1. Selection and Mutation

5.  If either of the following occurs then the population is responding to selection. 1. Some phenotypes allow greater survival to reproductive age. -or- 2. All individuals reach reproductive age but some individuals are able to produce more viable (reproductively successful) offspring. If these differences are heritable then evolution may occur over time.

6.  It needs to be mentioned that most phenotypes are not strictly the result of their genotypes.  Environmental plasticity and  interaction with other genes may also be involved.  In other words it is not as simple as we are making it here but we have to start somewhere. Caution

7. 1. Selection may alter allele frequencies or violate conclusion #1 2. Selection may upset the relationship between allele frequencies and genotype frequencies. Conclusion #1 is not violated but conclusion #2 is violated. In other words the allele frequencies remain stable but genotype frequencies change and can no longer be predicted accurately from allele frequencies.

9.  After random mating which produces 1000 zygotes we get:

10. Initial frequencies B 1 = 0.6; B 2 = 0.4 B1B1B1B1 B1B2B1B2 B2B2B2B2 1000 total

11. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 B1B2B1B2 B2B2B2B2 1000 total

12. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 B2B2B2B2 1000 total

13. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total

14. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes

15. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive

16. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive

17. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive

18. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving

19. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360

20. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360

21. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80

22. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total

23. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is

24. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45

25. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45

26. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45.10

27. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45.10 The resulting allelic frequencies in the new reproducing population is B 1 =B 2 =

28. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45.10 The resulting allelic frequencies in the new reproducing population is B 1 =.45+1/2(.45) = 0.675

29. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45.10 The resulting allelic frequencies in the new reproducing population is B 1 =.45+1/2(.45) = 0.675 B 2 =

30. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45.10 The resulting allelic frequencies in the new reproducing population is B 1 =.45+1/2(.45) = 0.675 B 2 = 1/2(.45)+0.10 = 0.325

31. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45.10 The resulting allelic frequencies in the new reproducing population is B 1 =.45+1/2(.45) = 0.675 B 2 = 1/2(.45)+0.10 = 0.325 an increase of.075 a decrease of.075

32. Initial frequencies B 1 = 0.6; B 2 = 0.4 360 B 1 B 1 480 B 1 B 2 160 B 2 B 2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % survive 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is.45.10 The resulting allelic frequencies in the new reproducing population is B 1 =.45+1/2(.45) = 0.675 B 2 = 1/2(.45)+0.10 = 0.325 an increase of.075 a decrease of.075 are evolving Thus, conclusion #1 is violated and the allele frequencies are changing; we are not in equilibrium. The population is evolving!

33.  analyze the population on the basis of the fitness of the offspring produced.  The fittest individuals will survive the selection process and leave offspring of their own.  We are going to define fitness as the survival rates of individuals which survive to reproduce.

34.  If : w 11 = fitness of allele #1 homozygote (exp B 1 B 1 ) w 12 = fitness of the heterozygote (exp B 1 B 2 ) w 22 = fitness of allele #2 homozygote exp (B 2 B 2 ) mean fitness of the population will be described by the formula: ŵ = p 2 w 11 + 2pqw 12 + q 2 w 22 MEAN FITNESS CAUTION! Use ONLY allele frequencies in these formulas NOT genotype frequencies!

35.  B 1 = 0.6 and B 2 = 0.4 and fitness of B 1 B 1 = 1.0 (100% survived) fitness of B 1 B 2 =.75 ( 75% survived) fitness of B 2 B 2 =.50 (50% survived)  Figure the mean fitness now.

36.  B 1 = 0.6 and B 2 = 0.4 and fitness of B 1 B 1 = 1.0 (100% survived) fitness of B 1 B 2 =.75 ( 75% survived) fitness of B 2 B 2 =.50 (50% survived)  Figure the mean fitness now.  ŵ= (.6) 2 (1)+

37.  B 1 = 0.6 and B 2 = 0.4 and fitness of B 1 B 1 = 1.0 (100% survived) fitness of B 1 B 2 =.75 ( 75% survived) fitness of B 2 B 2 =.50 (50% survived)  Figure the mean fitness now.  ŵ= (.6) 2 (1)+(2(.6)(.4)(.75)) +

38.  B 1 = 0.6 and B 2 = 0.4 and fitness of B 1 B 1 = 1.0 (100% survived) fitness of B 1 B 2 =.75 ( 75% survived) fitness of B 2 B 2 =.50 (50% survived)  Figure the mean fitness now. ŵ= (.6) 2 (1)+(2(.6)(.4)(.75)) + (.4) 2 (.5) =.80

39. B 1 B 1 = P 2 w 11 ŵ B 1 B 2 = 2pqw 12 ŵ B 2 B 2 = q 2 w 22 ŵ We can use these formulas which can calculate the new expected genotype frequencies based on the fitness of each genotype and the allele frequencies in the current generation.

40. B 1 = p 2 w 11 +pqw 12 B 2 = pqw 12 +q 2 w 22 ŵ ŵ

41. Δ B 1 = Δp = p (pw 11 +qw 12 – ŵ) ŵ Δ B 2 = Δq = q (pw 12 +qw 22 – ŵ) ŵ

42. new genotypeallele frequencies  Go back to the problem we did in class last time. Taking this current population that you have already analyzed, figure out what the new genotype and allele frequencies will be if the fitness of these individuals is actually as follows:  SS individuals 0.8 ; Ss individuals 1.0 and the ss individuals 0.6.

43. ŵ = p 2 w 11 + 2pqw 12 + q 2 w 22 ŵ= (.82) 2 (.8) + 2(.82)(.18)(1.0) + (.18) 2 (.6) ŵ =.537 +.295 +.019 =.85 B 1 B 1 = P 2 w 11 ŵ B 1 B 2 = 2pqw 12 ŵ B 2 B 2 = q 2 w 22 ŵ ; SS = (.82) 2 (.8) /.85 =.633 ;Ss = 2(.82)(.18)(1.0) /.85 =.347 ;ss = (.18) 2 (.6) /.85 = 0.023 Last time we calculated S =.82 and s =.18 Now we set the fitnesses at w 11 (SS)=.8;w 12 (Ss)=1;w 22 (ss)=.6 Calculate the ŵ and B 1 B 1; B 1 B 2; and B 2 B 2 values for the next generation now Hint: Do they add up to 1.0?

44. B 1 = p 2 w 11 +pqw 12 ŵ B 2 = pqw 12 +q 2 w 22 ŵ S = (.82) 2 (.8) + (.82)(.18)(1.0) =.806.85 s = (.82)(.18)(1.0) + (.18) 2 (.6) =.196.85 We an also calculate the new allele frequencies as well

45. So…… B 1 B 1 =.63 B 1 B 2 =.35 B 2 B 2 =.02 and B 1 =.80 B 2 =.20 Is this population in equilibrium? Have the allele frequencies changed? Can we predict the genotype frequencies from the allelic frequencies?

47. Fruit fly experiments of Cavener and Clegg

48.  Worked with fruit flies having two versions of the ADH (alcohol dehydrogenase) enzyme, F and S. (for fast and slow moving through an electrophoresis gel) Breeders for each generation were picked at random  Grew two experimental populations on food spiked with ethanol and two control populations on normal, non-spiked food. Breeders for each generation were picked at random.  Took random samples of flies every few generations and calculated the allele frequencies for Adh F and Adh S

49. Figure 6.13 pg 185

50.  only difference is ethanol in food no migration assured random mating population size, drift? no mutation.  Must be selection for the fast form of gene.  Indeed studies show that Adh F form breaks down alcohol at twice the rate as the Adh S form.  Therefore offspring carrying this allele are more fit and leave more offspring and the make-up of gene pool changes.

52.  Selection may upset the relationship between allele frequencies and genotype frequencies.  Conclusion #1 ( allele frequencies do not change) is not violated but conclusion #2 (that we can predict genotype frequencies from allele frequencies) is violated.

53. Initial B 1 = 0.5 Initial B 2 = 0.5 250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total

54. Initial B 1 = 0.5 Initial B 2 = 0.5250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive

55. Initial B 1 = 0.5 Initial B 2 = 0.5250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive 125 / 750 0.167

56. Initial B 1 = 0.5 Initial B 2 = 0.5250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667

57. Initial B 1 = 0.5 Initial B 2 = 0.5 250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167 The resulting allelic frequencies in new mating population

58. Initial B 1 = 0.5 Initial B 2 = 0.5250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167 The resulting allelic frequencies in new mating population B 1 =.167+1/2 (0.667) = 0.5

59. Initial B 1 = 0.5 Initial B 2 = 0.5 250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167 The resulting allelic frequencies in new mating population B 1 =.167+1/2 (0.667) = 0.5 B 2 = ½ (.667) +.167 = 0.5

60. Initial B 1 = 0.5 Initial B 2 = 0.5250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167 The resulting allelic frequencies in new mating population B 1 =.167+1/2 (0.667) = 0.5 B 2 = ½ (.667) +.167 = 0.5 No change

61. Initial B 1 = 0.5 Initial B 2 = 0.5250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167 The resulting allelic frequencies in new mating population B 1 =.167+1/2 (0.667) = 0.5 B 2 = ½ (.667) +.167 = 0.5 No change not Thus conclusion #1 is not violated therefore this population has not evolved…..but…..

62. Initial B 1 = 0.5 Initial B 2 = 0.5 250 B 1 B 1 500 B 1 B 2 250 B 2 B 2 1000 Total Differential fitness of the genotypes Fitness.50Fitness 1.0Fitness.50 Number of survivors to reproductive age 125500125750 total The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167 The resulting allelic frequencies in new mating population B 1 =.167+1/2 (0.667) = 0.5 B 2 = ½ (.667) +.167 = 0.5 Change in allele frequency No change Thus conclusion #1 is not violated therefore this population has not evolved….but….. Conclusion #2 is violated. We are not in equilibrium. Frequency of B 1 B 1 =.167 which is not equal to (.5) 2

63. Kuru example among the Foré in New Guinea ◦ Pg 188-191 ◦ Wanted to determine if there was a genetic basis to the resistance of kuru infection.  Ritualistic mortuary feasts, only young women ate the contaminated nervous system tissue leading to CJD (similar to mad cow disease)  Among young women who never participated he Met allele = 0.48 and the Val allele 0.52; Genotypes were: Met/Met 0.22; Met/Val 0.51 and Val/ Val 0.26 very close to the values expected for H-W.

64.  Met = 0.52 and Val = 0.48  The expected genotypes are  Met/Met 0.27 ; Met/ Val 0.5 and Val/Val 0.23  The actual were:  Met/Met 0.13 ; Met/ Val 0.77 and Val/Val 0.10  Appears homozygotes are susceptible but heterozygotes are protected.

65.  HIV example in book. pg 191  Two conditions must be met 1. Need a high enough frequency of the beneficial allele in the population gene pool 2. There must be high selection pressure for the allele in the same area. In this case a high incidence of HIV infection.

66.  If selection is acting, does the rate of evolution of a particular allele depend on whether it is…. heterozygote or homozygote? dominant or recessive?

67. Tribolium Beetle example

68.  Dawson’s Flour beetle example  Studied a gene locus that had a wild type (+) allele and a lethal allele.  +/+ or +/L are normal L/L is lethal.  Two experimental populations composed of all heterozygotes +/L  Therefore started with + = 0.5 and L =0.5.  Expected populations to evolve toward lower frequency of the L allele.

69. Results showed that the recessive lethal did drop rapidly at first but slowed down over successive generations. WHY?

70. As you go on there are less and less homozygous lethals for selection to act on and the lethal allele hides in the heterozygotes In each succeeding generation all LL are lost and ++ makes up a greater proportion of the survivors.

71. dominance frequencyinteract  Dawson showed that dominance and allele frequency interact to determine the rate of evolution when acted on by selection  If a recessive allele is common evolution is rapid because there are a lot of homozygotes that express the phenotype for selection to act on.  If recessive allele is rare, evolution is slow because the rare allele is hidden in the heterozygotes where selection cannot act.  His experiments also demonstrated that ◦ controlled lab situations can accurately predict the course of evolution ◦ populations do what you would expect if selection is occurring as predicted by the evolutionary theory.

72.  Normally in a recessive/ dominant gene, the fitness of the heterozygote will be equal to one of the homozygotes  Also, it is possible for the heterozygotes to have a fitness intermediate to the two homozygotes.  Thirdly we may find Heterozygote Superiority or Inferiority

73.  Studied a gene in which  Homozygotes for one allele are viable  Homozygotes for the other allele are not viable and are lethal.  Heterozygotes have a higher fitness than either homozygotes

74.  Started with all heterozygotes to establish a new population (each allele =.5) After several generations equilibrium was reached at.79 frequency for the viable allele maintained How could this be This means that the lethal allele was maintained at frequency of 0.21! How could this be? Started more populations beginning with frequency of.975 of viable allele. Expect the population to eliminate all lethal alleles and fix the viable allele at 1.0. But.....

75. Figure 6.18 pg 200 dropped The viable allele dropped in frequency and the same equilibrium around a frequency of.79 was reached for the viable allele!

76.  There is some advantage to the heterozygote condition and the heterozygote actually has a superior fitness to either homozygote.  Example in humans is sickle cell anemia balanced polymorphism  Leads to the maintenance of genetic diversity = balanced polymorphism

77.  Where the heterozygote condition is inferior to either of the homozygotes  What do you predict would happen to the allele frequencies here?  Leads to fixation of one allele in the population, while the other is lost.  Either allele may be fixed depending on conditions and beginning frequencies of each allele in the gene pool.

78.  Leads to a loss of genetic diversity  Although if different alleles are fixed in different populations can help maintain genetic diversity among populations

79.  When one allele is consistently favored it will be driven to fixation  When heterozygote is favored both alleles are maintained and at a stable equilibrium (balanced polymorphism) even though one of the alleles may be lethal in the homozygous state.

80.  The Elderflower orchid example in book  Population’s allele frequencies remain at or near an equilibrium but it is due to the direction of selection fluctuating.  First one allele is favored and then the other.  The population fluctuates around an equilibrium point.

81.  Bumblebees visit yellow and purple flowers alternately  The least frequent phenotype is visited more often and receives more pollination events.  In subsequent generations this color becomes more and more frequent until it becomes the dominant color.  Once this happens then the same color becomes less frequently visited and the other color becomes favored.  Oscillation between the two colors continues and the favored allele alternates over time around some mean equilibrium value.

83.  Mutation is the source of all new alleles  Mutation provides the raw material on which selection can act

84.  Mutation alone is a weak or nonexistent evolutionary force  If all mutations that happened, occurred in gametes so that they would be immediately passed on to their offspring and ….  the rate of mutation were high, say A  a at a rate of 1 in 10,000 per generation.  then the rates are very slow as shown in figure 6.23

85. Figure 6.23 pg. 211

86. In concert with selection, mutation becomes a potent evolutionary force.

87.  Used a strain of E. coli that cannot exchange DNA (conjugation) so the only possible source of genetic variation is mutation.  Showed steady increases in fitness and size over 10,000 generations in response to a demanding environment. (little over 4 years)  However, increases in fitness occurred in jumps when a beneficial mutation occurred and then spread rapidly through the population

88. Figure 6.25 pg 213

89.  When mutations are deleterious  Selection acts to eliminate them  Deleterious Mutations persist because they are created anew over and over again  When the rate at which deleterious mutations are formed exactly equals the rate at which they are eliminated by selection the allele is in equilibrium. = mutation-selection balance

90. mildlydeleterious selectionweakand Mutationhigh  If the mutation is only mildly deleterious and therefore selection against it is weak; and Mutation rate is high then mutated allelein the population ◦ The equilibrium frequency of the mutated allele will be relatively in the population. strong selection against mutation rate is low  If, on the other hand, there is strong selection against a mutation (the mutation is highly deleterious) and the mutation rate is low then mutated allele will be ◦ Equilibrium ratio of the mutated allele will be in the population high low

91.  Spinal muscular atrophy, second most common lethal autosomal recessive disease in humans. Selection coefficient is.9 against the disease mutations.  However, among Caucasians 1 in 100 people carry the disease causing allele.  Research shows that the mutation rate for this disease is quite high  Mutation selection balance is proposed explanation for persistence of mutant alleles.  http://www.smafoundation.org http://www.smafoundation.org

92.  Cystic fibrosis is the most common lethal autosomal recessive disease in Caucasians  Mutation-selection balance alone cannot account for the high frequency of the allele =.02  Appears to also be some heterozygote superiority involved  Heterozygotes are resistant to typhoid fever bacteria and have superior fitness during typhoid fever epidemic.  At the current time it is believed that CF is an example of heterosis and not mutation- selection balance

93.  An autosomal dominant allele  Is actually increasing in the human population.  Any ideas why?  May be because it increases the tumor supressor activity in cells dramatically lowering the incidence of cancer in those with the defective allele.  They survive through the reproductive years and leave more offspring than their unaffected siblings.