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marshall27 [118]
1 year ago
7

The process by which the gene in the nucleotide suddenly changes its position is called.

Physics
1 answer:
Ksenya-84 [330]1 year ago
8 0

Mutation is the process by which the gene in the nucleotide suddenly changes its position

Any change in the structure of a gen or in the usual DNA sequence of a cell is called mutation. It can be caused by mistakes during cell division when the DNA is copied or by the exposure to DNA damaging agents in the environment like UV light or cigarette smoke, and may be transmitted to subsequent generations.

A mutation is a permanent alteration of the nucleotide sequence of an organism, virus or other genetic elements.

There are three types of mutations:

  • Base substitutions
  • Deletions
  • Insertions

<h3>What is DNA?</h3>

The Deoxyribonucleic acid, also called DNA, is the hereditary material that carries genetic information for the development and functioning of humans and almost all other organisms. DNA is made of nucleotides.

Learn more about DNA at brainly.com/question/16099437

#SPJ4

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Which of the following expresses the answer to
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Answer: 17.95

Explanation:

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4 0
3 years ago
A 0.0427 kg racquet-ball is moving
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Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball (m_1) = 0.00427 kg

Velocity of racket-ball before collision (v_{1i}) = 22.3 m/s

Velocity of racket-ball after collision with box (v_{1f}) = -11.5 m/s

[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision v_{2i} = 0 m/s

<em>[Since the box is stationary, so velocity is taken zero]</em>

Velocity of box moving forward after collision v_{2f}= 1.53 m/s

To find the mas of the box m_2.

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

p_i=p_f

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

We can plugin the given value to find m_2

(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)

0.9522+0=-0.4911+1.53m_2

Adding both sides by 0.4911

0.9522+0.4911=-0.4911+0.4911+1.53m_2

1.4433=1.53m_2

Dividing both sides by 1.53.

\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}

0.9433=m_2

∴ m_2=0.9433 kg

Mass of the box = 0.9433 kg (Answer)

4 0
3 years ago
What is the concentration of H+ ions at a pH = 2? mol/L What is the concentration of OH– ions at a pH = 2? mol/L What is the rat
densk [106]

Answer:

In Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 2. Then divide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C.

What is the concentration of H+ ions at a pH = 2?

0.01   mol/L  

What is the concentration of OH– ions at a pH = 2?

0.000000000001   mol/L  

What is the ratio of H+ ions to OH– ions at a pH = 2?

10,000,000,000 : 1

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I LITERALLY spent 40 MINUTES trying to figure out this question, so please, use my VERY CORRECT answers!

I hope this helps!

4 0
3 years ago
The law of suggests that the orbit of planets is not circular but .
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One of Kepler's laws is that the orbits of planets are elliptical. It's not a suggestion. BTW, circles are ellipses too, but so special that their likelihood is close to zero.
3 0
3 years ago
Read 2 more answers
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
3 years ago
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