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.
Answer:
reproduction
Explanation:
reproduction, process by which organisms replicate themselves
Answer:
i think the first one is gravity and second one is rotation
Explanation:
Answer:
a. Moles of 
= 0.001643 moles
b. 0.296 g
c. 0.3098 g
d. Not acceptable
Explanation:
a.
Considering:
Or,
Given :
For :
Molarity = 0.1052 M
Volume = 15.62 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 15.62×10⁻³ L
Thus, moles of :
Moles of = 0.001643 moles
b.
The reaction of NaOH with the acetylsalicylic acid is in the ratio of 1:1.
Thus, Moles of NaOH = Moles of acetylsalicylic acid = 0.001643 moles
Molar mass of acetylsalicylic acid = 180.16 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass = Moles * Molar mass = 0.001643 moles * 180.16 g/mol = 0.296 g
c.
1.159 g of sample contains 0.296 g of acetylsalicylic acid
1.213 g of sample contains 
g of acetylsalicylic acid
Mass of acetylsalicylic acid = 0.3098 g = 309.8 mg
d. Sample contains = 309.8 mg
Manufacturer claiming = 315 mg to 335 mg
Thus , it is not acceptable.
