Basic concepts of population genetics - Bioversity International

... with a codominant gene having multiple
alleles (continued)
Locus A
Gel
Ind.1 Ind.2 Ind.3 Ind.4 Ind.5 Ind.6
A1 1 1 1 0 0 0
A2 0 1 0 1 1 0
A3 0 0 1 0 1 1
M
Genotypes
A 1 A 1
Individuals
1 2 3 4 5 6
A 1A 2
A 1 A 3
Scoring bands
Copyright: IPGRI and Cornell University, 2003 Population genetics 27
27
A 2A 2
Locus A
A 2 A 3
A 3A 3
M 1 2 3 4 5 6
(1,0,0) (1,1,0) (1,0,1) (0,1,0) (0,1,1) (0,0,1)
Again, with a codominant marker, the genotypes of the three genotypic classes can
be observed. In the drawing above, top centre, we see a gel image with the banding
pattern of a codominant marker with three alleles (A 1 , A 2 and A 3 ) in a diploid
sample. We score each band (each row) independently, and transform them to a
score of 1 if present or a score of 0 if not. We can do it by band (bottom left of slide)
or by genotype (bottom right corner). In the table below, we can see the calculations
of the expected and observed genotype frequencies, as well as the allele
frequencies (p 1 , p 2 and p 3 ). (M = size marker.)
Genotypes
Genotype
frequency
(exp.)
Number of
individuals
Genotype
frequency
(obs.)
A 1 A 1
p 1 2
n 11 =
4
P 11 =
n 11 /n
= 0.17
A 1 A 2
2p 1 p 2
n 12 =
6
P 12 =
n 12 /n
= 0.25
A 1 A 3
2p 1 p 3
n 13 =
0
P 13 =
n 13 /n
= 0
A 2 A 2
p 2 2
n 22 =
10
P 22 =
n 22 /n =
0.42
A 2 A 3
2p 2 p 3
n 23 =
2
P 23 =
n 23 /n
= 0.08
A 3 A 3
p 3 2
n 33 =
2
P 33 =
n 33 /n
= 0.08
p 1 = P 11 + ½P 12 + ½P 13 = P 11 + ½ j 1 P 1j = 0.17 + ½(0.25 + 0.00) = 0.30
p 2 = P 22 + ½P 21 + ½P 23 = P 22 + ½ j 2 P 2j = 0.42 + ½(0.25 + 0.08) = 0.59
p 3 = P 33 + ½P 31 + ½P 32 = P 33 + ½ j 3 P 3j = 0.08 + ½(0.00 + 0.08) = 0.12
Total
1
n = 24
1