Statistical BIOSTAT/STAT551, Genetics II: Quantitative Traits Properties of a Single Locus
Winter 2004 Handout 1
Reading: Chapter 1 and 4. Note that you should already be familiar with computing allele and genotype frequencies and testing for Hardy-Weinberg Disequilibrium.
DEFINITION From WWWebster Dictionary: Main Entry: quantitative inheritance Function: noun Date: circa 1929 : genic inheritance of a character controlled by polygenes From Lynch and Walsh (1997): . . . quantitative genetics, a statistical branch of genetics based upon fundamental Mendelian principles extended to polygenic (multilocus) characters . . . WHY STUDY QUANTITATIVE GENETICS? nature of quantitative trait variation
consequences of inbreeding and outcrossing
constraints on the evolutionary process
estimation of breeding values
development of predictive models for evolutionary change
examining quantitative traits in the study of multi-factorial disease
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BIOSTAT/STAT551, Statistical Genetics II: Quantitative Traits Properties of a Single Locus THE MASTER PLAN THE SET-UP single locus
Winter 2004 Handout 1
multiple loci
environment
resemblance between relatives
polygenes
FINDING THE GENES detecting major genes
usage of genetic markers
mapping and characterizing QTLs
ADVANCED TOPICS correlation among traits
multivariate linkage analysis
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BIOSTAT/STAT551, Statistical Genetics II: Quantitative Traits Properties of a Single Locus Characterizing the influence of a locus on phenotype z=phenotype G=genotypic value E=environmental deviation
Winter 2004 Handout 1
z=G+E
What is G? G is the expected phenotype that results from the joint expression of the genes that influence the phenotype. begin by assuming that we have a single autosomal locus that influences the phenotype
B1B1 B1B2 B2B2
0
(1+k)a
2a
(Figure 4.4 of text) a will be referred to as the homozygous effect and is a measure of additivity of alleles k measures departure from additivity and will be referred to as the dominance coefficient Genotype BiBj B1B1 B1B2 B2B2 Different kinds of behavior k=0 k=1 k=-1 k>1 k<-1 allelic effects are completely additive B2 is completely dominant to B1 B1 is completely dominant to B2 locus exhibits overdominance locus exhibits underdominance Genotypic Value Gij 0 (1+k)a 2a
Problem 1.1: The pygmy gene, denoted pg, in the mouse greatly reduces body size. An experiment reported the following means for body weight in grams: G+ + = 14, G+ pg = 12 and Gpg pg = 6. We will take these to be the expected phenotypes. Compute the homozygous effect and dominance coefficient. What does this tell us? (example from Falconer and Mackay 1996)
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BIOSTAT/STAT551, Statistical Genetics II: Quantitative Traits Properties of a Single Locus Fishers Decomposition of the genotypic value
Winter 2004 Handout 1
Gene content: number of copies of a particular allele, B1, in a genotype denoted by N1
Consider the relationship between phenotype and Ni. If k=0, then the alleles behave in a completely additive fashion. Thus the relationship between phenotype and N1 is linear. It may be of interest to quantify the deviation from this linear relationship that results from the presence of dominance.
Once we have expressed the relationship between the genotypic value and N2 in this form, we can describe the slope, , in terms of our population parameters (homozygous effect, dominance coefficient and allele frequencies). From regression theory, we know that
=
(G, N 2 ) 2 (N 2 )
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BIOSTAT/STAT551, Statistical Genetics II: Quantitative Traits Properties of a Single Locus
Winter 2004 Handout 1
With the following, we can then express both (G,N2) and 2(N2) in terms of a, k, p1 and p2 where pI is the population frequency for allele Bi.
Table 4.1 of text: Genotypic Gene Content Value (G) (N) Regression Value (G ) Dominance Deviation (G G )
Genotype B1B1 B1B2 B2B2
Freq.
GN
N2
We end up with, a[1+k(p1-p2)] = Now that we have an expression for , what is it? It is termed the average effect of allelic substitution. It is the expected change in phenotype given a substitution of allele B2 for an allele B1. If the alleles behave in a completely additive fashion, then =? If not, the relationship depends on both the level of dominance and allele frequencies.
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BIOSTAT/STAT551, Statistical Genetics II: Quantitative Traits Properties of a Single Locus
Winter 2004 Handout 1
What about the genetic variance?
We have the following relationship: G = G + . We can obtain the variance from our previous work:
2 The additive genetic variance, A , is the genetic variance associated with the average additive effects of alleles and can be shown to be: 2 A =
2 The dominance genetic variance, D , is the genetic variance associated with dominance effects and is equal to 2 D = (2p1p2ak)2
Note that BOTH of the variances are affected by the additive and dominance terms. Other terminology: Average Excess difference between the mean genotypic value of an individual carrying (at least one) copy of B2 and the mean genotypic value of a random individual
1* = -p2 and 2* = p1
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BIOSTAT/STAT551, Statistical Genetics II: Quantitative Traits Properties of a Single Locus
Winter 2004 Handout 1
Additive Effects correspond to the regression coefficients 1 and 2; HOWEVER, for a randomly mating population, these are equal to the average excesses 1 = -p2 and 2 = p1
Breeding Values the sum of the additive effects for an individual, i.e. for an individual with genotype Gij, their breeding value is Aij=i + j Recall, A2 =
Also note that, D2 =
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BIOSTAT/STAT551, Statistical Genetics II: Quantitative Traits Properties of a Single Locus
Winter 2004 Handout 1
Extensions for Multiple Alleles Consider a locus with n alleles. From the model for a diallelic locus, it is straightforward to construct a multi-allelic extension of the linear model for genotypic value: G = G + i N i +
i =1 n
Here,
the set of i are found through least-squares regression, i.e. the i that minimize E ( ij2 ) Ni is the number of copies of the ith allele.
Under a random-mating population, the additive effects can be written as a function similar to before:
i = p j Gij G
j =1
n
And the additive genetic variance can be written as
2 A = 2 pi i2 i =1 n
E...