Chapter 6: Exploring Genes

• You should review Chapter 5 in order to understand the nature of hereditary material and the flow of information from DNA to pro tein.

• The methods and techniques used to analyze and manipulate DNA are presented in this chapter.
• An overview of the tools that make it possible is what they begin with.
• The specificity of base pairs between nucleic acids and the enzymes is important.

• Specific DNA restriction fragments, by DNA ligase, are described in a more detailed description.
• The major method of determining the sequence of DNA and an automated method for synthesizing oligodeoxyribonucleotides are presented.
• The authors explain how specific fragments of genes can be amplified by the process of polymerase-catalyzed synthesis of DNA.
• A more detailed description of restriction and DNA ligase follows.
• Various self-replicating carriers of the target genes are discussed.
• The problems of locating specific genes in the genome and of expressing foreign genes in the eukaryotes are also considered.
• There is a special role in the production of cDNA.
• The pattern and level of gene expression can be monitored with the help of DNA chips.

• Chapter 6 was obtained from them.
• The authors show how site-specific mutagenesis can be used with cloned genes to produce a desired amino acid.
• They close the chapter with an overview of how the methods described allow the information to be manipulated.
• The authors talk about how the new biochemical technologies make possible previously unimaginable manipulation of living organisms.

• You should be able to complete the objectives once you have mastered this chapter.

• Explain the applications of the basic tools.

• The number of nucleotides in the genomes of lower and higher organisms was calculated.

• Indicate the roles of deprotection in the process.

• The practical applications of the technique are appreciated.

• Draw the ter mini of the fragments.

• Compare the properties and merits of some common vectors used in prokaryotes and eukaryotes.

• Explain how to use the genetic code to make a probe.

• The process for converting the information in mRNA into duplex DNA is outlined in 21.

How can this process be used to clone the DNA?

• oligonucleotides can be used to synthesise genes.

• High yields at each condensation step are required for the successful chemical syn thesis.

• The ends of the strands need to be moved so that a 3, OH is next to a 5, OH.

If each of the following is not a desired characteristic of the vectors, why not?

• It is possible to efficiently package very long DNA fragments from the eukaryotic genomes.

• The linker oligonucleotide d(GGAATTCC) and an isolated and purified DNA restriction fragment have been excised from a longer DNA molecule with a restriction endonuclease that produces blunt ends.

• Drug resistance transfer occurs when a long DNA fragment is inserted into the middle of a gene that specifies an antibiotic and is incorporated into a bacterium.

Explain how the presence of introns in eukaryotic genes complicates the production of theProtein products they encode when expression is attempted inbacteria

• The genes for a polypeptide hormone were isolated, cloned, and overexpressed in a bacterium.
• The polypeptide failed to function when it was subjected to a bioassay in the organisms from which it was isolated.
• Why was the product inactive?

• It depends on the sequence of the oligonucleotide and if it is different from the target gene.

• The steps needed to make a gene are outlined.

• You should be clear about the information and reagents that you need.

• The top strand has the same sequence as the bottom strand.
• Many restriction enzymes cut palindromic sequences.

• Ethidium bromide intercalates into the double strand and increases its quantum yield.
• It fluoresces with an intense orange color when it is irradiated.

• The reaction of the oligonucleotide with these reagents would yield.

• The dimethoxytrityl group must be removed from the 5,,-hydroxyl without removing the blocking groups on the exo.
• Although the synthesis can be carried out manually, it is slow and laborious.

• The steps are 3, 2, and 1.

• The double strand DNA molecule is completely separated by heating to 95degC.
• The excess primer is caused to hybridize to the parent strands by the rapid cooling.
• After the 95degC step, the Taq DNA polymerase is able to carry out DNA synthesis using the four dNTPs.
• Amplification of DNA can be achieved by repeating steps many times.

• If the two strands had a gap between their ends, they couldn't join.
• If the dNTPs were added to this structure, they could be used to convert the gapped, duplex DNA into a productive substrate for DNA ligase.

• Answers a, b, c, and d are desired characteristics.

• In the absence of host cell growth, am plification of the vector is possible.
• There are unique restriction sites that allow the cutting of the foreign DNA.
• Antibiotic resistance allows for the selection ofbacteria that carry a vaccine.
• Small size allows the introduction of long pieces of foreign DNA without interfering with the introduction of the molecule into the host bacterium.
• Answer e isn't an essential characteristic because they don't have to be circular to function effectively.

• The statements a and e are incorrect because large DNA fragments do not fit into a l capsid and are too big to be separated easily by a standard polyacrylamide gel electrophoresis.
• Large DNA fragments can be separated using a field of light.
• Answer c is incorrect because linkers and adapters can be used to clone noncohesive ends.

• The 5,phosphate group of the oligonucleotide must be used as a base for the DNA ligase.
• The palindromic (self-complementary) oligonucleotide would be joined to the blunt-end fragment with the help of the polynucleotides.
• The endonuclease will produce cohesive ends that match the cut.
• The fragment would be cut when it was treated with the enzyme.

• Drug resistance is caused by the production of the gene that confers drug resistance.
• A cell is sensitive to the antibiotic.
• Selecting cells that contain the inserted fragment can be done if the gene that confers resistance to a second drug is present.
• The cells that are resistant to the second antibiotic are likely to have the foreign DNA in them.

• A genomic library is composed of a collection of clones, each of which has its own genetic material.
• The entire collection should have all the sequence of the target organisms' genome.
• A collection of clones that contain the same sequence in the mRNA of the target organisms is called a cDNA library.
• Only a small portion of the genome is being transcribed into mRNA at any given time, so a cDNA library contains far fewer clones than a genomic library.
• The content of a library depends on the cells that were isolated.
• The type of cell, its state of development, and environmental factors affect the identity of the population.

• Trp and Met have one codon each and Cys has two, which makes this sequence the better choice for reverse translation.
• The Met-Trp-CysTrp sequence has different dodecameric coding sequences.
• Leu and Arg each have six codons, so for the (a) sequence, Met-Leu-Arg-Leu, there are over 200 different coding sequences.
• The probe for (b) would be simpler to build and give more results.

• A method of determining the sequence of long regions of DNA by subcloning pieces of a library is referred to as 18. chromosome walking.
• The overlaps can be used to place the shorter sequence into the correct order.

• The introns are removed from the primary transcript in eukaryotes.
• The translation product of the primary transcript wouldn't work because it wouldn't have the machinery to process it.
• The problem can be solved with the use of cDNA prepared from the genes that make the proteins, and the intron sequence will have been removed.

• Adding a radioactive antibody to lysed colonies could be used to screen the population for the cloned genes.

• Since reverse transcriptase makes DNA, dNTPs are required.
• There is a template for the synthesis of a DNA strand.
• The template for the synthesis of its complement is served by the DNA strand itself.
• A primer is needed to start the synthesis of a new DNA chain.

• It is possible that the polypeptide might not have undergone some posttranslational modification that is needed for it to function, as an obvious explanation as the destruction of the polypeptide during the bioassay.
• The polypeptide may need to be acetylated, methylated, or trimmed at the N- or C-terminus.
• If the bacterium that was produced did not have the necessary machinery to carry out these modifications, it would lack the ability to recognize the signals that direct them.
• It is1-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-6556

• If the sequence of the oligonucleotide was completely different from the target gene, it could not be used as a primer for DNA polymerase.
• Functional domains could be better identified by deletion mutagenesis, in which relatively large regions of the gene would be systematically removed and the resulting functional consequences tested.
• Although oligonucleotide-directed mutagenesis can be used to make deletions, it is not the method of choice for an initial survey to find functional domains because only a single, precisely defined deletion is produced with each oligonucleotide.

• Nucleases are used for deletion analysis.
• When one wishes to test a model or hypothesis regarding the function of one or a few amino acids, oligonucleotide-directed mutagenesis is better suited for the job.
• The perfect complement will be formed when the mutagenizing oligonucleotide forms a more stable hybrid with the newly produced Mutant sequence than with the original unmodified sequence.
• It can be used to distinguish the original and the Mutant.

• You need to know the sequence of the gene you want to make.

• The genetic code could be used to derive this from the amino acid sequence.
• You need to know what restriction sites you want to build into the synthetic sequence.
• You have to decide on the sequence of the different parts of the gene.
• The final desired sequence, the individual lengths (30 to 80 nucleotides long) that can be easily synthesised and purified, and the requirement for overlap of the cohesive ends of the partially duplex segments are some of the factors that would determine the sequence.
• The selfcomplementary oligomers would be mixed together to form duplex fragments.
• If appropriate ends have been designed into the synthesis, the product can be used for cloning.

• You are studying a newly isolated bacterial restriction enzyme that cleaves double-strand circles of plasmid pBR322 once to yield unit-length, linear, double-strand molecules.

• All the double-strand molecules are unit-length linears after they are denatured and allowed to reanneal.
• In another experiment, an enzyme that cleaves double-strand DNA at random sites is used at low concentration to cleave intact pBR322 molecule approximately once per molecule, again yielding unit-length, double-strand linear DNA molecule.
• There are double-strand circles with a single nick in each strand.

• Before the development of modern methods for analysis and manipulation of genes, many attempts were made to transform both prokaryotic and eukaryotic cells.

• Most of the experiments failed.
• The early attempts to transform cells largely failed.

• Pseudogenes are composed of nonfunctional (unexpressed) DNA sequences that are re lated by sequence similarity to actively expressed genes.
• Some researchers think that pseudogenes are copies of functional genes that have been inactivated.
• Suggestions on how the genes could have become nonfunctional.
• If you clone a number of closely related sequences, you may be able to code for a particularProtein.

• The method for determining a DNA sequence is limited by the number of bases that can be analyzed in one reaction.
• Suppose you want to sequence a newly isolated double-strand DNA tumor virus that has 5000 base pairs.

• The Sanger method is used for restriction fragments of the DNA.
• On average, you use an enzyme that makes a lot of cuts to give a small amount of fragments.

• It is possible to use the denaturation and reassociation of DNA strands as a tool for genetic analysis.
• Increasing the pH of the solution above 11 causes the dissociation of the strands.
• The strands will reanneal when the solution is cooled or the pH is lowered.

• Both double- and single-strand DNA can be visualized using a technique called Heteroduplex analysis.
• In a reannealing experiment, two types of doublestrand molecule, one containing the sequence for a single gene and the other containing the same sequence as well as an insertion of nonhomologous DNA, are mixed and used.

• A single-strand loop will be visible at the location of the deletion when a Heteroduplex is formed between a deletion strain and a wild-type strain.
• Suppose you are studying a deletion of about 200 base pairs, located at a unique site on the bacterium's chromosomes, which contains over 3000 genes.
• Why would it be a good idea?

• You want to know if the substitution of a glycine residue will change the structure of the polypeptide.
• You know the sequence of the protein because you cloned it.

• Two unique fragments, one 100 bases and the other 400 bases in length, can be obtained from a double-strand DNA fragment with 500 bases.

• There are three fragments, two containing 150 and one containing 200, that were cleaved.
• Two fragments 100 bases in length and two 150 bases in length are found when the 500-base fragment is in a double-digest.
• The 500-base fragment has two enzymes.
• The first 75 base pairs at the left end are deleted if you have a double-strand DNA fragment that is identical to the original fragment.

• To avoid translation, you should use cDNA.
• The level of expression of functional protein will be low if you only introduce the chicken ovalbumin cDNA.

• If the sample is to be used for amplification, it must be free of contaminating DNA.
• To see why, consider a procedure that begins with a small amount of DNA in a mixture of 100 mL.
• The sample can be amplified in 20 cycles.
• If 0.1 mL of DNA from the initial amplification cycle is inadvertently introduced into a different reaction mixture, what would happen?

• One method of analysis of evidence from cases of sexual assault is histo compatability.
• The sperm and cells from the victim can be found in the samples.
• The samples are first put into a mixture.
• The sperm heads are not lysed.
• The sperm heads are washed several times.
• Sperm heads are sensitive to the protease-detergent mixture when they are lysed with a reducing agent such as dithiothreitol.

• The products are used for analysis.
• Blood and hair samples can be found at the scene of the alleged crime.

• The inability to remove randomly incorporated bases from newly synthesized DNA strands is due to the lack of proof reading activity of the DNA polymerase I.

• The amplified sample will be challenged with a radioactive probe containing the oncogene sequence.

• There are two other catalytic activities and a 5, 3, polymerase activity.
• One of the exonuclease activity's functions is to remove from the growing 3, end of the chain some of the bases that were wrongly incorporated by the polymerase activity.
• A 5, 3, exonuclease removes both pairs of DNA stretches ahead of the polymerase.
• If the 5, 3, exonuclease acts with the polymerase in the same enzyme, new nucleotides are incorporated by the polymerase in place of the ones removed by the nuclease, that is, moved.

• A site-directed mutagenesis can introduce virtually any desired change in a specific gene.
• A very useful application of the technique is to modify a particularProtein and evaluate its effect on biological or chemical activity.
• Two methods have been used in the past to produce modified proteins.
• Chemical agents or ultraviolet light can be used to induce changes in the amino acid sequence.
• One example is the inactivation of a serine in the active site of chymotrypsin using diisopropylfluorophosphate.

• In one family with an affected son, a male fetus and several family members were analyzed to see if the fetus also carried the mutation.
• The intron 10 has a RI polymorphic site.
• The cleavage fragments were stained with ethidium bromide after they were separated by electrophoresis.
• The source of the blood sample for each lane is shown along with a diagram of the gel.
• The RI site will be cut into two pieces.

• The control for each member of the family is specified.

• Chapter 5 of the text describes hydrogen bond formation between bases as a funda mental feature.
• Discuss the role of base pairs in the context of the following: (a) the fidelity of messenger RNA synthesis, (b) the use of primers in the polymerase chain reaction, and (d) identifying a desired clone using a radioactive substance.

• The pBR322 is a double-strand circular DNA molecule.

• cloning experiments use 4 kilobase pairs.

• Assume that the average base pair in the plasmid is 660 and that each cell contains 100 plasmid molecules.

• The aim of the project is to develop an anti-malarial vaccine.
• It is important to consider the entire information flow pathway.

• I hydrolyzes the duplex sequence GCGC between the last G and C to cut a large double-strand plasmid DNA at the single site operator site.

• A circular molecule with a specific end-to-end base sequence can be cleaved at one specific site after denaturation.
• Double-strand linears will be formed by such molecules.
• Double-strand linears with a variety of end-to-end (or permuted) sequences can be produced by random single cleavages of the original intact molecule.
• The formation of double-strand linears with overlaps is a result of the random association of these linears.
• The circles are formed by the overlap of the ends of the molecules.

• During the early years of such experiments, it was difficult to determine what happened to the DNA during transformation attempts.
• The fate of the test DNA could not be determined.
• The failure of the cells to take up the DNA, the rapid degradation of the DNA inside the cell, and the lack of accurate transcription or translation are some of the reasons that these transformation attempts were not successful.

• A stop codon in the sequence is one of the ways that a gene could be inactivated.
• If you want to distinguish a functional and pseudogene, you have to determine the sequence of theProtein and compare it to the coding sequence for each of the genes.
• These types of analyses remind us that there is still a need for a tool in the field of biology.

• There is a chance that small fragments generated by the cleavages may not be detected if one tries to locate all the fragments using gel electrophoresis.
• The overall assignment of sequence is checked when determining the sequence of a second set of fragments.

• The bases are unable to partici pate in base pairs because of the high pH.
• The pKa of the protons on N-1 is 9.2 for guanine.
• The ability of guanine to pair with cytosine is disrupted by the removal of hydrogen.

• If you mix the two types of double-strand molecule, you would expect to see lin ear molecule that are double-strand all along their length as well as some molecule that are only partially double-strand.
• The partially double-stranded molecule will contain a single-strand loop that locates the position of the insertion; they are formed between one strand of the molecule containing the normal gene and one strand of the molecule containing the insertion.

• The formation of tact induplex heteroduplex molecule between the deletion and wild-type DNAs is difficult because the long single strands become entangled as they pair with each other, making them impossible to analyze by electron microscopy.
• The time required for reassociation of the strands is long.
• Since the deletion is located at a single unique site in the chromosomes, the probability of finding the desired molecule among the mixture of many Heteroduplex molecules is rather low.

• You can carry out reannealing experiments that yield a high concentration of Heteroduplex Molecules with the loop characteristic ofDeletions if you clone the deletion or its corresponding wild-type sequence.

• If you want the yeast gene to be joined efficiently by DNA ligase, you have to have both base pairs on the ends of each duplex.
• If both the yeast and l-phage are cleaved with the same enzyme, the yeast and l-phage will have different ends.
• You have to make sure that the sites of the cleavage are in the right places so that the gene to be cloned is intact, and that the fragment has not been fragmented by multiple cleavages.

• The codon for proline is 5, and the codon for glycine is 5.
• Suppose you know the proper codon for yourProtein.
• You can use the scheme outlined in Figure 6-36 of the text to make a primer that is compatible with the region of the gene that specifies the proline.
• After longation of the primer using DNA polymerase, the progeny plasmids will express a glycine substitution at the desired position.

• The 500-base fragment has one site that is cleaved.

• Since we can't distinguish between ends of the molecule, let's assume that the molecule is cut from the left end.

• Only one of the patterns for enzyme B, in which a cut occurs 200 bases from the left end, yields the results when the fragment is incubated with both enzymes.
• At least one 50-base fragment would be produced by the other patterns.

• The deletion fragment serves as a marker for the left-hand end of the molecule, and the double-digest technique allows us to establish which pattern is correct.
• If the pattern shown on the left is correct, the cleaved deletion molecule will yield four fragments, including one only 25 nucleotides in length.
• If you look at the pattern on the right, you can see that the smallest fragment will have a length of 75 nucleotides.

• When additional cleavages are involved, the complexity of the pattern increases greatly.
• The best way to use the double-digest technique is to separate the fragments from one another and then digest them together.
• You can determine the location of different sites within a fragment.
• It is important to remember that the sum of the fragment lengths generated by one and the other must be equal.

• The Pribnow box and the :35 region are two promoter sites that determine where transcription begins.
• You may need the stem-loop and GC-rich terminator sequence at the end of your cDNA in order to cause the messenger RNA to end at the correct site.
• The Shine-Dalgarno ribosome recognition sequence and the proper start and stop signals for translation are also present, so that the mRNA code is read in the proper frame.

• The signals make sure that the expressedProtein has the correct sequence and length.
• The desired signals can be built into a vector so that only the cDNA needs to be cloned.

• The analysis of the second sample could be complicated by the contaminant.
• If the 106 templates in the original sample are amplified one millionfold, the concentration of templates at the completion of 20 cycles is 1012/100 mL.
• The second sample contained 106 templates, compared with a contaminating volume of 0.1 mL.
• The identity of the DNA in the analyte sample could be masked.
• A number of precautions are taken to avoid the introduction of foreign DNA.
• The use of sterile containers and reaction solutions, disposable gloves, and separate work areas for preparing reaction mixtures, pipetting template samples, and analyzing the products are included.
• It's important to run reactions that don't contain added DNA in order to make sure that the solution containing the analyte isn't contaminated.

• It is common to use cells from the victim as well as sperm from the alleged rapist to generate DNA templates for amplification and analysis.
• The differential lysis procedure allows the two to be separated.
• It is necessary that the DNA templates are accessible to the DNA polymerase and the substrates.
• During the gathering of evidence from a crime scene, it is important to keep samples that contain large amounts of DNA, such as bloodstains, separate from samples that contain little DNA.

• The results of the experiments depend on the complement of relatively long sequences of DNA, and occasionally mismatches should not affect the ability of the probe to anneal with each other.
• The synthesis of new DNA strands continues even after the initial cycles of amplification have ended.
• If the mismatch is random, the chances are that any strand will have few base changes and still form base pairs with the radioactive probe.

• Random errors in the chain reaction can be altered in a non-controlled way by cloning.
• There are two ways to deal with this problem.
• Several of these are commercially available and can be used with a thermostable DNA polymerase.
• Clones can also be prepared and sequence.
• The procedure will make sure that the clone you want to examine has the sequence you want it to have.
• You should sequence the product to see that no changes were made.

• 14.Nick translation can be used with labeled nucleoside triphosphates to generate highly radioactive probes for use in Southern blot analysis as well as for other techniques that require such labeled DNA samples.

• CHAPTER 6 is extended.
• The stretches of radioactive sequences in the DNA molecule can be used for autoradiography.

• It would take a lot of work to find a organism that had the desired change in a protein.
• In addition to the one or more specific residues of interest, a chemical agent can be used to modify other amino acid residues.
• The ability of site-specific mutagenesis to target a specific region of a particular gene is different to the older approaches.

• More information about theProtein of interest would allow for more mutagenesis Some examples include the location of the active site, with allosteric interactions, or the association of the active site with the proteins-nucleic acid interactions.

• The amplified DNAs are cut into two fragments.
• The site has been lost and the son's genes are not cut.
• Two fragments and one larger fragment are given to the mother, daughter, and a Heterozygous control.
• The male fetus is normal because the amplified fetal DNA is cleaved by the enzyme.

• If the signal is weak, autoradiography can take a long time.
• Only small amounts of tissue or blood can be analysed with minimal sample preparation.
• One can use a DNA staining reagent such as ethidium bromide to detect amplified sequences instead of using radioactive probes.

• The order of ribonucleotide assembly is determined by the basis of base pairs.

• The order of deoxyribonucleotide assembly is guided by base pairs.
• Hydrogen bonding of base pairs establishes the order of deoxynucleotide addition, and reverse transcriptase can use cDNA as a template for the creation of a duplex DNA molecule.

• Adding deoxynucleotides to the primer is how DNA polymerase carries out chain extension.
• The strands are reannealed after the duplexes have been formed.
• The process can be done as many as 30 times.
• The specificity of the chain reaction for a particular segment is dependent on the base pairs between the primer and templates.

• The extent of hydrogen bonding is what determines the relationship between probe and DNA.
• Southern blotting uses this technique for resolving DNAs.
• The radioactive band in the gel is visualized by autoradiography.

• The degree of relatedness increases with the amount of duplex DNA formed.
• The amount of duplex DNA formation is determined by the number of hydrogen bonds between strands.

• There is no correlation between the levels of mRNA and the production of proteins.

• It might not be possible for an abundant mRNA to be used to synthesise the proteins it is used for.
• They have to be measured directly if they want to be interested in proteins.

• Proteomics is the science of looking at the levels of a single molecule in an animal.

• The mass of 1013 plasmids can be determined by using the weight of a base pair, length of a base pair, and Avogadro's number.

• 1013 plasmids have a mol bp of 4 8 10 5.

• In the reaction mixture 9 there is a 10-9 g DNA/8.6!101 g/mol bp and a 1.51!10-12 mol bp.

• You could use the same method to separate the DNA from the plasmids in which it had been cloned.

• Large amounts of mRNA would be tailored for the chosen translation system.
• The conditions have been developed that allow the production of large amounts of transcript.
• The potential secondary structure of the mRNA, the sequence preferred by the ribosomes, and the location of the start codon are some of the factors to consider.

• Chapter 6 was derived from Kain, K. C., Lanar, D. E., and Orlandi.
• There is a universal promoter for gene expression.

• You could take advantage of the fact that the binding of the repressor molecule to the operator DNA sequence prevents the action of the enzymes.
• The binding of the other to the other cannot be done at the same time.

• The sequence is read from the bottom to the top opposite of the direction of the movement.

• The sequence is 5.
• There is a spot in the autoradiogram that is the fastest moving because of the destruction of the guanine.
• The Pi spot is not shown in the example shown in the text.

• The strand shown in the figure is the complement of the template strand.

• The distance between the cleavage sites is dependent on the length of the strand.

• A small part of a complete gene is what one would get from frag ments.

• Exhaustive di gestion with a restriction enzyme causes nonoverlapping, short fragments.

• The normal and mutant genes would be determined by the II digest.
• Two fragments would be replaced by a single longer fragment if the restriction site was lost.
• It would not be possible to prove that GTG replaced GAG with other sequence changes at the restriction site.

• It would have been very difficult a few years ago.

• The ability of automated solid-phase chemical methods to synthesise DNA has made it possible.
• A simple strategy for generating many mutants is to make a group of oligonucleotides that differ only in the sequence of bases in one triplet.
• 16 different versions of the first triplet of the 30-mer will be provided.
• One can synthesise oligonucleotides in which two or more codons are different.

• There are a number of questions that could be asked about the nature of the original sample.
• It is1-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-65561-6556

• The analysis of the DNA sequence and sequence complexity could be revealing in relation to modern reptiles and other organisms.
• One can narrow the classification of the type of organisms from which the DNA originated with some confidence if sufficient length and number of DNA sequences are available.

• Higher temperatures require more base pairs between the primer and target DNAs.
• Sequence mismatch between the primer and the target can be caused by lower hybridization temperatures.
• Let us suppose that the yeast gene A has a slightly different sequence in humans.
• If the hybridization temperature is too high, no amplification will be observed, but a lower temperature would allow amplification of the target human DNA.

• If each strand is to be replicated, the polymerase must be active in both di rections.
• If the DNA is not linear, known sequence are needed on both sides of the portion to be amplified.
• One can digest and circularize the genomes of both sides of a single sequence.

• One can amplify only the fragments of the known sequence in CHAPTER 6.

• The results show that the DNA is composed of four repeating units.

• One would expect a molecule with a linear sequence composed of four repeating peptides if they were transcribed into mRNA.

• STSs can be a common framework for establishing the re lation between clones.
• The entire genome has a sequence of 200 to 500 bp long.
• The sequence and a pair of primers that can generate theSTS are in a database.
• Laboratory 1 states that it has a YAC.
• Laboratory 2 can learn if their YAC contains any of theseSTSs by synthesizing the corresponding PCR primers.
• The discussion of this strategy can be found in Watson, J. D., Gilman, M., Witkowski, J., and Zoller, M.

• Each new strand of DNA is synthesised from a strand made in the previous cycle.
• The entire process will be inefficient if the synthesis of one strand is inefficient.

• The experiment is performed in a single tube and the same temperature must be used for both strands.
• If the two primers have different values of Tm, one strand could amplify more quickly than the other, and the efficient doubling of DNA at each cycle might not occur.

• Individual B does not have any symptoms because he has one gene that works normally.
• The third restriction experiment might have one functional and one nonfunctional gene X.
• B has no symptoms and his YProtein that is produced from his shorter-than-normal mRNA is functional, even though it is of the normal size.

• Individuals C and D do not express their genes.
• They can't make Y without the mRNA.

• Individual E is able to express X but can't make Y with it.

• Individual F makes the correct size of X and Y.
• The YProtein ap fails to function properly.
• There is a point that makes a single change in the Y sequence that is critical for function.

• The gels should be read from top to bottom.
• The data shows the sequence of the strand.
• The normal sequence of codons from the first gel is Val-Leu-Ser-Pro-Ala-Asp-Lys.

• Hemoglobin Chongqing has CGG instead of CTG as the second codon.

• leucine changes to arginine.

• Hemoglobin Karachi has a different name than the fifth codon.
• From alanine to proline there is a corresponding change.

• Hemoglobin Swan River has GGC instead of GAC as the sixth codon.

• Aspartic acid changes to glycine.