E.coli:
Size of the genome 5 x 106 bp
Volume of a base pair: 1 nm3 / bp
Volume of genome: 5 x 106 nm3
Cell volume: 1 µm3 = 1000 nm3
P= (5 x 106 nm3)/ 103 nm3 = 5 x 103
Yeast
Size of the genome 12.1 x 106 bp
Volume of a base pair: 1 nm3 / bp
Volume of genome: 12.1 x 106 nm3
Volume of nucleus: 4 x 109 nm3
P= (12.1 x 106 nm3)/ 4 x 109 nm3 = 3.025 x 10-3
Lambda phage
Size of the genome 48,502 bp
Volume of a base pair: 1 nm3/bp
Volume of genome: 48,502 nm3
Volume of head: 9 x 10-5 µm3 = 9 x 10-2 nm3
P= (48,502 nm3)/ 9 x 10-2 nm3 = 538911.11
3. The DNA packing ratio is one of important parameters in biology. It is defined as...
En (2 points) You isolated your mitochondrial DNA in Part I. In step 6, you discard the supernatant, but keep the pellet. In step 15, you discard the pellet, but keep the supernatant. Explain why the pattern is different between the two steps and the consequence of mixing up these two steps. Procedure Part 1: mt DNA Isolation from your cheek cells. Lysis solution is used to breakdown the cells in this step, you will isolate MEONA from cheek cells....