Problem Set ——– Answer Key 1. The Genetic Code (a) Are all amino acids encoded by the same number of codons? no (b) Which amino acid(s) are encoded by the most codons? Leu, Ser and Arg are each encoded by 6 codons (c) Which are encoded by only one codon? Met and Trp (d) Which are encoded by only one codon? UAA, UAG, UGA (e) What are the tRNA anti–codons that can bind to the tyrosine codons? Note the 5’ and 3’ ends. If mRNA is 5’ UAU 3’, then anticodon will be 3’ AUA 5’ If mRNA or 5’ UAC3’, then anticodon will be 3’AUG 5’ (f) What is special about the amino acid methionine? Methionine is the first amino acid on the polypeptide chain. Its codon, AUG, serves as the start codon. 2. Regulation of Gene Expression Mutations can alter the function of an operon. Predict how the following mutations would affect the Lac Operon gene transcription (i) in the presence of lactose and (ii) in the absence of lactose? (a) Mutation of the Operator so that the repressor can no longer bind to it. (i) Lactose → Operon On (ii) No Lactose → Operon On (no represser bound to operator) (b) Mutation of the Promoter so that it is no longer seen by RNA polymerase. (i) Lactose → Operon Off (no polymerase bound to transcribe genes) (ii) No Lactose → Operon Off Answer Key (c) Mutation of the Promoter for the lacI repressor so that it is no longer bound by RNA polymerase. (i) Lactose → Operon On (ii) No Lactose → Operon On (no repressor made) 3. Gene expression You want to express a human keratin protein, so you need to integrate the human gene into a bacterial plasmid. If you cut out the human keratin gene directly from the human DNA genome and insert it into the plasmid, will the bacteria be able to express this gene for you? If not, why not? What additional steps would you need to do to get the bacteria to express your gene? No, the bacteria will not be able to express this gene because there are intron pieces inside the human genomic DNA for the gene. However, bacterial genomes do not contain introns and therefore do not have the machinery to splice introns. You will need to screen a cDNA library, or make a cDNA from the keratin mRNA using reverse transcriptase, to insert the keratin cDNA in the bacterial expression plasmid. 4. The Genetic Code II In studies of the evolution of amino acid sequences in Drosophila, the following changes have been observed. Using the genetic code in your text, determine a set of triplet codes in which only a single nucleotide change produces each amino acid change. (Each arrow indicates one amino acid change caused by a single nucleotide change from the previous codon). 3 5. Gene Analysis Below is a picture of human genomic DNA that contains a gene found to be abnormally regulated in skin cancer cells. There is a ruler below the genomic DNA that indicates every 100 base-pairs (the entire region shown is 1800 bp long). You want to study the gene further to learn about its function, and whether mutations in the gene are found in skin cancer patients. You identified four sequences that can be cut by the Restriction enzymes HindIII, AvaII and SmaI, as marked above. (Thursday and Tuesday) (a) If you cut with SmaI, how many fragments do you expect to get? two fragments. Answer Key (b) You separate the fragments based on size using gel electrophoresis. Draw a picture of the gel with what your fragments would look like on the gel. Label the approximate sizes of the fragments. Add in the + and – poles of the gel. (c) If you cut with HindIII, how many fragments will you get? three fragments. (d) If you cut with HindIII and SmaI how many fragments will you get? four fragments (e) Lets say you wanted to amplify the promoter region using PCR to identify transcription factor binding sites. Draw on the picture above to show where you would have your PCR primers placed to specifically amplify this region. Use arrows (→) to represent your primers and their required 5’ → 3’ orientation in the picture above. (f) Why is denaturation the first step during PCR? Primers can only anneal to a single strand DNA, which then can act as a template for DNA polymerases. For each round of replication the mixture needs to be heated above 90o C again to denature the newly formed DNA, ”melt” the hydrogen bonds, and separate the double helix. 6. Trp operon What would be the effect of a deletion of region 4 of trpL on regulation of transcription. Region 4 is involved in formation of the terminator sequence involved in attenuation of the trp operon. So, attenuation would not occur, giving higher levels of complete trp operon mRNA. 7. PCR Why is a heat-stable DNA polymerase used for the polymerase chain reaction (PCR)? Before the discovery of a heat-stable DNA polymerase, during each denaturation step of the PCR, the DNA Polymerase (Klenow fragment of E. coli DNA Polimerase I) was inactivated and needed to be added again before each new cycle. The use of a 5 thermostable DNA polymerase eliminates the need for having to add new enzyme to the PCR reaction during the thermocycling process. Exam Questions from 2008 8. Operon You are an undergrad research assistant in a UC Berkeley lab and are working in bacteria on the metabolism of the monosaccharide, theose. Metabolism of theose requires the enzymes X, Y, and Z. The genes encoding these enzymes are part of one operon, and the product of gene N (N protein) regulates the transcription of these three genes. In a normal cell, protein A is always produced. A diagram of this operon is shown below. (a) Give a brief definition of an operon. An operon is two or more genes that are regulated by the same control region(s). (b) In a cell where there is a high level of N protein, you detect no transcription of genes X, Y, and Z. You conclude that.... The N protein is (circle one) a repressor. (c) In further study, you discover that transcription of the N gene is controlled by protein A (protein A is the product of gene A shown in the diagram), and theose binds to the N protein. You examine the transcription of genes N, X, Y, and Z in cells where gene A is normal (A+) and where gene A is not functional (A-) and in the presence (+) and absence (-) of theose. The data was shown in a table. i. Given the results above, what does protein A do? Protein A promotes the transcription of gene N. ii. Given the results above, what does theose do? Theose binds to protein N and prevents it from binding to the O region. This prevents repression by N. Answer Key 9. Restriction digest You just isolated a novel recombinant clone and purified the desired insert (a 10,000 bp long linear duplex DNA) from the vector. Now you wish to map the recognition sequences for restriction endonucleases A and B. You cleave the DNA with these enzymes and fractionate the digestion products according to size by agarose gel elecrophoresis. In lane 1 you cut only with A, in 2 with B and in 3 with A+B. Comparison of the pattern of DNA fragments with DNAs of known sizes yields to the following result: Digestion with A alone gives 2 fragments: 3000bp and 7000bp. Dig. With B alone 3 fragments: 500bp, 1000bp, 8500bp. With both enzymes together you get 4 fragments: 500bp, 1000bp, 2000bp, 6500bp. The map looks a follows: In last years exam mirror maps got also full credit. Partial credit was given, if e.g. digestion with one of the enzymes was misinterpreted.