Answer:
The viral genome is covered by a nucleocapsid protein called N. Two other proteins in the virus are the large protein called L, and the phosphoprotein called P. Both of these are involved with making new copies of the measles virus.
Explanation:
I got this from the wikipedia
Would the correct answer be A
Answer:
in the long term the penny
Explanation:
A) The answer is 5.2 · 10⁻¹⁴ kg
It is known that the mass (m) is density (D) multiplied by volume (V):
m = D · V
D = 1.0 · 10³ kg/m³
m = ?
V = ?
Let's calculate the volume.
The volume (V) of a spherical cell is V = 4/3 π r³ (r - radius of a sphere)
r = ?
d = 1.0 um = 1.0 · 10⁻⁶ m
Since radius is a half of the diameter, then:
r = d ÷ 2 = 1.0 · 10⁻⁶ m ÷ 2 = 0.5 · 10⁻⁶ m
It is known than π = 3.14
Therefore:
V = 4/3 π r³ = 4/3 · 3.14 · (0.5 · 10⁻⁶)³ = 4/3 · 3.14 · 0.5³ · (10⁻⁶)³
= 4/3 · 3.14 · 0.125 · 10⁻⁶*³ = 4/3 · 3.14 · 0.125 · 10⁻¹⁸ = 0.52 · 10⁻¹⁸
= 5.2 · 10⁻¹⁷ m₃
So, now when we know D and V, it is easy to calculate m:
m = D · V = 1.0 · 10³ kg/m³ · 5.2 · 10⁻¹⁷ m³ = 5.2 · 10⁻¹⁷⁺³ kg = 5.2 · 10⁻¹⁴ kg
b) The answer is 12.56 · 10⁻⁶ kg
It is known that the mass (m) is density (D) multiplied by volume (V):
m = D · V
D = 1.0 · 10³ kg/m³
m = ?
V = ?
Let's calculate the volume.
The volume (V) of a fly in the shape of cylinder is V = h π r²
h = 4.0 mm = 4.0 · 10⁻³ m
r = ?
d = 2.0 mm = 2.0 · 10⁻³ m
Since radius is a half of the diameter, then:
r = d ÷ 2 = 2.0 · 10⁻³ m ÷ 2 = 1.0 · 10⁻³ m
It is known than π = 3.14
Therefore:
V = h π r³ = 4.0 · 10⁻³ · 3.14 · (1.0 · 10⁻³ )² = 4.0 · 10⁻³ · 3.14 · 1.0² · (10⁻³ )² =
= 4.0 · 10⁻³ · 3.14 · 1.0 · 10⁻³*² = 4.0 · 10⁻³ · 3.14 · 1.0 · 10⁻⁶ =
= 12.56 · 10⁻³⁻⁶ = 12.56 · 10⁻⁹ m³
So, now when we know D and V, it is easy to calculate m:
m = D · V = 1.0 · 10³ kg/m³ · 12.56 · 10⁻⁹ m³ = 12.56 · 10⁻⁹⁺³ kg = 12.56 · 10⁻⁶ kg
Answer:
False
Explanation:
The blood brain barrier (BBB) is a selective border that prevents the entry of foreign substances into the brain. It is made up of different components including endothelial cells, astrocytes, and pericytes etc.
Of all these mentioned three, the endothelial cells are directly responsible for the selectivity of the BBB. The endothelial cells are bound together by the tight junctions situated between them, impeding the passage of foreign materials e.g. solutes, pathogens etc.
Astrocytes help to maintain the BBB by surrounding the endothelial cells, providing biochemical strength.
The pericytes are located in the basement membrane, where they communicate with endothelial cells. They help to sustain the BBB by stabilizing and monitoring the maturation of endothelial cells via communicating directly between cell membrane as well as paracrine signalling.
