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Home , EcoRI, Helicase, Recombinant DNA, Reverse transcriptase

Bio 102 Practice Problems Recombinant DNA and Multiple choice: Unless otherwise directed, circle the one best answer: 1. Which of the following DNA sequences could be the recognition site for a restriction ? A. TGCCGT B. TGCGCA C. TGCTGC D. All of the above. E. None of the above. 2. Which of the following is not needed for DNA by the method we discussed in class? A. Radioactive primer B. DNA C. Fluorescent dideoxy D. Ordinary nucleotides (dNTPs) E. All of the above are required. 3. What is the key enzyme used in PCR? A. ATP synthase B. Taq DNA polymerase C. DNA D. Restriction E. Sigma factor

Short answer (show your work or thinking to get partial credit): 1. Recombinant human , produced by bacteria carrying a cloned insulin , is now the major form of insulin used to treat . The human insulin gene encodes an mRNA only 333 nucleotides long, but the entire gene spans more than 4000 nucleotides. There are three exons and two . a. If we were to clone this gene directly from the nuclear DNA, bacteria would not be able to express the insulin . Explain why this is true. This gene has introns; bacteria don't. If we cloned the gene directly, the bacteria would produce mRNA that includes the introns and be unable to splice it. Therefore, no functional protein could be made. b. What technique should be used instead in order to get a functional insulin coding sequence cloned into bacteria? Describe briefly how this technique works. We should use cDNA . In this technique, we isolate mRNA from the cytoplasm of a cell, so that it has already been spliced. We then use enzyme to make a DNA copy (cDNA) of the mRNA. This DNA can then be cloned, producing a bacteriareadable gene that lacks introns. c. Every cell in the human body has the same DNA, so every cell has an insulin gene. However, in order to use the technique you described in "b," you would have to start with cells from the pancreasthe only body cells that actually produce the insulin protein . Why are these the only cells that would work? We need insulin mRNA in order to do cDNA cloning. If the cells are the only ones that make insulin protein, then most likely they are also the only ones that make significant amounts of insulin mRNA.

2. Human remains a promising possibility but is still plagued by problems. In the table below are listed two possible vectors and two problems. For each combination, please briefly explain if the specific problem is expected to be encountered for the vector.

Retroviral Vector Liposome Could be a problem; someone who’s had a retroviral before could Should not be a problem. , Immune be immune to the vector, or if multiple not lipids, usually trigger the immune Response treatments are necessary, development response. of immunity could be a big drawback. Could be a problem, because the Liposomes usually deliver Insertional inserts DNA randomly into or other DNA that won’t integrate Mutagenesis the cell’s ; it could hit a gene into the genome. Shouldn’t be a by chance. problem.

3. The diagram below represents a section of the . The coding sequence of a gene, YFG, is shown by an arrow, and boxes indicate the locations of some regulatory sequences. Locations of recognition sequences (cut sites) for three common restriction enzymes (EcoRI, BamHI, and NcoI) are also marked.

You would like to clone this gene in E. coli for further study. You have available the expression vector () shown below:

a. Why is it important for this plasmid to be an expression vector? Eukaryotic have regulatory signals like enhancers and TATA boxes that are not recognized by bacteria and lack the 10 and 35 sequences that bacterial RNA polymerase needs to start . In addition, eukaryotic find the correct AUG codon by scanning from the cap to the first AUG, while bacteria rely on a ShineDalgarno sequence in the mRNA. These bacterial regulatory signals are provided by the expression vector. If we use a plasmid that did not have these signals, it's very unlikely that the bacteria would be able to transcribe and translate our gene. b. Why is it important for this plasmid to have an antibioticresistance gene? This gives a way to select for bacteria that acquire the plasmid. The of successful transformation is small, so we need a way to know that we got the clone into a cell. Only cells that acquire this plasmid will be able to grow in the presence of this antibiotic. c. What would you use to clone this gene? Explain your choice. The plasmid has sites for all three restriction enzymes. However, Eco RI cuts the plasmid twice , so we would wind up chopping out a piece of the plasmid, including the needed ShineDalgarno region. This is a bad choice. Bam HI is also a bad choice, because it cuts in the middle of the gene we want to clone. We want an enzyme that will leave our gene intact. Not I is the best choice. It will cut our gene out of the genome and makes a single cut for inserting it into the plasmid. Notice that it's not important that the enhancer and TATA sequences will be left behind; the plasmid provides the needed regulatory signals.

4. You would like to use PCR to amplify (make many copies of) the boxed section of the DNA sequence below: 5´ ACGACCGATAGACGACGTAGGACTTACTTACTTACGTAGGCA 3´ 3´ TGCTGGCTATCTGCTGCATCCTGAATGAATGAATGCATCCGT 5´ You ask your lab partner to order a pair of primers that can be used in the PCR reaction. The sequences of the primers he orders are: Primer #1: 5´ ATAGAC 3´ Primer #2: 5´ ACTTAC 3´ a. Oops! Looks like you shouldn’t have trusted your lab partner on this one. Which of the two primers is wrong, and why won’t it work? Primer #1 is OK: it will bind to the bottom strand, and its 3’ end will then be pointed in the right direction for Taq DNA polymerase to synthesize the desired DNA. Primer #2 won’t work, because it binds to other end of the same strand and so its 3’ end is pointed out away from the sequence to be copied: 5’ATAGAC-> 5´ ACTTAC-> 3´ TGCTGGCTATCTGCTGCATCCTGAATGAATGAATGCATCCGT 5´

b. Give the sequence of a primer that will work and could be used instead of the wrong one. Be sure to indicate the 5´ and 3´ ends. We need a primer that will bind to the top strand, with its 3’ end pointed toward the left so that the other strand can be copied. For example, 5’ GTAAGT 3’: 5´ ACGACCGATAGACGACGTAGGACTTACTTACTTACGTAGGCA 3´ <-TGAATG 5’ 5’ATAGAC-> 3´ TGCTGGCTATCTGCTGCATCCTGAATGAATGAATGCATCCGT 5´

5. You are trying to find the gene responsible for a human genetic disorder. You have mapped the gene to a particular region, and examining the human genome sequence for that region gives you the sequence below: 5’ CATACTTACTACTAGATTACGATTAGACGATTAGGATG|GCC |GAC |TCG |TGC |AGT |AAC | AGC |ATG |ACC |GAG |GCC |TAGACCAGATTAGGAGCCGGACCAGGACGGACCAGCGACT 3’

a. Assuming you are reading the noncoding strand and that there are no introns, find an open reading frame (ORF) in this region. Circle the point where will start, and put a box around the point where translation will stop. Then give the number of amino acids in the protein this gene would encode: 12 amino acids (start codon encodes an ; stop codon doesn’t) b. If you wanted to express this gene in E. coli , what would need to be present in your cloning vector to ensure that it will be transcribed and translated? The gene’s sequence won’t be recognized by E. coli’s sigma factor, so you’ll need a prokaryotic promoter (–10 and –35 sequences). E. coli ribosomes can only find the correct start codon by first binding the ShineDalgarno sequence, so the vector should also have this sequence positioned just before the desired start codon. c. How might the protein produced by E. coli differ from the protein produced from the same gene in a human cell? The aminoacid sequence (primary structure) will be the same. However, eukaryotic proteins are often modified in the ER or Golgi: carbohydrates added, phosphate groups added, etc. These modifications probably won’t happen in E. coli. Also, it’s possible that the protein won’t be correctly folded, if some specific protein or condition in the eukaryotic cell is needed for folding.

6. You have cloned a cDNA encoding a human hormone, and you hope to produce the hormone in bacteria in order to treat a severe genetic disorder. Unfortunately, when you insert this DNA into a plasmid and transform it into the bacteria, you get no hormone production. Give two valid reasons for your failure, and suggest a possible solution in each case. (1) Hormone must be modified after synthesis; bacteria lack ER and Golgi. Solution: determine the needed modification and try to reproduce it chemically, clone needed enzymes or use a eukaryotic host. (2) cDNA has no promoter and can’t be transcribed. Solution: insert it into an expression vector that includes a promoter. (3) cDNA from a eukaryotic cell doesn’t have a ShineDalgarno sequence needed for translation initiation in bacteria. Solution: insert it into an expression vector that includes a promoter. (4) cDNA is made from mRNA, and the mRNA for the hormone will only be present in a cell that normally produces the hormone. Could be that the source of the DNA needs to be changed. (5) The hormone might be toxic to bacteria. Solution: could try a different cell as the host.

7. Circle the DNA sequence(s) below which could potentially be a recognition site for a restriction enzyme.

5´ ATTTTA 3´ 5´ CGCG 3´ 3´ TAAAAT 5´ 3´ GCGC 5´

5´ GGATCC 3´ 5´ AGGAGG 3´ 3´ CCTAGG 5´ 3´ TCCTCC 5´

8. A Bio 102 student gets so excited about the tyrosinase experiment that she decides to try to clone the tyrosinase gene. She grinds up some potato, extracts the DNA from it and digests the DNA with two different restriction enzymes (separately, not together): Eco RI and Bam HI. She then obtains a cloning vector and digests it with the same two enzymes. She then runs a gel, which is shown at the right. a. Which enzyme would she want to use for cloning the potato DNA: Eco RI, or Bam HI? Explain why you made your choice. Bam HI, because it only cuts the plasmid once —if you cut the plasmid twice, then both pieces must go back together along with your insert in order to get a functional recombinant plasmid. b. Notice that the cloning vector made nice, tight bands on the gel, but the potato DNA just looks like a smear with no distinct bands. However, this is just what the student expected, so she’s not worried about it at all. Explain why this is the expected result. Plasmids are small and might have only one or two cut sites for a particular enzyme. The potato genome is huge and will have hundreds or thousands of sites for that enzyme. So we expect many, many more fragments, leading to the smeary appearance. That’s OK, though, because after cloning all the fragments (to make a ), we’ll have a way to identify the one correct clone. c. The student now mixes the potato DNA (digested with the enzyme you specified in part A) with cloning vector DNA (digested with the same enzyme). She then adds the mixture to E. coli cells that have been treated with CaCl 2, heats briefly to 42 °C, adds and incubates for an hour. What would be her next step? Be as specific as possible. Now the cells would be put on an antibioticcontaining plate. This will kill any cell that didn’t get a plasmid and allow those that did to grow into colonies. d. Unfortunately, after doing the next step as you specified, she doesn’t get a single bacterial colony. Not even one! When she reviews her procedure, she realizes she left out a critical step. What did she forget, and why would this be necessary? She forgot to add the ligase! Ligase enzyme is needed to join the potato DNA with the cloning vector to make a single, circular recombinant DNA molecule. 9. Suppose you want to clone the gene for human , so that you can use bacteria to produce the protein and obtain it in pure form for further study of its activity. a. Your first task is to isolate the hexokinase gene from the human genome. Assuming you have some human DNA on hand and access to the genomic databases on the Web, what technique might you use to obtain pure hexokinase DNA? Since you can use the human genome database to determine the sequence of the enzyme, you can design primers that could be used to amplify only the hexokinase gene from a human DNA sample by PCR. b. In your initial attempt, you succeed in obtaining hexokinase DNA and ligating it into a plasmid vector, but when you transform the recombinant plasmid into bacteria, you get no hexokinase protein produced at all. When you discuss your problem with your friend, she suggests that you might want to start with mRNA instead of with DNA. What problem can be overcome by starting with mRNA instead? The hexokinase gene, like most human genes, probably contains introns. Bacteria can’t splice out introns, so the bacteria couldn’t make the correctly processed mRNA (with an uninterrupted coding sequence), so they can’t make the hexokinase protein starting with the complete human gene. c. What will you need to do with your mRNA before you can make a new recombinant plasmid? Use reverse transcriptase to make a DNA copy of it (convert it to cDNA). d. What else would you need to provide in order for the bacteria to correctly transcribe and translate the human hexokinase gene? The cDNA won’t include a promoter (10 and 35 sequences, for bacteria) or a ShineDalgarno sequence, so the bacteria may not correctly recognize the gene unless these are added. An expression vector is commonly used for cloning cDNA for this reason. e. After consulting genome databases, you find that the human hexokinase aminoacid sequence begins with MetTrpLysTrp TrpMet. But the protein made from your plasmid begins with MetTrpMetTrpTrpMet! A must have occurred! Below, write the DNA sequence of the nontemplate strand for the beginning of the actual (unmutated ) hexokinase gene. Don’t forget 5′ and 3′ ends. There is only one codon for Met, AUG, and only one codon for Trp, UGG. So the mRNA sequence for the unmutated gene must be: 5′AUGUGG(Lys)UGGUGGAUG. But, there are two possible codons for Lys: AAA and AAG. How could we know which one it was? Well, in the mutated version, Lys is changed to Met (AUG). AAG could change to AUG with only one mutation, while AAA would need two to change to AUG. So it’s most likely that the Lys codon was AAG. Now we have 5′ AUGUGGAAG UGGUGGAUG for the mRNA. Now, the question asks us for the nontemplate strand of the DNA. The nontemplate strand looks like the mRNA, but of course it would have T ’s, not U ’s. So: 5′ATGTGGAAGTGGTGGATG. f. Perhaps you could salvage your recombinant protein by treating it with a mutagen in order to get a reversion mutation that restores the correct aminoacid sequence! Which of the following mutagens might work? Circle all that apply. 1) ethidium bromide (makes onebase insertions or deletions) 2) aminopurine (causes A→G substitutions) 3) imidazolecarboxamide (causes A→T substitutions) This could work because it could change the Met codon (TAC on the template strand) to TTC, which would be AAG (Lys) in the mRNA.

4) aflatoxin B 1 (allows any nucleotide to basepair with a T during replication) This could work because it could affect the T in the ATG Met codon; if T paired with T, we would get TTG on the template strand, producing an AAC Lys codon. 10. Explain one significant problem with using for gene therapy. (1) Because they insert DNA randomly into host DNA, they could produce a mutation (2) Because they contain a promoter, they could affect expression of other genes near the

Matching: 1. In our example of cloning insulin, we used several different enzymes. Match each enzyme below with its function in the cloning process.

F Reverse transcriptase a. Separating strands of DNA for PCR b. Causing “sticky ends” to G Restriction enzyme c. Copying a specific region of DNA d. Covalently joining a DNA fragment to a cloning vector D DNA ligase e. Transcribing the recombinant gene f. Making a DNA copy of RNA C Taq DNA polymerase g. Cutting DNA backbone at a specific sequence h. Replicating plasmid J DNA j. Not used in cloning

2. Matching. Match the enzymes on the left with their roles in gene cloning on the right. You may use the letters once, more than once or not at all, but only one letter per blank.

E DNA ligase a. Cut expression vector to produce “sticky ends” F DNA polymerase b. Produce cDNA from mRNA A Restriction enzyme c. Separate strands of doublestranded DNA B Reverse transcriptase d. Produce many copies of a specific DNA sequence F Helicase e. Make covalent bonds between two DNA molecules f. Not used in gene cloning