You are studying a strain of bacteria. You find that it has suddenly gained the

Question

You are studying a strain of bacteria. You find that it has suddenly gained the ability to survive antibiotics. After asking around, you find out that your partner did not fully clean out the test tubes after growing antibiotic-resistant bacteria, and that you then used the dirty tubes. Your partner claims that couldn't be why your bacteria changed, as his bacteria had been killed in the tubes. Could your bacteria have gotten a new trait from dead bacteria? If so, how? You infect a bacterial culture with two strains of bacteriophage. One is h*r and the other is hr^+. You obtain the following numbers of plaques of each type. How far apart are the h and r genes. Describe how the experiments of Griffith, and later Avery MacLeod and McCarty, helped show that DNA was the genetic material?

Explanation / Answer

Please find the answers below:

Part 1: According to the question, the native bacteria was made to grow or reside in a dirty tube which contained dead bacteria. Since the resident bacteria was dead, it might have released some cytotoxic materials in the periphery. These cytotoxic materials might have been acquired and assimilated by the native bacteria and helped them to transform. Gradually with cell division, these native bacteria might have undergone mutational transformation which made them resistant in nature. This is relevant to the classical transformation assay conducted initially by scientist. Thus, the dirty tube might have surely contained antibiotic which killed the initially inhabiting bacteria and helped the native new bacteria to undergo tranformation.

Part 2: According to the information, the frequency of the parental types is given by 846 and 852 whereas the frequency of the recombinant types is given by 73 and 75. Thus, since the recombinant took place due to exchange of genetic material, these will represent the actual recombinant types in the population. Thus, the frequency of the recombinant types and hence the distance between the genes will be given by:

The gene distance = number of recombinant types *100 / total progeny size

The gene distance = (73+75) * 100/73+75+846+852

Gene distance = 148 * 100/1846 = 8.01 cM.

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