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fenix001 [56]
3 years ago
7

What is responsible for morphogenesis

Chemistry
2 answers:
GaryK [48]3 years ago
8 0
Gene expression may be the correct answer...
Tcecarenko [31]3 years ago
4 0
Gene expression
Plz choose me for Brainest answer
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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
Write the Henderson-Hasselbalch equation for a solution of formic acid. Calculate the quotient [HCO2]/[HCO2H] at (a) pH 3.000; (
Elena L [17]

Answer:

a. 0.182

b. 1.009

c. 1.819

Explanation:

Henderson-Hasselbach equation is:

pH = pKa + log [salt / acid]

Let's replace the formula by the given values.

a. 3 = 3.74 + log [salt / acid]

3 - 3.74 = log [salt / acid]

-0.74 = log [salt / acid]

10⁻⁰'⁷⁴ = 0.182

b. 3.744 = 3.74 + log [salt / acid]

3.744 - 3.74 = log [salt / acid]

0.004 = log [salt / acid]

10⁰'⁰⁰⁴ = 1.009

c. 4 = 3.74 + log [salt / acid]

4 - 3.74 = log [salt / acid]

0.26 = log [salt / acid]

10⁰'²⁶ = 1.819

3 0
3 years ago
Help me on a and b pls
marusya05 [52]

Answer:

yes, it is increased by atomic mass

Explanation:

5 0
3 years ago
What is the oh−concentration and ph in each solution? (a) 0.225 m koh, (b) 0.0015 m sr(oh)2?
shepuryov [24]
Q1)
We have been given the OH⁻ concentration, therefore we first need to find the pOH value and then the pH value.
pOH = -log [OH⁻]
pOH = -log (0.225 M)
pOH =  0.65
pH + pOH = 14
pH = 14 - 0.65 = 13.35

Q2)
pOH = -log[OH⁻]
pOH = -log (0.0015 M)
pOH = 2.82
pH + pOH = 14
pH = 14 - 2.82 
pH = 11.18

7 0
3 years ago
Which is the correct Lewis dot structure of NH2?<br><br> A.A<br> B.B<br> C.C<br> D.D<br> E.E
Anarel [89]

I think D is your answer

5 0
2 years ago
Read 2 more answers
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