8.592: Solutions for Assignment 1
Problem 1
If the probability of having a stop is ps , then the probability of having a sequence of
DNA long N is:
PN = (1 ps )N ps
(1)
We have:
ps = 3/64
(2)
P600 = 1.4 1014
(3)
Setting N = 600, we get
Since there are 6 w
8.592: Solutions for Assignment 3
1
Spring 2005
Fibonaccis Rabbits
(a) The matrix relating the young and adult populations from one generation to the
next is
YN +1
AN +1
0
p
=
f
p
YN
AN
.
(1)
(b) To nd the behavior for large N , we can examine the eigenva
8.592: Solutions for Assignment 2
1. Number of gapped alignments
(a)
(A1 , A2 , , An )
(B1 , B2 , , Bm )
Let us to nd a recursive relationship for the number of ways of intercalating two
sequences of lengths n and m (F (n; m). The rst letter of the interc
8.592J: Solutions for Assignment 4
1
Spring 2005
Flory Theory
(a) For a charged polymer of length N , the dependence of R on Q and N , is estimated
2
by balancing the Coulomb energy Ec = RQ2 with the entropy associated with
d
conning the polymer to a size
8.592J: Solutions for Assignment 5
1
Spring 2005
Designed REM
(a) We have to compute the average energy of the system. As we saw in class, there
will be a freezing temperature Tf that separates two regimes of behaviour for this
system.
For T > Tf :
(E ) =
8.592J: Solutions for Assignment 8
Spring 2005
Problem 1
(a) The weight matrix is calculated as
1
fi (b)
+
,
f0 (b) n
wi (b) = ln
(1)
where fi (b) is the observed frequency of nucleotide b at position i, f0 (b) is the background frequency of b and the pse
8.592J: Solutions for Assignment 7
Spring 2005
Problem 1
(a) A lament of length 2 can be created by addition of a monomer to one of
length 1 (at rate a) or removal of a monomer from a lament of length + 1
(at rate b). Such laments change their length thro
8.592J: Solutions for Assignment 6
Spring 2005
Problem 1
a) For this part, we simply have to perform this computation:
3R2
(R) = (N )e 2Na2
gN
(N ) =
3
(2N a2 /r ) 2
(1)
(2)
(R)eE (R) =
Z=
(3)
R
= (N )
3R2
d3 Re 2Na2 eF Rcos() =
3
2 2 (N )
3 3 2 F 2 Na2