Ammissine 2018 biology(bacteriophage) Q

  1. Following the infection of a bacterial cell by a single bacteriophage, whose DNA is
    labeled with 32p, 100 new phage particles are produced. Unless recombination processes occur, how many of the new phages produced will have labeled DNA?
    A. 50
    B. 1
    C. 100
    D. 25
    E. 2

the correct answere here is E. i actually dont anderstand why isnt it zero. the new Virus DNA is formed from host’s nucleotids. So i would immagine non of the new phages have labeled DNA


the DNA is marked if it contains the ³²P
we assume that the bacteriophage replicates in an environment without radioactive P

replication is semi conservative so each time we keep 1 original DNA strand, and 1 new strand synthesised from the non radioactive/non marked elements found inside of the cell
say ³²X is a marked DNA strand and X is an unmarked DNA strand

initially we have the following DNA:
³²X ³²X

after replication an mitosis:
³²X X
³²X X

after replication and mitosis
³²X X
³²X X

this happens again and again until we produce 100 phages.
However since replication is semi conservative, only 2 phages will have the ³²P original strand

Hi, where does the non-radioactive strand come from? Theres someting im not understanding because I understand the semi-conservative replication but im confused by the rest.

the bacteriophage is radioactive, not the bacteria that it infects
the phage hacks the bacteria’s metabolism to do replication, so it will synthesize the new strand using C H O N P S atoms from the cell environment of the bacteria, that are non radioactive
So in semi conservative replication, the neo-synthesized strand made thanks to the bacteria’s building blocks and replication machinery will be non radioactive