Agriculture Reference
In-Depth Information
0to
z
) a function of probability (
p
). For the sake of
simplicity,
z
Would you expect all the potato plants to have exactly
the same weight of potatoes? - probably not. Indeed
the weights can be presented in the form of a histogram
(Figure 5.1) where 11 000 yields were divided into 17
weight classes. The variation in weight is obvious; some
plants produce less than 0.5 kg of tubers, while others
produce over 5 kg. Most, however, are grouped around
the average of 3.2 kg weight.
Yield in potato, as in other crops, is polygenically
inherited. Yield is therefore not controlled by a single
gene but by many genes, all acting collectively. Thus
yield has more chance of some of the processes these
many genes control being affected by differences in
the 'environment'. As the example above involved har-
vesting plants which are all indeed clones of the same
genotype, then the variation observed is due entirely
to the environmental conditions that each plant was
subjected to.
One of the major differences between single gene
inheritance and multiple gene inheritance is that the
former is less affected or influenced by the environ-
mental conditions compared to the later. When potato
yield are being recorded we are recording the expression
of the phenotype, but are interested in the genotype.
The two are related by the equation:
=
=
1.645 for
p
0.95 (i.e. 95% certain)
=
and
z
2.326 (for 99% certainty).
For example, consider the frequency of homozy-
gous recessive genotypes (
aabb
) resulting from the cross
AAbb
×
aaBB
at the F
3
stage. How many F
3
lines would
need to be evaluated to be 95% certain of obtaining 10
homozygous recessive genotypes? Here the probability
of
aabb
is 9/64 (i.e.
q
=
=
0.141),
p
0.95, therefore
=
z
1.645.
=
(
[
(
−
)
+
(
−
)
]
n
2
10
0.5
2.706
1
0.141
2
+
[
(
(
−
)
+
(
−
)
1.645
2.706
1
0.141
4
1
0.141
0.5
)
−
1
×
(
10
−
0.5
)
]
)(
2
×
0.141
110.69
Therefore 111 F
3
lines need to be grown.
=
QUANTITATIVE GENETICS
The basis of continuous variation
With qualitative inheritance, the segregating individ-
ual phenotypes are usually easily distinguished and fall
into a few phenotypic classes (i.e. tall or short). Char-
acters that are controlled by multiple genes do not fall
into such simple classifications. Consider a hectare field
of potatoes, all planted with one cultivar and so every
single plant in the field should be of identical geno-
type (ignoring mutation and errors). In that field there
are likely to be 11 000 plants. At harvest, you have been
given the task of harvesting all 11 000 plants and weigh-
ing the tubers that come from each plant separately.
2
e
P=G+E+GE
+
σ
where P is the phenotype, G is the genotypic effect, E is
the effect of the environment in which the genotype is
grown, GE is the interaction between the genotype and
the environment and
e
is a random error term.
Single gene characters are often less affected by envi-
ronment and usually do not show much influence
σ
600
500
400
300
200
100
0
0.4 0.7
1
1.3 1.6 1.6 2.2 2.7 3
Plant yield (kg)
3.3 3.6 3.9 4.2 4.5 4.8 5.1 5.4
Figure 5.1
Yield of tubers from individual potato plants taken at random from a single clonally reproduced potato cultivar.
