Pls help need it now as I don't know the answer

At constant temperature, when the volume of a gas is decreased, what happens to its pressure? Select one: a. It increases. b. It

natulia [17]

Answer:

The answer would be

a. it increases

3 0

3 years ago

Calnexin and calreticulin catalyze the removal of the final glucose residue from glycoproteins during the folding process. True

zloy xaker [14]

Answer:

Explanation:

A) False.

Glucosidase (not calnexin nor calreticulin) helps to remove glucose residue.

Both calnexin and calreticulin rather have an affinity for last glucose residue of misfolded protein (Only misfolded proteins are marked by glycosyltransferase by attaching glucose residue). They attach with misfolded protein and with the help of other proteins like ERp57 (a type of protein disulfide isomerase) and try to fold it properly. If protein is properly folded then glucosidase removes the glucose residue thereby releasing the properly folded protein from calnexin or calreticulin. and now protein is transported to the Golgi body. If folding is still not proper then the same cycle of glycosylation -binding of calnexin/calreticulin and effort to fold it properly is repeated.

B) True.

Transketolase is a key enzyme of the pentose phosphate pathway. It contains thiamine diphosphate (TPP) as a cofactor. it does transfer 2 carbon residue from a ketose to aldose. So, effectively it converts one ketose sugar to aldose with 2 carbonless and aldose to ketose with 2 carbon more.

C) True.

Theoretically, for the evolution of one molecule of oxygen, only 8 photons are required. But in practice, it is known that there are many variants like wavelength and the energy of the photon. The larger the wavelength, like the one which is used in PS1 (more than 700nM), the lesser the energy. Secondly, the energy of the photon is also wasted as heat energy. Because of these factors, more than 8 photons are needed in reality.

D) Wrong.

Fructose 2,6 bisphosphate is a key substrate and affects both the enzymes- phosphofructokinase and fructose bisphosphatase allosterically during gluconeogenesis. It strongly favors the breakdown of glucose during glycolysis by activating phosphofructokinase but it inhibits fructose bisphosphatase. Hence it activates the kinase enzyme while inhibiting the phosphatase and maintains a huge supply of glucose in the system.

E) Wrong.

The Calvin cycle shares similarity with the pentose phosphate pathway as both are involved in the synthesis of sugar (Triose and Ribose). However, it does not share similarity with enzymes of glycolysis (which is primarily focused on the breakdown of glucose) and gluconeogenesis.

8 0

3 years ago

PLEASE HELP!!! PLEASE.

adell [148]

Answer:

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄: The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl?

As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.

The normality of the NaOH and HCl = Their molarity.

∵ (NV)NaOH = (NV)HCl

∴ N of HCl = (NV)NaOH / (V)HCl = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?

As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.

∵ (NV)H₂SO₄ = (NV)KOH

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

∴ The concentration of KOH = 7.675 mN = 7.675 mM.

Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?

As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

The normality of NaOH = Molarity of NaOH = 0.5 N.

∵ (NV)H₂SO₄ = (NV)NaOH

∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄ = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.

Q₄: Explain the difference between an endpoint and equivalence point in a titration.

The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.

End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.

The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.

6 0

3 years ago

How has thermal energy, heat, and temperature change?

maria [59]

thermal energy leads to an increase in the kinetic energy also increasing in temperature hope this helped!

3 0

3 years ago

Read 2 more answers

Which two periods have the same number of elements

Stels [109]

Answer:

4 & 5 have the same number of elements