . 3 3 (2.26) 3.3 approx approx M 1.1ME.coli , 3 E.coli (2.27) This can be simplied to ME.coli which tells us that we make roughly a 10% error by using the density of water rather than the correct macromolecular density. 1 Problems and Solutions 7 Since the formula for glucose is C6 H12 O6 , the molecular mass is 180 Da. Hence, the number of sugars is # sugars 102 g 3 1019 glucose molecules. 180 g/6 1023 molecules (2.32) According to our estimate from part (a) (awed though it is because it emphasizes only the construction material cost of making a cell and ignores the energetic requirements), it takes 109 sugar molecules to make a bacterium and hence our 5 mL culture can support roughly 1010 bacteria. This is consistent with our intuition because a saturated culture has roughly 109 cells/mL. The molecular mass of NH4 Cl is about 50 Da. This means that in 5 mL of culture we have 6 1019 nitrogen atoms. As a result, if nitrogen was the limiting building block our culture would be able to support 3 1010 cell. From our estimations, when our culture starts running out of its carbon source it will also start running out of its nitrogen source. Just to reiterate, the point of this estimate is to get a sense of the cellular inventory and how it relates to the molecular contents of the growth medium that is used for those cells. The precise numbers should not be taken too seriously. 2.6 Atomic-level representations of biological molecules (a) Obtain coordinates for several of the following molecules: ATP, phosphatidylcholine, B-DNA, G-actin, the lambda repressor/DNA complex or Lac repressor/DNA complex, hemoglobin, myoglobin, HIV gp120, green uorescent protein (GFP), and RNA polymerase. You can nd the coordinates on the book's website or by searching in the Protein Data Bank and various other Internet resources. (b) Download a structural viewing code such as VMD (University of Illinois), Rasmol (University of Massachusetts), or DeepView (Swiss Institute of Bioinformatics) and create a plot of each of the molecules you downloaded above. Experiment with the orientation of the molecule and the dierent representations shown in Figure 2.32. (c) By looking at phosphatidylcholine, justify (or improve upon) the value of the area per lipid (0.5 nm2 ) used in the chapter. (d) Phosphoglycerate kinase is a key enzyme in the glycolysis pathway. One intriguing feature of such enzymes is their enormity in comparison with the sizes of the molecules upon which they act (their \"substrate\"). This statement is made clear in Figure 5.5 (p. 294). Obtain the coordinates for both phosphoglycerate kinase and glucose and examine the relative size of these molecules. The coordinates are provided on the book's website. (a) The following are the PDB accession number of the molecules used in this problem or the link to the relevant websites. All of them are also available on the book website. 8 Chapter 2. Biological Structures ATP: ATP.pdb from http://xray.bmc.uu.se/hicup. Phosphatidylcholine: stearyl-oleyl-phophacholine.pdb from http://faculty.gvsu.edu/carlsont/mm/lipids/pg.html. B-DNA: bdna.pdb from http://chemistry.gsu.edu/glactone/PDB/ pdb.html. G-actin: 1J6Z.pdb from the PDB. Lambda repressor/DNA complex: 1LMB.pdb from the PDB has the DNAbinding region of lambda repressor complexed with DNA. Lac repressor/DNA complex: 1lbg.PDB and 1tfl.pdb from the PDB. Hemoglobin: 1hga.pdb from the PDB. Myoglobin: 1mbo.pdb from the PDB. HIV gp120: 1GC1.PDB from the PDB. Green uorescent protein (GFP): 1GFL.pdb from the PDB. RNA polymerase: 1L9U.pdb from the PDB. (b) Crystal structures can be displayed in a variety of ways. All the gures here were generated using Jmol version 11.4. Figures 2.38, 2.39, and 2.42 are \"ball-and-stick\" representations. The hydrogens are not shown, which is often the case, and the rest of the atoms are color coded such that grey corresponds to carbon, red to oxygen, blue to nitrogen, and orange to phosphate (see g. 2.38). Figures 2.40, 2.43, 2.44, and 2.45 are generally called ribbon-diagrams or, in Jmole, the representation \"scheme\" is \"cartoon.\" Ribbon diagrams help make sense of complex protein structures because they can reduce the molecule to its secondary structural motifs, such as -helices and -sheets. Figures 2.48 and 2.46 are space-lling representations, in which each atom is marked by its equivalent Van der Waals sphere, the diameter of which are generally roughly 2.5 (see g. 2.48B). Space-lling models have the same color-coding scheme A as the ball-and-stick diagrams. The wiry representation in gure 2.41 is called a \"trace\" in Jmol. These and other kinds of representations are listed in Jmol under Schemes in the Style menu. (c) Fig. 2.48 shows that the eective cross section of a phosphatidylcholine polar head is roughly a square with a side of 0.5 nm. This results in an area of 0.25 nm2 , very close to the value of 0.5 nm2 used in the chapter. (d) See g. 2.49. Problems and Solutions 9 N O P C 1.5 nm Figure 2.38: A crystal structure of adenosine-triphosphate (atp.pdb). The three phosphates PO4 are on the right and the rest is adenosine. ATP is the energy currency of the cell. Most reactions that require energy obtain it from one of the bonds of this molecule, hydrolizing ATP to ADP. 2.7 Coin ips and partitioning of uorescent proteins In the estimate on cell-to-cell variability in the chapter, we learned that the standard deviation in the number of molecules partitioned to one of the daughter cells upon cell division is given by n2 n1 1 2 = N pq. (2.33) (a) Derive this result. (b) Derive the simple and elegant result that the average dierence in intensity between the two daughter cells is given by (I1 I2 )2 = Itot , (2.34) where I1 and I2 are the intensities of daughters 1 and 2, respectively, and Itot is the total uorescence intensity of the mother cell and assuming that there is a linear relation between intensity and number of uorophores of the form I = N . (a) The number of ways we can divide N indistinguishable molecules into two cells is given by: N! W = , (2.35) n1 !(N n1 )! where N is a total number of molecules and n1 is the number of molecules that went to daughter cell 1. The probability of a particular sequence of N molecules going to daughter cell 1 and daughter cell 2, where the probability of the molecule going to daughter cell 1 is p and to daughter cell 2 is q, is pn1 q N n1 , (2.36)