Answer:
The correct answer will be- 200
Explanation:
The cell cycle in eukaryotes mostly takes place in four phases: G₁, S, G₂ and M phase in sequence.
The DNA replication takes place during S phase which doubles the amount of DNA in a cell after which during M- phase the DNA divides into the daughter cell.
The amount of DNA present during G₂ phase is the 200 pg which will remain same until anaphase I as the segregation of chromosomes to the daughter cells takes place in the anaphase. Therefore, the amount of DNA observed after G₂ phase will remain 200 pg till metaphase I.
Thus, 200 is the correct answer.
Enzymes increase the rate of a given reaction by lowering Activation E<span>nergy. Hope that helps.</span>
Answer:
Explanation:
My best bet is DNA methylation at the site of Tweedledum's leptin gene or Histone Acetylation at the site of Tweedledee's gene.
B/c DNA methylation is a process by which methyl groups are added to the DNA molecule. Methylation can change the activity of a DNA segment without changing the sequence. When located in a gene promoter, DNA methylation typically acts to repress gene transcription. So this is probably repressing Tweedledum's leptin gene trancription which is not happening in Tweedledee.
Additionally, Histone Acetylation at site of Tweedledee's gene increases her trancription b/c Histone acetylation causes DNA to be more accessible and leads to more transcription factors being able to reach the DNA. Thus, acetylation of histones is known to increase the expression of genes through transcription activation.
The answer is a ...........................
