One E coli cell has a volume of 1 mu m3 and consists of abou
Solution
(a) It is assumed that amino acids have same density as water.
Density of water = 1g/cm3
= 10-12 g/ µm3
= 10-12 g/ µm3 (Converting to Daltons)
6.022* 1011 dalton/ µm3
The weight of proteins is 30% of the total weight of E.coli cell.
30% of the total weight= 1/3* 6.022* 1011
= 2* 1011 daltons
If an amino acid weighs 100 daltons then total amino acids in an E.coli cell can be calculated as:
100 daltons in 1 (AA) amino acid,
2* 1011 daltons in 2* 1011 /100
=2* 109 AAs
(b) No. of proteins in a cell = No. of AAs / AAs per protein
= 2* 109 / 300
= 2/3 * 107
= 0.67 * 107
(c) During translation, the total distance covered by ribosomes in units of codon is same as the total number of AAs (amino acids). Thus,
Rate of translation = Number of AAs
Number of ribosomes and time
= 2* 109/10000* 30* 60 sec
= 100/ sec or velocity
(d) According to the statement, it is assumed that transcription and translation occur at the same rate. It implies that DNA and RNA have same length i.e. no introns. Thus the number of mRNAs made depends upon velocity, cell generation time and number of RNAPs (RNA polymerases)i.e
No. of mRNA per cell= Velocity*time*no. of RNAPs
= 100* 30*60* 3000
= 54* 107
No. of proteins per mRNA = No. of AAs
No. of mRNA codons
= 2* 109
54* 107
= 3.7

