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jeka94
3 years ago
5

Why are dichotomous keys useful in classifying a new species

Biology
1 answer:
finlep [7]3 years ago
7 0
Helps build two main Characteristics out of two choices to classify and unidentified organism
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The concept of aging as a result of cellular duplication errors is based on the fact that the body's ability to make new cells t
babunello [35]

Answer:

Through Mutation

Explanation:

Since our cells are able to produce exact copies of themselves. Thus, until unless some mutation has occurred and affected the ability of the cells to produce exactly similar duplicates copies of the parent cell, there is no way that an individual gets aged.  

Since, mutation affects the mitochondrial DNA it is pretty sure that a mutated cell has affected the longevity of a normal cell and then that mutated cell has produced million other mutated aged cell by replicating itself

8 0
3 years ago
Read 2 more answers
Why does the DNA need to be extracted from a cell before it can be analyzed?
Arada [10]

Answer:

<em>To study the genetic causes of disease and for the development of diagnostics and drugs. And detecting bacteria and viruses in the environment and for determining paternity.</em>

<em></em>

5 0
2 years ago
The concentration of N2 in blood at 37 °C (body temperature) and atmospheric pressure (partial pressure of N2 = 0.80 atm) is 0.0
devlian [24]

Answer:

volume  = 0.285 L

Explanation:

Henry constant is given as

Henry constant  =  \frac{C}{P}

Where C  is concentration

P is atmospheric pressure

Henry\ constant = \frac{0.00056}{0.80} = 0.0007 M/atm

when atmospheric pressure 4 atm

solubility = K_H\times P

              = 0.0007 \times 4 = 0.0028 M

In 5 litere blood , moles of N_2 = Molarity \times volume

                                                     = 0.0028 \times 5 = 0.014

At surface moles  = 0.0056 \times 5 = 0.0028

Moles of N_2 release = 0.014 - 0.0028 = 0.0112

Mass of N_2 = 0.0112 \times 28 = 0.314 g

T = 37 + 273 = 310 K

VOLUME = \frac{nRT}{P}

= \frac{0.0112\times 0.0821\times 310}{1}

volume  = 0.285 L

6 0
3 years ago
Use the four population pyramids above to answer the question. Which country's population is
artcher [175]
I think it would be D hope this helps
8 0
1 year ago
Which of the following situations does diffusion not help, in our bodies?A. Oxygen entering the bloodstream.B. Food entering the
lubasha [3.4K]

Answer:

Glucose entering the intestines from the villus.This is by active transport.The latter is defined as the movement of molecules and  solutes from the region of lower concentration to the region of higher concentration against the concentration gradient.

Explanation:

All the above options depends on the diffusion gradients from one  higher region to another.e.g oxygen passing from high concentration from outside to lower concentration in blood stream,like wise Co2 from higher concentration from inside to outside,with passive diffusion lipids and water enter the small intestine(note amino acid and glucose,and fructose are different).

Active transport( through sodium dependent transporter) transports glucose to the intestine  from the villus.This is movements against concentration gradient because,glucose molecule are concentrated in intestine,and needed by the cells, in the body.Therefore they must be transported as a fast rate inn the blood .

However the absorption by the villi is very slow,therefore active transport is needed to move these against the concentration gradient into the intestine.And finally the blood.

If this were to be moved by  passive diffusion,the high concentration in the intestine will force this back into the villi

Therefore extra energy is needed by active transport for the glucose molecules to enter the intestine from the villus.

6 0
3 years ago
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