Quantitative characters · 2015-10-21 · ab aB Ab AB ab aB Ab AB 4 3 3 3 3 2 2 2 2 2 1 1 1 1 0 2 A...

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Transcript of Quantitative characters · 2015-10-21 · ab aB Ab AB ab aB Ab AB 4 3 3 3 3 2 2 2 2 2 1 1 1 1 0 2 A...

Quantitative characters

Joe Felsenstein

GENOME 453, Autumn 2015

Quantitative characters – p.1/38

ab

aB

Ab

AB

ab

aB

Ab

AB

4

3

3

3

3

2

2

2

2

2

1

1

1

1

0

2

A random−mating population with two genes having2 alleles each, at equal frequencies, symmetrically

affecting a quantitative character

Quantitative characters – p.2/38

4

3

3

3

3

1

1

1

1

0

2

2

2

2

2

2

Quantitative characters – p.3/38

4

3

3

3

3

1

1

1

1

0

2

2

2

2

2

2

Quantitative characters – p.4/38

43210

aabb

Aabb

aAbb

aaBb

aabB

AaBb

aABb

AAbb

AabB

aAbB

aaBB

AABb

AAbB

AaBB

aABB AABB

The distribution of the genotypes and the quantitative

character before artificial selection

Quantitative characters – p.5/38

An imaginary quantitative character with 4 loci

The character is a sum of effects including interaction and environmentaleffects:

AA

Aa

aa

−1

2

3

CC

Cc

cc

−2

0

3

+

+

BB

Bb

bb

DD Dd dd

0.85.24 −4.0

3.28 0.1 −3.08

0.56 −1.0 −2.2

+

startingvalue

two loci that add up

plus two that interact plus an environmental effect

12 +

0−4 4

What kind of distribution will this lead to when gene frequencies at the fourloci are pA = 0.4, pB = 0.5, pC = 0.3, and pD = 0.6 ?

Quantitative characters – p.6/38

The resulting distribution

0 5 10 15 20 25 30 35

Phenotype

freq

uenc

y

Quantitative characters – p.7/38

A similar distribution in a 5-locus character

0 1 2 3 4 5 6 7 8

Gene frequencies

A

BCDE

0.4

0.30.40.20.6

Phenotype:

Number of copiesof these alleles.No environmentaleffects

100 random

individuals:

Quantitative characters – p.8/38

Distributions of quantitative characters

Quantitative characters – p.9/38

Quantitative characters – p.10/38

Quantitative characters – p.11/38

Recall this distribution

Mean phenotype of

43210

aabb

Aabb

aAbb

aaBb

aabB

AaBb

aABb

AAbb

AabB

aAbB

aaBB

AABb

AAbB

AaBB

aABB AABB

The distribution of the genotypes and the quantitative

character before artificial selection

10

Before selection

frequency of A = 0.5

frequency of B = 0.5

population = 2.0

Quantitative characters – p.12/38

Truncation selection

43210

AaBb

aABb

AAbb

AabB

aAbB

aaBB

AABb

AAbB

AaBB

aABB AABB

The distributions after artificial selection which

saves only those individuals at or above 2

After selection:

0.6364

Before selection

frequency of A = 0.5

frequency of B = 0.5

Mean phenotype of

survivors = 2.545

frequency of A = 14/22 =

frequency of B = 14/22 = 0.6364

Quantitative characters – p.13/38

Offspring of truncation selection

43210

In fact, the offspring will have this distribution:

9

60

154

180

81

Mean phenotype

of offspring = 2.545

(484 copies in all)

Quantitative characters – p.14/38

Effect of truncation selection on one locus

phenotype

With a larger number of loci, focusing just on one locus

Quantitative characters – p.15/38

Effect of truncation selection on one locus

aa

Aa

AA

phenotype

With a larger number of loci, focusing just on one locus

Quantitative characters – p.16/38

Effect of truncation selection on one locus

Aa AA

phenotype

aa

The distribution of offspring at this locus

Quantitative characters – p.17/38

Heritability

V = V + V + V

Variance of a character = VT

T A D E

Total variance

Additive genetic variance

Dominance variance

Environmental variance

heritability = h = V

V

A

T

2

heritability in effect measures the fraction of

variations that are passed on to the next generation,

undisrupted by Mendelian segregation

assuming genes are additive and environments are independent)(

Quantitative characters – p.18/38

Heritabilities in a real-life example

Quantitative characters – p.19/38

Response to artificial selection

truncation point

breed from thesediscard these

mean of newborns mean of thosewho we breed from

S

mean of newborns

in next generation

Sh2

R = h S2

h = heritability2

Quantitative characters – p.20/38

Response to artificial selection

1000 lbs 1100 lbs

weight

If heritability = 0.4

This is the expected gain in one generation

Mean of selected individuals#

of in

divi

dual

s

S = selection differential

= mean of selected individuals − population mean = 100 lbs

R = gain

= 40 lbs2 = h S

Population mean

Quantitative characters – p.21/38

Some features of artificial selection experiments

replicate lines

relaxation of selection

reverse selection

a selection limit?

generations

mea

n ph

enot

ype

control line

selection up

selection down

Quantitative characters – p.22/38

Body weight in mice

Quantitative characters – p.23/38

Tooth decay in rats

Quantitative characters – p.24/38

White blood cell counts

Quantitative characters – p.25/38

Chicken legs

Quantitative characters – p.26/38

Thorax bristles in Drosophila

Quantitative characters – p.27/38

Abdominal bristles in Drisophila

Quantitative characters – p.28/38

Dobzhansky’s glass maze

Quantitative characters – p.29/38

Artificial selection on geotaxis

Quantitative characters – p.30/38

Artifical selection on phototaxis

Quantitative characters – p.31/38

Ken Weber’s “inebriometer”

diagram photograph

Quantitative characters – p.32/38

The Illinois corn selection experiment (oil content)

Quantitative characters – p.33/38

Quantitative characters – p.34/38

The Illinois corn selection experiment (protein, oil)

Stephen P. Moose, John W. Dudley and Torbert R. Rocheford . 2004.Maize selection passes the century mark: a unique resource for 21stcentury genomics. Trends in Plant Science 9 (7): 358-364.

Quantitative characters – p.35/38

Hermon Bumpus’s 1896 sparrows – lengths

length

Fre

quen

cy

150 155 160 165 1700

5

10

15

20

Histogram of female total lengths in millimeters

Quantitative characters – p.36/38

Sparrow survival data

length

Fre

quen

cy

150 155 160 165 1700

5

10

15

20

Histogram of total lengths of females, and those that survived

Quantitative characters – p.37/38

Least squares fit of estimated fitnesses

155 160 165

0.0

0.2

0.4

0.6

0.8

1.0

fitne

ss

total length of female (mm)

Fraction of surviving females, and estimate of (quadratic) fitness curve

Quantitative characters – p.38/38