Population genetics part 2 with Descriptions.pptx

Population genetics study

1. CS 105 GENETICS POPULATION GENETICS AND QUANTITATIVE GENETICS 1

2. Principles of Population genetics and Quantitative Genetics Definition/Terms: Population genetics is the study of how Mendel's Laws and other genetic principles apply to entire populations . Genes: A physical entity transmitted from parent to offspring during the reproductive process that influences hereditary traits. 2

3. Principles of Population genetics and Quantitative Genetics Alleles: Different forms of a gene for example a gene for haemoglobin may exist in normal form or in any one of a number of forms that produce haemoglobin molecules.   Locus: The position of a gene along a chromosome. In higher organism each individual has two copies of each type of chromosome one inherited from its mother through the egg and another one inherited from its father through the sperm. At any locus, hence every individual has two alleles - one at each corresponding position in the maternal and paternal chromosome. 3

4. Principles of Population genetics and Quantitative Genetics HOMOZYGOUS Is a state when two alleles at the same locus are identical the individual HETEROZYGOUS Is a state when two alleles at a locus are chemically different GENOTYPE Is the genetic constitution of an individual. Genotype refers to the specific alleles carried by an individual at all loci that affect the trait in question. 4

5. Principles of Population genetics and Quantitative Genetics PHENOTYPE Is the physical expression of a genotype 5

6. Phenotypic diversity and Genetic Variation Phenotypic diversity is one of the most important attributes in natural populations . For many traits, many differing phenotypes can be found. Population genetics deals with this phenotypic diversity and the portion of the diversity, which is caused by genotypic differences among individuals. 6

7. Phenotypic diversity and Genetic Variation Population genetic studies deals with determining how much genetic variation exists in natural populations and explaining this variation in terms of its origin, maintenance and evolutionary importance. 7

8. Phenotypic diversity and Genetic Variation In population genetics, the word population does not usually refer to the entire species, but it refers instead, to a group of individuals of the same species living within a sufficiently restricted geographical area that, any member of the group can potentially mate with any other member (provided of cause, they are of the opposite sex). 8

9. Genetic constitution of a population   A population in a genetics sense is not just a group of individuals but a breeding group , and the genetics of a population is concerned not only with the genetic constitution of the individuals, but also with the transmission of the genes from one generation to the next. 9

10. Genetic constitution of a population   In transmission, the genotypes of the parents are broken down and new sets of genotypes are formed in the progeny from genes transmitted in the gametes. Example: Parents: AA x aa Gametes: A A a a F1 A a 10

11. Genetic constitution of a population The genes carried by the population have continuity from generation to generation but the genotypes to which they appear do not. 11

12. Genetic constitution of a population   The genetic constitution of a population referring to the genes it carries is described by gene frequencies i.e. the specification of alleles present at each locus and the numbers on proportions of the different alleles at each locus. The frequency of all alleles at the locus must add up to unity or 100 percent. 12

13. Genetic constitution of a population   Example: Suppose A 1 is an allele at A 1 locus then the frequency of A 1 genes is the proportion of or percentage of all genes at this locus that are A 1 allele. Gene frequency at a particular locus among a group of individuals can be determined from knowledge of the genotype. 13

14. Genetic constitution of a population Genotype frequency – Is the proportion or percentage of a genotype among individuals in a group. Suppose there are three genotypes A 1 A 1 , A 1 A 2 , A 2 A 2. Genotypic frequency of A 1 A 1, will be the proportion of A 1 A 1 to all three genotypes i.e. A 1 A 1 A 1 A 1 + A 1 A 2 + A 2 A 2 14

15. Genetic constitution of a population For example if one quarter of individuals in the group is A 1 A 2 , the frequency of this genotype will be 0.25 or 25 percent. The frequency of the genotypes must add up to unity. 15

16. Genetic constitution of a population A 1 A 1 A 1 A 2 A 2 A 2 Total Number of individuals 30 60 10 100 Number of genes A 1 60 60 120 Number of genes A 2 60 20 80 16

17. Genetic constitution of a population The frequency of A 1 = 0.6 The frequency of A 2 = 0.4 Genotypes A 1 A 2 A 1 A 1 A 1 A 2 A 2 A 2 __________________________________________________ Freq p q P H Q ___________________________________________________   p + q = 1 and P + H + Q = 1   17

18. Genetic constitution of a population Since each individual contains two genes the frequency of A 1 is 1/2 (2P + H).   Thus p = P + 1/2H/ P+ H +Q q = Q + 1/2H/P + H + Q   Thus, A 1 = p = 30 + 1/2(60)/P + H + Q = 60/100 = 0.6   18

19. Genetic constitution of a population Freq. A 1 = p = 0.6 Freq. A 2 = q = 10 + 1/2 60/100 = 40/100 = 0.4 or q = 1-p q = 1 - 0.6 = 0.4 19

20. Agencies/Factors that influence transmission of genes Population size Differences in fertility and viability Immigration Mutation Mating system 20

21. Hardy - Weinberg equilibrium (1908) The Hardy-Weinberg Law - In a large random - mating population with no selection , mutation or migration the gene frequencies and the genotype frequencies are constant from generation to generation, and there is simple relationship between the gene frequencies and the genotype frequencies. 21

22. Hardy - Weinberg equilibrium (1908) A population with constant gene and genotype frequencies is said to be in Hardy - Weinberg equilibrium   P 2 (AA) + 2pq ( Aa ) + q 2 ( aa ) = 1 The relationship between gene frequencies and genotype frequencies is very important because many of the deductions about population genetics and quantitative genetics rest on it. 22

23. Hardy - Weinberg equilibrium (1908) If gene frequencies of the parents are p and q then the genotypes of the progeny are p 2 , 2pq and q 2 . ________________________________________________ Gene of parent Genotype in progenies   A 1 A 2 A 1 A 1 A 1 A 2 A 2 A 2 __________________________________________________ Freq. p q p 2 2pq q 2 23

24. Hardy - Weinberg equilibrium (1908) Other conditions for this law to hold are:   Gene segregate normally in gametogenesis Gene frequencies are the same in males and females 24

25. Hardy - Weinberg equilibrium (1908) The relationship between gene frequencies in the Hardy-Weinberg equilibrium can be shown graphically (see graph). 25

26. Hardy - Weinberg equilibrium (1908) These graphs show two important features:   1. The frequency of heterozygote individuals cannot be more than 50 percent and its maximum occurs when the gene frequencies are at p = q = 0.5.   2. When the frequency of an allele is low, the rare allele occurs predominantly in the heterozygotes and is very few in the homozygotes. This is important for effectiveness of selection.   26

27. Hardy - Weinberg equilibrium (1908) Applications of the Hardy-Weinberg Law   1. Gene frequency of recessive allele can be known without knowing the frequencies of all genotypes, if the genotypes are in Hardy-Weinberg (so as to use the equation p = P + 1/2H . For example of a is a recessive gene with a frequency of q then the frequency of aa is q 2 and the gene frequency is the square-root of the homozygote frequency. 27

28. Hardy - Weinberg equilibrium (1908) 2. If Hardy-Weinberg equilibrium is assumed it is possible to know the frequency of heterozygotes in the population especially if they are carriers of recessive abnormalities. The frequency of heterozygotes among normal individuals is denoted by H which is the ratio of genotype frequencies Aa /(AA + Aa ) where a is a recessive allele. If q is the frequency of a then: 28

29. Hardy - Weinberg equilibrium (1908) H = 2q (1-q) 2q ----------- = ---- ( 1-q) 2 + 2q(1-q) 1 + q 29

30. Hardy - Weinberg equilibrium (1908) Test of Hardy-Weinberg equilibrium   If data are available, the observed frequencies of the genotypes can be tested for agreement with a population in Hardy-Weinberg equilibrium. 30

31. Hardy - Weinberg equilibrium (1908) Example: Suppose gene R controls flower colour with genotypes RR being red, Rr purple and rr are white.   RR (red) Rr (purple) rr (white)   No. of 2857 2628 515 Individuals   Genotype Proportion 0.476 0.438 0.086 P H Q 31

32. Hardy - Weinberg equilibrium (1908) Steps: 1. Calculate R gene frequency = p = R p = P + 1/2H 2857 + 1/2 (2628) ---------- ---------------------- = 0.695 P + H + Q 6000 p = 0.695   32

33. Hardy - Weinberg equilibrium (1908) 2. Calculate q: = r q = Q + 1/2H = 515 + 1/2 (2628) ----------- -------------------- = 0.305 P + H + Q 6000   or 1 - p = 1 - 0.695 = 0.305 33

34. Hardy - Weinberg equilibrium (1908) 3. Calculate the expected proportions and number of individuals.   RR Rr rr p 2 2pq q 2 (0.695) 2 2(0.695 x 0.305) (0.305) 2 0.483 0.424 0.093 Expect. Genotype proportion    (0.483 x 6000) (0.424 x 6000) (0.093 x 6000) 2898 2544 558 Expected no. of individuals 34

35. Hardy - Weinberg equilibrium (1908) Test whether the observed and expected values differ significantly   X 2 =  ( obs - exp) 2 ------------------------- exp   X 2 = (2857 - 2898) 2 + (2628 - 2544) 2 + (515 - 558) 2 --------------- ----------------- -------------- 2898 2544 558   X 2 = 6.667 at 1 df 35

36. Hardy - Weinberg equilibrium (1908) Conclusion: At 5% (0.05) the tabulated value is 3.841. Therefore the observed proportions of individuals for each genotype is significantly different from the expected proportion, which means this population is not in Hardy-Weinberg equilibrium. 36

37. Random mating in case of multiple alleles i.e. three or more If there are three alleles, then these can be labelled as A 1 , A 2 , A 3 and the corresponding allele frequencies as p 1 , p 2 and p 3 . In this case then p 1 + p 2 + p 3 = 1. The genotypic frequencies of A 1 A 1 , A 1 A 2 , A 2 A 2 , A 1 A 3 , A 2 A 3 and A 3 A 3 are p 1 2 , 2p 1 p 2 , p 2 2 , 2p 1 p 3 , 2p 2 p 3 and p 3 2 respectively. 37

38. Random mating in case of multiple alleles Sperm Egg A 1 (p 1 ) A 2 (p 2 ) A 3 (p 3 ) A 1 (p 1 ) A 1 A 1 ( p 1 2 ) A 1 A 2 (p 1 p 2 ) A 1 A 3 (p 1 p 3 ) A 2 (p 2 ) A 1 A 2 (p 1 p 2 ) A 2 A 2 (p 2 2 ) A 2 A 3 (p 2 p 3 ) A 3 (p 2 ) A 1 A 3 ( p 1 p 3 ) A 2 A 3 (p 2 p 3 ) A 3 A 3 ( p 3 2 ) 38

39. Random mating in case of multiple alleles Frequencies of offspring A 1 A 1 = p 1 2 A 1 A 2 = p 1 p 2 + p 1 p 2 = 2p 1 p 2 A 1 A 3 = p 1 p 3 + p 1 p 3 = 2p 1 p 3 A 2 A 2 = p 2 2 A 2 A 3 = p 2 p 3 + p 2 p 3 = 2p 2 p 3 A 3 A 3 = p 3 2   39

40. Random mating in case of multiple alleles The blood groups (ABO) in humans are controlled by three alleles that are designated I o , I A and I B . Genotypes I A I A and I A I o have blood group A, genotypes I B I B and I B I o have blood type B, genotypes I o I o has blood type 0 and genotype I A I B has blood type AB. 40

41. Random mating in case of multiple alleles Example: From a test of 6313 people in one city the number of individuals were found to be as follows in blood group:   A type - 2625 B type - 570 O type - 2892 AB type - 226 41

42. Random mating in case of multiple alleles The allele frequency in this case is: p 1 =0. 2593 (I A ), p 2 = 0.0652 (I B ), p 3 = 0.6755 (I o ) 42

43. Random mating in case of multiple alleles The expected number of the four genotypes (blood groups) A type = [(0.2593) 2 + 2(.2593) (.6755)] (6313) = 2636.0   B type = [(0.0652) 2 + 2[.0652) (.6755)] (6313) = 582.9   O type = (0.6755) 2 (6313) = 2880.6   AB type = 2(.2593) (.0652) (6313) = 213.5   43

44. Random mating in case of multiple alleles Blood group Observed ` expected No of individuals No of individuals ______________________________________________ A 2625 2636.0 B 570 582.9 0 2892 2880.6 AB 226 213.5 ______________________________________ X 2 = 1.11 44

45. Random mating in case of multiple alleles There is one degree of freedom   4 - 1 - 1 (for estimating p 1 from the data) - 1 (for estimating p 2 from the data) = 1 (degree of freedom is not deducted for estimating p 3 because 1-p 1 -p 2 = p 3 ). Probability = 0.05. The conclusion is that, this population is in Hardy - Weinberg equilibrium. 45

46. Sex-linked genes Many higher animals have sex chromosomes, which determine characteristics associated with sex. The relationship between gene frequency and genotype frequency in the homogametic sex is the same as with autosomal gene, but heterogametic sex has only two genotypes and each individuals carries only one gene instead of two. 46

47. Sex-linked genes For this reason, two thirds (2/3) of the sex- linked genes in the population are carried by homogametic sex and one third by heterogametic. Heterogametic sex will be referred to as male. XX = Female XY = Male 47

48. Sex-linked genes Consider two alleles A 1 and A 2 , with frequencies p and q and let the genotypic frequencies be as shown in the table below. 48

49. Sex-linked genes Females Males A 1 A 1 A 1 A 2 A 2 A 2 A 1 A 2 Frequency P H Q R S 49

50. Sex-linked genes The frequency of A 1 among the females is then p f = P + ½ H, The frequency of A 1 among males is p m = R. The frequency of A 1 in the whole population is: p = 2/3 p f + 1/3 p m = 1/3 ( p f + p m ) = 1/3(2P + H + R)   50

51. Sex-linked genes If the gene frequencies among males and among females are different, the population is not in equilibrium. The gene frequency in the population as a whole does not change, but its distribution between the two sexes oscillates as the population approaches equilibrium. 51

52. Sex-linked genes The reason for this is that, males get their sex linked genes only from their mothers; therefore p m is equal to p f in the previous generation. Females get their sex linked genes equally from both parents, therefore p f is equal to the mean of p m and p f in the previous generations (Figure). 52

53. Sex-linked genes Progeny generation p’ m = p f p’ f = 1/2 (p m + p f ) The difference between the frequencies in the two sexes is P’ f – p’ m = 1/2 (p m + p f ) _ p f = -1/2 ( p f - p m )   That is half the difference in previous generation but in another direction. 53

54. Random mating two loci are involved Suppose there is locus with two alleles A 1 & A 2, and suppose that locus A is not sex- linked, if the frequency of A 1 and A 2 alleles are p 1 and p 2 respectively with p 1 + p 2 = 1. Then Hardy - Weinberg law tells us that genotypes A 1 A 1 , A 1 A 2 and A 2 A 2 will be found in the proportion p 1 2 , 2p 1 p 2 and p 2 2 respectively provided that mating is at random. 54

55. Random mating two loci are involved If we also consider another autosomal locus with alleles B 1 and B 2 at frequencies of q 1 and q 2 respectively where q 1 + q 2 = 1, then according to the Hardy Weinberg Law the genotypes B 1 B 1 , B 1 B 2 and B 2 B 2 will occur in the proportions q 1 2 2q 1 q 2 and q 2 2 respectively provided that mating is at random. 55

56. Random mating two loci are involved In this case, A 1 allele will be in random association with the A 2 allele, and B 1 allele will be in random association with the B 2 allele. The alleles in A locus may fail to be in random association with the alleles at the B locus.   56

57. Random mating two loci are involved A 1 (p 1 ) A 2 (p 2 ) B 1 (q 1 ) A 1 B 1 (p 1 q 1 ) A 2 B 1 (p 2 q 1 ) B 1 (q 2 ) A 1 B 2 (p 1 q 2 ) A 2 B 2 (p 2 q 2 ) 57

58. Random mating two loci are involved Expected gametic frequencies   Frequency of gametic types with linkage equilibrium   A 1 B 1 = (p 1 q 1 ) A 1 B 2 = (p 1 q 2 ) A 2 B 1 = (p 2 q 1 ) A 2 B 2 = (p 2 q 2 ) where p 1 + p 2 = 1 and q 1 + q 2 = 1 58

59. Random mating two loci are involved When the alleles at two loci are in random association, the frequency of a gamete carrying any particular combination of alleles is equal to the product of the allele frequencies, of those alleles. 59

60. Random mating two loci are involved Loci that are in random association are said to be in state of LINKAGE EQUILIBRIUM . Loci not in random association are said to be in LINKAGE DISEQUILIBRIUM .   The rate of approach to linkage equilibrium depends on the relative proportions of the various types of gametes that can be formed by an individual heterozygous for two loci. 60

61. Random mating two loci are involved Consider an individual of genotype A 1 A 2 B 1 B 2 – it is heterozygous for two loci. This genotype symbolism fails to distinguish between two cases.   The individual could have formed by the union of A 1 B 1 and A 2 B 2 gametes and this will produce A 1 B 1 /A 2 B 2 genotype.  The individual could have formed by union of an A 1 B 2 with an A 2 B 1 gamete – A 1 B 2 /A 2 B 1   61

62. Random mating two loci are involved Suppose the true genotype of individual is A 1 B 1 /A 2 B 2   This will produce four types of gametes.   1. A 1 B 1 2. A 2 B 2 3. A 1 B 2 4. A 2 B 1 Gametic types (1) and (2) are know as NON – RECOMBINANTS . Their alleles are associated in the same manner as in previous generation. Gametic types (3) and (4) are known as RECOMBINANT gametes – because their alleles are associated differently than in the previous generation. 62

63. Random mating two loci are involved The gametic frequency of non-recombinant types will be equal and that of recombinant types will also equal. The overall frequency of recombinant type (3) + (4)) will not be equal to the overall frequency of non - recombinant gametes type (1) + (2). 63

64. Random mating two loci are involved The proportion of recombinant gametes produced by an individual is termed as RECOMBINANT FRACTION   64

65. Random mating two loci are involved Example: A 1 B 1 / A 2 B 2 produces gametes A 1 B 1 , A 2 B 2, A 1 B 2 and, A 2 B 1 in the proportions 0.38, 0.38. 0.12 and 0 .12 respectively. The recombination fraction between loci is r = 0.12 + 0.12 = 0 .24. 65

66. Random mating two loci are involved The recombination fraction between two loci depends on the position of the loci on the chromosome. If they are on different chromosome the recombination fraction is r = 0.5 i.e. the four gametic types will be produced in equal frequency. 66

67. Random mating two loci are involved If the two are on the same chromosome the recombination fraction depends on their distance apart. The closer the two loci are the less likely that the breakage and reunion will occur between them, the further apart are the more likely that such event will occur. 67

68. Random mating two loci are involved If r = 0, this means that the two loci are very close together that the break never occur between them. The largest possible recombination fraction is r = 0.5 which occurs when loci are very far apart-on the same chromosome. 68

69. Non - random mating There are two major forms of non-random mating   When mated individuals are related to each other by ancestral descent. This increases the frequency of homozygotes at all loci. ( Inbreeding ).  When individuals tend to mate preferentially with respect to their genotypes at any particular locus under consideration. 69

70. Non - random mating There are two types of preferential mating .   ( i ) Assortative mating   Mated pairs are of the same phenotype more often than would occur by chance.   (ii) Disassortative mating   Mated pairs are of the same phenotype less often than would occur by chance. 70

71. Non - random mating The effect of assortative mating is to increase the frequencies of homozygotes and reduce that of heterozygotes.   If assortative mating is continued in successive generation, the population approaches on equilibrium of which genotypes frequencies remain constant. 71

72. Non - random mating Disassortative mating leads to an increase in heterozygotes and reduction of homozygotes. It also had an effect of changing gene frequency. If the mating is predominantly between unlike phenotypes, the rarer phenotype has a better chance of success in mating than has the commoner types. As a result the rarer alleles are favoured and gene frequency changes toward intermediate values at which the phenotypes are equal in frequency. 72

73. Effects of migration, mutation and selection on gene frequencies 1. Migration: Migration describes the effect of introducing previously non-interbreeding genotypes into breeders' populations. Migration can be responsible for introducing alleles previously absent from the population and for changing the frequency upwards or downward of alleles already present. 73

74. Effects of migration, mutation and selection on gene frequencies Suppose that a large population consist of a proportion of ‘m’ of new (individuals) immigrants in each generation 1-m being the native.   If the frequency of a certain gene is q m among the immigrants and q o among the natives. Then the frequency of the gene in a mixed population (q 1 ) will be: q 1 = mq m + (1 - m) q o = m ( q m - q o ) + q o 74

75. Effects of migration, mutation and selection on gene frequencies The change of gene frequency  q, brought about by one generation of immigration is the difference between the frequency before immigration and the frequency after immigration.    q = q 1 - q o = m( q m - q o ) 75

76. Effects of migration, mutation and selection on gene frequencies Change of gene frequency due to immigration will depend on: Rate of immigration (b) Difference in gene frequency between immigrants and natives . 76

77. Effects of migration, mutation and selection on gene frequencies 2. Mutation: Mutations that are very rare do not produce permanent change in a large population. But mutations that occur repeatedly produce permanent change in the population. Types of mutation:   Non recurrent mutation Recurrent mutation 77

78. Effects of migration, mutation and selection on gene frequencies Suppose gene A 1 mutates to A 2 with a frequency of u per generation. If the frequency of a A 1 in one generation is p o the frequency of new mutated A 2 genes in next generation will be up o .   New gene frequency is p o - up o and change of gene frequency is - up o   i.e. p o - ( p o - up o ) = p o - p o - up o = - up o 78

79. Effects of migration, mutation and selection on gene frequencies If a gene mutates in both directions  A 1 to A 2 A 2 to A 1   If the initial gene frequency A 1 = p o and A 2 = q o A 1 mutates to A 2 at a rate u per generation and A 2 to A1 at a rate v. Then after one generation there is a gain of A 2 genes equal to up o due to mutation in one direction and loss equal to vq o due to mutation in another direction. Mutation rate A 1 u A 1 A 2 A 1 A 2 v 79

80. Effects of migration, mutation and selection on gene frequencies Initial gene frequencies p o and q o The change of gene frequency in one generation is The change of q = up o - vq o This situation lead to an equilibrium in gene frequency at which no further changes takes place - This is so because if the frequency of one allele increases fewer of the other are left to mutate in that direction and more are available to mutate in other direction. 80

81. Effects of migration, mutation and selection on gene frequencies At equilibrium : pu = qv p/q = v/u q = u/u+ v   Mutation rates are generally very low about 10 -5 or 10 -5 per generation for most loci in most organisms. Therefore mutation alone can produce only very slow changes in gene frequency.   In evolutionary time scale they can be important   Reverse mutation is usually less frequent than forward mutation. 81

82. Effects of migration, mutation and selection on gene frequencies 3. Selection: The effect of selection on changing gene frequency is of primary interest to plant breeders.   In this case the fact that individuals differ in viability and that they contribute different numbers of offspring to the next generation have to be taken into account .   The contribution of offspring to the next generation is called fitness of individuals or adaptive value or selective value. 82

83. Effects of migration, mutation and selection on gene frequencies When a gene is subjected to selection its frequency in the offspring will not be the same as in the parents since parent of different genotypes pass on their genes unequally to the next generations. This is the reason why selection causes change in gene frequency and consequently also to genotype frequency. 83

84. Effects of migration, mutation and selection on gene frequencies The differences of fitness that result to the selection are the aspect of phenotype that is why the changes in gene frequency from selection are more complicated than that resulting from mutation.   The degree of dominance shown by gene is important i.e. dominance with respect to main effects of gene. 84

85. Effects of migration, mutation and selection on gene frequencies Suppose there are genes A 1 and A 2 and selection is against A 2   NO DOMINANCE    A 2 A 2 A 1 A 2 A 1 A 1 __________________________________________ 1-s 1-1/2 s 1 85

86. Effects of migration, mutation and selection on gene frequencies PARTIAL DOMINANCE A 2 A 2 A 1 A 2 A 1 A 1 _______________________________________ 1-s 1 – hs 1 86

87. Effects of migration, mutation and selection on gene frequencies COMPLETE DOMINANCE A 1 A 2 A 2 A 2 A 1 A 1 __________________________________________ 1-s 1 87

88. Effects of migration, mutation and selection on gene frequencies OVERDOMINANCE A 2 A 2 A 1 A 1 A 1 A 2 _______________________________________ 1-s 1 1-s 2 1 88

89. Effects of migration, mutation and selection on gene frequencies Selection against a particular gene - which may result elimination of one genotype of another operates through.   Reduced viability Reduced fertility   89

90. Effects of migration, mutation and selection on gene frequencies The strength of selection is expressed as the coefficient of selection (s) = A proportionate reduction in the gametic contribution of a particular genotype compared to standard genotype (the most favoured).   The contribution of favoured genotype = 1 The contribution of genotype selected against is 1-s   This expresses the fitness of one genotype relative to the other. 90

91. Effects of migration, mutation and selection on gene frequencies Example: If the coefficient of selection s = 0.2 the fitness then will be 0.8 which means for every 100 zygotes produces by favoured genotype only 80 are produced by genotype selected against. 91

92. Effects of migration, mutation and selection on gene frequencies The fitness of genotype with respect to any particular locus is not necessarily the same in all individuals. It depends on 1) Environmental situation in which the individuals live 2) Genotype with respect to genes at other loci.   92

93. Effects of migration, mutation and selection on gene frequencies Change of gene frequency under selection: 1. Selection against recessive Suppose there are genes A 1 & A 2 and the initial gene frequencies are under Hardy - Weinberg equilibrium. 93

94. Effects of migration, mutation and selection on gene frequencies _________________________________________________________ Genotypes   A 1 A 1 A 1 A 2 A 2 A 2 Total _________________________________________________________   Initial frequencies p 2 2pq q 2 1 Coefficient of selection 0 0 s Fitness 1 1 1-s Genetic contribution p 2 2pq q 2 (1-s) 1-sq 2 _________________________________________________________ 94

95. Effects of migration, mutation and selection on gene frequencies Multiplying initial gene freq. by the fitness value gives the frequency of genotypes after selection - Gametic contribution.   After selection the total frequency is not a unit (1) because there has been proportionate loss of sq 2 due to selection.   The interest of a plant breeder will be to find out the frequency of A 2 gametes produced and the frequency of A 2 genes in the progeny. 95

96. Effects of migration, mutation and selection on gene frequencies A 2 = A 2 A 2 + 1/2 A 1 A 2 ---------------------------- A 1 A 1 + A 1 A 2 + A 2 A 2 A 2 = q 1 = new gene frequency   q = q 2 (1-s) + pq ------------------= This is the new gene frequency 1 - sq 2 96

97. Effects of migration, mutation and selection on gene frequencies It can be further simplified by substituting p = (1 - q)   q 1 = q - sq 2 -------- 1 - sq 2   The change in gene frequency q after one generation of selection will be: q = q 1 – q   The change of q = sq 2 (1-q) ---------- 1 -sq 2 97

98. Effects of migration, mutation and selection on gene frequencies The effect of selection on gene frequency depends not only on the intensity of selections , but also on the initial gene frequency . Expressions for the new gene frequency and for the change of gene frequency with different conditions of dominance, are given in the table (handout). 98

99. Effects of migration, mutation and selection on gene frequencies Number of generations required for effective selection: Normally a plant breeder would like to know how many generations of selection would be needed to effect a specified change of gene frequency?   If we consider the case of selection against a recessive when elimination is complete i.e. s = 1 and there is no mutation ( eg . recessive lethal).  s = 1   Using equation q 1 = q – sq 2 -------- 1 – sq 2 99

100. Effects of migration, mutation and selection on gene frequencies and writing q o , q 1 , q 2 ---- qt for the gene frequency after 0, 1, 2 ---- t generation   q o q 1 = ----------- 1+ q o   q 1 q 2 = ----------- 1+ q 1     100

101. Effects of migration, mutation and selection on gene frequencies q o q t = ----------- 1+ tq o   t = number of generations q o , q t = The number of generations (t), required to change the gene frequency from q o to q t is:   t = q o - q t --------- q o q t   = 1/ q t – 1/ q o 101

102. Effects of migration, mutation and selection on gene frequencies Example: Suppose you want to eliminate disease susceptible plants ( rr ). If there is one out of four susceptible i.e. q 2 = ¼ therefore q = √ ¼ = ½ Objective is to reduce to one out of 16 i.e. q t 2 = 1/16 therefore qt = √1/16 = ¼ t = q – q t /q q t = 1/q t -1/q = 1/1/q t - 1/1/q = 1/ ¼ - 1/ ½ = 4- 2 = 2 t= 2 generations 102