Which of the following is NOT an effect of human development on the Everglades?

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

A log has a density of .8 g/cm³. What will happen to this log in freshwater, which has a density of 1.0 g/cm³?

Inga [223]

Answer:

The log will float on the water because its density is lower than the liquid, so it will stay at the top due to Archimedes' principle.

6 0

2 years ago

Read 2 more answers

Please help !!!!!!!!

Salsk061 [2.6K]

1) CH2 (gas) + Br (solid) -> BrC (solid) + H2 (gas)
2) a) CH4 + Br2 -> CH3Br + HBr
2) b) methane + bromine is substitution because one hydrogen atom from methane is replaced by one bromine atom. addition reaction takes place when one molecule combines with another to form a larger molecule so therefore a molecule from X and bromine combine to form XBr.

6 0

2 years ago

If we mix 25 grans if sodium oxide with a large amount of potassium chloride, how many grams of sodium chloride should be produc

amid [387]

Answer:

47.2 g

Explanation:

Data given:

mass of Sodium oxide (Na₂O) = 25 grams

mass of potassium chloride (KCl) = excess

amount of sodium chloride (NaCl) = ?

Solution:

First we look for reaction of sodium oxide with potassium chloride.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

As we know that potassium chloride is in excess amount and sodium oxide is 25 g so it means sodium oxide is a limiting reactant and amount of sodium chloride depends on the amount of sodium oxide.

So, now we will look for mole mole ration of Na₂O to NaCl.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol 2 mol

from above equation we come to know that 1 mole of Na₂O gives 2 moles of NaCl.

Now convert moles to mass

As we Know

Molar mass of Na₂O = 2(23) + 16 = 62 g/mol

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

So

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol (62 g/mol) 2 mol (58.5 g/mol)

62 g 117 g

So, it means that 62 g of Na₂O produce 117 g of NaCl then how many grams of NaCl will be produce by 25 g of Na₂O

Apply unity formula

62 g of Na₂O ≅ 117 g of NaCl

25 g of Na₂O ≅ X g of NaCl

Do cross multiplication

X g of NaCl = 117 g x 25 g / 62 g

X g of NaCl = 47.2 g

So,

25 g of Na₂O gives 47.2 grams of NaCl.

4 0

3 years ago

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0

3 years ago