Name: _________________________________ Genetics 314 - Spring 2005 Exam 1 – 100 points 1. You are given three samples of DNA from three different organisms: a virus, a bacteria and a pygmy owl. You are asked to identify the three samples just by heating and cooling the DNA. a) What information would heating the DNA tell you and would this help in identifying the source organism for each sample? By heating the DNA you could tell what temperature it would denature which is related to the amount of G – C pairs and A – T pairs in the DNA samples. Because of the greater number of hydrogen bonds shared between G – C pairs, DNA with a high proportion of G-C pairs will require a higher temperature to denature than DNA rich in A – T pairs. While this information may be useful, it will not give you any information that could be used to identify the source organism of each sample. b) What information could you gain by cooling the heated DNA and how would this information help in determining the source of each organism? The information you will get from cooling the DNA will be the size of the genome of the organism (number of unique sequences/genes) and the presence of highly repetitive and moderately repetitive sequences in the sample. This information could be used to determine the source organism for each sample because virus have a small genome size compared to bacteria so these two can be differentiated by the rate of renaturation (virus faster than bacteria) and pygmy owl could be differentiated from the virus and bacteria by the presence of repetitive sequences in the eukaryotic (pygmy owl) and by the time it takes for the whole sample to renature due to the much larger genome size and greater number of unique sequences found in eukaryotic DNA samples. c) Draw a set of Cot curves for how you would expect the three samples to behave, labeling which curve represented the virus, the bacteria and the pygmy owl. pygmy owl virus bacteria 1 Name: _________________________________ 2. DNA is said to be replicated in a semi-conservative manner. What is meant by the term semi-conservative replication and how was this determined? Semi-conservative replication of DNA means that the hydrogen bonds are broken during replication allowing the strands to separate but the phosphor-diester bonds remain intact the each individual strand of the original DNA is conserved. This results in the DNA after replication consisting of one new strand and one old strand of DNA. This was determined by using a radioisotope of nitrogen (N15) to label the nucleotides in the DNA prior to replication then allowing replication to occur with non-labeled nucleotides (N14) and look at the density of the DNA after successive rounds of replication. A single band consisting of a mix of N14 and N15 DNA was expected after the first replication and two bands one just N14 and the other a mix of N14/N15 after the second replication with semi-conservative replication. This is what was observed in the experiment, confirming DNA replicated in a semiconservative manner. 3. You are given the following sense strand of DNA that codes for a simple protein that could be added to food to control obesity. 3’ TAC CCC ACG GGC TTT GCA AAA TCG GCC AGC ACT 5’ a) What would be the mRNA sequence coded for by this DNA sequence? (For full credit label the 3’ and 5’ ends of the mRNA) 5’ AUG GGG UGC CCG AAA CGU UUU AGC CGG UCG UGA 3’ b) What would be the amino acid sequence coded for by your mRNA sequence? MET GLY CYS PRO LYS ARG PHE SER ARG SER STOP c) A friend says the following sequence codes for the same polypeptide but it does not have the exact base sequence of your DNA. 3’ TAC CCA ACA GGT TTC GCA AAG TCG GCT AGC ACT 5’ Is your friend correct and if so how is that possible? 5’ AUG GGU UGU CCA AAG CGU UUC AGC CGA UCG UGA MET GLY CYS PRO LYS ARG PHE SER ARG SER STOP Yes, because of the redundancy of the cod where more than one codon can code for the same amino acid. Note the deviation is usually in the third base of the codon. 2 Name: _________________________________ 4. You would like to have the obesity-preventing protein from question 3 mass produced so you want to insert it into an organism to get the DNA sequence transcribed and translated. a) What other DNA sequences would you need to add to your DNA sequence to insure proper transcription and translation in a prokaryotic organism? For full credit indicate the purpose of each sequence. promoter sequence leader sequence gene sequence termination sequence promoter sequence – binding site for RNA polymerase for the initiation of transcription leader sequence – binding site for the small subunit of the ribosome for the initiation of translation termination sequence – sequence that signals the end of transcription b) If you decided to move your gene from the prokaryotic organism to a eukaryotic organism would you need to change the additional DNA sequences you added to get expression in the prokaryotic organism? Briefly explain your answer. Yes, the promoter sequence, leader sequence and termination sequence are different between prokaryotic and eukaryotic organisms so they would need to be changed for the enzymes involved in transcription and translation to be able to recognize the initiation or termination sequences for transcription and the sequence needed for binding of the small ribosomal subunit in translation. 5. You have isolated a gene from an eukaryotic organism and get it expressed in a prokaryotic organism and discover you do not get a functional protein. If you go back to the eukaryotic organism and isolate the mRNA for this gene from the cytoplasm and then create a DNA sequence from the mRNA sequence, the synthesized gene will produce a functional protein when inserted in a prokaryote. Based on the information given answer the following questions: a) Why would the gene not produce the desired protein when it was placed in the prokaryotic organism the first time? The gene has introns that need to be processed out of the mRNA for the transcript to code for a functional protein. Because prokaryotes do not have introns they do not have the necessary systems needed to remove the introns before translation occurred. 3 Name: _________________________________ b) Why would synthesizing a gene based on the mRNA base sequence produce the desired protein? If the DNA sequence of the gene is based on the mRNA isolated from the cytoplasm of a eukaryotic cell the introns will not be presence in the sequence because mRNA in the cytoplasm has been processed so no introns would be present in the sequence. With no introns present the protein product produced by the prokaryote would be functional. 6. Being frustrated by the difference you find in gene expression using eukaryotic and prokaryotic cells you decide to take control by working with the gene in vitro. You buy a DNA replication kit from a USA company but notice the kit was really produced in Eastern Europe. a) You add your gene to the ingredients in the kit and discover that you get no replication of your DNA. Upon further inspection you find that the DNA had opened up but no DNA synthesis was occurring. What could be causing the problem and how could it be fixed. If the DNA strand has opened it means that helicase, topoisomerase and the single strand binding proteins are present and functioning but the DNA polymerases do not have an open 3’ end to initiate synthesis of DNA. The 3’ end is provided by RNA primers that are synthesized by the enzyme primase using RNA nucleotides. The absence of the primer indicates either primase is missing or the RNA nucleotides are missing. Solution, add RNA nucleotides and primase to produce RNA primers. b) You think you solve the problem and start the reaction again. This time you get DNA synthesis but it is much slower than expected. Now what is the problem and how could it be fixed? If this is based off of a prokaryotic model then the problem is that the DNA polymerase in the kit is DNA polymerase I not DNA polymerase III. The two enzymes differ in their rate of synthesis with DNA polymerase III being much faster that DNA polymerase I. Solution, add DNA polymerase III. 7. Now that you have your gene properly replicated you turn to doing in vitro transcription. You go to another American company for your transcription kit but discover it is now made in China. a) Undaunted, you combine the ingredients in your kit but discover that the transcripts that are produced are of variable lengths. What could be missing from your kit and how could you fix the problem? 4 Name: _________________________________ Again using a prokaryotic system as a model, the problem is that the RNA polymerase is not attaching at the promoter site but instead is randomly attaching on the DNA. This would be due to the lack of a sigma factor. The role of the sigma factor is to orientate the RNA polymerase (core enzyme) in the proper location on the gene on the promoter, Solution, add the correct sigma factor. b) Starting to get frustrated, you decide to read the instructions. You read one warning saying that the kit will only work for genes with rho independent termination. What is meant by this and why would it possibly be a concern? The only potential problem here is if you were using a termination sequence for your gene that required the rho factor (protein) to facilitate termination of transcription. The end result would be extra long transcripts. Solution, add the rho factor or make sure the termination sequence of your gene is a rho independent termination sequence. 8. You have now produced the transcripts you want, so you now are ready to attempt in vitro translation. You find a kit that is actually made in America but discover that the company to save money only included 45 of the possible 61 tRNAs in the kit. The disclaimer in the kit says this should not cause problem with translation. a) How could the company claim that its cost saving measures would not affect your ability to translate your gene? There is a redundancy in the code where more that one codon codes for a specific amino acid. Combine this with the wobble theory where the third base in the codon is not that necessary for matching the t-RNA anti-codon with a specific codon and you could have one anti-codon work for several codons meaning not all the t-RNA’s would be necessary. There would need to be at least one t-RNA for each amino acid. b) You try the kit and you get no translation. You discover you never see a complete ribosome attached to your DNA. You call the company and claim that their costsaving measures caused the problem. They say the problem is with your DNA sequence. What is the basis of your argument and what could the company think is wrong with your DNA sequence? If the small ribosome is attaching but the complete ribosome is not being assembled you could be missing the special methionine needed for initiation or missing the tRNA associated with methionine. If either are missing the second step of ribosome assembly can not occur so no complete ribosomes are assembled and translation can not begin. A second possibility is if the initiation factors for translation that facilitate assembly of the ribosome are missing then no ribosomes would be assembled and no translation would occur. 5 Name: _________________________________ If the DNA lacked a start codon AUG the second step of ribosome assembly would not occur again resulting in the lack of complete ribosomes and no translation. RNA Codons UUU UUC UUA UUG phe phe leu leu UCU UCC UCA UCG ser ser ser ser UAU UAC UAA UAG tyr tyr stop stop UGU cys UGC cys UGA stop UGG trp CUU CUC CUA CUG leu leu leu leu CCU CCC CCA CCG pro pro pro pro CAU CAC CAA CAG his his gln gln CGU CGC CGA CGG arg arg arg arg AUU AUC AUA AUG ile ile ile met ACU ACC ACA ACG thr thr thr thr AAU AAC AAA AAG asn asn lys lys AGU AGC AGA AGG ser ser arg arg GUU GUC GUA GUG val val val val GCU GCC GCA GCG ala ala ala ala GAU GAC GAA GAG asp asp glu glu GGU GGC GGA GGG gly gly gly gly 6