Pseudomonas sp. has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is 1.3 g/l (which is unusually high), and cell yield on acetate is 0.46 g cell/g acetate. If we operate a chemostat on a feed stream containing 38 g/l acetate, find the Cell concentration when the dilution rate is one-half of the maximum?

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#Question id: 14284

#SCPH05 I Biotechnology

The specific productivity (qp) of an enzyme production Is fitted linearly with specific growth rate p of a fungal organism according to the equation The estimated values of constants alpha and beta are 0.0006 and 25 respectively. The enzyme production kinetics is 

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#Question id: 23165

#SCPH06 I Botany

For visualizing two proteins which are localized in nucleus through fluorescence microscopy which two sets of dyes can be best used for clear visual?

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#Question id: 834

#SCPH01 Biochemistry

How the proteosome inhibitors  are potentially used in treatment of cancerous cells ?

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#Question id: 625

#SCPH05 I Biotechnology

For enzymes in which the slowest (rate-limiting) step is the reaction Km becomes equivalent to:

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#Question id: 13094

#SCPH28 | Zoology

You are studying a specific gene in yeast, and you want to express that yeast gene in E. coli. Your task is to design a strategy to insert the yeast gene into the bacterial plasmid. Below is a map of the area of the yeast genome surrounding the gene in which you are interested.

 

The distance between each tick mark placed on the line above is 100 bases in length

Below are the enzymes you can use, with their specific cut sites shown 5’-XXXXXX-3’ 3’-XXXXXX-5’

 

The plasmid is 5,000 bases long and the two farthest restriction enzyme sites are 200 bases apart. The plasmid has an ampicillin resistance gene somewhere on the plasmid distal from the restriction cut sites.

                                     

You transform your ligation planned in which two restriction enzymes would you use to design a way to get the insert into the vector if you had to use two enzymes simultaneously, into bacteria and plate the bacteria on Petri plates containing ampicillin. (You actually transform six different ligation mixtures, which are described below, into six different populations of cells, and plate each transformation onto a different plate, because you want to do all of the correct controls.) The next day you come in to lab to look at how many colonies of bacteria are on each plate. You are really excited, because the number of colonies you see on each plate tells you that the entire procedure worked! Which of the three following patterns of number of colonies did you see in order to conclude that you had a successful transformation?

In this table, DV = digested vector. DYG = digested yeast genome.

 

a) Pattern-1, DV only + Ligase→No colonies b/c you have digested with 2 different restriction enzymes that can’t ligate together 

b) Pattern-2, DYG only + Ligase→ No colonies because all you transformed is the digested, linear yeast DNA.

c) Pattern-3, Water + Ligase→ No plasmid with the ampicillin resistance gene (or any DNA) was transformed into the bacteria and so it won’t grow in the presence of ampicillin.

d)Pattern-3, DV + DYG + Ligase→Colonies. The plasmid and yeast gene can ligate together to form a functional plasmid that will express the ampicillin resistance gene.

e) Pattern-1 and 2 only, DV + DYG (No Ligase) →No colonies because, although you have both digested plasmid and a digested yeast gene with complementary sticky ends

Which of the following statements about these ligations and their pattern is correct?