| Contents | Back to Features | About the Program |
The equation
c
L = --------- + Lo
m - Mo
is used, where L is the size of the unknown fragement,
m is
the mobility (scan number) of the fragment, and c, Mo,
and
Lo are constants. The constants are determined in the
following
manner. Fragments of known size (from the size standard) are used as
known
values for L and m. For every set of three standard
points,
a curve is defined. Let the point (xi, yi) denote a
standard
point, where xi is the mobility of the known fragement and
yi is the size of the known fragement. These three points give
us
the following system of equations:
c
y1 = ------- + Lo (1)
x1 - Mo
c
y2 = ------- + Lo (2)
x2 - Mo
c
y3 = ------- + Lo (3)
x3 - Mo
Solving this system of equations yields:
-(-x3 + x2) (-x3 + x1) (x1 - x2) (-y3 + y2) (y1 - y3) (-y2 + y1)
c = ----------------------------------------------------------------
2
(-y1 x3 + y2 x3 - y2 x1 + x2 y1 - x2 y3 + x1 y3)
- (-y1 x1 y3 + y1 y3 x3 - y2 x2 y1 + y2 x2 y3 - y2 y3 x3 + y2 y1 x1)
Lo = --------------------------------------------------------------------
(-y1 x3 + y2 x3 - y2 x1 + x2 y1 - x2 y3 + x1 y3)
(-y2 x2 x1 + y2 x2 x3 - y1 x1 x3 + x1 y3 x3 - x2 y3 x3 + x2 y1 x1)
Mo = --------------------------------------------------------------------
(-y1 x3 + y2 x3 - y2 x1 + x2 y1 - x2 y3 + x1 y3)
However, a problem is encountered when all three of the standard
points
used are on a line. In this case, the denominator (which is the same
for all three constants) becomes 0. When that happens, the equation
equation suggested by E.M. Southern is abandoned and instead a
line is used. Note that in this case a line will pass through all
three standard points.
To size an unknown fragment, the following method is used. For most fragements, the average of two curves will be used to find the size. The first curve will be defined by two standard points below the unknown fragement and one above. The second curve will be defined by one standard point below and two above. (Above and below are determined by scan number.) Remember that one (or both) of these curves can be linear. In the case that the unknown fragement is near the beginning or the end of the size standard ranges, only one curve is used. For example, if the curve with only one size standard point below includes the first size standard point, the curve with two points below does not exists. For points smaller than the first size standard point, a line is used. The line is defined by the first size standard point and by assuming the size at scan number 0 is 0. For points greater than the largest size standard point, the curve for the last three points is simply extended. Note: This is not very accurate at all for points outside of the defined size standards! For those points inside, the sizing agrees with that of the ABI software and appears to be reliable. Therefore, it is recommended that only the points between defined size standards be used.
The program must also be able to convert a size into a scan number. This answer is not unique (see below). However, in almost every case, one of the answers is clearly out of bounds. The bounds are defined such that if a fragements size is between two standard points, it's scan number must also be between the two size standards. If the correct value cannot be determined, the program will display an error on the standard output and then take the smaller value.
The conversion back is broken down into three cases, a size derived from two curves, a line and a curve, and two lines. For two curves, the size s is defined as:
c1/(m - m1) + L1 + c2/(m - m2) + L2
s = -------------------------------------
2
solving for m yeilds:
-c1m2 - c2m1 2
m = f + sqrt(-------------- - m1m2 + f )
2s - L1 - L2
or
-c1m2 - c2m1 2
m = f - sqrt(-------------- - m1m2 + f )
2s - L1 - L2
where
1 c1 + c2
f = - ( m1m2 + -----------)
2 2s - L1 -L2
For a curve and a line, the size is determined by:
c/(m -mo) + Lo + am + b
s = --------------------------
2
solving yields:
g + sqrt(g^2 - 4a(mo*d + c)) g - sqrt(g^2 - 4a(mo*d + c))
m = -----------------------------, or m = ------------------------------
2a 2a
where
d = 2s - Lo - b g = amo + dFinally, for two lines, the solution is trivial and unique. This is because if points a, b, and c are on a line and points b, c, and d are on a line, then a and d must be on the same line.
| Contents | Back to Features | About the Program |