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3 years ago

NAme and size of ecoli clonning vector?

Engineering

1 answer:

elena-s [515]3 years ago

4 0

Answer:

pBR322 DNA is used in plasmid cloing vector in E.coli

molecule is a double-stranded circle 4,361 base pairs in lenght

pBR322 contains the genes for resistance to ampicillin and tetracycline, can be amplified with chloramphenicol

molecular weight is 2.38 x $10^{6}$ daltons

Explanation:

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A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pull

Nezavi [6.7K]

Find the complete solution in the given attachments

5 0

3 years ago

A continuous fiber-reinforced composite is to be produced by dispersing 60 vol% carbon fibers in a polycarbonate matrix. if the

devlian [24]

Answer:

Explanation:

A continuous fiber-reinforced composite is to be produced by dispersing 60 vol% carbon fibers in a polycarbonate matrix. if the stress in the polycarbonate matrix when the carbon fibers fail is 45 mpa, the longitudinal elastic modulus of the composite using rule of mixture

5 0

2 years ago

Two wafer sizes are to be compared: a 156-mm wafer with a processable area = 150mm diameter circle and a 312-mm wafer with a pro

Alja [10]

Answer:

a) 100%

b) 300%

c) 301 %

Explanation:

The first wafer has a diameter of 150 mm.

The second wafer has a diameter of 300 mm.

The second wafer has an increase in diameter respect of the first of:

((300 / 150) - 1) * 100 = 100%

The first wafers has a processable area of:

A1 = π/4 * D1^2

The scond wafer has a processable area of:

A2 = π/4 * D2^2

The seconf wafer has a increase in area respect of the first of:

(A2/A1 * - 1) * 100

((π/4 * D2^2) / (π/4 * D1^2) - 1) * 100

((D2^2) / (D1^2) - 1) * 100

((300^2) / (150^2) - 1) * 100 = 300%

The area of a chip is

Ac = Lc^2

So the chips that can be made from the first wafer are:

C1 = A1 / Ac

C1 = (π/4 * D1^2) / Lc^2

C1 = (π/4 * 150^2) / 10^2 = 176.7

Rounded down to 176

The chips that can be made from the second wafer are:

C2 = A2 / Ac

C2 = (π/4 * D2^2) / Lc^2

C2 = (π/4 * 300^2) / 10^2 = 706.8

Rounded down to 706

The second wafer has an increase of chips that can be made from it respect of the first wafer of:

(C2 / C1 - 1) * 100

(706 / 176 - 1) *100 = 301%

8 0

3 years ago

A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected

Pavlova-9 [17]

Answer:

the required diameter of the rod is 9.77 mm

Explanation:

Given:

Length = 1.5 m

Tension(P) = 3 kN = 3 × 10³ N

Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa

E = 70 GPa = 70 × 10⁹ Pa

δ = 1 mm = 1 × 10⁻³ m

The required diameter(d) = ?

a) for stress

The stress equation is given by:

$S = \frac{P}{A}$

A is the area = πd²/4 = (3.14 × d²)/4

$S = \frac{P}{(\frac{3.14*d^{2} }{4}) }$

$S = \frac{4P}{{3.14*d^{2} } }$

$3.14*S*{d^{2}} = {4P}$

${d^{2}} =\frac{4P}{3.14*S}$

$d= \sqrt{\frac{4P}{3.14*S} }$

Substituting the values, we get

$d= \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }$

$d= \sqrt{\frac{12000 }{125600000 } }$

$d= \sqrt{9.55*10^{-5} }$

d = (9.77 × 10⁻³) m

d = 9.77 mm

b) for deformation

δ = (P×L) / (A×E)

A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063

d² = (4 × A) / π = (0.000063 × 4) / 3.14

d² = 0.0000819

d = 9.05 × 10⁻³ m = 9.05 mm

We use the larger value of diameter = 9.77 mm

3 0

3 years ago

What does the plasma membrane do

aniked [119]

<u>Answer:</u>

The plasma membrane encloses specific structures.

<u>Explanation: </u>

plasma membrane is also called as cell membrane. Plasma membrane only allows the particle to get in and pass out of the cell by osmosis and diffusion method from the outside environment. It is responsible for the molecular traffic inside the cell.

It helps in maintaining the shape of the cell. It has many proteins in it. Therefore cell membrane are responsible for having specific structures.

7 0

3 years ago