Answer:
B. It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment.
Explanation:
Hershey and Martha Chase used radiolabeled the DNA of some of the bacteriophage cells with phosphorus (32P). They radiolabeled the sulfur (35S) of the coat protein in the second batch of the phage cells. They infected some of the bacterial cells with phage having radiolabeled DNA while the other <em>E. coli</em> cells were infected with the phage carrying radiolabeled coat protein. This allowed the clear identification of the radiolabelled molecule (DNA or protein) present in the host cell.
They observed that the <em>E. coli </em>cells infected with phage having radiolabeled DNA exhibited the radioactivity while the other batch of the host cell did not show it.
If B and b are the two alleles , B + b = 1 and B ^2 +2 Bb + b ^2 = 1.
If 16% of mice are homozygous black , B ^ 2 =0 .16, meaning B = 0 .4 and b = 1 - 0. 4 = 0 .6 .
Answer = Notice how you don't even need to know that 24% of the mice are heterozygous .
Answer: sorry I don't know but if I did I would help
Explanation:
Speed and precision have been one of the main differences between an older and a more modern genetically modified food. In addition, these genetically modified organisms have yielded crop efficiencies although it also involves in various debates because of its unethical nature.
