Answer:
Kc of reaction is 20.
Explanation:
The two proteins are X and Y.
The [X] = 1mM
[Y]=1mM
At equilibrium, [X] = 0.2mM [Y] = 0.2mM
we know that equilibrium constant is:
Kc=![\frac{[Products]}{[Reactants]}=\frac{[XY]}{[X][Y]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProducts%5D%7D%7B%5BReactants%5D%7D%3D%5Cfrac%7B%5BXY%5D%7D%7B%5BX%5D%5BY%5D%7D)
[XY]= 1-0.20=0.80 mM
putting values:
Kc=
Answer:
A. There were new technologies and new innovations to drive the sale of goods, since people could afford them
Explanation:
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
pH=8.5414
Procedure
The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.
pH = pKa + log₁₀ ([A⁻] / [HA])
Where
pH = acidity of a buffer solution
pKa = negative logarithm of Ka
Ka =acid disassociation constant
[HA]= concentration of an acid
[A⁻]= concentration of conjugate base
First, calculate the pKa
pKa=-log₁₀(Ka)= 8.6383
Then use the equation to get the pH (in this case the acid is HBrO)
Because K and Cl have such a large disparity in their electronegativities, KCl is a bipolar ionic molecule.
<h3>What exactly are polar and nonpolar bonds?</h3>
Polar covalent bonds develop when the distribution of electrons among atoms is uneven, whereas nonpolar side chains develop when the distribution of electrons is more even. The reason for the unequal sharing of electrons is because the atoms receiving them have various electronegativities.
<h3>How are polar bonds created?</h3>
Whenever a single pair of electrons is not shared equally, a polar molecule bond is created. This is caused by the electronegativity difference between the two elements. An unit of h as well as an unit of bromine share a pair of electrons, but not evenly.
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