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Rationalize each numerator. See Examples 5 and 6.A3 78

Intermediate Algebra | 6th Edition | ISBN: 9780321785046 | Authors: Elayn El Martin-Gay ISBN: 9780321785046 180

Solution for problem 67 Chapter 7.5

Intermediate Algebra | 6th Edition

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Intermediate Algebra | 6th Edition | ISBN: 9780321785046 | Authors: Elayn El Martin-Gay

Intermediate Algebra | 6th Edition

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Problem 67

Rationalize each numerator. See Examples 5 and 6.A3 78

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LIFE102 Week 12 Notes 4/4/16, Chapter 16, The Molecular Basis for Inheritance  The Structure of DNA  Chargaff’s Rules o The base composition (ATCG) of DNA varies between different species o In any species, the number of A and T bases are equal and the number of G and C bases are equal  DNA structure o Double-helix o Modeled by Watson and Crick o Watson and Crick’s model explained Chargaff’s rules (A=T and C=G) o 1 DNA nucleotide  Phosphate  Deoxyribose  N-base o A-T base pair: 2 bonds o G-C base pair: 3 bonds (more stable)  Structure of 2 DNA strands o Complementary (A-T & G-C) o Antiparallel (3’/5’  5’/3’)  Average chromosome: o 50 million base pairs (bp) o 5,000 genes o 1 gene: 10,000 bp  encodes for 1 protein o Before each cell division, all DNA is replicated  The basic principle of DNA replication: o Base pairing to a template strand o Since the two strands of DNA are complementary, each strand acts a template for building a new strand in replication o In DNA replication, the parent molecule unwinds and two new daughter strands are built based on base-pairing rules  DNA replication is “Semi-conservative” o DNA strands separate: each strand serves as a template for new strand o New strands are made from nucleotides o Each DNA molecule is ½ old and ½ new  Starting points of DNA replication: Origin of Replication (ori) o Bacterial chromosomes: 1 ori o Eukaryotic chromosomes: 100-1000 oris  DNA Replication o At the end of each replication bubble is a replication fork: a y shaped region where new DNA strands are elongating o Helicases are enzymes that untwist the double helix at the replication forks o RNA primer primes for replication (primase) o Always starts at the 3’ end of the double helix  Priming DNA synthesis o Enzymes called DNA polymerases elongate new DNA at a replication fork o DNA polymerases read the DNA and put down the matching enzyme’ 4/6/16, Chapter 16 cont.  The antiparallel structure of the double helix affects replication o DNA polymerase add nucleotides only to the free 3’ end of a growing strand; therefore, a new DNA strand can elongate only in the 5’ to 3’ direction o Along one template strand of DNA, the DNA polymerase synthesizes a leading strand continuously, moving toward the replication fork o To elongate the other new strand, called the lagging strand, DNA polymerase must work in the direction away from the replication fork 2 o The lagging strand is synthesized as a series of segments called Okazaki fragments, which are joined together by DNA ligase o DNA Polymerase 1 will later replace the RNA primer fragments with DNA o DNA Ligase connects DNA fragments  Makes the backbone of DNA continuous o SLIDE 23  Proofreading & Repairing DNA o DNA polymerases proofread newly-made DNA replacing any incorrect nucleotides o DNA can be damaged by exposure to harmful chemical or physical agents; it can also undergo spontaneous changes o In nucleotide excision repair, a nuclease cuts out damaged stretches of DNA, which are then replaced  Replicating the Ends of DNA molecules o Limitations of DNA polymerase create problems for the linear DNA of eukaryotic chromosomes o No way to complete the 5’ ends, so repeated rounds of replication produce shorter DNA molecules with uneven ends  Telomeres o Strand ends pose a problem for linear DNA:  Gaps remain at the end of the 5’ ends when RNA primers are removed o Telomeres: non-coding repetitive DNA segments o TTAGGG repeat in humans o Eukaryotic chromosomal DNA molecules have special nucleotide sequences at their ends called telomeres o Telomeres postpone the erosion of genes near the ends of DNA molecules o It has been proposed that the shortening of telomeres is connected to aging  A chromosome consist of DNA molecule packed together with proteins o Chromatin, a complex of DNA and protein is found in the nucleus of eukaryotic cells 3 o Chromosomes fit into the nucleus through an elaborate, multilevel system of packing 4/6/16, Chapter 17, From Gene to Protein  The Overview: The Flow of Genetic Information o The information content of DNA is in the form of specific sequences of nucleotides (ATCG) o The DNA inherited by an organism leads to specific traits by dictating the synthesis of proteins o Proteins are the links between genotype and phenotype o Gene expression, the process by which DNA directs protein synthesis, includes two stages: transcription and translation  Basic Princliples of Transcription and Translation o Transcription is the synthesis of RNA using information in DNA o Transcription produces messenger RNA (mRNA) o RNA is the bridge between genes and the proteins for which they code o Translation is the synthesis of a polypeptide, using information in the mRNA o Ribosomes are the sites of translation 4/8/16, Chapter 17, cont.  In a eukaryotic cell, the nuclear envelope separates transcription from translation o Eukaryotic RNA transcripts are modified through RNA processing to yield the finished mRNA o In prokaryotes, translation of mRNA can begin before transcription has finished  Can happen at the same time in the same place because DNA is located in the cytosol because it doesn’t have nucleus  Can make proteins very quickly  In Eukaryotes: o DNA: gene library in nucleus   Transcription 4 o Messenger RNA: copy made in nucleus brought to cytosol   Translation o Protein: made using mRNA info  The Central Dogma o Cells are governed by a cellular chain of command o DNA Replication  Transcription  Translation  Protein  The Genetic Code  How are the instructions for assembling amino acids into proteins encoded into DNA  There are 20 amino acids, but there are only 4 nucleotide bases in DNA o Codons: Triplets of Nucleotides  The flow of information from gene to protein is based on a triplet code:  A series of non-overlapping, three nucleotide words  During translation, the mRNA base triplets, called codons are read in the 5’ to 3’ direction  The words of a gene are transcribed into complementary nonoverlapping three-nucleotide words of mRNA (codons)  These codons are then translated into a chain of amino acids forming a polypeptide (a protein)  Transcription o During transcription, one of the two DNA strands, called the template strand, provides a template for ordering the sequence of complementary nucleotides in an RNA transcript o The template strand is always the same strand for a given gene o Each codon specifies the amino acid to be placed at the corresponding position along a polypeptide  Molecular components of transcription o RNA synthesis is catalyzed by RNA polymerase, which pries the DNA strands apart and hooks together the RNA nucleotides o RNA is complementary to the DNA template strand o RNA synthesis follows the same base pairing rules as DNA, except uracil substitutes for thymine 5  Messenger RNA (mRNA): o Made by RNA polymerase o Made from 5’3’ o Complementary and antiparallel with the template DNA  Transcription’s 3 stages: o Initiation  RNA polymerase binds to promoter  Transcriptions starts o Elongation o Termination:  RNA polymerase reaches the terminator:  Transcript released  Eukaryotic cells modify RNA after transcription o Enzymes in the eukaryotic nucleus modify pre-mRNA (RNA processing) before the genetic messages are dispatched to the cytoplasm o Each end of a pre-mRNA molecule is modified  The 5’ end reveices a modified nucleotide 5’ cap  Handle for a protein to grab on for transportation  The 3’ end gets a poly-A tail  Prevents RNA molecule from being degraded over time  Split Genes and RNA Splicing o Most eukaryotic genes and their RNA transcripts have long noncoding stretching of nucleotides that lie between coding regions o These are noncoding regions are called intervening sequences, or introns o The other regions are called exons becaue they eventually expressed, usually translated into amino acid sequences o RNA splicing removes introns and joins exons, creating an mRNA molecule with a continuous coding sequence  Translation 6 o Genetic information flows from mRNA to protein through the process of translation (in power point 3 times) o LOOK AT SLIDE 22 o A cell translates an mRNA message into protein with the help of transfer RNA (tRNA) o tRNAs transfer amino acids to the growing polypeptide in a ribosome  Genetic code o The genetic code is redundant (more than one codon may specify a particular amino acid) o Codons must be read in the correct reading frame in order for the right polypeptide to be produced o 7 LIFE 102, Week 11 3/28/16, Chapter 14 cont.  Environment affecting genetics can be something like tanning of skin color  Pedigree Analysis o A pedigree is a family tree that describes the interrelationships of parents and children across generations o Inheritance patters of particular traits can be traced and described using pedigrees o Pedigrees help us calculate the probability that a future child will have a particular genotype and phenotype  Recessively Inherited Disorders o Many genetic disorders are inherited in a recessive manner  The Behavior of Recessive Alleles o Recessively inherited disorders show up only in individuals homozygous for the allele o Carriers are heterozygous individuals who carry the recessive allele but are phenotypically normal; most individuals with recessive disorders are born to carrier parents 3/28/16, Chapter 15, The chromosomal Basis of Inheritance  Mendelian inheritance has its physical basis in the behavior of chromosomes o Mitosis & Meiosis were first described in the late 1800s o The chromosome theory of inheritance states:  Mendelian genes have specific loci (positions) on chromosomes  Chromosomes undergo segregation and independent assortment o The behavior of chromosomes during meiosis can account for Mendel’s law of segregation and independent assortment  The Chromosomal Basis of Mendel’s Laws o The behavior of nonhomologous chromosomes during meiosis accounts for the independent assortment of the alleles for 2 or more genes on different chromosomes o Starting with 2 true-breeding pea plants, let’s follow 2 genes through the F1 and F2 generations o The two genes specify seed color and seed shape  These 2 genes are on different chromosomes o The behavior of chromosomes during meiosis in the F1 generation and subsequent random fertilization gave rise to the F2 phenotypic ratio observed by Mendel  Morgan’s Experimental Evidence o Evidence associating a specific gene with a specific chromosome came from Thomas Hunt Morgan o Morgan’s experiments with fruit flies provided convincing evidence that chromosomes are the location of Mendel’s heritable factors o Morgan noted wild type (normal) phenotypes that were common in the fly populations o Traits alternative to the wild type are called mutant phenotypes  Correlating Behavior of a gene’s alleles with Behavior of a Chromosome Pair o Morgan mated male flies with white eyes with female flies with red eyes o The F1 generation had red eyes o F2 generation showed the 3:1 ratio but only males had white eyes o Morgan determined that the white-eyed mutant allele must be located on the X chromosome  Supported the chromosome theory of inheritance  Sex-linked genes exhibit unique patterns of inheritance o In humans and some other animals, there is a chromosomal basis of sex determination o There are two varieties of sex chromosomes: a larger X chromosome and a smaller Y chromosome  Females XX 2  Males XY o Each Ovum contains an X chromosome, while a sperm may contain either an X or a Y chromosome  Exceptional inheritance patterns… o X-linked/sex-linked inheritance: different for female and male o X-linked genes: on X chromosome o X & Y: sex chromosomes o X chromosomes: contains lots of genes o Y chromosomes: contain few genes o Men only need one recessive gene on their X instead of having two like women  X linked genes o Follow specific patterns of inheritance o For a recessive X-linked trait to be expressed:  A female needs two copies of the allele (homozygous)  A male only needs one cope of the allele (hemizygous, don’t need to know name) o X linked recessive disorders are much more common in males than in females  X Inactivation in Female Mammals o In mammalian females, one of the two X chromosomes in each cell is randomly inactivated during embryonic development  The inactive X condenses into a Barr body o If a female is heterozygous for a particular gene located on the X chromosome, she will be a mosaic for that character  Half of her cells will express one allele, while the others will express the alternate allele  Mottled coloration, tortoiseshell cats 3/30/16, Chapter 15 cont.  Linked genes tend to be inherited together because they are located near eachother on the same chromosome o Each chromosome has hundreds or thousands of genes (except the y) 3 o Genes located on the same chromosome tend to be inherited together are called linked genes o Morgan did other experiments with fruit flies to see how linkage affects inheritance of two characters  Morgan crossed flies that differed in traits of body color and wing size  He found that they must be inherited in specific combinations  These genes do not assort independently and are on the same chromosome  Exceptional inheritance patterns… o Genes on the same chromosome do not always segregate together o Cause was crossing over  Inheritance of Linked Genes o Crossing over-separates linked genes: recombination  Chromosomal basis for recombination of linked genes o Because crossing-over between the b and vg loci occurs in some, but not all, egg-producing cells, more eggs with parental-type chromosomes than with recombinant ones are produced in the mating females  Calculating Recombinant Frequency o 2 genes further apart on chromosome  More change of crossing over  Higher recombinant frequency  Genetic Alterations o Physical & Chemical disturbances o Mistakes in cell replication o Daughter cells get too much/too little DNA o Possible Mistakes  One chromosome too many/few (2n+1, 2n-1): Aneuploidy  Extra copies of entire genome ( 3n, 4n, 6n): Polyploidy  Meiotic Nondisjunction 4 o In nondisjunction pairs of homologous chromosomes do not separate normally during meiosis o As a result, one gamete receives two of the same type of chromosome and another gamete receives no copy  Aneuploidy: abnormal number of a particular chromosome o Down syndrome: 2n=47  Trisomy of chromosome 21: 3 copies of chromosome 21  Human disorders due to chromosomal alterations o Alterations of chromosome number and structure are associated with some serious disorders o Some types of aneuploidy appear to upset the genetic balance lss than others, resulting in individuals surviving to birth and beyond o Surviving individuals have a set of symptoms, or a syndrome, characteristic of the type of aneuploidy  Down Syndrome o Down syndrome is a aneuploidy condition that results from three copies of chromosome 21 o It affects about one of every 700 children born in the United States o The frequency of Down syndrome increases with age of the mother, a correlation that has not been explained  Polyploidy: extra copies of entire genome o Cause: DNA duplication without cell division o Polypoidy in animals: uncommon  Some amphibians and fish  Typically lethal in other animals o Polyploidy in plants: fairly common  Not lethal: asexual reproduction  Many crop plants (tobacco, bananas, strawberries, wheat) 5 LIFE102 Week 12 Notes 4/4/16, Chapter 16, The Molecular Basis for Inheritance  The Structure of DNA  Chargaff’s Rules o The base composition (ATCG) of DNA varies between different species o In any species, the number of A and T bases are equal and the number of G and C bases are equal  DNA structure o Double-helix o Modeled by Watson and Crick o Watson and Crick’s model explained Chargaff’s rules (A=T and C=G) o 1 DNA nucleotide  Phosphate  Deoxyribose  N-base o A-T base pair: 2 bonds o G-C base pair: 3 bonds (more stable)  Structure of 2 DNA strands o Complementary (A-T & G-C) o Antiparallel (3’/5’  5’/3’)  Average chromosome: o 50 million base pairs (bp) o 5,000 genes o 1 gene: 10,000 bp  encodes for 1 protein o Before each cell division, all DNA is replicated  The basic principle of DNA replication: o Base pairing to a template strand o Since the two strands of DNA are complementary, each strand acts a template for building a new strand in replication o In DNA replication, the parent molecule unwinds and two new daughter strands are built based on base-pairing rules  DNA replication is “Semi-conservative” o DNA strands separate: each strand serves as a template for new strand o New strands are made from nucleotides o Each DNA molecule is ½ old and ½ new  Starting points of DNA replication: Origin of Replication (ori) o Bacterial chromosomes: 1 ori o Eukaryotic chromosomes: 100-1000 oris  DNA Replication o At the end of each replication bubble is a replication fork: a y shaped region where new DNA strands are elongating o Helicases are enzymes that untwist the double helix at the replication forks o RNA primer primes for replication (primase) o Always starts at the 3’ end of the double helix  Priming DNA synthesis o Enzymes called DNA polymerases elongate new DNA at a replication fork o DNA polymerases read the DNA and put down the matching enzyme’ 4/6/16, Chapter 16 cont.  The antiparallel structure of the double helix affects replication o DNA polymerase add nucleotides only to the free 3’ end of a growing strand; therefore, a new DNA strand can elongate only in the 5’ to 3’ direction o Along one template strand of DNA, the DNA polymerase synthesizes a leading strand continuously, moving toward the replication fork o To elongate the other new strand, called the lagging strand, DNA polymerase must work in the direction away from the replication fork 2 o The lagging strand is synthesized as a series of segments called Okazaki fragments, which are joined together by DNA ligase o DNA Polymerase 1 will later replace the RNA primer fragments with DNA o DNA Ligase connects DNA fragments  Makes the backbone of DNA continuous o SLIDE 23  Proofreading & Repairing DNA o DNA polymerases proofread newly-made DNA replacing any incorrect nucleotides o DNA can be damaged by exposure to harmful chemical or physical agents; it can also undergo spontaneous changes o In nucleotide excision repair, a nuclease cuts out damaged stretches of DNA, which are then replaced  Replicating the Ends of DNA molecules o Limitations of DNA polymerase create problems for the linear DNA of eukaryotic chromosomes o No way to complete the 5’ ends, so repeated rounds of replication produce shorter DNA molecules with uneven ends  Telomeres o Strand ends pose a problem for linear DNA:  Gaps remain at the end of the 5’ ends when RNA primers are removed o Telomeres: non-coding repetitive DNA segments o TTAGGG repeat in humans o Eukaryotic chromosomal DNA molecules have special nucleotide sequences at their ends called telomeres o Telomeres postpone the erosion of genes near the ends of DNA molecules o It has been proposed that the shortening of telomeres is connected to aging  A chromosome consist of DNA molecule packed together with proteins o Chromatin, a complex of DNA and protein is found in the nucleus of eukaryotic cells 3 o Chromosomes fit into the nucleus through an elaborate, multilevel system of packing 4/6/16, Chapter 17, From Gene to Protein  The Overview: The Flow of Genetic Information o The information content of DNA is in the form of specific sequences of nucleotides (ATCG) o The DNA inherited by an organism leads to specific traits by dictating the synthesis of proteins o Proteins are the links between genotype and phenotype o Gene expression, the process by which DNA directs protein synthesis, includes two stages: transcription and translation  Basic Princliples of Transcription and Translation o Transcription is the synthesis of RNA using information in DNA o Transcription produces messenger RNA (mRNA) o RNA is the bridge between genes and the proteins for which they code o Translation is the synthesis of a polypeptide, using information in the mRNA o Ribosomes are the sites of translation 4/8/16, Chapter 17, cont.  In a eukaryotic cell, the nuclear envelope separates transcription from translation o Eukaryotic RNA transcripts are modified through RNA processing to yield the finished mRNA o In prokaryotes, translation of mRNA can begin before transcription has finished  Can happen at the same time in the same place because DNA is located in the cytosol because it doesn’t have nucleus  Can make proteins very quickly  In Eukaryotes: o DNA: gene library in nucleus   Transcription 4 o Messenger RNA: copy made in nucleus brought to cytosol   Translation o Protein: made using mRNA info  The Central Dogma o Cells are governed by a cellular chain of command o DNA Replication  Transcription  Translation  Protein  The Genetic Code  How are the instructions for assembling amino acids into proteins encoded into DNA  There are 20 amino acids, but there are only 4 nucleotide bases in DNA o Codons: Triplets of Nucleotides  The flow of information from gene to protein is based on a triplet code:  A series of non-overlapping, three nucleotide words  During translation, the mRNA base triplets, called codons are read in the 5’ to 3’ direction  The words of a gene are transcribed into complementary nonoverlapping three-nucleotide words of mRNA (codons)  These codons are then translated into a chain of amino acids forming a polypeptide (a protein)  Transcription o During transcription, one of the two DNA strands, called the template strand, provides a template for ordering the sequence of complementary nucleotides in an RNA transcript o The template strand is always the same strand for a given gene o Each codon specifies the amino acid to be placed at the corresponding position along a polypeptide  Molecular components of transcription o RNA synthesis is catalyzed by RNA polymerase, which pries the DNA strands apart and hooks together the RNA nucleotides o RNA is complementary to the DNA template strand o RNA synthesis follows the same base pairing rules as DNA, except uracil substitutes for thymine 5  Messenger RNA (mRNA): o Made by RNA polymerase o Made from 5’3’ o Complementary and antiparallel with the template DNA  Transcription’s 3 stages: o Initiation  RNA polymerase binds to promoter  Transcriptions starts o Elongation o Termination:  RNA polymerase reaches the terminator:  Transcript released  Eukaryotic cells modify RNA after transcription o Enzymes in the eukaryotic nucleus modify pre-mRNA (RNA processing) before the genetic messages are dispatched to the cytoplasm o Each end of a pre-mRNA molecule is modified  The 5’ end reveices a modified nucleotide 5’ cap  Handle for a protein to grab on for transportation  The 3’ end gets a poly-A tail  Prevents RNA molecule from being degraded over time  Split Genes and RNA Splicing o Most eukaryotic genes and their RNA transcripts have long noncoding stretching of nucleotides that lie between coding regions o These are noncoding regions are called intervening sequences, or introns o The other regions are called exons becaue they eventually expressed, usually translated into amino acid sequences o RNA splicing removes introns and joins exons, creating an mRNA molecule with a continuous coding sequence  Translation 6 o Genetic information flows from mRNA to protein through the process of translation (in power point 3 times) o LOOK AT SLIDE 22 o A cell translates an mRNA message into protein with the help of transfer RNA (tRNA) o tRNAs transfer amino acids to the growing polypeptide in a ribosome  Genetic code o The genetic code is redundant (more than one codon may specify a particular amino acid) o Codons must be read in the correct reading frame in order for the right polypeptide to be produced o 7

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Textbook: Intermediate Algebra
Edition: 6
Author: Elayn El Martin-Gay
ISBN: 9780321785046

This full solution covers the following key subjects: . This expansive textbook survival guide covers 90 chapters, and 8410 solutions. Intermediate Algebra was written by and is associated to the ISBN: 9780321785046. Since the solution to 67 from 7.5 chapter was answered, more than 242 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Intermediate Algebra, edition: 6. The full step-by-step solution to problem: 67 from chapter: 7.5 was answered by , our top Math solution expert on 12/23/17, 04:59PM. The answer to “Rationalize each numerator. See Examples 5 and 6.A3 78” is broken down into a number of easy to follow steps, and 9 words.

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Rationalize each numerator. See Examples 5 and 6.A3 78