a. sexual selection b. heritability c. linkage equilibrium d. fossil record e. hardy weinberg...
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A. Sexual selection
B. Heritability
C. Linkage Equilibrium
D. Fossil Record
E. Hardy Weinberg exceptions
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Q2. This graph represents….
• Heritability• Selection differential• Selection gradient• Directional selection• Stabilizing selection
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A. Flowers
B. Fruit flies
C. Beetles
D. Guppies
E. Newts
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Qualitative traits are all or none
• attached earlobes
• widows peak
• six fingers
• cystic fibrosis
•height
•skin color
•plant flower size
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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
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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
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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
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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
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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
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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 parentsparents that is due to variation in the parent’s genes
Looking at a hypothetical population…
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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Selection differential can be calculated from selection gradient
Divide the selection gradient by the variance. Explained in box 9.3 p. 340.
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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
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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)
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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
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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%
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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
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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
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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.
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pg 343 Fig 9.21
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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
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Fig 9.22 pg 343
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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%.
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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
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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
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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.
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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
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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
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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
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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.
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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
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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
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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
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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.
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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
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