Answer:
The correct answers are A. "crenation", B. "hemolysis", C. "hemolysis", D. "crenation" and E. "neither will occur".
Explanation:
0.9% (m/v) NaCl or 5.0% (m/v) glucose are isotonic solutions at which the cells will not suffer any harmful consequence. A solution with a higher concentration than the isotonic conditions would result in the cells crenation, while a solution with a lower concentration would result in the cells hemolysis. Therefore the consequences of putting the red cells to the solutions stated in the question are as following:
A: 3.21% (m/v) NaCl Solution = crenation (higher than 0.9% (m/v) NaCl)
B: 1.65% (m/v) glucose Solution = hemolysis (lower than 5.0% (m/v) glucose)
C: distilled H2O Solution = hemolysis (lower than 0.9% (m/v) NaCl or 5.0% (m/v) glucose)
D: 6.97% (m/v) glucose Solution = crenation (higher than 5.0% (m/v) glucose)
E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl = neither will occur (equal to 5.0% (m/v) glucose and 0.9% (m/v) NaCl)
Answer:
I would probably want to give my child one or two specific traits like intelligence or no inheritable diseases, because I would want my child to have a excellent chance of success in life.
Explanation:
<span>Fossil record, genetics, and examples of local adaptations all support the theory of evolution by natural selection.</span>
The plant belongs to the group Angiosperm. Vascular plants with seeds could be cycads, Gingko, conifers, or angiosperms. Among those groups, only angiosperms have flowers.
The plant they found have colored, scented flowers which suggests that it could be pollinated by insects or birds. Colored flowers attract birds and insects. Color serves as a guiding mark. Talking of scent, it does not attract bird, but attracts insects. It also serves as the guiding mark.
Answer:
0.153
Explanation:
We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.
So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.
ρVg = ρ'V'g + ρ"V"g
ρV = ρ'V'g + ρ"V"
Its new body volume = volume of water displaced, V = V' + V"
ρ(V' + V") = ρ'V' + ρ"V"
ρV' + ρV" = ρ'V' + ρ"V"
ρV' - ρ"V' = ρ'V" - ρV"
(ρ - ρ")V' = (ρ' - ρ)V"
V'/V" = (ρ - ρ")/(ρ' - ρ)
= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)
= (0.9988 g/cm³ ÷ 0.18 g/cm³)
V'/V" = 5.55
Since V = V' + V"
V' = V - V"
(V - V")/V" = 5.55
V/V" - V"/V" = 5.55
V/V" - 1 = 5.55
V/V" = 5.55 + 1
V/V" = 6.55
V"/V = 1/6.55
V"/V = 0.153
So, the fish must inflate its air sacs to 0.153 of its expanded body volume
