One codon has the instructions to make

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Answer 1

Answer:

1 protein

Explanation:


Related Questions

in chemical reactions, bonds blank in reactants.

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in chemical reactions, bonds break in reactants

The correct sequence for the general adaptation syndrome is:_________
a) Exhaustion phase, alarm phase and resistance phase
b) Alarm phase, resistance phase and exhaustion phase
c) Resistance phase, exhaustion phase and alarm phase
d) Alarm phase, exhaustion phase and resistance phase

Answers

Answer:

The correct sequence for the general adaptation syndrome is: b) Alarm phase, resistance phase and exhaustion phase

Explanation:

The general adaptation syndrome describes what happens to our body when it is under stress.

The first phase is the Alarm phase. In this stage, the body activates the fight or flight response, causing an increase in adrenaline and cortisol, blood pressure, respiratory frequency, amongst other things that prepare us to react in a situation that we perceive as dangerous.

The second phase is the resistance one. During this phase, the body undergoes a different process that adapts it to manage the stress, with an elevated amount of cortisol and blood pressure.

The third phase is the exhaustion phase. If the stress levels continue to be high, the body passes to this stage. The body is exhausted and cannot keep dealing with stress, which causes mental and physical problems such as anxiety, depression, fatigue, stress-related illnesses, among other symptoms.

What are the properties of water that you can observe when transpiration takes place?

Answers

Answer:

When water evaporates through the leaves, a pull is created through the xylem, and water moves back to the leaves. This is known as the transpiration pull.

dy÷dx=(x-1)(x+3) at x=2​

Answers

Answer:

[tex]\dfrac{dy}{dx}=\dfrac{4}{25}[/tex]

Explanation:

The given expression is :

[tex]y=\dfrac{(x-1)}{(x+3)}[/tex]

We need to find dy/dx at x = 2

[tex]\dfrac{dy}{dx}=\dfrac{d}{dx}(\dfrac{x-1}{x+3})\\\\=\dfrac{(x+3)\dfrac{d}{dx}(x-1)-(x-1)\dfrac{d}{dx}(x+3)}{(x+3)^2}\\\\=\dfrac{x+3-(x-1)}{(x+3)^2}\\\\=\dfrac{x+3-x+1}{(x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{4}{(x+3)^2}[/tex]

Put x = 2 in above expression

[tex]\dfrac{dy}{dx}|x=2=\dfrac{4}{(2+3)^2}\\\\=\dfrac{4}{25}[/tex]

Hence, the value at dy/dx is [tex]\dfrac{4}{25}[/tex]

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