A. Sexual selection

B. Heritability

C. Linkage Equilibrium

D. Fossil Record

E. Hardy Weinberg exceptions

Q2. This graph represents….

• Heritability• Selection differential• Selection gradient• Directional selection• Stabilizing selection

A. Flowers

B. Fruit flies

C. Beetles

D. Guppies

E. Newts

Qualitative traits are all or none

• attached earlobes

• widows peak

• six fingers

• cystic fibrosis

•height

•skin color

•plant flower size

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

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

VG

VP

VG

VG + VE

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

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 parentsparents that is due to variation in the parent’s genes

Looking at a hypothetical population…

If slope is near zero there is no resemblance

Evidence that the variation among parents is due to the environment.

Mid parent height

Mid

offs

prin

g he

ight

Figure 9.13a Pg 334

If this slope is near 1 then there is strong resemblance

Mid

offs

prin

g he

ight

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

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)

Figure 9.17 p. 339

1. Assign absolute fitness – fitness equals survival to reproductive age. Long tailed had a fitness of 1, short tailed a fitness of 0

2. 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.

3. Make a scatterplot of relative fitness as a function of tail length. Calculate the slope using best fit. The slope is the selection gradient

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

Figure 9.17 p. 339

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 differentialdifferential

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. Is selection by the bumblebees in the tundra responsible for the larger flower size?

2. 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. (h2 = VA / VP)

Therefore h2 lies somewhere between 0.2 and 1.0

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.

Examples of this type of selection are The Alpine Skypilot and the Finch beaks in times of drought. One extreme One extreme phenotypic phenotypic expression of expression of the trait the trait increases increases in fitness and the other extreme decreases. Slightly

reduces the variation in a population

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.

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.

Example in gall flies - Weis and Abrahamson 1986Example in gall flies - Weis and Abrahamson 1986Figure 9.26 p. 348

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

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.

Example of the black-bellied seed cracker

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 respondsresponds to the environment

This is a demonstration ofphenotypic phenotypic plasticityplasticity

Must always remember that variation has both a genetic and an environmental component.

Any estimate of heritability is specific to a particularparticular population living in a particularparticular environment.

High heritability within groups tell us nothing tell us nothing about the origin of the differencesabout the origin of the differences between groups

Cannot be usedCannot be used to determine the differences differences between populations of the same speciesbetween 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