Biology Reference
In-Depth Information
78
= S/L,
and
= T / L .
To derive these equations, refer to Figs. 3-3 and 3-4. First, we have:
[A+] = [B+] = [C+],
since they are all equivalent to each other. If we have three genetic markers
A, B and C in that order, instead of two genetic markers A and B shown in
Fig. 3-4, to get A+B+C+ transductants, we need a cross-over to the left of A
and another to the right of C. Thus, we have:
[A+B+C+]/[A+] = (1 - S / L - T/L)
3
3
.
= (1 -
-
If we consider A+B+ transductants, the third genetic marker can be C+ or
C-. Therefore, by definition, we have:
[A+B+] / [A+] = [A+B+C+] / [A+] + [A+B+C-] / [A+],
(1 - )
3
= ( 1 - - )
3
+ [A+B+C-] / [A+] .
or
[A+B+C-] / [A+] = (1 - )
3
- (1 - - )
3
.
or
Then among [A+], the sum of the four ratios must equal to one, with one of
the four being zero. Thus, we have:
