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Question 11 Multiple choice Question The table below summarizes the results of a replica plating experiment where a mixture of 347 E aanstrains of different genotypes were tested for their ability to growth on a variety of media. Medium Number of colonies that grew Complete medium (contains all amino acids, 347 vitamins, and a variety of carbon sources) Minimal medium (glucose as a carbon 261 source) Minimal medium ampicillin (an antibiotic) 51 303 Minimal medium +methionine (an amino acid) 301 Minimal medium +arginine (an amino acid) Minimal medium +methionine+ampicillin 57 Minimal medium arginine-ampicillin 64 Reference: Ref 9-1 Based on the data presented in the table above, how many colonies are ang? Answer

Solution

1. From the given data of replica plating, we can find out the number of colonies which are met arg.

Colonies which are met arg, are auxotrophic mutants which lack the ability to synthesize both the amino acid methionine and arginine. Their wild type prototrophic counterparts are met⁺arg⁺ .

It is given that in complete media (containing all supplements and amino acids), 347 colonies grew.

In minimal media, containing only glucose, 261 colonies grew.

Therefore, 261 colonies are met⁺arg⁺

Subtracting these number of colonies from the colonies that grew in complete media, 347-261 = 86 colonies are either met, arg or met-arg

Now, the number of colonies which grow in minimal media + methionine = 303

hence,number of met colonies = 347-303 = 44

The number of colonies which grow in minimal media + arginine = 301

hence, number of arg colonies = 347 - 301 = 46

Therefore, the number of colonies which are met and arg are = 44 + 46

= 90

We had found that 86 colonies are either met, arg or met-arg

Therefore number of colonies are met-arg = 90-86 = 4

2. The gene order for the given strain is b+a+d+

The distance between d and b marker can be found out in a two-step process.

The distance between b and a (from b+a-d-) = (58/400) x100 = 14.5 m.u

The distance between a and d (from b+a-d+ adn b+a+d-) = (16/400) x 100

= 4 m.u

Therefore the distance between b and d = 14,5 + 4 = 18.5 m.u (approximate to 19)

3. Adding up the series of markers (the last marker being continued): Hfr 1...DAF + Hfr 2.....EBF +Hfr 3....ECD

= DAFBECD

representing this on a circular chromosome with marker A showing continuity would be: ADCEBFA