A) 1:1:0
B) 1:4:0
C) 0:1:3
D) 0:1:7
Correct Answer: D
Solution :
[d] Option [d] is correct. Meselson and Stahl found that DNA of the first generation was hybrid or intermediate\[\left( ^{15}N\text{ }an{{d}^{15}}N \right)\]. It settled in cesium chloride at a level higher than the fully labelled DNA of parent bacteria \[\left( ^{15}N{{\,}^{15}}N \right)\]. The second generation of bacteria after 40 minutes, contained two types of DNA, 50% light \[\left( {{N}^{14}}\text{ }{{N}^{14}} \right)\] and 50% intermediate \[\left( {{N}^{15}}\text{ }{{N}^{14}} \right)\]. The third generation of bacteria after 60 minutes contained two types of DNA, 25% intermediate \[\left( {{N}^{15}}\text{ }{{N}^{14}} \right)\] and 75% light \[\left( {{N}^{14}}\text{ }{{N}^{14}} \right)\] in 1 : 3 ratio. It can be assumed that the fourth generation after 80 minutes would contain 12.5% \[{{N}^{15}}{{N}^{14}}\] and 87.5% \[{{N}^{14}}{{N}^{14}}\] DNA in 1:7 ratio. Thus, if Meselson and Stahl's experiment is continued for four generations in bacteria, the ratio of \[^{15}N{{/}^{15}}N{{:}^{15}}N{{/}^{14}}N\text{ }{{:}^{14}}N{{/}^{14}}N\] containing DNA in the fourth generation would be 0 : 1 : 7.
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