Breeding Schemes - An Introduction to Plant Breeding - page 50

The level of heterosis is measured by two methods
F 1 minus the mid-parent , called mid-parent heterosis,
or F 1 minus the best parent , called best parent hetero-
sis. The mid-parent heterosis need not detain us here
as it is of no direct interest. If best parent heterosis is
positive (i.e. the F 1 exceeds the performance of the best
parent), then generally heterosis has been ascribed to
over-dominance.
In Case 1 the best parent (P 1 ) has a phenotype of
10 yield units; and the F 1 has a phenotype of 10 yield
units; that is, they have identical yields and no heterosis
is detected .
In Case 2 (with the same alleles and effects but with
different starting arrangements of alleles between the
parents), the best parent (P 1 ) has a phenotype of 6 yield
units; and the F 1 has a phenotype of 10 yield units; that
is, heterosis would be 4 units and the F 1 is, in fact, 40%
more productive than the better parent.
It is however, clearly wrong to consider this to be over-
dominance, since none of the individual loci show such
an effect. It would certainly be possible (in fact Parent 1
in Case 1 shows this) to fix this level of performance in
homozygous, true breeding lines with no heterozygosity
(i.e. AABBCCDDEE).
Now let us consider the statistical probability under-
lying these combinations. With five loci the frequency,
assuming no linkage, no effects of selection and a ran-
dom assortment of gametes in the F 1 , after a number of
rounds of selfing, the probability (which is equivalent to
the frequency) of having a genotype which combines the
five dominant alleles as homozygotes would be (1/2) 5

wish to have some assurance that a breeding population
contains at least one of these genotype recombinants
amongst all possible inbred lines that will be produced,
the number of lines needed to be grown and screened
is given by the equation:

n = ln ( 1 − p )/ ln ( 1 − x )

where n is the number of inbred lines that would need
to be screened, p is the desired probability that at least
one genotype will exist in the population and x is the
frequency of the genotype of interest. In the example
with five loci, it would be necessary to screen at least
145 inbred lines from the progeny derived from the
cross between Parent 1 and Parent 2 (in both cases) to
be 99% certain that at least one example of the genotype
required will exist.
Of course, a quantitatively inherited trait like yield
is not controlled by five loci but more likely 50, 500 or
5000. To give some idea of the number of genotypes
needed to obtain a specific combination of alleles when
the number of loci increases is given in Table 4.1. In
this table the number of plants that need to be screened
to be 90%, 95%, 99%, 99.5% and 99.9% sure that
at least one genotype of desired combination exists is
shown for cases with 5, 6, 7, 8, 9, 10 and 20 loci.
With 20 loci, a modest number compared with the
possible real situation, a breeder would need to evaluate
almost 5 million inbred lines to be sure that the one he
wants is present. So hybrids do offer a better probability
of success in this instance, but not because they show
over-dominance at their loci!

=

/

1 32 (or 0.03125, just over 3%). If, however, you

Table 4.1 Number of recombinant inbred lines that would require evaluation for the
breeder to be 90%, 95%, 99%, 99.5% and 99.9% sure that a homozygous lines will
be a specific combination of alleles at each of 5, 6, 7, 8, 9, 10 and 20 loci.

Probability of obtaining at least one individual of the desired genotype

0.900 0.950 0.990 0.995 0.999

5 Loci 76 94 145 167 218

6 Loci 146 190 292 336 439

7 Loci 294 382 587 676 881

8 Loci 588 765 1 177 1 354 1 765

9 Loci 1 178 1 532 2 356 2 710 3 533

10 Loci 2 357 3 066 4 713 5 423 7 070

20 Loci 2 414 434 3 141 251 4 828 869 5 555 687 7 243 317