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distribution. We computed the variance
2
of the deviations of the asymptotic dis-
tribution and that of Eq. (15.10) for a rough landscape. As can be seen in Fig. 15.2,
for large genomes the transition to the mean-eld limit occurs for s!0.
15.3.3. Quasispecies, error threshold and Muller's ratchet
Before going in deep studying a general model of an evolving ecosystem that includes
the eect of competition (co-evolution), let us discuss a simple model [35] that
presents two possible mechanisms of escaping from a local optimum, i.e. the error
threshold and the Muller's ratchet.
We consider a sharp peak landscape: the phenotype u
0
= 0, corresponding to
the master sequence genotype x = 0(0; 0; : : : ) has higher tness A
0
= A(0), and
all other genotypes have the same, lower, tness A
. Due to the form of the tness
function, the dynamics of the population is fundamentally determined by the ttest
strains.
Let us indicate with n
0
= n(0) the number of individuals sharing the master
sequence, with n
1
= n(1) the number of individuals with phenotype u = 1 (only
one bad gene, i.e. a binary string with all zero, except a single 1), and with n
all
other individuals. We assume also non-overlapping generations,
During reproduction, individuals with phenotype u
0
can mutate, contributing
to n
1
, and those with phenotype u
1
can mutate, increasing n
. We disregard the
possibility of back mutations from u
to u
1
and from u
1
to u
0
. This last assumption
is equivalent to the limit L!1, which is the case for existing organisms. We
consider only short-range mutation with probability
s
. Due to the assumption of
large L, the multiplicity factor of mutations from u
1
to u
(i.e. L1) is almost the
same of that from u
0
to u
1
(i.e. L).
The evolution equation Eq (15.3) of the population becomes
1
N
K
n
0
0
=
(1
s
)A
0
n
0
;
1
N
K
n
0
1
=
((1
s
)A
n
1
+
s
A
0
n
0
) ;
(15.11)
1
N
K
n
0
=
A
(n
+
s
n
1
) :
and
A
0
n
0
+ A
(n
1
+ n
)
N
hAi=
is the average tness of the population.
The steady state of Eq. (15.11) is given by n
0
= n. There are three possible xed
points n
(i)
=
: n
(1)
= (0; 0; 0) (N
(1)
= 0), n
2
= (0; 0; K(11=hA
i))
n
(i)
0
; n
(i)
1
; n
(i)