Question

In the Meselson - Stahl experiment, if after the first round of replication, the DNA had been denatured at high temperature before its centrifugation in a column of cesium chloride, which two models would have shown the same sedimentation pattern?

Group of answer choices

a) semiconservative and dispersive

b) none of the three would have had the same sedimentation pattern

c) all three would have shown the same sedimentation pattern

d) conservative and dispersive

e) semiconservative and conservative

Answer 1

Ans. (e) semiconservative and conservative

Explaination: Denaturation leads to separation of two strands of a DNA molecule into separate single strands by breaking hydrogen bonds between bases on the compementary strands. In semiconservative mode of DNA replication, there is synthesis of a new DNA strand and with this newly synthesized strand along with an original strand will form the daughter molecule. Hence, denaturation will produce two types of strand (new strand with N14 and original strand with N15). Conservative mode of DNA replication results in the production of a new DNA molecule (both strands with N14) and the original DNA molecule is conserved ( DNA with N15 on both the strands). Denaturation of original and newly synthesized DNA molecule will result in strands with N14 and strands with N15 just like in case of denaturation of DNA molecules produced in semiconservative mode of DNA replication. Hence, both will show two bands on centrifugation - N14 containing DNA strand band at upper part of the tube and N15 containing DNA strand band at lower part of the tube. But dispersive mode produces DNA with strands which are combination of new and original DNA in a single strand i.e. in a single strand both N14 and N15 are present and hence it will produce single band on centrifugation.