Lecture 3 Gene Segregation and Interaction PDF

LECTURE 3 GENE SEGREGATION AND INTERACTION 1  MENDELIAN GENETICS Two Alleles Multiple Alleles A...

LECTURE 3 GENE SEGREGATION AND INTERACTION 1  MENDELIAN GENETICS Two Alleles Multiple Alleles ABO Blood Groups Law of Segregation Law of Independent Assortment (Monohybrid) (Dihybrid) Complete Modified ratios 9:3:3:1 Modified Ratios dominance (3:1) Non-Allelic Interactions Lethal Genes Pseudoalleles Allelic Interactions Novel Epistasis Complementary Duplicate Phenotype Gene Action Gene Action Incomplete dominance Over-dominance Co-dominance Dominant Recessive 2 Definition of Terms gene -inherited factor on the chromosome responsible for a trait locus -location of a gene on a chromosome 3 genotype - genetic constitution of an individual allele - alternative forms of a gene Ex. Gene for flower color - allele for purple color - allele for white color 4 phenotype - physical, physiological, biochemical and behavioral traits of an individual - determined by its genotype and its interaction with the environment 5 dominant - gene exerting full effect despite the presence of another allele of the same gene 6 recessive - gene not expressed in the presence of another allele homozygous - 2 copies of the same allele of a gene (e.g. YY, yy) 7 heterozygous - 2 different alleles of the same gene (e.g. Yy) 8 9 www.s-cool.co.uk 10 Hybridization - cross between 2 individuals with contrasting traits Ex. Purple x White Yellow x Green 11 F1 or first filial generation - first generation produced after mating between parents that are homozygous for different alleles 12 F2 generation or second filial generation - the generation produced by self fertilization or sib-mating of F1 individuals 13 Backcrossing - the cross of a heterozygote with one of its parents 14 Recall: Pre-Mendelian Heredity - thought of as a blending process - mixture of characteristics of both parents 15 …but blending process cannot explain if offspring is similar to one parent 16 All about Gregor Mendel 17  Mendel used self-pollinated plants, green peas (Pisum sativum) Mendel used pure line or true breeding parents with contrasting traits. homozygous plants 18 True breeding parentals 19  Mendel emasculated the female parent to prevent selfing. - Pollen is brushed on the stigma of the female flower. 20 Emasculation 21 https://www.google.com/imgres 22  presence of genetic determinant (or a hereditary factor) for a specific character passed from one generation to the next no blending with other genetic determinants 23  PP x pp Gametes P p F1 Pp To get F2: Pp x Pp Gametes: P P p p 24 Punnett Square Gametes P p P PP Pp p Pp pp 25 Genotypic Ratio (Pp x Pp): 1PP : 2 Pp : 1pp 1purple 2 purple 1 white Phenotypic Ratio: 3 purple : 1 white 26 Complete dominance - one dominant allele is enough to express the dominant trait - homozygous dominant and heterozygote have the same phenotype 27 28  Law of Segregation Alleles in a gene pair separate cleanly from each other during meiosis. AA Aa aa A A A a a a 29 Parentals: AA x Aa Gametes: A A A a Gametes A a A AA Aa A AA Aa 30 Genotypic Ratio: 2 AA : 2 Aa or 1 AA : 1 Aa 31  AA x Aa A 1 AA A a 1 Aa 32 Parentals: Aa x aa Gametes: A a a a Gametes a a A Aa Aa a aa aa 33 Genotypic Ratio: 2 Aa : 2 aa or 1 Aa : 1 aa 34 https://www.google.com/imgres?i 35 round vs. wrinkled seeds 36  Mendel’s Observation (Fairbanks, D.J. and W.R. Anderson, 1999) Phenotypes of F1 F2 Phenotypes F2 True breeding Phenotypes Ratio Parents Round Wrinkled All round 5472 round 2.96 : 1 seeds seeds 1850 wrinkled Yellow Green All yellow 6022 yellow 3.01 : 1 seeds seeds 2001 green Purple White All purple 705 purple 3.15 :1 224 white Inflated Constricted All inflated 882 inflated 2.95: 1 pods pods 299 constricted 37 Phenotypes of F1 F2 F2 true breeding phenotypes phenotypes Ratio parents Green Yellow All green 428 green 2.82 : pods pods 152 yellow 1 Axial Terminal All axial 651 axial 3.14 : flowers flowers 207 1 terminal Tall Dwarf All tall 787 tall 2.84 : plants plants 277 dwarf 1 38  Dihybrid Cross Consider two traits at the same time seed shape - round vs. wrinkled seed color – yellow vs. green 39 True breeding parentals round, yellow x wrinkled, green RRYY rryy RY ry F1 RrYy F2 RrYy x RrYy 40 Mendel’s Results: yellow, round 315 yellow, wrinkled 101 green, round 108 green, wrinkled 32 Total 556 41 Take each trait independently: Yellow 315 green 108 101 32 Total 416 Total 140 416/140 = 2.97 : 1 42 Round 315 wrinkled 101 108 32 Total 423 Total 133 423/133 = 3.18 43 Parentals: RRYY x rryy Gametes: RY ry F1: RrYy 44 To get F2 : RrYy x RrYy Gametes: Y R RY y Ry Y r rY y ry 45  Punnett Square Gametes RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy 46 47 F2 genotypic and phenotypic ratios: Genotypic ratio Phenotypic Ratio 1 RRYY 1 round, yellow 2 RRYy 2 round, yellow 1 RRyy 1 round, green 2 RrYY 2 round, yellow 4 RrYy 4 round, yellow 2 Rr yy 2 round, green 1 rrYY 1 wrinkled, yellow 2 rrYy 2 wrinkled, yellow 1 rryy 1 wrinkled, green 48  Summarized form of genotypic ratio 9 R_Y_ 9 round, yellow 3 R_yy 3 round, green 3 rr Y_ 3 wrinkled, yellow 1 rr yy 1 wrinkled, green 49 Round, yellow round, green RRYY RRyy RRYy Rryy RrYY RrYy wrinkled, yellow wrinkled, green rrYY rryy rrYy 50  Phenotypic Ratio 9 round, yellow 3 round, green 3 wrinkled, yellow 1 wrinkled, green Total 16 Phenotypic Ratio: 9 3 3 1 16 16 16 16 51 Law of Independent Assortment Alleles of different gene pairs separate independently from each other and randomly combine during meiosis 52 Genotypes AABB aabb Anaphase I A A a a B B b b Gametes: AB ab 53  Obtaining Gametes: AABB aabb B --- AB b --- ab A a B --- AB b --- ab B --- AB b --- ab A a B --- AB b --- ab 54  AaBb Anaphase I A a A a B b b B 55 Alleles randomly combine B ---- AB A b ---- Ab B ---- aB a b ---- ab 56  Dichotomous Branching RrYy x RrYy Take each gene pair independently (TEGI): Rr x Rr --- 1 RR : 2 Rr : 1 rr Yy x Yy --- 1 YY : 2 Yy : 1 yy 57  1YY --- 1RRYY 2Yy --- 2RRYy 1 RR 1yy --- 1RRyy 1YY --- 2 RrYY 2Yy --- 4 RrYy 2Rr 1yy --- 2 Rryy 1YY--- 1 rrYY 2Yy --- 2 rrYy 1rr 1yy --- 1 rryy 58  Short cut in finding phenotypic ratio: TEGI: Rr x Rr = 1RR : 2 Rr : 1rr 3 round : 1 wrinkled Yy x Yy = 1YY : 2 Yy :1yy 3 yellow :1 green 59  3 yellow- 9 round, yellow 3 round 1 green - 3 round, green 3 yellow- 3 wrinkled,yellow 1 wrinkled 1 green - 1 wrinkled, green 60 Gene Segregation in Haploids 61 Consider one gene: c x c+ c : c : c :c : c+ : c+ : c+ : c+ 4 : 4 + c c 8 8 1 : 1 + c c 2 2 62 63  For two genes mt+c+ x mt-c Take each gene pair independently mt+ x mt- c+ x c ½ mt+ : ½ mt- ½ c+ : ½ c 64  ¼ mt +c+ ½ c+ -- ½ mt+ ½c -- ¼ mt+c ½ c+ -- ¼ mt-c+ ½ mt- ¼ mt -c ½ c -- 65  Exercises: 1. Enumerate the gametes that can be derived from the ff: a. AaDd b. AaDdHh c. AAHHGgLL 66 2. Provide the genotypic ratios. Use dichotomous branching method. a. AaBb x aaBb b. BBAadd x BbaaDD 67 3. Provide the phenotypic ratios. Use dichotomous branching method. Given: P-- purple vs. pp -- white T-- tall vs. tt -- dwarf a. PpTt x PPTt b. ppTT x PpTt 68  If two pairs of contrasting traits are inherited independently, to predict the frequencies of F2 phenotypes - apply Product Law of Probabilities 69 For simultaneous occurrence of two independent events, the combined probability of the two outcomes is equal to the product of their individual probabilities 70  RrYy x RrYy TEGI: Rr x Rr --1RR : 2Rr : 1rr 3/4 round : 1/4 wrinkled Yy x Yy --1YY : 2 Yy : 1yy 3/4 yellow :1/4 green 71  ¾ yellow - 9/16 round, yellow ¾ round ¼ green - 3/16 round, green ¾ yellow - 3/16 wrinkled, yellow ¼ wrinkled ¼ green - 1/16 wrinkled, green 72 Example: Determine the probability that a plant with genotype CcAa will be produced by the given cross. Cc Aa x CcAA 73 TEGI: Cc x Cc = 1/4 CC : 2/4 Cc : 1/4 cc 1/2 Cc Aa x AA = 1/2 AA : 1/2 Aa p = (1/2 Cc) (1/2 Aa) p = 1/4 CcAa 74 If two events are not independent  the likelihood of an outcome is referred to as - Conditional Probability 75  What is the probability that one outcome will occur given the specific condition upon which this outcome is dependent? 76 Use the formula: n! P= w! x! pw qx n = total no. of progeny w = no. of progeny with genotype or phenotype p x = no. of progeny with genotype or phenotype q p = probability of genotype/phenotype w q = probability of genotype/ phenotype x 77  In a sibship of 8 children, what is the probability of having 5 boys and 3 girls? 8! 1 5 1 3 = 5! 3! 2 2 8.7.6.5.4.3.2.1 1 1 = 5.4.3.2.1.3.2.1 32 8 336 1 1 = 6 32 8 56 1 1 56 7 = 32 8 = 256 or 32 78 Albinism is due to the presence of homozygous recessive genes. Two parents heterozygous for the gene have 5 children, what is probability that 3 will be normal? 79  Aa x Aa = 1 AA : 2 Aa : 1 aa 3/4 normal : 1/4 albino = 5! 3 3 1 2 3! 2! 4 4 = 5.4.3.2.1 27 1 3.2.1.2.1 64 16 27 270 = 10 = 1024 1024 80  Take Note! Please review the concepts of Independent Segregation and Assortment Practice obtaining gametes, deriving genotypic and phenotypic ratios from a cross (monohybrid or dihybrid) Solve problems on product law of probabilities, and conditional probabilities 81  Discoveries before Chromosomal Theory of Inheritance (before 1903) 1865 Mendel’s Principles 1871 Friedrich Miescher isolated nuclein from nuclei of pus cells 82  1875 O. Hertwig nucleus required in cell division and fertilization 1882-1885 E. Strassburger and Walter Fleming the chromosomes are in the nucleus 83  1900 Hugo de Vries, Carl Correns, Erich von Tschermak confirmed Mendel’s principles in plants 1902 William Bateson, E.Rebecca Saunders, Lucien Cuenot confirmed Mendel’s principles in animals 84 1903 Walter Sutton and Theodor Boveri Chromosome Theory Inheritance resemblance between Mendelian factors and chromosomes 85 Correlations between Chromosomes and Mendelian Factors 1. Chromosomes exist in pairs. Mendelian factors exist in pairs. maternal and paternal origin 2. Homologous chromosomes separate at anaphase I. Mendelian factors separate at anaphase I. 86 Correlations between Chromosomes and Mendelian Factors… 3. Fertilization restores the diploid chromosome number. Alleles of a gene also pair up. 2n x 2n AA x aa n n A a 2n Aa 87 88 89 90 91 92  Types of Dominance Relationships 1. Complete dominance AA and Aa have the same phenotype the presence of a dominant allele is enough to express the dominant trait 93 1. Complete dominance… F2 genotypic ratio: 1AA : 2Aa : 1aa F2 phenotypic ratio 3 : 1 94 2. Incomplete dominance F1 phenotype is intermediate F2 genotypic ratio: 1AA : 2Aa : 1aa F2 phenotypic ratio: 1 : 2 : 1 95 To illustrate incomplete dominance: Red x white Parents: RR rr F1 : Rr pink F2 1RR : 2 Rr : 1 rr 1red : 2 pink : 1 white 96 Incomplete dominance 97 (https://slideplayer.com/slide/15142999/) 98 3. Overdominance Aa is superior compared to AA and aa. heterosis or hybrid vigor (https://bohatala.com/what-is-heterosis/) 99 4. Co-dominance the products of the two alleles in the heterozygote are present (https://slideplayer.com/slide/8802115) 100 101  Multiple Alleles More than two alleles at a single locus In a diploid individual: Maximum of two alleles One on each of the homologous chromosomes 102 Example: ABO blood groups - discovered by Karl Landsteiner - early 1900s Blood A B AB O Type Genotype AA BB AB OO AO BO Antigen A B A and B H 103  Glycolipid (oligosaccharide + lipid) is present on the surface of RBC. The difference in oligosaccharide is the reason behind ABO blood types. 104  oligosaccharides (antigen) synthesis of oligosaccharides involves many steps 105  Last two steps in oligosaccharide synthesis H antigen (fucose)5th A antigen B antigen (N-acetyl glucosamine)6th (galactosamine)6th 106 107  Blood Surface Blood Type of Type antigen on Compatible of RBC Donor Recipient Type A A antigen Type A or Type O Type B B antigen Type B or Type O Type AB A+B Type A, B, AB or antigen O Type O H antigen Type O 108 Supposing… a paternity suit was filed in court. Father Mother Type B Type A Son Type O Is this possible? 109 Father x Mother Type B Type A BB AA BO AO (Thus, they can beget: AB, BO, AO, OO) 110  Lethal Genes genes that can cause death Two types: Recessive lethal Dominant lethal 111  A. Recessive lethal gene lethal when homozygous recessive could result to a recognizable phenotype when heterozygous 112 Example 1: wild type x yellow YY Yy 1YY : 1Yy 1 wild : 1 yellow 113 Cross between two carriers: yellow x yellow Yy Yy 1 YY : 2Yy : 1yy 1 wild : 2 yellow : die 1 : 2 : 0 114 Example 2: Manx allele (Mnl) abnormal spine development extreme development abnormality causes death of the embryo Manx cat (tailless) Heterozygous MnLMnl 115 Example 3: Tay-Sachs disease homozygous recessive; normal at birth deterioration of the central nervous system starts before one year old Loss of neuromuscular control; blindness 116  Lack of hexosaminidase A Accumulation of GM2 gangliosides (lipids in the brain and nerve cells) Usually fatal at three to four years old Matsushita et al., Neuroscience 414, 21 August 2019, Pages 128-140 117 Ex. 4: Xeroderma pigmentosum lacks DNA repair enzyme photosensitive if exposed to light: intense pigmentation freckling warty growths (may become malignant) 118 (http://www.pcds.org.uk/clinical-guidance/xeroderma-pigmentosum) 119  A. Dominant lethal gene (0:1) lethal when homozygous dominant or heterozygous example: Huntington’s disease the dominant allele codes for an abnormal huntingtin protein 120  Described by George Huntington in 1872 progressive degeneration of the CNS; involuntary movements onset of symptoms at 30 y.o. or earlier death at 40-50 y.o. 121  Modifier Genes gene changes phenotypic effect of other genes in a quantitative fashion dilution or enhancement effect 122 Example in tomato genotype color lycopene: -carotene R_T_ red 10 : 1 R_T_Mo orange 1 : 1 123  Red Orange tomatoes tomatoes 124  Gene Interaction non allelic interaction of two or more genes which results in a modified phenotypic ratio the interaction between two or more genes determine a single phenotype 125  Epistasis interaction of two or more genes determined by observing certain phenotypic ratios in the progeny of heterozygous parents 126  Types of Non-allelic Interactions 1. Novel phenotypes F2 ratio: 9:3:3:1 Example 1: comb type in poultry R_ = rose is dominant to non-rose (rr) P_ = pea is dominant to non-pea (pp) 127 F1 RrPp F2 9 R_P_ 9 walnut 3 R_pp 3 rose 3 rrP_ 3 pea 1 rrpp 1 single 128 129 Example 2: Fruit color in bell pepper Y_ = elimination of chlorophyll (no green color) yy = chlorophyll formation (green color) R_ = red color rr = yellow carotenoids 130 F1 YyRr F2 9 Y_R_ 9 red 3 Y_rr 3 yellow 3 yyR_ 3 brown (green+red) 1 yyrr 1 green 131  2. Recessive epistasis F2 ratio: 9 : 3 : 4 homozygous recessive gene hides the effect of the other gene Example 1: Coat color in mouse 132 A_ = agouti is dominant to black (aa) C_ = color expression is dominant to color inhibition (cc) 133  F1 AaCc F2 9 A_C_ 9 agouti 9 3 A_cc 3 albino 3 aaC_ 3 black 3 4 1aacc 1 albino 134 Black Agouti (alternate dark and light bands) Albino 135  Example 2: Coat color in Labrador (black, chocolate and yellow) B_ = codes for enzyme for melanin pigment synthesis bb = reduced pigment synthesis E_ = high pigment deposition in the hair ee = low pigment deposition 136  F1 BbEe F2 9 B_E_ 9 black 9 3 B_ee 3 yellow 3 bbE_ 3 chocolate 3 4 1 bbee 1 yellow 137  Coat color in Labrador Black Chocolate Yellow 138 www.cubocube.com 139  3. Dominant epistasis Case 1 F2 ratio 12 : 3 : 1 dominant gene masks the expression of the other gene 140 Example: Fruit color in summer squash W_ = white is dominant to color (ww) Y_ = yellow is dominant to green (yy) 141  F1 WwYy F2 9 W_Y_ 9 white 3 W_yy 3 white 12 3 wwY_ 3 yellow 3 1 wwyy 1 green 1 142 (Holisticfitnesslondon.co.uk) 143 Case 2: Dominant epistasis ( F2 13 :3) one gene when dominant is epistatic to the second; the second gene when homozygous recessive is epistatic to the first 144 Example: Feather color in poultry I_ = color inhibition is dominant to color appearance (ii) C_ = color is dominant to white (cc) 145  F1 IiCc F2 9 I_C_ 9 white 3 I_cc 3 white 13 3 ii C_ 3 colored 3 1 iicc 1 white 146 4. Complementary gene action F2 9:7 either gene when homozygous recessive is epistatic to the other gene 147 Example: Flower color in pea P_ = purple is dominant to white (pp) C_ = color is dominant to non-color (cc) 148  F1 PpCc 9 P_C_ 9 purple 9 3 P_cc 3 white 3 ppC_ 3 white 7 1 ppcc 1 white 149 5. Duplicate gene action F2 15 :1 either gene when dominant is epistatic to the other gene 150 Ex. shape of seed capsule in Bursa A_ = triangular is dominant to ovoid (aa) B_ = triangular is dominant to ovoid (bb) 151  F1 AaBb F2 9 A_ B_ 9 triangular 3 A_bb 3 triangular 15 3 aaB_ 3 triangular 1 aabb 1 ovoid 1 152 Capsella bursa-pastoris (L.) Medik. http://www.friendsofthewildflowergarden.org/ 153 https://www.healthbenefitstimes.com 154 Sample Problem: Give the genotypes and phenotypes of parents in the cross: 99 gray coat, brown eyes, normal 44 white coat, brown eyes, dwarf 33 black coat, brown eyes, normal 155 Take each phenotype independently: Coat color gray coat 99 white coat 44 black coat 33 Ratio 9 : 3 : 4 thus AaBb parents 156  Eye color brown eyes 99 44 33 176 All brown eyes, thus CC parents 157  Height normal 99 dwarf 44 33 132 Ratio 132 : 44 or 3:1 Thus Nn parents 158 In summary… Genotype of the parentals: AaBb CC Nn Phenotype: gray coat, brown eyes, normal 159  Pseudoalleles Lewis 1951 Star-asteroid in Drosophila Star and star recessive (ast) two different mutants located on the same chromosome 160 S AST Cis Normal s ast eyes (blog.wellcome.ac.uk) S ast Trans Reduced s AST eyes 161  Lewis effect or position effect phenotype is not only dependent on the genotype but also on the position of the genes on the chromosome 162  Environmental Influence on Gene Expression Phenotype = Genotype + Environment or P = G + E different genes different factors development 163 Penetrance proportion of genotype that shows the expected phenotype 164 a. complete penetrance - all will show the trait (100%) b. incomplete penetrance - not all will show the trait 165 Expressivity degree in which a particular phenotypic effect is exhibited by an individual a. constant expressivity b. variable expressivity 166 Variable expressivity in polydactyly 167 (or Constant expressivity) (https://courses.lumenlearning.com/wm- biology1/chapter/reading-penetrance-and-expressivity/) 168 Pleiotropy one gene has multiple phenotypic effects example: Sickle cell anemia 169 170  Sickle cell anemia  genetically based human disease normal sickled RBC RBC 171 172 Phenocopy an environmental mimic of gene action an environmental factor induces a particular phenotype that resembles a genetically determined phenotype 173 Thalidomide – drug to cure morning sickness Phocomelia – underdeveloped limbs 174 175 176 BEFORE AFTER 177 Genetics will tell the truth. 178 Environmental factors responsible for differences in penetrance & expressivity 1. external environmental a. temperature b. Light c. nutrition d. maternal relations 179 2. Internal environment a. age b. sex - sex limited - sex influenced c. substrates 180 Twin studies in humans – help determine the effect of genes and environment: Identical twins - identical genotype Fraternal twins - dissimilar genotypes 181 Greater phenotypic similarities among identical than fraternal twins - due to similar genotypes of IT 182 Phenotypes are the same for IT and FT - greater environmental influence than genotypic influence 183  Role of environment and heredity Concordant both members or twins show or don’t show the trait Discordant only one member shows the trait 184  High hereditary influence - high concordance in identical twins - low concordance in fraternal twins 185  Low hereditary influence & high environmental influence - equal concordance and discordance between identical and fraternal twins 186 Genetic concordance among identical and fraternal twins Trait % Concordance Identical Fraternal age of sitting 82 76 Hair color 89 22 Feeble 94 47 mindedness Schizophrenia 80 13 Criminality 68 28 187