INTRODUCTION TO PLANT BREEDING
AGRONOMY 815 / COURSE NOTES

P. STEPHEN BAENZIGER, 338 Keim Hall, 472-1538

DEPARTMENT OF AGRONOMY / UNIVERSITY OF NEBRASKA

SELECTION FOR QUALITATIVE TRAITS:
CHANGES IN GENE FREQUENCY

Fehr, Chapt. 3.
Falconer, Chapt. 2 — Required reading.
Allard, p.170-175 — Required reading.

Gene frequency will not change in a large, random mating population unless there is  migration, mutation and selection. Also remember that one generation of random mating will return a population to Hardy-Weinberg Equilibrium.

SELECTION — DIFFERENTIAL REPRODUCTIVE RATE

COEFFICIENT OF SELECTION (S)= THE PROPORTION REDUCTION IN THE GAMETIC CONTRIBUTION
OF A PARTICULAR GENOTYPE COMPARED WITH THE STANDARD GENOTYPE.

e.g.,s=.1For every 100 zygotes produced by the favored genotype, only 90 are produced by the other.

Fitness is 1 - s = w.

Consider complete dominance with selection against the recessive.
(Note . . . selection is often only partially effective.)

Genotypes	A A	A a	a a	Total
Initial Frequency	p²	2pq	q²	1.0
Fitness	1	1	1 - s
Genetic contribution	p²	2pq	(1 - s)q²	1 - sq²

Total =

=1 - q²s; or 1 - sq²

q₁=

delta q=

multiply by -1

Note if s and q are small values, then delta q ~ -sq²p.

CHANGES IN GENE FREQUENCY DUE TO SELECTION DEPENDS UPON

SELECTION COEFFICIENT (S);
INITIAL GENE FREQUENCY.

and also influenced by type of gene action.

Fitness	A₁A₁	A₁A₂	A₂A₂
No dominance (selecting against A₂)	1	1 - 1/2s	1 - s
Complete dominance (selecting against A₁)	1 - s	1 - s	1
Over dominance (selection for A₁A₂)	1 - s₁	1	1 - s₂
Partial dominance (selection against A₂)	1	1 - hs	1 - s

Note: when q is very small, 1 - sq and 1 - q² are approximately one and the formulae can be greatly simplified.

EFFECTIVENESS OF SELECTION

Selection is most effective at intermediate gene frequencies. (Maximum variability when gene frequency .5, therefore have maximum selection differential and maximum heritability.)

Selection for or against a recessive gene is extremely ineffective when the recessive allele is rare. (When an allele is rare it is represented almost entirely in heterozygotes.)

e.g., Complete dominance, selection against the recessive with s = 1 (i.e., complete elimination of A₂A₂ genotype, and gene frequencies in the initial population = .5).

A₁A₁A₁A₂A₂A₂

p =.252pq =.50q = .25

Fitness110

Gametic contrib.p2pq0

In generalq₁=q₀/(1 + q₀)

q₂=q₀/(1 + 2q₀)

andq_t=q₀/(1 + tq₀)

Thereforet=(q₀ -q_t)/(q₀q_t)

For example: You have a recessive color gene that is showing up in your cross pollinated corn variety at a level of 1 in 100. The meet certification standards, you need to reduce the number to 1 in 1000. How many generations will it take you to reduce the gene frequency?

sq root (0.001) = 0.0316

or 22 generations

Can you think of a quicker way to reduce the gene in the population?

Answer:

The problem in reducing the gene frequency is that one cannot identify the carrier heterozygotes. The heterozygotes can be identified by either a test cross or a progeny test. Plants would be harvested individually. The plant progenies would be selfed and/or crossed onto the tester. Any parental lines whose self or test cross segregates for the color marker will be eliminated and the frequency of the gene will become 0 in the reconstituted population made by intermating the selfed homozygous progeny. Note the key is the ability to self or test cross. Self incompatible plants would require a more complicated mating system coupled with the test cross. COMMENT ON INBREEDING VS. RANDOM MATING, ALSO EUGENETICS.

Suppose there is a deleterious recessive trait within the human population. The affected individuals occur 1 in 20,000 (gene freq. = 0.0070711). The government decides to try to reduce the affected individuals to 1 in 40,000 (gene freq. = 0.005).

How many generations of selection would it take to reduce the gene frequency?

t = (0.0070711 - 0.005)/(0.0070711)(0.005) = 58.6 generations. Assuming 20 years per generation, this selection would require 1172 years.