Re: [ccp4bb] How to compare B-factors between structures?
James, Thank you for your help. I appreciate the very thorough explanation. I have never heard of noise although I have produced a 'kicked' map in phenix and refined the structure using those maps..though I guess this is different. I will try it. Thanks. -Yarrow 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
Re: [ccp4bb] How to compare B-factors between structures?
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.
Re: [ccp4bb] How to compare B-factors between structures?
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 jmhol...@lbl.gov 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. --
Re: [ccp4bb] How to compare B-factors between structures?
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 jmhol...@lbl.govmailto:jmhol...@lbl.gov 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
Re: [ccp4bb] How to compare B-factors between structures?
Good point; I've tested this (n=1) in the past with a high-resolution dataset (synchrotron data) and low-resolution dataset (in-house) of crystals of the same protein grown in the same drop. Same space group, same unit cell. B-factors for the low-resolution dataset were higher. After dividing every individual B-factor by the average B-factor of each, the normalized-B-factor-versus-residue plot was identical for both structures. Adding or subtracting a constant value didn't do that. As I pointed out, this is only n=1, but comparing the high-and low-resolution structures of the same condition should give the answer as to which B-factor normalization is the most appropriate. Filip Van Petegem On Mon, Mar 4, 2013 at 11:16 AM, Jacob Keller j-kell...@fsm.northwestern.edu 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 jmhol...@lbl.gov 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).
Re: [ccp4bb] How to compare B-factors between structures?
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 jmhol...@lbl.gov mailto:jmhol...@lbl.gov 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
Re: [ccp4bb] How to compare B-factors between structures?
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 jmhol...@lbl.gov 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 jmhol...@lbl.gov 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
Re: [ccp4bb] How to compare B-factors between structures?
Perhaps I was a bit gruff, but what I was trying to do was throw out a voice of caution. In crystallography, it is very tempting to normalize everything by setting the mean to zero and the RMS deviation to one, but this is a linear transformation, and linear transformations only work when the things you are comparing are linear. True, everything is linear to a first-order approximation, and you may well be okay comparing B factors between two crystals that are not all that different in average B, but the assumption of linearity with something fundamentally non-linear like B factors will break down at some point. I'm not really sure where, but I would suspect that normalizing B factors from a 5 A structure to those of a 2.0 A structure might be ill advised. My opinion is that multiplying and dividing B factors makes no physical sense. If you must normalize something, I'd suggest normalizing the half-occupancy volume (proportional to sqrt(B)^3), or the width of the half-occupancy volume (proportional to sqrt(B)). If you want to try to make a leap to entropy, then if you assume the half-occupancy volume contains an ideal gas, then the entropy should be proportional to log(B). But B itself is proportional to the surface area of the half-occupancy volume, and I can't think of too many physical-chemical properties that would be proportional to that. Nevertheless, purely on theory, I predict you will much more easily hit pitfalls (or quicksand) by trying to normalize B directly instead of first converting it into something more tangible (like a volume). That's all I'm sayin' -James Holton MAD Scientist On Mon, Mar 4, 2013 at 12:31 PM, John Fisher johncfishe...@gmail.com wrote: 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 jmhol...@lbl.gov 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 jmhol...@lbl.gov 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
Re: [ccp4bb] How to compare B-factors between structures?
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. -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697 -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697
Re: [ccp4bb] How to compare B-factors between structures?
Why not normalize each to its average b-factor? And...maybe they are not significantly different in the first place? JPK On Mon, Feb 25, 2013 at 3:08 PM, Yarrow Madrona amadr...@uci.edu 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. -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697 -- *** Jacob Pearson Keller, PhD Postdoctoral Associate HHMI Janelia Farms Research Campus email: j-kell...@northwestern.edu ***
Re: [ccp4bb] How to compare B-factors between structures?
Thanks Nat, I was planning on plotting B-factors vs. residue anyway, so maybe this will save me time. I will take a look. -Yarrow Not a CCP4 solution, but the structure comparison program in the Phenix GUI will plot B-factors for different structures of the same protein. (I am happy to make additions or modifications to this, but so far I haven't received much feedback.) -Nat On Mon, Feb 25, 2013 at 12:08 PM, Yarrow Madrona amadr...@uci.edu 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. -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697 -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697
Re: [ccp4bb] How to compare B-factors between structures?
Thank you Manfred, To be honest I have tried to understand the Pearson linear CC since reading the nature crystallography methods paper by Karplus and diederichs, but I am having trouble. I do not clearly understand what it represents. Do you know any good resources that could help me out? -Yarrow If you have two sets of numbers which correspond ideally, you should calculate a correlation coefficient, to be precise a Pearson linear CC. This is independent of scaling. If you want to compare mutant vs native, you probably have to calculate residue average B-factors, because you won't have the exact same number of atoms. Best, Manfred On 25.02.2013 21:08, 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. -- Dr. Manfred. S. Weiss Helmholtz-Zentrum Berlin für Materialien und Energie Macromolecular Crystallography (HZB-MX) Albert-Einstein-Str. 15 D-12489 Berlin GERMANY Fon: +49-30-806213149 Fax: +49-30-806214975 Web: http://www.helmholtz-berlin.de/bessy-mx Email: mswe...@helmholtz-berlin.de Helmholtz-Zentrum Berlin für Materialien und Energie GmbH Mitglied der Hermann von Helmholtz-Gemeinschaft Deutscher Forschungszentren e.V. Aufsichtsrat: Vorsitzender Prof. Dr. Dr. h.c. mult. Joachim Treusch, stv. Vorsitzende Dr. Beatrix Vierkorn-Rudolph Geschäftsführung: Prof. Dr. Anke Rita Kaysser-Pyzalla, Thomas Frederking Sitz Berlin, AG Charlottenburg, 89 HRB 5583 Postadresse: Hahn-Meitner-Platz 1 D-14109 Berlin http://www.helmholtz-berlin.de -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697
Re: [ccp4bb] How to compare B-factors between structures?
Hi Yarrow, I'm sure other visualizing tools can do this, but I just wanted to share that our free Discovery Studio Visualizerhttp://accelrys.com/products/discovery-studio/visualization-download.php can do this quite easily. Contact me off the list and we can set up a time when I can show you how to do this. I could also send you some instructions as well. Cheers, Francisco Sr. Product Manager Accelrys Software, Inc -Original Message- From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On Behalf Of Yarrow Madrona Sent: Monday, February 25, 2013 1:52 PM To: CCP4BB@JISCMAIL.AC.UK Subject: Re: [ccp4bb] How to compare B-factors between structures? Thanks Nat, I was planning on plotting B-factors vs. residue anyway, so maybe this will save me time. I will take a look. -Yarrow Not a CCP4 solution, but the structure comparison program in the Phenix GUI will plot B-factors for different structures of the same protein. (I am happy to make additions or modifications to this, but so far I haven't received much feedback.) -Nat On Mon, Feb 25, 2013 at 12:08 PM, Yarrow Madrona amadr...@uci.edumailto:amadr...@uci.edu 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. -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697 -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697
