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3 years ago

During genetic engineering, how do Restriction enzymes know what base pairs to act on?

Engineering

1 answer:

MaRussiya [10]3 years ago

6 0

Answer:

The correct/closest option is b

Explanation:

Restriction enzymes are enzymes (endonucleases) that cut short DNA strands at specific sites. Hence, each restriction enzyme has it's own specific site (between two bases) it cuts at. There are two types of end that can be produced by this cut; the blunt end and the sticky end.

A restriction enzyme recognizes (palindromic sequence) and cut in it's own specific end.

For example, if a restriction enzyme cuts between a guanine (G) and an adenine (A), and it cuts a palindromic double stranded DNA in the manner below, it produces a sticky end.

G║AATTC

CTTAA║G

And if a restriction enzyme cuts between guanine (G) and cytosine (C) in the manner below, it produces a blunt end.

GGG║CCC

CCC║GGG

Hence, from the question, restriction enzymes (although chosen by the scientist based on desired sequence to be cut) recognize the sticky or blunt ends itself.

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Determine if the following errors are systematic or random. Justify your response. (a) Effect of temperature on the circuitry of

k0ka [10]

Answer:

a) temperature: random error

b) parallax: systematic error

c) using incorrect value: systematic error

Explanation:

Systematic errors are associated with faulty calibration or reading of the equipments used and they could be avoided refining your method.

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3 years ago

Whose job is it to ensure that a group stays focused and on schedule?

Sedbober [7]

Answer:

the leader or organizer. you should have assigned jobs, so one of them should be in charge. if you did not assign jobs, assign them now, and one person has a job, and they must be held accountable. if they cannot do their job, someone might have to take over, but then you tell your prof. that they could not do their part, so hopefully you will get the credit you deserve and they will not.

Explanation:

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3 years ago

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A material has the following properties: Sut = 275 MPa and n = 0.40. Calculate its strength coefficient, K.

Tems11 [23]

Answer:

The strength coefficient is K = 591.87 MPa

Explanation:

We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

$S_{ut}=K \left(\cfrac ne \right)^n$

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.

Solving for strength coefficient

From the strain hardening equation we can solve for K

$K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}$

And we can replace values

$K = \cfrac{275}{\left(\cfrac {0.4}e \right)^{0.4}}\\K=591.87$

Thus we get that the strength coefficient is K = 591.87 MPa

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4 years ago

Pls follow me in brainly

Sergio [31]

Answer:

sure

Add on:

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3 years ago

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6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right

dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0

3 years ago