Best Answer: Well, you have to look at the onion slide under the microscope and looks at each cell and tell what stage (interphase, prophase, etc.) they are in. and then for the description probably tell which one there is the most of and maybe why you think that is. You can find pictures of each phase in your book. I did a lab like this not too long ago in my bio class lol. And, i believe that you will get a lot in interphase (when cells are not currently dividing), just to let you know.
Answer:
Endosymbiosis of an oxygen-using scavenger bacterium in a larger host cell−the endosymbiont evolved into lysosomes.
Explanation:
The evolution of eukaryotic cells in all probability included: endosymbiosis of an oxygen-utilizing bacterium in a bigger host cell−the endosymbiont developed into mitochondria. development of an endomembrane framework and consequent advancement of mitochondria from a segment of the Golgi. endosymbiosis of an oxygen-utilizing photosynthetic bacterium in a bigger host cell−the endosymbiont advanced into mitochondria. endosymbiosis of an oxygen-utilizing scrounger bacterium in a bigger host cell−the endosymbiont developed into lysosomes.
Answer:
I think it is B
Explanation:
They aren't very clear about how acupuncture will stop pain channels
<span>Connective tissue supports the framework of the liver, and epithelial tissue protects the liver.
Connective tissue holds the liver in place during movement, and epithelial tissue forms the lining of the liver.
Connective tissues are mainly used in forming support networks within tissues and Epithelial tissue tend to line the organs and form protective cell layers.</span>
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!
