Evolution Lab:
Evolution - Change in allelic frequencies in a population
Anagenesis – change in genetic lineage
(3 ½ billion years – unbroken chain of DNA transfer; this chain has not
branched; no net gain of species).
Cladogenesis – creation of new genetic lineages (net gain of species)
One gene with two alleles (Aa)
Count all individuals in population and see how many of each is there:
# AA = 30, # Aa = 50, # aa = 20 (N=100).
How many A’s  60 + 50 = 110
How many a’s  40 + 50 = 90
Total alleles = 200; proportion of A’s = 110/200 = 0.55; proportion of a’s = 90/200
= 0.45.
F(A) = 0.550  0.549; a loss of A’s in the population = anagenesis.
How anagenesis can occur:
1) mutation: A  a or a  A; rate of change due to mutation ranges from
1X10-5 to 1 X 10–7; not very common.
2) Natural selection: differential survival or reproduction that results in
differential differences in representation in succeeding generations.
(Survival of the Fittest).
3) Genetic Drift: Change in allele frequency due to unrepresentative sampling
from one generation to the next.
Unrepresentative sampling:
Shuffle six decks of cards and deal out 150. What is the chance of all six being
hearts? Not much. Should be about 25% hearts – a representative sample. If
almost all of the cards were hearts, that would be considered unrepresentative
sampling. If I deal out only 4 cards, would you expect to get one of each suit?
This small sample in analogous to genetic drift. Perfectly fit individuals are
removed from the population.
Suppose:
#A’s = 1,000,000 f(A) = 0.5
#a’s = 1,000,000 f(a) = 0.5
We only pick 3 individuals to reproduce.
Could get 2 AA and one Aa  AA Aa, AA, or Aa.
Hardy=Weinberg equilibrium (HWE): f(A) = p and f(a) = q
Assumptions required for a population to be in HWE (Table 1 page 47):
Large population size, diploid organism, sexual reproduction, random
mating, generations do not overlap, no mutation, no immigration or
emigration, no natural selection.
With random mating  AAxAA, AAxAa, AAxaa, AaxAa, Aaxaa, aaxaa
In the next generation and all that follow, the frequency of:
AA=p2; Aa=2pq; aa=q2
Suppose f(A) = 0.2 and f(a) = 0.8 then HWE 
f(AA) = p2 = 0.22 = 0.04
f(Aa) = 2pq = 2x0.2x0.8 = 0.32
f(aa) = q2 = 0.82 = 0.64
(p + q) = p2 + 2pq + q2
p + q = 1; 0.2 + 0.8 = 1
If the frequency of albinism in North America is 0.00005, then what is the
frequency of carriers?
Take the square root of 0.00005 (=q2) to get q, then to get p=1-q. Heyerozygot
frequency = 2xpxq.
Fitness:
Fitness – the ability of an organism to survive and reproduce (make offspring)
Absolute fitness – How many of each genotype survive. For example – if 70%
of AA and Aa survive and reproduce, while only 35% of Aa survive and
reproduce, then AA and Aa genotypes have a higher fitness.
Relative fitness – The comparison among genotype fitness. For example – AA
= 0.7/0.7=1; Aa = 0.7/0.7=1; aa = 0.35/0.7= 0.5.
Fitness values can be used to predict hoe allelic frequencies will change from
one generation to the next.
Suppose f(A) = 0.5 and f(a) = 0.5, then:
f(AA)=0.52=0.25, f(Aa) = 2x0.5x0.5=0.5, f(aa) = 0.52=0.25
Genotype
Genotypic
frequency
X
Relative
=
Preliminary value
AA
Aa
aa
0.25
0.5
0.25
X
X
X
fitness
1.0
1.0
0.5
Genotype of frequencies of adults after selection are:
f(AA) = 0.25/0.875 = 0.29 (was 0.25; increased)
f(Aa) = 0.5/0.875 = 0.5 (was 0.5; decreased)
f(aa) = 0.125/0.875 = 0.14 (was 0.25; increased)
If this continues, what will happen to f(aa)?
=
=
=
Sum=
0.25
0.5
0.125
0.875