* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Inbreeding 1
Survey
Document related concepts
Polymorphism (biology) wikipedia , lookup
Pharmacogenomics wikipedia , lookup
Transgenerational epigenetic inheritance wikipedia , lookup
Genetic engineering wikipedia , lookup
Designer baby wikipedia , lookup
Genome (book) wikipedia , lookup
Human genetic variation wikipedia , lookup
Gene expression programming wikipedia , lookup
Koinophilia wikipedia , lookup
Dominance (genetics) wikipedia , lookup
Microevolution wikipedia , lookup
Population genetics wikipedia , lookup
Genetic drift wikipedia , lookup
Transcript
Inbreeding depression in corn Inbreeding Alan R. Rogers November 16, 2015 (Jones 1924) Inbreeding depression in humans I Two plants on left are from inbred homozygous strains. I Next: the F1 offspring of these strains I Then offspring (F2 ) of two F1 s. I Then F3 I 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 I Heterozygotes tend to have high fitness I 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 .. . 1 .. . 3 N 8 .. . 1 4 Decay of heterozygosity under selfing A1 A2 N .. . .. . . . . 1 2. .. 1 N 2 . . .. .. .. . A2 A2 1 4 .. Half of heterozygosity is lost each generation. . 1 N 4 1 2. 1 4 .. 1 N 4 .. . .. . 1 .. . 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 ...................... .......... ..... ..... .... .... ... ... .. .... .. . . . .. . . . .. ...... .... .. . . . ... .... .. . ...... .... ...... . . ........ . . . ... ...... .................................. .... .... .... ... .... . . . .... .... .... ......................... ............................... ...... . . ....... . . . . . .... .... .. .......... .... . . ..... .... ....... .. . ... .. ... .... ...................... .. ................ .. .... . .. ... ... . . . ... .. . .. . .. .. . .. . ... ... . .. . . .... . . . .... . . . . ...... . . . . . . . . ........ . ............ ............... ........................... ...... ◦• ◦ •·· ◦ •·· a b ·· ·· ··· · ·· ·· ·· ··· · ·· Identical in state, not by descent ...................... .......... ..... ..... .... .... ... ... .. .... .. . . . . .. . . .... .... .. ..... . .... .. ... ... ...... ....... .. ....... .... ....... .... ................................. .... .... .... .... .... . . . .... .. . .... .................... . . ........ . ... ............................... ..... . . . . . . .... .. . .. .......... .... ..... .... ....... .. ............. ... ... .... ................ .............. ... .. .. .... . .. ... . . .. ... .. . .. . .. .. . .. . ... ... . .. . . .... . . . .... . . . . ...... . . . . . . . . ........ . ............ ............... ........................... ...... •• ◦ •·· ◦ •·· a b ·· ·· ··· · ·· ·· ·· ··· · ·· 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. ....................... ......... ..... ..... .... ... ... ... .. ... .. ... . . . . .... .. .. ...... . .... .... ... ... . . . . ..... ...... . . . . . . .. ..... .... ....... .... ............................... .... .... .... ... .... .... .... . . . .... ......................... ......................... ..... . . . . ....... . .... . . ..... ... .... .... . ........ . . . . . ... . .. .... ........... .................... .... .. ............. ... .. .... ... ... ... ... ... ... ... ... .. .. .. . . ... ... .. . . . . . .... .... .. .. . . . . ...... . . . ....... ............ ................ ................................ ...... ◦• ·· ·· ··· · ·· ·· ·· ··· · ·· a F , is the probability that they are IBD I What is the probability that they both carry A1 ? Event IBD from A1 -bearing ancestor Descend from two random ancestors who both carry A1 ◦ •·· • ◦·· I 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 ...................... .. .................... ....................... ... .................... ................................... ... ... . . . . . . . . . . . . . . . . . . . .................... ....... ............ .................... .... .......... ............................................ . . . .... ............. Susannah Wedgewood Josiah Wedgewood II ... ... .. ......... .... ... ... .. ......... .... Charles Darwin Emma Wedgewood B A ...................... .. .................... ....................... ... .................... .................................... ... ... . . . . . . . . . . . . . . . . . . . . . .................... ........ ........... ........................ ........... ............................................. . . .... ............ C D ...................... ... .................... ... ... .................... ... ... . . . . . . . . . . . . . . .................... ........ . . ......................... ......... .... ........... E F ........... ........ ........... .......... ........... .......... . . . ........... . . . . . . ........... .. .. .............. ........................... ........... ......... ........... .......... ........... .......... . . . ........... . . . . . . ........... .. .. .............. .......................... George Darwin G Mating between full siblings B A ...................... . .................... ........................ ... ................... ................................... ... ... . . . . . . . . . . . . . . . . . . . . . . . .................... ........ ........... .................................... ............................................ .... ............ C Inbreeding and Genetic Drift Alan R. Rogers D ........... ........ ........... .......... ........... .......... . . . ........... . . . . . . ........... .. .. .............. .......................... 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. Â = 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.