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Transcript
Extra credit question
The genes (loci) that influence
quantitative traits are known as:
_________________________
Uses of heritability
• The degree to which offspring resemble
their parents is determined by the
narrow-sense heritability h2
• The efficacy of natural and artificial
selection is also determined by h2
h2 = 1
h2 = 0
VA/VP = 1
VA/VP = 0
Efficacy of artificial selection:
size of Labradors
Breeder’s Question
Q: A horse breeder wants to
win the Kentucky Derby. If
she breeds her mare to a
really fast stallion, how
likely is it that the colt will
be faster than all the other
three-year-olds when it runs
in the Derby?
A: It depends on the heritability of running
speed!
Breeder’s Equation
• R = h2 S
• S = Selection differential
difference between selected parents and the population
as a whole (within a generation)
• R = response to selection
difference between selected offspring and the unselected
population (across generations)
Breeder’s Equation
R = h2 S
A dog breeder chooses his largest dogs to
breed together. The average height of the
breed is 60 cm (at the shoulder), and the
dogs he chooses to breed average 70 cm
tall.
He knows from previous work that the
heritability of height is 0.5.
How big can he expect the offspring to be?
R = h2 S = 0.5 * 10cm = 5cm
Breeder’s Equation
R = h2 S = 0.5 * 10 cm = 5 cm
If the response to selection is 5 cm, he
can expect his puppies to grow to be
60 cm + 5 cm = 65 cm tall
Exactly the same equation
can be used to understand
natural selection!
Efficacy of natural selection:
Darwin’s finches
h2
= 0.8
If large bills
are favored in
drought years,
what effect will
an El Nino
year have on
the
population?
R = h2 S
Before El Nino mean bill depth = 10 cm. Birds
that survived the drought had bills that were
2 mm deeper (on average) than the
population mean.
Q: What happened to the average bill depth
in the next generation?
A: R = 0.8 * 2mm = 1.6 mm. Bill depth next
generation = 10 + 1.6 = 11.6 mm.
How do you measure the heritability?
h2 = 1
h2 = 0
Another way to measure h2
if R = h2 S, then h2 = R/S
A corn breeder chooses to breed from plants that
have large cobs. The average cob length in his
crop is 15 cm, but he breeds from plants that
average 18 cm cobs.
Next year, he measures cob length in the offspring of
the selected plants and discovers that the mean is
16 cm
What is the heritability of cob length in this
population?
h2 = R/S = 1 cm/3 cm = 0.33
Most quantitative traits have
substantial random environmental
variance
2
h
= VA/VP = VA/(VA+VD+VE)
2
h =0.5,
If
then 50% of
the phenotypic variance
is additive genetic.
What’s the other 50%?
Hydrangeas
Environmental effects on
Quantitative Traits:
Monozygotic Twins
Heritability measured in one
population does not tell you
anything about differences
between populations
Environmental variation (VE) is not the
same within and between populations
Jones’s Farm
h2
1.0
milk yield
3 qts/day
Smith’s Farm
1.0
2 qts/day
Genetic and Environmental Effects on
Quantitative Traits
Can these methods be applied
to humans?
Environmental variation cannot be
controlled
Twin Studies
• Identical twins share 100% of
alleles
• Fraternal twins share 50% of
alleles
• Assume both kinds of twins
share VE to the same extent
Correlations between identical
co-twins
Complete genetic
determination
No genetic
determination
176
174
r=1.0
172
r=0.0
185
180
Trait in Twin 2
170
Trait in Twin 2
190
168
166
164
162
160
158
175
170
165
160
155
150
145
156
140
154
150
155
160
165
Trait in Twin 1
170
175
150
155
160
165
Trait in Twin 1
170
175
Correlations between co-twins
for adult height
Monozygotic (identical)
Dizygotic (fraternal)
180
190
r=0.91
r=0.46
185
180
Twin 2 Height
Twin 2 Height
175
170
165
160
175
170
165
160
155
150
155
145
150
140
150
155
160
165
Twin 1 Height
170
175
150
155
160
165
Twin 1 Height
170
175
Heritability estimates from
correlations between co-twins
• Identical twins share 100% of alleles
• Fraternal twins share 50% of alleles
H2 ≈ 2* (ri - rf)
Height example: ri = 0.91, rf = 0.46
H2 = ?
H2 = 0.90
ri
Identical
Fingerprint ridges 0.96
Height
0.90
IQ score
0.83
Social maturity
0.97
rf
Fraternal
0.47
0.57
0.66
0.89
H2
0.98
0.66
0.34
0.16
Heritabilty estimates from human
twin studies are biased
1. H2≈2* (ri - rf) overestimates true H2 by
0.5* VD/VP
2. H2 includes all genetic effects, not just
additive genetic
3. GxE interaction increases variability
among fraternal twins (it lowers rf) but
does not affect ri, so H2 is
overestimated
Heritabilty estimates from human
twin studies are biased
4. Estimates assume identical and
fraternal twins share environments to
the same extent.
Do you think this is true?
a) Identical twins share embryonic
membrane.
b) Identical twins share more similar
social environment.
Heritabilty estimates from human
twin studies are biased
5. Studies often based on small sample
sizes, and therefore estimates are not
very precise (large standard errors)
6. Some studies include male-female
fraternal twins, whereas identical
twins are always the same sex.
Conclusions
Heritabiltiy estimates derived
from human twin studies
should be considered very
approximate, and probably
too high.
Molecular Biology and
Quantitative Genetics
QTL Mapping
• Use DNA-based markers
• Marker is anything that differs among
strains (RFLPs, microsatellites,
single-base pair differnces)
• Markers close to genes will tend to
be inherited along with genes
(recombination rare)
Small
m
s
Large
m
s
M
S
M
S
F1
m
s
M
S
Markers that are close to the QTL are inherited with it
(recombination is rare)
m
s
F1:
M
S
Markers that are close to the QTL are inherited with it
(recombination is rare)
F2:
m
s
m
s
Small
m
s
M
S
Intermediate
M
S
Large
M
S
If a marker is far from any gene
affecting fruit weight, there will be
recombination between the fruitweight and the marker
A marker far from a gene will not show
a statistical association with the
phenotype
Many Markers
Results
• 28 QTLs responsible for fruit-weight
variation in tomato
• One gene has been cloned and
transferred between plants
• When “small” allele is transferred to
normally large-fruited variety, the fruit
weight is reduced by 30%
• Proof that the gene is really a QTL