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tankabanditka [31]
3 years ago
9

What is FALSE about E coli and the lac operon? A. The lac Y and lac Z genes are turned off when lactose is present.B. The lac Y

and lac Z genes are turned off when glucose is present. C. Lactose is a disaccharide which requires more energy to eat than the monosaccharide glucose. D. The lac Y gene encodes the permease protein. E. B-galactosidase breaks the covalent bond between the glucose and galactose of the lactose molecule.
Physics
1 answer:
rjkz [21]3 years ago
6 0

Answer: The lac Y and lac Z genes are turned off when lactose is present. Is a false statement about E coli and the lac operon. Therefore the correct option is A.

Explanation:

Escherichia coli is an enteric bacteria that makes use of lactose operon (or lac operon) for its lactose metabolism. The lac operon is made up of three genes which includes lacZ, lacY and Lac A. These genes are expressed only when lactose is present and glucose is absent. Also the lac repressor acts as a lactose sensor while the catabolite activation protein (CAP) acts as a glucose sensor. They help in turning on and off the genes with respect to lactose and glucose levels.

The lac Y and lac Z genes are turned off (not expressed) when glucose is present and lactose absent, but are expressed when glucose is absent and lactose present. Therefore option A is a false statement.

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Answer:

His launching angle was 14.72°

Explanation:

Please, see the figure for a graphic representation of the problem.

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v = (vx, vy)

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In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:

sin angle = v0y / v0  then: v0y = v0 * sin angle

In the same way for vx:

vx = v0 * cos angle

Using the equation for velocity in the x-axis we can find the equation for the horizontal position:

dx / dt = v0 * cos angle

dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)

x - x0 = v0 t cos angle

x = x0 + v0 t cos angle

For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:

dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)

vy -v0y = g t

vy = v0y + g t

vy = v0 * sin angle + g t

The position will be:

dy/dt = v0 * sin angle + g t

dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)

y = y0 + v0 t sin angle + 1/2 g t²

The displacement vector at a time "t" will be:

r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)

If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)

r = (x0 + v0 t cos angle, 0)

The module of this vector will be the the total displacement (65 m)

module of r = \sqrt{(x0 + v0* t* cos angle)^{2} }  

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65 m / v0 cos angle = t

Then, using the equation for the position in the y-axis:

y = y0 + v0 t sin angle + 1/2 g t²

0 =  y0 + v0 t sin angle + 1/2 g t²

replacing t =  65 m / v0 cos angle and y0 = 0

0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²  

cancelating v0:

0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)

-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)  

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-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²

sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²

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sin (2* angle) /2 = 0.25

sin (2* angle) = 0.49

2 * angle = 29.44

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Answer:

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