Inbreeding

Alan R. Rogers

November 10, 2017

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Inbreeding depression in corn

(Jones 1924)

I Two plants on left arefrom inbredhomozygous strains.

I Next: the F1 offspringof these strains

I Then offspring (F2)of two F1s.

I Then F3I And so on.

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Inbreeding depression in humans

Offspring of cousinmarriages are less likely tosurvive.(Bittles and Neel 1994)

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Genotype frequencies without random mating

Genotype Frequency

A1A1 p2 + pqFA1A2 2pq(1− F )A2A2 q2 + pqF

Describes any bi-allelic locus.

F = 0 under random mating.Reduces to Hardy-Weinberg.

F > 0 under inbreeding. Givesexcess of homozygotes.

F is the coefficient of inbreeding.

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Example

Assume p = 1/2Frequency

Genotype F = 0 F = 0.1

A1A1 0.25 0.275A1A2 0.50 0.450A2A2 0.25 0.275

Inbred population has morehomozygotes.Suffers if either

I Heterozygotes tend to havehigh fitness

I Deleterious alleles tend tobe recessive.

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Outline of theory

Inbreeding increases F , which increases homozygosity, whichdecreases fitness.

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Decay of heterozygosity under selfing

Gen. A1A1 A1A2 A2A2

0 N

. .. ...

. . .14

12

14

. .. ...

. . .

1 14N 1

2N 1

4N

... . .. ...

. . ....

1 14

12

14

1... . .

. .... . .

...2 3

8N 1

4N 3

8N

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Decay of heterozygosity under selfing

Half of heterozygosity is lost each generation.

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What about cousin mating, or mating between sibs?

It is extremely difficult to work this out, using the method we justused.

Between 1903 and 1915, no one could get it right.

Pearl (1913): Only for brother-sister matings does inbreedingreduce heterozygosity. [Wrong!]

Solution: build theory looking backwards in time, not forwards.

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Kinds of gene identity

There are two senses in which a pair of gene copies may be“identical:”

identity in state : copies of same alleleidentity by descent : copies of same gene copy in an ancestor

Abbreviation: IBD = Identity by Descent

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Gametes a and b are identical by descent

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Identical in state, not by descent

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a b

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May be IBD relativeto an earliergeneration.

Not IBD relative tothe pedigree shownhere.

IBD is alwaysrelative to aparticulargeneration.

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Identical neither in state nor by descent

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a b

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Uniting gametes

Consider the two gametes that unite to form an individual.

I F , is the probability that they are IBD

I What is the probability that they both carry A1?

Event Probability

IBD from A1-bearing ancestor FpDescend from two random ancestors who bothcarry A1

(1− F )p2

P11 = Fp + (1− F )p2

= p2 + pqF

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All three genotypes

Genotype Frequency

A1A1 p2 + pqFA1A2 2pq(1− F )A2A2 q2 + pqF

Same formulas as before.

F is no longer arbitrary.

F is probability that unitinggametes are IBD.

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For blackboard

Calculating F from pedigrees

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Darwin-Wedgewood Genealogy

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Josiah Wedgewood Sarah Wedgewood

Susannah Wedgewood Josiah Wedgewood II

Charles Darwin Emma Wedgewood

George Darwin

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A complex pedigree

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A B

C D

E F

G

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Mating between full siblings

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A B

C D

E

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Inbreeding and Genetic Drift

Alan R. Rogers

November 10, 2017

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Inbreeding and drift

Even under random mating, there is inbreeding in any finitepopulation.

This “random inbreeding” is the same thing as genetic drift.

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Number of ancestors

generation year ancestors

0 1994 11 1965 22 1936 43 1907 84 1878 165 1849 326 1820 647 1791 1288 1762 2569 1733 512

10 1704 1,02411 1675 2,04812 1646 4,09613 1617 8,19214 1588 16,384

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Number of ancestors: II

generation year ancestors

15 1559 32,76816 1530 65,53617 1501 131,07218 1472 262,14419 1443 524,28820 1414 1,048,57621 1385 2,097,15222 1356 4,194,30423 1327 8,388,60824 1298 16,777,21625 1269 33,554,43226 1240 67,108,86427 1211 134,217,72828 1182 268,435,456

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Number of ancestors: III

generation year ancestors

29 1153 536,870,91230 1124 1,073,741,82431 1095 2,147,483,64832 1066 4,294,967,296

If you were born in 1994, then you had over 4 billion ancestors in1066.

But there were not that many people on the planet.

Many of your ancestors in 1066 were the same people—we are allinbred.

Let us build a model of this inbreeding.

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Drift and inbreeding

Drift After t generations of genetic drift, the expectedheterozygosity is

E [H(t)|p0] = 2p0q0(1− 1/2N)t

Inbreeding If Ft is the average inbreeding coefficient in generationt, relative to generation 0,

E [H(t)|p0] = 2p0q0(1− Ft)

Equating these expressions gives

Ft = 1− (1− 1/2N)t

Inbreeding is genetic drift.

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Fitness and Inbreeding

Alan R. Rogers

November 10, 2017

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Genotype frequencies and fitnesses

Genotype Frequency Fitness

A1A1 Fp + (1− F )p2 1A1A2 2pq(1− F ) 1− hsA2A2 Fq + (1− F )q2 1− s

Mean fitness:

w = 1− 2pq(1− F )hs − [Fq + (1− F )q2]s

= 1− a− bF ⇐ linear func of F

where

a = 2pqsh + q2s

b = 2pqs(1/2− h)

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Why is inbreeding harmful?

We have just seen that mean fitness is

w = 1− a− bF

whereb = 2pqs(1/2− h)

Fitness decreases with inbreeding if b > 0, which is true if s > 0and h < 1/2, or in other words, if deleterious alleles are at leastpartially recessive.

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What inbreeding depression tells us

(Jones 1924)

Inbreeding depression iswidespread in Nature.

Implies that deleteriousalleles tend to be partiallyrecessive.

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Morton and Crow: how large is the effect?

Model

S = Pr[survival]

=L∏

i=1

1− ai − biF

≈∏

i=1

e−ai−biF

≈ e−A−BF

where A =∑

ai and B =∑

bi .

Estimates

A = 0.1612

B = 1.734

ExampleFor mating between full sibs,F = 1/4, and

S = exp{−0.1612− 1.734/4}= 0.85

So we expect 15% mortality inthe offspring of full-sib matings.

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