Consider two fragments of DNA of radius r1 31 nm and r2 38

Consider two fragments of DNA of radius r1 = 31 nm and r2 = 38 nm when they are coiled up as they move through a gel. An electrophoresis analysis (see figure below) is carried out to separate these two fragments. Suppose the drag factor C in the equation Fdrag = Crv is C = 5.0 10-5 kg/(m · s), and assume each fragment has a charge q = 14e.

(a) If the electric field has a magnitude of 1.5 N/C, what is the speed of each fragment in the gel?

(b) How long will it take the smaller fragment to move a distance 1 cm (a typical distance used in DNA fingerprinting)?

(c) After the time in part (b) has elapsed, what is the separation of the two different-size fragments in the electrophoresis tube?

Solution

we know E=F/q

here F is force and q is charge

given Fdrag = Crv is C = 5.0 10-5 kg/(m · s

for first particle r1=31nm=31*10^-9 m

given q=-14e

(a) E=1.5N/C

so 1.5= (-5.0* 10^-5 *31*10^-9 v1)/(-14*1.6*10^-19 )

v1=0.216*10^-5 m/sec

similarly for second particle

v2=0.176*10^-5 m/sec

(b)smaller fragment radius= 31nm

velocity= 0.216*10^-5 m/sec

distance=1cm

velocity=distance/time

time= 10^-2 /0.216*10^-5 m/sec= 4.6*10³ sec