Lecture 26: Quantitative Traits VI Date: 11/21/02  Parent-offspring finished  Sib pairs  Twins

Review: Parent-Offspring Regression

 Assume (1) insubstantial epistasis, (2) equal phenotypic variation in sexes and generations, (3) no general environmental effects contributing to resemblance, and (4) no assortative mating.

  When regressing offspring phenotype on parent phenotype, the regression slope is an estimate of 0.5 2 times the

narrow-sense heritability

h

2   

A

2 .

z

When regressing offspring phenotype on the midparent phentoype, the regression slope is a direct estimate of

h

2 .

Multiple Offspring

Suppose there are

n

offspring in each family.

The dependent variable becomes the average offspring phenotype.

Making all previous assumptions, we have    

j n

  1

n z oj

,

z p

   1

n j n

  1  

z oj

,

z p

   

z oj

,

z p

 E  

op

       

j n

  1

z oj n

  

p

,

z p

         

z

,

z o

 

p p



Unequal Family Size

 When there are unequal numbers of offspring in each family, we would like to weight each family’s contribution to the analysis in such a way as to minimize the sampling error in the heritability estimate.

 The weight should naturally take into consideration the number of offspring sampled. Presumably families with more offspring “count” more (are weighted heavier).

Within/Between Family Phenotypic Variance

  The

among-family phenotypic variance

t

Var

 

 Cov

 

is where Cov(

S

) is the phenotypic covariance of sibs and

t

is called the

intraclass correlation

.

An estimate of the

within-family phenotypic variance

(due to differences among individuals of the same family) is given by:

   

 Var Cov

 

Residual Variance

  The linear regression we consider is:

z oi

   

op z pi



e i

The residual error variance consists of  the variance attributable deviations of the “true” family means from the linear prediction (independent of sample size)  sampling variance (dependent on sample size).

Components of Residual Variance

 Sampling variance is estimated by:

n

 The variance of true offspring family means is:

t

Var    Variance accounted for by linear regression: 2

r op

Var  2

b op

Var For midparent, there is factor of 0.5 on right.

Total Residual Variance

 Thus, the variance of true family means from  the regression is given by:

t

  2

b op

 Var So the residual variance is the sum: Var  

i

  

t



b

2

op

 1 

t n i

  Var

Weighted Least Squares

 So the weights for families that minimize the sampling variance of the regression coefficient are:

w i



n i



t

 2

b op n i

   And the weighted least-squares regression coefficient is:

b



i N

  1

w i



z oi



z o i N

  1

w i



z pi

  

z z p pi

 2 

z p



z p



i N

  1

i N

  1

w i z pi w i

,

z o



i N

  1

w i z oi i N

  1

w i

Unequal Family Size – Conclusions

 The weight approaches a limiting value as

n i

becomes large. After a while, there is little gain by sampling more offspring.

 When

t

is large, this limit is achieved more rapidly. When there is little within family variation, you need measure only a few offspring to get a good estimate.

 Since the weights are a function of the regression coefficient, it would seem impossible to perform the analysis. It’s OK, just iterate.

Relaxing Assumptions

 If the mean or variance of traits differ between the sexes, then the data can be normalized (subtract mean and divide by standard deviation), such that both now have mean 0 and variance 1.

 Begin analysis.

Precision of Estimates

 Assume the parent and offspring phenotypes are bivariate normal.

 Then the sampling distribution of the regression coefficient is normal with Var  

op

  1  2

r op



N

Var Var  

p

 

o

number of parent/offspring pairs

Precision of Estimates

Var  

p

 1 Var  

p

When midparent value is used, 2 make the following substitutions: When multiple offspring used, make the following substitution: Var  

o r op

 

t r o p

 

n b o p

1 

t

2 Var  

o

When unequal family sizes used, the slope estimate becomes: Var 

i N

  1

w i

Var  

o



z



pi z p

 2

Sib Analysis – Intro

 Sometimes parent data is not available  Can’t identify the parent  Parent no longer alive  Three types of analysis  full sib   half sib full and half sib  Assume random sample of families and random mating of parents.

Half-Sib

 If epistatic effects are small and common environmental effects do not contribute to resemblance of sibs, then 4 times the covariance of half sibs is an estimate of additive genetic variance and approximately twice the parent-offspring covariance.

Relationship Parent-offspring Half sibs A ½ ¼ D AA ¼ 1/16 AD DD

Half-Sib – Common Environment

 A shared environment is the biggest potential problem with half-sib analysis.

 One solution is to use paternal half-sibs that are raised by multiple mothers.

 A common design is one sib from each mother.

 The analysis is ANOVA-based. Regression causes problems because of non-independence between pairs of sibs.

Half-Sib – ANOVA

  The linear model used is:

z

  

s



e ij i ij

The residual error (

e ij

) is caused by: environmental variance     genetic variance among mothers segregation in father dominance  Assume

e

variance (

ij

are uncorrelated and have common

within family variance

) 

e

2

Half-Sib – Sire Effects

    The

s i

are called

sire effects

and are assumed to have mean 0 (true when fathers are random sample).

Let the variance among sire effects be  2

s

Critical ANOVA assumption is that the random  

s i

,

e ij

  0 Also note that the phenotypic covariance between members of a group equals the variance among groups.

Half-Sib Covariation

  Paternal half sibs       

z ij

  ,

z ik s i

 

e ij

,  

s i



e ik

  

s

2 When epistatic effects are minimal:  2

A

 4 

s

2

General Approach

    Specify the linear model.

State assumptions of model and write an expression for the total phenotypic variance in terms of the components.

Write the components of variance associated with the model in terms of the covariance between classes of relatives.

Using the resemblance between relatives results, write these covariances between relatives as sums of causal components of variance (e.g. additive, dominance, epistatic, etc).

Paternal Half-Sibs ANOVA Table

E(MS) Factor Among families Within families Total df SS MS

N

-1

T

-

N

SS

s



i N

  1

n i



z i



z

 2 SS

e



i n i N

   1

j

1 

z ij



z i

 2 SS

s N

 1

T

SS

e



N T

-1 SS

e



i n i N

   1

j

1 

z ij



z

 2

T



i N

  1

n i

,

n

0 

T



T N

1 

i N

  1 1

n i

2 SS

T T

 1 

e

2 

n

0 

s

2 

e

2 

s

2

Method of Moments Estimators of Variance Components



s

2 

e

2   E  MS

s

 E  MS

e



n

0 MS

e

 Plug in sample estimates: Var  MS

s



n

0 MS

e

Var  MS

e

Heritability Estimator

intraclass correlation

t

PHS  Var Var    

h

2  4

t

PHS bias PHS  2   1   

n nN

 1    1  unimportant for

N

>20

Hypothesis Testing

  Hypothesis tests assume normal distribution and constant variance.

 Estimates of variance components are unbiased regardless of the distribution of the linear effects (no normality assumption require).

Normality of observed data does not guarantee linear effects are normally distributed.

Testing Heritability>0

 Assuming normality and constant variance.

E E  MS  MS

s e

    

e e

2 2 

n

0 

s

2

F

 MS MS

e s

~

F



N

 1 ,

T



N

  A significant result implies: 

s

2  0

Meaning of Significant Result

 A lack of significance could merely mean that there was insufficient power.

 Standard errors can provide more information by indicating the accuracy of the estimates.

 To calculate standard errors we need the sampling variances of the observed mean squares.

 2  MS

x

  MS df 

x x

2  2

Sampling Variance of Variance Components

Var  Var Var  Var Cov  Var   , Var      Var  MS

e

 

T

2  MS

e



N

  2 2 Var 2

n

2   MS

s

    MS

N



s

1  2 

n

MS

e

  

T

 MS

e



N

  2 2        Cov     MS

s

 MS

e

Var  MS

e



n n n

 2

T

 MS

e



N

  2 2    , MS

e

  Assume equal family size.

Standard errors obtained by substituting MSE and taking square roots.

These are LARGE SAMPLE estimators.

Sampling Variance of Heritability

Var     2

Nn

  1 

n

 

n

1   1 

t

 2

s

ˆ

h

2  4 Var -

t

is the intraclass correlation.

- Obtained by delta method.

Again, a large sample estimator.

Confidence Intervals

  Assume normally distributed effects and homogenous variance.

s e

~ ~

N N

  0 , , 

s e

Often, researchers simply assume parameter estimates are approximately normally distributed. Then standard errors are used to calculate confidence intervals +/ 1.96*standard error.

Confidence Interval – Within Family Variance

E SS  MS

s s

  

e

2 SS 

s n

0 

s

2   2

N

 1 E SS

e

 MS

e

  SS

e



e

2  

T

2 

N

  2

T

SS

e



N

   / 2   Var  SS

e

  2

T



N

1   / 2 

Confidence Interval – Among Family Variance

MS 

n e

Var    

F

  

N

 1

F

 / 2   1 

F



N

 1  MS

e n

   

FF

 , 

N

 1

T



N F

 / 2   1 

F



T



N

1  / 2     

F



N

 1

F



N

 1

T



N

 / 2   / 2   1        

N

 1  / 2      1 

F F

 , 

N



T



N

 1  / 2 

N

 1    / 2          MS

s

/MS

e

Assume equal size sibships.

Confidence Interval Heritability

4    

F F F



F



N

 1

N

 1

T



N T



N

 / 2   / 2   

n

1  1     

h

2  4    

F F F



T F



T



N



N N

 1

N

 1  / 2   / 2   

n

1  1     Assume equal size sibships.

Example

    The wild parsnip cannot self-fertilize. Assume that all its seeds produced by one plant are therefore half-sibs. Two flowers are unlikely to be pollinated with pollen from the same plant when there is lots of pollen from multiple sources floating around.

Select 20 random plants.

Collect 10 seeds from each and plant.

Analyze the mature plants.

Example – Data

     Some plants died prior to assay.

5 plants had 10 offspring assayed.

8 plants had 9 offspring assayed.

4 plants had 8 offspring assayed 1 plant each had 7, 6, and 4 offspring assayed.

 UNBALANCED DESIGN

Example - Calculations

        

t

MS

e

MS

s

Var( Var(

e

) = 0.0370

Var(

z

) = 0.0462

h

2

s

= 0.1156

= Var( = 4

t

SE(

h

2 = 0.0370

) = 0.0092

s

)/Var( = 0.80 (no epistatic effects) ) = 4SE(

t z

) = 0.20

) = 0.32

F

= 3.12 with 19 and 151 degrees of freedom => p value < 0.01.

Example - Analysis

 Maternal half-sibs used.  It is actually quite conceivable that many of the seeds were full sibs.

 Both of these possibilities affect heritability estimates in the same way. Increase or decrease?

Resampling Procedures

 Since the hypothesis tests assume normality and equal variance and since many of the results were contingent on a large sample, numerical procedures that do not make these assumptions have been employed.



jacknife:

delete one family from dataset at a time and recalculate estimates.

 

bootstrap:

draws samples of same size with replacement from dataset.

permutation:

randomize the half-sibs among families which generates null distribution under assumption 

s

2  0

Full-Sib

 Twice the inraclass correlation gives an estimate of heritability if dominance, environmental, and epistatic effects are minimal. These are BIG assumptions for full sibs, so full sibs are not often used to estimate heritibility.

 Combined with half-sib analysis, full sib analysis can provide information on the relative role of dominance and environmental effects.

Half-Sib/Full-Sib Design

 Mate each of

N

randomly selected fathers with several different females and sample multiple offspring from each female.

 The offspring of one female are full sibs.

 The offspring of different females with same father are half-sibs.

Linear Model

z ijk

  

s i



d ij



e ijk j i

is the

i

th sire is the

j

th “dam” mated to the

i

th sire

k

is the

k

th offspring of the

i

th sire and

j

th dam 

z

2  

s

2  

d

2  

e

2

Derivation

 We now convert observable components of variance to covariance between relatives.

 The residual variance is the variance within full-sib families:  

e s

2 2     2  

z

 PHS  Covariance between full sibs = variance between families

Total Variance



d

2      PHS  

s

2 

d

2 

e

2      4  4 2

A

2 2

A

2

A

    2

AA

 16 2

D

3 4  4 2

D

  3  16 3  2

AA

4 2

AA

   8 2

AD

7  8 2 

AD

 2

DD

  16  15  16 2

DD

2

E c

  2

E s

common environment special environment

Estimation

 Thus, four times the sire component provides and estimates of the additive genetic variance.

 The difference between the dam component and the sire component provides an estimate of the impact of the dam and common environment. The latter two cannot be separated.

HS/FS ANOVA

Factor df SS MS E(MS) Sires

N

-1 Dams (sires) Sibs (dams) Total

N

(

M

-1)

T-NM T

-1

i

1

M N

  

j i

1

n ij



z i i

1

M N

  

j i

1

n ij



z ij



z

 2 

z i

 2

i N

   

j M i k

1 1

n ij

1 

z ijk



z ij

 2 SS

s

/df

s

SS

d

/df

d

SS

e

/df

e



e

2 

k

2  2

d



k

3 

s

2 

e

2 

k

1 

d

2 

e

2

i N

   

j M i k

1 1

n ij

1 

z ijk



z

 2

Sample Size Specifics

k

1

k

2

k

3   

N



M

1  1       

T



i j M N

   1

i

 1

n i n ij

2      

N

1  1      

i N



M j

  1

i

 1

n i n ij

2

N

1  1      

T



i N

  1

n i

2

T

      

i

 1

j M N



i

 1

n ij

2

T

     

T

is total number of individuals in experiment.

M

is the mean number of dams per sire.

n i

is the total number of offspring for the

i

th sire.

Hypothesis Testing

 Assume normality and balanced design.

 A test for significant sire effects:

F

 MS MS

d s

~

F



N

 1 ,

N



M

 A test for significant dam effects:  1  

F

 MS

d

MS

e

~

F



N



M

 1  ,

T



N M



Unbalanced Hypothesis Testing

F



k

1 MS

s

 

k

2 

k

1  MS

e k

2 MS

d r

numerator 

i m

  1 2 

c i Q

MS

i

 2 df

i

where

Q

=

c

1 MS 1 +

c

2 MS 2 + ...+

c m

MS

m

Interpretation

 Since variance of sires is an estimate of genetic additive effect, a significant sire effect indicates significant additive effect.

Twins

 Monozygotic twins are genetically identical (MZ).

 Dizygotic twins are full sibs (DZ).

 A greater amount of phenotypic variability is assumed between DZ vs. MZ twins when genetics underly the trait.

Twins – Linear Model

 We define a simple linear model for both types of twins (MZ and DZ):

z ij

  

b i



e ij



e ij

is a result of segregation (in DZ) and environmental variance (in both DZ and MZ).

 Assume DZ and MZ twin pairs are a random sample such that E(

b i

) = 0.

Among-Twin Variance

 Define the following:    

b b e e

2 2 2 2         to be variance to be variance to be variance to be variance among among within within MZ pairs DZ pairs DZ pairs MZ pairs  By the linear model, 

b

2   

z i

1 ,

z i

2   

b

2

Causal Components

   In terms of causal components, the variance among MZ pairs is: 

b

2  

G

2   2

E c

Among DZ pairs it is: 

b

2  1 2  2

A

 1 4 

D

2  1 4  2

AA

   DZ and MZ may not have same common environment.

 2

E c

Within Pair Component

 Since 

z

2  

b

2  

e

2  The remaining component is within-pair component: 

e

2 

e

2   2

E s

 1 2  2

A

 3 4 

D

2  3 4  2

AA

    2

E s

Among-Pair Variance Estimates

Var  MS

b

 MS

e

2 Var  Var   Var  MS

b

  Var  MS

e

4   MS

b

  MS

e

 2 2

N

Within-Pair Variance

Var  MS

e

Var  Var   2  MS

e

 2

N

Total Variance

Var  MS

b

 MS

e

2  Same sample variance as for Var(

b

).

Statistical Test



F

= MS

b

/MS

e

is a test statistic for evaluating significant of among-pair component of variance.

Estimating Heritability

H

2 

t

MZ  MS

b

MS

b

      MS

e

MS

e

    SE  1 

t

MZ

N

   

F F F



N

 1

N

 / 2 

F



N

 1    / 2    1 1     

H

2     

F F F N

, 

N

 1

F N

, 

N

 1  / 2   / 2    1 1    

Example

      

N

=260 pairs of MZ twins.

MS MS

H F

2

b e

= 3619.2

= 82.9

= 0.955

SE(

H

2 ) = 0.005.

= 43.66

Highly significant estimate of heritability.

MZ Analysis – Difficulties

 Heritability is inflated by common environmental effects.

 Check by comparing twins raised together vs. those raised apart.