cloning a gene that has secondary structure

January Weiner nospam_jweiner1 at ix.urz.uni-heidelberg.de
Fri Sep 25 10:56:12 EST 1998


[ Rich Dudley ]
> I really want to know how.  Bacterial expression vectors usually can't handle such
> large inserts--they're physically too small, and the high copy number ori's can be
> lethal to the host cell due to depletion of resources (think about it--200 copies
> of what would be approx. 40 kDa is the equvalent of 8Mbp--8x the E. coli
> chromosome!)

	What??? 40kDa is 8Mbp=8000000bp=8*10^6. Now, 8e6/200= 40000 bp??
You got it all wrong. Lookup any biochemistry book for the definitions of
molecular mass and the chemical formulas for the aminoacids. 
	In M.pneumoniae I am working with the homolog of the P1 protein,
170 kDa, is encoded by 1241 bp.

> A 34 kDa ORF?  What in God's name could this code for?  The protein would be
> approx.1,258,000 Da (or 1258 kDa for those who can't do the math).  My guess is
> also that any RE site in an MCS probably shows up somewhere in a 34 kDa span.

	Ah! That explains all.
	Well, take a look at that:
	
Exp	Factor			Abbreviation (example)
9	1000000000		giga (GB)
6	1000000			mega (MB)
3	1000			kilo (1 kDa = 1000 Da)
1	10			decy (dl)
0	1			-
-2	0,01			centy (cm)
-3	0,001			mili (mM)
-6	0,000001		mikro (uF)
-9	0,000000001		nano  (ng)
-12	0,000000000001		pico (pM)
-15	0,000000000000001	femto (fM)

> Good call!  I forgot about this.
> rich

	Erm...
	Regards,
			January


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