Notes on Population Genetics: Understanding Mutation and Its Role in Evolution, Study notes of Human Genetics

Download Notes on Population Genetics: Understanding Mutation and Its Role in Evolution and more Study notes Human Genetics in PDF only on Docsity! “Coarse” Notes Population Genetics III-1 MUTATION INTRODUCTION TO MUTATION READING: Hedrick pp. 357–367 and 376–385 – Mutation plays two key roles in evolution: (1) It is an evolutionary force that changes gene frequencies; (2) It is the ultimate source of all genetic variation • Note: any complete theory of evolution must explain processes that create mutations. – Most mutations are rare. • Alleles are rarely incorrectly replicated. • Ballpark figures for spontaneous mutation rates: – Most mutations are deleterious. • Most common mutations involve loss of function. • Deleterious mutations are generally recessive. • Mutation is a destructive force. • Mutation is a creative force. MUTATION AS A FORCE OF EVOLUTIONARY CHANGE – Simplest case: two alleles A and a • Reality: consider an exon that is 100 bp long. • Approximation: alleles refer to 2 classes of alleles, e.g., T/not T (=G, C, or A) – Will consider recurrent mutation. – How do allele frequencies change given recurrent mutation? • Rate of mutation = probability a gene copy will mutate per generation “Coarse” Notes Population Genetics III-2 – Let u = rate of mutation from A to a (“forward” mutation rate). v = rate of mutation from a to A (“backward” mutation rate). • Equations of genetic change: ! p = p 1" u( ) + 1" p( )v !p = " p # p = #up + v 1 # p( ) = # u + v( )p + v $ % & ' ˆ p = v u + v • Observations: (1) Since u, v are small, !p due to mutation is typically small. – Mutation is a "weak" evolutionary force. (2) !p is linear in p. – I.e., in contrast to selection no term like p 1 ! p( ) . – rate of creation of new mutations depends on both • the frequency of "mutable" copies (e.g., p). • the mutation rate per copy (u). (3) ˆ p makes sense. – proportional to relative mutation rates. – at equilibrium, the number of A's converted to a's = number of a's to A's. (4) Can rewrite above rate equation as !p = " u + v( ) p " ˆ p ( ) – Shows that the rate of evolution towards equilibrium is proportional to • the total mutation rate; • the deviation of the current allele frequency from the equilibrium, ˆ p = v u + v( ) . (5) The rate of mutation does not depend on genetic variance. – No “variance” factor in the equation for !p . – Mutation as an evolutionary force does not require pre-existing variation • in fact, it's most effective as an evolutionary force when p = 0 or p = 1! • How fast is equilibrium approached? – pt ! ˆ p = 1! u ! v( ) t p 0 ! ˆ p ( ) – Answer: very slowly. • e.g., when u = v = 10!6 , it takes almost 350,000 generations to evolve half way to the equilibrium ˆ p = ˆ q = 0.5 . “Coarse” Notes Population Genetics III-5 • As t —> ∞, f t —> ˆ f ! 1 4Nu +1 = eventual probability of homozygosity (or the average frequency of homozygotes). – The quantity 4Nu is often denoted by ! so that ˆ f = 1 1 +! – With k < ! alleles, ˆ f = 1 + 4Nu k !1( ) 1+ 4Nu k k ! 1 = 1+ " k !1( ) 1 +" k k !1 # 1 1 +" as k →∞. • ˆ f gives a good idea of homozygosity, but how many different alleles are maintained in a population of size 2N? – Minimum number = 1; Maximum, 2N – Define 1 ˆ f = "effective number of alleles" in the population; this is directly related to the amount of heterozygosity. • Rationale: –Consider a population in H-W proportions with 3 equally frequent alleles. • Freq.[homozygotes] = 1 3( )2 + 1 3( )2 + 1 3( )2 = 1 3 . • Then, 1/Freq. of homozygotes = 1/(1/3) = 3 = # alleles! – Consider a pop. whose alleles have unequal freqs., 1/4, 1/4, 1/2 • Freq.[homozygotes] = 1 2( )2 + 1 4( )2 + 1 4( )2 = 5 8 => 8/3 =2.6 "equally frequent" of "effective" alleles • At equilibrium, effective number of alleles is 1 ˆ f ≈ 4Nu + 1 = ! +1 . – Punchline: Substantial genetic variability is possible with mutation if 2Nu > 1 • i.e., at least one mutant/locus/generation • Dynamics of Mutation & Drift – Q: What is the rate of substitution of selectively neutral mutations? 1) How frequently does a new neutral mutation arise? Ans. 2Nu, where u is the mutation rate to selectively neutral alleles per generation. “Coarse” Notes Population Genetics III-6 2) How likely is this new mutant to be fixed? Ans. Since alleles are "neutral", 1/2N 3) How frequently will one neutral allele replace another? Ans. (how often they arise) x (how often they fix) = 2Nu x (1/2N) = u per generation. – Conclude Rate of neutral substitution = neutral mutation rate. • Note: the substitution rate is independent of N! – Why? While neutral mutations are more likely to fix in a smaller population, the rate at which they arise (2Nu) is small. Just the opposite is true for larger pops. The result is that substitution rate is independent of a population's size. – Property is consistent with a central observation motivating the "molecular clock" hypothesis – Property is also central to the "neutral theory" of molecular evolution advanced by Kimura. • Problem: Some observed substitution rates are constant “per year” rather than “per generation.”