Agriculture Reference
In-Depth Information
=
(
/
)
×
So percentage of recombinants
178
1195
it is impossible to detect double recombination events,
and as a result map distances are always going to provide
under estimates of actual recombination frequencies.
To avoid such discrepancies, recombination frequen-
cies, which are not additive, are usually converted to a
cM scale using the function published by J.B.S. Haldane
in 1919, being:
=
100
14.9%
For
A
-
C
loci we have:
Parental types
A
_
C
_
500
+
35
=
535
aacc
510
+
38
=
548
Recombinants
aaC
_50
+
3
=
53
cM
=
(
−
ln
(
1
−
2
R
))
×
50
A
_
cc
55
+
4
=
59
=
(
/
)
×
where
R
is the recombination frequency. From this we
see that the map distances transform to:
So percentage of recombinants
112
1195
=
100
9.4%.
For the
B
-
C
loci we have:
Parental types
A
------------17.6
cM
------------
B
A
------10.4
cM
-----
C
- - -7.2
cM
---
B
B
_
C
_
500
+
50
=
550
bbcc
510
+
55
=
565
The standard error of these map distances can be
calculated using the equation:
+
=
Recombinants
bbC
_35
3
38
+
=
B
_
cc
38
4
42
s
.
e
.
=[
R
/(
1
−
R
)
]
/
n
=
(
/
)
So
percentage
of
recombinants
80
1195
where
R
is the recombination frequency, and n is the
number of plants observed in the three-way test cross.
×
6.7%.
We obtain the
map distance
as being equivalent to
the percentage recombinants and so we have:
100
=
Pleiotropy
µ
A
------------14.9 m
-------------
B
µ
µ
---
B
From the map, it is clear that the map distance
A
to
B
(14.9 m
A
------9.4 m
-------
C
- - -6.7
m
Very tight linkage between two loci can be confused
with pleiotropy, the control of two or more characters
by a single gene. For example, the linkage of resistance to
the soybean cyst nematode and seed coat colour seems
to be a case of pleiotropy because the two characters
were always inherited together. The only way link-
age and pleiotropy can be distinguished, is effectively
the negative way, that is, to find a crossover product,
such as a progeny homozygous for resistance and yellow
seed coat. Resistance and yellow seed coat could never
occur in a true breeding individual if true pleiotropy
was present. Thousands of individuals may have to be
grown to break a tight linkage that appears to be one of
pleiotropy.
is less than the added distance A-C
plus C-B. The method of calculation used to estimate
these distances assumes that only the non-parental types
are included, as would be the case where the link-
age between two loci is considered. Where three loci
are involved it is possible to include the double cross
over recombination events in estimating the distance
between the furthest two loci. In this case we could
calculate the
A
-
B
distance from:
µ)
=
Non recombinant types
A_B_C_
500
500
aabb
510
=
510
Recombinant events
aaB_
50
+
38
=
88
A_bb
55
+
35
=
90
Epistasis
Twice double cross over
2
(
3
+
4
)
=
14
recombinants
Different loci may be independent of each other in their
segregation and recombination patterns. Independence
of gene transmission, however, does not necessarily
imply independence of gene action or expression. In
fact, in terms of its final expression in the phenotype of
the individual,
no gene acts by itself
.
=
(
/
)
×
So percentage of recombinants
192
1195
=
100
16.1%, which is the same distance that would
be estimated by simple adding
A
-
C
to
C
-
B
.
This indicates a general flaw in linkage map distance
estimation in that where there is no 'centre gene' locus,
