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Transcript
Evolution &
Microevolution
Tutorial
Introduction
Microevolution
Hardy Weinberg Equilibrium
Practice!
In this tutorial, you will learn:
 The
difference between macroevolution &
microevolution.
 How Hardy-Weinberg equilibrium works as
well as factors that can upset this
equilibrium.
 How to use the equation, p2 + 2pq + q2 = 1,
to calculate allele frequencies in a
population.
Credits:
Figures and images by N. Wheat unless otherwise noted.
Lesser ball python image used with permission from Tim Bailey, Bailey & Bailey Reptiles.
Funded by Title V-STEM grant P031S090007.
Introduction
 Evolution
– includes all of the changes in
the characteristics and diversity of life that
occur throughout time.

Evolution can occur on both large and
small scales.
Macroevolution – Evolution on a
Large Scale
 Macroevolution
a grand scale.



– evolutionary change on
Origin of novel designs
Evolutionary trends
Adaptive radiation
Microevolution – Evolution on a
Small Scale
 Microevolution
- a change in the genetic
composition of a population over time.

A change in the frequency of certain alleles
in a population over several generations.
Polymorphism
 Polymorphism
occurs when there are
different allelic forms of a gene in a
population.

Mojave (left) and Lesser (middle) are
different alleles of the same gene. Wild type
ball python is shown on the right.
Photo courtesy of Bailey & Bailey Reptiles
Gene Pool
 All
of the alleles of all of the genes
possessed by all of the members of the
population are contained in the gene
pool of the population.
 We can measure the relative frequency
of a particular allele in a population.

Allelic frequency
Population Genetics
 Population
Genetics – the study of how
populations change over time.


Dependent on both Darwin’s theory of
natural selection and Mendel’s laws of
inheritance.
All heritable traits have a genetic basis,
some are controlled by multiple genes – not
as simple as in Mendel’s studies.
Genetic Equilibrium
 According
to Hardy-Weinberg
equilibrium, the hereditary process alone
does not produce evolutionary change.

Allelic frequency will remain constant
generation to generation unless disturbed
by mutation, natural selection, migration,
nonrandom mating, or genetic drift.
 These
are sources of microevolutionary
change.
Frequency of Alleles
 Each
allele has a frequency (proportion)
in the population.
 Example population of 500 wildflowers.

CRCR = red; CRCW = pink; CWCW = white
 250
red, 100 pink, 200 white
Frequency of CR =
(250 x 2) + 100 / 1000 = 600/1000 =.6 = 60%

Frequency of Alleles
p
is the frequency of the most common
allele (CR in this case).

q
p = 0.6 or 60%
is the frequency of the less common
allele (CW in this case).
p+q=1
 q = 1- p = 1 – 0.6 = 0.4 or 40%
Hardy-Weinberg Theorem
 Populations
that are not evolving are said to
be in Hardy-Weinberg equilibrium.
Hardy-Weinberg Theorem
 As
long as Mendel’s laws are at work, the
frequency of alleles will remain
unchanged.
Review Punnett squares in the genetics tutorial.
Hardy-Weinberg Theorem
 The
Hardy-Weinberg theorem assumes
random mating.
 Generation after generation allele
frequencies are the same.
Hardy-Weinberg Theorem
 Conditions
required for Hardy-Weinberg
equilibrium to hold true:





Very large population
No gene flow into or out of the population
No mutations
Random mating
No natural selection
Hardy-Weinberg Theorem
 Departure
from these conditions
results in a change in allele
frequencies in the population.
 Evolution
has occurred!
Practice with Hardy Weinberg
 Frequency
– the proportion of individuals
in a category in relation to the total
number of individuals.
 100 cats, 75 black, 25 white – frequency of
black = 75/100 = 0.75, white =0.25.

Two alleles: p is common, q is less common.
 p+q
=1
Question 1
The frequency of black cats is:
 0.75
 75
 0.25
 25
 100
Question 1
Sorry!
 That
is incorrect.
 Try again!
Question 1
Congratulations!
 You
are correct!
Question 2
What would the frequency of black
cats be if the population size was
80 instead of 100 (still 75 black)?
 0.75
 75
 0.94
1
(75/80)
Question 2
Sorry!
 That
is incorrect.
 Try again!
Question 2
Congratulations!
 You
are correct!
Hardy-Weinberg Theorem
 At
a locus with two alleles, the three
genotypes will appear in the following
proportions:
 (p + q) x (p + q) = p2 + 2pq + q2 = 1
Practice with Hardy Weinberg

(p + q)2 =
p2
+
Individuals
homozygous
for allele B
2pq +
Individuals
heterozygous
for alleles B
&b
q2
Individuals
homozygous
for allele b
Practice with Hardy Weinberg
 We
will use a population of 100 cats as a
practice example.


84 of the 100 cats are black.
16 are white.
Practice with Hardy Weinberg
 We
can use the equation and our color
observations to calculate allele
frequencies in our population of 100 cats.


p2 + 2pq + q2 = 1
100 = population size
Practice with Hardy Weinberg
 84

of our 100 cats are black.
Black is the dominant phenotype.
 Cats
with the genotype Bb or BB will be black.
 The frequency of black cats is 84/100, but we
can’t yet say anything about the B allele.
 See
the genetics tutorial to review these
terms.
Practice with Hardy Weinberg
 16



of our 100 cats are white.
White is recessive (bb) and is represented
by q2 in our equation: p2 + 2pq + q2 = 1
So, q2 = 16/100 = 0.16
q = square root of 0.16 = 0.40.
Practice with Hardy Weinberg
q
= square root of 0.16 = 0.40.
 Since p + q = 1; p = 1 – q = 0.60.
 p2 = 0.36


p2 represents the proportion of individuals in
the population with the homozygous
dominant phenotype (BB).
Remember population size = 100
Question 3
So, the number of cats in our
population that have the BB
genotype would be:
 0.36
cats
 0.36 x 100 = 36 cats
 0.16 x 100 =16 cats
 84 cats
Question 3
Sorry!
 That
is incorrect.
 Try again!
Question 3
Congratulations!
 You
are correct!
Practice with Hardy Weinberg
 Now
we know how many of our cats have
the BB genotype and the bb genotype.


We can find the number of Bb cats using our
equation: p2 + 2pq + q2 = 1.
2pq represents the proportion of cats with Bb.
2
x 0.6(p) x 0.4(q) = 0.48
 0.48 x 100 = 48 cats with Bb genotype.
Question 4
Let’s try another! In our population of
100 cats, 75 are black & 25 are white.
Where do we start?
 75
black cats = p2.
 75/100 = 0.75 black cats = p2.
 25 white cats = q2.
 25/100 = 0.25 white cats = q2.
 Need more information.
Question 4
Sorry!
 That
is incorrect.
 Try again!
Question 4
Congratulations!
 You
are correct!
Question 5
If q2 = 0.25, q=
 0.05
5
 0.5
 50
Question 5
Sorry!
 That
is incorrect.
 Try again!
Question 5
Congratulations!
 You
are correct!
Question 6
If q=0.5, p=
 0.5
5
 0.6
 0.1
Question 6
Sorry!
 That
is incorrect.
 Try again!
Question 6
Congratulations!
 You
are correct!
Question 7
So, if p=0.5, and p2=0.25, how
many of our cats have the BB
genotype?
 0.25
 25
 50
 75
Question 7
Sorry!
 That
is incorrect.
 Try again!
Question 7
Congratulations!
 You
are correct!
Question 8
Now, how many of the cats
are heterozygous (Bb)?
 48
 100
 .5
 50
Question 8
Sorry!
 That
is incorrect.
 Try again!
Question 8
Congratulations!
 You
are correct!
Question 9
If we measure allele frequency one year at
p=0.8 & q=0.2 and then go back 5
generations later to find p=0.5 & q=0.5, what
has happened?




The population has remained in
Hardy-Weinberg equilibrium.
The population has doubled in
size.
There has been a change in
allele frequencies: evolution has
occurred.
Nothing has changed.
Question 9
Sorry!
 That
is incorrect.
 Try again!
Question 9
Congratulations!
 You
are correct!