Today I'll be discussing the improvement of
populations through mass selection. The organism
I chose is spiders, and I'll be selecting
the top 10% of the population to breed for
more legs. This will allow the spare legs
to be released when the spider is caught by
a predator. Our original population is a data
pool of 8 spiders, a broad sense heritability
of 0.85, and a realized heritability of 0.67.
This chart describes the variables and equations
used to make calculations and predictions.
The selection intensity and standardized selection
point will be used in calculations. These
values are derived from the chart at the bottom.
Since we are taking the top 10%, our proportion
selected is 0.1 and we will use the corresponding
values. The mean original population is calculated
by adding the amount of legs each spider has
then dividing by the total number of individuals.
This comes out to 8. The standard deviation
is found by adding the squared differences
of the population and the population average,
then dividing by the total number of individuals,
which is 0.5. The mean of the selected parents
is found by multiplying the selection intensity
from the chart by the standard deviation,
then adding the average of the original population,
which comes to 8.88. The selection differential
is equal to the difference of the mean of
the selected parents and the population average,
which is 0.88. The truncation point is found
by adding the mean of the original population
to the product of the standardized selection
point and standard deviation. This equals
8.64. The response from selection is calculated
by multiplying the realized heritability by
the selection differential, which comes out
to be 1.55. And the mean of the offspring
is found by adding the original population
average to the response from selection, which
comes out to be 9.55. Here is a histogram
summarizing our findings. As you can see,
the mean of the offspring is moving towards
our selected mean and away from the original
mean, meaning the trait for more legs is being
passed down. I repeated these calculations
for the offspring population. The difference
is that our new mean of this F1 generation
is 9.55 and the standard deviation is 0.11.
The mean of the selected parents is 9.74.
The selection differential is 0.45. The truncation
point is 9.69. The response from selection
is 0.31. And the mean of the offspring is
9.86. Here's another histogram to summarize
the results. As you can see the mean offspring
of this group will have 9.86 legs, which is
greater than the offspring of the parental
group. A difference to note between the two
graphs is that the response to selection has
decreased in this one. Let's discuss why this
has happened. The response to selection is
defined as the extent to which the characteristic
changes over time and can be used to predict
heritability. If this mass selection was repeated
indefinitely, the response would decrease
in later rounds of selection. One reason this
may happen is because the extreme type may
not be healthy. Breeding spiders for more
legs may not be to their environmental advantage,
and could cause them to die prematurely before
passing on the trait. Another possible reason
for this is that two traits are negatively
correlated. In this example, that could mean
that the trait for more legs could be negatively
correlated to the ability to spin a web. Without
spinning a web, the spider will not make the
trap to catch its prey.
