You created a serial dilution of 10-1 to 10-6 of a sample containing phages. Assume that you properly diluted your samples so that you had 10 times fewer phages in each test tube before adding your bacterial growth and then pouring over your agar for incubation.
What quantitative difference would expect in phage plaque formation after bacterial growth following incubation?

Explanation:

According to the question the difference between the dilution of every two successive tubes is 10. The second tube is diluted by a factor of 10 as compared to the first tube, the third tube is diluted by a factor of 10 as compared to the second tube and so on.

It means the amount of phages in two successive tubes is also different by a factor of 10. The amount of phages in the second tube is 10 times lower than the amount of phages in the first tube, the amount of phages in the third tube is 10 times lower than the amount of phages in the second tube and so on.

So we can expect that if we plate these phages into a bacterial lawn and they lead to the formation of plaques, then the number of plaques between two successive plates will be different by a factor of 10. The number of plaques in the second plate will be 10 times lower as compared to the first plate and the number of plaques will be 10 times lower in the the third plate as compared to the second plate and so on.

answer: dna

because all cells have genes