Answer:
lower power= 20 x 10= 200x
higher power= 20 x 43= 860x
Answer:
A)100mL B)50mL C)The second option D)Hypoosmotic Environment
Explanation:
The average Na concentration in the seas and oceans of the world is around 3,5% which mean that in 100 ml of sea water, there is around 3,5 grams of Na.
The weight of one mol of NaCl is 58,44 grams. For 3,5 grams of NaCl, we get 3,5/58,44 = 0,060 mol of NaCl which is 0,060x1000 = 60 mmol/100ml. According to this and the information given in the question about the secretion of the salt glands', if the average sodium concentration is 600mmol/L, we have 60*10 = 600mmol/L so it would take 100 mililiters of water to excrete.
If the average Na concentration of the salt gland's secretion were 300 mmol/L, only 50 mililiters of water would be needed to excrete the same sodium load.
The second option of secretion is hyperosmotic to seawater because the concentration is higher.
Osmoregulation is the process of balancing the amount of water and salt between the body of the organism and its surrounding environment. For salt glands to be advantageous for osmoregulation, they need to be in a hypoosmotic environment.
I hope this answer helps.
Respiration takes place in the mitochondria
The mass is increased as two things are added together so the two masses of the two substances are added together.
Answer:
1. Allele frequency of b = 0.09 (or 9%)
2. Allele frequency of B = 0.91 (0.91%)
3. Genotype frequency of BB = 0.8281 (or 82.81%)
4. Genotype frequency of Bb = 0.1638 (or 16.38%)
Explanation:
Given that:
p = the frequency of the dominant allele (represented here by B) = 0.91
q = the frequency of the recessive allele (represented here by b) = 0.09
For a population in genetic equilibrium:
p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)
(p + q)^2 = 1
Therefore:
p^2 + 2pq + q^2 = 1
in which:
p^2 = frequency of BB (homozygous dominant)
2pq = frequency of Bb (heterozygous)
q^2 = frequency of bb (homozygous recessive)
p^2 = 0.91^2 = 0.8281
2pq = 2(0.91)(0.9) = 0.1638
