<span>Since alkylating agents work to prevent the cell from replicating its genetic material, the cell would most likely stop at the interphase checkpoint, which is the phase before mitosis (which consists of prophase, metaphase, anaphase, and telophase). During interphase, the cell replicates its genetic material (DNA), but this would be prohibited by alkylating agents.</span>
Answer:Explanation: Viruses are acellular organisms and although they do not have cells, they are extremely dependent on these structures, since they do not have their own metabolism and do not have any organelles. Viruses can only reproduce within a host cell other than fungi and bacteria
Explanation:
Hey hi your answer would be A<span> glycolysis </span>
Answer:
There will be few if any complications.
Explanation:
Just took the unit test review
This is the DNA. I'm going to only use the upper strand to demonstrate what this strand would code for before and after a single bp deletion (so write it as mRNA). I will also write it how it's easier to see this which is to split them up into the 3 base codon system. Note that you don't need to know the amino acid code - you use a table to find these.
ORIGINAL (mRNA on top, Amino Acid (AA) on bottom:
5'-AGC GGG AUG AGC GCA UGU GGC GCA UAA CUG-3'
SER GLY MET SER ALA CYS GLY ALA STOP LEU
Note that the protein would stop being made at the stop codon and the LEU wouldn't matter at the end...
Now, I will remove one bp...(I bolded it up top). Rewrite the mRNA and find the corresponding AA...
NEW
5'-AGC GGG AUG GCG CAU GTG GCG CAU AAC UG-3'
SER GLY MET ALA HIS VAL ALA HIS ASN .....
Completely different amino acid sequence after the methionine (MET). The stop codon is gone...the protein would continue being translated until it reaches another stop codon...so not what was supposed to be made!