Problem_Set9_Answer

Problem_Set9_Answer - RNA sequences for GU followed by AG...

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Genetics 380 : Spring 2008 Problem Set #9 Answers Q. 1. (a) 100 + 50 + 25 + 200 = 375 nucleotides (b) 100 +75 + 50 + 25 + 200 = 450 nucleotides Q. 2. In the β -globin DNA: β -globin mRNA hybrid, the introns will loop out (giving double-stranded loops). At the exons, one strand of DNA will be hybridized to the mRNA and one strand will be displaced out of the duplex, giving ss-loops. Nothing will hybridize to the phage Lambda portion of DNA duplex. So, it will remain as DNA duplex. (see diagram below for details). Q. 3 . RNA with 2 nucleotides (75% C and 25% A) will have 8 codons in the proportions given below- CCC (Pro) = ¾ x ¾ x ¾ = 27/64 CCA (Pro), CAC (His), ACC (Thr) = ¾ x ¾ x ¼ = 9/64 CAA (Gln), ACA (Thr), AAC (Asn) = ¾ x ¼ x ¼ = 3/64 AAA (Lys) = ¼ x ¼ x ¼ = 1/64
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Q. 4. Introns begin with GU and end in AG (there are other finer details). An examination of the 5
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Unformatted text preview: RNA sequences for GU followed by AG reveals that-- (1) No GU in this RNA. (2) There is a GU (nucleotides 3 &4) but no AG after the GU. (3) There are several GU in this RNA, but no AG after the first GU. (4) There are several GU in this RNA and one AG near the 3'-end. Therefore, there is a potential intron here. (5) There are multiple GU in this RNA, and two AG downstream from the GU. Therefore, there is a potential intron here. A search for the branch-point sequence (YNCURAY; N= any nucleotide; Y=Pyrimidine; R=Purine) reveals that only RNA #5 has such a sequence (UACUAAC). Therefore, probably only RNA #5 has an intron. In this RNA, only one of the 2 AG is downstream from the branch point sequence. After splicing the intron out, the RNA sequence will be either 5- UAG/UCUCA-3 or 5'-UAGGUUCGCAUUGAC/UCUCA-3' (/ indicates splice point)...
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This note was uploaded on 04/29/2008 for the course BIOLOGY 447:380 taught by Professor Sinha during the Spring '08 term at Rutgers.

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Problem_Set9_Answer - RNA sequences for GU followed by AG...

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