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a. Hotchkiss and Marmur noted that the percentage of cotransformation was higher than would be expected on a random basis. For example, the results show that 2.6% of the cells were transformed into M and 4% were transformed into S. If the M and S traits were inherited independently, the expected probability of cotransformation of M and S (M S) would be 0.026 x 0.04 = 0.001, or 0.1%. However, they observed o.41% M S cotransformants, four times more than they expected. What accounts for the relatively high frequency of cotransformation of the traits they observed? b. On the basis of the results, what conclusion can you draw about the order of the M, S, and F genes on the bacterial chromosome?

a. Hotchkiss and Marmur noted that the percentage of cotransformation was higher than would be expected on a random basis. For example, the results show that 2.6% of the cells were transformed into M and 4% were transformed into S. If the M and S traits were inherited independently, the expected probability of cotransformation of M and S (M S) would be 0.026 x 0.04 = 0.001, or 0.1%. However, they observed o.41% M S cotransformants, four times more than they expected. What accounts for the relatively high frequency of cotransformation of the traits they observed? b. On the basis of the results, what conclusion can you draw about the order of the M, S, and F genes on the bacterial chromosome?

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
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DATA 24. MA Rollin Hotchkiss and Julius Marmur studied transformation in the ANALYSIS bacterium Streptococcus pneumoniae (R. D. Hotchkiss and J. Marmur. 1954. Proceedings of the National Academy of Sciences of the United States of America 40:55-60). They examined four mutations in this bacterium: penicillin resistance (P), streptomycin resistance (S), sulfanilamide resistance (F), and the ability to use mannitol (M). They extracted DNA from strains of bacteria with different combinations of different mutations and used this DNA to transform wild-type bacterial cells (P* S* F* M*). The results from one of their transformation experiments are shown here: Donor Recipient Transformants Percentage DNA DNA of all cells MSF M* S* F M* SF 4.0 M* S* F 4.0 MS' F 2.6 MSF 0.41 M* SF 0.22 MS* F 0.0058 MSF 0.0071 a. Hotchkiss and Marmur noted that the percentage of cotransformation was higher than would be expected on a random basis. For example, the results show that 2.6% of the cells were transformed into M and 4% were transformed into S. If the M and S traits were inherited independently, the expected probability of cotransformation of M and S (M S) would be o.026 x 0.04 = 0.001, or 0.1%. However, they observed o.41% M S cotransformants, four times more than they expected. What accounts for the relatively high frequency of cotransformation of the traits they observed? b. On the basis of the results, what conclusion can you draw about the order of the M, S, and F genes on the bacterial chromosome?
Transcribed Image Text:DATA 24. MA Rollin Hotchkiss and Julius Marmur studied transformation in the ANALYSIS bacterium Streptococcus pneumoniae (R. D. Hotchkiss and J. Marmur. 1954. Proceedings of the National Academy of Sciences of the United States of America 40:55-60). They examined four mutations in this bacterium: penicillin resistance (P), streptomycin resistance (S), sulfanilamide resistance (F), and the ability to use mannitol (M). They extracted DNA from strains of bacteria with different combinations of different mutations and used this DNA to transform wild-type bacterial cells (P* S* F* M*). The results from one of their transformation experiments are shown here: Donor Recipient Transformants Percentage DNA DNA of all cells MSF M* S* F M* SF 4.0 M* S* F 4.0 MS' F 2.6 MSF 0.41 M* SF 0.22 MS* F 0.0058 MSF 0.0071 a. Hotchkiss and Marmur noted that the percentage of cotransformation was higher than would be expected on a random basis. For example, the results show that 2.6% of the cells were transformed into M and 4% were transformed into S. If the M and S traits were inherited independently, the expected probability of cotransformation of M and S (M S) would be o.026 x 0.04 = 0.001, or 0.1%. However, they observed o.41% M S cotransformants, four times more than they expected. What accounts for the relatively high frequency of cotransformation of the traits they observed? b. On the basis of the results, what conclusion can you draw about the order of the M, S, and F genes on the bacterial chromosome?
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