Lecture 4: Heritability Flashcards
Variability in phenotype is due to…
variability in genetics + variability in environment
Vp = Vg + Ve
Broad sense heritability
proportion of Vp that is attributed to Vg
H2 = Vg/Vp
- lowest is 0, highest is 1
- higher the value, the more heritable
Vg =
additive genetic variance + dominance genetic variance + epistatic genetic variance
Vg = Vga + Vgd + Vgi
Additive genetic variance
alleles for a gene sum together to create an additive effect
- would give an intermediate phenotype for heterozygote
Dominance genetic variance
some alleles may be stronger than others (within the same gene)
- ex if you have an allele for brown eyes, your eyes are likely brown
Epistatic genetic variance
one allele may affect another (between diff genes)
- lab coat colour. The dog could have the brown coat colour allele but if it’s missing the gene for placing pigment in the hair shaft it’ll be blonde anyway
Narrow sense heritability
takes out dominance and epistatic genetic variability to only look at additive
h2 = Vga / Vp
- so the variabil attributed to genetic additive inheritance as a proportion of the total phenotypic variability
Environmental variance
Ve = Ves + Vens
- Ves = shared env variance (concerned with shared environments that increase trait similarity)
- Vens = diff environments that decrease trait similarity
technically encompasses everything that is not genetic (including error)
What is the main source of environmental variance for behavioural traits
Non-shared environments
As environmental similarity changes over time…
h2 will also change over time too
- example two siblings might start in the same home but end up in different parts of the world thus affecting the heritability estimate for a certain trait (h2 = Vga/Vp and Vp = Vg + Ve)
After leaving home, DZ twins become… and MZ…
less correlated and remain strongly correlated
Explain all the terms in the equation:
1 = h2 + c2 + e2
- h 2 is narrow sense heritability (proportion of Vg due to additive effects)
- c2 = proportion of phenotypic variance due to shared environment
- e2 = proportion of p variance due to non shared environment
Heritability of IQ … with age
increases
Hallmark of heritable trait
Strong relationship between genetic similarity and trait similarity
Ways to study heritability
- mid-parent offspring regression
- twin sibling studies
- Adopted sibling studies
- hybrid studies
Mid-parent offspring regression
- collect data on a quantitative trait like IQ
- calculate the average value for the two parents and average value for the children
- plot parent and child value on each axis
- regress
- slope of the line is an estimate of narrow sense heritability
Equal assumptions environment
DZ twins are subject to approx the same environmental similarity as MZ twins
- not rlllllllyyy true
Twin sibling studies
- Collect data from large set of of MZ and DZ twins
- plot twins as X and Y values (separate plots for MZ and DZ)
- determine x y correlation
Falconer Model and equations
uses trait similarity between mono and dizygotic twins
2(Rmz - Rdz) = h2
and
Rmz = h2 + c2
2(Rmz - Rdz) = h2
and then using herit estimate you can find variation due to shared env. with first eq
Rmz
trait resemblance
- the correlation between monozygotic twins on a particular trait
- can be attributed to heritability and shared environmental effects (c2)
R_mz =h2 + c2
Rdz
if dizygotic twins are 50% similar and MZ twins are 100% similar then
Rdz = 0.5h2 +c2
(because they share half the genes and the same env)
Key assumption of the falconer model
MZ and DZ twins share the same env
THREE KEY EQUATIONS FOR TEST
- 2(Rmz - Rdz) = h2
- Rmz = h2 +c2
= 1 = h2 + c2 +e2
Adoptee studies
studies shared env - assumes genetic differences are random
- children raised together are more similar to each other than children raised apart (role of env)
- Adopted children raised in the same env are less similar to each other than genetically related siblings raised in the same env (role of genetics)
Hybrid studies
MZ twins raised apart have same IQ basically
Heritability of ASD and SZ and IQ
around 0.6 to 0.7 and 0.4-0.8 depending on age
Heritability across all traits is
49% ish
Genetic mediation of Env: Passive, evocative, active
passive - receive an env correlated with genotype (artist parents, artistic household)
evocative - genetically-driven behaviours bring about certain environments (ex. aggression elicits punishment)
(behaviour evokes response thus modifying env)
Active - people will seek environments correlated with their genetic predisposition (actively looking for env)
Limitations to heritability
- its a characteristic of a specific population (does not apply to other popns or other individuals)
- estimate is based on specific genetic background and type and frequency of genes may vary in diff sub pops.
- environmental circumstances may vary in different subpopulations and also vary over time - tells us nothing about actual genes
- GWAS doesn’t rlly help - calculations can be misleading in certain situations
Gene innovation and amplification
ways in which genetic effects can vary with time
- take on new roles
- accumulate over time
If there is no variability in phenotype, heritability will always come out as…
zero. Can’t measure effect of different genes because all the traits are the same
Concordance study
useful for dichotomous traits like ASD, SZ, MDD, BPD.
not useful for continuous traits.
- determine presence of disorder in large pop of MZ and DZ twins
- in each twin pop, count the cases in whcih both twins have the disorder and the cases where only one has it
Concordance rate
Calculated for both twin populations so CRmz and CRdz
= cases of both twins affected / cases of both twins affected and case of one twin affected
Compare the two rates (ignoring env variance because assuming twins raised in sim env)
The greater the difference between the concordance rate of MZ and DZ, the larger the
heritability
