BMMAG_BMMSG 525 Microbiology I Problem Set 2. The problem set is worth 20 points
ID: 187195 • Letter: B
Question
BMMAG_BMMSG 525 Microbiology I Problem Set 2. The problem set is worth 20 points, divided as indicated below for each question. There is partial credit possible only for question 4c 1. Create a line diagram of a gene with the following elements: promoter, operator, ribosome binding site, start codon, end codon, open reading frame, start of transcription, rho-independent transcription termination. Start with the arrow provided. Label where appropriate. (4 pts). 2. Transcribe the following sequence, and indicate all of the start and stop codons in all six frames (2 pts). 5'AATGTTACTCGAGGGCTACTTAGCCACTAGGCTTTAGCCAGCG 3 3' TTACAATGAGCTCCCGATGAATCGGTGATCCGAAATCGGTCGC 5' 3. Translate the transcription (mRNA) sequence of the longest ORF in problem 2 (2 pts). 4. Assume a substitution mutation occurred near the beginning of the sequence, so that the DNA sequence is now: 5' A GTGTTACTCGAGGGCTACTTAGCCACTAGGCTTTAGCCAGCG3 3' TCACAATGAGCTCCCGATGAATCGGTGATCCGAAATCGGTCGC 5 a. What type of mutation has occurred (e.g. nonsense or missense, transition or transversion) (1 pt)? b. Does this sequence still have an ORF (1 pts)? c. Explain why or why not. (2 pts)? 5. Assume a second mutation occurred, such that the DNA sequence is now: 5 AGTGTTACTCGAGGGCTACTAGCCACTAGGCTTTAGCCAGCG 3' 3' TCACAATGAGCTCCCGATGATCGGTGATCCGAAATCGGTCGC 5 a. What type of mutation has occurred (hint: it may not be the same as in problem 4) (1 b. If a peptide can be produced after this mutation, what is its amino acid sequence (2 pt)? pts)? The lac operon produces three proteins. From the genotypes below, state whether or not LacZ will be produced in the presence of allolactose and the absence of glucose (lacO is the primary lac operator). For a. through e. explain why LacZ will or will not be expressed. (1 pt each): 6. a. Iacl., 1aco, lacZ-, lacY-, lacA+ b. lad», 1aco, lacZ+, lacY-, lacA. c lacl+,laco, lacz+, lacY, lacA- d. laci-, lacO, lacz., lacY+, lacA- e. Iacl+, 1aco, lacZ+, lacY-, lacA+Explanation / Answer
5' AATGTT.........GCG 3' => Coding strand
3' TTACAA........CGC 5' => Template strand
The mRNA transcribed from this DNA will be 5' AAUGUU........GCG 3' which contains a start codon.
A mutation changed A to G in the second postion in the coding strand. So the resultant DNA is as follows.
5' AGTGTT.........GCG 3' => Coding strand
3' TCACAA........CGC 5' => Template strand
The mRNA transcribed from this DNA will be 5' AGUGUU........GCG 3' which does not contain the start codon. Thus the resultant protein is incomplete. So this is a nonsense mutation and transition.
Nonsense mutation: Mutation produces a stop codon which terminates the transcription before it reaches the original stop codon, resulting in incomplete protein.
Missense mutation: Mutation that changes one amino acid to another.
Transition: Purine to purine or pyrimidine to pyrimidine.
Transversion: Purine to pyrimidine or pyrimidine to purine
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