Given Hfr: his+ phe+ (amp = ampicillin; str = streptomycin; r = resistant; s = s

Question

Given Hfr: his+ phe+ (amp = ampicillin; str = streptomycin; r = resistant; s = susceptible) F: his phe amp^5 sir^5 Hfr and F^- were crossed and the transfer of genes happened in the order given with his^+ entering first. Plating of colonies happened on four types of media containing str, amp, histidine, and phenylalanine. Results were as follows: Medial: 415 strains grew on media with ampicillin Media2: 400 strains grew on media with amp + his Media3: 477 strains grew on media with amp + phe Media4: 500 strains grew on media with amp + his + phe What was the purpose of growing the strains on streptomycin media? What do the surviving colonies represent? Based on the media growths, give the different surviving genotypes and the number of individuals in each. Draw a map of the three genes (his, phe, and amp) based on the frequencies of recombination.

Explanation / Answer

A . The purpose of the bacteria growing on the media containing the streptomycin shows which colonies are streptomycin resistant and which are sensitive . The surviving colonies represent that they are having the streptomycin resistant gene and conjugation has occured between the Hfr strain and F plasmid .

B. Genotypes of the bacteria grown on different media ;

Ampicillin ; ampR

Ampicillin and histidine ; HIS+ ,AmpR ,

Ampicillin and Phe ; HIS +,pHE +AmpR

Ampicillin and His and phe ; AmpR ,His+ , Phe +

C. map of three genes ;

total number of colonioes ; 1792

colonies with amp and phe + total number of colonies with amp ,phe and his 477+500 , 977

colonies with amp and histidine + number of colonies with amp, phe ,his , 415 +500 ,915

Map distance between the amp and phe is 977/1792*100 which is equal to 54.2cM .

Map distance between the amp and histidine is 915/1792 *100 which is equal to 51.0cM.

So the distance betwen the HIstindine and phe is   400 +500 ,900 , 900/1792*100 equal to 50.2cM

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