Lecture 4: Heritability

Question 1

Variability in phenotype is due to…

Answer 1

variability in genetics + variability in environment

Vp = Vg + Ve

Question 2

Broad sense heritability

Answer 2

proportion of Vp that is attributed to Vg
H2 = Vg/Vp
- lowest is 0, highest is 1
- higher the value, the more heritable

Question 3

Vg =

Answer 3

additive genetic variance + dominance genetic variance + epistatic genetic variance

Vg = Vga + Vgd + Vgi

Question 4

Additive genetic variance

Answer 4

alleles for a gene sum together to create an additive effect

- would give an intermediate phenotype for heterozygote

Question 5

Dominance genetic variance

Answer 5

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

Question 6

Epistatic genetic variance

Answer 6

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

Question 7

Narrow sense heritability

Answer 7

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

Question 8

Environmental variance

Answer 8

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)

Question 9

What is the main source of environmental variance for behavioural traits

Answer 9

Non-shared environments

Question 10

As environmental similarity changes over time…

Answer 10

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)

Question 11

After leaving home, DZ twins become… and MZ…

Answer 11

less correlated and remain strongly correlated

Question 12

Explain all the terms in the equation:

1 = h2 + c2 + e2

Answer 12

• 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

Question 13

Heritability of IQ … with age

Answer 13

increases

Question 14

Hallmark of heritable trait

Answer 14

Strong relationship between genetic similarity and trait similarity

Question 15

Ways to study heritability

Answer 15

1. mid-parent offspring regression
2. twin sibling studies
3. Adopted sibling studies
4. hybrid studies

Question 16

Mid-parent offspring regression

Answer 16

• 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

Question 17

Equal assumptions environment

Answer 17

DZ twins are subject to approx the same environmental similarity as MZ twins
- not rlllllllyyy true

Question 18

Twin sibling studies

Answer 18

• 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

Question 19

Falconer Model and equations

Answer 19

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

Question 20

Rmz

Answer 20

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

Question 21

Rdz

Answer 21

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)

Question 22

Key assumption of the falconer model

Answer 22

MZ and DZ twins share the same env

Question 23

THREE KEY EQUATIONS FOR TEST

Answer 23

• 2(Rmz - Rdz) = h2
• Rmz = h2 +c2
  = 1 = h2 + c2 +e2

Question 24

Adoptee studies

Answer 24

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)

Question 25

Hybrid studies

Answer 25

MZ twins raised apart have same IQ basically

Question 26

Heritability of ASD and SZ and IQ

Answer 26

around 0.6 to 0.7 and 0.4-0.8 depending on age

Question 27

Heritability across all traits is

Answer 27

49% ish

Question 28

Genetic mediation of Env: Passive, evocative, active

Answer 28

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)

Question 29

Limitations to heritability

Answer 29

1. 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
2. tells us nothing about actual genes
   - GWAS doesn’t rlly help
3. calculations can be misleading in certain situations

Question 30

Gene innovation and amplification

Answer 30

ways in which genetic effects can vary with time

• take on new roles
• accumulate over time

Question 31

If there is no variability in phenotype, heritability will always come out as…

Answer 31

zero. Can’t measure effect of different genes because all the traits are the same

Question 32

Concordance study

Answer 32

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

Question 33

Concordance rate

Answer 33

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)

Question 34

The greater the difference between the concordance rate of MZ and DZ, the larger the

Answer 34

heritability