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Two independent random samples were selected from normally

Statistics for Business and Economics | 12th Edition | ISBN: 9780321826237 | Authors: James T. McClave, P. George Benson, Terry T Sincich ISBN: 9780321826237 51

Solution for problem 90SE Chapter 8

Statistics for Business and Economics | 12th Edition

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Statistics for Business and Economics | 12th Edition | ISBN: 9780321826237 | Authors: James T. McClave, P. George Benson, Terry T Sincich

Statistics for Business and Economics | 12th Edition

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Problem 90SE

Problem 90SE

Two independent random samples were selected from normally distributed populations with means and variances and , respectively. The sample sizes, means, and variances are shown in the following table.

a. Test against  . Use a = .05.

b. Would you be willing to use a small sample t-test to test the null hypothesis against the alternative hypothesis Why?

Step-by-Step Solution:
Step 1 of 3

February 7, 2016  Reproduction: Problems to be solved o Transfer of Information  Between parent and daughter cells  Between parent and offspring (in multicellular organisms) o Apportionment of biological materials (e.g., organelles)  Types of Reproduction o 1. Asexual (“clonal”) Reproduction  Offspring (cell, individual) are genetically identical* to THE parent  All prokaryotes**, many eukaryotes  Occurs by binary fission in prokaryotes, mitosis in eukaryotes o 2. Sexual Reproduction  TWO parents combine their genes in a stereotypical way (i.e., meiosis)  Leads to recombination, i.e., offspring are genetically different from either parent  ONLY occurs in eukaryotes  Interphase (G1) o Each chromosome is a single, unreplicated double strand of DNA o One chromosome from each parent (Male, Female) forms a Homologous Pair (=“homologs”) o Chromosomes in nucleus, surrounded by nuclear membrane o Single centrosome  S-Phase o After DNA replication, TWO “sister chromatids” are present for each homolog o Each sister chromatid is the SAME double-stranded DNA molecule o The two sister chromatids are attached by proteins o Centrosomes duplicate  Early Prophase o Chromosomes condense o Mitotic spindle forms from centrosome o Centrosomes begin to migrate to poles of cell o Nucleoli disappear  Mid-Prophase o Chromosomes fully condensed o Centrosomes complete migration to the poles o Nuclear envelope begins to degrade o Spindle fibers enter nuclear area from the pole o Kinetochores form at centromeres  Prometaphase o Nuclear envelope completely degraded o Spindle fibers begin to attach at kinetochores o Sister chromatids attached to opposite poles  Metaphase o Spindle fibers attached to centromere at kinetochore o All sister chromatids attached to opposite poles o Chromosomes migrate to center plane of cell, “metaphase plate”  Anaphase o Protein bond between sister chromatids degrades o Sister chromatids separate, migrate toward opposite poles o Poles move farther apart as non-kinetochore spindle fibers lengthen  Telophase o Non-kinetochore spindle fibers continue to elongate cell o Nuclear envelopes begin to form at poles o Chromosomes de-condense back into chromatin o Nucleoli re-form, cytokinesis begins  Eukaryotic Cell Cycle Control o Cell replication must be under precise control. o If unicellular organisms had no control over reproduction, they would exhaust their resources and starve to death. o Signals from external environment o In multicellular organisms, cell reproduction must be controlled for proper development. o Internal signals, e.g., growth factors February 11, 2016  Cell control o 1. CDKs act on cell-cycle regulators o 2. Example: G1-S checkpoint  Sexual Life Cycles – Human Example o 46 Chromosomes o 22 Homologous Pairs of autosomes  Same length  • Same centromere position  • Same sequence (+/-)  • Same set of genes o • One pair of sex chromosomes  • Females XX, Males XY  • Only small region of sequence homology  Meiosis o • RECALL: Function of MITOSIS is to faithfully replicate the parental genome in each daughter cell with no change in information content o • Proximate function of MEIOSIS is to produce haploid cells from diploid cells o • Ultimate function of MEIOSIS is to generate genetic variation upon which natural selection can act  Sexual Life Cycles – Diplontic (e.g., animals) o Free-living stage is diploid o Gametes formed by meiosis o Haploid gametes merge genomes to form diploid zygote (“syngamy”)  Sexual Life Cycles – Alternation of Generations (e.g., Plants) o Diploid sporophyte forms haploid spores by meiosis o Spores form gametophyte by mitosis o Gametophyte forms gametes by mitosis o Gametes merge to form diploid zygote  Sexual Life Cycles – Haplontic (e.g., Fungi) o Free-living, multicellular organism is haploid o Gametes formed by mitosis o Gametes merge to form diploid zygote o Zygote undergoes meiosis to form haploid cells  Meiosis – an overview o Interphase 1: G1 phase –  • Begin with two homologous chromosomes (out of n homologous pairs)  • DNA content = 2C  • Ploidy = 2n (diploid) o Interphase 1: S-phase •  Chromosomes replicate  • DNA content = 4C  • Ploidy = 2n o “Meiosis I”  • Homologous chromosomes separate  • Cell Division* #1  • Result is TWO haploid (ploidy = n) cells with TWO SISTER CHROMATIDS of ONE of the two homologs o “Meiosis II”  • Sister chromatids separate  • Cell Division # 2  • Result is FOUR haploid daughter cells, each with a single unreplicated chromosome (= 1C) o • Each cell contains ONE member of each homologous pair of chromosomes  Meiosis I – early Prophase I o Homologous chromosomes pair o Synaptonemal complex (proteins) attaches homologs  • “synapsis” o Homologs form tetrad  Meiosis I – late Prophase I o Chromosomes cross over, form “chiasmata” o Exchange of DNA between homologs occurs at chiasma o Spindles form and attach to kinetochores as in mitosis  Meiosis I – Metaphase I o Chromosomes lined up on metaphase plate in homologous pairs o Spindles from one pole attach to one chromosome of each pair o Spindles from the other pole attach to the other chromosome of the pair  Meiosis I – Anaphase I o Homologous chromosomes separate and move along spindle fibers toward pole o Sister chromatids remain attached at centromeres o Note that recombination has occurred  Meiosis I – Telophase and cytokinesis o Homologous chromosomes reach (opposite) poles o Each pole has complete haploid complement of chromosomes o Each chromosome consists of two sister chromatids  Meiosis II – Prophase II o Spindle forms o Chromosomes move toward metaphase plate  Meiosis II – Metaphase II o Chromosomes reach metaphase plate, as in mitosis o Kinetochores of sister chromatids attach to spindle fibers from opposite poles  Meiosis II – Anaphase II o Centromeres of sister chromatids separate o Sister chromatids move toward opposite poles  Meiosis II – Telophase and cytokinesis o Mechanism as before o Note that now FOUR HAPLOID DAUGHTER CELLS formed from each parent cell o Note that some chromosomes are recombinant, some are no  Apoptosis - Programmed Cell Death o At certain points in development, cells are programmed to die o Failure of apoptosis leads to developmental abnormalities (e.g., webbed hands and feet in humans) o Cells with irreparable DNA damage (can lead to cancer, abnormal development) also subjected to apoptosis o "Suicide" proteins (caspases) are always present in inactive form  Regulation is post-translational, not of transcription February 17, 2016  Dihybrid cross- dependent assortment o Predict ¾ round and yellow, ¼ wrinkled and green o Not what Mendel observed  Independent assortment o Four combos of alleles in gametes o All equally likely o 9:3:3:1 ratio expected o This is what Mendel observed  Mendel’s two laws o 1. Law of segregation  If the locus is heterozygous, half of the gametes get one allele, half get the other allele o 2. Law of independent assortment- multiple loci  Alleles at each locus segregate independently of one another  Probability theory: the chance of something happening, relative to all possible outcomes o Independent: outcome of one event has no effect on the outcome of the other  Ex. Rolls of a die, flips of a coin  Probability of both events  Pr(A and B)= Pr(A)x Pr(B)  Probablility of either or two events happening  Pr(A or B) = Pr(A)+ Pr(B)  Incomplete dominance o Ex. Snapdragons o Two alleles: C^R and C^W o Heterozygotes are intermediate between two phenotypes  Co-dominance o One locus, two alleles o Ex. M/N blood groups o M/N locus encodes glycoprotein antigen that occurs on the surface of the human red blood cells o Heterozygotes show both phenotypes  Multiple alleles o Most genes have more than 2 alleles o ABO blood type  One gene, 3 alleles: I^A, I^B, i (i is null)  I^A and I^B are co-dominant, i is recessive  Pleitrophy: one gene effects multiple characters o Ex. Tyrosinase (Tyr-albinism) o Ex. Sry (sex- determining locus in mammals)  Epistasis: an allele at one locus affects the phenotypic effects of an allele at the other locus o Ex. Mouse coat color  One locus controls hair pigment color  Black, brown; black is dominant (BB, Bb, bb)  Second locus controls pigment deposition  Pigment dominant to no pigment (CC, Cc, cc)  Dihybrid cross (BbCc x BbCc)  Polygenic inheritance o Most traits do not have a simple genetic basis o Called “quantitative traits” or “complex traits” o Often continuous traits  Height, weight, skin color  Ex. Many chronic diseases o Consider the offspring of a mating between two individuals heterozygous at all loci- AaBbCc x AaBbCc  What fraction of gametes will be ABC 1/8  What fraction of gametes will be AABBCC 1/64  Environment effects on phenotypes o Many traits not completely determined by genes  Height, weight, skin color  Many chronic human diseases, ex. Coronary heart diseases o Ex. Light intensity on plant height o Called “genotypes by environment interaction”  Human genetics- pedigree analysis o Can’t do controlled crosses with humans o We track phenotypes on “pedigree”  Genetics of disease o Recessive disorders  Most diseases are recessive  Homozygous recessive have disease  Heterozygous are carriers March 7, 2016  Mutation: change in the composition of the genome from parent to daughter strand o (total genomic damage-repaired damage) = mutation o Occur in both somatic and germline cells (only considering heritable mutations- which occur in germline cells) o Is the ultimate source of genetic variation (!)  Categories of mutations o Nucleotide sequence mutations o Chromosomal rearrangements o Transposable elements  Types of nucleotide sequence mutations o Point mutation (+ base substitution) o Insertions/deletions (“indels”) o Duplications  Point mutations o Substitution of one nucleotide for another at a homologous site o Rates vary- generally around 10^-8 to 10^-10 per site per generation, varies by taxon o Occur anywhere in the DNA sequence  Rate may (and does) vary with genomic text Chemical types of point mutations o o Transitions are more common- magnitude depends on gene and taxon The genetic code o Mutation- “silent” or “synonymous” substitution o Substitute a purine for pyrimidine or pyrimidine for purine Mutation- “Replacement” substitution o Substitute purine for purine or pyrimidine for pyrimidine Mutation- insertion/deletion (“indels”) o Deletion: remove a pyrimidine or purine o Insertion: dd a pyrimidine or purine o Can be one or more nucleotides Mutations occur because o Spontaneous errors in DNA replication (mismatches) o Mutagenic damage to DNA (replicating or non-replicating) o “Selfish” elements replicate themselves, often to the detriment of the host genome Causes of mismatches o Polymerase base misincorporation  DNA polymerases have 3’->5’ and/or 5’->3’ exonuclease (proofreading) function o Tautomerization o Spontaneous deamination (e.g., deaminated cytosine is uracil, leads to C/G-> U/A -> T/A DNA repair o First line of defense: DNA polymerase  DNA polymerase III has a “proofreading” capability  If the wrong nucleotide is entered, DNA synthesis stops, the offending nucleotide is removed, and synthesis resumes Mechanisms of mutagenic damage o Base analogs: purines/pyrimidines that mimic legitimate bases pair differently o Direct damage to DNA: e.g. Deamination, alkylation, intercalating agents, UV o Indirect damage to DNA: e.g. Agents that generate oxygen free radicals  Oxygen free radicals predominantly cause G/C->A/T via 5- hydroxyuracil from cytosine deamination and G/C-> T/A via 8- oxoguanine Mechanisms of DNA repair o Direct repair: damage repaired directly o Excision repair: damaged DNA excised and repaired using undamaged strand as template; base excision, nucleotide excision o Mismatch repair: recognizes misincorporated bases, removes the mispaired base, and repairs using the parent strand as a template o Recombination repair: (double stranded break repair): employs the recominational machinery o Transcription- coupled repair: (ER, specific to transcribed strand during transcription) Insertions/Deletions (indels) o Insertions  Addition of nucleotides to a sequence o Deletions  Deletion of nucleotides from a sequence o Result from  “replication slippage” when replicating DNA reanneals at short tandem repeats  Non-homologous (ectopic) recombination Chromosomal aberrations and their consequences o Non-disjunction  Homologous chromosomes do not separate at meiosis I  Sister chromatids do not separate at meiosis II  Consequences If fertilization with a gamete with abnormal ploidy occurs, result is offspring with abnormal chromosome number, called “aneuploidy” Polyploidy o Sometimes organisms get an entire extra (haploid) set of chromosomes o 3 sets are called “triploidy”  Can occur by fertilization of a diploid egg in which all chromosomes underwent non-disjunction o 4 sets are called “tetraploidy”  E.g. Zygote fails to divide after DNA replication; mitosis leads to a 4n organism Duplications o Result from non-homologous recombination o Probably very important for evolution of new genes Chromosomal rearrangements o Typically from non-homologous recombination or double-stranded breaks in the chromosome o 2 types  Inversions Once inversion occurs, recombination is usually prevented because recombinant gametes cannot get a full complement of genes  Translocations Pieces of non-homologous chromosomes break and fuse together Often cause improper segregation of chromosomes at meiosis Transposable elements: “jumping genes” o Transposable elements are genetic elements that contain the information for their own replication o Insert into new locations in the genome  Conservative transposition (cut and paste)  Replicative transposition (copy and paste) o Unlike episomes or viruses, TEs never exist independent of the host genome o Most recombination requires complementary base-pairing of homologous (sequence-similar) regions of DNA  Meiosis (in eukaryotes)  Transformation  Generalized transduction  Conjugation o Transposition is a form of non-homologous recombination o TEs can move to novel sites in the genome o The simplest TEs consist of  A gene that encodes an enzyme for excision an insertion of the TE sequence; transposase in DNA TEs  A recognition sequence that the enzyme recognizes as the boundary of the TE o In prokaryotes  Insertion sequences A transposase gene A recognition sequence that is an inverted repeat  Transposase recognizes boundaries of TE inverted repeats and cuts the TE out of the donor site  Transposase cleaves chromosome at target side via endonuclease  Transposon is joined to the single-stranded ends at the target site  Gaps are filled in by DNA polymerase I and DNA ligase  Result is a new copy (copy and paste) or moved copy (cut and paste) of the TE o Composite transposable elements  Contain additional genetic material besides transposase, IRs  Structure consistent with two ISs close together in genome  One transposase may not be functional March 9, 2016  Transposable elements (TEs) o DNA transposons  Composite transposable elements  Contains additional genetic material besides transposase, IRs  Structure consistent with two ISs close together in genome  One transposase may not be functional  Prokaryotic gene regulation: metabolic control o Metabolism can be defined as the total of the chemical reactions of a cell (or organism)  Must be controlled, or chaos ensues o Anabolic metabolism: building large molecules out of small molecules; requires energy input o Catabolic metabolism: breaking down large molecules into small molecules; releases energy o Direction of flux through metabolic pathway depends on cellular needs o Flux through a metabolic pathway can be regulated in two ways  Activity of the enzymes themselves  Regulation of enzyme activity o Competitive inhibition: inhibitor binds to active site o Non-competitive inhibition: inhibitor binds to some other part of the cell o Allosteric regulation: multiple subunits o Feedback inhibition  Metabolic pathway switched off by the end- product o Cooperativity  Active form stabilized by substrate  Expression of genes that encode enzymes  Regulation of gene expression: operon o In prokaryotes, functionally related genes often are clustered together in the genome, called “operons” o Under the control of a single promoter sequence, so are transcribed together o Single “switch” sequence controls transcription of the whole operon; called an operator o Operator controls access of RNA Pol to genes o Repressor proteins controls access of Pol to operator o Repressor protein encoded by a “regulatory” gene o Presence or absence of an end-product or precursor controls the operon “on-off switch” o Example: the E. coli trp operon o Trp operon is repressible because transcription is repressed by the presence of an end-product  Repressible operons usually function in anabolic metabolism o An inducible operon becomes transcriptionally active by the presence of a precursor  Example: E. coli lac operon  Negative v. positive o Trp and lac operons are negative regulation: transcription switched off by active repressor o Positive regulation: transcription switched on by activator molecule  Operon control: repressible v. inducible o Repressible operon: end-product turns the operon off o Inducible operon: substrate turns the operon on March 11, 2016  Eukaryotic genomes: physical structure o Eukaryotes face different organizational challenges tan prokaryotes  Cells are much larger and more complex  Many eukaryotes are multicellular; cells have to differentiate  Genomes are larger and more complex o Chromatin: the complex of DNA and proteins that together constitute the chromosome o Problem: how to pack an enormous amount of DNA into a cell o Chromatin is sequentially packaged o Level 1: Chromatin fibers (10 nm)  Histones are positively charged proteins that bind to chromosomal DNA  Remain complexed with DNA throughout the cell cycle  Changes in shape and orientation of nucleosomes influence accessibility of DNA to RNA polymerase o Level 2: 30 nm fiber  Histone tails interact with linker DNA to form loops in the 10 nm fiber o Further folding of the chromatin results in mitotic condensation of chromosomes o Heterochromatin remains densely coiled during interphase o Euchromatin uncoils during interphase  Control of gene expression o All cells in a multicellular organism contain the exact same genetic compliment o Any cell expresses only a small fraction of its genes at any given time o Regulation of the expression of specific genes most common at the level of transcription o Gene expression typically depends on the cell receiving some signal o Upon receipt of a signal, regulation can potentially occur at many steps o 1. Regulation of chromatin structure- modification of histones  Acetylated histones: -COCH3 groups attached to lysine in histone tails  Acetylated histones cannot bind nucleosomes, chromatin de- condenses  Conversely, methylation of histone tails causes chromatin to condense  Each histone tail contains several lysine residues that can be acetylated or methylated  Information stored in pattern of acetylation/methylation o 2. DNA methylation  DNA of most higher eukaryotes is methylated in place  DNA methyltransferases, other demethylating enzymes  Most common is methylated cytosine  Methylation usually represses expression  Can be preserved across replication; methylation enzymes methylate daughter strand  Information contained in methylation can be passed to offspring; “epigenetic” inheritance  Genomic imprinting  Basic assumption of Mendelian genetics is that the effects of an allele do not depend on whether the allele was inherited from Mom or Dad  Sometimes, effects of an allele depend on the parent from which the allele was inherited  Imprint is often due to methylation of one allele  Inherited imprint is removed in gamete producing cells  Now, sex specific imprint is applied in the germ cells  Gametes have appropriate imprint  Dosage compensation of sex-linked genes  In organisms with chromosomal sex determination, the homogametic sex typically has two copies of sex-linked genes and the heterogametic sex only has one copy  This can cause a problem, because all else equal, the homogametic sex will have twice as much gene product expressed as the heterogametic sex  Dosage compensation has evolved to equalize gene expression between the sexes  In placental mammals, one X chromosome in somatic cells in females is inactivated, X inactivation  Mechanism of X inactivation is expression of a functional RNA gene called Xist  Xist RNA coats the X to be inactivated, causing the whole X to be heterochromatin  The targeted X is random o 3. Regulation of initiation of transcription  Transcription factors associated w/ RNA Pol are necessary for transcription of all genes  The general transcription machinery usually transcribes genes at a low rate  Efficient transcription requires specific transcription factors  The same gene may be controlled by different enhancers in different cell types, developmental stages, etc  Some transcription factors act as activators  Activators recruit proteins that acetylate nearby histones  Other TFs act as repressors  Repressors recruit proteins that deacetylate histones  1. Activator protein binds to enhancer  2. Binding of activator causes DNA bending, activators brought in proximity to promoter  3. Activators bind to other TFs, allow formation of an active transcription complex on the promoter  Coordinated control of gene expression o In general, many genes need to be co-regulated  Eukaryotes typically do not have operons  Relatively few sequences in control elements  Enhancers typically composed of about 10 control elements  Specificity of control elements depends on the particular combination rather than a single unique sequence o Co-regulation usually depends on association of a specific set of control elements with each gene in the co-regulated group  Post transcriptional processing o Gene expression ultimately depends on the amount of gene product o Gene expression can be regulated post-transcriptionally by  Alternative splicing  mRNA degradation  microRNA, small interfering RNA  Regulating translation  “Maternal effect” genes lack a poly-A tail  Post-translational control  Why introns o Allow a single gene to encode >1 polypeptide March 14, 2016  Viruses o Viruses are obligate parasites of cellular organisms o Consist of nucleic acid genome (DNA or RNA) enclosed in protein o Use the host’s cellular machinery to replicate o Viral genome encodes the information necessary to replicate  Phage reproduction: lytic and lysogenic o Bacteriophage are viruses that attack bacteria o Lytic reproduction kills host by lysis  Phage that reproduce only by lytic reproduction are called virulent  1. Phage binds to host’s surface receptor  2. Phage injects DNA into the cell  3. Host’s DNA is hydrolyzed (degraded)  4. Host’s metabolic machinery used to make phage protein and DNA from free nucleotides and amino acids  5. Phage encodes lysozyme production, cell lyses, new phage released o Lysogenic reproduction replicates phage genome along with host genome without killing host  Phage that can reproduce by lytic or lysogenic reproduction are called temperate  1. Phage attaches, injects DNA  2. Phage DNA circularizes  3. Phage DNA integrates by crossing over via phage integrase. Integrated phage called prophage: lives for a long time  4. Occasionally, prophage is removed from host genome via phage excisionase; lytic cycle ensues  Genetics of prokaryotes o Bacteria have a single circular chromosome o Many bacteria have plasmids: self-replicating elements that can integrate into the chromosome and which contain only a few genes o Bacterial genomes contain between 500 and several thousand genes o Prokaryotic genome is much smaller than eukaryotic genomes o Very few introns; no splicosomes  Prokaryotic recombination o Bacteria reproduce by clonal fission (asexually) o Yet, recombination is common  Transformation  Uptake of naked DNA  Transduction  Phage mediated DNA transfer  Conjugation  “mating” between bacteria  Inference of recombination o 2 strains of E. coli; arg- and trp- o Plate on minimal medium o Neither strain grows by itself, but a mixture grows o Must be recombination not mutation because the mutation rate is known  Recombination o Transformation  Takes up foreign DNA double stranded RNA  Integrated into bacterial genome at a site of sequence homology  Many species of bacteria have receptor proteins on the cell surface that function in DNA uptake  Uses homologous sequence to replace a strand of host DNA  One daughter has the donor strand, one has the host strand o Phage-mediated exchange  1. Phage infects bacteria  2. DNA hydrolyzed, phage DNA replicated  3. Fragment of bacterial DNA incorporated into phage  4. Phage infects new host  5. Crossing over can occur between bacterial DNA fragment and homologous region of chromosome  6. Recombinant genotype results o Recall lysogenic phage  Prophage exits host genome before entry into the lytic cycle  Occasionally the prophage takes some of the host genome with it  Results as before, recombination can occur  Comparative genomes: sequence organization o Prokaryotes most of the genome encodes either protein, functional RNA, or regulatory sequence o In multicellular eukaryotes, the vast majority of DNA does not code on functional gene product o Much non-coding DNA is not “junk DNA” (but some may be) o Rate of increase of amount of DNA from prokaryotes to single- celled eukaryotes to multicellular eukaryotes is much greater than the increase in number of gametes  TE related sequences o Many TE related sequences are TEs that have lost the ability to transpose  Called non- autonomous elements  Can sometimes be mobilized by enzymatic machinery of “live” TEs  Other repetitive DNA o ≈15% of human genome is repetitive DNA not related to TEs o Mostly due to errors in DNA replication and/or recombination  Large-segment duplications  Multi-gene families o Most eukaryotic genes present in only one copy o Some genes present in >1 copy (eg. rRNA) o Some genes closely resemble each other in sequence  Genes that closely resemble each other probably resulted from an ancestral gene duplication  May be clustered or dispersed in genome  Fuctional genomics: quantitative trait locus (QTL) mapping o Hypothetical example: there is a genetic variation among inbreeding lines of mice in the susceptibility to infection by staphylocus aureus o Question: how to find the genes that confer susceptibility o Cross a susceptible line to a resistant line o F1s will be heterozygous at all loci o F2s will segregate different combinations of alleles o Genotype the lines at many marker loci  Functional genomics: genome- wide association study (GWAS) o Example: susceptibility of schizophrenia in humans o Schizophrenia seems to happen in families more often than expected by chance, but it’s obviously not due to one or a few Mendelian loci o How to track down genetic causes You can’t do a controlled breeding with humans o Solution: Let nature provide the meiosis for you! o Genotype a large number of individuals at a large number of marker loci, “single nucleotide polymorphisms” (SNPs) March 16, 2016 Purposes of biotechnology o 1. Establish the genetic basis of phenotype  Eg. Find mutations that cause colon cancer o 2. Manipulate the genetic basis of phenotype  Eg. Fix mutations that cause colon cancer Methods o Restriction enzymes/gene cloning  Restriction enzymes allow for the manipulation of DNA in a repeatable predictable fashion  Cloning vector – DNA or RNA molecule (e.g., bacterial plasmid or virus) capable of independent replication  Plasmid – Small, double-stranded ring of DNA, found in the cytoplasm of bacteria and capable of independent replication  Basic strategy is to ligate (link) target DNA to vector DNA, so that when the vector replicates, so does the linked target DNA o Polymerase chain reaction (PCR)  Isolate and amplify a specific DNA sequence o Transformation/transfection/transduction o Expression vectors o Genome editing with CRISPR/Cas9

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Chapter 8, Problem 90SE is Solved
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Textbook: Statistics for Business and Economics
Edition: 12
Author: James T. McClave, P. George Benson, Terry T Sincich
ISBN: 9780321826237

Since the solution to 90SE from 8 chapter was answered, more than 273 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 90SE from chapter: 8 was answered by , our top Business solution expert on 07/21/17, 05:42AM. This textbook survival guide was created for the textbook: Statistics for Business and Economics , edition: 12. This full solution covers the following key subjects: test, hypothesis, use, variances, against. This expansive textbook survival guide covers 15 chapters, and 1631 solutions. Statistics for Business and Economics was written by and is associated to the ISBN: 9780321826237. The answer to “Two independent random samples were selected from normally distributed populations with means and variances and , respectively. The sample sizes, means, and variances are shown in the following table. a. Test against . Use a = .05.b. Would you be willing to use a small sample t-test to test the null hypothesis against the alternative hypothesis Why?” is broken down into a number of easy to follow steps, and 57 words.

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Two independent random samples were selected from normally