Phosphorous is required to synthesize the deoxyribonucleoside triphosphates used in dna replication. a geneticist grows some e. coli in a medium containing nonradioactive phosphorous for many generations. a sample of the bacteria is then transferred to a medium that contains a radioactive isotope of phosphorus (32p). samples of the bacteria are removed immediately after the transfer and after one and two rounds of replication. assume that newly synthesized dna contains 32p and the original dna contains nonradioactive phosphorous. what will be the distribution of radioactivity in the dna of the bacteria in each sample?

First sample is removed immediately after the transfer hence replication has not taken place and 32P wont be incorporated into DNA strands. Radioactivity will be almost zero.

Second sample is removed after first round of replication. The new strand synthesised will have 32P. The DNA strands will separate and replication will take place according to the conservative model. In all DNA molecules, out of the two strands one strand will have 32 P and show radioactivity.

Third sample is removed after second round of replication. The two strands will separate again and new strand will incorporate 32P. 50% of DNA molecules will have one old phosphorus containing strand and one new 32P containing strand. Rest 50% of DNA will have 32P in both strands.

the correct answer is sex-linked recessive.

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