2. A. Sexual selection B. Heritability C. Linkage Equilibrium D. Fossil Record E. Hardy Weinberg exceptions

3. Q2. This graph represents…. Heritability Selection differential Selection gradient Directional selection Stabilizing selection

4. A. Flowers B. Fruit flies C. Beetles D. Guppies E. Newts

5. Qualitative traits are all or none attached earlobes widows peak six fingers cystic fibrosis height skin color plant flower size

6.  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

7.  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

8.  Phenotypic variation (V P ) is the total variation in a trait (V E + V G )  Environmental variation. (V E ) is the variation among individuals that is due to their environment  Genetic variation (V G ) is the variation among individuals that is due to their genes

9.  Additive Genetic Variation (V A ) = Variation among individuals due to additive effects of genes  Dominance Genetic Variation (V D ) = Variation among individuals due to gene interactions such as dominance  V G = V A + V D

10.  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 VGVPVGVP V G V G + V E

11.  Broad sense heritability = V G / V P  Narrow sense heritability = V A / V P  We will deal only with narrow sense heritability = h 2  Use of narrow sense heritability allows us to predict how a population will respond to selection

12.  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 h 2 parents  h 2 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…

13. If slope is near zero there is no resemblance Evidence that the variation among parents is due to the environment. Mid parent height Midoffspring height Figure 9.13a Pg 334

14. If this slope is near 1 then there is strong resemblance Midoffspring height Evidence the variation among parents is due to genes

15.  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.

16. 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

17.  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

18.  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

19.  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

20. 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

21. 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

22. 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

23.  Selection differential can be calculated from selection gradient  Divide the selection gradient by the variance. Explained in box 9.3 p. 340.

24.  Once we know the heritability and the strength of selection we can predict response to selection  R = h 2 S * R = predicted response * h 2 = heritability differential * S = selection differential

25.  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)

27.  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

28. 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%

29. 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

30.  results provided a best fit number of 0.5 for heritability. Actual calculations give h 2 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. (h 2 = V A / V P ) Therefore h 2 lies somewhere between 0.2 and 1.0

31.  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.

32. pg 343 Fig 9.21

33.  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

34. Fig 9.22 pg 343

35.  using the low end h 2 of.2 and an S of.05 R = h 2 S =.2 (.05) =.01  using a high end for h 2 of 1.0 and S =.05 R = h 2 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%.

36. 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

39.  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 phenotypic expression of the trait increases in fitness and the other extreme decreases. Slightly reduces the variation in a population

40.  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.

41.  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 1986 Figure 9.26 p. 348

42.  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

43.  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

44.  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.

46. 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

47. 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

48. 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 responds  Apparently there are genetic differences that control how each responds to the environment phenotypic plasticity  This is a demonstration of phenotypic plasticity

49. The upper part of the illustration shows the different appearances of Achillea lanulosa-populations like the variance in the plants' heights cultivated under identical standard conditions in climatic chambers. The lower part of the illustration gives the natural geographic origin of the single populations by way of a profile of a west-to-east cross-section through California; to the left: Sierra Nevada, to the right: Great Basin (J. CLAUSEN, D. D. KECK, W. M. HIESEY, 1948)

50.  Must always remember that variation has both a genetic and an environmental component. particularparticular  Any estimate of heritability is specific to a particular population living in a particular environment. tell us nothing about the origin of the differences  High heritability within groups tell us nothing about the origin of the differences between groups  Cannot be useddifferences between populations of the same species  Cannot be used to determine the differences between populations of the same species that live in different environments.

51. 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