Seriously? I believe this specific forum has become quicksand rather than a useful tool. Normalization based simply on ratios of one struct to the other should allow normalization. Correct?, or have I just simply lost my mind here? J
John Fisher, M.D./PhD St. Jude Children's Research Hospital Department of Oncology Department of Structural Biology W: 901-595-6193 C: 901-409-5699 On Mar 4, 2013, at 1:26 PM, James Holton <[email protected]> wrote: > > No, you can only add and subtract B values because that is mathematically > equivalent to multiplication in reciprocal space (which is equivalent to > convolution in real space): > > exp(-B1*s^2) * exp(-B2*s^2) = exp(-(B1+B2)*s^2) > > Multiplying and dividing B values is mathematically equivalent to applying > fractional power-law or fractional root functions in reciprocal space (and I > don't even want to think about what that does in real space). > > exp(-B1*B2*s^2) = ??? > > -James Holton > MAD Scientist > > > On 3/4/2013 11:19 AM, Bosch, Juergen wrote: >> Yep, I agree calculate the average B per structure and divide each B by this >> value, then multiply it by any value that is reasonable so you can visualize >> color differences :-) >> Jürgen >> >> On Mar 4, 2013, at 2:16 PM, Jacob Keller wrote: >> >>> You only entertain addition+subtraction--why not use >>> multiplication/division to normalize the b-factors? >>> >>> JPK >>> >>> On Mon, Mar 4, 2013 at 2:04 PM, James Holton <[email protected]> wrote: >>>> Formally, the "best" way to compare B factors in two structures with >>>> different average B is to add a constant to all the B factors in the low-B >>>> structure until the average B factor is the same in both structures. Then >>>> you can compare "apples to apples" as it were. The >>>> "extra B" being added is equivalent to "blurring" the more well-ordered >>>> map to make it match the less-ordered one. Subtracting a B factor from the >>>> less-ordered structure is "sharpening", and the reason why you shouldn't >>>> do that here is because you'd be assuming that a sharpened map has just as >>>> much structural information as the better diffracting crystal, and that's >>>> obviously no true (not as many spots). In reality, your comparison will >>>> always be limited by the worst-resolution data you have. >>>> >>>> Another reason to add rather than subtract a B factor is because B factors >>>> are not really "linear" with anything sensible. Yes, B=50 is "more >>>> disordered" than B=25, but is it "twice as disordered"? That depends on >>>> what you mean by "disorder", but no matter how you look at it, the answer >>>> is generally "no". >>>> >>>> One way to define the "degree of disorder" is the volume swept out by the >>>> atom's nucleus as it "vibrates" (or otherwise varies from cell to cell). >>>> This is NOT proportional to the B-factor, but rather the 3/2 power of the >>>> B factor. Yes, 3/2 power. The value of "B", is proportional to the >>>> SQUARE of the width of the probability distribution of the nucleus, so to >>>> get the volume of space swept out by it you have to take the square root >>>> to get something proportional the the width and then you take the 3rd >>>> power to get something proportional to the volume. >>>> >>>> An then, of course, if you want to talk about the electron cloud (which is >>>> what x-rays "see") and not the nuclear position (which you can only see if >>>> you are a neutron person), then you have to "add" a B factor of about 8 to >>>> every atom to account for the intrinsic width of the electron cloud. >>>> Formally, the B factor is "convoluted" with the intrinsic atomic form >>>> factor, but a "native" B factor of 8 is pretty close for most atoms. >>>> >>>> For those of you who are interested in something more exact than >>>> "proportional" the equation for the nuclear probability distribution >>>> generated by a given B factor is: >>>> kernel_B(r) = (4*pi/B)^1.5*exp(-4*pi^2/B*r^2) >>>> where "r" is the distance from the "average position" (aka the x-y-z >>>> coordinates in the PDB file). Note that the width of this distribution of >>>> atomic positions is not really an "error bar", it is a "range". There's a >>>> difference between an atom actually being located in a variety of places >>>> vs not knowing the centroid of all these locations. Remember, you're >>>> averaging over trillions of unit cells. If you collect a different >>>> dataset from a similar crystal and re-refine the structure the final x-y-z >>>> coordinate assigned to the atom will not change all that much. >>>> >>>> The full-width at half-maximum (FWHM) of this kernel_B distribution is: >>>> fwhm = 0.1325*sqrt(B) >>>> and the probability of finding the nucleus within this radius is actually >>>> only about 29%. The radius that contains the nucleus half the time is >>>> about 1.3 times wider, or: >>>> r_half = 0.1731*sqrt(B) >>>> >>>> That is, for B=25, the atomic nucleus is within 0.87 A of its average >>>> position 50% of the time (a volume of 2.7 A^3). Whereas for B=50, it is >>>> within 1.22 A 50% of the time (7.7 A^3). Note that although B=50 is twice >>>> as big as B=25, the half-occupancy radius 0.87 A is not half as big as >>>> 1.22 A, nor are the volumes 2.7 and 7.7 A^3 related by a factor of two. >>>> >>>> Why is this important for comparing two structures? Since the B factor >>>> is non-linear with disorder, it is important to have a common reference >>>> point when comparing them. If the low-B structure has two atoms with B=10 >>>> and B=15 with average overall B=12, that might seem to be "significant" >>>> (almost a factor of two in the half-occupancy volume) but if the other >>>> structure has an average B factor of 80, then suddenly 78 vs 83 doesn't >>>> seem all that different (only a 10% change). Basically, a difference that >>>> would be "significant" in a high-resolution structure is "washed out" by >>>> the overall crystallographic B factor of the low-resolution structure in >>>> this case. >>>> >>>> Whether or not a 10% difference is "significant" depends on how accurate >>>> you think your B factors are. If you "kick" your coordinates (aka using >>>> "noise" in PDBSET) and re-refine, how much do the final >>>> B factors change? >>>> >>>> -James Holton >>>> MAD Scientist >>>> >>>> >>>> On 2/25/2013 12:08 PM, Yarrow Madrona wrote: >>>>> Hello, >>>>> >>>>> Does anyone know a good method to compare B-factors between structures? I >>>>> would like to compare mutants to a wild-type structure. >>>>> >>>>> For example, structure2 has a higher B-factor for residue X but how can I >>>>> show that this is significant if the average B-factor is also higher? >>>>> Thank you for your help. >>> >>> >>> >>> -- >>> ******************************************* >>> Jacob Pearson Keller, PhD >>> Postdoctoral Associate >>> HHMI Janelia Farms Research Campus >>> email: [email protected] >>> ******************************************* >> >> ...................... >> Jürgen Bosch >> Johns Hopkins University >> Bloomberg School of Public Health >> Department of Biochemistry & Molecular Biology >> Johns Hopkins Malaria Research Institute >> 615 North Wolfe Street, W8708 >> Baltimore, MD 21205 >> Office: +1-410-614-4742 >> Lab: +1-410-614-4894 >> Fax: +1-410-955-2926 >> http://lupo.jhsph.edu >
