For rock pocket mice, fur color is a phenotype
<h3>What are phenotypes?</h3>
Phenotypes refer to the morphological characters of organisms.
In other words, phenotypes refer to the physical appearance of living organisms.
In this case, the appearance of the rock pocket mice varied from darker fur on the black rock to lighter-colored fur on the lighter-colored rock. This means that the phenotype varies with the environment.
More on phenotypes can be found here: brainly.com/question/20730322
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:
C
Explanation:
Ribosomes are organelles which produce protein inside a cell. They do the same even if they are on an ER membrane.
Hope this helps! :)
Different steps of the oxidative decarboxylation of pyruvate by the pyruvate dehydrogenase pdh complex are given by placing them in the order as followed.
<h3>What is Oxidative decarboxylation?</h3>
The Oxidative decarboxylation reactions are oxidation reactions wherein a carboxylate institution is removed, forming carbon dioxide. They regularly arise in organic systems: there are numerous examples withinside the citric acid cycle. This sort of response probable began out early on the starting place of life.
- Pyruvate reacts with TPP and is decarboxylated, forming hydroxyethyl-TPP.
- The lipoamide arm movements to the energetic E underline three in which the decreased lipoamide is oxidized through FAD, forming the energetic lipoamide and triangle down ADH2 .
- ADH_ is reoxidized to FAD, lowering NAD to NADH.every so often known as thiamine.
- The acetyl lipoamide arm of E_ movements to the energetic of E_ , in which the acetyl institution is transferred to CoA forming acetyl-CoA and the decreased shape of lipoamide.
Read more about the lipoamide:
brainly.com/question/25870256
#SPJ2
Answer:
both eukaryotes and prokaryotes generate between 36-38 ATP during cellular respiration depends on the need.
some textbook says the net yield is 36 beacuse of the usage of 2 ATP during the process. but the net yield is same for both prokaryotes and Eukaryotes. .
