Biology Reference
In-Depth Information
Example 3-2
.......................
Let us assume a population of 100 individuals of genotype Aa. The
initial frequencies of the A allele, p
0
, and the a allele, q
0
, are both 0.5. This
initial population reproduces, producing 100 offspring in the following
proportions:
2
p
0
¼ð
P
ð
AA
Þ¼
0
:
5
Þ
¼
0
:
25
:
P
ð
Aa
Þ¼
2p
0
q
0
¼
2
ð
0
:
5
Þð
0
:
5
Þ¼
0
:
50
:
2
q
0
¼ð
P
ð
aa
Þ¼
0
:
5
Þ
¼
0
:
25
:
Because there are 100 offspring, we can calculate that 100
0.25
¼
25 of
them will have genotype AA; 100
0.5
¼
50 will have genotype Aa; and
100
25 will have genotype aa. However, let us also assume the
aa allele is associated with a selective disadvantage of 0.2, such that 20%
of the aa offspring will not survive to reproduce. Thus, only 80% of the
25 aa will be able to pass their genes on to the next generation, and,
therefore, the proportions of the A and a alleles will be different from the
initial generation. Let us look at the genes that will be passed on:
0.25
¼
The 25 AA individuals will contribute 2
25 or 50 A alleles.
The 50 Aa individuals will contribute 50 A alleles and 50 a alleles.
The 0.8
25
¼
20 aa individuals will contribute 2
20 or 40 a alleles.
Therefore, there will be 95 individuals contributing their alleles to the
next generation, and the total gene pool will be twice that number or 190
alleles. Of those, 100 are A alleles and 90 a alleles. If we denote the allele
frequencies for A and a by p
1
and q
1
, respectively, we can calculate that:
p
1
¼
:
Notice the proportions of the A and a alleles have changed from p
0
¼
100
=
190
¼
0
:
5263 and
q
1
¼
90
=
190
¼
0
:
4737
0.5
and q
0
¼
0.4737 in
the population surviving until reproduction. To see how these
proportions will change even further, suppose now that the 95
individuals reproduce, and 100 offspring result. What would be the
allele frequencies in the next generation?
0.5 in the initial population to p
1
¼
0.5263 and q
1
¼
The AA individuals would be represented by Np
1
, where N is the
population size (in our case N
(0.5263)
2
¼
27.7 or, rounding up, 28 people. The Aa individuals would be N(2p
1
q
1
)
or 100
¼
100). This would be 100
2
(0.5263)
(0.4737)
¼
49.8, or 50 people. The aa individuals
would be Nq
1
or 100
(0.4737)
2
22.4, or 22 people. We can see there
are fewer aa individuals in this generation than in the previous one.
¼
The 28 AA individuals will contribute 2
28 or 56 A alleles.
The 50 Aa individuals will contribute 50 A alleles and 50 a alleles.
