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sweet-ann [11.9K]
3 years ago
11

How much DNA can be extracted from an onion

Chemistry
1 answer:
konstantin123 [22]3 years ago
4 0
Since the onion (Allium cepa) is a diploid organism having a haploid genome size of 15.9 Gb, it has 4.9x as much DNA as does a human genome (3.2 Gb). Other species in the genus Allium vary hugely in DNA content without changing their ploidy.
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Examples of type of matter that_ through the environment are carbon, nitrogen, and water​
oee [108]

Answer:

examples of type of matter that <u>c</u><u>i</u><u>r</u><u>c</u><u>u</u><u>l</u><u>a</u><u>t</u><u>e</u><u> </u>through the environment are carbon, nitrogen, and water

8 0
3 years ago
Which of the following best decribes the fluoride content of bottled water? unpredictable, high, low, moderate
Anton [14]
The fluoride content of bottled water is unpredictable.  
8 0
3 years ago
Why is a cell membrane like a security guard because.......<br> Why?????
White raven [17]

The cell membrane allows certain items into a cell and blocks others from entering, just like a security guard protects a real factory.

5 0
3 years ago
I'm kinda stumped here :/
maria [59]

Answer:

To prepare 20,0 mL of the liquid mixture you should mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃

Explanation:

Here you have two variables: The volume of both CHCl₃ (X) and CHBr₃ (Y). To find these two variables you must have, at least, two equations.

You know total volume is 20,0 mL. Thus:

X + Y = 20,0 mL (1)

The other equation  is:

\frac{X}{20,0mL\\} × 1,492 g/mL + \frac{Y}{20,0 mL} × 2,890 g/mL = 1,82 g/mL (2)

If you replace (1) in (2):

\frac{X}{20,0mL\\} × 1,492 g/mL + \frac{20,0 mL - X}{20,0 mL} × 2,890 g/mL = 1,82 g/mL

Solving:

X = 15,3 mL

Thus, using (1):

20,0 mL - 15,3 mL = Y = 4,7 mL

Thus, to prepare 20,0 mL of the liquid mixture you must mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃.

I hope it helps!

4 0
3 years ago
Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydrat
Afina-wow [57]

Explanation:

According to the given data, the turnover number can be calculated as follows.

      Turnover number = K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}

     V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}

Therefore, moles of CO_{2} hydrolyzed is as follows.

Moles of CO_{2} hydrolyzed = \frac{Mass of CO_{2}}{Molar mass of CO_{2}}

                 = \frac{0.30}{44}

                 = 0.00682 moles

Now, moles of CO_{2} hydolyzed per second is calculated as follows.

Moles of CO_{2} hydolyzed per second = \frac{0.00682}{60}

             = 1.137 \times 10^{-4} moles/second = V_{max}

And,

Moles of enzyme = \frac{Mass}{\text{Molar mass}}

                       = \frac{10.0 \mu g}{30000}

                       = 3.33 \times 10^{-10} moles

Therefore, the value of K_{cat} is as follows.

    K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}

               = 0.3411 \times 10^{6} per second

               = 0.3411 \times 60 \times 10^{6} per minute

               = 20.466 \times 10^{6} per minute

Thus, we can conclude that the turnover number (K_{cat}) of carbonic anhydrase (in units of min^{-1}) is 20.466 \times 10^{6} per minute.

6 0
4 years ago
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