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BMB400Test_1 - Total = 68 BMB 400 Keith Hutchinson...

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Total = 68 BMB 400 Keith Hutchinson Hutchison 10/23/07 Exam 1 1) 5pts My DNA coded for the desert hedgehog homolog (DHH) from Homo Sapiens. It’s on Chromosome 12 at location 12q12-q13.1. the following is a description from the NCBI results: “This gene encodes a member of the Hedgehog family. The hedgehog gene family encodes signaling molecules that play an important role in regulating morphogenesis. This protein is predicted to be made as a precursor that is autocatalytically cleaved; the N-terminal portion is soluble and contains the signaling activity while the C-terminal portion is involved in precursor processing. More importantly, the C-terminal product covalently attaches a cholesterol moiety to the N-terminal product, restricting the N-terminal product to the cell surface and preventing it from freely diffusing throughout the organism. Defects in this protein have been associated with partial gonadal dysgenesis (PGD) accompanied by minifascicular polyneuropathy. This protein may be involved in both male gonadal differentiation and perineurial development.” 2a) 12pts Tm is the temperature at which a strand of dsDNA is in the double helix form and half in the single strand form. The hydrogen bonds between base pairs are a major factor in what the Tm for a given strand will be. Since GC bonds have 3 hydrogen bonds and AT bonds only have two a strand with a higher GC content will have a higher Tm then one with more AT contents. This also leads to the fact that a longer strand of DNA will have a higher Tm then a smaller one since it has more hydrogen bonds that need more energy to break (thus a higher Tm). Up to a point. As I pointed out with the 2 nd homework assignment once one gets above 75 to 100 bp it becomes primarily dependent on GC content alone. The stacking interactions of GC bonds also help to increase the Tm of a given strand. The interactions of GC base pairs with neighboring base pairs also increases the Tm since these interactions are more favorable then AT stacking interactions.
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2b) This is an alignment for: Macaca mulatta similar to desert hedgehog preproprotein. Here is the alignment comparison of my DNA to this one with my DNA on top and the heterologue on the bottom: Query 1 GCGCCGGGCTTCATTTGTGGCTGTGGAGACCGAG T GGCC T CCACGCAAACTGTTGCTCAC 60 |||||||||||||||||||||||||||||||||| |||| |||||||||||||||||||| GCGCCGGGCTTCATTTGTGGCTGTGGAGACCGAG C GGCC C CCACGCAAACTGTTGCTCAC 512 Query 61 G CCCTGGCACCTGGTGTTTGC C GCTCGAGGGCCGGCGCCC G CGCCAGGCGACTTTGCACC 120 |||||||||||||||||||| |||||||||||||||||| ||||||||||||||||||| A CCCTGGCACCTGGTGTTTGC G GCTCGAGGGCCGGCGCCC T CGCCAGGCGACTTTGCACC 572 Query 121 GGTGTTCGCGCGCCGGCTACGCGCTGGGGA C TCGGTGCTGGCGCCCGGCGGGGA T GCGCT 180 |||||||||||||||||||||||||||||| ||||||||||||||||||||||| ||||| GGTGTTCGCGCGCCGGCTACGCGCTGGGGA T TCGGTGCTGGCGCCCGGCGGGGA C GCGCT 632 Query 181 TCGGCCA G CGCGCGTGGCCCGTGTGGCGCG 210 ||||||| |||||||||||||||||||||| TCGGCCA A CGCGCGTGGCCCGTGTGGCGCG 662 The differences in the strands would cause a difference in the Tm from my DNA to that of the heterologue. From the base changes (which are underlined and bold above) the Tm of the heterologue strand would be lower then that of my DNA.
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