Solute potential of a solution is calculated using the formula,
Ψ
Where,
Ψ
is the solute potential of the solution,
<em>i</em> is the degree to which the solute ionizes(ionization constant) in solution = 1, as sucrose is a nonelectrolyte.
C is the concentration of the solution in molarity = 0.5 M
R is the gas constant or the pressure constant = 0.0831 L.bar/(mol.K)
T is the temperature in Kelvin scale = 
Calculating the solute potential of the surrounding sucrose solution:
Ψ
= -(1 * 0.5 M * (0.0831 L.bar/(mol.K))* 303 K)
= 12.6 bar
Therefore, the solute potential of the surrounding solution is 12.6 bar
Answer:
Both biotic and abiotic factors affect the survival and reproductive success rate of fishes in an ecosystem.
The number of predators that a particular type of fish will have will influence the survival and reproductive rate of that fish. An increased number of predators will mean that the fishes will not be able to survive in that ecosystem and will be eaten up.
The number of preys will also affect the survival rate of fishes in an environment. The lesser the number of preys, the more the competition for food among species.
Abiotic factors like the concentration of salts in water will also influence the survival and reproduction of fishes. If a fish is not adapted to live in saline water conditions, then accumulation of salts in the ecosystem might destroy the whole fish species in that ecosystem.
Human activities, like the throwing of wastes into the aquatic ecosystem, drainage of fertilizers in the aquatic system can also destroy the survival and reproduction rate of a particular kind of aquatic animal species.
Explanation:
If purple is dominant, the genotype of the purple flower could be PP or Pp, as both would give a purple colour. However, if the crossed flower turned out to be white, we could say with certainty that the genotype of the purple flower is Pp as the genotype of the white crossed flower would be pp, which would not be possible if the purple flower's genotype was PP.
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!
