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.
It’s most likely missing water
plants need to be watered in order to survive
- 407.4 kJ of heat is released.
<u>Explanation:</u>
We have to write the balanced equation as,
2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O
Here 2 moles of ethane reacts in this reaction.
Now we have to find out the amount of ethane reacted using its given mass and molar mass as,
2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆
Heat released = ΔH × given mass / 60.14
= - 1560. 7 kj ×15.7 g / 60. 14 g = -407. 4 kJ
Her computer will start to heat up, and the temp. would be 60 degrees celcius.
I hope this helps you ᕕ( ᐛ )ᕗ
Answer:
is there a table or an equation for that ??
