Answer:
where the is the picture?
Answer:
The third sentence is not correct. The correct statement would be as follows:
because the contents of the lumen of ER (or any other compartment) in the secretory or endocytic pathways never mix with the cytosol, Proteins that enter these pathways is never imported again.
Explanation:
During mitosis, specifically during prophase, when the nuclear envelope breaks down, or in other words, retracts from the chromatin, its content that is the membrane protein intermix with the ER membrane protein. However its content will always remain separated from that of the cytosol because of the presence of an intact membrane.
those are nucleotides
since all three of them contain deoxyribose (because there's only one hydroxil group) they are DNA nucleotides
the first nucleotide has cytosine as it's nitrogenous base
the second nucleotide has adenine as it's nitrogenous base
the third nucleotide has thymine as it's nitrogenous base
Answer:
1. G° = -RT ln (G1P/P)
3.1 = 8.314 × 310 × ln (G1P/P)
3.1 / 2577.34 = ln (G1P/P)
0.0012 = ln (G1P/P)
0.0012 = (log G1P/P)/log 2.71828
0.4342 × 0.0012 = log G1P/P
0.00052 = log G1P/P
G1P/P = 10^0.00052 = 1.0012
P/G1P = 1/1.0012 = 0.9988
2. The cleavage of glycogen phosphorolytically is beneficial for the cell to conduct the process as the discharged glucose is phosphorylated. A general hydrolytic cleavage would give rise to only a glucose, which has to be phosphorylated again with the help of ATP.
Another merit of phosphorylated glucose is that it comprises the negative charge and cannot diffuse out of the muscle cell. Thus, the reaction will not be at equilibrium under the physiological conditions and always encourages the generation of the products. The formation of products will amend the change in free energy in such a manner that the reaction will always carry in the forward direction.
3. Greater the ratio of [Pi]/[glucose-1-phosphate], higher will be the relative rate of glycogen phosphorylase in comparison to the phosphoglucomutase as the transformation of Glu-1-P becomes slow because of lesser accessibility of substrate.
Answer: B. The bacteria must adjust to the nutrient content in the new medium, synthesizing necessary amino acids, growth factors, and enzymes.
Explanation: The lag phase is a time for adjustement to new environments, in this case new mediums. In that phase, bacteria sense the available nutrients to synthesize those required and grow as preparing for division, metabolizing to produce energy, making proteins, fatty acids, etc.
