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!
Answer:
Spotted Eagle Rays have two sets of five gills on each side of their body on the ventral side. In order to breathe, A. narinari must be in continuous swimming motion so that oxygen can be absorbed from the water passing through the gills
Explanation:
Percent error is calculated by the expression:
%error = |actual value - observed value| / actual value x 100
We calculate the error of the values as follows:
<span>a. 23.487 cm
% error = |</span><span>23.490 - 23.487| / 23.490 x 100 = 0.013%
</span><span>
b. 23.493 cm
</span>% error = |23.490 - 23.493| / 23.490 x 100 = 0.013%
<span>
c. 23.516 cm
</span>% error = |23.490 - 23.516| / 23.490 x 100 = 0.11%<span>
d. 23.501 cm
</span>% error = |23.490 - 23.501| / 23.490 x 100 = 0.05%<span>
e. 23.477 cm
</span>% error = |23.490 - 23.477| / 23.490 x 100 = 0.055%
The altered protein affected the function of the cell membrane. The cell membrane which contain a bilayer control the movement of substance in and out of the cell and organelles. Due to this it is selective permeable to ion and organic molecules.
Answer:
C- it help to show that some bacteria are just one cell.
Explanation:
Anton Leeuwenhoek was the first scientist who observed the presence of bacteria in his tooth scratch under the microscope made by himself. At that time there was no concept of bacteria and he named those small creatures as animalcules which were later termed as bacteria.
Leeuwenhoek during 1670s sent the details of his experimentation on algae and bacteria to Royal Society of London.
It was due to his that primitive experiment that made the discovery of bacteria and make it known to other scientists. Therefore option C is the best option that demonstrates that bacteria are single celled organisms that are very minute. When we look at cell theory there is a postulate that shows that cell is the fundamental unit of function and structure of all living organisms. So, the work of Anton Leeuwenhoek holds substantial importance in the formulation of cell theory.
Hope it helps!
