Help in chemistry !!!!!!!

Plasmid DNA and a gene of interest are cut with the enzyme PpuMI. Write a possible sequence of bases for the sticky end of the g

nasty-shy [4]

Answer:

5' RG GWCCY 3'

3' YCCWG GR 5'

Explanation:

The enzyme PpuMI is a restriction endonuclease enzyme, it has a specific recognition site where it cut the DNA. The source of the enzyme is from an E. coli strain that carries the PpuMI gene from Pseudomonas putida (R. Morgan).

The enzyme PpuMI recognizes specific sequence with palindrome arrangement. It target the sequence 5' RGGWCCY 3'

target Sequence: 5' RGGWCCY 3'

3' YCCWGGR 5'

The enzyme cleavage point is at:

5' RG^GWCCY 3'

3' YCCWG^GR 5'

The product of the cleavage will give a sticky end Cleavage:

5' RG GWCCY 3'

3' YCCWG GR 5'

Note: R stands for purines (adenine and guanine). Y stands for pyrimidines (cytosine, thymine, and uracil). And W represents adenine or thymine.

5 0

2 years ago

A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atmatm at 20°C.Calculate the van't Hoff factor iii for AlCl

Alja [10]

Answer:

The actual Van't Hoff factor for AlCl3 is 3.20

Explanation:

Step 1: Data given

Molarity of AlCl3 = 0.050 M

osmotic pressure = 3.85 atm

Temperature = 20 °C

Step 2: Calculate the Van't Hoff factor

AlCl3(aq) → Al^3+(aq) + 3Cl^-(aq)

The theoretical value is 4 ( because 1 Al^3+ ion + 3 Cl- ions) BUT due to the interionic atractions the actual value will be less

Osmotic pressure depends on the molar concentration of the solute but not on its identity., and is calculated by:

π = i.M.R.T

⇒ with π = the osmotic pressure = 3.85 atm

⇒ with i = the van't Hoff factor

⇒ with M = the molar concentration of the solution = 0.050 M

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 20 °C = 293.15 Kelvin

i = π /(M*R*T )

i = (3.85) / (0.050*0.08206*293.15)

i = 3.20

The actual Van't Hoff factor is 3.20

6 0

2 years ago

When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C

UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

$[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M$

Equilibrium constant of the reaction :

$K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}$

K = 4

$CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)$

Concentration at eq'm:

0.24 M 0.24 M 0.48 M 0.48 M

After addition of 0.34 moles per liter of $CO_2$ and $H_2$ are added.

(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

$K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}$

$4=\frac{(0.48+x)^2}{(0.24+0.34)^2}$

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0

2 years ago

PLEAS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! DUE TONIGHT

Jobisdone [24]

Answer:

It is required answer.

Explanation:

Given that :

1. using balanced chemical equation:

ammonium acetate:

The balanced equation is:

NH₃ + H₂O ===> NH₄OH

when ammonia gas dissolves in water then we get the base in form of ammonium hydroxide.

When NH₄OH reacts with CH₃COOH then we get ammonium acetate and water

NH₄OH + CH₃COOH ===> [CH₃COO]- & NH₄+ & H₂O

So, we can say that,

when we are adding an acid and a base together then we get the product of H₂O and given elements.

2. addition of barium hydroxide to sulfuric acid:

the balanced equation is

H₂SO4+ Ba(OH)₂--> BaSO₄+ 2H₂O

when acid and base reacts together than we get barium sulphate and water

when sulfuric acid and barium hydroxide.

Hence, it is required answer.

8 0

2 years ago

When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro

Bas_tet [7]

<span>Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3. .0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>

7 0

2 years ago