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Chapter 2

PSYCH 2TT3 Chapter Notes - Chapter 2: Common Descent, Termite, Eusociality


Department
Psychology
Course Code
PSYCH 2TT3
Professor
Ayesha Khan
Chapter
2

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Textbook notes – chapter 2 (The Evolution of Behaviour)
ARTIFICIAL SELECTION:
Behaviour is hereditary. We can perform artificial selection by choosing the species that portray our desired behaviours
and breed them together, and eventually we will get a population of species that are optimal for the selected trait. This
reduces variation in population. Each generation becomes more optimal.
NATURAL SELECTION:
Behavioural traits that affected reproductive success were passed from parents to offspring.
Selective advantage of a trait:
When nature is the selective agent, traits, including behavioural traits, increase or decrease in frequency as a function of
how well they suit organisms to their environments.
If a trait increases fitness and increases survival, it will be passed on to the next generation and increase in frequency in
the population.
Phenotype – observable properties of an organism
Which phenotype is a result of genotype – genetic makeup
Even a fitness advantage of 1 percent per generation is sufficient for one behaviour to replace another over evolutionary
time.
allele— is a gene variant, one of two or more alternative forms of a gene
Example: Individuals that hunt in packs may have high foraging success rates for large prey. Natural selection would
then favour pack hunting.
Over evolutionary time, small differences in fitness can accumulate into large changes in gene frequencies.
HOW NATURAL SELECTION OPERATES:
Three requirements for natural selection that include:
1. Variation in the trait—different varieties of the trait.
2. Fitness consequences of the trait—different varieties of the trait must affect reproductive success differently.
3. A mode of inheritance—a means by which the trait is passed on to the next generation.
Variation in trait can be caused by environmental factors or genetic factors.
Genetic variation in populations that be generated by mutation which is a random change in genetic structure.
Addition and deletion mutations occur when a single nucleotide is either added or deleted from a stretch of DNA. This
type of mutation typically causes the production of an inactive enzyme, which may, in turn, affect an animal’s
behaviour.
Base mutations occur when one base in a nucleotide replaces another. Silent mutation which does not change the amino
acid made is also a type of base mutation.
Genetic variation is also caused by genetic recombination by sexual reproduction.
In sexually reproducing organisms when pairs of chromosomes (one of which comes from the mother and one from the
father) line up during cell division, sections of one chromosome may “cross-over” and swap positions with sections of
the other chromosome. This swapping creates new genetic variation. Crossing-over points are essentially random, and
so virtually any crossover between a pair of chromosomes is possible in principle.
Migration can increase genetic diversity in a given population because individuals coming from other populations can
introduce new trait variants.
Variation alone is insufficient to allow natural selection to operate. Not only must there be behavioural variation but
that variation must have fitness consequences.
The fitness consequences of a trait refer to the effect of a trait on an individuals reproductive success.
Reproductive success - refers to the mean number of reproductively viable offspring an individual produces.
For natural selection to act on a trait, that trait must be passed down from one generation to the next.

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One way to study genes as a mode of transmission is by calculating narrow-sense heritability—a measure of the
proportion of variance in a trait that is due to genetic variance.
Behavioural differences can also be the result of genetic differences. It is this genetic variance that is measured in
heritability experiments.
TRUNACATION SELECTION:
One means for measuring narrow-sense heritability is by designing a truncation selection experiment.
In step 1 we measure the approach score of every bird in our population when it reaches twelve months of age. Suppose
this gives us a mean approach score of 60 seconds for
generation 1. Lets label this mean value x0. Step 2 of our
experiment is to truncate, or cut off, the population
variation in approach scores by allowing only those
individuals with approach scores greater than some value—
for example, 80 seconds—to breed. We then calculate the
mean approach score of those individuals that we have
allowed to breed. Lets label that mean as x1 and say that x1
equals an approach score of 90 seconds in our population.
The difference between x1 and x0 is referred to as the
selection differential, or S. In our case, S = 30 seconds.
One way to think about S is as the maximal amount we
could expect natural selection to change approach scores—
the amount of change that we might expect if all the
variation in approach score was genetic variation upon
which natural selection could act.
In step 3 of our truncation, we raise the
offspring produced by generation 1 birds, under
identical conditions to those experienced by
their parent, until they have reached twelve
months of age, and then we measure their
approach scores. Lets label the mean approach
score of these generation 2 individuals as x2, and imagine this value to be 70 seconds. The difference between this mean
(x2) and the mean of the entire population in the first generation (x0) is referred to as the response to selection, or R. It
is a measure of how much truncation selection has changed approach scores across generations 1 and 2. In our case, R
= 10 seconds.
Heritability is defined as R/S, so in our population of birds, the heritability of approach is 10/30 or about 0.33. In other
words, one-third of all the variance in approach is due to genetic variance upon which natural selection can act
So we have to find the mean of parent distribution, then truncate and say above a certain mean can only breed, then find
the mean of that population, then subtract the two mean values to give S, then get the mean of the F1 population,
subtract that by the mean of parents, that gives you response to selection.
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