Solution to PS 1 #2

a) since the average E.coli gene is 1kb (1000 bp )  long, 4000 genes account for 4000 kb of DNA.  S the perecentage not transcribed is:
600/4600 x 100%=  13%

Most of the non-transcribed DNA consist of promotors and regions that regulate transcription initiation.

b)  Since gene products in mammals and  bacteria are similair in size, the amount of DNA in the exons of a typical mammalian gene must also be about 1000 bp.  The total amount of DNA in the exons is therefore:

5 x 104 genes x 1 kb/gene= 5 x 104 kb

This DNA represents

(5 x 104 kb/ 3 x 106 kb) x 100%= 1.7 % of the mammalian genome.

The remaining 98.3% of the DNA consists of introns and other sequences.



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