Agriculture Reference
In-Depth Information
resistant. Multiplex breeding can be used in a similar
way to develop parents in hybrid cross combinations.
The statistical test most commonly used when such
a problems arises, is simple in design and application.
Each deviation is
squared,
and the expected number
in its class then divides each squared deviation. The
resulting quotients are then all added together to give
a single value, called the chi-square (
The chi-square test
2
χ
)
. To substitute
symbols for words, let
d
represent the respective devi-
ations (observed minus expected),
e
the corresponding
expected values, then:
If plant breeders have an understanding of the inher-
itance of simply inherited characters it is possible
to predict the frequency of desired genes and geno-
types in a breeding population. Breeders are often
asking questions relating to the nature of inher-
itance as well as the number of alleles or loci
involved in the inheritance of a particular charac-
ter of importance. For example, it is often valuable
to determine whether a single gene is dominant or
additive in inheritance. In the dominant case the
F
2
family will segregate in a 3 : 1 dominant : recessive
ratio, while in the latter a 1:2:1 homozygous
dominant : heterozygous : homozygous recessive ratio.
In segregating families such as those noted above, the
ratios actually observed will not be exactly as predicted
due to sampling error. For example a coined tossed 10
times does not always result in 5 heads and 5 tails. So
the breeder is faced with interpreting the ratios that are
observed in terms of what is expected, so it might be
necessary to decode if a particular ratio is 1 : 2 or 1 : 3
or 3 : 4 or 9 : 7. How can this be done objectively? The
answer is to use chi-square tests.
It should be clear that the significance of a given devi-
ation is related to the size of the sample. If we expect a
1 : 1 ratio in a test involving six individuals, an observed
ratio of 4 : 2 is not at all bad. But if the test involves 600
individuals, an observed ratio of 400 : 200 is clearly a
long way off. Similarly, if we test 40 individuals and
find a deviation of 10 in each class, this deviation seems
serious:
2
d
2
χ
=
(
/
)
e
We can calculate chi-square for the two arbitrary
examples above to show how this value relates the
magnitude of the deviation to the size of the sample.
Sample of 40
Sample of 200
individuals
individuals
Observed (obs)
30
10
90
110
Expected (exp)
20
20
100
100
obs
−
exp(
d
)
10
10
10
10
d
2
100
100
100
100
d
2
/exp
5
5
1
1
2
10
2
χ
You will note that the value of chi-square is much
larger for the smaller population, even though devia-
tions in the two populations are numerically the same.
In view of our earlier common-sense comparison of the
two, this is a practical demonstration that the calcu-
lated value of chi-square is related to the significance of
a deviation. It has the virtue of reducing many different
samples, of different sizes and with different numerical
deviations, to a common scale for comparison.
The chi-square test can also be applied to samples
including more than two classes. For example, the table
below shows the chi-square analyses of the tall, 6-row
×
observed
30
10
expected
20
20
−
obs
exp
10
10
short 2-row barley example earlier. Suppose that a
test cross was carried out where the heterozygous F
1
(
TtSs
) is crossed to a genotype with the recessive alle-
les at these loci. When this was actually done, and
400 test cross progeny were grown out, there were
96 tall and 2-row, 107 tall and 6-row, 97 short and
6-row, and 102 short and 2-row. Inspection of this
would show that there are a higher proportion of the
original parental types (tall/6-row and short/2-two)
But if we test 200 individuals, the same numerical
deviation seems reasonably enough explained as a purely
chance effect:
observed
90
110
expected
100
100
−
obs
exp
10
10
