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http://en.wikipedia.org/wiki/Bach_family#Family_tree
Lecture 22:
Inbreeding
(or, the origin
of P. D. Q.
Bach?)
Lecture 22 reading
Primary reading - DF - Chapter 9, pp. 197-
Background - C & H - Chapter 2, pp. 37-43; Chapter 7, Fig. 7.4, pp. 238
Revised lecture syllabus 10 Oct Lecture 20 - Genetic Variation and random mating 12 Oct Lecture 21 - Random mating, assortative mating, linkage disequilibrium 15 Oct Lecture 22 - Inbreeding 17 Oct Lecture 23 - Mutation and migration 19 Oct Lecture 24 - Genetic Drift 22 Oct Lecture 25 - Drift-migration equilibrium and F-stats 24 Oct Lecture 26 - Quantitative Genetics I 26 Oct Lecture 27 - Quantitative Genetics II 29 Oct SECOND MID-TERM EXAM (lectures 14-25) 31 Oct Lecture 28 - Fitness 2 Nov Lecture 29 - Artificial selection 5 Nov Lecture 30 - Natural Selection 7 Nov Lecture 31 - Sexual Selection 9 Nov Lecture 32 - Geographic variation 12 Nov Lecture 33 - Molecular Evolution I 14 Nov Lecture 34 - Molecular Evolution II 16 Nov Lecture 35 - Species and Speciation 26 Nov Lecture 36 - Allopatric speciation 28 Nov Lecture 37 - Sympatric speciation 30 Nov Lecture 38 - The fossil record 3 Dec Lecture 39 - Macroevolution 5 Dec Lecture 40 - Macroevolution 7 Dec Lecture 36 - Human population structure
Positive assortative mating
Genotype AA Aa aa Frequency P H Q AA P P^2 PH zero Aa H PH H^2 zero aa Q zero zero Q^2
What are frequencies of each type of mating? No longer random mating!
Thought example: if 25 red males, 25 blue males, 25 red females, and 25 blue females (n = 100, freq. red organisms = 0.50, freq. blue = 0.50) start a new population, then the frequency of blue x blue matings (out of 50 matings total) is 25/50 = 0.50. In other words, blue x blue mating frequency is not Q 2 but Q! Genotype (^) AA Aa aa Frequency (^) P H Q AA P P 2 /(1-Q) PH/(1-Q) zero Aa H PH/(1-Q) H^2 /(1-Q) zero aa Q zero zero Q
This is wrong
This is correct
Positive assortative mating
Hedrick, P. 2005. Genetics of populations, Jones and Bartlet, p.
So, positive assortative mating reduces H and increases P and Q - as you would expect intuitively. No change in allele frequencies. Negative assortative mating does the opposite.
So far we have only talked about one gene (one locus) at a time. Expanding to two genes (two loci) brings in some complications.
http://www.countrysideinfo. co.uk/devon_bap/images/
pin (^) thrum
What is D for pins and thrums?
haplotypes GA ga frequen c y0.3 0. GA 0.3 0.09 0. ga 0.7 0.21 0.
G A
g a
haplotypes
D = (fGA)(fga)-(fGa)(fgA)
D = (x11)(x22)-(x12)(x21)
D = (0.3)(0.7)-(0.0)(0.0)
D = 0.
An important point - single genes come to HWE in one generation, but two genes come to an equilibrium very slowly if linkage is tight.
The four different forms of the butterfly Papilio memnon (the black and green individual is the only male) are all the result of ~ genes located near one another, but all are in linkage disequilibrium so there are fewer morphs than if D were 0.
Inbreeding: mating with relatives
Genetic effect on individuals: causes them to contain alleles that are identical - identical by descent from a common ancestor allele - also being autozygous. Genetic effect on populations: causes a loss of heterozygotes - heterozygosity (H). Phenotypic effect (the good): causes individuals to be uniform.
Phenotypic effect (the bad): causes inbreeding depression (loss of vigor, longevity, fertility).
Inbreeding acts on all genes simultaneously, unlike assortative mating.
B C
I
X A Y
Z W
D E
A B C
D E
I
a1a a1 a
a1 (^) a
a1 a a1a
How to calculate the effect of inbreeding on individual genotypes - use the inbreeding coefficient F.
pedigree gamete pathway
A B C
D E
I
a1a a1 a
a1 (^) a
a1 a a1a
Probability of I being autozygous for a 1 , f = (1/2)^6
But could be autozygous for a 2 also - so equation needs to be f = 2(1/2)^6 = (1/2)^5. Thus the algorithm for calculating f for individual I is to count ancestors up to the common ancestor, the down the other side: DBACE and calculate f (^) I = (1/2)count.
A B
C D
E I
F
ECADF so (1/2) 5 = 0. ECBDF so (1/2) 5 = 0.
OR a total of 0.0625.
Multiple paths for autozygosity - 1st cousin mating:
Now, what happens with inbreeding at the population level? Let’s do a thought-experiment on a self-fertilizing organism.
HWE allozygous (1-F) autozygous F A 1 A 1 = p 2 p^2 (1-F) pF A 1 A 2 = 2pq 2pq(1-F) - A 2 A 2 = q 2 q 2 (1-F) qF
P (^) inb = p 2 (1-F) + pF Hinb = 2pq (1-F) Q (^) inb = q 2 (1-F) + qF
How do we calculate f for a whole population? Two ways - either measure individual f for many individuals and take a mean. Or (much easier), measure heterozygote reduction.
(1.)
(2.)
Hinb= 2pq (1-F) Hinb= H (^) HWE (1-F) F = (H (^) HWE - H (^) inb)/HHWE F = (H (^) HWE - H (^) obs)/H (^) HWE
(3.)
Fpop = ( H^^ HWE^ ^ H^ OBS ) H (^) HWE
Hinb= 2pq (1-F) Hinb= H (^) HWE (1-F) F = (H (^) HWE - H (^) inb)/HHWE F = (H (^) HWE - H (^) obs)/H (^) HWE
(3.)
Phenotypic effect (the good): causes individuals to be uniform.
Family 25 Family 30
Sewall Wright, 1919 (^) http://www.equine- world.co.uk/about_horses/horse_images/thoroughbred_horse_1.jpg
“All of us thoroughbreds are descended from Byerley Turk (1680-1696), Darley Arabian (1700-1733) or Godolphin Arabian (1724-1753)”
Phenotypic effect (the bad): causes inbreeding depression (loss of vigor, longevity, fertility).
Phenotypic effect (the bad): causes inbreeding depression (loss of vigor, longevity, fertility).
