INTRODUCTION TO PLANT BREEDING

AGRONOMY 815 / COURSE NOTES

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

DEPARTMENT OF AGRONOMY / UNIVERSITY OF NEBRASKA

GENE FREQUENCIES IN POPULATIONS

Fehr, Chapt. 3
Briggs & Knowles,. Chapt. 14
Falconer, Chapt. 1 & 2.

In the previous discuss, we have looked at the genetics of the progeny from selfed lines and single crosses having defined parents. Those examples best describe what occurs in self pollinated crops. In cross pollinated crops, population genetics are most important in that any female may be pollinated by many different male lines.

HARDY-WEINBERG EQUILIBRIUM:

In a large, random-mating population (MEANING NO GENETIC DRIFT) with no SELECTION, MUTATION, or MIGRATION, the gene frequencies and the genotype frequencies will remain the same from generation to generation. Furthermore there is a simple relationship between the gene frequencies and the genotype frequencies.

In a population with two alleles at a locus (A,a) and each allele has a frequency of
F(A) = p and F(a) = q then F(AA) = p2, F(Aa) = 2pq, F(aa) = q2. Note p + q = 1 and q = (1-p).

Instead of using the gene frequencies, can also use the genotype frequencies
F(AA) = P, F(Aa) = H, and F(aa) = Q. In this case, F(A) = P + 1/2H, F(a) = Q + 1/2H. Note: 1 = P + H + Q.

In proving the Hardy-Weinberg equilibrium, you must show:

  1. The gene frequency in the parents is identical to the gene frequency in their gametes. Equal transmission.

  2. The gene frequency in the gametes explains the genotype frequency in the zygotes. Equal fertilization.

  3. The genotype frequency of the zygotes and of the adults are the same. Equal survival.

  4. The gene frequency in the adults is identical to the gene frequency in their gametes. Equal transmission.

Proofs of the Hardy-Weinberg Equilibrium:

GAMETE/GENE PROOF
Genes in parentsGenotypes in Progeny

AaAAAaaa

Frequencypqp22pqq2
pHQ




Note that p = P + 1/2H = p2 + 2 (2pq) = p (p + q) = p.



Female gene frequency

Aa
pq


Male gene Frequency Ap p2pq
aq pqq2





Genotype and frequency of female parent
MATING PROOF


AAAaaa
PHQ


Male


AAPP2PHPQ
AaHPHH2QH
AaQPQQHQ2





Genotype and Frequency of Progency


Mating TypeFrequencyAAAaaa


AAxAAP2 P2__________

AAxAa2PH PHPH_____

AAxaa2PQ _____2PQ_____

AaxAaH2 1/4H21/2H21/4H2

Aaxaa2HQ _____HQHQ

aaxaaQ2 __________Q2



SUMS

EQUALS



How is gene frequency determined in a population? If you know which gene is recessive then:

q2=(# recessive individuals) / (total # individuals)

q=square root of above,p=1-q

The proportion of the population that carries a masked recessive allele is:

heterozygotes=

upon substitution of1 - q=p.

Key point to note is that with cross pollination, proportion of population that is homozygous is entirely dependent upon gene frequency and not generation of crossing. (Exception is when population size is small.)

ALSO, IT IS IMPORTANT TO REMEMBER THAT HARDY-WEINBERG EQUILIBRIUM IS ATTAINED AFTER ONE GENERATION OF RANDOM MATING (EXCEPTION IS DUE TO LINKAGE).

Gene frequencies can change due to:

  1. Population size C random drift;

  2. Differences in fertility and viability (selection);

  3. Migration and mutation (for plants C inadequate isolation of population and/or deliberate adding of genes to the population);

  4. Non-random mating (i.e., assortive mating).


Change in inbreeding =1/2Ne=delta F where Ne is the effective population size.
(See Allard, p. 200-203, Fehr, Chapt. 8).

If 50 corn plants are intermated then delta F=1/(2(50))=.01

If self pollination is excluded then Ne = N + 1/2 (approximate) where is the number of individuals in the population.

If exclude sib matings then Ne = N + 2

If have different numbers of males and females then:

1/Ne = 1/4Nf + 1/4Nmor

Ne = (4NfNm)/(Nf + Nm).

NOTE:THIS MEANS THAT WITH ONLY ONE MALE AND AN INFINITE NUMBER
OF FEMALES, THE EFFECTIVE POPULATION SIZE IS ONLY 4.

Ft = 1 - (1 - (delta F))t where t is the generation.

With inbreeding the Hardy-Weinberg equilibrium becomes:

p2 (1 - F)+Fp:2pq(1-F):q2(1-F)+Fq.

If F = 0 then the formula is unchanged. If F = 1 then the formula becomes:

1 p:1 q

NOTE:WHILE THIS IS POPULATION GENETICS, IT IS STILL FOR QUALITATIVE
GENES AND NOT FOR QUANTITATIVE GENES.