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by: Ms. Griffin Franecki

Genetics BIOL 222

Ms. Griffin Franecki
Penn State
GPA 3.99


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This 0 page Study Guide was uploaded by Ms. Griffin Franecki on Sunday November 1, 2015. The Study Guide belongs to BIOL 222 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 45 views. For similar materials see /class/233054/biol-222-pennsylvania-state-university in Biology at Pennsylvania State University.

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Date Created: 11/01/15
Chapter 1Genetics and the Organism 1831051 Genetics Study of gene function from molecules of DNA to the gene pool of a population DNA contains ALL of the information for building an organism Central Dogma of Molecular Biology Information Flow DNA9 RNA9 Protein Genes Functional units of DNA Individuals inherit genes Individuals contribute to the gene pool of a population Alleles Different but related forms of the same gene Genotype The set of genes an organism inherits Very little change over time Phenotype Characteristics of an organism Phenotypes change throughout the life of an organism as its genes interact with the environment Phenotype is determined by the genotype as it interacts with the environment Interactions between genes and the environment determine what organisms become Geneticists study partial genotypes and partial phenotypes of organisms ie study the effect of one or a few genes A particular phenotype can be caused by more than one genotype AA dominant phenotype Aa dominant phenotype aa recessive phenotype Wild Type WT9 genotypic or phenotypic characteristics of natural populations or standard laboratory strains An individual that is not W is a variant or mutant Mutation is the basis for variation within a species and is the raw material for evolution Model Organisms Genetic analysis is greatly simplified by using an easily followed phenotype in an organism that is practical to study e g mice vs elephants 1 Bacteriophage and Viruses 2 Bacteria E coli and B subtilis 3 Yeast Bread and Beer 4 Caenorhabditis elegans round worm 5 Drosophila melanogaster fruit y 6 Maize corn 7 Arabidopsis thaliana small plant 8 Mus musculus mice Developmental Noise Random molecular events in cells of a developing organism that cause phenotypic variations Thus the genotype determines the phenotype however expression of the information in the genotype is in uenced by the environment developmental noise and other genes For Genetic Analysis 1 A phenotype that is easy to identify and monitor 2 A phenotype that is not highly in uenced by the environment control the environment 3 Try to determine the in uence of a single gene Genetic Dissection Identify a process to study Isolate mutants that interfere with the process Identify clone the WT genes Determine the function of the gene products Chapter 29Patterns of Inheritance 19205 1 Mendelian Analysis Analyzing hereditary information using Mendel s techniques and principles Gregor Mende191860s Conducted quantitative and systematic studies of inheritance Proposed the concept of the gene Particulate Inheritance Characteristics determined by discrete units that are inherited intact Mendel s Peas An individual pea plant produces both pollen and eggs Peas can self pollinate or be cross pollinated Each pea individual results from a separate fertilization event Selfpollination self9Pollen and egg from the same ower Both parents have the same genotype Crosspollination cross9Pollen from one plant fertilizes an egg from another plant Parents have different genotypes Pure Line All offspring produced by selfing or crossing individuals within the same line have the same phenotype P9Parental generation Individuals in the first cross of a particular series of experiments Fl First Filial Generation Progeny from the first cross Fz second Filial Generation Progeny from selfing individuals from the F1 generation D0minant9The phenotype that is expressed in the F1 progeny when two pure lines are crossed Recessive The phenotype that is covered up by the presence of the dominant phenotype Reappears in F2 Mendel used reciprocal crosses Figures 2 5 and 2 6 Experiment 1 P pure breeding X pure breeding Cross purple owers white owers xl F1 All purple flowers Reciprocal crosses gave the same result Purple is dominant and white is recessive F1 X F1 Self xl F2 31 ratio of purplewhite Recessive phenotype remained as a separate heritable trait Mendel studied 7 distinct characteristics or phenotypes Figure 2 4 Table 2 1 Experiment 2 P yellow X green cross xl F1 All yellow self xl F2 31 yellow green self 13 yellow F2 always gave rise to yellow 23 yellow F2 gave 31 yellow green ratio 1 1 green F2 always gave rise to green Underlying the 31 F2 phenotypic ratio is a 121 genotypic ratio of YYYyyy Mendel proposed that each gene is present twice gene pair Figure 2 7 Experiment 2 Again P YY yellow X yy green cross xl F1 All Yy yellow self xl F2 1 YY yellow 2Yy yellow lyy green Test Cross Crossing to a recessive individual F1 Yy X W xl 11 Yy Dominant Recessive yy Autosomes All chromosomes other than the sex chromosomes Alleles Different forms of the same gene Y and y Homozygous Both alleles of a gene pair are the same YY dominant or yy recessive Heterozygous Different alleles of a gene pair Yy dominant YY and Yy have different genotypes but the same phenotype Flower Color AA active enzyme purple pigment anthocyanin Aa active enzyme purple pigment aa no active enzyme white Mendel s First Law The two members of a gene pair segregate from each other into gametes so that one half of the gametes carry one member of the gene pair and the other half of the gametes carry the other member of the gene pair Chapter 29Patterns of Inheritance 19705 1 Monohybrid Cross Analyzing one characteristic gene Dihybrid Cross Analyzing two genes simultaneously Trihybrid Cross etc Dihybrid Cross Usually has 4 possible phenotypes 1 Both dominant 2 One dominant other recessive 3 Other recessive one dominant 4 Both recessive Phenotypic ratio of F2 progeny is 9331 Figure 2 10 Independent Assortment Different genes assort independently from one another Mendel s Second Law During gamete formation the segregation of alleles of one gene is independent of the segregation of alleles of another gene ie gene pairs on different chromosomes assort independently at meiosis Probability The number of times an event is expected to happen divided by the number of opportunities for an event to happen ie p xy Dice pgofa 3 16 Gametes if heterozygous Aa pA 12 pa 12 Product Rule The probability that two independent events will occur simultaneously is the product of their probabilities pof two 4s 1616 136 RrYy pgamete with 2 recessive alleles 121 2 14 Sum Rule The probability of either one of two independent events is the sum of the probabilities pof two 4s or two 5s 136 1362118 RrYy pgamete with 2 recessive or 2 dominant alleles 1212121214 14 212 Probability or chance governs the transmission of genes Punnett Square Figure 2 1 1 P RRYY X rryy cross 1 Rround peas F1 RrYy sell rwrinkled peas 1 Yyellow peas F2 ygreen peas 1 List the gametes from one parent on one side 2 List the gametes from other parent on the other side 3 Fill in gametic combinations in the squares 3 Determine genetic ratios phenotypic or genotypic by counting squares Branch Diagram Use this method 1 List probabilities for one event 2 List probabilities of second event next to those of the first 3 Use the product rule to determine genetic ratios Composition of F72 Product rule result 34 Y 3434 916 R Y round yellow 34 R YYYyyY RRRrrR 14 yy 3414 3 16 R yy round green 34 Y 1434 316 rrY wrinkled yellow 14 rr YYYyyY 14 yy 1414 116 rryy wrinkled green F2 phenotypic ratio of a dihybrid cross is 933 1 Use the product rule directly for trihybrid etc F1 AaBchDdEe X AaBchDdEe To calculate genotypic freg uency pAAbchDDEe14141214121256 To calculate phenotypic frequency pA bbC D E 341434343481 1024 Pedigree Analysis Used in human genetics Family tree Symbols used in pedigree analysis Figure 2 12 Autosomal Recessive Disorders Rare Figure 2 13 Mendelian inheritance of an autosomal recessive disorder is revealed by the appearance of the phenotype in male and female progeny of unaffected individuals Inbreeding increases the chances of children being affected e g PKU cystic fibrosis albinism Dominant disorders Figure 2 16 Mendelian inheritance of an autosomal dominant disorder show affected males and females in each generation and also show affected males and females transmitting the condition to sons and daughters in equal proportions e g achondroplasia quotdwarfismquot D dwarfism dd normal Interferes with bone growth during development eg Huntington s disease Late onset after having children Chapter 29Patterns of Inheritance 19905 1 Sex Linkage lnheritance pattern of genes on sex chromosomes In many organisms the gender is determined by the combination of sex chromosomes Sex chromosomes also carry genes unrelated to male and female development Xlinked recessive inheritance Many more males than females Drosophila eye color phenotype correlated with chromosome differences in microscope Thomas Hunt Morgan Nobel prize 1934 Experiment 19Drosophila eye color Figure 2 24 F XWX X XWY Red WT female white eyed male xl F1 All females and males red red dominant X WXW X WY white recessive xl F2 31 redwhite BUT Red 2 female 1 male White all male WHY Experiment 29Test Cross F1 Red female X WXW X white male XWY xl 1 1 11 red female white female red male white male XWXW XWXW XW Y XWY Experiment 39Reciprocal cross of experiment 1 Figure 2 24 F Xwa white female X X WY red male xl F1 11 red female white male X Xw XWY xl F2 1 1 11 red female white female red male white male XMXW XWXW XW Y XWY Morgan s Explanation XX92 chromosomes with eye color genes XY91 chromosome with eye color genes Experiment 3 revisited P X Xw X XWY white female red male xl F1 X Xw amp XWY red female white male PLUS rare exceptional progeny 12000 X XWY amp XWO white female red male Exceptional chromosome arrangements were observed in a microscope N0n Disjuncti0n9 Failure of chromosomes or sister chromatids to separate in meiosis Exceptional progeny arise due to non disjunction during meiosis Figure 3 7 The number of X chromosomes dictates the gender of Drosophila The presence of the Y chromosome dictates maleness in humans Xw09 sterile Drosophila male Xw09 sterile human female Table 2 3 Human Sex Linkage Pedigree Analysis XLinked Recessive Figure 2 25 Many more males than females show the recessive phenotype If rare almost all affected people are male If rare none of the offspring of an affected male are affected but all daughters are carriers Female carriers will pass the condition to 12 of her sons and 12 of her daughters will be carriers eg red green color blindness hemophilia XLinked Dominant Figure 2 28 Affected males pass condition to all daughters never to sons Affected females pass condition to 12 of sons and 12 of daughters eg hypophosphatemia Vitamin D deficient rickets YLinked Inheritance Other than maleness very few human phenotypes are known to be Y linked XInactivation p 3239Female mammals inherit 2 X chromosomes Early in development one of the 2 Xs is inactivated If the functional X has a recessive allele the recessive phenotype is expressed Barr Body The inactive X microscope Since the inactivation process is random all females are genetic mosaics ie a mixture of cells having 2 genotypes corresponding to inactivation alternatives e g sweat glands in humans calico cats Chapter 39The Nature of Chromosomes 1912051 Each chromosome contains one DNA molecule Cvtological features of CI Chromosome number92 to 100s depending on species Chromosome size Large variation within a genome Centromere Structure that attaches to spindle fibers during mitosis and meiosis constricted appearance at metaphase Metacentric Centromere in middle Acrocentric Centromere of f center Telocentric Centromere at one end Telomeres Chromosome ends Euchromatin Active DNA lightly packed Heterochromatin Inactive DNA tightly packed A Constitutive Always inactive permanent feature of chromosomal region e g centromeres telomeres inactive X chromosome B Facultative Present or absent at various times active or inactive Banding Patterns Stains result in characteristic banding patterns Figure 3 17a eg G bands Polytene Chromosomes Unusual feature of some ies Figure 3 18c Many rounds of replication without separation leads to a characteristic banding pattern Balbiani rings or puffs are active regions of RNA synthesis ie active DNA Chromosome Structure E coli 4600 kb946 million bp 13 mm Human ce1196 million kb96 billion bp 2m 2 In of DNA must be packaged into a 6 pm nucleus Human body 1013 cells9sun and back 65 times 500000 times around earth Chromatin Chromosomal DNA and protein Nucleosomes DNA and chromosomal proteins called histones 10 nm fiber diameter Figure 3 20b First level of packaging beads on a string Each nucleosome contains an octamer of 2 subunits each of the histones H2A H2B H3 and H4 DNA wrapped twice around histone octamer Solenoid A coil of nucleosomes 30 nm fiber Figure 3 20b Stabilized by histone H1 that runs down the center Supercoil Seen during mitosis and meiosis 700 nm fiber Figures 3 22 Held together by a scaffold at scaffold attachment regions SARs Figure 3 21 Replication amp transcription of Chromatin Nucleosomes do not dissociate Loosen up to allow transcription and replication Chromatin remodeling complexes control gene expression Chromosome Structure Movie Organelle Chromosomes Chloroplasts chNA and Mitochondria mtDNA contain unique genes in their own chromosome Inheritance patterns differ from nuclear genes White female X green male9 all white Green female X white male9 all green Maternal inheritance Phenotype of all progeny is identical to the female parent II I 1 a I J Inheritance comes from egg pl andor 39 ia Mitochondria and chloroplasts arose in eukaryotic cells by the engulfment of primitive prokaryotic cells Evolution led to their present function Mitochondria Genome Figure 3 41a Mitochondria function oxidative phosphorylation Contain genes involved in translation and oxidative phosphorylation Chloroplast Genome Figure 3 41b Chloroplast function photosynthesis Contain genes involved in transcription translation and photosynthesis Requires communication between nuclear and mitochondrialchloroplast genomes because several organelle proteins are nuclear encoded Communication is necessary to coordinate expression of all genes necessary for organelle function Nuclear encoded proteins are transported into the organelle Chapter 49Eukaryotic Linkage Mappipg by Recombination 1914051 Meiotic Recombination Figures 4 6 4 7 Generates haploid genotypes gametes that differ from the haploid parental genotypes Interchromosomal Recombination Figure 4 8 If genes assort independently crosses between a heterozygote and a tester strain generate 50 Parental type gametes 50 Recombinant gametes 1 1 1 1 phenotypic ratio Interchromosomal recombination movie Gene Linkage Figure 4 2 Each chromosome can contain 1000s of genes Genes close together on the same chromosome do not assort independently Linked genes segregate together during meiosis Mendelian ratios are not observed Experiment 1 P prprvgvg X prprvgvg xl F1 prpr vgvg pr pr vg vg X pr pr vg vg Test Cross xl pr vg 1339 pr red eye normal pr v g 1195 pr purple eye pr v g 151 vg normal wings pr vg E v g vestigial wings 2839 This is not a 1111 ratio Pr and v g are linked on the same chromosome Parental Classes Original arrangement of alleles on the two chromosomes Two most frequent classes Recombinant Classes Observed less frequently than parentals Figure 4 10 Intrachromosomal Recombination New gene combinations are generated during meiosis when non sister chromatids cross over X over between the genes under study Figures 4 3 cover 4 9 X overs generate two reciprocal products or classes that are about equal in frequency Recombinant freguency RF significantly lt 50 suggests linkage RF 50 suggests that the genes are unlinked on separate chromosomes Intrachromosomal recombination movie Linkage Mapping To separate linked genes the X over must occur between them As the distance between linked genes increase the chance of a X over between the genes increase The RF between linked genes is used to map their relative distance apart on a chromosome RF 001 1 1 map unit mu Gene Locus Region on a chromosome where alleles of a certain gene are found Problem 1 evisited pr Vg pr Vg X pr Vg pr Vg Test Cross xl pr Vg pr Vg 1339 prVgprvg 1195 pr Vgprvg 151 pr Vg lpr Vg 2839 305 recombinant X 100 107 107 mu 2839 total 107 lt 50 pr and Vg are linked Problem 2 Given a genetic distance in mu we can predict frequencies of progeny from a test cross a b a b X a b a b Test Cross If 20 mu apart abab 40 abab 40 ablab 10 ablab 10 Chanter 49Eukarv0tic CIquot 39 bv 39 9160 ThreePoint Testcross Map units are additive but to order 3 genes we need to perform a three point testcross Figure 4 1 1 Double Xovers9 Smallest class because two X overs required Compare parental types with double X overs The gene that is quotswitchedquot is in the middle Figures 4 12 and 4 13 P VV39CVCV39CtCt X VV39CVCV39CtCt xl F1 V V CV CV Ct ct Test Cross xl F2 V CV ct 580 V CV ct 592 V CV ct 45 V CV ct 40 V CV ct 89 V CV ct 94 V CV ct V CV ct 5 1448 1 Consider V and CV neglect ct for now V ct amp V ct parentals V Ct amp V ct recombinants RF89943 5 14480132 132 2 Now consider V and ct neglect CV CV ct amp CV ct parentals CV ct amp CV ct recombinants RF 45 40 3 51448 0064 64 3 Now consider CV and Ct neglect V V CV amp V CV parentals V CV amp V CV recombinants RF45408994335514480196196 Now construct the linkage map V 132 ct 64 cV Interference A X over in one region of the chromosome decreases the chances that a X over will occur in an adjacent region Calculate the frequency and number of double recombinants expected if there is no interference ie use the product rule Expected frequency 01320064 00084 Expected number 000841448 12 Interference I1 coc coefficient of coincidence I 1 observed number of double X overs expected number of double X overs I1 8124121303333 This is another example of a three point testcross aabbcc X aabbcc xl a b c 72 a b c 68 a b c 128 a b c 122 a b c 9 a b c A 2000 Construct a linkage map and calculate interference First determine the middle gene a 1 Consider a and b neglect c a b2 a b gtparentals a b a b gtrecombinants RF726891 1 2000 160 20000088 2 Consider a and c neglect b a c a c gtparentals a c a c gtrecombinants RF128122911200027020000135135 3 Consider b and c neglect a b c b c gtparentals b c2 b c gtrecombinants RF7268 1 28 122991 1 1 1200043 02000 0215215 c 135 a 680 b 11 OE E00801352000216 11 202161 0926007474 20 Chapter 59Bacterial and Bacteriophage Genetics 91905 Prokaryotes Eubacteria and Archaebacteria Divide by binary fission resulting in genetically identical progeny Figure 5 2 Single circular chromosome No mitosis meiosis nucleus Haploid Bacteriophage Phage9bacterial Virus Not free living parasitize bacteria to replicate Conjugation Transfer of DNA from one cell to another Via direct cell to cell contact Figure 5 7 Requires the fertility 1F factor small circular DNA element that can replicate mini chromosome Cells with F are P cells without F are F F donor cells produce 1g which attach to F recipient cells Donor transfers a copy of F to the recipient A copy of F always remains in donor ie replication F can integrate into the chromosome by X ing over thereby generating an Hfr strain Figure 5 8 Hfr Strains high quot of 39 39 3 Can transfer chromosomal genes to F recipient DNA fragment can recombine with chromosome and generate recombinants Figure 5 10 Conjugation Movie 21 Determining Linkage Interrupted Mating Figure 5 1 1 Donor Recipient Hfr a b c d strs X F a b c d StI R Time course remove and plate samples on rich medium containing streptomycin str an antibiotic that prevents survival of donor cells Screen surviving cells for transferred genes by replica plating Origin Fixed point on donor chromosome where gene transfer starts Hfr transferred last Orientation of integrated F determines polarity direction of transfer First gene transferred depends on orientation and position of F origin terminus Many Hfr strains exist depending on position amp orientation of F Figure 5 13 Bacterial Mapping 1E coll Bacterial matings result in a partial diploid merodiploid Transferred genes must recombine with recipient chromosome Results in exchange of genetic material requires double X over Not a reciprocal change Figure 5 15 Figure 5 14 22 Determining gene order from interrupted mating Hfr leu met arg strS X F leu met arg st1R 1 Determine the gene that is closest to the origin by performing an interrupted mating experiment met1 2 Then with a natural gradient of transfer experiment plate on minimal all supplements except methionine ie select for met recombinants 3 Screen for other genes by replica plating 100 met Gene order leu arg met 60 arg gt 20 leu High Resolution Mapping Figure 5 16 Hfr leu arg met strS X F leu arg met st1R 1 select for last gene to enter ie leu Thus every transferred fragment has all 3 genes 2 screen for the other genes by replica plating 1 map unit 2 1 X over in the interval leu arg met 8 8200 2 4 4 map units leu arg met 18 18200 9 9 map units leu arg met 174200 2 87 200 And rare quadruple X overs leu arg met leu arg met determines the middle gene 4 9 Product rule 0050 05000250 25 23 Bacterial Transformation Conversion of one genotype to another by introducing chromosomal DNA Requires recombination similar to Hfr X F cross If two genes are close to each other both can be transferred on the same DNA fragment Limited use for genetic mapping Bacteriophages Virulent Phages Always lytic Temperate Phages Lytic or lysogenic LysogenyPhage infected bacteria not leading to lysis Lysogenic strains carry a prophage phage DNA with lytic functions turned off Can be integrated or as a separate circle UV light or chemicals induce prophage to enter the lytic cycle Transduction Phage can transfer bacterial genes from one cell to another by packaging bacterial DNA by mistake Generalized Transducing Phage Can carry any part of the bacterial chromosome Specialized Transducing Phage Can only carry a specific chromosomal region Generalized Transduction When a cell is infected by a non lysogenic phage the bacterial chromosome is broken up Phage can package pieces of bacterial DNA instead of phage DNA eg P1 Figure 5 27 Transducing phage ejects bacterial donor DNA into a recipient strain merodiploid recombination 24 Linkage data from P1 transduction Genes must be close enough for P1 to transduce them in a single piece of DNA 1 Donor P1 met arg X met arg Recipient 2 Select for met plate on minimal with arginine 3 Screen for the of met that are also arg Strains transduced to met arg are cotransductants The greater the cotransduction frequency the closer the genes are to each other Donor P1 a b c X a b c recipient 7 Select for a then screen for b and c Genotype Colonies a b c 7 a b c 20 a b c 46 a b c 27 100 The cotransduction frequency of a and b is 27 The cotransduction frequency of a and c is 47 25 Specialized Transduction eg bacteriophage lambda M Integration by a single X over between gal and bio Useful for moving specific genes ie gal or bio Figures 5 30 5 3 1a 2enes on E coli CI 1 First use Hfr strains to map genes within about 10 15 minutes min 2 Then use P1 transduction within the interval established by Hfr conjugation Conjugation transformation and transduction require an equal number of X overs to replace DNA in the bacterial chromosome 26 Chapter 69From Gene to Phenotype 1923051 Bac Igground Information Genes specify the structure of proteins Genes do not act in isolation ie the gene products protein interact with one another Different alleles of a gene specify different forms of the corresponding protein e g active inactive etc Cellular processes occur by pathways in which enzymes act in a series gene A gene B gene C A gt B gt C gt D enzyme A enzyme B enzyme C Incomplete Dominance Heterozygotes have intermediate phenotypes of the two homozygotes eg plant ower color P AA X aa Cross Red White xl F1 Aa Self Pink xl F2 1 AA Red 2 Aa Pink 1 aa White Codominance Heterozygote has phenotype of both homozygotes e g Human Hemoglobin HbAHbA Normal Red Blood Cells RBCs HbSHbS Abnormal sickle shaped RBCs anemia often fatal HbAHbS No anemia RBCs form sickles at low oxygen 1 Regarding anemia HbA is dominant 2 Regarding cell shape HbA is incompletely dominant 3 Regarding hemoglobin molecules HbA andeS are codominant 27 Multiple Alleles A gene with more than two alleles eg Human ABO Blood Groups 3 alleles Genotype Blood Group IA IA IA i A IB 1B IB i B IA IB AB i i 0 IA and IB are fully dominant in IAi and IBi and codominant in IAIB AB Universal acceptor both A and B Ag on cell surface O Universal donor no antigen on cell surface Lethal Alleles Alleles of a gene that cause lethality May be lethal if homozygous or even heterozygous Lethality may occur early during embryonic development after birth or later in life Many human lethal genes result in spontaneous abortions Mouse coat color Dark coat Normal WT Yellow coat Variant P Dark AA X Yellow AYA xl F1 1 1 DarkYellow P Yellow AYA X Yellow AYA xl F1 21 Yellow Dark Suggests no homozygous yellow 14 AA Dark 12 AYA Yellow 14 AYAY Die Before Birth 28 Pleiotropic Allele that causes more than one phenotype eg mouse coat color Yellow when heterozygous lethal when homozygous eg manx cats Tailess when heterozygous lethal when homozygous eg human blood clotting mutations Complementation Wild type phenotype requires at least one dominant allele of each of two genes Homozygosity for the recessive allele of either gene results in the recessive phenotype eg Pea petal color biochemical pathway enzyme A enzyme B P AAbb white line 1 X white line 2 aaBB xl F1 AaBb purple xl A B purple A bb white aaB white aabb white HWWO 97 F2 ratio if different genes all recessive if the same gene 29 Epistasis The ability of a mutation at one locus to override a mutation at another in a double mutant Overriding mutation is epistatic overridden mutation is hypostatic gene A gene B white gt magenta gt blue enzyme A enzyme B P aaBB white X AAbb magenta xl F1 AaBb blue xl A B blue A bb magenta aaB white aabb white HWWO 934 F2 ratio if gene A is epistatic to gene B Suppressors eg an enzyme consisting of two separate polypeptides bb 2 second site suppressor that suppresses the mutant aa phenotype P aaBB purple X AAbb red xl F1 AaBb red xl A B red A bb red aaB purple aabb red HWWO 133 F2 ratio if gene s suppresses the mutant a phenotype 30


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