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Lecture 6

# PSYC 2P25 Lecture Notes - Lecture 6: Trait Theory, Heritability, Epistasis

Department
Psychology
Course Code
PSYC 2P25
Professor
Michael Ashton
Lecture
6

This preview shows page 1. to view the full 4 pages of the document. PSYC 2P25 Oct. 22, 2012
Heredity and Environment
What proportion of the variation in a personality trait is due to inherited (genetic) variation?
Environmental variation)
o Why are people different in their personality? [variation]
Researchers want to find the heritability (h2) of a trait the proportion of the trait variation that
is due to heritable (genetic) variation in a give population
Of course, 0 h2 1
To find the heritability of a personality trait, we need to know whether (genetic) relatives have
similar trait levels beyond any similarity due to their common environments
Ways to estimate heritability (h2)
1) study many set of identical (monozygotic; MZ) twins raised apart
o MZ twins share 100% of genes. If they’re raised apart, then similarity between them
can’t be due to being raised together
o Measure trait, find correlation (r) between twins’ levels (if twin A is “high” in trait, is
twin B also?)
o For MZ twins raised apart, r indications heritability (h2):
E.g., r= .60 -> h2= .60
2) study many sets of fraternal (dizygotic; DZ) twins (or even non-twin siblings) raised apart
o DZ twins share 50% of genes (same as for non-twin siblings). If they’re raised apart,
then similarity between them can’t be due to being raised together
o Measure trait, find r between twins’ levels, then double r to get h2 (need to double r
because it represents the effect of our 50% genetic similarity)
E.g. r= 30 -> h2 = 2(.30) = .60
3) study many sets of MZ twins raised together AND may sets of DZ twins raised together, and
compare them
o Measure trait, find r between MZ twins’ levels, and r between DZ twins’ levels
o Find difference between the r’s, then double it (difference in rs is due to the extra 50%
genetic similarity of MZ twins; need to double it to get the full effect of heredity)
E.g., MZ r= .60, DZ r= .30 -> h2 = 2 (.60 - .30) = 2 (.30) = .60
o This method of comparing similarity of MZ twins with similarity of DZ twins is the most
common way of estimating trait heritability
o Note: that to find h2, you can’t just take the r between MZ twins raised together (or 2r
between DZ twins raised together) because r might be due to [ADD FROM NOTES]
In example given above, the r for MZ twins together r(.60) was exactly twice as
big as the r [ADD FROM NOTES]
First, it’s possible that the r for MZ twins together could be more than twice as
big as the r for DZ twins together
E.g. MZ r= .70 and DZ r= .30