BIOE 109 Summer 2009 Lecture 7- Part II Selection on quantitative characters.

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Transcript of BIOE 109 Summer 2009 Lecture 7- Part II Selection on quantitative characters.

BIOE 109Summer 2009

Lecture 7- Part IISelection on quantitative characters

Selection on quantitative characters

What is a quantitative (continuous) character?

Selection on quantitative characters

What is a quantitative character?

• quantitative characters exhibit continuous variation among individuals.

Selection on quantitative characters

What is a quantitative character?

• quantitative characters exhibit continuous variation among individuals.

• unlike discrete characters, it is not possible to assign phenotypes to discrete groups.

Examples of discrete characters

Example of a continuous character

Two characteristics of quantitative traits:

1. Controlled by many genetic loci

Two characteristics of quantitative traits:

1. Controlled by many genetic loci

2. Exhibit variation due to both genetic and environmental effects

Two characteristics of quantitative traits:

1. Controlled by many genetic loci

2. Exhibit variation due to both genetic and environmental effects

• the genes that influence quantitative traits are now called quantitative trait loci or QTLs.

What are QTLs?

What are QTLs?

• QTLs possess multiple alleles, exhibit varying degrees of dominance, and experience selection and drift.

What are QTLs?

• QTLs possess multiple alleles, exhibit varying degrees of dominance, and experience selection and drift.

• some QTLs exhibit stronger effects than others – these are called major effect and minor effect genes, respectively.

What are QTLs?

• QTLs possess multiple alleles, exhibit varying degrees of dominance, and experience selection and drift.

• some QTLs exhibit stronger effects than others – these are called major effect and minor effect genes, respectively.

• the number and relative contributions of major effect and minor effect genes underlies the genetic architecture of the trait.

Mapping QTLs is expensive, labor intensive, and fraught with statistical problems!

Mapping QTLs is expensive, labor intensive, and fraught with statistical problems!

QTL mapping can reveal:1. Number of loci that influence a QT2. Magnitude of their effects on phenotype3. Their location on genome

Mapping QTLs is expensive, labor intensive, and fraught with statistical problems!

QTL mapping can reveal:1. Number of loci that influence a QT2. Magnitude of their effects on phenotype3. Their location on genome

QTL mapping CANNOT reveal:1. Identity of loci2. Proteins they encode

What is heritability?

What is heritability?

• heritability is the proportion of the total phenotypic variation controlled by genetic rather than environmental factors.

What is heritability?

• heritability is the proportion of the total phenotypic variation controlled by genetic rather than environmental factors.

The total phenotypic variance may be decomposed:

VP = total phenotypic variance

The total phenotypic variance may be decomposed:

VP = total phenotypic varianceVG = total genetic variance

The total phenotypic variance may be decomposed:

VP = total phenotypic varianceVG = total genetic varianceVE = environmental variance

The total phenotypic variance may be decomposed:

VP = total phenotypic varianceVG = total genetic varianceVE = environmental variance

VP = VG + VE

The total phenotypic variance may be decomposed:

VP = total phenotypic varianceVG = total genetic varianceVE = environmental variance

heritability = VG/VP (broad-sense)

The total genetic variance (VG) may be decomposed:

The total genetic variance (VG) may be decomposed:

VA = additive genetic variance

The total genetic variance (VG) may be decomposed:

VA = additive genetic varianceVD = dominance genetic variance

The total genetic variance (VG) may be decomposed:

VA = additive genetic varianceVD = dominance genetic varianceVI = epistatic (interactive) genetic variance

The total genetic variance (VG) may be decomposed:

VA = additive genetic varianceVD = dominance genetic varianceVI = epistatic (interactive) genetic variance

VG = VA + VD + VI

The total genetic variance (VG) may be decomposed:

VA = additive genetic varianceVD = dominance genetic varianceVI = epistatic (interactive) genetic variance

heritability = h2 = VA/VP (narrow sense)

Estimating heritability

Estimating heritability

• one common approach is to compare phenotypic scores of parents and their offspring:

Estimating heritability

• one common approach is to compare phenotypic scores of parents and their offspring:

Junco tarsus length (cm)

Cross Midparent value Offspring value

Estimating heritability

• one common approach is to compare phenotypic scores of parents and their offspring:

Junco tarsus length (cm)

Cross Midparent value Offspring value

F1 x M1 4.34 4.73

Estimating heritability

• one common approach is to compare phenotypic scores of parents and their offspring:

Junco tarsus length (cm)

Cross Midparent value Offspring value

F1 x M1 4.34 4.73

F2 x M2 5.56 5.31

Estimating heritability

• one common approach is to compare phenotypic scores of parents and their offspring:

Junco tarsus length (cm)

Cross Midparent value Offspring value

F1 x M1 4.34 4.73

F2 x M2 5.56 5.31

F3 x M3 3.88 4.02

Slope = h2

Regress offspring value on midparent value

Heritability estimates from other regression analyses

Comparison Slope

Heritability estimates from other regression analyses

Comparison Slope

Midparent-offspring h2

Heritability estimates from other regression analyses

Comparison Slope

Midparent-offspring h2

Parent-offspring 1/2h2

Heritability estimates from other regression analyses

Comparison Slope

Midparent-offspring h2

Parent-offspring 1/2h2

Half-sibs 1/4h2

Heritability estimates from other regression analyses

Comparison Slope

Midparent-offspring h2

Parent-offspring 1/2h2

Half-sibs 1/4h2

First cousins 1/8h2

Heritability estimates from other regression analyses

Comparison Slope

Midparent-offspring h2

Parent-offspring 1/2h2

Half-sibs 1/4h2

First cousins 1/8h2

• as the groups become less related, the precision of the h2 estimate is reduced.

Heritabilities vary between 0 and 1

Cross-fostering is a common approach

Heritability of beak size in song sparrows

Q: Why is knowing heritability important?

Q: Why is knowing heritability important?

A: Because it allows us to predict a trait’s response to selection

Q: Why is knowing heritability important?

A: Because it allows us to predict a trait’s response to selection

Let S = selection differential

Q: Why is knowing heritability important?

A: Because it allows us to predict a trait’s response to selection

Let S = selection differential

Let h2 = heritability

Q: Why is knowing heritability important?

A: Because it allows us to predict a trait’s response to selection

Let S = selection differential

Let h2 = heritability

Let R = response to selection

Q: Why is knowing heritability important?

A: Because it allows us to predict a trait’s response to selection

Let S = selection differential

Let h2 = heritability

Let R = response to selection

R = h2S

Predicting the response to selection

Example: the large ground finch, Geospiza magnirostris

Predicting the response to selection

Example: the large ground finch, Geospiza magnirostris

Mean beak depth of survivors = 10.11 mm

Predicting the response to selection

Example: the large ground finch, Geospiza magnirostris

Mean beak depth of survivors = 10.11 mm

Mean beak depth of initial pop = 8.82 mm

Predicting the response to selection

Example: the large ground finch, Geospiza magnirostris

Mean beak depth of survivors = 10.11 mm

Mean beak depth of initial pop = 8.82 mm

S = 10.11 – 8.82 = 1.29

Predicting the response to selection

Example: the large ground finch, Geospiza magnirostris

Mean beak depth of survivors = 10.11 mm

Mean beak depth of initial pop = 8.82 mm

S = 10.11 – 8.82 = 1.29

h2 = 0.72

Predicting the response to selection

Example: the large ground finch, Geospiza magnirostris

Mean beak depth of survivors = 10.11 mm

Mean beak depth of initial pop = 8.82 mm

S = 10.11 – 8.82 = 1.29

h2 = 0.72

R = h2S = (1.29)(0.72) = 0.93

Predicting the response to selection

Example: the large ground finch, Geospiza magnirostris

Mean beak depth of survivors = 10.11 mm

Mean beak depth of initial pop = 8.82 mm

S = 10.11 – 8.82 = 1.29

h2 = 0.72

R = h2S = (1.29)(0.72) = 0.93

Beak depth next generation = 10.11 + 0.93 = 11.04 mm

Modes of selection on quantitative traits

Directional selection on oil content in corn

Modes of selection on quantitative traits

Modes of selection on quantitative traits