ted07exam1 - Name: page 1 Genetics 320 Fall 2007, Midterm...

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Name: page 1 1. Expression, mutation, consequences (25pt) 2. Gene expression/domains (25pt) 3. Mitotic and meiotic recombination (25pt) 4. T/F short answer (25pt) TOTAL Genetics 320 Fall 2007, Midterm Exam 1. Sept 21
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Gene expression, mutation and consequences. The top strand of a DNA sequence of a gene is shown. The complementary sequence on the bottom strand is not shown. The promoter sequence is in capital letters and boxed in. +1 of the mRNA is a C. There is no splicing. The Atg is the start codon and is circled. Each amino acid encoded in the protein is essential for protein function. gcgTATAttgaccgctaCgtagg c tttAtgccggttaaatttcgcgggtagcccgcaag +11 Amount of Transcription Functioning Protein (per cell) d. TATAcc to TggAcc less less e. Delete t at +6 same same f. Insert g between 10g and 11g same less g. Change cgc to cgt same same h. Explain briefly the change, if any in # and kind of amino acids for parts f and g. (6pt) In f, inserting a g introduces a frameshift mutation, and a stop codon after a total of 3 amino acids. In g, the basepair change generates a syonomyous codon--they both encode arginine. No change to amino acid sequence. Name: page 2 a. Label (5’, 3’) the ends of the strands with the correct polarity. (4pt) b. Put an arrow to the template strand used by RNA polymerase to make mRNA.(2pt) c. Circle the stop codon. How many amino acids are encoded in this gene? (4 pt) 7 d-g. Fill in the table below. Use less, same or more to indicate the AMOUNT PER CELL in the mutant compared to the original gene.
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This note was uploaded on 06/16/2008 for the course ECOL 320 taught by Professor Weinert during the Spring '07 term at Arizona.

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ted07exam1 - Name: page 1 Genetics 320 Fall 2007, Midterm...

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