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3 years ago

Why the recombinant DNA is reintroduced into the bacterium?

1 answer:

8 0

Bacterial cells can pick up the DNA through the process of transformation. Lambda phages are used by eliminating the middle of its liner genome and adding the foreign DNA in the created space. The phage is then introduced into the bacterial cell where it replicated itself via the lysogenic cycle.

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what bodyy system Filters waste materials (esp. nitrogenous wastes) from blood; maintain water, salt, acid/base, & ion balan

Arada [10]

Please don’t listen to those people who always put links those are scams. The answer is the kidney, ureters, bladder, and urethra. This system filters your blood,removing waste and excess water. Your welcome!

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2 years ago

What are the differences between sand and potting soil? Are they both mixtures? How do you know?

katen-ka-za [31]

What class is this for because it depends

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3 years ago

Calculate the number of moles of C2H6 in 6.29×1023 molecules of C2H6.

Kipish [7]

1.05moles

Explanation:

Given parameters:

Number of molecules of C₂H₆ = 6.29 x 10²³molecules

Unknown:

Number of moles = ?

Solution:

The mole is the amount of substances that contains Avogadro's number of particles i.e 6.02 x 10²³

To find the number of moles:

number of moles = $\frac{number of particles}{Avogadro's number}$

number of moles = $\frac{6.29 x 10^{23} }{6.02 x 10^{23} }$

number of moles = 1.05moles

Learn more:

moles brainly.com/question/1841136

#learnwithBrainly

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2 years ago

What is the enthalpy change (in kj) of a chemical reaction that raises the temperature of 250.0 ml of solution having a density

Pepsi [2]

<span>250 ml * 1.25 g/ml * 3.74 j/g-K * 9.2 K = 10.752 kJ Pretty much, all you need to do here is multiply all of these out to get your final answer. Not all questions are this easy, but this is certainly one of them.</span>

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3 years ago

Read 2 more answers

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

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2 years ago