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Wewaii [24]
3 years ago
13

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

or False Transketolase transfers two carbon units in the pentose phosphate pathway using a thiamine pyrophosphate cofactor. True or False In theory, chloroplasts should only absorb 8 photons per oxygen molecule evolved during photosynthesis, but they actually absorb more than 8 photons per evolved oxygen molecule. True or False Fructose-2,6-bisphosphate activates both phosphofructokinase and fructose bisphosphatase to increase flux through gluconeogenesis. True or False The Calvin cycle shares enzymes that are homologous to enzymes involved in glycolysis, gluconeogenesis, and the pentose phosphate pathway with only three enzymes unique to the Calvin cycle. True or False
Chemistry
1 answer:
zloy xaker [14]3 years ago
8 0

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.

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The carbon-14 content of a wooden harpoon handle found in an Inuit archaeological site was found to be 61.9% of the carbon-14 co
astraxan [27]

Answer:

3,964 years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
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  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

  • The half-life of the element is 5,730 years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 100%).

[A] is the remaining concentration of the sample ([A] = 61.9%).

∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.

7 0
4 years ago
PLEASE I NEED HELP ASAP
Murljashka [212]

Answer:

Because the density of water is one

4 0
3 years ago
Answer right = Brainliest!!
amid [387]
A b and e is are the answers
8 0
3 years ago
How many molecules are in 18moles of CH.​
-Dominant- [34]

Answer:

1.08 x 10²⁵molecules

Explanation:

From the mole concept we know that ;

      1 mole of any substance contains 6.02 x 10²³ molecules

This number is the Avogadro's number.

 So;

  18 mole of CH will contain:

       Number of molecules of CH  = 18 x 6.02 x 10²³ = 1.08 x 10²⁵molecules

The number of molecules is therefore 1.08 x 10²⁵molecules

8 0
2 years ago
If one of the impurities in diesel fuel has the formula c2h6s, what products will be formed when it burns? enter a balanced chem
inna [77]
<span>If one of the impurities in diesel fuel has the formula c2h6s, then the products that will form would be carbon dioxide, water and sulfur dioxide. The balanced chemical reaction would be as follows:

</span>C2H6S(l)<span> + 9/2O2(g) = 2CO2(g) + 3H2O(v) + SO2(g)
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Hope this answers the question.
7 0
3 years ago
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