Agriculture Reference
In-Depth Information
These three generations are typified by genetic seg-
regation. It is therefore necessary to derive the pro-
portions of the different genotypes, and their relative
contributions to the means, in the various generations.
The
Aa
gene segregates with gamete formation in
the F
1
thus giving, in the F
2
generation, the genotypes
AA
,
Aa
and
aa
in the ratio 1
AA
:2
Aa
:1
aa
or, as
proportions,
explain the inheritance of quantitative (continuously
varying) characters exclusively in terms of the additive
and dominance properties of the single gene differ-
ences which underlie it. It is now necessary to consider
whether models based on only additive and domi-
nance genetic differences are adequate. From a plant
breeding standpoint it is important to know if the inher-
itance of a character under selection is controlled by an
additive-dominance model because many assumptions,
notably the response to selection, are based on heritabil-
ity, and estimating heritability is usually assumed that
this model is suitable. Testing the additive- dominance
model in quantitative genetics should be regarded as
directly comparable to testing for the absence of linkage
or epistasis with qualitative inheritance. However, with
genes showing qualitative differences the most common
testing method to compare frequencies of genotypes or
phenotypes is a
1
4
1
1
AA
,
2
Aa
and
4
aa
. The mean expression
+
of the
AA
plants is
m
a
. But, since only one quarter of
the F
2
generation is of the
AA
genotype, it contributes
1
to the F
2
generation mean. Similarly, since
the mean expression of the
aa
plants is
m
/
4
(
m
+
a
)
−
a
and they
1
4
make up
of the F
2
generation,
aa
plants contribute
1
4
−
)
to the F
2
generation mean. Finally, since
half the F
2
generation has the genotype
Aa
which has a
mean expression of
m
(
m
a
+
d
, the heterozygotes contribute
1
2
+
)
[
]
is
a summation of additive effects over all loci therefore,
for simplicity we assume (without proof ):
(
m
d
to the F
2
generation mean. The term
a
2
test. In quantitative genetics geno-
types (or phenotypes) do not fall into distinct classes
and hence frequency
χ
2
tests are not appropriate. So
we need an equivalent but appropriate test.
First you should recall that the components
m
,
χ
1
4
P
1
+
1
2
F
1
+
1
4
P
2
F
2
=
[
]
a
1
4
(
1
2
(
1
4
(
[
]
=
+[
]
)
+
+[
]
)
+
−[
]
)
and
are derived from the generation means of the
P
1
, P
2
and
d
m
a
m
d
m
a
F
1
generations:
1
4
m
1
4
[
1
2
m
1
2
[
1
4
m
1
4
[
=
+
a
]+
+
d
]+
−
a
]
P
1
=
P
2
=
F
1
=
m
+[
a
]
:
m
−[
a
]
:
m
+[
d
]
1
2
[
and from these similar equations can be formulated for
the generation means of the F
2
,
=
+
]
m
d
B
1
and
B
2
generations:
Considering a single gene model again, we have the
B
1
generation is composed of
1
2
[
1
2
[
1
2
[
F
2
=
B
1
=
+
]
+
]+
]
m
d
:
m
a
d
:
1
2
1
2
AA
and
Aa
. Again
assuming without proof that
[
a
]
is an accumulation
1
2
[
1
2
[
B
2
=
m
−
a
]+
d
]
of all additive effects and
[
d
]
an accumulation of all
dominant effects. We have:
B
1
=
These six equations lead to various
predictive
rela-
tionships between the means of different combinations
of the generations.
1
1
2
P
1
+
2
F
1
For example,
if the additive-
1
2
(
1
2
(
=
+[
]
)
+
+[
]
)
dominance
model
is
correct
then
it
can
be
pre-
m
a
m
d
dicted that:
1
2
[
1
2
[
=
+
]+
]
2 B
1
−
F
1
−
P
1
=
m
a
d
0
Similarly the B
2
generation would be:
B
2
=
This relationship is easily proved by the substitution
of the appropriate combinations of
m
,
[
]
[
]
a
and
d
for
1
2
[
1
2
[
m
−
a
]+
d
]
the three generation means:
2
m
1
2
[
1
2
[
+
]+
]
−
(
+[
]
)
−
(
+[
]
)
=
a
d
m
d
m
a
0
TESTING THE MODELS
(
+[
]+[
]
)
−
(
+[
]
)
−
(
+[
]
)
=
2
m
a
d
m
d
m
a
0
Earlier, a model which we shall now call the
additive-
dominance model
, was put forward that purports to
+[
]+[
]−
−[
]−
−[
]=
2
m
a
d
m
d
m
a
0
