BIL 250 Exam 1 Notes
BIL 250 Exam 1 Notes BIL 250
Popular in Genetics
Popular in Biology
This 69 page Bundle was uploaded by Caitlyn Traenkle on Monday September 28, 2015. The Bundle belongs to BIL 250 at University of Miami taught by Dr. Wang in Spring 2015. Since its upload, it has received 19 views. For similar materials see Genetics in Biology at University of Miami.
Reviews for BIL 250 Exam 1 Notes
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 09/28/15
Ch 1 The Genetics Revolution in the Life Sciences Introduction DNA molecule that encodes the genetic instructions Genetics the study of all aspects of genes includes molecular and genomics Genes fundamental units of biological information o Composed of DNA 0 Molecular genetics the study of the molecular processes underlying gene structure and function 0 Studies 1 or a few genes at a time Genomics The cloning and molecular characterization of entire genomes o the study of complete gene sets genomes 11 The Nature of Biological Information 0 life regenerates itself every generation from a single cell zygote o zygote must contain genetic instructions 0 Chromosomes wormshaped densely staining bodies found win the nucleus chromosomes suspected as info carriers bc they re passed intact from generation to generation via nuclear divisions meiosis and mitosis 0 DNA molecule that carries biological info within chromosomes 0 Structure contains a genetic code set of correspondences bW nucleotide triplets in RNA and amino acids in protein 0 DNA is made from a linear series of 4 molecular building blocks nucleotides Nucleotides the basic building block of nucleic acids a molecule composed of a nitrogen base a sugar and a phosphate group 0 speci c sequence of nucleotides constitutes the language of the code 0 Figure 12 0 DNA is a double helix 0 Each chromosome is 1 long DNA molecule Genes are functional regions of this DNA 0 Each chromosome has a pair 0 Each cell nucleus has an identical complement of chromosomes in 2 copies Each copy is a genome 0 Human body made up of trillions of cells 0 Cells make up tissues which make up the organism The Molecular Structure of DNA 0 DNA molecule made up of 2 long strands of nucleotides wound in a double helix 0 4 types of nucleotides in DNA 0 each nucleotide has a deoxyribose sugar a phosphate group and a nitrogenous base 0 4 different bases adenine A purine base pairs with T thymine T pyrimidine base pairs with A guanine G purine base pairs with C cytosine C pyrimidine base pairs with G o phos groups and sugars form chain sides of a ladder o bases face center and each is H bonded to base of other strand rungs of the ladder sequence of nucleotides represents the coded info carried by DNA DNA is organized into genes and chromosomes Genome an organism s complete set of genetic info encoded in its DNA 0 Eukaryotes DNA found in nucleus 0 Individual chromosome composed of 1 highly coiled DNA double helix of chromosomes in genome is speciesspeci c Diploid 2n A cell having 2 chromosome sets or an individual organism having two chromosome sets in each of its cells 0 Nuclei contains 2 complete copies of the genome 0 Ex humans have 2 copies of 23 distinct chromosomes N23and2n46 Haploid n A cell having 1 chromosome set or an organism composed of such cells 0 Nuclei contains just 1 chromosome set Homologous chromosomes Chromosomes that pair with each other at meiosis OR chromosomes in different species that have retained most of the same genes during their evolution from a common ancestor o Homologs A member of a pair of homologous chromosomes identical to other homolog Genes found in segments along 1 DNA molecule 0 Primary carries of info in genome 0 Variation in chromosomal landscape and size of genes among species 0 Exons coding regions in gene 0 Introns noncoding segments in gene affects gene size Gene pair 2 copies of a particular type of gene present in a diploid cell one in each chromosome set 0 Nucleus in a somatic body cell contains pairs of chromosomes but they re not physically paired physical pairing of homologs takes place in meiosis Electrophoresis technique used to separate a DNA molecule in a genome by size 0 DNA bands haploid chromosome number DNA packages very efficiently in a chromosome 0 Nucleosome basic unit of eukaryotic chromosome structure a ball of 8 histone molecules proteins that is wrapped by two coils of DNA DNAnucleosome chain further coiled and folded o Chromatin DNA and associated nucleosomes together the substance of chromosomes 0 Centromere region of chromosome that acts as a site for the binding of kinetochore proteins used to move chromosome during cell division Teleomeres tips of the chromosomes contain specialized DNA sequences needed for chromosomal division Extranuclear genome small specialized fraction of eukaryotic genomes found in mitochondria or chloroplasts Prokaryotes bacteria no nuclei genome found unbound in cytoplasm o Genome a single noncoiled chromosome usually circular o Often have plasmids small circular chromosomes in addition to main chromosome 0 Genomes of viruses smaller usually linear 12 How Information Becomes Biological Form Proteins main elements of form in organisms 3 basic types of proteins 0 structural physical structure hair muscle cytoskeleton o enzymatic catalyze rxns going on win cells rxns that make molecules proteins nucleic acids carbs fats 0 regulatory turn onoff gene activity 0 proteinsynthesis includes transcription and translation 0 DNA l RNA l protein Transcription The synthesis of RNA from a DNA template 0 RNA A singlestranded nucleic acid similar to DNA but having ribose sugar rather than deoxyribose sugar and uracil rather than thymine as one of the bases 0 Initial transcript is modi ed by excising the introns o Messenger RNA mRNA An RNA molecule transcribed from the DNA of a gene 0 mRNA transported to cytoplasm ribosomes translate the mRNA into proteins Translation The ribosome and tRNAmediated production of a polypeptide whose amino acid sequence is derived from the codon sequence of an mRNA molecule Polypeptide a protein chain of linked amino acids 0 20 main amino acids various combos give protein its speci c shapefunction nucleotide sequence in mRNA translated into amino acid sequence in protein by reading consecutive codons o codons section of RNA three nucleotides in length that encodes a single amino acid 0 Functional RNAs RNA type that plays a role without being translated include o ribosomal RNA rRNA part of the ribosome 0 transfer RNA tRNA carry amino acids to translational system 0 how is cell differentiation achieved 0 mRNA synthesized at speci c developmental stages 0 gene transcription controlled by regulatory proteins How does life replicate itself 0 DNA replication 0 2 strands of DNA separate and newly synthesized nucleotides are deposited on the old strands each paired with its appropriate partner AT and GC 0 new array of nucleotides then ligated joined while held in place by old strand 0 results in 2 DNA molecules each with 1 of the old strands and a newly synthesized strand 0 DNA replication takes place every time somatic cells divide and when gametes are formed Change at the DNA level 0 Species have established characteristics that de ne them 0 But win a species there is neutral variation does not affect that individual s survival 0 Variation possible bc DNA in a genome can be changed 0 Mutation a change to the DNA sequence Occur naturally or from exposure to environment Most detrimental but some have no effect or can be advantageous lf mutation in germ cells it can be passed onto progeny and contribute to variation among individuals Ex albinism in humans Heredity diseases disease that are passed on from 1 generation to the next mutation in gonads o Mutations not in germ cells only impact that organism Can just kill that cell Or if mutation affects regulatory proteins that control cell divisiongrowth can caused cancer tumor 0 Many human diseases caused by mutation in a single gene 0 Epigenetic nongenetic chemical changes in histones or DNA that alter gene function without altering the DNA sequence 0 Histones coil the DNA for chromosome packing but also have regulatory function by restricting access of regulatory proteins thereby silencing them 0 Hereditary change caused mostly by mutations in DNA but also by epigenetic effects 0 Some natural variation caused by environmental effects not acting on DNA 0 Ex dietary differences among individuals not heritable 13 Genetics and Evolution similarities bw species due to common ancestry and differences due to natural selection in different habitats Natural Selection the process whereby individuals with a particular characteristic may reproduce better than others in a given environment 0 These individuals have more offspring l abundance of individuals with that characteristic will increase Homology similarity due to shared ancestry from a common ancestor 0 Theory of evolution notion of natural selection acting on variation Genetics is the cause of variation for natural selection to act on Constructing evolutionary lineages Evolutionary tree treelike branched diagram that shows descent of various modern and fossil species thru intermediate ancestral forms over time o Quanti es differences in DNA sequences species with similar sequences closer in tree of relatedness 0 Ex DNA and amino acid sequence of gene for electron transport protein cytochrome c homologous across range of organisms on the planet 0 Compare chimpanzee and human genome shows that chimps are our closest living relative both evolving from apes Comparing human genomes from around the world conclude that Homo sapiens evolved in Africa 0 Can obtain insight into different species with similar structure to well established experimental organisms 14 Genetics Has Provided a Powerful New Approach to Biological Research 0 Both forward and reverse genetics work by analyzing mutations and their effect 0 by showing how a gene goes wrong we deduce its normal func on Forward genetics classical approach to genetic analysis in which genes are rst identi ed by mutant alleles and mutant phenotypes and later cloned and subjected to molecular analysis Mutation l gene discovery l DNA sequence and function 0 1 are these properties inherited as a single mutated gene 0 Answer by crossing mutant organism with wildtype then inspecting ratios of wildtype to mutant progeny o 2 determine the function of each gene identi ed 0 once gene for speci c property mapped in genomic sequence if that gene studied in experimental organisms bc of evolutionary homology likely that a function already known Reverse genetics experimental procedure that begins with a cloned segment of DNA or a protein sequence and uses it through directed mutagenesis to introduce programmed mutations back into the genome to investigate function 0 Gene DNA sequence l mutation l function 0 Start with gene sequence attempt to nd it s function Directed mutagenesis experimental approached used to target mutations in an individual gene 0 Ex knock out approach knock out gene s function by eliminating the gene then look for effects on organism s func on Manipulating DNA Genome very large many nucleotide pairs DNA manipulation usually done by focusing on parts of the genome often single genes 0 DNA cloning A section of DNA that has been inserted into a vector molecule such as a plasmid or a phage and then replicated to produce many copies 0 Use to isolate a small segment of DNA 0 Take a DNA fragment and replicate it many times 0 Amplifying replicating a DNA sequence 0 Fragments of genome obtained by cutting DNA in some way 0 Fragments inserted into small selfreplicating chromosome vector 0 Vectors then introduced into separate live bacterial cells 0 Vector replicated as cell divides 0 DNA clone used in many ways 0 Modi ed and reintroduced back into original organism 0 Or into a different organism l creates a transgenic organism 0 Or sequences and assembled with other cloned sequences to produce a genomic sequence Detecting speci c sequences of DNA RNA and protein Probing Labeled nucleic acid segment that can be used to identify speci c DNA molecules bearing the complementary sequence usually through autoradiography or uorescence Probing for a speci c DNA 0 A cloned gene can act as a probe for nding segments of DNA that have the same or a similar homologous sequence 0 Probe works bc its nucleotide sequence is complementary to its target 0 Southern blot The transfer of electrophoretically separated fragments of DNA from a gel to an absorbent sheet such as paper this sheet is then immersed in a solution containing a labeled probe that will bind to a fragment of interest 0 Finding sets of genes using DNA microarrays o Microarray array of DNA fragments representing all the genes in a genome glued onto a glass slide Bathed in labeled probe Detecting and amplifying sequences using the polymerase chain reaction PCR 0 Polymerase chain reaction PCR An in vitro method for amplifying a speci c DNA segment that uses two primers that hybridize to opposite ends of the segment in opposite polarity and over successive cycles prime exponential replication of that segment only 0 Probing for a speci c RNA 0 Northern blot The transfer of electrophoretically separated RNA molecules from a gel onto an absorbent sheet which is then immersed in a labeled probe that will bind to the RNA of interest mRNA extracted from tissue separated into fragements using electrophoresis blotted on membrane 0 Probing for a speci c protein 0 Use antibodies as probes antibodies bind to antigen 0 Western blot to detect protein protein mixture extracted from cells is separated into bands of distinct proteins by electrophoresis and then blotted onto a membrane 15 Model Organisms have been crucial in the genetics revolution Model organisms species used in scienti c research must be suitable for the biological question under investigation small each and cheap to maintain grow quickly can see many generations Common model organisms E coli bacteria 0 Bacteria used often bc Small size many can be sampled permits detection of rare genetic events Bacteriophages bacterial viruses can be used as vectors to carry DNA from 1 bacteria cell to another Ascomycete fungi yeast Arabidopsis small owing plant model for plants nematodes Drosophila fruit y used bc chromosomes have wellmarked pattern of banding possible to observe largescale chromosomal alterations Mus musculus mouse model for vertebrates 16 Genetics Changes Society agriculture most crops genetically modi ed to produce more and be more resistant to diseasepests o transgenic organism produced by introducing a foreign gene from a different species medicine many diseases identi ed that are caused by mutations in single genes 0 ex gene for PHU discovered can alleviate disease through diet gene therapy correcting genetic disease at the DNA level crime investigation 0 DNA ngerprinting from body uids 0 Amplify the DNA using PCR 17 Genetics and the Future 0 further understanding of genetic disease 0 increase population l increase need for food clothing etc o Eugenics improving the quality of human births O 0 Gene therapy Designing babies 18 Summary genetics involved in all areas of biology biological info encoded in DNA heritable O O 0 info provides directions for an organism s proteins DNA divided into functional units called genes Genes encode for a speci c protein Proteins synthesis Transcription DNA l RNA Translation RNA l protein a Nucleotides read in groups of 3 codons each corresponds to an amino acid in that gene s protein DNA structure allows it to make copies of itself DNA replicated every time cellorganism reproduces Info can be passed on 0 DNA structure can undergo random change mutation O O O Mutation is main source of variation bw individuals of a species Via natural selection mutation can produce new species evolution DNA sequences in species show considerable similarilty homology DNA homology used to apply what s learned in model species to others and used to make evolutionary trees 0 Genetic dissection a biological function can be picked apart via mutations o Genomics has allowed complete gene sets genomes to be analyzed Genetic approach to understanding a biological property nd the subset of genes in the genome that in uence that property gene discovery 0 After genes identi ed observed the ways the genes act to determine the biological property 0 Gene discovery process whereby geneticists nd a set of genes affecting some biological process of interest by the singlegene inheritance patterns of their mutant alleles or by genomic analysis Singlegene inheritance patterns analytical approach to gene discovery recognized in progeny of controlled matings o crosses deliberate mating of two parental types of organisms in genetic analysis 0 mutants have altered form of a normal property have a mutation 0 wild type have the normal form of a property singlegene inheritance patterns produced bw genes are parts of chromosomes and chromosomes partitioned precisely through generations Genetic dissection The use of recombination and mutation to piece together the various components of a given biological function 0 Mate a WT individual with a MT and observe the ratio of progeny with and without the mutation o The ratio reveals whether a single gene controls that property 0 Each mutant potentially identi es a separate gene affecting that property 21 SingleGene Inheritance Patterns Mendel s pioneering experiments 0 Used garden pea as model organism Investigated the inheritance of 7 properties aka charactertrait o For each character studied 2 contrasting phenotypes o Phenotype 1 The form taken by some character or group of characters in a speci c individual 2 The detectable outward manifestations of a speci c genotype 0 All lines used were pure lines a population of individuals all bearing the identical fully homozygous genotype O Matings were bw identical plants for that phenotype Made use of O O Crosses pollen transferred from anthers of 1 plants to stigmata of another no self fertilization Self To fertilize eggs with sperms from the same individual Crossed 2 pure lines of opposing phenotypes green and yellow seeds 0 O 0 Parental generation The two strains or individual organisms that constitute the start of a genetic breeding experiment their progeny constitute the F1 generation First lial generation F1 progeny individuals from a cross of two homozygous diploid lines Progeny peas a yellow regardless of which parent was malefemale Second lial generation F2 progeny of a cross between two individuals from the F1 generation Got 34 yellow and 14 green Shows that genetic determinants fro green were still present in yellow F1 even tho unexpressed Sefed F2 green seed plants progeny have only green seeds Sefed F2 yellow seed plants 13 pure breeding for yellow 23 gave progeny ratio of 34 yellow and 14 green same as F1 crossed F1 with any green seed plant l 12 yellow 12 green Results from all traits O 0 31 phenotypic ratio in selfpollination of F1 11 phenotypic ratio in cross of F1 yellow with green Mendel s law of equal segregation 1 a hereditary factor gene necessary for producing pea color 2 each plant has a pair of this type of gene 3 gene comes in 2 forms alleles Y yellow or y green 4 plant can be YY yy or Yy 5 in Yy plant Y allele dominates l phenotype yellow 0 Y dominant allele 0 y recessive allele 6 Mendel s rst law law of equal segregation The two members of a gene pair segregate from each other in meiosis each gamete has an equal probability of obtaining either member of the gene pair 7 a single gamete contains only 1 member of the gene pair 8 at fertilization gametes fuse randomly Zygote A cell formed by the fusion of an egg and a sperm diploid cell will divide mitotically to create a differentiated diploid organism o Homozygote organism with a pair of identical alleles Homozygous dominant YY Homozygous recessive yy 0 Heterogygote organism with a pair of different alleles Heterozygous Yy Monohybrid heterozygote for one gene 0 Genotypes aeic composition of an individual or of a cell YYYyyy Parental generation was pure lines YY or yy when crossed together the two different lines produced F1 generation of all heterozygous individuals Yy o Sel ng F1 Yy x Yy a monohybrid cross F2 31 ratio 0 Equal segregation of the Y and y aees results in gametes that are 12 Y and 12 y All 11 31 and 121 ratios are diagnostic of singlegene inher ance 0 Based on equal segregation in a heterozygote In de ning the allele pairs for that phenotype Mendel identifies a gene that affects pea coor 22 The Chromosomal Basis of SingleGene Inheritance Patterns Equal segregation members of a gene pair segregate equally in gamete formation 0 happens bc gene pairs located on chromosome pairs and chromosomes segregate carrying the genes Singlegene inheritance in diploids When somatic body cells divide accompanying nuclear division is called mitosis type of nuclear division occurring at cell division that produces two daughter nuclei identical with the parent nucleus 0 2n2n2nORnnn In sexual cycle meiocytes specialized diploid cells divide via meiosis to produce sex cells 0 Meiosis 2 successive nuclear divisions with corresponding cell divisions that produce gametes in animals or sexual spores in plants and fungi that have onehalf of the genetic material of the original cell Only happens in diploid cells Gametes formed are haploid 2n l n n n n o Mitosis o 1 cell division 0 produces 2 daughter cells have same genomic content as mother cell 0 1 replication the chromosomes replicate interphase 2n l prophase 4n produces pairs of identical sister chromatids o 2 segregation when cell divides each member of sister chromatids pulled into a daughter cell metaphase sister chromatids align linearly anaphase chromatids begin to be pulled apart telophase cell begins to divide o 3 results in 2 identical daughter cells 2n 2n Meiosis o 2 cell divisions 0 produces 4 haploid gametes o 1 replication chromosomes replicate to form sister chromatids interphase 2n prophase l 4n o 2 pairing homologous pairs of sister chromatids unite to form a bundle of 4 homologous chromatids synapsis prophase forms 2 dyads n dyad pair of sister chromatids joined at the centromere metaphase forms pair of synapsed dyads bivalent n bivalent 2 homologous chromosomes paired at meiosis n tetrad 1 Four homologous chromatids in a bundle in the rst meiotic prophase and metaphase 2 The four haploid product cells from a single meiosis crossing over takes place at this stage 0 3 segregation cell division 1 metaphase n bivalents move to cell s equator anaphase I teophase u when ce divides 1 dyad moves to each new cell 0 4 segregation cell division 2 prophase metaphase anaphase II n centromeres divide the dyad each member of a pair of chromatids moves into a daughter ce teophase II 0 products of meiosis 4 haploid cells each cell gets 1 member of the tetrad o summary 2 homologs l replication 2 dyads l pairing tetrad I 1st division 1 dyad to each daughter cell l 2nCI division 1 chromatid to each daughter cell synapsis joining of homologous pairs relies on synaptonemal complex SC macromolecular assemblage that runs down center of the pair Spindle bers that pull apart chromosomes are polymers of tubulin o Pulling apart caused by depolarization that shorten the bers Singlegene inheritance in haploids In haploid organism yeast equal segregation can be observed directly Cross mutant with a wild type 0 Mutant allele r 0 Wild type allele r or 2 haploid cells fuse l diploid meiocyte rr replication and segregation l tetrad of 2 meiotic products spores of genotype r and 2 of r haploid genetics more simple bc 0 cross requires only 1 meiosis while diploid cross must consider meiosis in male and female parent 0 all alleles expressed in phenotype no masking of recessives by dominant alleles 23 The Molecular Basis of Mendalian Inheritance Patterns Structural differences between alleles at the molecular level Alleles different forms of the same gene 0 A and a differ only at a few nucleotides in their entire sequence mostly identical Mutation can occur anywhere along the nucleotide sequence of a gene creating a new allele 0 Could change identity of a single nucleotide or deletion of 1 nucleotides or ad 1 nucleotides 0 Mutation can occur at many different sites Alleles with mutation are usually recessive o bc takes only 1 copy of a wildtype to provide normal function 0 so a can represent a recessive allele with a mutation at different places Molecular aspects of gene transmission Replication of alleles during the S phase 0 First step is the formation of sister chromatids O Preludes mitosis and meiosis bb l 4b bb l 2b and 2b Replication duplicates all genetic info wild type or mutant Homolog GC replication l 2 chromatids GC Meiosis and mitosis at the molecular level 0 O S phase produces 2 copies of each allele A and a they can now be segregated into separate cells Figure 212 Mitotic divisions haploid and diploid conserve the genotype of the parent cell daughters identical to parent have both A and a Meiosis has 2 successive divisions that half the number of chromosomes producing gametes with either A or a o Demonstrating chromosome segregation at the molecular level 0 Phenotypic segregation shows singlegene phenotypic inheritance patterns in relation to the segregation of chromosomal DNA at meiosis get either a or A Sequence alleles A and a in parents and the meiotic products Results in 12 products with A sequence and 12 with a sequence Restriction fragment length polymorphism RFLP restriction enzymes cut DNA at speci c base sequences target site sometimes missing or there s an extra site Southern hybridization reveals an RFLP Sites might be within a gene or not But sometimes mutant phenotype also introduces a new target site for a restriction enzyme new site provides molecular tag for the alleles Polymerase chain reaction PCR ampli es DNA sequence to reveal products of different sizes RFLP and PCR provide molecular markers A site of DNA heterozygosity not necessarily associated with phenotypic variation used as a tag for a particular chromosomal locus RFLP and PCR allele are tags that aren t associated with a speci c biological function but can be used to track the inheritance of a segment of a chromosome at a speci c position Alleles at the molecular level 0 Primary phenotype of a gene is the protein it produces Functional differences bw proteins that explain the different effects of WT and mutant alleles on the organism Ex human disease PKU caused by defective allele on gene encoding for liver enzyme PAH 0 Normally enzyme converts phenylanine in food into the amino acid tyrosine o Mutation makes enzyme unable to bind to phenylanine which builds up and is converted instead to phenylpyruvic acid interferes with nervous system development Mutation may alter amino acid sequence in vicinity of the enzyme s active site 0 PAH enzyme made from a single type of protein but mutant alleles show many mutations at different sites mainly in the exons proteinencoding region 0 Effect of the mutation on function of the gene depends on where within the gene the mutation occurs 0 Exons very sensitive to mutation o Mutations in introns have slightly different effect Null alleles An allele whose effect is the absence either of normal gene product at the molecular level or of normal function at the phenotypic level 0 Proteins encoded by them completely lack PAH function 0 Leaky mutations A mutation that confers a mutant phenotype but still retains a low but detectable level of wildtype function 0 Reduce the level of enzyme function 0 Most mutations alter the amino acid sequence of the gene s protein product resulting in reduced or absent function Dominance and recessiveness o Recessiveness observed in mutations in genes that are functionally haplosufficient a gene that in a diploid cell can promote wildtype function in only one copy dose 1 copy of haplosufficient gene provides enough gene product protein to carry out normal function in heterozygote m m is a null remaining copy of allele provides enough protein product 0 Haploinsufficient genes null mutant allele will be dominant M and the single WT allele cannot provide enough product for normal function 0 Dominant mutations can result in a new function for the gene bc WT allele cannot mask new function 24 Some Genes Discovered by Observing Segregation Ratios 0 Goal of genetic analysis is to dissect a property by discovering the set of single genes that affect it 0 Can identify by phenotypic segregation ratios generated by their mutations 11 and 31 0 Based on equal segregation Mutant alleles can be either dominant or recessive 0 To determine which cross mutant with WT A gene active in the development of ower color 0 Cross white owered plant mutant with red owered plants WT All F1 plants are red 0 F2 has 31 ratio 0 Ratio indicates singlegene inheritance l can conclude that mutant caused by a recessive alteration in a single gene 0 P x albalb 0 F1 all ab 0 F2 14 12 ab14 albalb A gene for wing development 0 Cross short winged y mutant with long winged y WT 0 F1 has 11 ratio 0 Ratio same within males and females 0 Short wings likely produced by a dominant mutation 0 With dominant mutation will be expressed when heterozygous recessive mutation only expressed in homozygous state 0 P x S 0 F1 12 12 S o Longwinged progeny interbred all progeny longwinged 0 Expected for recessive WT allele 0 x 0 F2 all Shortwinged progeny interbred l 34 short and 14 long 0 S x S 0 F2 14 55 12 S 14 A gene for hyphal branching Cross hyperbranching fungus mutant with normal branching WT F1 has 11 ratio l singlegene inheritance In haploids dominance not usually possible Gene s WT aee essential for normal control of branching Investigate mutant to see location in normal developmental sequence where mutant produces block l reveals time and place in cells where normal allele acts In diploids sterile recessive mutant can be propagated as heterozygote that can then be selfed In haploids sterile dominant mutant cannot propagate sexually but can be asexually If cross doesn t have 11 or 31 ratio can be due to interactions of several genes or environmental effect Forward genetics Forward genetics approach to understanding biological function starting with random singlegene mutants and ending with detailed cellbiochemical analysis genomic analysis Choose property of interest nd mutants affecting that property check mutants for singlegene inheritance l identify timeplace of action of genes l zero in on molecular nature of gene by genomic DNA analysis Predicting progeny proportions or parental genotypes by applying the principles of singlegene inheritance Gene discovery 0 Observe phenotypic ratios in progeny l deduce genotypes of parents AA Aa or aa Equal segregation 0 Cross parents of known genotypes l predict phenotypic ratios in progeny 0 NA x Aa AA x aa etc If singlegene inheritance individual with dominant phenotype but of unknown genotype Al can be tested to see if genotype homozygous or heterozygous 0 Al x aa o Heterozygous 11 12 Na 12 aa Homozygous all dominant phenotype Na 0 Testcross cross of an individual organism of unknown genotype or a heterozygote or a multiple heterozygote with a tester recessive individual o If tester not available self the unknown lf heterozygous 31 0 Principles of inheritance such as the law of equal segregation can be applied in two directions 0 1 inferring genotypes from phenotypic ratios 0 2 predicting phenotypic ratios from parents of known genotypes O 25 Sexlinked SingleGene Inheritance Patterns chromosomes analyzed thus far are autosomes regular chromosomes that form most of genomic set 0 sex chromosomes also segregate equally o phenotypic ratios different from autosomal ratios Sex chromosomes chromosome whose presence or absence is correlated with the sex of the bearer a chromosome that plays a role in sex determination Humans have 46 chromosomes 22 homologous pairs of autosomes 2 sex chromosomes 0 Females have pair of identical sex chromosomes X chromosomes 0 Males have nonidentical pair X and Y shorter than X Meiosis in females 2 X chromosome pair segregates each egg gets 1 X chromosome homogametic sex Meiosis in males X and Y chromosome pair segregates half sperm get X and other half get Y heterogametic sex 0 Inheritance patterns of genes on sex chromosomes different from autosomal genes 0 Drosophila has 3 pairs of autosomes 1 pair of sex chromosomes 0 Number of X chromosomes in relation to autosomes determines sex 2 Xs l female 1 lemale 0 similar inheritance patterns of genes on sex chromosomes to humans vascular plants show variety of sexual arrangements 0 those with sexual dimorphism most have nonidentical pair of chromosomes associated with its sex 0 others have no visibly different pair of chromosomes Sexlinked patterns of inheritance Divide X and Y chromosomes into homologous and differential regions 0 Hemizygous A gene present in only one copy in a diploid organism Ex an Xlinked gene in a male mammal 0 Genes in differential regions show inheritance patterns called sex linkage location of a gene on a sex chromosome 0 X linkage inheritance pattern of genes found on the X chromosome but not on the Y chromosome 0 Y linkage inheritance pattern of genes found on the Y chromosome but not on the X chromosome rare Contrasts autosome inheritance pattern which are in the same sex 0 If genomic location unknown sexlinked inheritance pattern indicates that the gene lies on a sex chromosome Pseudoautosomal regions 1 and 2 Small regions at the ends of the X and Y sex chromosomes they are homologous and undergo pairing and crossing over at meiosis o X and Y can act as a pair and segregate into equal numbers of sperm Xlinked inheritance Drosophila eye color determined by 2 alleles of gene on differential region of X chromosome 0 Mutant white eye w wild type red eye w 0 White males crossed with red females all progeny red l white is recessive 0 white F1 males crossed with red females l 31 redwhite ratio but all white are males 0 All F1 receive WT allele from mom but F1 females also received white allele from dad All F1 females are heterozygous WT ww F1 female pass white allele to 12 sons and 12 to daughters who don t express it Must inherit WT from dad 0 white F1 females x red males all females red all males white 0 every female inherited dominant w from father s X chromosome 0 males inherited recessive w from mother 0 F2 was half and half for both sexes Sex linkage different ratios in different sexes and different bw reciprocal crosses Abnormal allele associated with white recessive but abnormal alleles of genes on X chromosome that are dominant arise The chromosome theory of inheritance genes located on chromosomes proven by inheritance pattern correlated with 1 speci c chromosome pair 26 Human Pedigree Analysis Pedigree analysis Deducing singlegene inheritance of human phenotypes by a study of the progeny of matings within a family often stretching back several generations 0 Propositus the person who rst came to the attention of the geneticist has some medical disorder 0 Male square 0 Female circle 0 Filled in black affected individual Singlegene inheritance ratios usually not seen bc humans have few children 0 Approach depends on whether one of the contrasting phenotypes is a rare disorder or both phenotypes of a pair are common polymorphism o 2 phenotypes presence and absence of the disorder 0 4 patterns of singlegene inheritance revealed Autosomal recessive disorders Affected phenotype inherited as a recessive allele p o Sufferers pp o Unaffected Pp or PP Patterns in pedigree that would reveal autosomal recessive inher ance o 1 generally the disorder appears in progeny of unaffected parents 0 2 affected progeny include both males and females infers simple Mendelian inheritance of gene on an autosome gene not on a sex chromosome pedigrees of autosomal recessive disorders tend to have few black symbols 0 recessive condition shows up in groups of affected siblings people in earlierlater generations tend not to be affected 0 bc most people who carry the abnormal allele are heterozygous o heterozygotes more common than recessive homozygotes birth of affected person depends on rare chance of unrelated heterozygous parents 0 inbreeding increase chance that 2 heterozygotes will mate Autosomal dominant disorders Pedigree patterns expected 0 Phenotype tends to appear in every generation 0 Affected fathers or mothers transmit the phenotype to both sons and daughters Aa much more common than AA so most affected people are heterozygotes o Matings usually Aa x Aa 11 ratio Autosomal polymorphisms Polymorphism morphs the coexistence of 2 or more common phenotypes of a character 0 Often inherited as alleles of a single autosomal gene in Mendelian manner 0 Ex eye color hair color 0 Dimorphism polymorphism with only 2 forms most common Xlinked recessive disorders Pedigrees will show 0 1 More males than females with the rare phenotype bc female can only inherit the genotype if both mom and dad have that allele male can inherit phenotype only when the mom carries the allele 0 2 no offspring of affected male show the phenotype but all daughters are carriers heterozygotes in next generation 12 sons of carrier daughters show phenotype o 3 no sons on an affected male show the phenotype AA will not pass on condition bc sons obtains his Y chromosome from father cannot inherit father s X chromosome too 0 normal female of unknown genotype assumed to be homozygous ex redgreen color blindness 0 genetic determinants for red and green cone cells on the X chromosome color blind people have mutation in 1 of these 2 genes 0 affects more males than females 0 ex hemophilia failure of blood to clot ex Duchenne muscular dystrophy fatal phenotype is a wasting and atrophy of muscles 0 ex testicular feminization syndrome chromosomally males 44 autosomes and an X and Y chromosome but they develop as females 0 condition not reversed with male hormone therapy bc of mutation in androgen receptor gene Xlinked dominant disorders Pedigree characteristics 0 1 affected males pass condition on to all daughters but none of their sons 0 2 affected heterozygous females married to unaffected males pass condition to 12 sons and daughters 0 not common mode of inheritance Ylinked inheritance 0 Only males inherit genes in differential region of the Y chromosome 0 Fathers transmit genes to their sons SRY gene testisdetermining factor plays primary role in maleness maleness gene residing on the Y chromosome 0 Maleness itself is Y linked 0 Shows expected pattern of exclusively maletomale transmission 0 Male sterility can be caused by deletions of Y chromosome but sterility not heritable o Fathers have normal Y chromosomes shows deletions are new 0 inheritance patterns with unequal representation of phenotypes in malesfemales can locate the genes concerned to 1 of the sex chromosomes Calculating risks in pedigree analysis When disorder with well documented singlegene inheritance known win a family transmission patterns used to calculate probability of prospective parents having kid with the disorder Wifehusband both hav uncle with TaySachs disease autosomal recessive 0 Neither have the disease so either 1139 or Tt 0 First calculate prob of both being heterozygotes then if so the prob of passing deleterious allele to kid 1 husbands gparents both Tt bc had son tt o gparents a monohybrid cross Tt x Tt 14 TT 12 Tt 14 tt o husbands father TT 14 or Tt 12 0 prob father heterozygote 12 divided by 34 23 2 husbands mother assumed to be TT o husbands momdad TT x Tt 12 39IT and 12 Tt 3 overall probability of husband being heterozygous calculated using the product rule probability of two independent events occurring simultaneously is the product of the individual probabilities o prob father being heterozygous x prob father having heterozygous son 0 23 x 12 13 4 13 for wife also 5 if both Tt would be monohydrib cross prob of kid with tt 14 6 prob of having affected kids prob of them both being heterozygous and then both transmitting the recessive allele 0 independent events 0 13 x 13 x 14 136 27 Summary Somatic cell division genome transmitted by mitosis nuclear division 0 Mitosis haploid or diploid each chromosome replicates l pair of chromatids chromatids pulled apart l 2 identical daughter cells 0 Sexual cell division takes place in sexual cycle in meiocytes o Meiosis diploid Each homolog replicates l dyad of chromatids Dyads pair l tetrad Tetrad segregates at each of the 2 cell divisions l 4 haploid cells gametes o Haploid organism unite l diploid meiocyte Genetic dissection begins with a collection of mutants 0 Each mutant tested for singlegene inheritance 0 Analysis based on observing speci c phenotypic ratios in progeny of controlled crosses 0 11 31 and 121 ratios stem from equal segregation principle haploid products of meiosis from Aa will be 12 A and 12 a 0 molecular force of segregation is the depolymerization and subsequent shortening of microtubules attached to centromeres o recessive mutations usually in haplosufficient genes 0 dominant mutations usually in haploinsufficient genes 0 sex determined chromosomally in many organisms o XX female XY male 0 Genes on X chromosome Xlinked genes have no counterparts on Y chromosome 0 show singlegene inheritance that differs in the 2 sexes 0 results in different ratios for malefemale progeny singlegene segregation used to identify mutant allele 0 analysis of pedigrees reveals autosomal or Xlinked disorders of both dominant and recessive types 0 31 Mendel39s Law of Independent Assortment Aa Bb I 2 genes on different chromosomes ABab l genes on the same chromosome Aa Bb l unknown if genes on same or different chromosome Monohybrid heterozygote for a single gene Aa o Dihybrid heterozygote Aa Bb o Dihybrid cross Aa Bb x Aa Bb Mendel started with 2 pure parental lines wrinkledyellow and roundgreen rr YY x RR yy 0 Produced gametes rY and Ry 0 F1 a dihybrid Rr Yy seeds were round and yellow 0 Selfed F1 to get F2 Rr Yy x Rr Yy 0 F2 gave 4 different types of proportions 9331 from 2 different 31 ratios combined at random Mendel s second law principle of independent assortment gene pairs on different chromosome pairs assort independently at meiosis gamete formation 0 For 2 heterozygous gene pairs Aa and Bb the b allele is just as likely to end up in gamete with a as with A o Applies to genes on different chromosomes 0 1111 gametic ratio produced by dihybrid RrYy ratios of 1111 and 9331 are diagnostic of independent assortment in 1 and 2 dihybrid meiocytes 32 Working with Independent Assortment Predicting progeny ratios Genetics can work in 2 directions 0 1 predicting the genotypes of parents by using ratios of progeny o 2 predicting progeny ratios from parents of known genotype Punnett squares used to show hereditary patterns based on 1 gene pairs Branch diagram better for many gene pairs To calculate probabilities of speci c phenotypes coming from cross 0 Product rule the probability of independent events both occurring together I the product of their individual probabilities 0 Sum rule the probability of either of 2 mutually exclusive events occurring is the sum of their individual probabilities What proportion of progeny will be of a speci c genotype 0 Have 2 plants of genotypes AabbCcDdEe AaBbCcddEe Assume all gene pairs assort independently 0 Cross each gene to obtain 5 individual probabilities Aa x Aa 0 Use product rule to nd overall probability multiply the 5 individual probabilities together 39 14 X 12 X 14 X 12 X 14 1256 How many progeny do we need to grow 0 Want to estimate how many progeny need to be grown to have a 95 chance of obtaining desired genotype aabbccddee Average probability of success 1256 0 Probability of failure obtaining no progeny with the desired genotype 1 1256 255256 probability of no successes in sample of n 255256quotn probability of at least 1 success 1 255256quotn o to satisfy 95 con dence 095 1 255256quotn n 765 number of progeny needed to guarantee success How many distinct genotypes will a cross produce 0 Self the tetrahybrid AaBbCcDd o 3 genotypes for each gene pair Aa AA aa 0 O OO 0 3quot4 81 different genotypes o testcross of this tetrahybrid 2 genotypes for each gene pair Aa and aa 2A4 16 genotypes in progeny assuming all genes on different chromosomes testcross genotypes occur at equal frequency of 116 Using the chisquare test on monohybrid and dihybrid ratios Often get results that are close to expected ratio but not identical want to know how close to expected is close enough Chisquare Xquot2 test determines the probability of obtaining observed proportions by chance 0 Xquot2 test is a way of quantifying the various deviations expected by chance if a hypothesis is true Xquot2 sum ofOEquot2 E for all classes 0 E expected number in a class 0 O observed number in a class 0 Look up Xquot2 value in table to get the probability 0 Rows list degrees of freedom df number of independent variables in the data of phenotypic classes minus 1 o If Plt005 l reject hypothesis Synthesizing pure lines 0 Pure lines essential tools of genetics bc 0 Only pure lines fully homozygous lines will express recessive alleles 0 Pure lines can act as a constant source of the genotype for experiments if left to interbreed Pure lines made through repeated generations of sel ng 0 Start with population of heterozygotes and allow them to self 0 next generation has half the heterozygosity proportion of heterozygotes 0 repeat sel ng for another generation all progeny of homozygotes will be homozygous but again heterozygosity will be halved 0 keep sel ng for numerous generations to reduce heterozygosity o to create a pure line with traits from 2 different plants 0 cross the 2 plants with the desired traits 0 then self F1 with desired trait 0 select the homozygote of each trait an cross self and select homozygote again 0 when a multiple heterozygote is selfed a range of different homozygotes produced 0 each distinct homozygote can be the start of a new pure line repeated sel ng leads to an increased proportion of homozygotes a process that can be used to create pure lines for research or other appHcanns Hybrid vigor situation in which an F1 is larger or healthier than its two different pure parental lines 0 When 2 disparate lines of plants united in an F1 hybrid presumed heterozygote hybrid shows greater size and vigor than the 2 contributing lines 0 Negative aspects 0 every season the 2 parental line must be grown separately and then intercrossed to make hybrid seed With pure lines just can let them self 0 Seeds from hybrid cannot be expected to have equal vigor the next year When hybrid undergoes meiosis independent assortment forms many different allelic combinations very few combos will be that of the hybrid Some hybrids bw genetically different lines show hybrid vigor 0 But gene assortment when the hybrid undergoes meiosis breaks up the favorable allelic combination thus few members of the next generation have it 33 The Chromosomal Basis of Independent Assortment like equal segregation independent assortment of gene pairs on different chromosomes explained by behavior of chromosomes during meiosis o homologs of 2 different chromosomes can be pulled in same or opposite directions segregate independently Independent assortment in diploid organisms law of independent assortment separate behavior of 2 different chromosome pairs gives rise to 1111 ratio of gametic types 0 random fusion of these gametes results in the 9331 F2 phenotypic ratio Independent assortment in haploid organisms 0 Products of meiosis in fungi are sexual spores 0 When they come together form diploid meiocyte In Neurospora nuclear spindles of meiosis l and II don t overlap so 4 products of a single meiocyte lie in straight row 0 ln heterozygous meiocyte Aa with no crossovers bw the gene and its centromere get 2 adjacent blocks of ascospores 4 A and 4 a 0 Examine a dihybrid cross bw 2 distinct mutants having mutations in different genes on different chromosomes 0 First mutant is albino a contrasts pink wild type a 0 Second is biscuit b compact colony WT b o ab x ab o bc of spindle attachment get equal frequency demonstrated independent assortment occurring in individual meiocytes Independent assortment of combinations of autosomal and Xlinked genes Autosomes and sex chromosomes moved independently by spindle bers attached randomly to their centromeres lf autosomal and Xlinked genes combined F2 phenotypic ratios will reveal elements of both autosomal and Xlinked inheritance Recombination Recombination production of new allele combinations 0 1 any process in a diploid or partly diploid cell that generates new gene or chromosomal combinations not previously found in that cell or in its progenitors o 2 At meiosis the process that generates a haploid product of meiosis whose genotype is different from either of the two haploid genotypes that constituted the meiotic diploid o recombination provides variation raw material for natural selection meiotic recombination any meiotic process that generates a haploid product with new combinations of alleles carried by the haploid genotypes that unite to form the meiocyte 0 input 2 haploid genotypes than combine to form meiocyte diploid cell that undergoes meiosis 0 output genotypes that are the haploid products of meiosis detection of recombinants in organisms with haploid life cycles 0 inputoutputs are the genotypes of individuals rather than gametes 0 can be inferred directly from phenotypes detection of recombinants in organisms with diploid life cycles O inputoutputs are gametes 0 must know genotypes of both gametes to detect recombinants 0 cannot detect directly use techniques to know input gametes use pure breeding diploid parents bc they can only produce 1 gametic types to detect recombinant output gamete testcross diploid individuals observe progeny recombinant frequency a test for whether 2 genes are on different chromosomes 0 recombinants produced by 2 different cellular processes independent assortment of genes on different chromosomes crossing over bw genes on the same chromosome 0 genes on separate chromosomes recombinants produced by independent assortment 1111 proportion of recombinants is 50 of total progeny independent assortment at meiosis produced recombinant frequency of 50 genes most likely on different chromosomes 34 Polygenic Inheritance 0 large proportion of variation in natural populations takes form of continuous variation opposed to singlegene differences 0 found in characters than can take measureable value bw 2 extremes height weight 0 when character quanti ed and plotted against frequency in population distribution is a bell curve avg in middle most common extremes more rare 0 many cases of continuous variation have purely environmental basis plant height with water availability 0 other cases are genetic skin color polygenesquantitative trait loci QTLs interacting genes underlying hereditary continuous variation 0 distributed throughout genome 0 often on different chromosomes show independent assortment if alleles at each gene pair approximately equal frequency in population then dihybrid cross can represent avg cross for a population in which 2 polygenes are segregating O variation and assortment of polygenes can contribute to continuous variation in a population polygenes no special class of genes just are alleles that contribute to continuous variation 35 Organelle Genes Inheritance Independent of the Nucleus nucleus contains most of eukaryote s gene distinct and specialized subset of genome found in mitochondria and chloroplasts O O inherited independently of nuclear genome constitute special case of independent inheritance extranuclear inheritance mitochondriachloroplasts specialized organelle in cytoplasm O O O 0 contain small circular chromosomes carrying a de ned subset of total cell genome mitochondrial genes concerned with mito s task of energy production oxidative phosphorylation chloroplast genes carry out function of photosynthesis neither function autonomously rely on nuclear genes many organelle genes present in a cell bc many organelles per cell and each cell has many copies of its chromosome 0 DNA sometimes packaged into suborganellar structures nucleoids 0 Assume all copies of an organelle chromosome within a cell are identical Organelle genes very closely spaced some organisms have introns O Mitochondrial DNA mtDNA subset of the genome found in the mitochondrion specializing in providing some of the organelle s functions Chloroplast DNA chNA small genomic component found in the chloroplasts of plants concerned with photosynthesis and other functions taking place within that organelle Patterns of inheritance in organelles Uniparental inheritance progeny inherit organelle genes exclusively from 1 parent but not the other 0 In most cases that parent is the mother pattern called maternal inheritance Occurs bc organelle chromosomes located in cytoplasm and malefemale gametes do not contribute cytoplasm equally to the zygote egg contributed the majority 0 Nuclear genes both parent contribute equally Organelle inheritance pattern 0 Mutant female x WT male l progeny all mutant 0 WT female x WT male l progeny all WT o Variant phenotypes by mutations in cytoplasmic organelle DNA are generally inherited maternally and independent of the Mendelian patterns shown by nuclear genes Cytoplasmic segregation Cytohetesheteroplasmons cells that contain mixtures of mutant and normal organelles o Cytoplasmic segregation detected in cytohets the 2 types apportion themselves into different daughter cells 0 Most likely stems from chance partitioning in cell division Variegated zygotes demonstrate cytoplasmic segregation o Progeny come from eggs that are cytohets 0 When zygote divides white and green chloroplasts segregate dividing themselves into separate cells yielding green and white sections 0 How is it possible to obtain a pure mutant cell given that a cell is a population of organelle molecules 0 Pure mutants created in asexual cells Variants arise by mutation of a single gene in a single chromosome Mutationbearing chromosome may by change increase in frequency in population within the cell random genetic drift 0 Cytoplasmic segregation NOT a mitotic process although it does take place in dividing asexual cells Organelle populations that contain mixtures of 2 genetically distinct chromosomes often show segregation of the 2 types into the daughter cells at cell division cytoplasmic segregation Alleles on organelle chromosomes 0 In sexual crosses inherited from 1 parent only usually mother No segregation ratios 0 In asexual cells show cytoplasmic segregation can occasionally show processes analogous to crossing over when cytohets are dihybrid Cytoplasmic mutations in humans Some human pedigrees show transmission of rare disorders only thru females 0 Patterns suggests cytoplasmic inheritance and mutation in mtDNA as reason for the phenotype Some cases cells of sufferer contain mix of normal and mutant chromosomes 0 proportions of each passed on to progeny can vary as result of cytoplasmic segregation o proportions in 1 person can vary in different tissues over time MtDNA in evolutionary studies Differencessimilarities of homologous mtDNA sequences bw species used to construct evolutionary trees Possible to introduce extinct organisms into evolutionary trees using mtDNA from remains MtDNA evolved rapidly useful in plotting recent evolution 0 Suggest Homo sapiens originated in Africa 36 Summary synthesis of genotypes are complex combinations of alleles from different genes 0 genes can be on same or different chromosomes dihybrid 2 gene pairs on different chromosome pairs each individual gene pair shows equal segregation at meiosis o bc nuclear spindle bers attach randomly to centromeres at meiosis 2 gene pairs partitions independently into meiotic products from a dihybrid AaBb 3 genotypes products AB Ab Ba ab all at equal frequency of 25 o testcross of dihybrid with double recessive phenotypic proportions of progeny also 25 1111 o 1111 and 9331 ratios diagnostic of independent assortment more complex genotypes composed of independently assorting genes treated as extensions of singlegene inheritance 0 product rule multiplying proportions of relevant individual genes 0 sum rule probability of the occurrence of any of several categories of progeny o Xquot2 test used to test whether observed proportions conform to expectations of a genetic hypothesis Plt05 l hypothesis rejected Sequential generations of sel ng increases proportions of homozygotes genes on different chromosomes 0 Sel ng used to create complex pure lines with combinations of desired mutations Independent assortment of chromosomes at meiosis can be observed cytological by using heteromorphic chromosome pairs show structural difference 0 Ex X and Y chromosomes Main function of meiosis is to produce recombinants new combinations of alleles of haploid genotypes that united to form the meiocyte o Dihybrid testcross showing independent assortment recombinant frequency 50 Continuous distributions can be based on environmental variation or on variant alleles of multiple genes or combo of both 0 Polygenes active alleles of several genes that contribute moreless to the character 0 Produces bell curve with plotted against frequency Small subset of genome found in mitochondriachloroplasts o Inherited independently of nuclear genome o Marternal inheritance o Genetically mixed cytoplasm cytohets the 2 genotypes often sort themselves into different daughter cells by cytoplasmic segregation 0 Mapping used to nd the location of genes on a chromosome Reasons mapping important Strain building Gene position crucial for building complex genotypes needed for experimentscommercial applications Discovering gene s function Knowing gene position provides way of zeroing in on its structure and function Interpreting evolutionary mechanisms Genes present and their location on chromosomes often slightly different in related species by comparing differences can deduce evolutionary genetic mechanisms through with genomes diverged Chromosome map representation of all chromosomes in the genome as lines marked with the positions of genes known from their mutant phenotypes plus molecular markers Based on analysis of recombinant frequency Loci locus gene position Distances bw loci based on a scale Recombinationbased maps map loci of genes that have been identi ed by mutant phenotypes showing singlegene inheritance 0 Physical maps show genes as segments arranged along the DNA molecule that constitutes a chromosome 41 Diagnostics of Linkage o Recombination maps chromosome map in which the positions of loci shown are based on recombinant frequencies 0 Assembled 23 genes at a time via linkage analysis 0 Linked genes loci of the genes on the same chromosome Alleles on any 1 homolog are physically joined linked by the DNA bw them Using recombinant frequency to recognize linkage o In a self of a dihybrid F1 F2 did not show 9331 ratio independent assortment found that certain combos of alleles showed up more often than expected 0 Morgan proposed linkage as hypothesis to explain apparent allele association 0 Morgan performs cross to get dihybrids then did testcross only recessive alleles o phenotypes of offspring reveal alleles contributed by gametes of dihybrid parent 0 found that 2 allele combinations showed up much more often than the others they re linked can assess testcross results with percentage of recombinants in progeny o recombinants are the types that are not the genotypes contributed by F1 dihybrid o recombinant frequency total recombinants total progeny x 100 o linked genes on same chromosome so parental allelic combinations held together in majority of progeny Morgan did another cross each parent homozygous for WT allele of 1 gene and MT allele of the other 0 Get different ratio but similar recombinant frequency as before 0 General pattern of dihybrid testcross results 0 2 equally frequent nonrecombinant classes totaling in excess of 50 o 2 equally frequent recombinant classes totaling less than 50 0 when 2 genes close together on same chromosome pair linked they don t assort independently produce a recombinant frequency lt50 0 recombinant freq lt50 l linkage How crossovers produce recombinants for linked genes 0 linkage hypothesis explains why allele combinations from the parental generations remain together genes are physically attached by segment of chromosome bw them crossing over exchange of corresponding chromosome parts between homologs by breakage and reunion 0 when homologous chromosomes pair at meiosis chromosomes occasionally break and exchange parts 0 crossover products 2 new combinations from crossover at meiosis when duplicated homologous chromosomes pair with each other 2 dyads unite as a bivalent crossshaped chiasma often forms bw 2 nonsister chromatids for linked genes recombinants produced by crossovers o chiasmata are visible manifestations of crossovers Linkage symbolism and terminology o Linked genes in a dihybrid present in 1 of 2 conformations o Cis conformation 2 dominant or WT alleles present on same homolog A1A2a1a2 0 Trans conformation they re on different homologs a1 a2 Alleles on same homolog have no punctuation bw them Slash separates the 2 homologs Genes on different chromosomes unlinked genes separated by semicolon Aa Cc Unknown linkage separated by dot Aa Dd Evidence that crossing over is a breakage and rejoining process Recombinants produced by some kind of exchange of material bw homologous chromosomes 0 Plants was a dihybrid in cis conformation o Progeny of testcross compared recombinants and parental genotypes 0 found recombinants inherited 1 or the other of the 2 chromosomes crossover results from breakage and reunion of DNA 0 2 parental chromosomes break at same position then each piece joins with neighboring piece from other chromosome Evidence that crossing over takes place at the fourchromatid stage crossover takes place at the 4chromatid stage of meiosis crossovers are bw nonsister chromatids crossing over cannot take place at the 2strand stage before DNA replication bc more than 2 different products of a single meiosis can be seen in some tetrads o if crossovers at 2chromosome stage could only ever be a max of 2 different genotypes in an individual tetrad Multiple crossovers can include more than 2 chromatids Tetrad analysis shows 2 features of crossing over 0 In some individual meiocytes several crossovers can occur along a chromosome pair 0 In any 1 meiocyte multiple crossovers can exchange material bw more than 2 chromatids But single crossover is bw 2 chromatids 42 Mapping by Recombinant Frequency fungal tetrad shows that for any 2 speci c linked genes crossovers take place bw them in some but not all meiocytes o farther apart genes are more likely crossover will take place and higher proportion of recombinant products 0 proportion of recombinants clue to distance separating 2 gene loci on chromosome map recombinant freq lt50 has linkage closer to 50 the father apart genes are Map units 0 Linkage map abstract map of chromosomal loci that is based on recombinant frequencies 0 Greater distance bw the linked genes l greater chance of crossovers in region bw the genes l greater proportion of recombinants produced 0 Genetic map unit centimorgan cM distance on the chromosome map corresponding to 1 percent recombinant frequency 0 Recombinant frequency RF proportion or percentage of recombinant cells or individuals 0 Genes arranged linearly making map distances additive Distance on a linkage map is a physical distance along a chromosome 0 Physical mapping has shown that genetic distances are roughly proportional to recombination distances 0 Exceptions caused by recombination hotspots places in genome where crossing over takes place more frequently than normal expansion of some regions and recombination blocks opposite effect Crossovers occur randomly along chromosome pair o Longer regions avg number of crossovers higher more recombinants produced longer map distance Threepoint testcross Threepoint testcross testcross in which one parent has three heterozygous gene pairs 0 Used to deduce whether 3 genes are linked and if they are their order and distances bw them 0 Testcross trihybrid female with triple recessive males 0 2quot3 8 gamete genotypes possible 0 R genotypes are recombinant for the loci taken 2 at a time Any combination different from the 2 parent genotypes constitutes a recombinant 0 Add up total recombinants R and divide by total progeny RF 0 Largest RF genes farthest apart can deduce which gene in middle 0 Linkage maps just the loci in relation to one anther don t know where loci are on a chromosome 0 Can also identify 2 rarest classes double recombinants Deducing gene order by inspection 0 For trihybrids of linked genes gene order can be deduced without recombinant frequency analysis 0 For linked genes have 8 genotypes with frequencies 0 2 at high freq 0 2 at intermediate freq 0 2 at a different intermediate freq 0 2 rare low freq 0 only 3 gene orders possible each with different gene in middle position 0 to deduce gene in middle allele pair that has ipped position in double recombinant classes lowest freq Interference Interference interaction bw crossovers that inhibit each other 0 measure of the independence of crossovers from each other calculated by subtracting the coefficient of coincidence from 1 0 double recombinant classes can be used to deduce extent of interference how to measure interference o if crossovers in 2 regions independent product rule l predicts freq of double recombinants l equal to product of recombinant frequencies in adjacent regions 0 if continually get different number of double recombinants than what s actually observed l interference l crossover reduces probability of crossover in adjacent region coefficient of coincidence coc ratio of the observed number of double recombinants to the expected number 0 interference 1 coc drosophilia males show no crossing over in humans women show higher recombinant freq than males Using ratios as diagnostic 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 assort 11111111 000000 43 Mapping with Molecular Markers ways to map gene loci 0 RF values by counting visible phenotypes produced But some differences in DNA bw 2 chromosomes don t produce visibly different phenotypes 0 Molecular markers site of DNA heterozygosity not necessarily associated with phenotypic variation used as a tag for a particular chromosomal locus 0 Useful as genomic landmarks that can be used to locate genes of interest 2 types of molecular markers o singlenucleotide polymorphisms o simplesequencelength polymorphisms Single nucleotide polymorphisms Genomic sequences of individuals of same species 999 identical 0 01 difference based on single nucleotide differences a change in just 1 nucleotide 0 both molecular allele common in population 0 Single nucleotide polymorphisms SNPs nucleotidepair difference at a given location in the genomes of two or more naturally occurring individuals 0 Can lie win genes or not 0 Change in single nucleotide pair could produce a new allele causing mutant phenotype 0 Most SNPs don t produce different phenotypes bc Don t lie in a gene or Lie in gene but both versions produce same protein 0 2 ways to detect a SNP 0 sequence a segment of DNA in homologous chromosomes and compare that homologous segments 0 Restriction fragment length polymorphisms RFLPs difference in DNA sequence between individuals or haplotypes that is recognized as different restriction fragment lengths for SNPs located at a restriction enzyme s target site there will be 2 RFLP alleles morphs 1 of which has the restriction enzyme target and other doesn t enzyme will cut DNA at SNP with the target SNPs then detected as different bands in electrophoresis Simple sequence length polymorphisms o Genomes contain repetitive DNA in different individuals there s often different numbers of copies 0 Simple sequence length polymorphisms SSLPs existence in the population of individuals showing different numbers of copies of a short simple DNA sequence at one chromosomal locus aka VNTRs 0 Variable number tandem repeats VNTRs A chromosomal locus at which a particular repetitive sequence is present in different numbers in different individuals or in the two different homologs in one diploid individual 0 2 types of SSLPS o minisatellite markers based on variation in the number of tandem repeats of a repeating unit from 15100 nucleotides long 0 microsatellite markers based on variable numbers of tandem repeats of an even simpler sequence dinucleotide most common type repeat of CA and its complement GT Detecting simple sequence length polymorphisms Homologous regions with different numbers of tandem repeats will be different lengths Use PCR to amplify these differences then electrophoresis In case of minisatellites patterns produced on gel referred to as DNA ngerprints Recombination analysis using molecular markers 0 Individual has base pair GC at position 5658 on 1 chromosome and AT at same position on other chromosome 0 Individual is molecular heterozygote ATGC for that position 0 Mapped just like phenotypic heterozygote Na 0 Locus inserted into chromosomal map by analyzing recombinant freq 0 Molecular markers useful in locating a gene of interest 0 bc tell you how close you are to a destination 0 to zero in on disease gene do a number of crosses 0 cross individual with disease and individual with 1 range of molecular markers whose positions are known 0 using PCR parentsprogeny scored for markers then recombination analysis used to see if gene of interest linked to any of the markers 0 different markers can be mapped to each other creating map that can act like a series of stepping stone on way to some gene loci of any DNA heterozygosity can be mapped and used as molecular chromosome markers or milestones 44 Centromere Mapping with Linear Tetrads in eukaryotes recombination analysis cannot be used to map loci of centromeres bc they show no heterozygotsity enabling them to be used as markers 0 but in fungi that produce linear tetrads centromeres can be mapped o in fungi haploid meiotic divisions take place along axis of the ascus so each meiocyte produces linear array of 8 ascospores an octad plus a postmeiotic mitosis centromere mapping considers a gene locus and asks how far this locus is from its centromere 0 based on fact that different pattern of alleles will appear in linear tetrad or octad that arises from meiosis with a crossover bw a gene and its centromere o if no crossover in region bw Ala and centromere l 2 adjacent blocks of 4 ascospores in linear octad rstdivision segregation patterns M1 patterns 0 with a crossover l 1 of 4 different patterns in octad each pattern showing blocks of 2 adjacent identical alleles seconddivision segregation patterns M2 difference is that chromatids move in different directions at second division 0 frequency of octads with M2 pattern proportional to size of the centromereAa region used as measure of size of region 0 M2 freq 42300 14 o Crossover in meiosis results in only 50 recombinant chromatids 0 To convert freq of meiosis to freq of recombinant chromatids 142 7 region is 7 map units in length 45 Using the ChiSquare Test for Testing Linkage Analysis 0 Xquot2 test provides way of deciding if 2 genes are linked o 2 genes linked on same chromosome if RFlt50 0 but if RF close to 50 use Xquot2 test 0 rst calculate expectations E for each class 0 but cannot get E without knowing linkage distance 0 instead can use independent assortment absence of linkage 0 null hypothesis hypothesis that proposes no difference between two or more data sets if observed results cause us to reject hypothesis of no linkage can infer linkage observed values 0 A 255500 a 254500 B 254500 b 246500 0 E for A B 255500 x 254500 x 500 12954 Do for entire grid 0 Expected values 0 Xquot2 OEquot2E 497 Do for each combination 0 Degrees of freedom classes in rows 1 x of classes in columns 1 Df 21x211 0 Look at table 497 very close to 5021 so p value close to 025 or 25 plt5 so hypothesis of independent assortment rejected no linkage l loci probably linked 46 Accounting for Unseen Multiple Crossovers o in 3point testcross some parental nonrecombinant chromatids resulted from double crossovers o initially could not be counted in recombinant freq skewing results A mapping function a formula that related an observed recombinant frequency value to a map distance corrected for multiple crossovers o relate RF to mean number of crossovers m that must have taken place in that chromosomal segment per meiosis 0 then deduce map distance this m value should have produced 0 to nd relation of RF to m think of outcomes of various crossover possibilities 0 true determinant of RF is the relative sizes of the classes with no crossovers zero class compared with classes with any nonzero number of crossovers now calculate size of zero class 0 Poisson distribution distribution giving the probability of observing various numbers of a particular event in a sample when the mean probability of an event on any one trial is very small Described occurrence of crossovers in a speci c chromosomal region Described distribution of successes in samples when avg probability of success is low Proportion of classes with different numbers of crossovers n Fi equotm x mquotii El e base of natural logarithms n m mean successes in sample size i actual succeses in sample of that size freq of samples with l successes in them factorial symbol 0 freq of the i0 class is equotm o freq of class with any nonzero number of crossovers 1 equotm in meiosis 50 of products will be recombinant so RF 121equotm To get corrected map distance 0 Multiply calculated avg crossover freq by 50 The Perkins formula 0 For fungi and other tetradproducing organisms another way of compensating for multiple crossovers Corrected map distance 50T 6NPD o T tetratype Produced from crossover bw either of the 2 loci and their centromeres Size of T class depends on total size of 2 regions bw locus and centromere o NPD nonparent ditype El El El 47 Using RecombinationBased Maps in Conjunction with Physical Maps Recombinantion maps show the loci of genes for which mutant alleles and their mutant phenotypes have been found 0 Determined by freq of recombinants at meiosis o Freq assumed to be proportional to distance bw 2 loci on chromosome 0 Sites of molecular heterozygosity not associated with mutant phenotypes can be incorporate into maps via molecular markers 0 Like any heterozygous site markers mapped by recombination then used to zero in on gene of interest Recombination maps are hypothetical physical maps are closer to real genome map Physical maps ordered and oriented map of cloned DNA fragments on the genome 0 Units of distance are numbers of DNA bases kilobase Union of recombination and physical maps can ascribe biochemical function to a gene identi ed by its mutant phenotype 48 The Molecular Mechanism of Crossing Over heteroduplex DNA DNA in which there is one or more mismatched nucleotide pairs in a gene under study heteroduplex DNA and crossovers outcomes of doublestranded break DNA break cleaving the sugarphosphate backbones of both strands of the DNA double helix 0 O O O O 1 both chromatids of a pair break in same location 2 DNA eroded in 5 strand of each broken end leaving both 3 ends single stranded 3 1 of the single stands invaded DNA of other chromatid 4 tip of invading strand uses adjacent sequence as template for new polymerization proceeds by forcing the 2 resident strands of helix apart displaced single stranded bonds with other single strand if invasion and strand displacement spans site of heterozygosity region of heteroduplex DNA formed 5 Holliday junctions 49 Summary gene linkage 2 genes located on same pair of homologous chromosomes explains why parental gene combinations stay together but not how recombinant nonparental combos arise crossing over physical exchange of chromosome parts in meiosis 0 result of physical breakage and reunion of chromosome parts 0 takes place at 4chromoatid stafge of meiosis 2 types of meiotic recombination o recombination by independent assortment results in recombinant freq of 50 o crossing over results in recombinant freq of lt50 RF corresponds to distances bw genes on chromosome l linkage map In ascomycete fungi centromeres can be located on map by measuring seconddivision segregation freq SNPs single nucleotide differences in DNA sequence SSLPs difference in number of repeating units 0 SNP and SSLP used as molecular markers Xquot2 test tells how often observations deviate from expectations purely by chance 0 Used in determining whether loci are linked Multiple crossovers can result in nonrecombinant chromatids leads to underestimation of map distance based on RF 0 Mapping function corrects for this 0 Perkins formula has same use in fungal tetrad analysis Recombination maps of loci conferring mutant phenotypes used in conjunction with physical maps shows genelike sequences 0 Enable melding of cellular function with gene s effect on phenotype Mechanism of crossing over starts with double stranded break in 1 chromatid o Erosion leaves ends single stranded o 1 single strand invaded double helix of other chromatid 0 leads to heteroduplex DNA 0 gaps lled by polymerization Introduction Bacteria are prokaryotes DNA not enclosed in a membranebound nucleus 0 DNA arranged in long series on a chromosome 0 Genome is a single molecule of double stranded DNA in form of a closed circle 0 Contain extra DNA elements plasmids Bacteria can be parasitized by speci c viruses called bacteriophages phages 0 Virus particle consisting of nucleic acid and protein that must infect a living cell to replicate and reproduce Hereditary processes observed in prokaryotes o Asexual division DNA replicated but partitioning of the new copies into daughter cells accomplished by mechanisms different from mitosis o Mutation does occur permits genetic dissection of bacterial funcUon To detect sexlike fusion based on detection of recombinants o If different genome ever do get together in same cell should occasionally produce recombinants o If recombinant detected must have been some type of quotsexualquot union 0 Bacteria different from eukaryotes in that union is of 1 complete chromosome plus a fragment of another Not 2 complete chromosomes brought together Conjugation process of gene exchange from contact and fusion of 2 different cells 0 After fusion donor cell sometimes transfers DNA to other cell 0 Transferred DNA may be part or rarely all of the genome 0 Sometimes plasmids transferred 0 Any genomic fragment transferred may recombine with recipients chromosome after entry Transformation bacterial cell takes up a piece of DNA from external environment and incorporates it into its own chromosome Transduction phase picks up piece of DNA from 1 bacterial cell and injects it into another where it can be incorporated into the chromosome Phage recombination production of recombinant phage genotypes as a result of doubly infecting a bacterial cell with different parental phage genotypes 51 Working with Microorganisms Bacteria good model organisms bc they re fastdividing and small 0 Can be cultured in liquid or solid medium agar gel Bacterial cell divides asexually until nutrients depleted or until toxic waste products accumulate Plating Spreading the cells of a microorganism bacteria fungi on a dish of nutritive medium to allow each cell to form a visible colony 0 Cells divide but cannot travel in gel form colonies visible clones of cells 0 Each colony from a single original cell 0 Cell clone Members of a colony that have a single genetic ancestor Bacterial mutants easy to obtain 0 Ex WT bacteria are prototrophic can growdivide on minimal medium substrate containing only inorganic salts carbon source for energy and water Auxotrophic mutants will not grow unless medium contains 1 speci c cellular building blocks Resistant mutants mutant that can grow in a normally toxic environment 0 These types of mutants allow differentiation of individual strains providing genetic markers An allele used as an experimental probe to keep track of an individual organism a tissue a cell a nucleus a chromosome or a gene 52 Bacterial Conjugation Discovery of conjugation Bacteria possess processes similar to sexual reproduction and recombination 0 Original experiment with E coli 0 2 auxotrophic cultures A and B mixed yielded prototrophic WT 0 cells of type A or type B cannot grow on unsupplemented minimal medium MM bc A and B each carry mutation that causes the inability to synthesize constituents needed for cell growth 0 when A and B mixed and plated however a few colonies appear on agar plate 0 these colonies derive from single cells in which genetic material has been exchanged 0 thus are capable of synthesizing all the required constituents of metabolism ruled out quotcrossfeeding cells of the 2 strains not really exchanging genes but leak substances that other cells can absorb and use for growing 0 physical contact bw the 2 strains required for WT cells to form Conjugation the union of two bacterial cells during which chromosomal material is transferred from the donor to the recipient cell Discovery of the fertility factor F quotCrossesquot via conjugation unequal 1 parent transferred someall of its genome into another cell 0 Donor bacterial cell used in studies of unidirectional DNA transmission to other cells 0 Recipient bacterial cell that receives DNA in a unilateral transfer between cells Donor ability is a hereditary state 0 Fertility factor F A bacterial episome whose presence confers donor ability maleness F cell donor In E coli a cell having a free fertility factor n a male cell F cell recipient In E coli a cell having no fertility factor n a female cell 0 F is a plasmid a small nonessential circular DNA molecule that can replicate in the cytoplasm independent of the host chromosome 0 F plasmid transfer during conjugation Pilus pulls 2 bacteria together A pilus forms bw the 2 cells A singlestranded copy of plasmid DNA produced in donor cell by rolling circle replication a mode of replication used by some circular DNA molecules in bacteria such as plasmids in which the circle seems to rotate as it reels out one continuous leading strand This copy then passes into the recipient bacterium where the single strand serving as a template is converted into the doublestranded helix A copy of F remains in the donor and another appears in the recipient Hfr strains 0 Discovery of a derivative of an F strain with 2 unusual properties 0 1 crossing with F strains new strain produced 1000x more recombinants as normal F strain derivative an Hfr strain ability to promote higher frequency of recombination High frequency of recombination Hfr cell In E coli a cell having its fertility factor integrated into the bacterial chromosome n a donor male cell 0 2 in Hfr x F crosses no F parents converted into F or Hfr in contrast with F x F crosses infectious transfer of F results in large proportion of F parents being converted into F Hfr strain results from integration of F factor into the chromosome 0 O 0 During conjugation F inserted in chromosome drives partall of that chromosome into the F cell Chromosomal fragment can then engage in recombination with recipient chromosome Rare recombinants observed in F x F crosses due to formation of Hfr cells in F culture Hfr chromosome replicates and transfers a single strand to F cell during conjugation O O O O Ensures complete chromosome for donor after mating Transferred stranded converted into double helix in recipient cell Donor gene become incorporated in recipient s chromosome thru crossovers l creating a recombinant cell No recombination l transferred fragments lost Linear transmission of the Hfr genes from a xed point 0 O O Investigated pattern of transmission of Hfr genes to F cells during cross Used interrupted mating technique used to map bacterial genes by determining the sequence in which donor genes enter recipient cells Plated in medium that killed Hfr donors had allele str Surviving str cells tested those bearing a donor allele must have taken part in conjugation Exconjugant A female bacterial cell that has just been in conjugation with a male and contains a fragment of male DNA Key elements in results 1 Each donor allele rst appears in F recipients at speci c time after mating began 2 donor alleles appear in speci c sequence 3 later donor alleles present in fewer recipient cells in conjugating Hfr singlestranded DNA transfer beings from xed point on donor chromosome origin 0 and continues in a linear fashion point 0 site the F plasmid inserted farther gene from 0 later it s transferred to F D these genes included in fewer exconjugants in Hfr crosses F exconjugants rarely converted into Hfr or F bc the inserted F transmitted as the last element of the linear chromosome transmission process usually stops before getting this far the Hfr chromosome circular unwinds a copy of itself that s transferred to the F cell in a linear fashion with the F factor entering last o Inferring integration sites of F and chromosome circularity o How and where F plasmid integrates to form an Hfr l deduced that chromosome circular If F in a ring insertion might be a simple crossover bw F and bacterial chromosome Any linear Hfr chromosomes could be generated by insertion of F into ring in appropriate placeorientation 1 end of integrated F the origin terminus at other end 0 orientation F inserted determines order of entry of donor alleles o insertion sequences regions of homology that allows 1 to act as pairing region that could be followed by crossover o F factor exists in 2 states 1 plasmid state free cytoplasmic element easily transferred to F recipients 2 integrated state contiguous part of circular chromosome F transmitted late in conjugation Mapping of bacterial chromosomes Broadscale chromosome mapping by using time of entry 0 Construct linkage maps from interruptedmating results 0 Distance measured in mins donor alleles rst appear after mating Finescale chromosome mapping by using recombinant frequency 0 For exconjugant to acquire donor genes permanently donor fragment must recombine with the recipient chromosome 0 Recombination bw 1 complete genome from F endogenote and an incomplete 1 from the Hfr donor exogenote 0 At this stage cell is a merozygote A partly diploid E coli cell formed from a complete chromosome the endogenote plus a fragment the exogenote 0 Single crossover would break ring to keep ring intact must be even number of crossovers o Other product of quotdouble crossoverquot generally lost only 1 of the reciprocal products of recombination survives 0 Want to calculate map distance separating 3 close loci Use interruptedmating to show their order To get merozygote select stable exconjugants bearing the last donor allele Count frequencies of crossovers at different locations See which of the other markers were integrated O unselected marker In a bacterial recombination experiment an allele scored in progeny for the frequency of its cosegregation with a linked selected allele F plasmids that carry genomic fragments F in Hfr strains usually stable but F can exit from chromosome by reversal of recombination process 0 Plasmid can carry with it a part of the bacterial chromosome 0 F plasmid F plasmid carrying bacterial genomic DNA Faulty excision occurs bc there s another homologous region nearby that pairs with the original 0 The DNA of an F plasmid is part F factor and part bacterial genome 0 Like F plasmids F plasmids transfer rapidly 0 They can be used to establish partial diploids for studies of bacterial dominance and allele interaction R plasmids a plasmid containing one or several transposons that bear resistance genes 0 Transferred rapidly on cell conjugation like the F plasmid pBR 332 and pUC preferred vectors for molecular cloning 0 gene on R plasmid that confer resistance can be used as markers to track movement of vectors bw cells 0 alleles for antibiotic resistance often contained within a unit called a transposon unique segments of DNA that can move around to different sites in genome transposition thus thru plasmids antibiotic resistance alleles can spread rapidly throughout bacteria population 53 Bacterial Transformation Transformation The directed modi cation of a genome by the external application of DNA from a cell of different genotype The nature of transformation 0 Transforming DNA incorporated into bacterial chromosome by process analogous to doublerecombination events in Hfr x F crosses a Conjugation DNA transferred from 1 living cell to another thru close contact 0 Transformation isolated pieces of external DNA taken up by cell Chromosome mapping using transformation 0 Transformation can be used to measure how closely 2 genes are linked on bacterial chromosome When DNA extracted some breakage into smaller pieces likely o If 2 donor genes located close good chance they will be carried on same piece of transforming DNA double transformation Simultaneous transformation by two different donor markers 0 if genes linked proportion of double transformants greater than the product of singletransformant frequencies Bacteria can take up DNA fragments from the surrounding medium 0 Inside the cell these fragments can integrate into the chromosome 54 Bacteriophage Genetics Bacteriophage phages bacterial virus 0 parasitize and kill bacteria bacteriophages can be used in 2 different types of genetic analysis 0 1 2 distinct phage genotypes can be crosses to measure recombination thus map viral genome o 2 phages can be used as a way of brining genes together for linkage and other genetic studies can also be used in DNA technology as carriers or vectors of foreign DNA Infection of bacteria by phages Phage consists of a nucleic acid chromosome DNA or RNA surrounded by coat of protein molecules During infection phage attaches to bacterium and injects its genetic material into bacterial cytoplasm o Phage genetic info takes of machinery of bacterial cell by turning off synthesis of bacterial components and redirecting to make phage components 0 Many phage descendants made and released 0 Lysis The rupture and death of a bacterial cell on the release of phage progeny o Lysate Population of phage progeny Phages too small to see but can produce visible manifestation of a phage by using several phage characteristics 0 Plaque clear area on a bacterial lawn left by lysis of the bacteria through progressive infections by a phage and its descendants o plaque morphology is a phage character that can be analyzed at the genetic level o host range phage phenotype bc phages differ in spectra of bacterial strains they can infectlyse Mapping phage chromosomes by using phage crosses genotypes of 2 parental strains crossed hr x hr o h can infect 2 different E coli strains 0 h can infect only 1 strain 0 r rapidly lyses cells produces large plaques o r slowing lyses cells produces small plaques 0 mixed double infection infection of a bacterial culture with two different phage genotypes 4 genotypes classi ed as parental hr and Hr and recombinant hr and hr 0 RF hrhr total plaques Recombining phage chromosomes linear single crossovers produce viable reciprocal products Phage crosses subject to analytical complications 0 Several rounds of exchange take place within host 0 Recombination can take place bw genetically similar phages as well as bw different types 0 So recombinants consequence of a population of events 0 Large number of phages used l rare crossover events can be detected 0 Selective system only the desired rare event can produce a certain visible outcome 0 Screen system in which large numbers of individuals are visually scanned to seek the rare Recombination between phage chromosomes can be studied by bringing the parental chromosomes together in one host cell through mixed infection 0 Progeny phages can be examined for both parental and recombinant genotypes 55 Transduction Transduction The movement of genes from a bacterial donor to a bacterial recipient with a phage as the vector Discovery of transduction Recovered recombinants from a Utube experiment in which conjugation prevented by lter 0 By varying size of pores in lter found that agent responsible for gene transfer was same size as a phage o Virulent phages those that immediately lyse and kill host 0 Cannot become prophages Temperate phages those that can remain win host cell for period without killing it o Prophage phage integrated into the bacterial genome o Lysogenic bacterium harboring a inert prophage lysogen o Capable of occasional bacterial lysis 2 kinds of transduction o generalized transducing phages can carry any part of the bacterial chromosome 0 specialized transducing phages carry only certain speci c parts Generalized transduction ability of certain phages to transduce any gene in the bacterial chromosome 0 when donor cell lysed bacterial chromosome broken up into small pieces and occasionally newly forming phages mistakenly incorporate a piece of bacterial DNA in phage head 0 origin of transducing phage o phage carrying bacterial DNA can infect another cell that bacterial DNA can then be incorporated into recipients cell s chromosome by recombination Generalized transduction can be used to obtain bacterial linkage info when genes close enough that phage can pick them up and transduce them in a single piece of DNA 0 Cotransductants 2 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 can estimate size of piece of host chromosome that phage can pick up Specialized transduction situation in which a particular phage will transduce only speci c regions of the bacterial chromosome 0 lnserts into bacterial chromosome at only 1 position 0 Behavior of the prophage o A prophage behaves as a bacterial gene locus behaves as part of the bacterial chromosome 0 prophage enters F cell at speci c time corresponding to its position in chromosome o earlier gene recovered bw they enter before the prophage 0 later genes not recovered bc lysis destroys recipient cell in HfrA x F cross entry of A prophage into cell immediately triggers lytic cycle 0 zygotic induction The sudden release of a lysogenic phage from an Hfr chromosome when the prophage enters the F cell followed by the subsequent lysis of the recipient cell in cross of 2 lysogenic cells HfrA x FA no zygotic induction 0 prophage produces cytoplasmic factors that represses multiplication of the virus A insertion o A prophage part of lysogenic bacterium s chromosome 0 inserts by single crossover bw circular A phage chromosome and circular E coli chromosome 0 attachment site region at which prophage integrates on bacterial chromosome bw genes gal and bio Mechanism of specialized transduction Recombination bw specific regions of A and bacterial chromosome catalyzed by specialized phageencoded enzyme system that used A attachment site as substrate 0 Enzyme dictates that A integrates only at specific point bw gal and bio During lysis A prophage normally excises at correct point to produce normal circular A chromosome Rarely excision abnormal due to faulty outlooping o Outlooping phage DNA can pick up nearby gne and leave behind some phage genes 0 Resulting phage genome defective but also gained bacterial gene gal or bio o Abnormal DNA packaged into phage heads that can infect other bacteria Adgal or Adbio ln presence of a second normal phage particle in double infection Adgal can integrate into chromosome at A attachment site 0 Gal genes transduced into second host 56 Physical Maps and Linkage Maps Compared physical map can be useful in mapping new mutations o insertional mutagenesis situation when a mutation arises by the interruption of a gene by foreign DNA such as from a transgenic construct or a transposable element 0 causes mutations thru random insertion of foreign DNA fragments o transposons useful inserts bc their sequence known mutant gene can be located and sequenced 57 Summary 0 gene transfer and recombination take place bw different strains of bacteria 0 in bacteria genetic material passed in only 1 direction donor F or Hfr to recipient F o donor ability determined by presence of fertility factor F a type of plasmid o occasionally F factor in F cells can integrate into E coli chromosome to form an Hfr cell fragment of donor chromosome can transfer into recipient cell and recombine with recipient chromosome 0 interrupted mating method for constructing linkage map of single chromosome of bacteria map unit is mins 0 RF bw markers known to have entered recipient can provide nerscale map distance 0 R plasmids carry antibioticresistance alleles often win mobile element transposon 0 Transformation genetic traits transferred from 1 bacterial cell to another in form of pieces of DNA taken into cell from extracellular environment 0 DNA must be taken into recipient cell and recombination must take place bw recipient chromosome and incorporated DNA 0 Lysogeny infection injected phage lies dormant in bacterial cell 0 Prophage dormant phage can incorporate into host chromosome 0 Can leave its dormant state and lyse host cell Transduction phage carries bacterial genes from donor to recipient o Generalized random host DNA incorporated into phage head during lysis o Specialized faulty excision of prophage from unique locus results in inclusion of speci c host genes along with phage DNA in phage head Mutations can be precisely located o Mutations produced by insertion of transposons insertional mutagenesis 0 Then DNA sequence surrounding inserted transposon obtained and match to a sequence in physical map 0 Provides locus sequence and possibly function of the gene 0 Introduction 0 Forward genetics used to identify individual genes 0 Begin with set of mutants then cross each mutant with WT to see if mutant shows singlegene inheritance How genes in a set interact to in uence phenotype o Analyze protein interactions directly in vitro by using 1 protein as bait and observe which other proteins attach to it Proteins that bind are candidates for interaction 0 Analyze mRNA transcripts Genes that collaborate in speci c developmental process can be de ned by set of RNA transcripts present when process going on 0 Genetic analysis 0 Gene interactions classi ed in 2 categories 0 1 interactions bw alleles of 1 locus variations on dominance o 2 interactions bw 2 loci reveal number and types of genes underlying a function 61 Interactions bw the Alleles of a Single Gene Variations on Dominance many ways to alter sequence of gene each producing a mutant allele but only some mutant alleles appear in population 0 multiple alleles allelic series set of forms of one gene differing in their DNA sequence or expression or both dominance or recessiveness of new mutation gives insight to gene func ons o dominance is a manifestation of how the alleles of a single gene interact in a heterozygote o interacting alleles may be wild and mutant m or 2 different mutants mlm2 Complete dominance and recessiveness Full complete dominance an allele that expresses itself the same in single copy heterozygote as in double copy homozygote 0 WT alleles recessive in heterozygotes Haplosufficient genes a single copy of WT adequate for full expression 0 Null mutations fully recessive Haploinsufficient genes WT does not adequate for full expression 0 Heterozygote WTnull mutant and mutation is dominant Dominant negative mutant allele that in single dose a heterozygote wipes out gene function by a spoiler effect on the protein Incomplete dominance situation in which a heterozygote shows a phenotype quantitatively but not exactly intermediate between the corresponding homozygote phenotypes Each WT allele produces set does of its protein product 0 Number odes in WT allele determine concentration of pigment o 2 doses red 0 1 does l pink 0 0 doeses l white Codominance situation in which a heterozygote shows the phenotypic effects of both alleles equally Ex human ABO blood groups 3 major alleles i la lb but person can only have 2 o Genotypes lai and lbi l la and lb fully dominant overi o Genotype lalb l A and B allele codominant Type of dominance complete incomplete codominance determined by the molecular functions of the alleles of a gene and by the investigative level of analysis Recessive lethal alleles Lethal allele allele that is capable of causing death of an organism 0 Shows that new gene of unknown function essential to organism s operation 0 Useful in determining developmental stage at which the gene normally acts 0 Phenotype associated with death informative to gene function 0 Can explain unusual ratio if death at zygote o Pleiotropic allele that affects several different properties of an organism 0 Whether an allele lethal or not often depends on the environment 0 Some alleles viable in 1 environment but lethal in another Recessive allele sublethal when lethality expressed in only some of the homozygotes o Lethality range from 0100 depending on gene itself rest of genome and environment 0 ln diploids recessive lethal alleles maintained as heterozygotes ln halploids heatsensitive lethal alleles useful 0 Temperaturesensitive ts mutations conditional mutation that produces the mutant phenotype in one temperature range and the wildtype phenotype in another temperature range 0 Permissive temperature temp at which a ts mutant allele is expressed the same as the wildtype allele 0 Restrictive temperature temp at which a ts mutation expresses the mutant phenotype Null alleles for genes identi ed thru genomic sequencing can be made by using quotreverse geneticquot procedures that speci cally knock out the function of that gene 62 Interaction of Genes in Pathways Genes act by controlling cellular chemistry Biosynthetic pathways in Neurospora Neurospora a haploid fungus Forward genetics 0 Irradiated fungus cells to produce mutations 0 Tested cultures for interesting mutant phenotypes relevant to biochemical function 0 Found numerous mutants that had defective nutrition Auxotrophic mutants cannot synthesize all nutrients Con rmed each mutation inherited as a singlegene mutation 11 ratio when crossed with WT Classi ed speci c nutritional requirement of each auxotroph 0 Found mutated genes mapped to 3 different loci on 3 separate chromosomes Auxotrophs for each of 3 loci differed in their response to structurally related compounds 0 Cellular enzymes known to interconvert such related compounds 0 Mutation at particular gene assumed to interfere with production of a single enzyme onegeneonepolypeptide hypothesis hypothesis that originally proposed that each gene nucleotide sequence encodes a polypeptide sequence 0 generally true with the exception of untranslated functional RNA RNA type that plays a role without being translated 0 clear that gene encodes the physical structure of a protein which in turn dictates its function Chemical synthesis in cells is by pathways of sequential steps catalyzed by enzymes 0 The genes encoding the enzymes of a speci c pathway constitute a functionally interacting subset of the genome Gene interaction in other types of pathways Synthetic pathway chain of enzymatic conversions that synthesizes essential nutrients 0 ex Neurospora arginine pathway Signaltransduction pathway chain of complex signals from the environment to the genome and from 1 gene to another 0 Ex mating response in yeast Developmental pathways comprises the steps by which a zygote becomes an adult organism 0 Steps require gene regulation and signal transduction 63 Inferring Gene Interactions genetic approach that reveal interacting genes for particular biological property 0 1 obtain many singlegene mutants and test for dominance o 2 test mutants for allelism are they at 1 or several loci o 3 combine mutants in pairs to form double mutants genotype with mutant alleles of two different genes to see if genes interact gene interation inferred from phenotype of double mutant 0 if genes interact then phenotype differs from simple combination of both singlegene mutant phenotypes o if mutant alleles from different genes interact then WT genes interact normally as well 0 when 2 mutants interact modi ed 9331 ratio observed 0 mutant screen could have unintentionally favored genes 0 set of gene loci needs to be de ned Sorting mutants using the complementation test 0 How to decide whther 2 mutations belong to same gene 0 Map each mutant allele if 2 mutations map to 2 different chromosomal loci likely of different genes Time consuming for large set of mutations o Complementation test test for determining whether two mutations are in different genes they complement or the same gene they do not complement o In diploid complementation test performed by intercrossing 2 individuals that are homozygous for different recessive mutations o Observes whether progeny have the WT phenotype o If progeny WT 2 recessive mutations must be in different genes l mutations complemented o If progeny not WT recessive mutations must be alleles of same gene l mutations don t complement o Complementation the production of a wildtype phenotype when 2 haploid genomes bearing different recessive mutations are united in the same cell Result of the cooperative interaction of the WT alleles of the 2 genes 0 In haploid complementation test cannot be performed by intercrossing o In fungi fusion resulting in a heterokaryon culture of cells composed of 2 different nuclear types in a common cytoplasm o Fungal cells fuse readily when 2 different strains fuse haploid nuclei from different strains occupy 1 cell heterokaryon nuclei don t fuse o in different strains there are mutations in 2 different genes conferring the same phenotype fuse 2 strains to form heterokaryon with 2 nuclei in shared cytoplasm gene products in same cytoplasm 2 WT alleles can exert dominance and coorperate to produce heterokaryon of WT phenotype n 2 mutations complement 0 When 2 independently derived recessive mutant alleles producing similar recessive phenotypes fail to complement they must be alleles of the same gene Analyzing double mutants of random mutations To learn if 2 genes interact l assess phenotype of the double mutant to see if anything other than the combo of both single mutations 0 Double mutant obtained by intercrossing 0 Assume complementation in F1 different genes 0 F1 selfed to get F2 F2 should have the double mutant 0 Double mutant identi ed by ratios 9331 is the null hypothesis any modi ed ratio informative The 9331 ratio no gene interaction The 97 ratio l genes in the same pathway 0 Ratio only possible if double mutant has same phenotype as the 2 single mutants Modi ed ratio gives way of identifying double mutant s phenotype o Identical phenotypes of single and double mutants l each mutant allele controls a different step in the same pathway 0 absence of either gene function leads to absence of the end product of the pathway The 934 ratio l recessive epistasis o Epistasis situation in which the differential phenotypic expression of a genotype at 1 locus depends on the genotype at another locus a mutation that exerts its expression while canceling the expression of the alleles of another gene overriding mutation is epistatic and overridden one if hypostatic results from genes being on same pathway n epistatic mutation carried by gene that is farther upstream than gene of overridden mutation 0 double mutant expresses only 1 of the 2 mutant phenotypes o 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 0 ln fungi tetrad analysis useful in identifying double mutant The 1231 ratio l dominant epistasis o 2 genes act in a common developmental pathway Suppressor mutant allele of a gene that reverses the effect of a mutation of another gene resulting in a wildtype or nearwildtype phenotype o lmplies that target gene and suppressor gene normally interact at some functional level in WT states 0 Sometimes have no effect in absences of the other mutation o Other times suppressor allele produces its own abnormal phenotype Screening for suppressors 0 Start with mutant in process of interest expose to mutation causing agents 0 ln haploids plate mutagenized cells and look for colonies with WT phenotypes Revertants WT arising from reversals of original mutational event Pseudorevertants double mutants in which 1 of the mutations is a suppressor Appearance of original mutant phenotype identi es parent as a suppressed mutant 0 ln diploids suppressors produce various modi ed F2 ratios 1316 must include double mutant WT phenotype I expected from recessive suppressor that has no detachable phenotype If recessive suppressor has same phenotype as target mutation then F2 ratio 106 wildmutant Difference bw suppressor and epistasis o Suppressor cancels expression of mutant allele and restores corresponding WT phenotype o Often only 2 phenotypes segregate rather than 3 in epistasis suppressor mutation that causes a compensatory shape change in the second protein can restore t and hence normal function 0 From suppressor ratios interacting proteins often can be deduced in situations in which a mutation causes a block in a metabolic pathway the suppressor nds some way of bypassing the block nonsense suppressors mutations in tRNA genes resulting in an anticodon that will bind to a premature stop codon within a mutant coding sequence 0 little effect on phenotype other than in suppression Modi ers mutation at a second locus changes the degree of expression of a mutated gene at the rst locus o mutation in a regulatory protein can downregulate or up regulate the transcribed gene o synthetic lethals double mutant that is lethal whereas the component single mutations are not 64 Penetrance and Expressivity o Penetrance proportion of individuals with a speci c genotype that manifest that genotype at the phenotype level 0 Reasons for organism having genotype but not expressing corresponding phenotype o 1 in uence of environment individuals with the same genotype may show a range of phenotypes depending on the environment 0 2 in uence of other interacting genes modi ers epistatic genes or suppressors may act to prevent the expression of the typical phenotype o 3 sublety of the mutant phenotype subtle effects brought about by the absence of a gene function may be difficult to measure Expressivity degree to which a particular genotype is expressed in the phenotype penetrance and expressivity quantify the modi cation of gene expression by varying environment and genetic background 0 they measure respectively the percentage of cases in which the gene is expressed and the level of expression 65 Summary 0 genes act in concert with many other genes in genome 0 forward genetic analysis deduces these interactions 0 individual mutations tested rst for dominance allelic interaction 0 recessive mutations often result of haplosuf ciency of WT allele 0 dominant mutations result of either haploinsuf ciency of WT or of mutant acting as a dominant negative 0 some mutations can cause death lethal mutations lethality of homozygous recessive mutation way to asses if gene essential to genome 0 interaction of different genes result of their participation in same or connecting pathways synthetic signal transduction developmental complementation test determine whether 2 distinct recessive mutation are of 1 gene or 2 different genes mutant genotypes brought together in F1 if phenotype mutant no complementation l 2 alleles of the same gene if phenotype wild l complementation l alleles of different genes interaction of different genes detected by testing double mutants bc allele interaction implies interaction of gene products at functional level 0 O 0 key types of interaction epistasis suppression synthetic lethality epistasis replacement of mutant phenotype produce by 1 mutation with mutant phenotype from mutation of another gene common developmental or chemical pathway suppressor mutation of 1 gene that can restore WT phenotype to a mutation at another gene physically interacting proteins or nucleic acids synthetic lethality some combos of viable mutants are lethal different types of gene interactions produce F2 dihybrid ratios that are modi cations of 9331 gene interaction and geneenvironment interaction revealed by variable penetrance ability of genotype to express itself and expressivity quantitative degree of expression of a genotype
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'