1Solutions to 7.014 Problem Set 5Question 1a) Which of the following molecules functions directly to transfer information from thenucleus to the cytoplasm? Circle all that apply.DNA mRNAtRNA transporter vesicles proteinsb) Translation is the synthesis of proteins on ribosomes, where the information for an aminoacid sequence is encoded in the mRNA. This information is originally encoded by DNA,transcribed by a protein complex called RNA polymerase that binds to the promoter region of thegene.c) Why are retroviruses considered unusual with respect to the central dogma?In general, the idea of the central dogma is that information flow in the cell isDNA RNA protein. Retroviruses carry their genetic material as RNA that when inserted into thecell is copied into DNA. In effect, in retroviruses, RNA DNAd) From first principles, why do you need at least three RNA bases (a codon) to code for 1amino acid?DNA is composed of 4 different nucleotides but the proteins encoded by DNA are composed of 20different amino acids. 42 = 16, not sufficient to give 20 different amino acids, whereas 43 = 64.Spring 2002 7.0142Question 2After the double stranded model for DNA structure was proposed, two possible mechanismsfor DNA replication were envisioned:Semi-conservative +Conservative +Each of these possibilities has distinct predictions with respect to the daughter molecules.Notice that in the semi-conservative case, each daughter duplex in the first generation willhave one strand from the original and one new strand. In the conservative case, each daughterduplex will only have new strands. To distinguish between the two possibilities, the followingexperiment was carried out. Cells were grown in a medium containing “heavy” nitrogen (15N)[REMEMBER: nucleic acids contain nitrogen in their rings!). When all the DNA in the cellscontained the “heavy” nitrogen, they were transferred to a medium containing the normal,“light” nitrogen (14N) isotope. Samples of cells were removed at different times after additionof “light” nitrogen. The DNA was then analyzed by density-gradient centrifugation, which canseparate heavy-heavy, heavy-light, and light-light chains of DNA.This is what you would observe if you grow cells in either “heavy” or “light” nitrogen: Grown in Grown in Light Nitrogen heavy nitrogen Tubea) Given what you know about DNA replication, where and how many bands would youexpect to find in your tube after the cells had undergone exactly one round of DNA replicationgrowing in “light” nitrogen? Draw the DNA bands in the empty tube taking into account theposition for the “heavy” and “light” bands. Light Heavy EmptyOriginal DNADaughter DNASpring 2002 7.0143Question 2, continuedb) If you let the cells undergo exactly two rounds of DNA replication, how many bands wouldyou find? Where would you find them with respect to the “light” or heavy” bands? Draw yourprediction in the empty tube. Light Heavy EmptyFrustrated with your career as a biologist, you become an astronaut and travel to space. On adistant planet you are surprised to find life, in the form of chewing-gum like creatures.Curious to see if this form of life replicates DNA the same way as life on Earth does, you senda sample to your 7.014 TA’s. After some time, they send you an email saying that theirexperiments suggest that the new life form uses “random dispersive” replication. In randomdispersive replication the DNA is copied so that each daughter duplex will end up withroughly 50% of the original DNA and 50% new DNA (See below). They do not tell you whatexperiments they did to come up with this model.Random dispersive +c) If your TA’s did the “light”, “heavy” nitrogen experiment, what would they have observed?Draw your predictions in the empty tubes label below.Light Heavy After 1 round After 2 roundsd) Why would your TA's need to examine the cells after two rounds of replication to propose amodel like this?After one round of DNA replication, the results for semi-conservative and random dispersive replicationare the same. The difference only appears after two rounds.Spring 2002 7.0144Question 3E.coli DNA polymerase I is an enzyme that has three activities:1. 5'-to-3' DNA Polymerase activity2. 3'-to-5' exonuclease activity3. 5'-to-3' exonuclease activitya) What would be the effect of removing the 3’ to 5’ exonuclease activity on DNA replication?Removing this activity would affect the proofreading and fidelity of DNA replication.b) In 1969, John Cairns isolated a viable E. coli mutant that lacked DNA polymerase I (Pol I )activity.i) Do you think that an organism that cannot replicate its DNA would be viable?No, any organism that cannot replicate DNA would die.ii) What was the conclusion from Cairns’ experiment?From the viable E. coli mutant, one must conclude that DNA polymerase I is not essential forDNA replication. There must be another DNA polymerase in E. Coli cells.c) You find an E. coli temperature-sensitive (conditional lethal) mutant lacking only the 5'-to-3'exonuclease activity of DNA polymerase I. This mutant is not viable at non-permissivetemperatures. This is interesting because you know from part b above that DNA polymerase Iactivity is dispensable.i) What can you say about the 5’ to 3’ exonuclease activity of Pol I?You can conclude that, if Pol I is present in the cells, then the 5’ to 3’ exonuclease activity isessential. If Pol I is completely absent, some other DNA polymerase can take over the job ofremoving and replacing the RNA primers. If a defective Pol I is present, it likely prevents theother DNA polymerase from taking over the job.ii) Given what you know about the steps involved in DNA replication, what do youthink the function of the 5’ to 3’ exonuclease activity is during this process?E.coli DNA polymerase I is primarily responsible for removing the RNA primer of the newlymade Okazaki fragments, and replacing it with DNA. Without this activity, the completion ofthe lagging strand would be inhibited.Spring 2002 7.0145Question 4Shown here is the double-stranded DNA sequence coding for human hemoglobin. Below the 2strands is the one letter abbreviation for the amino acids of the hemoglobin protein. 20 40 60 5' ACACTCGCTTCTGGAACGTCTGAGATTATCAATAAGCTCCTAGTCCAGACGCCATGGGTC 3'
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