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kherson [118]
3 years ago
13

In the gene TATTCATTGTTA—TGATTT—ATTCG, CATTGTTA encodes for pepsin, a digestive enzyme. The rest of the sequence doesn’t code fo

r any protein. Which sequence contains a mutation that will affect the formation of pepsin?
TATTCATTCATTA—TGATTT—ATTCG



TATTCATTGTTA—TGACTTT—ATTCG



TATTCATTGTTA—TGATTT—ATTGGCG



TATTCATTGTTA—TGAT—ATTCG



TGCATTCATTGTTA—TGATTT—ATTCG
Physics
1 answer:
MatroZZZ [7]3 years ago
8 0

Answer:

TATTCATTCATTA—TGATTT—ATTCG

Explanation:

A mutation is a permanent change in the nucleotide sequence of DNA. A mutation occurs during replication or recombination. It may be due to base substitutions, deletions and insertions. As per the question, DNA segment forms encodes for the enzyme pepsin is CATTGTTA.

Option TATTCATTCATTA—TGATTT—ATTCG is the correct answer. In the DNA segment which encodes pepsin, a purine base (G) guanine is replaced by another purine (A) adenine. This type of mutation is called a transition type point mutation.

Due to base substitution, the mutated segment CATTCATTA will nor encode pepsin.

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The possible magnitude for the force of static friction on the stationary cart is 72.1 N.

The given parameters:

  • <em>Applied force on the cart, F = 72.1 N</em>

<em />

Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.

F = ma

Static frictional force is the force resisting the motion of an object at rest.

\Sigma F = 0\\\\F -F_f = 0

where;

F_f is the frictional force

F= F_f \\\\72.1 = F_f\\\\F_f = 72.1\  N

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.

Learn more about Newton's second law of motion: brainly.com/question/25307325

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2 years ago
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4 0
3 years ago
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jek_recluse [69]

Answer:

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Explanation:

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8 0
3 years ago
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A policeman is chasing a criminal across a rooftop at 10 m/s. He decides to jump to the next building which is 2 meters across f
valina [46]

Answer:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, u_{x} = 10 m/s

The distance between the buildings, d_{x} = 2.0 m

The building is lower by a height, h = 2.5 m

Solution:

Now,

When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.

Thus

From the second eqn of motion, we can write:

h = ut + \frac{1}{2}gt^{2}

h = \frac{1}{2}gt^{2}

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Now,

When the policeman was chasing across:

d_{x} = u_{x}t + \frac{1}{2}gt^{2}

d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m

The distance they will meet at:

9.57 - 2.0 = 7.57 m

   

8 0
3 years ago
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7 0
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