Answer:
d. ATP; Fructose-2,6-bisphosphate
Explanation:
Phosphofructokinase-1 is the enzyme that catalyzes the formation of fructose-1,6-bisphosphate from fructose-6-phosphate and ATP. The phosphofructokinase step is the first rate-limiting step of glycolysis.
Phosphofructokinase-1 activity is allosterically regulated. Its activity is increased whenever the cell's ATP supply is depleted or when its breakdown products, ADP and AMP accuulates in he cell. However, it is inhibited when the cell is amply supplied with ATP.
The activity of phosphofructokinase-1 is restored by fructose-2,6-bisphosphate, its most potent activator.
From the given options:
a. AMP :::: citrate is wrong because AMP increases the activity of phosphofructokinase-1 and citrate is not its activator but an inhibitor
b. AMP :::: Fru-2,6-P2 is wrong because AMP increases its activity same as fructose-2,6-bisphosphate
c. ATP :::: citrate is wrong because both citrate and ATP are inhibitors
d. ATP :::: Fru-2,6-P2 is correct as explained above
e. All of these is wrong because of the other wrong options above.
Answer:
The mentioned parental types are c+m- and c-m+. Thus, the recombinants will be c+m+ and c-m-.
Now, the given distance between c and m is 8 map units. Thus, the recombinant frequency is 8% or 0.08.
The total recombinants from 1000 plaques will come out to be 80,
Thus, the recombinants of each type will be 40.
Total parental type will be 920, and therefore, each parental type count will be 460.
Thus, expected c+m- = 460, expected c-m+ = 460, expected c+m+ = 40 and expected c-m- = 40.
Answer:
The correct answer is the third option- the large ribosomal unit.
Explanation:
The translation is the second process of the protein synthesis in which transcribed mRNA molecule and transfer RNA or tRNA and ribosomes assemble together and complete synthesis of peptide chain or protein.
The assembly of initiator tRNA to ribosome subunits at the start codon of the mRNA is the initiation complex of the translation. The initiator tRNA is basically a met-tRNA molecule.
The initiator tRNA is bound to small subunit (30S) at 5' cap and scan for the start codon of mRNA.
Start codon bind to initiator RNA and in the end larger ribosomal unit assemble to this complex to complete the initiation complex of translation.
Thus, the correct answer is option - the large ribosomal subunit
Answer:
It is explain about how is the cell division.and what are the examples for it.
Thera are two type of cell division.
1. Mitosis.
2.Meiosis
Answer:
The most appropriate answer would be The cell will not make functional proteins from that mRNA strand.
The amino acid sequence of the proteins is derived from the nucleotide sequence of the mRNA (messenger ribonucleotide).
If the sequence of mRNA is miscopied, it will change the sequence of amino acids of the protein.
Consequently, the protein may become non-functional.
