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max2010maxim [7]
3 years ago
5

Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /m

l final concentration) liquid media. [Include things like how many grams of each component that you use, how much antibiotic (in ml) to add (stock solutions – 100 mg/ml ampicillin, 25 mg/ml kanamycin), and in what type of container you perform the sterilization step.] Show your calculations.
Chemistry
1 answer:
mario62 [17]3 years ago
5 0

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

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Mumz [18]

Answer:

Option 2= Glucose

Explanation:

Cell membrane is made up of two phospholipid layers and each contain phosphate head and fatty acid or lipid tails. the head is present between the outer and inner boundaries and tail is present in between. The small non- polar molecules can pass the membrane through simple diffusion. This lipid tail restrict the passage of polar molecules including water soluble substances like glucose. However, transmembranes are present that allow the molecules to inter that are blocked by the tails.

Facilitated diffusion:

it is a type of diffusion in which caries protein without using the cellular energy shuttle the molecules to the cell membrane. Glucose is bind on the carrier protein ,change the shape and transport it from one to another side of membrane. In order to absorb the glucose red blood cells use this kind of diffusion.

Primary active transport:

The cells that are present along small intestine use this type of transport to pump the glucose inside the cell. The primary active transport require energy to transport the glucose inside.

Secondary active transport:

It is another method of transport of glucose into the cell. This method can not use ATP but it is based on concentration gradient of the sodium that provide electro chemical energy for the glucose transport.

5 0
3 years ago
The table below gives the equilibrium concentrations for this reaction at a certain temperature : N2(g) + O2(g) 2NO(g) what is t
makkiz [27]
D. 100 I hope this help :)
4 0
2 years ago
Use the unbalanced equation NH3+O2=NO+H2O to find the mole ratio between NO and H2O
Leona [35]

Answer:

For every 4 moles of NO created, 6 moles of H2O are created so the ratio is 4:6

Explanation:

You just need to balance the equation.

NH3 + O2 -> NO + H2O

1. I started with hydrogen; there's 3 on the left and 2 on the right. Multiply them together to find a number they both go into (3×2=6, but in this case 6 hydrogen on each side does not work so I doubled it so there is 12 hydrogen on each side).

This will bring you to this:

4NH3 + O2 -> NO + 6H2O

2. Now get equal amounts of nitrogen on each side. There's 4 nitrogen on the left side, and 1 on the right. Multiply the right by 4. Then you will have this:

4NH3 + O2 -> 4NO + 6H2O

3. Last thing you need to do is have the same amount of oxygen on both sides. On the left you have 2 and on the right you have 10. Get the left to 10 by multiplying it by 5.

Balanced: 4NH3 + 5O2 -> 4NO + 6H2O

In word form, for every reaction between 4 moles of ammonia and 5 moles of oxygen, 4 moles of nitric oxide and 6 moles of water will be created.

I hope this helps!

5 0
3 years ago
What is another name for a coefficient? this is for chemistry
krok68 [10]

Answer:

synergistic

Explanation:

Synergistic means: relating to the interaction or cooperation of two or more organizations, substances, or other agents to produce a combined effect greater than the sum of their separate effects

7 0
2 years ago
Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0
2 years ago
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