Agriculture Reference
In-Depth Information
Table 7.10 Relative ranking of eight potato crosses
(C1 to C8) based on cross prediction of Breeder's
Preference that genotypes chosen at random from
each cross will exceed a Breeder's Preference rating
greater than 5, on the 1 to 9 scale and the number
of selected breeding lines from each cross that was
selected in the 4th, 5th and 6th selection stage in
the breeding program at the Scottish Crop Research
Institute.
Therefore given the above genetic parameters we
would expect that 30.85% of all possible recombinants
from the cross will have a greater (or equal) value than
the target value.
Consider the same set of parameters ( m
σ g =
4) but now with a target value of 11 (i.e. a target value
less than the progeny mean). We now have:
=
12,
T
m
σ g =
11
12
=−
0.25
Cross
Rank
Selected to stage
4
Four
Five
Six
Looking this value up from the tables we have a prob-
ability value of 0.5987. In the example above we then
subtracted this value from one to obtain the correct
probability. In this case however, this
C1
1
15
3
2
C2
3
9
3
2
C3
6
1
0
0
)/σ g has a
negative value as so our required probability is 1
(
T
m
C4
4
2
0
0
table value) which is in fact, simply the value obtained
from tables.
In summary, four possibilities exist:
(
1
C5
8
1
0
0
C6
5
11
6
1
C7
2
12
7
3
C8
7
0
0
0
If the probability that a recombinant is to exceed a
target value and
This has prompted several other breeding organizations
to change the means by which they reduce clonal num-
bers in the early generations of potato and sugarcane
breeding schemes.
(
)/σ g is positive then the
T
m
probability is 1
the tabulated value
If the probability that a recombinant is to exceed a
target value and
)/σ g is negative then the
probability is simply the tabulated value
(
T
m
Use of normal distribution function tables
The area under a unit normal distribution (a normal
distribution with mean of zero and variance of one) is
frequently tabulated in statistical tables. It is common
for the area to be given from
If the probability that a recombinant is less than the
target value and
)/σ g is positive then the
probability is simply the tabulated value
(
m
T
If the probability that a recombinant is less than the
target value and
−∞
to the required target
(
m
T
)/σ g is negative , then the
−∞
+∞
value ( T ). The whole area from
to
is of course
probability is 1
tabulated value
+∞
equal to 1, so the area from T to
can be obtained
by subtracting the table value from 1.
For example, from a given cross (A
×
B) estimates
Univariate cross prediction example
To consider further possible problems involved in selec-
tion of the 'best' cross combinations consider the
following example.
Below are shown the means and genetic variances of
crop yield (t/ha) of four barley crosses (A, B, C and
D). Also grown in this prediction trial were five control
lines. These controls are all commercially grown culti-
vars predominating over the region where new varieties
are to be grown. The average yield of the controls was
21 t/ha and the variance of the controls was 7.5 t/ha.
Which of the crosses should have greatest emphasis in
=
were obtained of the mean ( m
12.0), genetic vari-
σ g = σ
= 16
g
σ g =
=
ance (
4).
What would be the probability that a recombinant
will exceed a set target value of 14?
16.0, therefore
To solve this
we have:
T
m
σ g =
14
12
=
0.5
4
Using the unit normal distribution tables and a value
of 0.50 we have the probability of
−∞
to T to be equal
to 0.6915 (from table of unit normal distribution). The
actual probability we want is 1
=
0.6915
0.3085.
 
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