Lecture V

S318 - Honors Evolution

Lecture V: Testing the Syllogism II

Natural Selection continued

1) What do we want to know?

a) Does evolution natural selection explain the fit between organism and environment?

b) Does evolution by natural selection explain the splitting of lineages into new species? Adaptive radiation?

2) Where are we now with respect to answering these questions?

a) with respect to 1a, we are making progress but we need to apply these concepts to REAL species

b) with respect to 1b, we have a long way to go.

3) Review of the card game

4) Evolution in Darwin's finches

5) Groups: Design a "lab" for high school teachers to explain heritability and the response to selection.

Review of the card game

(remember, we will also post a separate link after Lab III showing the data we collected and highlighting the main points of each exercise [Games I-III])

WITHOUT environmental variance (Game I)

Here is a copy of the figure we handed out in class. I've drawn some extra things on it.

Several things to notice here

1) The mean after selection minus the mean before selection is equal to the selection differential

2) The mean of the offspring distribution minus the mean of the distribution before selection is equal to the response to selection

3) The response to selection is also equal to the product of heritability and the selection differential

4) In this game, all of the phenotypic variation in our population was due to the additive effects of alleles. We found that the heritability of this trait was 1.0. Therefore, we expect the response to selection (point 2) to be equal to the selection differential (point 1).

5) The mean after selecton (17.93) minus the mean before selection (14.73) equals the selection differential (3.20). The mean of the offspring distribution (17.93) minus the mean of the distribution before selection (14.73) is equal to the response to selection (3.20). These results match our predictions.

WITH environmental variance (Game II)

Here is a copy of the figure we handed out in class. Again, I've drawn some extra things on it.

1) In this game, we found that the heritibility of our trait was less than 1, but greater than 0. Therefore, we expect the response to selection to be less than the selection differential.

2) The mean after selection (17.57) minus the mean before selection (13.45) equals the selection differential (4.12). The mean of the offspring distribution (15.86) minus the mean of the distribution before selection (13.45) is equal to the response to selection (2.41). These results match our predictions.

Evolution in Darwin's finches

See section 3.2 in the text (pages 49-57).

Some other stuff to think about...

Remember that heritability = the proportion of total phenotypic variation in a population that is due to the additive effects of alleles. It also represents the proportion of phenotypic variation in a population that can respond to selection.

Selection occurs within generations, a response to selection (i.e., evolution by natural selection) occurs across generations

You can think of heritability as "scaling" the response to selection in the next generation. If all of the phenotypic variation in a population is due to the additive effects of alleles, then the response to directional selection will be equal to the selection differential. The new trait mean in the next generation will be equal to the old trait mean (before selection) plus the selection differential.

An example using the Breeder's equation

Frank has a population of beetles and knows there is phenotypic variation in his population for exoskeleton thickness (he has an exciting life...). These beetles breed at the same time and die immediately following the mating period. Curious about the evolution course he is taking at the time, Frank measures the heritability of exoskeleton thickness in his population using a mid-parent / offspring regression technique (carefully controlling for environmenal factors) and gets a value of 0.60. He also notes that the mean thickness in his population (just the offspring now) is about 0.50 mm.

After going home for an extended weekend, Frank returns to find his roomate has thrown a party and somehow his beloved beetles were used in some kind of squashing / crunching game. He quickly locates the survivors and returns them to their home. Furious, but sensing the opportunity to test the theory of evolution by natural selection, Frank quickly (there aren't that many left) measures the exoskeleton thickness of the survivors and gets a mean value of 0.68mm.

He can predict the mean exoskeleton value in the next beetle generation as follows:

selection differential = 0.68mm - 0.50mm = 0.18mm

heritability = 0.60

response to selection = 0.60 * 0.18mm = 0.108mm

Therefore, there should be a 0.108mm increase in mean exoskeleton thickness in the next offspring generation (relative to the original population before selection).

The mean exoskeleton thickness should be 0.50mm + 0.108mm = 0.608mm

Notice that this value is between the mean before selection (0.50mm) and the mean after selection (0.68mm) because the heritability is less than 1.

You can also think about this stuff graphically as follows:

Consider a scatter-plot of mid-parent trait values vs. offspring trait values. Recall that the slope of the best-fit line through those data points (which equals COV(o.p)/Var(p)) is the heritability of that trait.

Now, let us assume that the heritability of this trait is 1 (the slope of the green line above is 1) and select on the parents such that the mean after selection is greater than the mean before selection. It is this difference that is the selection differential (s).

Next, let us map the mean trait value of the parental generation onto the offspring axis (dotted inner square in the figure below) - this is equal to the mean value of the trait in the kids.

We can determine the mean trait value of the kids after selection (remember, selection was in the parental generation & the kids are the survivors' offspring) by drawing a line from the mean of the parents after selection up to the "heritability line" and across to the offspring axis (outer dotted lines in the figure below).

The response to selection (R) is the difference between the mean of the kids after selection in the parental generation and the mean of the parents before selection (which is equal to the mean of the kids before selection).

Notice that R = s when the heritability is 1

In the second example, consider the exact same situation, only now the heritability is less than one. I've lightly drawn in a line showing a heritability of 1 so you can see the comparison.

Again, we can map the mean trait value of the parental generation onto the offspring axis (dotted inner square in the figure below) - this is equal to the mean value of the trait in the kids.

Next, we can determine the mean trait value of the kids after selection by drawing a line from the mean of the parents after selection up to the "heritability line" and across to the offspring axis (outer dotted lines in the figure below).

Notice how the response to selection is smaller than the selection differential (distance between the dotted lines above the "heritability line" is smaller than the distance between the dotted lines below the "heritability line")

Also notice how the relationship between R and s is scaled by the heritability. Looking at the figure below, imagine twisting the "heritability line" clockwise as a dial about its mean (point where the inner square lines intersect it) and consider how further lowering the heritability also lowers the response to selection.

If the "heritability line" becomes completely horizontal, then no matter what the selection differential, the response to selection will always be 0 (the mean of the kids after selection will always be the same as the mean of the kids before selection on the y-axis above). In other words, if none of the phenotypic variation in the population is due to the additive effects of alleles, then there can be no response to selection.