Molecular Drive (Dover)

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1 Molecular Drive (Dover) The nuclear genomes of eukaryotes are subject to a continual turnover through unequal exchange, gene conversion, and DNA transposition. Both stochastic and directional processes of turnover occur within nuclear genomes. Double-Strand Break Model Gene Conversion Holliday Model Walsh (Genetics 105: , 1983) 1 Locus, 2 Allele Model (A and a) Such That: γ= the probability of an unequal gene conversion event β= the conditional probability that a converts to A given an unequal conversion occurs 1-γ=the probability of getting a 1:1 ratio of Mendelian Segregation γβ=probability of segregation yielding only A alleles γ(1 β)=probability of segregation yielding only a alleles Then the segregation ratio A:a in Aa heterozygotes is: 1 / 2 (1- γ)+ γβ: 1 / 2 (1- γ)+ γ(1 β) or k:1-k where k= 1 / 2 (1- γ)+γβ Unequal gene conversion Equal gene conversion A is fixed if k> 1 / 2, and a is fixed if k< 1 / 2, at a rate dependent upon the frequency of heterozygotes in the population 1

2 Extension of Walsh s Model To Include Molecular Drive (γ,β), Drift (N ev ), and System of Mating/Population Subdivision (f) Transposition Probability of Fixation of a Neutral Mutation = 1/(2N) Probability of Fixation of a Mutation With Biased Gene Conversion = 2(2k "1)N ev (1" f ) N when 4N ev (1" f )(2k "1) >>1 (2k "1)N ev (1" f )e 4 N ev (2k"1)(1" f ) N when 4N ev (1" f )(2k "1) <<1 Thus, the evolutionary impact of gene conversion interacts with and is modulated by traditional evolutionary forces. For example, as f goes down, the probability of fixation of a biased gene conversion allele goes up. Molecular Factors Do Not Override Traditional Evolutionary Forces; Rather, They Strongly Interact With Them Transposition: Mutator Transposition: Direct Phenotypic Effects A flower of Petunia hybrida transposon genotype derived from the inbred line W138 showing a large number of white-pink sectors (see Ramulu et al., ANL10: 19-21, 1998). 2

3 Transposition: Horizontal & Vertical Transfer Transposition: Horizontal Transfer Anxolabehere et al. Mol. Biol. Evol. 5: , 1988 Comparison of a) species and b) P-element phylogenetic histories. Diagonal lines unite P-element clades with the species from which they were sampled. From Silva, J.C. and M.G. Kidwell. Horizontal Transfer and Selection in the Evolution of P Elements. Mol Biol Evol 17(10): , Unequal Exchange Unequal Exchange Can Also Create A New Types of Genes 3

4 Unequal Exchange: Concerted Evolution Unequal Exchange: Concerted Evolution Gene Duplication Without Concerted Evolution Gene Duplication With Concerted Evolution Gentile, K.L., W.D. Burke and T.H. Eickbush. Multiple Lineages of R1 Retrotransposable Elements Can Coexist in the r DNA Loci of Drosophila. Mol Biol Evol 18(2): , Unequal Exchange: Concerted Evolution Model of Weir et al. J. Theor. Biol. 116: 1-8, 1985 n=number of repeats in a multigene family N=ideal population size µ=neutral mutation rate per repeat per generation 1/(2N)=probability of fixation of new mutant at homologous sites α=probability of a repeat converting a paralogous repeat to its state (Molecular drive exists such that a neutral mutant will eventually go to fixation at all paralogous sites as well) 1/(2Nn)=probability of fixation of a new mutant at all homologous and paralogous sites 2Nnµ=expected number of new mutants per generation Rate of neutral evolution in multigene family evolving in concert = (2Nnµ) 1/(2Nn)=µ= Same neutral rate as if it were a single locus! Unequal Exchange: Concerted Evolution Recall that the time to neutral coalescence of all homologous copies of a gene to a common ancestral form = 4N The time to neutral coalescence of all homologous and paralogous copies in a multi-gene family to a common ancestral form = 2/(1-τ) where τ is the Maximum of one of two forms: /(2N) or 2. 1-α In the first case [α>1/(2n); that is molecular drive is more powerful than drift], then t= 2/{1-[1-1/(2N)]} = 2/[1/(2N)] = 4N = the same rate of coalescence as a single locus and no effect of α! In the second case (α<1/(2n); that is molecular drive is weak compared to drift), α dominates the coalescence process! Therefore, molecular drive has its biggest evolutionary impact when it is Weak compared to drift. Under these conditions, the multigene family will have much diversity among paralogous copies within a chromosome. 4

5 Abnormal Abdomen in Drosophila mercatorum R1 & R2 Retrotransposons In rdna of D. mercatorum A through E Can All Be Found In Different rdna Repeats On A Single X Chromosome, Although Only Types A and B Are Common In Natural Populations The Phenotypic Expression of aa Requires > 1/3 of rdna Repeats Have an Insert That Functionally Inactivates The Repeat Modes of Molecular Drive for aa R1 transposition leaves a molecular footprint - these are observed at a low frequency, implying a limited amount of transposition Almost all R1 bearing repeats are physically clustered in the rdna - this implies that unequal exchange is important Therefore, the primary source of molecular drive in this system is unequal exchange. 5

6 Under-Replication of Ribosomal DNA in Drosophila Polytene X Chromosomes The Phenotypic Expression of aa Requires The Absence of Selective Under-Replication of Inserted Repeats (Controlled by ur locus with r=0.004 to the rdna Cluster) Distribution of Insert Proportions in X Chromosomes with ur + versus ur aa Frequency X aa X The ur aa Allele Is in Linkage Disequilibrium with the rdna Cluster Such that Virtually All X Chromosomes With ur aa Also Have > 1/3 of the rdna Repeats with the R1 Insert. This Creates the aa Supergene Proportion of Inserted 28S Genes on X 6

7 The High Amount of Diversity Both Within and Among Chromosomes for the Number of R1 Bearing Repeats Implies that Molecular Drive Is Weak Compared To Other Evolutionary Forces, and Therefore Molecular Drive Has An Important Evolutionary Impact On Shaping Variation In This System aa Genotype to aa Phenotype X aa X Proportion of Inserted 28S Genes on X Sufficient Potential for ribosome ribosomes in all deficiency in cells cells Compensatory response in diploid cells insures sufficient ribosomes Note CR is a normal developmental control system that dampens out the phenotypic impact of variation in the number of functional ribosomal repeats In polytene tissues (such as the larval fat body), CR does not work and the amount of cellular rdna is controlled by the under-replication system. aa Genotype to aa Phenotype Polytene Tissues Produce Large Quantities of Certain Proteins Over Short Time Periods A Functional Deficiency of rdna in aa Flies Leads to Translational Limits On The Rate of Production of Such Proteins Among Such Proteins Produced in the Fat Body is Juvenile Hormone (JH) Esterase aa Flies Produce JH Esterase At A Slower Rate Than non-aa Flies, Resulting in Higher Titers of JH During Certain Critical Developmental Time Periods. aa Genotype to aa Phenotype 25 o C The reduced rate of JH degradation by aa flies during the late third larval instar stage results in the failure of the abdominal histoblast cells to undergo full adult differentiation. This phenotypic effect disappears with lower larval rearing temperatures that slow down the rate of larval development. It is not important in the natural populations. 7

8 aa Genotype to aa Phenotype Prolonged Larval Phase JH is reactivated in adults after pupation, where it controls critical aspects of egg production. Number of Females /+ Females aa/- Females 1 st 2 nd 3rd 4th Quartile Overall, aa/- females take 1.44 days longer than +/+ females to emerge from bagged natural rots. The amount of JH bound in ovariole tissue is deficient the first week after eclosion in aa flies, but is normal by the second week Precocene II JH Titer (& 30 analogs such as 20 precocene) Is 10 Linearly Related To Ovarian Output Vitellogenic Oocytes/Female Precocene I -x Precocene, 10 M Mean number of eggs lain by aa and + females in a laboratory environment + Females aa Females Time Interval (from Eclosion) Eggs/Female Sample Size Eggs/Female Sample Size Days Days Days Days

9 aa Genotype to aa Phenotype Determining Age Structure in Natural Populations aa Genotype to aa Phenotype Cactus Site Year Age Structure Freq. aa in Freq. aa in Males a aa = p/p Sons from rots B 1982 Old IV 1982 Old C 1983 Young F 1984 Young IV 1984 Old B 1987 Young B 1989 Old Pooled Age Structures: for Old Pooled for Young Age Structures: aa Genotype to aa Phenotype High Titers of JH In Late Third Instar Larvae Cause a Prolongation of the Larval Phase and Juvenilized Adult Abdominal Cuticle High Titers of JH in Early Adult Females Cause Earlier Egg Laying and Increased Fecundity ALMOST ALL OF THE PLEIOTROPIC TRAITS OF aa ARE CONSEQUENCES OF THE NORMAL RESPONSES TO HIGH JH TITER AND CAN BE PRODUCED AS PHENOCOPIES BY APPLICATION OF JH AT THE APPROPRIATE TIMES. 9

10 Variation for aa is strongly affected by molecular drive - how about other evolutionary forces such as natural selection? Natural Selection & Adaptation Need To Consider How The aa Genotype Interacts With The Environment. 10

11 10 8 Si te A IV >15 Site B 6 Site C > >15 6 Site IV Cactus B >15 Estimated Surviorship Curves From Observed Daily Mortality Rates At Wet and Dry Sites. Let Note The Critical Role Of Pleiotropy In Determining the Reversal of Fitnesses As A Functional Of The Ecological Conditions Affecting Age Structure Then: n "r (a+ e) l e # l a b a e = 1 $ r % a=1 "Dry" age structure: r(aa/-) = r(+/+) = x = Age from Hatching e = Age at Eclosion a = x - e = Adult Age from Eclosion l x = Surviorship to Age x l a = l a+e /l e = Adult Surviorship to Age a b a = Expected Number of Offspring Born to an Adult of Age a r = Malthusian Parameter (Fitness) n # a=1 n # a=1 l a b a "1/ l e l a b a (a + e) "Humid" age structure: r(aa/-) = r(+/+) =

12 Response of aa to Spatial Heterogeneity In Humidity, But Not of X-Linked G-6-PD Response of aa to Spatial Heterogeneity In Humidity, But Not of X-Linked G-6-PD Temporal Variation Over 12 Years For example, 1980 was normal, 1981 had the third worst drought in recorded history, and 1982 (due to El Chichón eruption) was the wettest in recorded history. Frequency of aa X Response of aa to Spatial and Temporal Variation Over 12 Years Normal Average Wet (1982) Dry Average Kona Average 0.1 A & B C & D E & F High Site Low 12

13 Response of aa to Temporal Variation At Site IV (Weather Often Discordant With Kohala Transect) 0.5 The Impact of Epistasis Between aa Insert Number and Underreplication Upon Adaptive Response 0.58 X chromosomes, 1986 (Kona Year) Frequency of aa X Dry ('80,'81,'83) Years Wet ('82,'84-87) Insert Proportion Upper (A-D) Sites Lower (F, IV) ur + X's ur aa X's The Adaptive Response of aa To Environmental Variation in Humidity and Other Factors Influencing Adult Surviorship Is Modulated Through Pleiotropy and Epistasis. Weak Molecular Drive Provides Critical Variation In This System Upon Which Natural Selection Acts Molecular Drive Does NOT Negate The Importance of Other Evolutionary Forces. Molecular Drive INTERACTS With Other Evolutionary Forces In Determining the Path of Evolution. 13

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