Bio 382-1 Homework Assignment #6 Name: The complete ovalbumin gene (7,700 bp) is
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Bio 382-1 Homework Assignment #6 Name: The complete ovalbumin gene (7,700 bp) is picture below. The introns are labeled A-G, and the size of each intron is listed. The exons are labeled 1-7, and the size of each exon is labeled on the figure (in bp)? The L region is the Leader Exon, which is present in mature transcripts,but translation does not begin until cxonI A-1 ,428bp B-350 bp C-400 bp D-300 bp E-1,500 bp F-250 bp G 1,600 bp ?7,700 BASE PAIRS 5 6 BCD 47 185 51 129 118 143 156 1,043 On the agarose gel below, the size marker lane (M) shows the corresponding sizes of the ladder bands in base pairs. Lanes 1-5 contain different alternatively spliced transeripts of the ovalbumin gene pictured above. For each lane, you want to include the leader exon and use only exons whenever possible. If you include introns, you want to minimize the total number of them used when possible. One of the lanes contains unspliced pre-mRNA (including ALL exons and introns), and one lane is spliced normally. To get from the components that you list for each lane. In the ur band sizes, add up the base pairs corresponding spaces below, list the size of each band, and your best guess as to which components of the gene are present (L, A, B, C, D, E, F, G, 1, 2,3,4, 5, 6, and/or 7). 05 points/band size, 0.5 points/correct listed 7500 5000 3500 1800 1000 800Explanation / Answer
Band Sizes of different ovalbumin fragments are as follows (Approximately):
Lane 1: 825 bp (greater than 800 bp)
Lane 2: 1000 bp
Lane 3: Higher than 7500 bp (Approx 7700 bp)
Lane 4: Higher than 1800 bp
Lane 5: Slightly less than 3500 bp
The lane 3 band is the pre mRNA as its size is equal to the entire transcript of 7700 bp. It contains the components L, A, B, C, D, E, F, G, H, G, 1, 2, 3, 4, 5, 6, 7.
Lane 5 is the mature transcript lacking the introns. It has the components A, B, C, D, E, F, G, H and G.
Lane 4 is the mature transcript.
Lane 4 = (Pre mRNA)- (Sum of all introns)= 7700- 5828= 1872
Lane 2 = (Pre mRNA)- (Sum of all introns)- (Sum of exon 1,2,3,4,5,6)= 1872-782= 1090
Lane 1= (Pre mRNA)- (Sum of all introns)- (Exon 7)= 1872- 1043= 829
Lane 3= (Pre mRNA) –(Sum of all introns except intron G)= 7700-4228= 3472
Hence,
Lane 1= Transcript with exons L, 1, 2,3,4,5,6
Lane 2= Transcript with exons L,7
Lane 3= Transcript with L, 1,2,3,4,5,6,7 and intron G
Lane 4= Mature transcript with L,1,2,3,4,5,6,7
Land 5= Pre mRNA with L, A, B, C, D, E, F, G, 1, 2, 3,4,5,6,7
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