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MICB 201 Chapter #4 Answers Problem #1 (a) First 10: N-Met-Lys-Ile-Thr-Leu-Leu-Val-Thr-Leu-Leu. There are many potential TAC sequences which could encode the start codon AUG. The question is: How does one decide which one is the right one? Last 3: Lys-Ala-Ala-C. There are many potential sequences before the sequence encoding the transcription terminator (b) which could encode a stop codon (UAA, UGA or UAG). The question is: How does one decide which one is the right one? (b) A C G G G U G A G C U G C C C A C U C G U 5'- U-U 3'- A U U A C G A A A A U G Problem #2 75/(5000 + 75) • 100 = 1.48% not 75/(5000) • 100 = 1.50% Problem #3 The average protein is specified by the average gene which possesses a size of 1 kb or 1000 bp One average gene has a mass of 6.6 x 10 5 Daltons 2000 nucleotides • 330 Daltons/nucleotide = 6.6 x 10 5 Daltons In total, 11% of the genome does not specify protein. Thus the number of average genes is 1 x 10 9 Daltons • 0.89 • 1 gene/6.6 x 10 5 Daltons = 1348 genes

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Problem #4 Protein #1: N-Met-Asn-Ser-Ser-Thr- .......
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