Dear Netters,
in reply to a more or less simple question (everybody should be able to
calculate himself) I was curious to see the numbers which were requested.
Glen (gshearer at whale.st.usm.edu) asked for:
1. GC content of the yeast nuclear genome
The genomic sequences currently available at MIPS contain 12,067,280 nt
and the GC content of this portion is 38.30 % .
Assumed that there are 120 repeats of rDNA with 9138 nt each the GC content
changes to 38.751 % because the rDNA repeats have a GC % of 43.751 .
2. amount of repetitive sequences
The rDNA repeats are already mentioned and there might be some more
including the Y' sequences, the CUP repeats, the ENA repeats.
My personal estimation therefore is (again assuming 120 rDNA repeats):
120 * 9138 (rDNA)
+ ~30 * ~6000 (Y')
being a total of 1,276,560 nt that is close to 10% of the total genome!
(The CUP and ENA repeats don't count too much to the total genome)
This does not cover what is called redundancy in the yeast genome!!!
(eg. pairs of ribosomal proteins, families of sugar transporters, etc.)
3. fg DNA/nucleus
assuming 1bp = 660 g/Mol
then you end up with
13163840 bp * 660 g = 8,6881 Gg/Mol yeast nuclei
which is 14.425 fg/nucleus (haploid) (1 Mol = 6.023E+23)
Have fun!
Karl
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