Transcript Chapter 9 - Heritability

A.
B.
C.
D.
E.
Sexual selection
Heritability
Linkage Equilibrium
Fossil Record
Hardy Weinberg exceptions
Q2. This graph represents….
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Heritability
Selection differential
Selection gradient
Directional selection
Stabilizing selection
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B.
C.
D.
E.
Flowers
Fruit flies
Beetles
Guppies
Newts
•height
•skin color
•plant flower size
Qualitative traits are all or none
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attached earlobes
widows peak
six fingers
cystic fibrosis
provides tools to analyze genetics and
evolution of continuously variable traits
 Provides tools for:
1. measuring heritable variation
2. measuring survival and reproductive
success
3. predicting response to selection

When assessing heritability we need to
make comparisons among individuals.
Cannot assess a continuous trait’s
heritability within one individual
 Need to differentiate whether the
variability we see is due to environmental
or genetic differences
 Heritability = The fraction of the total
variation which is due to variation in
genes

Phenotypic variation (VP) is the total
variation in a trait (VE + VG)
 Environmental variation. (VE) is the
variation among individuals that is due to
their environment
 Genetic variation (VG) is the variation
among individuals that is due to their genes

Additive Genetic Variation (VA) = Variation
among individuals due to additive effects of
genes
 Dominance Genetic Variation (VD) =
Variation among individuals due to gene
interactions such as dominance
 VG = VA + VD

VG
VP =
VG
VG + V E

heritability =

Heritability is always between 0 and 1

If the variability is due to genes then it makes
sense to evaluate the resemblance of
offspring to their parents
 Broad
sense heritability = VG / VP
 Narrow
sense heritability = VA / VP
 We
will deal only with narrow sense
heritability = h2
 Use
of narrow sense heritability allows
us to predict how a population will
respond to selection
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
Plot midpoint value for the 2 parents on x axis
and mid-offspring value on y axis and draw a
best fit line.
This slope which is calculated by least squares
linear regression is a measure of heritability
called narrow-sense heritability or h2
h2 is an estimate of the fraction of the variation
among the parents that is due to variation in the
parent’s genes
Looking at a hypothetical population…
Figure 9.13a Pg 334
If slope is near zero there is no resemblance
Mid parent height
Evidence that the variation among parents is due to the
environment.
If this slope is near 1 then there is strong resemblance
Evidence the variation among parents is due to genes
Any study of heritability needs to account for
possible environmental causes of similarity
between parent and offspring.
 Take young offspring and assign them
randomly to parents to be raised
 In plants, randomly plant seeds in a given
field
 Example in text Song Sparrows studied by
James Smith and Andre Dhondt.

Showed song
sparrow chicks
(eggs or
hatchlings)
raised by foster
parents
resembled
their biological
parents
strongly and
their foster
parents not at
all
Figure 9.14 p. 335
Done by measuring the strength of
selection by looking at the differences in
reproductive success.
 Basically we measure who survives, who
doesn’t, and then quantify the difference
 Example breeding mice with longer tails

DiMasso and colleagues bred mice in order
to select for longer tails
 Each generation they picked the 1/3 of the
mice who had the longest tails and allowed
them to interbreed
 Did this for 18 generations
 Calculated the strength of selection

Selection differential (S) = difference
between mean tail length of breeders (those
that survive long enough to breed) and the
mean tail length of the entire population.
 Selection gradient = slope of a best fit line
on a scatter plot of relative fitness as a
function of tail length

Figure 9.17 p. 339
Selection differential (S)
Average tail
length of the
breeders only
minus the
average tail
length of the
entire
population
entire population
breeders (survivors)
Only the 1/3 of mice
with the longest tails
allowed to breed
(survive)
1.
2.
3.
Assign absolute fitness – fitness equals
survival to reproductive age. Long tailed had a
fitness of 1, short tailed a fitness of 0
Convert absolute fitness to relative fitness.
Figure the mean fitness of the population.
Then divide the absolute fitness by the mean
fitness . (Mean fitness = .67(0) + .33(1) = .33).
So relative fitness of breeders = 1/.33 = 3.0 and
relative fitness of non-breeders = 0/.33 = 0.
Make a scatterplot of relative fitness as a
function of tail length. Calculate the slope
using best fit. The slope is the selection
gradient
Figure 9.17 p. 339
Selection gradient
1. Calculate relative fitness
for each mouse, then plot
relative fitness of each as a
function of tail length
2. the slope of the best fit
line is the selection gradient
Selection differential can be calculated
from selection gradient
 Divide the selection gradient by the
variance. Explained in box 9.3 p. 340.

Once we know the heritability and the
strength of selection we can predict
response to selection
 R = h2 S

* R = predicted response
* h2 = heritability
* S = selection differential
We can estimate how much variation in a trait is
due to the variation in a gene (heritability)
 Quantify the strength of selection that results
from differences in survival or reproduction.
(selection differential)
 Predict how much a population will change from
one generation to the next. (predicted response
to selection)

Candace Galen (1966) studied the effect of
selection pressure by bumblebees on flower
diameter
 Worked with alpine skypilots from two
elevations, timberline and tundra

› Tundra flowers are larger and are pollinated
exclusively by bumblebees
› Timberline flowers are pollinated by a mixture of
insects and are smaller
1.
2.
Is selection by the bumblebees in the
tundra responsible for the larger flower
size?
How long would it take for selection
pressure to increase flower size by 15%
1.
Determine heritability
• measured flower diameters
• collected seeds germinated them and
transplanted seedlings to random locations in
the same habitat as the parents
• seven years later measured the flowers from
the 58 plants which had matured enough to
flower
• plotted offspring flower diameter as a function
of maternal (seed bearing parent) flower
diameter


results provided a best fit
number of 0.5 for
heritability. Actual
calculations give h2 of
1.0 (because multiple
offspring with only one
parent [female]).
Scatter (fig 9.20)
necessitated a statistical
analysis which showed
she could only be certain
that at least 20% of the
phenotypic variation was
due to additive genetic
variation.
Therefore h2 lies somewhere
(h2 = VA / VP)
between 0.2 and 1.0

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
caged some about-to-flower Skypilots with
bumblebees
measured flower size when Skypilots bloomed and
later collected their seeds
planted seedlings back out in the original parental
habitat
Six years later she counted the number of surviving
offspring produced by each of the parent plants
She used the number of surviving 6 year old
offspring as her measure of fitness
Plotted relative fitness (# of surviving 6 year old
offspring / total number planted) as a function of
maternal flower size.
pg 343 Fig 9.21
Calculated the selection differential (S) ( by
dividing selection gradient by variance in flower
size)
 Her S value told her that, on average, the
flowers visited by bumblebees were 5% larger
than the average flower size.
 Control experiments from random hand
pollination and by a mixture of pollinators other
than bumblebees, showed no relationship
between flower size and fitness

Fig 9.22 pg 343
using the low end h2 of .2 and an S of .05
• R = h2S = .2 (.05) = .01
 using a high end for h2 of 1.0 and S = .05
• R = h2 S = 1(.05) = .05
 Means that a single generation of selection
should produce an increase in the size of the
average flower by from 1% to 5%.

Observations of a population of timberline
flowers pollinated exclusively by
bumblebees showed that on average
flowers that were produced by bumblebee
pollination were 9% larger than those
pollinated randomly by hand.
Galen’s prediction that response was
rapid was verified

Fitness of a
phenotype
increase or
decreases
with the
value of a
trait.
Slightly
reduces the
variation in a
population
Examples of this
type of selection
are The Alpine
Skypilot and the
Finch beaks in
times of drought.
One extreme
phenotypic
expression of
the trait
increases in
fitness and the
other extreme
decreases.



Those individuals with
intermediate values
are favored at the
expense of both
extremes.
The average value of
a trait remains the
same but the variation
is reduced
The tails of the
distribution are cut off.
Example in gall flies - Weis and Abrahamson 1986
Figure 9.26 p. 348
 A fly lays eggs in
Goldenrod bud.
 Plant produces a gall in
response to the fly larva
 Wasps lay eggs in galls
that eat fly larva
 Birds also eat galls.
 Pressure from wasps
selects for larger galls
and
 Pressure from birds
selects for smaller galls
 The result is selection
for mid sized galls.
Selects for
individuals with
extreme values for a
trait
 Does not change
AVERAGE value but
INCREASES
phenotypic variance
 Result far fewer
individuals at the
middle of the
continuum for the
trait

Example of the black-bellied seed cracker
Breeding Populations
have birds with
EITHER large OR
small beaks
 Juveniles show the full
spectrum of beak size
 But only the large OR
small beaked birds
survive to reproduce.

Fiogure 9.27 p. 349
Unlike our example of the moths and other
ONE gene traits….
 We are talking here about quantitative traits
determined by multiple genes:

› As phenotypic variation decreases so should
genetic variation
› However in most populations substantial genetic
variation continues to be exhibited.
› A satisfactory explanation for this unexpected
outcome is under debate and no acceptable
hypothesis is yet agreed upon.
Clausen Keck and Hiesey 1948
• Worked with Achillea
lanulosa
• On average plants from
the low altitude
Populations produce
slightly more stems than
those native to higher
elevations. (30.20 to
28.32)
Figure 9.31 p. 354
When grown together at low elevation,
low elevation plants produced more stems
This is consistent with the idea that high-altitude
plants are genetically programmed to produce fewer
stems
When the two source plants were
grown together at high altitude ….
 High altitude plants
had more stems!
(19.89 vs 28.32)
 Each population
was superior in its
own environment
 Apparently there
are genetic
differences that
control how each
responds to the
environment
 This is a
demonstration of
phenotypic
plasticity
Must always remember that variation has both a
genetic and an environmental component.
 Any estimate of heritability is specific to a
particular population living in a particular
environment.
 High heritability within groups tell us nothing
about the origin of the differences between
groups
 Cannot be used to determine the differences
between populations of the same species
that live in different environments.

All that we can really gain by
measuring heritability is the ability to
predict whether selection on the trait
will cause a population to evolve