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BIL 360 Exam 3 Study Guide Part 2

by: Ashi Ma

BIL 360 Exam 3 Study Guide Part 2 BIL360

Marketplace > University of Miami > Biology > BIL360 > BIL 360 Exam 3 Study Guide Part 2
Ashi Ma
GPA 3.2
Comparative Physiology
Dr. DuBois

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Comparative Physiology
Dr. DuBois
Class Notes
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This 29 page Class Notes was uploaded by Ashi Ma on Saturday January 31, 2015. The Class Notes belongs to BIL360 at University of Miami taught by Dr. DuBois in Fall. Since its upload, it has received 32 views. For similar materials see Comparative Physiology in Biology at University of Miami.

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Date Created: 01/31/15
BIL250 Unit 1 Notes Chapter 4 Mapping Eukaryote Chromosomes by Recombination 41 Diagnostics of Linkage Recombination maps chromosomes assembled two or three genes at a time with the use of a method called linkage analysis Linked genes loci of the genes are on the same chromosomes and the alleles on one homolog are physically joined by the DNA between them Tester parent contributes gametes carrying only the recessive alleles offspring directly reveal gametes of the dihybrid parent Meiosis in other parent can be ignored only focus on dihybrid When two allele combinations have a high frequency then they are associated via linkage Two equally frequent nonrecombinant classes totaling IN EXCESS of 50 percent Two equally frequent recombinant classes totaling LESS than 50 percent 0 Two genes are close together on the same chromosome pair they do not assort independently but produce a recombinant frequency of less than 50 percent Recombinant frequency of less than 50 is diagnostic for linkage Crossing over when homologous chromosomes pair at meiosis the chromosomes break and exchange parts in this process Crossover products two new recombinants Chiasma two dyads unit as a bivalent crossshaped structure often forms two nonsister chromatids For linked genes recombinants are produced by crossovers Chiasmata are the visible manifestations of crossovers Cis conformation alleles are present on the same homolog or adjacent Trans conformation alleles are on different homologs or opposite Linkage symbolism Alleles on the same homolog have no punctuation between them A slash symbolically separates the two homologs Alleles are always written in the same order on each homolog Genes known to be on different chromosomes are shown sparated by a semicolon Aa Cc 5 Genes of unknown linkage are separated by a dot Aa Dd PWF Knob large densely staining element A crossover is the breakage of two DNA molecules at the same position and their rejoining in two reciprocal recombinant combinations If crossing over takes place at the twochromosome stage then there can only be a maximum of two genotypes on an individual tetrad Four chromosome stage explains four chromosomes on a tetrad Tetrad analysis shows two important features of crossing over 1 In some individual meiocytes several crossovers can occur along a chromosome pair 2 In any one meiocyte these multiple crossovers can exchange material between more than two chromatids Any single crossover is between two chromatids Crossovers between sister chromatids happens rarely they do not produce new allele combinations and are not considered 42 Mapping by Recombinant Frequency The farther apart genes are the more likely crossover will take place and the higher proportion of recombinant products will be 0 Proportion of recombinants is a clue to the distance separating two gene loci on a chromosome map Farther apart the genes the frequency approaches 50 Genetic map unit mu distance between genes for which 1 product of meiosis in 100 is recombinant Recombinant frequency RF 0 Presence of hotspots causes proportional expansion of some regions of the map Recombination blocks have the opposite effect Longer regions the average number of chromosomes is higher recombinants are more frequently obtained also Recombination between linked genes can be used to map their distance apart on a chromosome Trihybrid triple heterozygote with triple recessive tester Threepoint testcross or a threefactor test cross 0 Goal whether or not the three genes are linked and if so deduce their order and map distances between them For trihybrids of linked genes gene order can usually be deduced by inspection without a recombinant frequency analysis Linked genes have 8 genotypes of the following frequencies 0 2 at high frequency 0 2 at intermediate frequency 0 2 at a different intermediate frequency 0 2 rare Double recombinant classes are the smallest ones Only one order is compatible Only three gene orders are possible with a different gene in the middle position To determine middle gene allele pair has ipped position in the double recombinant classes 1 v ct cv 2 ct v cv 3 ct cv v Interference crossovers inhibit each other or from another happening 0 Double recombinants can be used to deduce the extent of this interference If the crossovers in the two regions are independent the product rule can be used to predict the frequency of double recombinants Frequency would equal the product of the recombinant frequencies in the adjacent regions vctcv o vct RF value 0132 ctcv RF value 0064 0 01320064 00084 084 l if there is no interference double recombinants might be expected at this frequency 0 1448 ies 000841448 12 ies 0 However only 8 are observed the lack shows that the two regions are not independent and suggest that the distribution of crossovers favors singles at the expense of doubles There is some kind of interference crossover does reduce probability of a crossover in the adjacent region Coef cient of coincidence coc ratio of observed to expected double recombinants Interference I 1 coc Interference I 1 observed frequencyexpected frequency In example I 1 812 412 0 412 13 33 IfI 1 then interference is complete Heterozygous females are used for testcrosses in Drosophila because the men don39t show crossing over females have higher recombinant frequencies Ratios as diagnostics phenotypic ratios Monohybrid testcrossed 11 Monohybrid selfed 31 Dihybrid testcrossed independent assortment 1111 Dihybrid selfed independent assortment 9331 Dihybrid testcrossed linked PRRP Trihybrid testcrossed independent assortment 11111111 Trihybrid testcrossed all linked PPSCOSCOSCOSCODCODCO 43 Mapping with Molecular Markers Molecular makers differences in DNA between two chromosomes that do not produce visibly different phenotypes either because DNA differences are not located in genes or located on genes but do not alter protein These sequence differences are called molecular alleles or molecular markers Singlenucleotide polymorphisms SimplesequenceIength polymorphisms Comparisons of different individual sequences are 99 identical the 01 percent difference is due to singlenucleotide differences 0 Large portion are polymorphic meaning molecular alleles are quite common in population Singlenucleotide polymorphisms SNPs also called snips o 3 million SNPs 1 in every 3001000 bases 0 Some lie within genes some do not Single nucleotide pair could produce a new allele producing a mutant phenotype Two nucleotide pairs wild type and mutant are an example of an SNP Two ways to detect an SNP 1 Sequence a segment of DNA in homologous chromosomes and compare homologous segments to spot differences 2 In case of SNPs located at restriction enzyme39s target site these SNPs are restriction fragment length polymorphisms RFLPs a Two RFLP alleles one which has restriction enzyme target and one that does not b Restriction enzyme will cut DNA at SNP containing the target and ignore the other c SNPs detected as different bands on an electrophoretic gel Simple sequence length polymorphisms SSLPs type of repetitive DNA different individuals have different number of copies 0 Sometimes called variable number tandem repeats or VNTRs SSLPs have multiple alleles 15 on one locus two types 1 Minisatellite Markers variation in the number of tandem repeats of a repeating unit from 15100 nucleotides long human length 1 to 5kb 2 Microsatellite Markers variable numbers of tandem repeats of an even simpler sequence small number of nucleotides such as dinucleotide Satellite genomic DNA is isolated and fractionated with the use of physical techniques repetitive sequences often form a fraction that is physically separate from the rest It is the satellite fraction Detecting Simple Sequence Length Polymorphisms Detected by taking advantage of the fact that homologous regions bearing different numbers of tandem repeats will be different lengths PCR is used for this technique 0 DNA Fingerprints patterns produced on the gel in the case minisatellites Molecular heterozygote ATGC individual has all base pairs on the same position Loci of any DNA heterozygosity can be mapped and used as molecular chromosome markers or milestones 44 Centromere Mapping with Linear Tetrads Centromeres not genes regions of DNA on which the orderly reproduction of living organisms absolutely depends on Fungi that produce linear tetrads have centromeres that can be mapped Octad meiocyte produces a linear array of eight ascospores 8 ascospores constitute the 4 products of meiosis a tetrad plus a postmeiotic mitosis Mendel39s law of equal segregation dictates that an octad will have four ascospores of genotype A and four of a o If there is crossing over there will be four different patterns in the octad each pattern showing blocks of two adjacent identical alleles Firstdivision segregation patterns M1 patterns two different alleles segregate into the two daughter nuclei at the rst division of meiosis no crossover Seconddivision segregation patterns Mll patterns as a result of crossing over in the centromeretolocus region the A and a alleles are still together in the nuclei at the end of the rst division of meiosis Map units de ned as the percentage of recombinant chromatids issuing from meiosis divide by 2 for the Mll segregation 45 Using the ChiSquare Test for Testing Linkage Analysis Degrees of freedom for larger grids number of classes in rows 1 X number of classes in column 1 Df 21 X 21 1 46 Accounting for Unseen Multiple Crossovers Mapping function formula that relates an observed recombinantfrequency value to a map distance corrected for multiple crossovers m mean number of crossovers that must have taken place in chromosomal segment per meiosis True determinant of RF is the relative sizes of the classes with no crossovers the zero class compared with the classes with any nonzero number of crossovers Poisson Distribution occurrence of crossovers in a speci c chromosomal region described by statistical distribution describes the successes Perkins Formula In a tetrad analysis of a dihybrid there are only three possible tetrads possible 1 Parental ditype PD 2 Tetratype T 3 Nonparental ditype NPD If genes are linked then the formula used is Map distance RF 100NPD 12T Should always yield 050 re ecting independent assortment Formula that compensates for the effects of double crossovers Corrected map distance 50T 6NPD The inherent tendency of multiple crossovers to lead to an underestimation of map distance can be circumvented by the use of map functions in any organism and by the Perkins formula in tetradproducing organisms such as fungD 47 Using RecombinationBased Maps in Conjunction with Physical Maps Physical map map of the actual genomic DNA a very long DNA nucloeitde sequence showing where genes are how big they are what is in between them and other landmarks Units of distance numbers of DNA bases kilobase Physical map shows a gene39s possible action at the cellular level Recombination map contains information related to the effect of the gene at the phenotypic level The union of recombination and physical maps can ascribe biochemical function to a gene identi ed by its mutant phenotype 48 Molecular Mechanism of Crossing Over In a rare octad ratio there may be to many spores of one allele and one few of the other 0 There is a nonidentical sister spore pair this suggests that the DNA of one of the nal four meiotic homologs contains heteroduplex DNA Heteroduplex DNA DNA in which there is a mismatched nucleotide pair in the gene under study lnvolves only one nucleotide position Nonidentical sister spores were found to be correlated with crossing over in the region of the gene concerned crossing over might be based on the formation of heteroduplex DNA Doublestranded break the outcome of heteroduplex DNA and a crossover Broken ends of DNA will promote recombination between different DNAs 0 If the invasion and strand displacement spans a site of heterozygosity such as Aa then a region of heteroduplex DNA is formed HoIidayjunctions the replicated ends are sealed and the net is result is a strange structure with two singlestranded junctions These are potential sites of singlestrand breakage and reunion Chapter 4 Learning Curve Questions A DNA sequence with a either a GT or an AC base pair is heteroduplex The recombinant ie nonparental genotypes seen in the progeny of Morgan39s y testcrosses occur in approximately equal frequency and make up of the progeny39s genotypes 0 Less than 50 percent A genetic map unit as de ned by Sturtevant is the distance between genes that produces 1 percent recombinants during meiosis Genomic DNA is nearly identical between individuals however at a random location in the genome sequences from different people can differ by one nucleotide For example at a speci c site one person can have a G nucleotide whereas another has a T nucleotide These sequences differing in only one nucleotide are commonly found in the population These differences are called 0 Single nucleotide polymorphisms A researcher has crossed two types of Neurospora one with an Eallele and another with an e allele She observes a resulting octad that has ascospores with genotypes in the following arrangement from top to bottom E E e e E E e e This is an example of an pattern Mll and seconddivision segregation To investigate the deviation from expected genotype ratios in dihybrid crosses Morgan used a testcross In other words he crossed a dihybrid y with o A homozygous recessive y In a threepoint testcross gamete genotypes are possible 0 Eight A researcher has mated two types of Neurospora one with a D allele and the other with a dallele He observes four different octads with ascospores in the following orders octad 1 DDddDDddDD octad 2 DDDDdddd octad 3 ddDDDDdd and octad 4 DDddddDD Which of these octads does NOT represent an M segregation pattern Octad 2 A researcher has just drawn a map for a chromosome region in cats On this map she has noted that genes D and Tare 80 kb apart and that each of these genes is approximately 15 kb in size She has also included the positions of two molecular markers M1 and M2 This researcher has drawn a map 0 Physical In the progeny of a threepoint testcross the rarest classes of genotypes are usuaHy Double recombinants A researcher is studying two types of DNA The rst has a 100nucleotide sequence repeated 10 times consecutively This is a The second sequence has a CA dinucleotide repeated 10 times This is a Minisatellitemicrosatellite When mapping centromeres in Neurospora researchers often look at the genotypes of the eight ascospores constituting an All of these ascospores result from the same set of meiotic divisions and can be evaluated for M or M segregation patterns refer Octad The formuas can be used to calculate the distance between two genes on the same chromosome theyit takes into consideration double crossover events that can produce progeny with genotypes identical to the parents39 Poisson amp Perkins The proportion of progeny that have undergone crossing over and have combinations of alleles that are different from those of the parents is called the Recombinant frequency You are studying clovers and are interested in locating a gene involved in fungal resistance Luckily the clover genome has many molecular markers already mapped to chromosomes From your preliminary crosses you believe that the resistance gene is on chromosome 5 You would next 0 Cross clovers that are dihybrids for resistance and a chromosome 5 speci c molecular maker with plants that are homozygous for fungal susceptibility chromosome 5specific molecular marker A researcher has mated two Neurospora with different genotypes resulting in a ZW X ZW cross She examines a resulting tetrad which contains cells with the genotypes ZW and 2W This tetrad would be classi ed as a Nonparental ditype You are interested in the inheritance of two autosomal genes in sun owers These genes determine seed color black and white governed by the alleles b and b and plant height short and tall determined by the alleles 5 and 5 You cross two pure lines of sun owers and let the F1 generation bb 55 self The F2 progeny deviate from the expected Mendelian ratio so you cross a dihybrid F1 female with a tester male Of the 500 progeny 198 are b 5 are b 5 48 are b 5 and 57 are b 5 o 197 A cross of parents with unknown genotypes produces progeny with a phenotypic ratio of 11111111 What type of cross is this Trihybrid testcross unlinked genes During the course of his experiments Morgan studied ies with various phenotypes which helped him to determine whether genes were linked lf Morgan were inspecting male ies with the genotype prpr vgvg which he used as testers what would be the phenotype of these ies 0 Purple eyes short wings A researcher is trying to calculate the crossover interference for a particular region in human chromosome 2 He determines that the coef cient of coincidence for this region is 0 What can he conclude The interference for this region of chromosome 2 is 1 and double crossovers tend not to occur in this chromosomal region Simple sequence length polymorphisms can be easily studied by designing polymerase chain reaction PCR primers that ank the simple sequence length polymorphism region The PCR products are then run on an agarose gel that separates the SSLPs based on their 0 Length The hypothesis that researchers typically evaluate when using a x2 test to determine whether two genes are linked assumes that the two genes are c On different chromosomes and not linked You are interested in seeing if two genes in the clown sh are linked You do a dihybrid testcross and nd that the progeny do not have a 1111 genotype ratio Are the genes linked Yes These genes are on the same chromosome A researcher wants to use molecular markers to help map a gene involved in type I diabetes To this end what types of markers could she use 0 Single nucleotide polymorphisms restriction fragment length polymorphisms simple sequence length polymorphisms variable number tandem repeats You are working with two genes in banana plants fruit color 9 and plant height t Yellow fruit color and tallness are dominant and are the result of wildtype 9 and t alleles respectively You cross two pure lines and then perform a testcross with the F1 dihybrid females The 700 F2 progeny show the following genotypes 265 are g t 278 are 9 t 76 are g t and 81 are 9 t What is the recombinant frequency of the two genes 0 224 percent Single nucleotide polymorphisms can be located when they remove or insert a target for a restriction enzyme These changes also known as can be seen when digestion products are run on an agarose gel 0 Restriction fragment length polymorphisms Parental genotypes which would be classi ed as nonrecombinant were seen in of the progeny resulting from Morgan39s testcrosses in ies as certain alleles tend to be inherited together 0 More than 50 percent What does it mean when two genes are linked The two genes are on the same chromosome Linked genes with a trans conformation can be represented as AbaB Chapter 5 The Genetics of Bacteria amp Their Viruses 51 Working with Microorganisms Prokaryotes bacteria belong to this class as well as bluegreen algae cyanobacteria 0 DNA is not enclosed in a membranebound nucleus 0 Form of a closedcircle NOT LINEAR 0 Extra DNA elements called plasmids in shape of a circle Bacteriophagesphages bacteria can be parasitized Genetic material can be DNA or RNA Categorized as nonliving because cannot reproduce alone need molecular machinery of cells Viruses must be propagated in the cells of their host organisms Reproduce asexually ln bacteria rarely are two complete chromosomes brought together usually the union is of one complete chromosome a fragment of another 10 Conjugation process of gene exchange which is the contact and fusion of two different cells Transformation bacterial cell takes up a piece of DNA from the external environment and incorporate it in its own chromosome Transduction certain phages pick up a piece of DNA from one bacterial cell and inject it into another where it can be incorporated into the chromosome Phage recombination phages themselves can undergo recombination when two different genotypes both infect the same bacterial cell Each bacterial cell divides asexually 124D816 Plating when a small amount of liquid culture can be pipetted onto a petri plate containing solid agar medium and spread evenly on the surface with a sterile spreader Colony when mass in plating reaches 10quot7 cells Cell clones members of a colony that have a single genetic ancestor Prototrophic wildtype bacteria they can grown and divide on minimal medium substrate containing only inorganic salts a carbon source for energy and water Auxotrophic mutants obtained from prototrophic culture these mutants are cells that will not grow unless the medium contains one or more speci c cellular building blocks such as adenine threonine or biotin Wild type lac can use lactose to grow Mutant type lac cannot use lactose to grow Resistant mutants divide and form colonies in the presence of an inhibitor Genetic markers different types of mutants that allow a geneticist to distinguish different individual strains to keep track of genomes and cells in experiments 52 Bacterial Conjugation Strain A and Strain B by themselves did not show any phototrophic colonies Strain A and Strain B mixed produced phototrophic colonies from the plating Some form of recombination of genes had taken place between the genomes of the two strains to produce prototrophs Conjugation physical union of bacterial cells can be con rmed under an electron microscope o Conjugating parents act unequally only one parent transfers some or all of its genome to another 0 One acts as a donor and one as a recipient The transfer of genetic material in ECoi conjugation is not reciprocal One cell the donor transfers part of its genome to the other cell which acts as the recipient 11 Fertility factor F donor ability is itself a hereditary state sterile donors can regain ability by association with another donor strains F strains that can donate 0 F strains cannot donate and are recipients Plasmid nonessential circular DNA molecule that can replicate in the cytoplasm independent of the host chromosome Rolling circle replication F DNA in the donor cell makes a singlestranded copy of itself circlular plasmid quotrollsquot and it reels out the singlestranded copy 0 F copy remains in the donor and another copy in the recipient Derivative of F strains with two unusual properties 1 Crossing the F strains the new strain produces 1000 times as many recombinants as a normal F strain Hfr is the name of this ability to promote high frequency of recombination Hfr chromosome replicates and transfers a single strand to the F cell during conjugation If there is no recombination the transferred fragments of DNA are lost in course of cell division 2 Hfr X F crosses none of the F parents were converted into F or Hfr This is the opposite of the above property Linear transmission of the Hfr genes from a xed point Superscripts quotrquot and quot5quot stand for resistant and sensitive Interrupted mating a cross is mixed and put into a kitchen blender at speci c times to separate mating cell pairs Exoconjugants any str cell bearing a donor allele must have taken part in conjugation Key elements 1 Each donor allele rst appears in the F recipients at a speci c time after mating began 2 The donor alleles appear in a speci c sequence 3 Later donor alleles are present in fewer recipient cells Origin 0 the conjugating Hfr singlestranded DNA begins from a xed point on the donor chromosome and continues in a linear fashion 0 Point O is known to be the site at which the F plasmid is inserted Farther a gene is from 0 later it is transferred to the F 12 The Hfr chromosome originally circular unwinds a copy of itself that is transferred to the F cell in a linear fashion with the F factor entering last 1 One end of the integrated F factor would be the origin where transfer of the Hfr chromosome begins The terminus would be at the other end of the F 2 The orientation F is inserted would determine the order of entry of donor alleles f circle contains genes A B C D then insertion between A and D would give order ABCD or DCBA Insertion sequences regions of homology are known to be mainly segments of transposable elements Fertility Factor Exists in 2 States 1 The plasmid state free cytoplasmic element F is easily transferred to F recipients 2 The integrated state as a contiguous part of a circular chromosome F is transmitted only very late in conjugation Using time of entry for chromosomal mapping b begins to enter the F cell 10 minutes after a begins to enter then a and b are 10 units apart Finescale chromosome mapping by using recombinant frequency Recombination between bacteria takes place between one complete genome from the F called endogenote and incomplete one derived from the Hfr donor called exogenote Merozygote partial diploid cell has two copies of one segment one is endogenote and other is exogenote There must be an even number of crossovers to keep the chromosome intact Recombination during conjugation results from a doublecrossover Iike event which gives rise to reciprocal recombinants of which only one survives F plasmids that carrv oenomic fragments F F prime an F plasmid carrying bacterial genomic DNA The DNA of an F plasmid is part F factor and part bacterial genome Like F plasmids F39 plasmids transfer rapidly They can be used to establish partial diploids for studies of bacterial dominance and aee interaction R plasmids vectors carrying multiple resistances transferred rapidly on cell conjugation Transposons alleles for antibiotic resistance contained within this unit unique segment of DNA that can move around to different sites in the genome process called transposition 53 Bacterial Transformation 13 Transformation Bacteria takes fragments of DNA from external medium and integrates into the recipient s chromosome lf DNA is a different genotype than recipient the genotype of recipient can become permanently changed Double transformation if two donor genes are located close together on a chromosome there is a strong chance that they will be carried on the same piece of transforming DNA Bacteria can take up DNA fragments from the surrounding medium Inside the cell these fragments can integrate into the chromosome 54 Bacteriophage Genetics Viruses can be used in two different types of genetic analysis 1 Two distinct phage genotypes can be crossed to measure recombination and map the viral genome 2 It can be used as a way of bringing bacterial genes together for linkage and other genetic studies Phages are used in DNA technology as carriers or vectors of foreign DNA Phage Consists of a nucleic acid quotchromosomequot DNA or RNA surrounded by a coat of protein molecules Identi ed by symbols 0 During infection phage attaches to bacterium to inject genetic material into the bacterial cytoplasm Phage genetic information turns off synthesis of bacterial components and makes phage components instead 0 Many phage descendants are made and released when bacterial cell wall breaks open Lysis breaking of the bacterial cell wall because of the phages Lysate population of phage progeny Plaque opaque lawn of bacteria covering the surface of a plate of solid medium happens when phage infects surrounding bacterial cells Mapping phage chromosomes bv usino phage crosses Mixed infectiondouble infection infect both parental T2 phage genotypes The infection of a bacterial culture with two different phage genotypes Several rounds of exchange can take place within the host 1 Recombinant produced shortly after infection may undergo further recombination in the same small or in later infection cycles 2 Recombination can take place between genetically similar phages as well as between different types Recombinants from phage crosses are a consequence of a population of events rather than singlestep exchange events All other things being equal RF calculation does not represent a valid index of map distance in phages 14 As distance between two mutant sites increase a cross over event is much more likely Selective system only the desired rare event can produce a certain visible outcome 0 A mutational selection technique that enriches the frequency of speci c usually rare genotypes by establishing environmental conditions that prevent the growth or survival of other genotypes Screen system in which large numbers of individuals are visually scanned to seek the rare quotneedle in the haystackquot in contrast to selective system Mutagenesis procedure in which essentially all mutagenized progeny are recovered and are individually evaluated for mutant phenotype often the desired phenotype is marked in some way to enable its detection Recombination between phage chromosomes can be studied by bringing the parental chromosomes together in one host cell through mixed infection Progeny phages can be examined for both parental and recombinant genotypes 55 Transduction Transduction some phages are able to pick up bacterial genes and carry them from one bacterial cell to another 0 Another mode of gene transfer Virulent phages those that immediately lyse and kill the host Temperate phages can remain within the host cell for a period without killing it 0 DNA either integrates into the host chromosome to replicate with it or replicates separately in the cytoplasm as does a plasmid Prophage a phage integrated into the bacterial genome Lysogenic a bacterium harboring a quiescent phage itself is called a ysogen Quiescent phage in a ysogenic bacterium becomes active replicates itself and causes the spontaneous ysis of its host cell A resident temperate phage confers resistance to infection by other phages ofthattype Two kinds of transduction 1 Generalized transducing phages can carry any part of the bacterial chromosome 2 Specialized transducing phages carry only CERTAIN speci c parts Viruent phages cannot become prophages they always yse a cell immediately on entry Temperate phages can exist within the bacterial cell as prophages allowing their hosts to survive as ysogenic bacteria39 they are also capable of occasional bacterial ysis Generalized Transduction 15 Transducing phage donor cell is lysed bacterial chromosome is broken up into small pieces Newly forming phage particles mistakenly incorporate a piece of the bacterial DNA into a phage head in place of phage DNA 0 Genes on any of the cutup parts of the host genome can be transduced this type of transduction is by necessity of the generalized type Generalized transduction Used to obtain bacterial linkage information when genes are close enough that the phage can pick them up and transduce them in a single piece of DNA Cotransductants strains that produce both types of colonies two donor alleles that simultaneously transduce a bacterial cell their frequency is used as a measure of closeness of the donor genes on the chromosome map 0 The greater cotransduction frequency the closer two genetic markers must be opposite of mapping measurements 0 Linkage values are usually expressed as cotransduction frequencies Specialized Transduction Specialized transducer inserts into the bacterial chromosome at one position only It exists with a n outlooping similar to F plasmids It can pick up and transduce only genes that are close by Zygotic induction entry of the prophage into the cell immediately triggers the prophage into a lytic cell o In the cross of two lysogenic cells there is no zygotic induction 0 Presence of any prophage prevents another infecting virus from causing lysis Prophage produces a cytoplasmic factor that represses the multiplication of the virus Attachment site crossover point would be a speci c site prophage inserted into bacterial genome Region at which prophage integrates O Lysogeny does increase timeofentry or recombination distances between the bacterial genomes Mechanism of Specialized Transduction Transduction occurs when newly forming phages acquire host genes and transfer them to other bacterial cells Generalized transduction can transfer any host gene It occurs when phage packing accidently incorporates bacterial DNA instead of phage DNA Specialized transduction is due to faulty outlooping of the prophage from the bacterial chromosome and so the new phage includes both phage and bacterial genes The transducing phage can transfer only speci c hos genes 56 Physical Maps amp Linkage Maps Compared 16 Insertional mutagenesis rapidly zero in on a mutation s position on a known physical map 0 Technique causes mutations through the random insertion of quotforeignquot DNA fragments Inserts inactivate any gene in which they land by interrupting the transcriptional unit 0 Usually insert transposons Transformation prototrophic auxotrophic Chapter 5 Learning Curve Questions Which type of DNA exchange between bacteria relies on viral infection Transduction n Hfr mapping using an interrupted mating strategy the quotlaterquot donor alleles are present in what type of recipient cells 0 The minority of Which of the following would be components of a bacteriophage such as the T4 phage A head DNA or RNA tail bers a sheath Cell clones such as those that constitute a bacterial colony are Cells originating from the same quotfounderquot cell Which of the following researchers was the rst to describe bacterial transformation 0 Frederick Griffith bacteria are considered to be and they can grow and divide on medium Prototrophic wild type minimal In 1953 William Hayes conducted a quotcrossingquot experiment between two conjugating bacterial cells and found the transfer of genetic material in conjugation is Unequal Bacteria can acquire new DNA by quotpicking upquot pieces of DNA from the external environment This process is known as 0 Transformation The immediate source for an F plasmid is 0 An Hfr bacterial cell A researcher has purposefully infected bacteria with a phage Following culture the researcher notices that that majority of his bacteria have ysed 17 and produced many quotphage progenyquot which are now oating in the culture media All of these progeny constitute the Lysate An undergraduate is watching one of her laboratory mates grow bacteria He takes a liquid sample of bacteria and places this liquid onto an agar containing plate He then uses a glass rod to spread this liquid culture over the agar The undergraduate has just observed 0 Bacterial plating A scientist has added Hfr cells to Fquot cells and these cells have undergone conjugation The resulting exconjugants formerly Fquot cells would be classi ed as as they contain both an endogenote and an exogenote and as a result are quotpartiallyquot diploid Merozygotes If a bacterium carries a prophage this cell will be Lysogenic A researcher has identi ed what appears to be an F plasmid occurring independently of the chromosome in a new species of bacteria On closer inspection she determines that this plasmid actually contains two additional bacterial genes that are normally found on the bacterial chromosome not on a plasmid However neither of these genes appears to be involved in drug resistance The researcher has likely discovered an F plasmid Which of the following statements is true regarding the work of Norton Zinder and Joshua Lederberg These researchers employed an experimental method similar to that developed by Bernard Davis and used a Ushaped tube and lter Which of the following statements is true regarding the work of Alfred Hershey Hershey performed a mixed infection by introducing bacteria to two types of T2 phages having the genotypes hr or hr The organization of the genetic material in a bacterium is different from the organization of genetic material in a eukaryote in all of the following ways except o The bacterial genome can be a single molecule of singlestranded RNA in the form of a closed circle A researcher crosses an Hfr strain carrying the alleles a U C and dF to an Fquot strain carrying the alleles aquot bquot C and d She observes that the rst resulting recombinant bacteria have the genotype aquot b C and d Which Hfr allele is likely to be closest to the site at which the F plasmid is inserted ie the origin b 18 A researcher is staining lac and lac bacteria with a red dye She notices that after she performs her staining protocol the majority of the bacterial cells stain red The researcher can conclude That the majority of her bacteria can utilize actose Lac symbolizes a mutant type of bacteria that 0 Cannot use a speci c energy source Bernard Davis39s experiment with the Ushaped tube demonstrated that 0 Physical contact between cells is required for the exchange of DNA between two bacterial cell lines If a bacterium carries a prophage this cell will be Lysogenic Mapping genes using Hfr cells relies on the fact that 0 During conjugation DNA moves unidirectionally from the donor to the recipient at a constant rate Bacteria designated as arg are Auxotrophic bacteria and bacteria that will not grow normally in minimal media 19 Chapter 6 Gene Interaction Two categories 1 Interactions between alleles of one locus does not address the range of genes affecting a function 2 Interactions between two or more loci 61 Interactions between the Alleles of a Single Gene Variations on Dominance Multiple allelesallelic series known mutant alleles of a gene and its wild type allele Dominance is a manifestation of how the alleles ofa single gene interact in a heterozygote Fully dominant allele will be expressed when only one copy is present as in a heterozygote where as the alternative allele will be fully recessive Homozygous dominant cannot be distinguished from heterozygote AA Aa Haploinsufficient one wildtype dose is NOT enough to achieve normal levels of function Null mutation produces a nonfunctional protein Heterozygote wild typenull null is dominant Dominant negative polypeptides with this type of mutation act as quotspoilersquot or quotroguesquot Gene product is a unit of homodimeric protein a protein composed of two units of the same type Heterodimer composed of polypeptides of different genes 20 0 Example of dominant negative collagen For most genes a single copy is adequate for full expression such genes are haplosuf cient and their null mutations are fully recessive Harmful mutations of haplosuf cient genes are often dominant Mutations in genes that encode units in homo or heterodimers can behave as dominant negatives acting through quotspoilerquot proteins Incomplete dominance occurrence of the intermediate phenotype used to describe the general case in which the phenotype of heterozygote is intermediate between those of the two homozygotes pink petal Codominance the expression of both alleles of a heterozygote blood type Alleles determine the presence and form of a complex sugar on the surface of red blood cells 0 Sugar molecule is an antigen cellsurface molecule that can be recognized in the immune system c A and B are two different forms 0 i no cellsurface molecule null allele Sicklecell anemia illustrates the arbitrariness of the terms dominance incomplete dominance and codominance Type of dominance inferred depends on the phenotypic level at which the assay is made organismal cellular or molecular Type of dominance is determined by the molecular functions of the alleles of a gene and by the investigative level of analysis Allele can show dominance with one partner but not with another Recessive Lethal Alleles Lethal aee aee that is capable of causing the death of an organism Useful in determining the developmental stage at which the gene normay acts Pleiotropic any allele that affects several properties of an organism Most recessive ethas are silent in heterozygote 25 of progeny die at some stage of development Lethal alleles depend on the environment In diploids recessive etha aees can be maintained as heterozygotes Haploids heat sensitive etha alleles are useful Temperaturesensitive ts mutations lethal alleles in haploids Phenotype is wild type at the permissive temperature often room temperature Mutant is higher at restrictive temperature Temperature sensitive alleles are said to be caused by mutations that make the protein prone to twist or bend its shape to an inactive conformation at the restrictive temperature Nu alleles for genes identi ed through genomic sequencing can be made by using a variety of quotreverse geneticquot procedures that speci cally known out the function of the gene 21 To see if a gene is essential a null allele is tested for lethality 62 Interaction of Genes in Pathways Onegeneonepolypeptide hypothesis 0 Originally proposed that each gene nucleotide sequence encodes a polypeptide sequence 0 Generally true with the exception of untranslated functional DNA Functional RNAs an RNA type that plays a role without being translated Chemical synthesis in cells is by pathways of sequential steps catalyzed by enzymes The genes encoding the enzymes of a speci c pathway constitute a functionally interacting subset of genome Signal transduction pathway chain of complex signals from the environment to the genome and from one gene to another Crucial to the proper function of an organism Developmental pathways comprise the steps by which a zygote becomes an adult organism 63 Inferring Gene Interactions 1 Obtain many singlegene mutants and test for dominance 2 Test the mutants for allelism are they at one or several loci 3 Combine the mutants in pairs to form double mutants to see if the genes interact Double mutant o If the genes interact then the phenotype differs from the simple combination of both singlegene mutant phenotypes lf mutant alleles from different genes interact then we infer that the wildtype genes interact normally as well 0 Two mutants interact 9331 Sorting mutants Complemntation test performed by intercrossing two individuals that are homozygous for differnet recessive mutations Observe whether the progeny have the wildtype phenotype lf wild type two recessive mutations must be DIFFERENT genes 0 Two mutations are said to have complemented lf NOT wild type recessive mutations must be alleles of the same gene 22 Complementation is the production of a wildtype phenotype when two haploid genomes bearing different recessive mutations are united in the same cell In a haploid organism the complementation test cannot be performed by intercrossing ln fungi an alternative way brings mutant alleles together to test complementation fusion resulting in a heterokarvon When two differnet strains fuse haploid nuclei from the different strains occupy one cell which is the heterokaryon When two independently derived recessive mutant alleles producing similar recessive phenotypes fail to complement they must be alleles of the same gene Analvzino double mutants of random mutations 9331 ratio no gene interaction 0 Ratio is produced because the two pigments act independently at the cellular level 97 ratio genes in the same pathway 0 Possible that double mutant has the same phenotypes as the two single mutans Identical phenotypes of the single and double mutants suggest that each mutant allele controls a different step in the SAME pathway 97 F2 ratio suggests interacting genes in the same pathway absence of either function leads to absence of the end of produce of the pathway 934 ratio recessive epistasis Epistasis referring to the situation in which a double mutant shows the phenotype of one mutation but not the other Overriding mutation is epistatic and overridden one is hypostatic Epistasis results from genes being in the same pathway Epistasis is inferred when a mutant allele of one gene masks the expression of a mutant allele of another gene and expresses its own phenotype instead 1231 ratio dominant epistasis Suppressor mutant allele of a gene that reverses the effect of mutation of another gene resulting in a wildtype or nearwildtype phenotype lmplies that the target gene and the suppressor gene normally interact at some functional level in their wildtype states 0 Sometimes have no effect in the absence of other mutation In other cases suppressor allele produces its own abnormal phenotype Suppression is sometimes confused with epistasis The key difference is that suppressor cancels the expression of a mutant allele and restores the corresponding wildtype phenotype Only two phenotypes segregate rather than three in epistasis 23 Revertants an allele with wildtype function arising by mutation of a mutant allele caused either by a complete reversal of the original event or by a compensatory secondsite mutation Pseudorevertants doubIe mutants in which one of the mutations is the suppressor Modi ers mutation at a second Iocus changes the degree of expression of a mutated gene at the rst Iocus Synthetic lethals speci c category of gene interaction 0 It can point to speci c types of interactions of gene products A range of modi ed 9331 F1 ratios can reveal speci c types of gene interaction 64 Penetrance and Expressivity Penetrance the percentage of individuals with a given allele who exhibit the phenotype associated within the allele Reasons to have genotype but not express phenotype In uence of the environment In uence of other interacting genes Uncharacterized modi ers Epistatic genes Suppressors in the rest of the genome may act to prevent the expression of a typical phenotype 3 Subtlety of the mutant phenotype oooNH Expressivity measures the degree to which a given allele is expressed at the phenotypic level measures the intensity of the phenotype Incomplete penetrance Often diseasecausing aIIeIe that is not fully penetrant The terms penetrance and expressivity quantify the modi cation of gene expression by varying environment and genetic background they measure the percentage of cases in which the gene is expressed and the level of expression Chapter 6 Learning Curve Questions A researcher is studying mice that carry a mutant allele of a gene caIIed Pax3 Animals heterozygous for this mutation demonstrate a malformed nervous system a cleft palate and abnormal fur pigmentation As this mutant Pax3 aIIeIe affects severaI mouse traits this allele is PIeiotropic 24 Which of the following statements is true All of them a Genes can encode proteins that act together in a biosynthetic pathway like the arginine p Genes involved in a developmental pathway can encode proteins that act together to drive speci c organ such as the heart Genes involved in a signaltransduction pathway can produce proteins that determine how environmental signal and can result in gene transcription Not all genes encode proteins some encode only RNAs A 934 phenotypic ratio in the progeny of a dihybrid cross is diagnostic for Recessive Epistasis A beagle carrying an allele for piebald spotting can have a single large black spot or several small black spots The determining factors for the kind of spots that a beagle has are the alleles at multiple other genes in its genome This demonstrates Variable Expressivity When two alleles at a single locus are both expressed in the heterozygote they are said to be Codominant A researcher is studying a group of genes that all appear to function in the same pathway These genes encode proteins that are involved in multiple phosphorylation events and when this pathway is activated the result is the transcription of a speci c target gene These genes are most likely involved in a pathway Signaltransduction pathway The progeny from a dihybrid cross of corn snakes both with the genotype 0 70 bb have a phenotypic ratio of 9331 What is the interaction of these two genes 0 No interaction The de nition of is the percentage of individuals who have an allele and express the phenotype associated with that allele Penetrance A doctor isolates hemoglobin from a patient and runs this sample on an electrophoretic gel She notices that two hemoglobin bands are present one near the top of the gel and the other near the bottom The doctor concludes that the patient39s genotype is HbAHbS To determine the step at which a nonfunctional enzyme blocks a biosynthetic pathway can be supplied to cells The cells will be able to produce the 25 nal product from any step after the block or where the nonfunctional enzyme interferes with the pathway 0 Products that occur at different stages of the pathway The test allows a researcher to determine whether two recessive mutations are from different genes or are alleles of the same gene Complemenation Variable expressivity can be caused by c Other alleles in the genome at different loci and environmental factors In the ABO blood groups the blood type AB is an example of Codominance A gene encodes the physical of a protein which in turn dictates its Structure function A dihybrid cross of harebell plants with the genotype WlW1W2W2 produces plants with blue wildtype and white mutant owers in a ratio of 97 indicating that the two genes function in the same pathway Which plants make up the 7 in the 97 ratio 0 Both single mutants and the double mutants A species of owering plant includes plants with blooms that range from bright golden yellow to very pale yellow no owers carrying the yellow pigment allele appear white The allele for yellow ower color Demonstrates variable expressivity Two pure lines of fouro39clock plants one with red owers and the other with white owers are crossed The F1 progeny all are pink When these are selfed the F2 progeny are 25 percent red owers 50 percent pink owers and 25 percent white owers This is an example of Incomplete dominance Beadle and Tatum used fungi that were unable to produce an essential nutrient arginine to determine the cellular pathway that synthesizes this nutrient The discovery of biosynthetic pathways that were controlled by the sequential action of enzymes led to the formulation of the hypothesis Onegeneonepolypeptide The complementation test is a cross of individuals homozygous for different recessive mutations If all of the F1 progeny of this cross have a wildtype phenotype the mutations Are recessive alleles of different genes and they complement one another 26 You have completed a complementation test for three recessive alleles that produce white instead of blue wildtype owers in harebells The results of the crosses are as follows white 1 x white 2 all blue white 1 x white 3 all blue white 2 x white 3 all blue What is the interpretation of these results 0 The three mutations affect different genes In foxgloves the dominant mutant for petal color D produces a dark red color The recessive wildtype allele 039 produces light red owers A second mutant W restricts pigmentation only to the throat spots of a petal while the wildtype allele W distributes pigment throughout the entire petal If a dihybrid Dd WW foxglove is selfed the phenotypic ratio of the progeny is 1231 white petals with spots to dark red petals to light red petals What does the double mutant look like 0 White with throat spots The arginine biosynthetic pathway is as follows precursor gt ornithine gt citrulline gt arginine The three arginine mutants of Beadle and Tatum are named argl arg2 and arg3 The arg2 mutant does not grow with ornithine treatment but does grow with citrulline or arginine treatment Where in the pathway does the defective enzyme of arg 2 mutants occur Ornithine l citrulline The arginine biosynthetic pathway is as follows precursor gt ornithine gt citrulline gt arginine The three arginine mutants of Beadle and Tatum are named arg l arg 2 and arg 3 The argl mutant grows with treatment of ornithine citrulline or arginine The wildtype argl gene codes for an enzyme that Converts a precursor molecule to ornithine Beadle and Tatum irradiated the fungus Neurospora to produce mutants that were defective in some aspect of nutrition What step did they take next 0 They tested for singlegene inheritance of the mutation Which of the following is a dominant negative mutation A mutation in the gene encoding collagen which produces a misshapen protein that spoils a collagen trimer Incomplete dominance is the result of a dose effect by the wildtype allele Imagine that it takes 3 units of pigment to produce red owers in peonies If the wildtype allele produces 2 units of pigment and a mutant allele produces no pigment what color of owers will homozygous wildtype heterozygous and homozygous mutant peonies respectively have 0 Red pink white 27 You are a botanist searching for mutant wheat plants resistant to a speci c fungus susceptibility to this fungus is the wildtype phenotype You have found three strains that are resistant to the fungus The strains are available as pure lines but you don39t know if they carry mutations affecting the same gene or different genes You also know that susceptibility to the fungus is dominant so the strains must be homozygous recessives You complete a complementation test of the mutants and nd the following resistant 1 x resistant 2 all susceptible resistant 1 x resistant 3 all susceptible resistant 2 x resistant 3 all resistant Based on this data mutants occur in the same gene and mutant occurs in a different gene 2and31 You have completed a complementation test for three recessive alleles that produce white owers instead of blue wildtype owers in harebells The results are as follows white 1 x white 2 all white white 1 x white 3 all white white 2 x white 3 all white What is the interpretation of these results 0 The three mutations all affect the same gene ie are different alleles Gene interaction can result in a phenotypic ratio in the progeny 97 which shows that two genes are involved in the same pathway 934 which demonstrates recessive epistasis 1231 which demonstrates dominant epistasis 933 which demonstrates synthetic lethality i All of the answers are correct Two pure lines of the blueeyed Mary plant one with white petals and one with magenta petals are crossed Both of these petal colors are the result of recessive mutations the wildtype petal color is blue The resulting dihybrids are crossed resulting in a phenotypic ratio of 934 blue to magenta to white If the petal pigment pathway can be depicted as white gt magenta gt blue where in the pathway does the epistatic mutation have its effect 28 e White D magneta Complementation tests are typically carried out in diploid organisms What is TRUE of complementation tests in haploid organisms Complementation can be assessed in a haploid organism provided that fusion can occur in this organism and a heterokaryon can be formed Random Midterm I Notes bbA and aabb give you the same phenotype know what recombination is used forto map genes locate the position of the gene Look at picture in chapter 4 and understand this Polydactyly 6 ngers result of incomplete penetrance 29


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