keyHW#1 - Homework Assignment #1 (72.5 points), IB 302 Due...

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Homework Assignment #1 (72.5 points), IB 302 Due at the beginning of class on February 10, 2010 Please print out this assignment and write your answers in the spaces provided unless otherwise noted. 1. (24 points total) The following DNA sequence represents the beginning of the coding region of the alcohol dehydrogenase ( Adh ) gene of Drosophila simulans (Bodmer and Ashburner 1984), arranged into codons: CCC ACG ACA GAA CAG TAT TTA AGG AGC TGC GAA GGT (a) Find the corresponding mRNA sequence, and use Figure 8.2 to find the amino acid sequence. Write the aa sequence in the space below. (2 points) mRNA sequence: CCC ACG ACA GAA CAG UAU UUA AGG AGC UGC GAA GGU Amino acid sequence: Pro-Thr-Thr-Glu-Gln-Tyr-Leu-Arg-Ser-Cys-Glu-Gly. (b) Again using Figure 8.2, and assuming that the mRNA sequence is translated, determine for each site how many possible mutations (changes of individual nucleotides) would cause an amino acid change, and how many would not. Synonymous (S) or nonsynonymous (N) mutational changes are listed below: © 2009 Sinauer Associates, Inc. 1
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© 2009 Sinauer Associates, Inc. Site Original  nucleotide Base  position A C G U 1 C 1 N - N N 2 C 2 N - N N 3 C 3 S - S S 4 A 1 - N N N 5 C 2 N - N N 6 G 3 S S - S 7 A 1 - N N N 8 C 2 N - N N 9 A 3 - S S S 10 G 1 N N - N 11 A 2 - N N N 12 A 3 - N S N 13 C 1 N - N N 14 A 2 - N N N 15 G 3 S N - N 16 U 1 N N N - 17 A 2 - N N N 18 U 3 N S N - 19 U 1 N S N - 20 U 2 N N N - 21 A 3 - N S N 22 A 1 - S N N 23 G 2 N N - N 24 G 3 S N - N 25 A 1 - N N N 26 G 2 N N - N 27 C 3 N - N S 28 U 1 N N N - 29 G 2 N N - N 30 C 3 N - N S 31 G 1 N N - N 32 A 2 - N N N 33 A 3 - N S N 34 G 1 N N - N 35 G 2 N N - N 36 U 3 S S S - 2
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For the entire sequence, what proportion of possible mutations are synonymous? (3 points) 22/108=20.4% What proportion of possible mutations are nonsynonymous? (3 points) 86/108=79.6% (c) What proportion of the possible mutations at first, second, and third base positions within codons are synonymous? First: 5.6% (3 points) Second: 0% (3 points) Third: 55.6% (3 points) (d) What would be the effect on the amino acid sequence of inserting a single base, G, between sites 10 and 11 in the DNA sequence? (2 points) Insertion of a C between sites 10 and 11 will result in a frameshift mutation. The resulting protein will have a vastly different amino acid sequence downstream of the insertion site. (e) What would be the effect of deleting nucleotide 16? (2 points) Deleting nucleotide 16 also causes a frame-shift mutation. If this frameshift were to occur, we would end up with a shortened protein (as the UAA stop codon appears close to nucleotide 16). 2. (23 points total) Table 10.1 in Futuyma reports the results of a very influential study by
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keyHW#1 - Homework Assignment #1 (72.5 points), IB 302 Due...

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