Natural Selection as the Mechanism
for Evolution
3 Mechanisms for selection and an
introduction to Hardy-Weinberg
Lyell’s Influence
In attempt to explain the past in terms of present
day processes, Darwin went to local farmers and
animal breeders and observed…
 Variation in the organisms that could be
inherited
 After many generations, organisms appear very
different from ancestors
Darwin called this process ARTIFICIAL SELECTION
Artificial Selection in Dogs
130,000 years
Artificial Selection in Pigeons

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It was pigeons, not finches, that
Darwin used to argue natural
selection
The pigeon (better termed the Rock
Dove) domesticated 5,000-10,000
years ago
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Food? Fun? Why pigeons??
Pigeon fanciers – began breeding
these pigeons for aesthetic
qualities.
By the time Darwin became
interested in them, there
were several hundred varieties of
domestic pigeon available.
NATURAL SELECTION
Variation in Nature
1.

No two organisms are exactly alike!
Struggle for Existence (Malthus)
2.

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High birth rates and limited resources will cause organisms to
compete
Selective pressures (predation, competition, parasitism,
disease, pesticides, etc)
Survival of the Fittest
3.

Organisms with the best adaptations survive and reproduce
more often
Reproduction of Viable Offspring
4.

Offspring must be fertile and reproduce
Understanding Histograms


A histogram is a graphical representation of
the distribution of data. It is an estimate of the probability
distribution of a continuous variable (quantitative variable)
Helps to understand how data is distributed. Shape of many
distributions in natural systems are bell curves.
Three mechanisms for selection
Natural Selection on Populations
STABILIZING SELECTION:
Individuals in the center of the curve
have higher levels of fitness than the
individuals at either end
Example: Birth weight in humans or
marine creatures on ocean floor
Natural Selection on Populations
Why would
stabilizing
selection occur in
human babies?
Natural Selection on Populations
DIRECTIONAL SELECTION:
Individuals at one end of curve have
higher levels of fitness than the
individuals at the other end of the curve
Ex: Peppered Moths in England post
Industrial Revolution
Directional Selection


Would you
expect the same
amount of each
phenotype in
different
environments?
In which
environment
would you find
the most black
moths? Why
Natural Selection on Populations
DISRUPTIVE SELECTION:
Individuals on the ends of the
curve have higher fitness levels
than individuals in the center of
the curve
Example: Beak size in Finches
Natural Selection on Populations
Microevolution: A Population’s Gene
Pool


A gene pool is all the
alleles available in all
of the individuals in a
population
Changing allele
frequencies means
the population is
evolving →
microevolution
Hardy-Weinberg Equilibrium
It is a condition in which no change in
alleles occur
 The equation used to determine allele
frequencies:
p2 + 2pq + q2 = 1
p = frequency of dominant allele (A)
q = frequency of recessive allele (a)
p2 = AA
q2 = aa
2pq = Aa

Hardy-Weinberg Equilibrium
Hardy-Weinberg Equilibrium

Example 1: In one hypothetical Zebra Mussel
(Dreissena polymorpha) population, most of the
individuals have dark, zebra-striped shells
(below left). However, solid light-colored shells
(below right, caused by a homozygous recessive
gene, aa) occur in 1 of every 10,000 individuals.
Hardy-Weinberg Equilibrium

Solution 1:

frequency of aa = q2 = 1/10,000 = 0.0001, so q = 0.01
number of aa = 0.0001 x 10,000 = 1 individual
p + q = 1, so p = 0.99
frequency of AA = p2 = 0.9801
number of AA = 0.9801 x 10,000 = 9,801 individuals

How many heterozygous individuals?
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Hardy-Weinberg Equilibrium

Example 2: The Coquina Clam (Donax variabilis) is
highly polychromic: (Polymorphism expressed as existing in
several different colors. (adj. polychromic)) (with shells of
many different colors.) In a population of 2,000 clams, 1,920
are solid colored, whereas the remainder has radiating color
bands. Solid color occurs in homozygous dominant (BB) and
heterozygotes (Bb); color banding only occurs in
homozygous recessive individuals (bb).
 Problem: Calculate gene frequencies
and numbers of BB and Bb.
Hardy-Weinberg Equilibrium
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Solution 2:
1,920 are solid (BB and Bb), so 80 banded are recessive (bb)
frequency of bb = q2 = 80/2000 = 0.04, so q = 0.20
p + q = 1, so p = 0.80
number of BB: p2 = 0.64, so BB in population of 2,000 =
0.64 x 2,000 = 1,280 individuals
number of Bb: 2pq [frequency of Bb] = 2 x 0.2 x 0.8 = 0.32,
so Bb = 0.32 x 2,000 = 640 individuals