Unit 2 - Unit 2 Information Transfer and Mobile Genetic...

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Unformatted text preview: Unit 2 Information Transfer and Mobile Genetic Elements 27 UNIT 2: INFORMATION TRANSFER AND MOBILE GENETIC ELEMENTS Introduction: A cell's DNA contains all the information needed for the cell's growth and division into two similar cells. In active cells some of the genetic information is used to generate "working" molecules, such as the enzymes that catalyze the reactions leading to accumulation of cell material. We have already learned that most enzymes are proteins, and now we shall see how the information in the cell's DNA is used to make its proteins-including DNA polymerase, the enzyme that helps to replicate the DNA itself. In the replication of DNA, the entire molecule is copied exactly. A cell’s DNA typically codes for a much larger diversity of proteins than those that are regularly used by the cell, so synthesis of proteins following the instructions contained in DNA is selective. For instance, bacteria grown on one type of food source will usually not contain the enzymes for using an alternative food source that is not present. Consider also that DNA in the nucleus of any mammalian cell contains the information for making the oxygen carrier protein hemoglobin, although hemoglobin is normally made only in cells destined to become red blood cells. Transfer of the information contained in DNA into functional proteins is indirect, involving the production of a "middle-molecule," or messenger ribonucleic acid (mRNA). This messenger molecule in turn must interact with a complex organelle, the ribosome, made up of proteins and ribonucleic acid, before amino acids can be correctly assembled into a functional polypeptide. Furthermore, translation requires adaptor molecules, tRNAs, to carry activated amino acids to the correct position on the ribosomemRNA complex. In Units 3 and 4 last semester we focused on aspects of metabolism in which ATP was synthesized. In this unit we will confront some of the primary processes in which ATP is utilized. Macromolecules, which make up most of the dry matter of cells, are built up from monomer subunits. These monomers are assembled into the cell's DNA, RNA, and proteins by way of condensation reactions requiring an energy input in the form of ATP. By far the greatest part of the ATP used in synthesizing macromolecules-about 70% of the total--goes into making protein. This is partly because proteins are costly to make in terms of ATP used/gram of protein produced, and also because proteins make up the greatest part of the cell's dry weight. Readings: Chapter 17 Chapter 19 (pages 381-385) Chapter 27 (pages 561-563) Chapter 21 (page 435) Chapter 20 Chapter 38 (pages 812-819) To Do This Unit: 1. 2. 3. Read through the objectives. Read the text assignments, focusing on the material covered in the objectives. Examine carefully the demonstration materials online and in the Study Center. Unit 2 4. 5. Information Transfer and Mobile Genetic Elements 28 Go back over the text and demonstration materials and write out answers to the objectives. Take an examination on this unit. You will be given a sequence of DNA with the direction of transcription, and asked to replicate it and to transcribe and translate it, using the information presented in this unit and the last. KEY CONCEPTS AND OBJECTIVES: After you have studied the material in this unit you should understand the following concepts and you should be able to carry out the objectives for each. The sequence of bases in DNA determines the sequence in which amino acids will be linked in protein synthesis. DNA transcription mRNA translation protein 1. Using Fig. 17.2 (p. 327) explain the "one gene-one polypeptide hypothesis." the functin of a gene is to dictate the production of a specific enzyme. supported by experiments on mutant bacteria that has a singe-gene defect. Later changed to accomodate proteins that require two genes to synthesize. During transcription, a faithful copy of the information in DNA is transformed into mRNA. transcription mRNA DNA 2. a) In the diagrams below, label the 5 carbons in each sugar, indicate which sugar would be found in DNA and which in RNA, and explain the structural difference between the two sugars. (See Fig. 5.27, p. 87.) Unit 2 Information Transfer and Mobile Genetic Elements RNA 29 DNA lacks an oxygen in the second carbon. DNA P O CH 2 HH OH P O O BASE HH OH CH 2 HH OH O BASE P = phosphate group HH H Nucleotides b) List three differences between DNA and RNA. 1. RNA has ribose instead of dyoxyribose 2. RNA has uracil instead of thymine 3. RNA is usually single stranded. c) Which nucleotides are found in DNA and which in RNA? Which are purines and which are pyrimidines? (See Fig. 5.27, p. 87 and p. 88.) DNA: A,T,G,C RNA: A,U,G,C Purine: A,G Pyrimidines: T,C Purine: A,G pyrimidines: U,C In DNA, the strand that is being transcribed into mRNA is called the template strand and the non-template strand is referred to as the complementary strand. (Note that transcription of the template produces an RNA version of the complementary strand). The site at which RNA polymerase binds is called a promoter. The promoter sequence specifies which strand is to be transcribed and orients the RNA polymerase in the right direction. 3. The following information on transcription in bacterial cells is given to help you understand how the RNA polymerase complex "knows" where to bind on the DNA and how it is oriented to move along the DNA in a particular direction. (1) In bacteria, the promoter sequence consists of two DNA sequences, each six nucleotides long, that are separated from one another by 17 other base pairs (bp): 5' ---TTGACA ---(17 bp)--- TATAAT---> 3' (2) RNA polymerase recognizes and binds to the promoter on the complementary strand. The promoter is read in the 5' to 3' direction. Unit 2 Information Transfer and Mobile Genetic Elements 30 Template Strand 3' 5' TTGACA -17 bpTATAAT RNA polymerase Complementary Strand mRNA 5' 3' (3) Once bound to the complementary strand, the RNA polymerase is oriented to move along the DNA in one direction (direction of the arrowhead, in this case, left to right) and will move "downstream," toward the 3' end of the complementary strand. (4) The RNA polymerase will transcribe the other (template) strand (the 3' to 5' strand--in this case, the top strand) because nucleic acid synthesis can only occur in the 5' to 3' direction. In other words, the polymerase complex recognizes the promoter on the 5' to 3' complementary strand, but must transcribe the 3' to 5' template strand in order to synthesize 5' to 3'. REMEMBER: ! " # $ RNA polymerase recognizes and binds to the promoter on the complementary strand. The template strand = the strand to be transcribed. Transcription always goes 3' ! 5' on template strand (because mRNA synthesis is 5' ! 3'). The mRNA synthesized has the same sequence and polarity as the complementary strand (with U instead of T). (5) In bacteria, the start signal for transcription on the complementary strand is 5' CAT 3'. What would it be on the template strand? __________ GTA a) Locate the promoter sequence, and determine the direction of transcription and which strand will be transcribed in the following sequence of bacterial DNA. Circle the start signal on the complementary strand. 5' 3' GTCCACAGATGTAGGCATTATA..17.bp..TGTCAACGATGGCCTGCATGA CAGGTGTCTACATCCGTAATAT..17.bp..ACAGTTGCTACCGGACGTACT 3' 5' b) Note: The polymerase may use different strands for different genes; it is the promoter sequences that determine the direction of transcription and the strand used. Unit 2 Information Transfer and Mobile Genetic Elements 31 Arrows represent direction of transcription. c) The sequence of nucleotides constituting a promoter is not the same for all promoters--rather, each promoter's nucleotide sequence is a variation on a theme (theme = consensus sequence). Given a group of related sequences, derive a consensus sequence from them. Example: 5' 5' 5' 5' 5' 5' 5' TTGACG TAGACA CTTACA TTGAAA TTGACC TTGTCA TTGACA Consensus sequence: What do you think determines whether a given promoter is strong or weak (i.e., whether initiation of transcription occurs at that promoter with high or low frequency)? Why is this important? In thinking about this question, consider that E. coli has only a single RNA polymerase that transcribes all genes. Would it be advantageous for an organism to transcribe all of its genes at the same frequency? Why or why not? See demo for additional information. The promoters are a variation of a theme(consensus sequence). The more similar the promoter is to the consensus sequence the stronger it is. This determines how often the RNA is transcribed. Since it's stronger, the RNA polymerase will attach to the promoter site more often and will be, in a sense, given p riority over transcription of other genes. It will not be advantageous for an organism to transcribe all its genes at the same frequency because some genes may be required more than others; then transcription of what is more than necessary will be a waste. It would be more effective if the time and energy was taken to transcribe what it actually needs. d) Using information in the demo, explain how the RNA polymerase knows where on the DNA to stop transcription in bacteria? The RNA polymerase will continue transcription until it runs into a termination signal which is composed of a h airpin loop and a following repeated sequence of adenine. When the polymerase runs into the adenine run, the hairin loop forms in the mRNA just produced by the ribosone. This structure then causes stress on complex and the transcription is halted. When the hairpin loop forms, the DNA is able to rewind back into its d ouble helix putting more strain in the structure. Second, since adenine has weak hydrogen bonding (only two) it can't hold the mRNA to the DNA template for long and so the mRNA is released. In eukaryotes, the information in most genes is split into coding and non-coding sequences. The mRNA is modified after transcription. Unit 2 4. Information Transfer and Mobile Genetic Elements 32 Describe the process of transcription in eukaryotic cells using Figs. 17.3 (p. 329), 17.7 (p. 332), and 17.8 (p. 333). Then describe the mRNA processing that occurs in eukaryotic cells (see Fig. 17.9, p. 334). In doing so, identify and define the following. Are the same processes found in prokaryotes as well? a) binding of transcription factors b) TATA box c) sites where transcription is initiated1 d) sites where transcription is terminated (What is the signal in eukaryotes?) e) the primary transcript (or pre-mRNA molecule) f) the exons and introns; Which are excised? g) the end of the primary transcript that is capped (5' vs. 3') h) the end of the primary transcript that carries a poly A tail (5' vs. 3') i) the mechanism by which editing occurs (Figs. 17.10-17.11, p. 335). (Note: you are not responsible for the details of the editing process; strive for an overall understanding of the process.) 5. a) In your written exam, you will be given a sequence of DNA, together with the polarity of the chains (5' vs. 3') and the direction of transcription (arrow). For example: _________________________Direction of transcription_____________________> 5' - AUCAGGAGGUCUACGGAAUGUUCCCACUGACAUGAUCAACCACGU - 3' complimentary DNA strand 5'-ATCAGGAGGTCTACGGAATGTTCCCACTGACATGATCAACCACGT -3' 3'-TAGTCCTCCAGATGCCTTACAAGGGTGACTGTACTAGTTGGTGCA -5' * Template DNA strand From this information, you are to determine which strand is being transcribed (i.e., top or bottom) and, starting at the end of the correct strand, write the sequence of nucleotides in the mRNA that would be transcribed from this DNA. Indicate the 5' and 3' ends of the transcript. (Note: On your test we will not put the promoter sequence in; assume that the polymerase has bound and begun transcription.) 1Prokaryotes have but one type of RNA polymerase but transcription in eukaryotes is more complex and requires three kinds of RNA polymerase. Like the promoter sequences in bacteria, eukaryotic promoters have two recognition sites, one between 40 and 110 bases upstream of the start site for transcription, and a second is a TATA sequence on the complementary strand about 25 bases upstream of the start site. (The TATA box was discovered by Professor Michael Goldberg here at Cornell.) Unit 2 Information Transfer and Mobile Genetic Elements 33 b) What is the start signal (i.e., initiation codon) for translation in both prokaryotes and eukaryotes? Using the sequence of mRNA that you transcribed, find the proper start signal for translation, and, using the genetic code on page 330, write the sequence of amino acids that might be translated from this mRNA. The start signal for translation in both prokaryotes and eukaryotes is AUG on the mRNA c) What are the termination signals for translation? (See Fig. 17.5, p. 330.) The termination signal for translation is UAA, UGA, UAG. Which codes for water molecule instead of amino acid on the tRNA that when the water molecule is added to the polypeptide 6. a) In Fig. 17.14b (p. 338) point out the amino acid attachment site and the anticodon. Using Fig. 17.5 (p. 330), determine the codon for tryptophan (trp) and give the anticodon. (Label the 3' and 5' ends!) b) What are ribosomes composed of? Using Fig. 17.16 (p. 339), explain the process of translation in bacteria. The demo model will clarify the process (see objective 6d). ribosomes are composed of small and large unit rRNA. During translation, fi rst the tRNA, mRNA and the small ribosome unit binds together. the tRNA is already bound to the small unit where the small unit binds to the attachment site. Then once the mRNA is bound to the small unit, the tRNA binds to the start c odon. *(uses a GTP as energy for recognition of codon) Then the large ribosomal subunit binds to the latter to form the translation initiation complex. The i nitiator tRNA is placed at site P. The next tRNA is palced at site A. The amino acid of the site A tRNA forms a peptide bond with the amino acid of the initial tRNA (N-terminus). Once the bond forms, the initiator tRNA lets go of the amino acids and so the chain is passed on to the tRNA in site A. Then the site A tRNA is translocated to site P, which nees GTP for energy, as the initator tRNA or the tRNA, former occupant of site P is moved to site E then let go. c) Using Fig. 17.17 (p. 340), describe the initiation of translation. see above d) Using the demo, cardboard models of ribosomes, mRNA, tRNA, and amino acids, and the information provided with the model, describe the individual steps in polypeptide synthesis including: • attachment of amino acids to tRNAs to form activated amino acids (Fig. 17.15, p. 338). What compound provides the energy to attach the amino acids? formation of ribosome-mRNA complex and initiation. binding of charged tRNAs to P and A sites. What is the initiation codon for translation? formation of peptide bond. What happens to the met (or fmet) in most proteins? • • • • Unit 2 Information Transfer and Mobile Genetic Elements 34 e) Use Fig. 17.18 (p. 341) to describe the elongation cycle. see 6c) f) Use Fig. 17.19 (p. 342) to show how translation is terminated. translation is terminated once the ribosomal units read the terminal codon which could be UAA or UAG a nd such. These codons have water molecules intead of amino acids attached that when the water forms a bond with the peptide chain, the ribosome is forced to let go. g) Where does protein synthesis take place in eukaryotic cells? What special roles do signal recognition particles (SRPs) and endoplasmic reticulum, play in protein synthesis? eukaryotic cells produce proteins in both cytoplasm or the rough ER. mRNAs that translate for ribosomal p roteins code for signal peptide which the SRP attaches to and brings it to the receptor protein, part of the translocation complex on the rough ER. Then the signal peptide is removed as the rest of the polypeptide leaves the ribosome into the rough ER and folds into its shape. Then the polypeptide chain can be altered b y adding phosphates, lipids and such, which is then packaged, sent to golgi apparatus to be sent to a specific location, such as other organelles. 7. a) Using Fig. 17.3 (p. 329), 17.24 (p. 347), and 17.25 (p. 348) compare the processes of transcription and translation in prokaryotes and eukaryotes. Completing the following chart may help you to summarize the differences. Prokaryotes Are transcription factors required? (Y/N) Does transcription proceed past the termination signal? (Y/N) Is the mRNA processed to remove introns? (Y/N) Is the transcript capped? (Y/N) Does the transcript receive a poly-A tail? (Y/N) Can transcription and translation take place at same time? (Y/N) Are the ribosomes large or small? Eukaryotes b) Summary objective: Describe the flow of information from the gene to the synthesis of a protein in (1) prokaryotic cells and (2) eukaryotic cells. Unit 2 Information Transfer and Mobile Genetic Elements 35 A gene determines primary structure of a protein, and primary structure determines protein conformation. 8. a) Find the asterisk in the sequence of nucleotides shown in Objective 5a. What is the consequence (in terms of the polypeptide that would be translated from this mRNA) if adenine were substituted for the guanine at the asterisk and the GC base pair were changed to an AT base pair? (Hint: Check the codons in Fig. 17.5, p. 330.) that would cause a silent subsititution mutation. the codon would cange from UUC to UUU on the mRNA where both code for phe. b) What is the consequence (in terms of the polypeptide that would be translated from this mRNA) of making the same base substitution in the next GC pair to the right after the asterisk? that would cause a missense subsitution mutation. the codon would change from CCA to UCA changing the amino acid from phe to ser. depending on how crucial this amino acid is to the folding of the polypeptide, the seriousness of the result varies. c) What is the consequence (in terms of the polypeptide that would be translated from this mRNA) of deleting the GC base pair at the asterisk? Of deleting the GC base pair and the next two base pairs? Which of these changes (mutations) would have the most detrimental effect on the structure and function of the polypeptide in question? Why are mutations in which one or two nucleotides are inserted or deleted called frameshift mutations? It will result in shift in the reading frame and thus is called frameshift mutations. It's pretty serious since it will cause the following base codons to be read in the wrong groups and thus may code form completely different a mino acid and alter the secondary and tertiary structure of the protein product. which ever one that codes for the most different amino acids will cause more detrimental effect. more likely the one with three deletion since it will definitely change two of the amino acids as supposed to one (?) BULLLSHITTTT Difference: missense mutation=> one base pair becomes substituted for another. unlike the silent mutation, the change causes change in amino acid sequence. nonsense mutation => The substitution of a base pair coincidentally changes the amino acid codon to termination signal codon and cause the polypeptide to cut short. Usually nonsense mutations are more serious because it causes the polypeptide to be terminated prematurely, leading to a nonfunctional protein. Missense may cause alteration in the shape but will not affect the activity of the p rotein as much as the nonsense mutation. d) Give the difference between missense mutations and nonsense mutations. Which is generally the more serious? e) What are some ways that such mutations might be caused? 1. The template itself may have errors. DNA often have errors during replication. If the error is not corrected, the error will be passed on genetically = spontaneous mutations. 2. physical chemical agents, mutagens, cause mutations. = base analogs - chemicals that are similar to DNA base and pair inccoretly during DNA replication = others interfere with correct DNA replication by inserting themselves into the DNA and distorting the double helix. Unit 2 9. Information Transfer and Mobile Genetic Elements 36 Use diagrams such as Figs. 19.5-19.6 (pp. 385-386) to describe the lytic and lysogenic cycles of replication of a temperate bacteriophage. In doing so: a) Use the following terms correctly. • • • • • virulent versus temperate v irulent virus - lytic, causes lysis of the host. Temperate - lysogenic, does not kill host, but reproduces and is passed on as the host divides. may b ecome virulent at environmental signals when the gene is to be expelled lysogenic strain of bacteriophage In lysogenic cycle the virus undergoes replication in the host without destorying it. the viral proteins break the DNA and rejoin it with the viral gene incorporated into it creating a prophage (the lysogenic cycle v iral DNA). the gene codes for a protein that prevents t ranscription of most of teh other prophage genes and so is able prophage (provirus) t o remain silent in the host. So the virus is carried in the prophage form. It can transform into lytic cycle when the virus genome is to be expelled. so then it breaks out of the host in t he virus form. For lytic cycle, the virus induces the host to produce the materials to replicate itself and create copies of itself. Then it breaks out of the host causing lysis. lytic cycle b) Explain how the lytic cycle can be induced in lysogenic bacteria. lytic cycle can be induced by giving environmental signal such as chemicals or high-energy radiation, causing the genome to exit the host. 10. a) What is bacterial transformation? Relate this to Griffith's pneumococci experiments that were covered in Unit 1. transformation: bacteria takes up forein DNA to transform its phenotype or genotype. Pneumococci t ook up the DNA from the surrounding dead cells to become fatal. -_- something like that. b) Using Fig. 27.11 (p. 562), explain how bacteriophages can be used to transfer genes from one bacterial cell to another. What is this process called? How does this process differ from bacterial transformation? (These processes will be further described in the Recombinant DNA objectives that follow.) The process is called transduction where the bacteriophage carry bacterial genes from one host to another. The virus carrying the bacterial DNA may not be able to reproduce because its own DNA has been replaced with the bacterial DNA. But still, the bacteriophage can attach itself to another host and release its content, but it won't affect the bacterial virulently. The bacterial DNA can then be taken up by t he bacteria and do an exchange of homologs. It differs from transformation in that transduction is more accidental and is the result of an error during the phage reproductive cycle. 11. a) Using Fig. 27.13 (p. 563), explain what a plasmid is. What types of genes may be found in plasmids that are medically important? plasmid is a small circular piece of DNA that is replicated within a bacteria apart from the c hromosome. It can be used to transfer genetic info from one cell to another during conjugation and such others. R plasmids that contain resistance gene against medicine or antibiotics. Unit 2 Information Transfer and Mobile Genetic Elements 37 b) Using diagrams such as Figs. 27.12-27.13 (pp. 562-563), describe the process of conjugation in bacteria. What is the difference between a F-, F+, and Hfr cell? c onjugation is when a donor reaches out to another cell using sex pili. Through the sex pili, the plasmid can be exchanged. Cells that are F+ means they have the F factor usually in form of F plasmid. F- do not have F plasmid and therefore cannot generate a sex pili. Hfr cells have F factor in its chromosome. It also acts as the donor. 12. a) Explain what transposons are and indicate whether they are characteristic of prokaryotes or eukaryotes, or both. (Note: Barbara McClintock, who discovered transposons, received her Ph.D. at Cornell and did much of her work here.) t ransposons: a transposable element that moves within a genome by a means of DNA intermediate. Both pro and eukaryotes have transposons b) Explain how transpositions are fundamentally different from other mechanisms of genetic shuffling. How does “cut-and-paste” transposition differ from “copyand-paste” transposition? the cut-and-paste does not leave the DNA in the original location. the DNA is essentially cut out. In c opy-and-paste, the DNA is copied, not cut out, then transferred to new location. so the DNA is still t here in the original spot as well as the new. c) What would be the result (in terms of the protein that would be synthesized) if a transposon inserts itself into the middle of a functional gene? ...its promoter? then the structure of the protein would change and therefore become useless to its original function. if it was in t he promoter, then the polypeptide may become elongated, combo of both the transposon and the gene that the promoter was supporting. 13. a) Explain what is meant by "cloning" genes, and give reasons why scientists are interested in cloning genes. cloning is the production of multiple copies of a single gene. It can be useful in promoting expression of what is considered the good gene. It can be used to cut out and incorporate into an organism to give the wanted phenotype. Its protein product is what is usually useful (obviously). It can also result in large harvest of wanted proteins such as growth hormone. b) Using Fig. 20.3 (p. 398), explain how large molecules of DNA can be broken down into smaller pieces of DNA that can be manipulated in the laboratory. What are the characteristics of a restriction site? What is the importance of "sticky ends?" The DNA can be broken down by the use of restriction enzyme that can identify a certain specifi c sequence in the DNA and cut. The restriction site are symmetrical when both sides are read in the 5' to 3' direction. the cuts will result in restriction fragments. sticky ends are the one single stranded end on the double stranded restriction fragment. the s hort extensions can form hydrogen bonds with complementary sticy ends on other DNA molecules cut with the same enzyme. As long as the sticky ends fi t, you can create different combo of DNA segments. Association without the help of a ligase is temporary. Ligase strengthens the bond and makes it permanent. Unit 2 Information Transfer and Mobile Genetic Elements 38 c) Using Fig. 20.4 (p. 399), describe the procedure by which eukaryotic DNA can be inserted into plasmids (vectors). What is the role of ligase? 1. isolate plasmid DNA from bacterial cells and DNA from organism. using the same restriction enzyme, cut both DNA s amples. Then mix the fragments of wanted gene and cut plasmids. the use of ligase will make the recombinant DNA permanent. some are recombinant some are not. the ligase forms covalent bonds along the sugar backbone. the cells can be separate with the use amp(r). the Amp(r) is included in the plasmid along with the gene of interest. Then after transformation occurs, the bacteria is cultured in a agar with ampicillin. Those that have transformed will have amp(r) that will resist the ampicillin and will therefore survive; the nontransformed bacteria will have no resistance and therefore will die out. d) Explain how the cloning vectors can be taken up into bacterial cells. How can the cells that successfully incorporated new DNA be separated from the millions of cells that received no new DNA? e) How can the host cell (or cells) that received the gene of interest be isolated from the vast majority of cells in the culture medium? (See Fig. 20.7, p. 402.) the use of lacZ can help. in bacteria that have nonrecombinant plasmid will have lacZ gene intact that when the bacteria is cultured in x-gal, the nonrecombinant bacteria will turn blue due to the hydroltic enzyme it produces, where as the others will be white. Problem 1: difference in promoters and other DNA control sequence. 2: introns in eukaryotes 3: (?) the problems can be solved using expression vector, a cloning bector that contains ahighly active bacterial promoter just upstream of a restirction site where the eukaryotic gene can be inserted in the correct reading frame. also the intron problem can be solved using cDNA which uses mRNA and reverse transcriptase. using an mRNA the transcriptase will produce DNA, still in the 5' to 3' direction. (mRNA is in the form of 5' to 3') then the mRNA is broken down by enzymes. Then DNA polyerase synthesizes the other DNA strand in place of mRNA. f) Describe three problems in getting eukaryotic genes expressed in prokaryotic cells and explain two procedures that have been used to surmount these problems. g) Explain the different advantages of cDNA approaches and those using genomic DNA. See p. 401 and the demo. if interested in knowing what cell type expresses the gene , genomic library is more useful. it's also useful for getting regualtory sequences or introns associated with a gene. If you want the coding DNA only, then cDNA is more useful. in genomic: no single gene is targeted for cloning. In cDNA, it has only part of the entire genome. 14. Describe the PCR procedure and give one practical application of PCR. See demo for additional information. 1: heat to denature and separate DNA strands. 2. then cool to anneal (hydrogenbonding with single stranded DNA primers complementary to sequences on opposite strands at each end of the target sequence) 3. add heat stable polymerase to extend DNA. practical: can amplify small amount of DNA. Therefore can be used in crime scenes where small amounts of blood can yield a lot of DNA through PCR 15. a) Explain how gel electrophoresis works (see Fig. 20.9) and, using Fig. 20.11 (p. 407), explain the procedure of Southern blotting. electrophoresis creates poles on the ends of the gel. Since DNA is negatively charged, the DNA is put in the negative side and when the electrophoresis is let on, then the positive magnetic pull will move the DNA out of its location. Then, due to the size of the DNA, the speed and distance differs by the size. The larger it is, the less distance it will cover. electrophoresis can be used to determine the lengths of DNA, where with reference, it can tell you what genes are in it. Southern blotting fuses electrophoresis and nucleic acid hybridization. After electrophoresiss, the gel is pressed on alkaline solution and nitrocellulose membrane and so the pattern is transferred to the blot. Then the blot is put into a solution with the radioactively labeled probe. Then the blot will show the bands thats been bound to the probe. b) What are RFLPs (see page 417) and how can they be used? RFLP: restriction fragment length polymorphism: changes in the restriction fragment length due to change in DNA sequence caused by SNP: sincle nucleotide polymorphism: a single base-pair site where variation is found in at least 1% of the population. They can be used by comparing locus of the SNP. If one person unaffected has ex. C on a loci and an affected person has T in this loci, then that shows that the affected person has disease causing allele. Unit 2 16. Information Transfer and Mobile Genetic Elements 39 a) Explain how the following techniques can be used to help map the human genome: nucleic acid hybridization, genome analysis (analyzing DNA sequences, studying gene expression, and determining gene function), linkage mapping, physical mapping, and DNA sequencing. (analyzing DNA sequences, studying gene expression, and determining gene function). Note: You need only understand the general principles, not the details. b) Give an example of how DNA technology can be used for (1) diagnosis of disease in humans; (2) human gene therapy (What are some of the problems faced?); and (3) producing pharmaceutical products. 17. a) Explain how DNA technology can be used in the process of DNA profiling (DNA fingerprinting) and explain what STRs are and why they are useful. b) Give one example of how DNA technology can be used in environmental work or agriculture. How has DNA technology been used to make products for animal husbandry? What are transgenic organisms? c) Give reasons why plants have proven to be easier to engineer than animals (see also page 814). Using Figs. 20.25 (p. 421) explain how foreign genes can be inserted into plants using the Ti plasmid? What are some of the uses of this technology in crop plants? d) Describe how stem cells can be used to treat human diseases. Unit 2 Information Transfer and Mobile Genetic Elements 40 Below are summary questions relating to important concepts in this unit. The TA may use these questions in his or her oral test or you may see one of them as an essay question on the final exam. Take a few moments now to formulate your answers. Describe the DNA molecule and relate the concept of "genes" to its structure. Explain how genetic material is replicated and how genes control cellular functioning. Include some detail regarding the mechanism of protein synthesis. Describe the steps involved in recombinant DNA technology, and explain how this technology could be used for "DNA fingerprinting." Unit 2 Information Transfer and Mobile Genetic Elements 41 Unit 2 Evaluation: PLEASE FILL OUT THIS EVALUATION SHEET AFTER YOU HAVE COMPLETED THE UNIT, AND PLACE IT IN THE WOODEN BOX IN THE STUDY CENTER. THESE FORMS WILL PROVIDE FEEDBACK NECESSARY FOR IMPROVING THE UNITS. How many hours did you spend studying for this unit? _________ Do you consider the time you spent studying for this unit to be: A. too little? B. about right? C. excessive? Was the majority of the material in this unit: A. new to you? B. somewhat familiar? How many times have you taken this unit test? C. very familiar? _________ How effective were the objectives in organizing your study? A. ineffective B. effective C. very effective Did the objectives clearly indicate what you were expected to know about the unit material? A. Yes B. No What objectives did you have the most problems with? ___________________________ Was the demo helpful in learning the unit material? A. not at all B. moderately C. very helpful; essential The amount of information in this unit is: A. too little B. about right The level of difficulty of this unit is: A. too easy B. about right C. excessive C. too hard How crowded was the Study Center when you took your test? A. not crowded B. crowded, but OK C. very crowded; a long wait Indicate day of the week and time of day that you took this exam: __________________ Did you need any special help from TAs in learning this unit material? A. Yes B. No If you needed help from a staff member, was he/she willing and able to provide the help you needed? A. unwilling/unable B. satisfactory C. very helpful Name of TA who tested you on this unit: ______________________________________ Did you feel the TA who tested you was well preprared? A. poorly prepared B. satisfactorily prepared C. very well prepared Do you feel the TA who tested you gave you a fair and impartial test? A. Yes B. No Any comments or suggestions? ...
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This note was uploaded on 12/06/2010 for the course BIOG 1106 taught by Professor Don'tknow during the Fall '10 term at Cornell University (Engineering School).

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