Biology Reference
In-Depth Information
be the sum of those possessing one copy of mutant and two copies of
wild type, 3
pq
2
, and those possessing three copies of wild-type mutant,
q
3
, that is, 3
pq
2
21.6%. If only
one mutant pRNA per procapsid was sufficient to render the procap-
sid unable to package DNA, then only those procapsids with three
wild-type pRNAs bound (0 mutant pRNA) would package DNA.
As another example, if
Z
+
q
3
=
3(0.7)(0.3)2
+
(0.3)3
=
0.216
=
6, but only five of the six need to be
wild type in order to produce final products, the relative yield will be
the sum of: (1) the probability of each receptor containing five wild
type; and (2) the probability of each receptor containing six wild type.
The probability that a given receptor will have
M
mutant and
W
wild
type bound at increasing percentages of mutant can be calculated
under the conditions that
Z
=
=
6 and 12, respectively, using Eq. (3)
(Table 1 and 2);
Z
!
M
!
W
!
p
M
q
W
(3)
Table 3 shows the prediction of the probability of sound reac-
tions (products) in the presence of various ratios of mutant and wild
type. In this case, it is assumed that the number of mutant,
X
, needed
to block the reaction is one, and
Z
varies from one to six. That is, one
mutant per single reaction is sufficient to kill the reaction, notwith-
standing that the total number of said component in each single
reaction is a constant, one, two, three, four, five, or six. The simpli-
fied formula
q
Z
was used to predict the probability of the dominant
productive reaction, since, if one mutant will block the reaction, the
only reaction that would be able to occur would be the probability
when
W
=
Z
and
M
=
0. An example of such a probability assessment,
when
Z
1, has been shown for pRNA of phi29.
107
Since
it was assumed that six copies are used and one mutant is sufficient
to block, all reactions involving one, two, three, four, or five copies
of mutant, respectively, would have been inactive. Only those involv-
ing six copies of wild type would be competent. The chance that a
receptor will catch six copies of wild type and no mutant pRNA in a
mixture containing
q
wild-type and
p
mutant pRNA is
q
6
.
=
6 and
X
=
