Molecular Biology

Question 4 a) Below is a schematic of gene Y, which encodes protein Y. The promoter region is indicated by the dotted box. Transcription begins immediately following the promoter. 5’ 3’ Promoter Exon 1 Intron 1 Intron 2 1 1100 2100 Exon 2 3100 4100 5100 base pairs Positions 200-203 = Start codon 100 Positions 4800-4802 = Stop codon Transcriptional terminator Exon 3 • The transcript first produced by this gene would be approximately how many nucleotides long? The transcript first produced by this gene would be approximately 5000 nucleotides long. • Two different transcripts are produced from this gene, one is approximately 2000 nucleotides long, the other is approximately 3000 nucleotides long. Explain how two different transcripts can be produced from this gene. Alternative splicing could produce two different transcripts. One transcript could contain exons 1, 2 and 3, the other could contain only exons 1 and 3. • Assume each transcript produces a protein. Given your answer above and the diagram, what is the approximate size of the two proteins produced from this gene? The transcript contain all three exons would encode a protein beginning at the ATG at position 200 and end at the stop codon. This is about 2600 nucleotides, which represents about 866 codons. SO the two proteins expected would be about 866 amino acids long and about 533 amino acids long. b) Assume gene Y was mutated such that the base pair found at position 200 was changed from an A/T to a G/C. Transcription and translation of the altered gene Y still occurs. • Would the first transcript be the same length, shorter or longer than the first transcript produced from the wild type gene Y? Give all possible answers and explain your thinking. This mutation changes a single base pair well after the promoter, so it is unlikely to affect transcription. The first transcript is expected to be the same length. • Would the protein produced be the same length, shorter or longer than the protein produced from the wild type gene Y? Give all possible answers and explain your thinking. This mutation changes a single base pair, eliminating the start codon. It is impossible to tell what affect this will have on the protein. If there is another start codon that maintains the original reading frame, then the protein will be shorter. If there is another start codon that changes the reading frame, then the protein will certainly be different. Because the location of the stop codon is also altered in this alternative reading frame, the resulting protein might be longer or shorter or even the same. c) Now assume gene Y was mutated such that only the base pair found at position 2100 was changed from a C/G to an A/T. Transcription and translation of gene Y still occur. • Would the first transcript be the same length, shorter or longer than the first transcript produced from the wild type gene Y? Give all possible answers and explain your thinking. This mutation changes a single base pair well after the promoter, so it is unlikely to affect transcription. The first transcript is expected to be the same length. • Would the protein produced be the same length, shorter or longer than the protein produced from the wild type gene Y? Give all possible answers and explain your thinking. This mutation changes a single base pair of a single codon. There are two possibilities. If the new codon still encodes an amino acid, then the resulting protein will be the same length. If the new codon creates a stop codon, the resulting protein will be the shorter.

MIT7 | 01 | Molecular Biology | exam | Question 4 a) Below is a schem

MIT7_01SCF11_2.5prob-lac_operon.pdf-Practice Problems for Molecular Biology, Session

Question 4 a) Below is a schematic of gene Y, which encodes protein Y. The promoter region is indicated by the dotted box. Transcription begins immediately following the promoter. Promoter Transcriptional terminator 5’ Exon 1 Intron 1 Intron 2 Exon 2 Exon 3 Positions 200-203 = Start codon Positions 4800-4802 = Stop codon 3’ 1 100 1100 2100 3100 4100 5100 base pairs • The transcript first produced by this gene would be approximately how many nucleotides long? • Two different transcripts are produced from this gene, one is approximately 2000 nucleotides long, the other is approximately 3000 nucleotides long. Explain how two different transcripts can be produced from this gene. • Assume each transcript produces a protein. Given your answer above and the diagram, what is the approximate size of the two proteins produced from this gene? b) Assume gene Y was mutated such that the base pair found at position 200 was changed from an A/T to a G/C. Transcription and translation of the altered gene Y still occurs. • Would the first transcript be the same length, shorter or longer than the first transcript produced from the wild type gene Y? Give all possible answers and explain your thinking. • Would the protein produced be the same length, shorter or longer than the protein produced from the wild type gene Y? Give all possible answers and explain your thinking. c) Now assume gene Y was mutated such that only the base pair found at position 2100 was changed from a C/G to an A/T. Transcription and translation of gene Y still occur. • Would the first transcript be the same length, shorter or longer than the first transcript produced from the wild type gene Y? Give all possible answers and explain your thinking. • Would the protein produced be the same length, shorter or longer than the protein produced from the wild type gene Y? Give all possible answers and explain your thinking. Question 5 You design a summer class where you recreate experiments studying the lac operon in E. coli (see schematic below). In your experiments, the activity of the enzyme β -galactosidase ( β -gal) is measured by including X- gal and IPTG in the growth media. X-gal is a lactose analog that turns blue when metabolized by β -gal, but it does not induce the lac operon. IPTG is another analog of lactose that is an inducer of the lac operon but is not metabolized by β -gal. lacZ I P lac P i Promoter for O Binding site for CAP activator protein Gene encoding β -gal Repressor (I) a) Which of the following would you expect to bind to b -galactosidase? Circle all that apply. Lactose (or allolactose) X-gal IPTG b) Which of the following would you expect to bind to the lac repressor? Circle all that apply. Lactose (or allolactose) X-gal IPTG After mutagenesis you find 7 mutants that never turn blue as shown in the table below. Each mutant has a single loss-of-function mutation. Cell Type Media + glucose No lactose + X-gal + glucose + lactose + X-gal No glucose No lactose + X-gal No glucose + lactose + X-gal Wild type White colonies White colonies White colonies Dark blue colonies Mutants 1-7 White colonies White colonies White colonies White colonies c) A single loss-of-function mutation in which component or components could produce the phenotype seen in these mutants? List all that apply. Choose from: I, P i , CAP binding site, O, lacZ, or Pl ac You also find three mutants with the following phenotype. Each mutant has a single loss-of-function mutation. Cell Type Media + glucose No lactose + X-gal + glucose + lactose + X- gal No glucose No lactose + X-gal No glucose + lactose + X-gal Wild type White colonies White colonies White colonies Dark blue colonies Mutants White colonies White colonies Dark blue colonies Dark blue colonies d) A single loss-of-function mutation in which component or components could produce the phenotype seen in these mutants? List all that apply. Choose from: I, P i , CAP binding site, O, lacZ, or Pl ac

Solutions to Practice Problems for Molecular Biology, Session 5 : Gene Regulation and the Lac Operon Question 1 a) How does lactose (allolactose) promote transcription of LacZ? 1) Lactose binds to the polymerase and increases efficiency. 2) Lactose binds to a repressor protein, and alters its conformation to prevent it from binding to the DNA and interfering with the binding of RNA polymerase. 3) Lactose binds to an activator protein, which can then help the RNA polymerase bind to the promoter and begin transcription. 4) Lactose prevents premature termination of transcription by directly binding to and bending the DNA. Solution: 2) Lactose binds to a repressor protein, and alters its conformation to prevent it from binding to the DNA and interfering with the binding of RNA polymerase. b) What molecule is used to signal low glucose levels to the Lac operon regulatory system? 1) Cyclic AMP 2) Calcium 3) Lactose 4) Pyruvate Solution: 1) Cyclic AMP . Question 2 You design a summer class where you recreate experiments studying the lac operon in E. coli (see schematic below). In your experiments, the activity of the enzyme b-galactosidase ( β -gal) is measured by including X-gal and IPTG in the growth media. X-gal is a lactose analog that turns blue when metabolisize by b-gal, but it does not induce the lac operon. IPTG is an inducer of the lac operon but is not metabolized by b-gal. a) Which of the following would you expect to bind to β -galactosidase? Circle all that apply. Lactose (or allolactose) X-gal IPTG b) Which of the following would you expect to bind to the lac repressor? Circle all that apply. Lactose (or allolactose) X-gal IPTG lacZ I P lac P i Promoter for Repressor (I) O Binding site for CAP activator protein Gene encoding β gal Question 2, continued After mutagenesis you find 7 mutants that never turn blue as shown in the table below. *Note each mutant has a single loss-of-function mutation. Cell Type Media + glucose No lactose + X-gal + glucose + lactose + X-gal No No glucose lactose + X-gal No glucose + lactose + X-gal Wild type White colonies White colonies White colonies Dark blue colonies Mutants 1-7 White colonies White colonies White colonies White colonies c) A single loss-of-function mutation in which component or components (I, Pi, CAP binding site, Plac, O, lacZ) could produce the unidicible phenotype seen in these mutants? List all that apply. CAP binding site, Plac, lacZ You also find three mutants with the following phenotype. *Note that each mutant has a single loss-of-function mutation. d) A loss-of-function mutation in which component or components (I, Pi, CAP binding site, Plac, O, lacZ) could produce the constitutive phenotype seen in these mutants? I, Pi, O e) You introduce a piece of DNA into the constitutive mutant 8 bacterial cells that contains the following: LacI and its promoter, LacZ, LacA, and LacY and their promoter. (Introducing additional DNA is a common laboratory technique in bacteria, and all added genes and regulatory regions can be expected to act as if they were a part of the genome.) This does not rescue the mutant phenotype observed in mutant 8; that is, these bacteria are still constitutive. Does this additional information allow you to narrow your options as to possible causative mutations? Explain your answer. Yes. Because this extra piece of DNA does not rescue the constitutive mutant phenotype we know the mutation in mutant 8 is in the O region. Mutations in I or P i would be complemented “in trans” because having a single functioning repressor protein could regulate expression from the genomic lac operon and the inserted lac operon sequences. If the O region on the genomic lac operon is mutated such that it can not bind to the repressor protein, then the added DNA can not stop the transcription of the genes adjacent to the mutated O region. Cell Type Media + glucose No lactose + X-ga + glucose + lactose + X-gal No No glucose lactose + X-gal No glucose + lactose + X-gal Wild type White colonies White colonies White colonies Dark blue colonies Mutants 8-10 White colonies White colonies Dark blue colonies Dark blue colonies

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