While working for Dr. Strap you isolate a mutant of E. coli; this time defective
ID: 218609 • Letter: W
Question
While working for Dr. Strap you isolate a mutant of E. coli; this time defective in aromatic amino acid biosynthesis (aro-). You decide to map this relative to other markers by crossing an Hfr strain which is aro+ pyr+ cmlR strs to F- aro- pyr- cmls strR.
Note: cml is for chloramphenicol resistance and pyr is pyrimidine biosynthesis.
To kill the Hfr strain you plate the exconjugants on streptomycin. From these plates you isolate 300 colonies and then test them on different media.
Media ---> Number of colonies
Cml only ---> 263
Cml + Pyr ---> 264
Cml + Aro ---> 290
Cml + Pyr + Aro ---> 300
a) Indicate the number of individuals that are present for each of the genotypes recovered.
b) What are the recombinant frequencies?
Explanation / Answer
Strain giver (Hfr) is: aro+ pyr+ cmlR strs
Strain receptor (f-) is : aro- pyr- cmls strR
Given enoughf time for conjugation, we kill the parents with streptomycin. 300 will be our total, but in order to know which genotype they have we need to grow them in media containing different nutrients. This happens because the conjugation is not complete, some genes may have passed and some may not.
In the first plate containg only clm, we are seeing the colonies that incorporated aro and pyr as well as cmlR. This genotype is aro+ pyr+ cmlR strR and has 263 colonies. recombinant frequency? well, we do 263/300 = 0.87 .
Second media has clm + pyr, so, we are seeing those genotypes that can grow in this (only aro+). But pyr could be both pyr + or - . Anyways this genotype has 264 colonies. RF: 264/300 = 0.88.
Third grows colonies that are pyr+. 290 colonies. RF: 290/300 = 0.97
And last one contains all the colonies that only got the cmlR from the Hfr strain: 300.
Remember that higher frequencies means genes are closer. (In this case cml and pyr are closer than cml and aro).
Hope this helps!
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