Download Five Drivers of Evolution

BIOLOGY 3B
Natural Selection
Week 3
Darlak
WHICH BEAR IS THE “MOST FIT”?
EXPLAIN YOUR ANSWER
NAME
George
Ben
Spot
Sam
SIZE
10 ft
1,200 lb
8.5 ft
1,000 lb
9 ft
1,100 lb
9 ft
1,000 lb
# OF CUBS
FATHERED
19
25
20
20
# OF CUBS
SURVIVING
TO
ADULTHOOD
2
0
1
5
COMMENTS
George is
very large,
very
healthy.
The
strongest
bear.
Ben
mated
with the
greatest
number of
females.
When the area that
Spot lived in was
destroyed by fire,
Spot was able to
move to a new area
and change his
feeding habits.
Sandy was
killed by an
infection
resulting
from a cut in
his foot.
AGE AT
DEATH
13 yr
16 yr
12 yr
9 yr
AGENDA 04/20
Modes of Selection Check
 Selection and Survival Wrap
 5 Drivers of Evolution


Quiz Tomorrow!
Types of Adaptations
 Steps of Selection
 Modes of Selection

2B Essay Due Wednesday
 Standard 2B Friday 04/24

RECORD YOUR GROUP DATA HERE P3
Group
Children
Short
Long
Neck
Neck
Allele (N) Allele (n)
1
2
3
4
5
6
7
8
9
10
Grand Children
Short
Neck
Allele (N)
Long
Neck
Allele (n)
Great-Grand Children
Short
Neck
Allele (N)
Long
Neck
Allele (n)
Using the class generated data below:
1. Explain in your own words what happened to the
Tortoise population.
2. Did evolution occur? If so, which of the 5 evolutionary
forces was at work?
3. How do you know if evolution occurred? Use data below
to support your answer.
4. What type of adaptation was modified? Biotic or abiotic
Pressure?
Class Average
Allele Frequencies
Children
Grand
Children
Great-Grand
Children
Short Neck (N)
.53
.30
.12
Long Neck (n)
.47
.70
.88
SELECTION & SURVIVAL OF THE GALAPAGOS
Selection & Survival of the Galapagos P1
0.900
0.800
0.700
Children Short Neck Allele (N)
0.600
Children Long Neck Allele (n)
0.500
Grand Children Short Neck Allele (N)
0.400
Grand Children Long Neck Allele (n)
0.300
0.200
Great-Grand Children Short Neck Allele
(N)
0.100
Great-Grand Children Long Neck Allele
(n)
0.000
Allele (N)
Short Neck
Allele (n)
Allele (N)
Long Neck Short Neck
Children
Allele (n)
Allele (N)
Long Neck Short Neck
Grand Children
Allele (n)
Long Neck
Great-Grand Children
SELECTION & SURVIVAL OF GALAPAGOS TORTOISE
Selection & Survival of Galapagos Tortoise p4
1.000
0.900
0.800
Children Short Neck Allele (N)
0.700
Children Long Neck Allele (n)
0.600
Grand Children Short Neck Allele (N)
0.500
Grand Children Long Neck Allele (n)
0.400
Great-Grand Children Short Neck Allele
(N)
0.300
Great-Grand Children Long Neck Allele
(n)
0.200
0.100
0.000
Allele (N)
Allele (n)
Allele (N)
Allele (n)
Allele (N)
Allele (n)
Short Neck
Long Neck
Short Neck
Long Neck
Short Neck
Long Neck
Children
Grand Children
Great-Grand Children
1. WHAT TYPE OF SELECTION?
Example:
Birds with varying sizes
of beaks, after a
drought, the large beak
birds have selective
advantage
2. WHAT TYPE OF SELECTION?
Example:
Tourists in the desert like to
pick up a souvenir from their
travels, they pick up the
medium spine cactus.
Leaving the “homelier” low
spine and avoiding the
“prickly” high spine cactus.
3. WHAT TYPE OF SELECTION?
Example:
Birth weight in humans is
highly variable, however, a
child that is of average weight
has a higher chance of being
born and of good health, as to
mature and reproduce. Nature
selects against the extreme
phenotypes.
1. WHAT TYPE OF SELECTION?
Directional Selection
Selection Favors One
Extreme Variation of a Trait
Example:
Birds with varying sizes
of beaks, after a
drought, the large beak
birds have selective
advantage
2. WHAT TYPE OF SELECTION?
Disruptive Selection
Individuals with either
extreme trait
are selected for.
Example:
Tourists in the desert like to
pick up a souvenir from their
travels, they pick up the
medium spine cactus.
Leaving the “homelier” low
spine and avoiding the
“prickly” high spine cactus.
3. WHAT TYPE OF SELECTION?
Stabilizing Selection
Selection favors average
individuals of the
population.
Example:
Birth weight in humans is
highly variable, however, a
child that is of average weight
has a higher chance of being
born and of good health, as to
mature and reproduce. Nature
selects against the extreme
phenotypes.
AGENDA 04/21
Quiz
 5 Drivers of Evolution
 Hardy-Weinberg

2B Essay Due Wednesday
 Standard 2B Friday 04/24

WHAT DRIVES EVOLUTION? P. 16
Changes in Allelic Frequency
or the Gene Pool size
Natural Selection
 Genetic Drift (small Pop)
 Gene Flow
 Sexual Selection
 Mutations

5 DRIVERS
NATURAL SELECTION OF FAVORABLE TRAITS
5 drivers of Evolution
GENETIC DRIFT SMALL POPULATIONS
Founder Effect

New small population
Bottle neck effect

Small # of survivors of a
large population
GENETIC DRIFT
Impact on small
Populations
Decreasing Variation

Decreasing Variation

small populations
GENETIC DRIFT EXAMPLES
1. BLOOD TYPE DISTRIBUTION
Race
United States
Blackfoot
Navajo
Mexico
Chinese
India
O
45%
24%
76%
84%
34%
33%
A
42%
76%
24%
11%
31%
24%
B
10%
--------4%
28%
35%
AB
3%
--------1%
7%
8%
GENETIC DRIFT - SMALL POPULATIONS
2. FOUNDER EFFECT
COMMON COCKLEBUR (XANTHIUM
STRUMARIUM)
Spread by
“hitchhiking” as a bur
 Few plants colonizes
new area, low genetic
variability
 One trait (allele) may
show up more
in new population than
larger ancestral
population

GENETIC DRIFT OF SMALL POPULATION
3. BOTTLE NECK EFFECT
SEAL HUNTING
Large Population reduced to very small in 1800’s
 Surviving population had different allele frequency
and little genetic diversity

5 Drivers of Evolution
GENE FLOW
 Immigration-
to come into a population
 Emigration- to leave a population
5 drivers of Evolution
SEXUAL SELECTION
Strong force of evolution
 Females/Males only choose mates
based on a certain trait.
 Males/Females with the trait are “fit”

SEXUAL SELECTION = NON-RANDOM

Gentleman, which woman would you select as a
perspective mate?
SEXUAL SELECTION = NON-RANDOM

Ladies, which man would you select as a perspective
mate?
5 Drivers of Evolution
MUTATIONS
 Mutations
can
introduce new
phenotypes
 May be beneficial or
detrimental
 New genetic material
for selection to work
with
WHAT DRIVES EVOLUTION? P. 16
Changes in Allelic Frequency
or the Gene Pool size
Natural Selection
 Genetic Drift (small Pop)
 Gene Flow
 Sexual Selection
 Mutations
Mechanisms of Evolution

HARDY-WEINBERG

http://youtu.be/xPkOAnK20kw
THE HARDY-WEINBERG EQUATION IS USED TO
TEST WHETHER A POPULATION IS EVOLVING

For a population to remain in Hardy-Weinberg equilibrium
for a specific trait, it must satisfy five conditions:
1.
2.
3.
4.
5.
No Genetic Drift (Very large population)
No gene flow (immigration/ emigration)
No mutations
Random mating (no Sexual selection)
No natural selection
EQUILIBRIUM EQUATION
p2 + 2pq + q2 = 1 and p + q = 1





p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
HOW TO REMEMBER?

p = frequency of the dominant allele in the population


q = frequency of the recessive allele in the population


PP = individuals have 2 alleles so p2
q2 = percentage of homozygous recessive individuals


q= 1 allele
p2 = percentage of homozygous dominant individuals


p= 1 allele
QQ= individuals have 2 alleles so q2
2pq = percentage of heterozygous individuals

PQ = individuals have 1 of each allele so 2pq
The Hardy-Weinberg equation can be used to
test whether a population is evolving

Imagine that there are two alleles in a
blue-footed booby population: W and w
– W is a dominant allele for a nonwebbed booby foot
– w is a recessive allele for a webbed
booby foot
Copyright © 2009 Pearson Education, Inc.
W
Webbing
w
No webbing
The Hardy-Weinberg equation is used to
test whether a population is evolving

Consider the gene pool of a population of
500 boobies
–
–
–
320 (64%) are homozygous dominant (WW)
160 (32%) are heterozygous (Ww)
20 (4%) are homozygous recessive (ww)
Copyright © 2009 Pearson Education, Inc.
p2
2pq
q2
Phenotypes
Genotypes
WW
Ww
ww
Number of animals
(total = 500)
320
160
20
Genotype frequencies
320
–––
500
Number of alleles
in gene pool
(total = 1,000)
Allele frequencies (p, q)
160
–––
500
= 0.64
160 W + 160 w
640 W
800
1,000
20
–––
500 = 0.04
= 0.32
= 0.8 W
p
200
1,000
40 w
= 0.2 w
q
The Hardy-Weinberg equation can be
used to test whether a population is
evolving
Frequency of dominant allele
 (W) = 0.80 = p
– 80% of alleles in booby population are W
 Frequency of recessive allele
 (w) = 0.20 = q
– 20% of alleles in booby population are w

Copyright © 2009 Pearson Education, Inc.
THE HARDY-WEINBERG EQUATION CAN BE USED
TO TEST WHETHER A POPULATION IS EVOLVING

What about the next generation
of boobies?
– Probability that a booby sperm
or egg carries W = 0.8 or 80%
– Probability that a sperm or
egg carries w = 0.2 or 20%
Copyright © 2009 Pearson Education, Inc.
Gametes reflect
allele frequencies
of parental gene pool
W egg
p = 0.8
Eggs
w egg
q = 0.2
Next generation:
Genotype frequencies
Allele frequencies
Sperm
W sperm
p = 0.8
w sperm
q = 0.2
WW
p2 = 0.64
Ww
pq = 0.16
wW
qp = 0.16
ww
q2 = 0.04
p2 + 2pq + q2
0.64 WW
0.32 Ww
0.8 W
0.04 ww
0.2 w
THE HARDY-WEINBERG EQUATION CAN BE USED
TO TEST WHETHER A POPULATION IS EVOLVING

Frequency of all three genotypes is 100% or 1.0
–
–
p2 + 2pq + q2 = 100% = 1.0
homozygous dominant + heterozygous +
homozygous recessive = 100%
Copyright © 2009 Pearson Education, Inc.
WARM-UP 04/21

1.
2.
3.
Based on the condition given, is the population
Evolving or in Equilibrium (Hardy-Weinberg).
Random Mating EQUILIBRIUM no sexual selection
Large Population EQUILIBRIUM no small population
No Natural Selection
EQUILIBRIUM
All traits are beneficial
6.
Many mutations EVOLVING Adding new alleles
No immigration EQUILIBRIUM No mixing
Peahens choosing Peacocks for healthy tails EVOLVING
7.
Propeller Seeds traveling 100’s of miles EVOLVING
4.
5.
8.
Sexual Selection
Cactus plant in stable desert conditions
EQUILIBRIUM No Natural Selection
Mixing
AGENDA 04/21
Quiz Back/Take
 2B Essay Due
 Hardy-Weinberg Problems p. 17
 Homework: pp. 17-18


Standard 2B FRIDAY!

Be ready to create NS scenario for standard
2B ESSAY PEER-REVIEW & GRADE

4 Steps (10 pts each, if explained correctly)
40 pts
Overpopulation
 Variation
 Selection
 Adaptation


Type of Selection
20 pts
Do they explain which one? Is it correct? 10 pts
 Correct Graph provided.
10 pts


Type of Adaptation

Explanation of adaptation and typed of pressure
Conventions
 Intro & Conclusion

20 pts
10 pts
10 pts
0.488 or 48.8%
WINGED TRAIT IS DOMINANT
7/21= 0.333
√ 0.33 = 0.577
1- 0.57 = 0.423
0.4232 = 0.179
2 (.577)(0.423) = 0.488
PRACTICE
A population of birds contains 16 animals with red tail
feathers and 34 animals with blue tail feathers. Blue tail
feathers are the dominant trait.
 What is the frequency of the red allele?
q2 = 16/50 = 0.32 so q =

What is the frequency of the blue allele?
p = 1 - 0.566 =

√.32 = 0.566
0.434
What is the frequency of heterozygotes?
2 (0.566)(0.434) = 0.491

What is the frequency of birds homozygous for the blue
allele?
P2 = (0.434)2 = 0.188
2. You have sampled a population in which you know that
the percentage of the homozygous recessive genotype
(aa) is 36%. Using that 36%, calculate the following:
A. The frequency of the "aa" genotype.
Given as 36% = 0.36 = q2
B. The frequency of the "a" allele.
q = √ 0.36 = 0.6
C. The frequency of the "A" allele.
p + q = 1, so p = 1- 0.6 = 0.4
D. The frequencies of the genotypes "AA" and "Aa."
AA = p2 = (0.4)2 = 0.16
Aa = 2pq = 2(.6)(.4) = 0.48
E. The frequencies of the two possible phenotypes if "A" is
completely dominant over "a."
p2 + 2pq = (0.4)2 + 2(.6)(.4) = 0.64
Or 0.16 + 0.48 = all Dominant = 64%
HOMEWORK #2-9
WARM-UP 04/23

1.
2.
The yellow bellied three-toed skink (Saiphos equalis) is
a lizard of New South Wales, in Australia, that appears
to be undergoing the change from laying eggs to live
birth as the result of a series of mutations. Skinks
living on the coast tend to lay eggs, probably because
the warm weather is predictable and sufficient for
embryonic development. Those skinks living in the
cooler mountains tend to give birth to live young, the
mother’s body providing a more stable temperature. It
is to be predicted that these two populations will at
some point separate into different species.
Physiological
What type of Adaptation?
Biotic or Abiotic pressure? Abiotic - temperature
AGENDA 04/23

Homework Check pp.16-17

2B Review
Retake Pre-test on Clickers
I can Statements
2B Essay

3. There are 100 students in a class. Ninety-six did well
in the course whereas four blew it totally and received
a grade of F. Sorry. In the highly unlikely event that
these traits are genetic rather than environmental, if
these traits involve dominant and recessive alleles,
and if the four (4%) represent the frequency of the
homozygous recessive condition, please calculate the
following:
A. The frequency of the recessive allele.
q2 is given as 4% = 0.04 so q = √ 0.04 = 0.2
B. The frequency of the dominant allele.
q = 0.2 so p = 1- 0.2 = 0.8
C. The frequency of heterozygous individuals.
2pq = 2(0.2)(0.8) = 0.32
4. Within a population of butterflies, the color brown (B)
is dominant over the color white (b). And, 40% of all
butterflies are white. Given this simple information,
which is something that is very likely to be on an
exam, calculate the following:
A. The percentage of butterflies in the population that
are heterozygous.
if q2 = 0.40 so q = √0.40 = 0.63 and p = 1 - 0.63 = 0.37
So, 2pq = 2(0.63)(0.37)= 0.46 or 46%
B. The frequency of homozygous dominant individuals.
P2 = (0.37)2 = 0.14
or 40% (bb) + 46% (Bb)= 86%
so 100% - 86% = 14% (BB)
5. After graduation, you and 19 of your closest friends (lets
say 10 males and 10 females) charter a plane to go on a
round-the-world tour. Unfortunately, you all crash land
(safely) on a deserted island. No one finds you and you
start a new population totally isolated from the rest of
the world. Two of your friends carry (i.e. are
heterozygous for) the recessive cystic fibrosis allele (c).
Assuming that the frequency of this allele does not change
as the population grows, what will be the incidence of
cystic fibrosis on your island? ______
2pq = 2/20 = 0.1
q = 2 (c alleles)/40 total alleles = 0.05 or 5%
p = 1- 0.05 = 0.95
Let’s just check our answer
2pq = 2(.05)(.95) = 0.095 (~ 0.1)
6. Cystic fibrosis is a recessive condition that affects about
1 in 2,500 babies in the Caucasian population of the
United States. Please calculate the following.
The frequency of the recessive allele in the population.
q2 is given as 1/2500 = .0004 so q = √ 0.0004 = 0.02
The frequency of the dominant allele in the population.
q = 0.02 so p = 1 - 0.02 = 0.98
The percentage of heterozygous individuals (carriers) in
the population. ____
q = 0.02, p = 0.98, so 2pq = 0.039 or 3.9%
7. This is a classic data set on wing coloration in the
scarlet tiger moth (Panaxia dominula). Coloration in
this species had been previously shown to behave as a
single-locus, two-allele system with incomplete
dominance. Data for 1612 individuals are given below:
White-spotted (AA) = 1469
Intermediate (Aa) = 138
Little spotting (aa) = 5
Total = 1612
Calculate the allele frequencies ( p and q )
aa = q2 = 5/1612 = 0.0031
q2 = 0.0031, q = √0.0031 = 0.055
so p = 1 – 0.055 = 0.945
p2 = (0.945)2 = 0.893
2pq = 2(.945)(.055) =.103
Check It…..
AA = 89.3%
Aa = 10.3%
aa = 0.31%
8. The allele for a widow's peak (hairline) is dominant
over the allele for a straight hairline. In a population
of 500 individuals, 25% show the recessive phenotype.
How many individuals would you expect to be
homozygous dominant and heterozygous for the trait?
Given q2 = 0.25 q = 0.5, so 1 – 0.5 = 0.5 = p
So p2 = 0.25, or 25% of 500 = 125 homozygous dominant
So 2pq= 2 (0.5)(0.5) = .5 or 50% of 500 = 250 heterozygous
9. The allele for a hitchhiker's thumb is recessive
compared to straight thumbs, which are dominant. In
a population of 1000 individuals, 510 show the
dominant phenotype.
How many individuals would you expect for each of
the three possible genotypes for this trait.
1000 – 510 = 490 recessive individuals
Given q2 = 490 recessive/1000 = 0.49 q = 0.7
so 1 – 0.7 = 0.3 = p
So p2 = 0.09, or 9% of 1000 = 90 homozygous dominant
So 2pq= 2 (0.7)(0.3) = .42 or 42% of 1000 = 420
420 heterozygous individuals
EXTRA CREDIT
10. REAL WORLD APPLICATION PROBLEM
Choose a human trait to study and survey a population at your
school. (Aim for at least a sample size of 50 to get meaningful
results). Use your sample to determine the allele frequencies in
the human population.
Traits (dominant listed first)
Hitchhiker's Thumb vs Straight Thumbs
Widow's peak vs straight hairline
PTC taster vs non-taster
Short Big toe vs long big toe
Free earlobes vs attached earlobes
Tongue rolling vs non-rolling
Bent little fingers vs straight little fingers
Arm crossing (left over right) vs right over left
Ear points vs no ear points
Warm-up 04/24
FIND FOUR INCORRECT WORDS:
1.
2.
3.
4.
Organisms produce less offspring than
can survive.
In any population, mutations exist.
Individuals with certain “useful” or
beneficial variations perish and pass
on their variations to the next
generation.
Overtime, offspring with “beneficial”
variations will make up half of the
population.
1.
2.
3.
4.
LESS offspring
Organisms produce MORE
than can survive.
MUTATIONS exist.
In any population, VARIATIONS
Individuals with certain “useful” or
beneficial variations SURVIVE
PERISH and pass
on their variations to the next
generation.
Overtime, offspring with “beneficial”
variations will make up HALF
MOST of the
population.
AGENDA 04/24

Standard 2B






Intro to Engineering


Steps of Natural Selection
3 Types of Adatations
3 Modes of Selection
5 Drivers of Evolution
Hardy-Weinberg
I Spy ……..
Midterm Next Week

(no passes or improve grade)