Download Inbreeding 1

Inbreeding depression in corn
Inbreeding
Alan R. Rogers
November 16, 2015
(Jones 1924)
Inbreeding depression in humans
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Two plants on left are
from inbred
homozygous strains.
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Next: the F1 offspring
of these strains
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Then offspring (F2 )
of two F1 s.
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Then F3
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And so on.
Genotype frequencies without random mating
Describes any bi-allelic locus.
Offspring of cousin
marriages are less likely to
survive.
(Bittles and Neel 1994)
Genotype
A1 A1
A1 A2
A2 A2
Frequency
p 2 + pqF
2pq(1 − F )
q 2 + pqF
F = 0 under random mating.
Reduces to Hardy-Weinberg.
F > 0 under inbreeding. Gives
excess of homozygotes.
F is the coefficient of inbreeding.
Example
Outline of theory
Assume p = 1/2
Genotype
A1 A1
A1 A2
A2 A2
Frequency
F = 0 F = 0.1
0.25
0.275
0.50
0.450
0.25
0.275
Inbred population has more
homozygotes.
Suffers if either
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Heterozygotes tend to have
high fitness
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Deleterious alleles tend to
be recessive.
Inbreeding increases F , which increases homozygosity, which
decreases fitness.
Decay of heterozygosity under selfing
Gen. A1 A1
0
.
1
2
..
1
4
1
N
4
..
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1
..
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3
N
8
..
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1
4
Decay of heterozygosity under selfing
A1 A2
N
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1
2.
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1
N
2
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A2 A2
1
4
..
Half of heterozygosity is lost each generation.
.
1
N
4
1
2.
1
4
..
1
N
4
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..
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1
..
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3
N
8
What about cousin mating, or mating between sibs?
It is extremely difficult to work this out, using the method we just
used.
Between 1903 and 1915, no one could get it right.
Pearl (1913): Only for brother-sister matings does inbreeding
reduce heterozygosity. [Wrong!]
Kinds of gene identity
There are two senses in which a pair of gene copies may be
“identical:”
identity in state
: copies of same allele
identity by descent : copies of same gene copy in an ancestor
Abbreviation: IBD = Identity by Descent
Solution: build theory looking backwards in time, not forwards.
Gametes a and b are identical by descent
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Identical in state, not by descent
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May be IBD relative
to an earlier
generation.
Not IBD relative to
the pedigree shown
here.
IBD is always
relative to a
particular
generation.
Identical neither in state nor by descent
Uniting gametes
Consider the two gametes that unite to form an individual.
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a
F , is the probability that they are IBD
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What is the probability that they both carry A1 ?
Event
IBD from A1 -bearing ancestor
Descend from two random ancestors who both
carry A1
◦ •··
• ◦··
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Probability
Fp
(1 − F )p 2
P11 = Fp + (1 − F )p 2
= p 2 + pqF
b
All three genotypes
For blackboard
Same formulas as before.
Genotype
A1 A1
A1 A2
A2 A2
Frequency
p 2 + pqF
2pq(1 − F )
q 2 + pqF
F is no longer arbitrary.
Calculating F from pedigrees
F is probability that uniting
gametes are IBD.
Darwin-Wedgewood Genealogy
Josiah Wedgewood
A complex pedigree
Sarah Wedgewood
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Susannah Wedgewood
Josiah Wedgewood II
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Charles Darwin
Emma Wedgewood
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George Darwin
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Mating between full siblings
B
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Inbreeding and Genetic Drift
Alan R. Rogers
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November 16, 2015
E
Inbreeding and drift
Number of ancestors
Even under random mating, there is inbreeding in any finite
population.
This “random inbreeding” is the same thing as genetic drift.
Number of ancestors: II
generation
15
16
17
18
19
20
21
22
23
24
25
26
27
28
generation
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
year
1994
1965
1936
1907
1878
1849
1820
1791
1762
1733
1704
1675
1646
1617
1588
ancestors
1
2
4
8
16
32
64
128
256
512
1,024
2,048
4,096
8,192
16,384
Number of ancestors: III
year
1559
1530
1501
1472
1443
1414
1385
1356
1327
1298
1269
1240
1211
1182
ancestors
32,768
65,536
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
8,388,608
16,777,216
33,554,432
67,108,864
134,217,728
268,435,456
generation
29
30
31
32
year
1153
1124
1095
1066
ancestors
536,870,912
1,073,741,824
2,147,483,648
4,294,967,296
If you were born in 1994, then you had over 4 billion ancestors in
1066.
But there were not that many people on the planet.
Many of your ancestors in 1066 were the same people—we are all
inbred.
Let us build a model of this inbreeding.
Drift and inbreeding
Drift After t generations of genetic drift, the expected
heterozygosity is
Fitness and Inbreeding
E [H(t)|p0 ] = 2p0 q0 (1 − 1/2N)t
Inbreeding If Ft is the average inbreeding coefficient in generation
t, relative to generation 0,
Alan R. Rogers
E [H(t)|p0 ] = 2p0 q0 (1 − Ft )
November 16, 2015
Equating these expressions gives
Ft = 1 − (1 − 1/2N)t
Inbreeding is genetic drift.
Genotype frequencies and fitnesses
Genotype
A1 A1
A1 A2
A2 A2
Why is inbreeding harmful?
Frequency
Fp + (1 − F )p 2
2pq(1 − F )
Fq + (1 − F )q 2
Fitness
1
1 − hs
1−s
We have just seen that mean fitness is
w̄ = 1 − a − bF
Mean fitness:
w̄
= 1 − 2pq(1 − F )hs − [Fq + (1 − F )q 2 ]s
= 1 − a − bF
⇐
linear func of F
where
where
b = 2pqs(1/2 − h)
Fitness decreases with inbreeding if b > 0, which is true if h < 1/2,
or in other words, if the deleterious allele is at least partially
recessive.
a = 2pqsh + q 2 s
b = 2pqs(1/2 − h)
What inbreeding depression tells us
Morton and Crow: how large is the effect?
Estimates
Model
S
Inbreeding depression is
widespread in Nature.
 = 0.1612
= Pr[survival]
=
L
Y
i=1
Implies that deleterious
alleles tend to be partially
recessive.
≈
Y
B̂ = 1.734
1 − ai − b i F
Example
For mating between full sibs,
F = 1/4, and
e −ai −bi F
i=1
−A−BF
≈ e
S
= exp{−0.1612 − 1.734/4}
= 0.85
(Jones 1924)
where A =
P
ai and B =
P
bi .
So we expect 15% mortality in
the offspring of full-sib matings.