Physics 450 - Problem Set 2 Solutions - February 13 2001

1. (a) Note that
(2/x2) e-ax2 = (/x) (-2ax) e-ax2 = (-2 a+ 4 a2 x2) e-ax2
 
and so plugging in a = -1/(4Dt) gives:
(2/x2) e-x2/(4Dt) = [ -1/(2Dt) + x2/(4D2t2) ] e-x2/(4Dt)
 
Next compute
(/t) t-1/2 e-x2/(4Dt) = [-t-3/2/2 + x2 / (4Dt5/2) ] e-x2/(4Dt) 
 
A bit more algebra shows that
(/ t - D 2 / x2)t-1/2 e-x2/(4Dt) = 0 
 
(b) We need to compute
ó 
õ 		 -¥ 

-¥ 

 dx t-1/2 e-x2/(4Dt) = t-1/2 (2 p×2Dt)1/2 = 2 D1/2
 
i.e. constant in time. Here we have used the most important integral in physics,
ó 
õ 		 ¥ 

-¥ 

 dx e-x2/(2s2) = (2p)1/2 s
 
(c) This is just 3 times the algebra of (a), i.e.
(2/x2 + 2 / y2 + 2 / z2 ) e-(x2+y2+z2)/(4Dt) = [-3/(2Dt) + (x2 + y2 + z2)/(4D2 t2)]e-(x2+y2+z2)/(4Dt)
 
and
(/t) t-3/2 e-(x2+y2+z2)/(4Dt) = [-(3/2) t-5/2 + (x2+y2+z2)/(4Dt7/2)]e-(x2+y2+z2)/(4Dt) 
 
so that
(/ t - D [2 / x2+ 2 / y2+ 2 / z2] )t-3/2 e-x2/(4Dt) = 0 
 
(d) The integral of interest,
ó 
õ 		 ¥ 

-¥ 

 dx  ó 
õ 		 ¥ 

-¥ 

 dy  ó 
õ 		 ¥ 

-¥ 

 dze-(x2+y2+z2)/(4Dt)
 
is just the cube of the integral of (b), since eA+B = eA eB, i.e.
=  ó 
õ 		 ¥ 

-¥ 

 dx e-x2/(4Dt)  ó 
õ 		 ¥ 

-¥ 

 dy e-y2/(4Dt)  ó 
õ 		 ¥ 

-¥ 

 dz e-z2/(4Dt) =  é 
ë 		 ó 
õ 		 ¥ 

-¥ 

 dx e-x2/(4Dt)  ù 
û 		 3 
 
 
which by (b) is independent of t.

2. (a) A good way to solve this problem would be to look up the sequence on Genbank, and immediately find that it is a protein from the E. coli lambda virus, and that this sequence starts at base position 7202. Referring to the Genbank map at the start of the file, we find that the protein amino-acid sequence is

MTKDELIARLRSLGEQLNRDVSLTGTKEELALRVAELKEELDDTDETAGQDTPLSRENVLTGHENE VGSAQPDTVILDTSELVTVVALVKLHTDALHATRDEPVAFVLPGTAFRVSAGVAAEMTERGLARMQ

Note that this is actually only 132 amino acids since the final taa codon is a STOP signal. It is worth noting that the leading methionine (M) is always the first amino acid of a protein.

(b) 5'-ctggccggacct-3' (remember that hybridizing strands pair in opposite directions)

(c) There are of course many possibilities. One is 5'-attggccatt-gcgatatgcg-aatggccaat-3'

3. As discussed in class, all of the solutions are constructed from the basic law < r2 > = 6 D t

(a) Obviously c = 6D, so that D = c/6

(b) We can write

< (r1 - r2)2 > = < (r1 - r2)·(r1-r2) > 
 
 
= < r12 + r22 + 2 r1 ·r2 > 
 
Now, we use the linearity of the average, < A + B > = < A > + < B > to break this into
= < r12 > + < r22 > +2 < r1·r2 > = 6 D t + 6 D t + < r1 ·r2 > 
 
The final term is zero, since the Brownian motions of the two particles are independent of one another - on average they will each independently average to zero, so their averaged dot product must also be zero.

Thus, ct = 12 Dt giving D = c/12 for each particle.

(c) Actually not so hard. All you do is include the initial displacement, expand the dot-products as above, and find D = c/12 again.

4. (a) The concentration as a function of height is proportional to the probability distribution, and must be given by the Boltzmann distribution
c(z) = A e-Dm gz/(kB T)
 
where A must be set by requiring that the mean concentration be c0.

The mean concentration is just

h-1  ó 
õ 		 h 

0 

 dz c(z) = A  kB T 
Dmg h		 [1 - e-Delta mgh/(kB T)] 
 
Since this must be equal to c0, we find that
A = c0  Dm g h /(kB T) 
1 - e-Dmgh/(kB T) 
 
and therefore that the concentration profile is
c(z) = c0  Dm g h /(kB T) 
1 - e-Dm g h / (kB T) 		 e-Dmgz/(kB T) 
 
(b) For particles which are 20% heavier than water and radius r, we should use Dm = 4 p r3 ×0.2 ×rwater /3.
The factor   4 p ×0.2 ×rwater / 3 =  840 kg/m3, so that for the 1 nm and 1 micron particles,
Dm = 8.4×10-25 kg and 8.4×10-16 kg, respectively.
Now consider the combination T/(Dm g) = 5×102 m and 5×10-7 m, respectively.   This is the decay length of the exponential distribution, and therefore is the height that the particles can be excited up to thermally.

Thus, the 1 nm particles would be suspended up to  a height of 500 m, so in the 1 cm tube they will be at essentially constant concentration with height - this is the size of a small protein, so they are `well suspended'.

By contrast, the 1 micron particles will only be suspended up to a height of 0.5 microns, i.e. they will be in a very narrow layer at the bottom of a 1 cm tube.

The figure shows the concentration profile with height for the 1 nm and 1 micron particles;  the 1 nm particles have an almost constant concentration profile, while the 1 micron particles are all at the bottom of the tube (I have grossly exaggerated the width of the 1 micron particle distribution).

I also show the result for 10 nm radius particles (a not-too-large protein) which is now suspended in equilibrium with a distribution which is a fraction of the height of the tube.  These particles are marginally suspended in the tube.


File translated from TEX by TTH, version 2.53.
On 13 Feb 2001, 22:32.