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.
Cl is highly electronegative and will actually pull away 1 electron from sodium, forming an ionic bond.
<u>Answer:</u> The 
for the reaction is -1052.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
The intermediate balanced chemical reaction are:
(1) 
(2) 
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B1%5Ctimes%20%5CDelta%20H_1%5D%2B%5B1%5Ctimes%20%28-%5CDelta%20H_2%29%5D)
Putting values in above equation, we get:
Hence, the for the reaction is -1052.8 kJ.
Answer:
2.01 M
Explanation:
Step 1: Calculate the moles of acetic acid (HC₂H₃O₂)
The molar mass of acetic acid is 60.05 g/mol. We will use this data to calculate the moles corresponding to 36.2 g of acetic acid.
Step 2: Convert the volume of solution to liters
We will use the relation 1000 mL = 1 L. We assume that the volume of solution is that of water (300 mL)
Step 3: Calculate the molarity of the solution
The molarity is equal to the moles of solute (acetic acid) divided by the liters of solution
Answer:
B) exothermic.
Explanation:
Hello!
In this case, we need to keep in mind that exothermic reactions release heat, so they increase the temperature as the final energy is less than the initial energy; in contrast, endothermic reactions absorb heat, so they decrease the temperature as the final energy is greater than the initial energy.
In such a way, when a dissolution process shows off a negative enthalpy of dissolution, we infer it is an exothermic process due to the aforementioned; therefore, the answer is:
B) exothermic
.
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