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Masteriza [31]
3 years ago
15

In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the

antibiotic ampicillin (ampr). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized above. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria.Plates I and III were included in the experimental design in order toa. demonstrate that the E. coli cultures were viable.b. demonstrate that the plasmid can lose its ampr gene.c. demonstrate that the plasmid is needed for E. coli growth.d. prepare the E. coli for transformation.
Biology
1 answer:
brilliants [131]3 years ago
7 0

Answer:

The correct answer is option A.  demonstrates that the E. coli cultures were viable.

Explanation:

In this lab experiment or culture,  Plate I and plate III demonstrate, that E.coli bacteria can grow both in the presence and absence of plasmid DNA if ampicillin is not there.

So, which means that plasmid DNA is not required for the growth of E.coli in absence of ampicillin. The presence of growth in wild type (plate 1) and a plasmid containing bacteria (plate 3)  in media without ampicillin shows that bacteria are viable in nature.

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A corn plant known to be heterozygous at three loci is testcrossed. The progeny phenotypes and numbers are as follows:+ + + 455a
astraxan [27]

Answer:

The progeny phenotypes and numbers are as follows:

+ + + 455-------- parental

a b c 470-------- parental

+ b c 35----------- recombinant

a + + 33---------- recombinant

+ + c 37------------ recombinant

a b + 35--------------- recombinant

+ b + 460---------- parental

a + c 475----------- parental

Total 2,000

+ + + 455-------- parental

a b c 470-------- parental

+ b + 460---------- parental

a + c 475----------- parental

It is two point testcross

So the parental ++ (455+460); ac (470+475)------------------++/ac

Gene order is ------------- ++/ac

+ b c 35----------- recombinant

a + + 33---------- recombinant

+ + c 37------------ recombinant

a b + 35--------------- recombinant

sum of recombinants--------------140

Frequency of recombintion= recombinants/total x100

Map distances between a-c = (140/2000) x 100 = 7 mu

a------------------7 mu -----------------c

Explanation:

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Answer:

Sympathetic Nervous System

Explanation:

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