1) There's no way to do this without knowing the volume of the myocyte, so I would recommend you to search the book or your notes, should be somewhere in there. Basically, you would approach this question by finding the volume of the actin molecule first. V = 4/3*pi*(1.8nm)^3 = 24.42nm^3. Then your next step is just take the volume of the myocyte divide by 24.42nm^3 and you will get the number of molecules of actin inside a single myocyte (make sure the volume for myocyte is in nm^3, if not, you will need to do some unit conversions so the units can cancel out).
And just a hint or you can call it a cheat, lol. Since you already gotten the answer, if you don't want to go through the hassle of digging through your book and notes for the volume of the myocyte, you can just go 2.7x10^12 = x/24.42nm^3. Solve for x and that's the volume of the myocyte in nm^3. However, you wouldn't be able to do this if you do not know the answer ahead of time, so it's up to you. It doesn't really matter because in an exam, I'm sure your teacher would provide that value for you and won't make you memorize it.
Yeah for the rest of these, you just need to know the volume of the cells to do the math.
please explain how to get these answers (answers provided):
1. If this cell were a muscle cell (myocyte), how many molecules of actin could it hold? (Assume the cell is spherical and no other cellular components are present; actin molecules are spherical, with a diameter of 3.6 nm. The volume of a sphere is 4/3 πr3
Answer: 2.7 X 10^12 actin molecules
2. If this were a liver cell (hepatocyte) of the same dimensions, how many mitochondria could it hold? (Assume the cell is spherical; no othe cellular components are present; and the mitochondria are spherical, with a diameter of 1.5 μm).
Answer: 36000 motochondria
3. Glucose is a major energy-yielding nutrient for most cells. Assuming a cellular concentration of 0f 1 mM, calculate how many molecules of glucose would be present in our hypothetical (and spherical) eukaryotic cell. (Avogadro's number, the number of molecules in 1 mol of a nonionized substance, is 6.02 X 10^23).
Answer: 3.9 X 10^10 glucose molecules
4. Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase in our eukaryotic cell is 20 μM, how many glucose molecules are present per hexokinase molecule?
Answer:50 glucose molecules per hexokinase molecule
5. E. coli cells are rod-shaped, about 2 μm long and 0.8 μm in diameter. The volume of a cylinder is πr2h, where h is the height of the cylinder.
A) if the average density of E. coli (mostly water) is 1.1 X 103 g/L, what is the mass of a single cell?
Answer: 1 X 10^-12 g = 1 pg
B) E. coli has a protective cell envelope 10 nm thick. What percentage of the total of the bacterium does the cell envelope occupy?
Answer: 10%
C) E. coli is capable of growing and multiplying rapidly because it contains some 15000 spherical ribosomes (diameter 18 nm), which carry out protein synthesis. What percentage of the cell volume do the ribosomes occupy?
Answer: 5%
2 answers
6th textbook
2. Components of E. coli E. coli cells are rod-shaped, about 2 _m long and 0.8 _m in diameter. The volume of a cylinder is _r2h, where h is the height of the cylinder.
(a) If the average density of E. coli (mostly water) is 1.1 3 103 g/L, what is the mass of a single cell?
Ans：It’s diameter is 0.8 μm so radium is 0.4 μm. Volume is pir^2h.
V=πr^2 h=π×〖0.4〗^2×2=0.32π 〖μm〗^3
The number of E. coli bacteria is 1L/V.
N=(1 L)/(0.32π 〖μm〗^3 )=(1 〖dm〗^3)/(0.32π 〖μm〗^3 )=(〖10〗^15 〖dm〗^3)/(0.32π 〖μm〗^3 )=9.94718 × 〖10〗^14
The mass of single cell is M⁄N=(1.1×〖10〗^3 g)⁄(9.94718 × 〖10〗^14 )≈〖10〗^(-12) g.
(b) E. coli has a protective cell envelope 10 nm thick. What percentage of the total volume of the bacterium does the cell envelope occupy?
Ans: The volume of a E. coli bacterium except cell envelope should use smaller radium as 0.39μm.
V'=π〖r^'〗^2 h=π×〖0.39〗^2×1.98=0.301158π 〖μm〗^3
Cell envelope occupies 1-0.301158π⁄0.32π=5.89%.
(c) E. coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical ribosomes (diameter 18 nm), which carry out protein synthesis. What percentage of the cell volume do the ribosomes occupy?
Ans: We regard ribosome as ball. One ribosome volume is V=4/3 πr^3=4/3 π〖(9nm)〗^3=972π 〖nm〗^3.
The ribosomes volume is 15,000V=15,000×972π 〖nm〗^3=0.0458 〖μm〗^3 . They occupy 15,000V⁄0.32π=4.55625%
2. Components of E. coli E. coli cells are rod-shaped, about 2 _m long and 0.8 _m in diameter. The volume of a cylinder is _r2h, where h is the height of the cylinder.
(a) If the average density of E. coli (mostly water) is 1.1 3 103 g/L, what is the mass of a single cell?
Ans：It’s diameter is 0.8 μm so radium is 0.4 μm. Volume is pir^2h.
V=πr^2 h=π×〖0.4〗^2×2=0.32π 〖μm〗^3
The number of E. coli bacteria is 1L/V.
N=(1 L)/(0.32π 〖μm〗^3 )=(1 〖dm〗^3)/(0.32π 〖μm〗^3 )=(〖10〗^15 〖dm〗^3)/(0.32π 〖μm〗^3 )=9.94718 × 〖10〗^14
The mass of single cell is M⁄N=(1.1×〖10〗^3 g)⁄(9.94718 × 〖10〗^14 )≈〖10〗^(-12) g.
(b) E. coli has a protective cell envelope 10 nm thick. What percentage of the total volume of the bacterium does the cell envelope occupy?
Ans: The volume of a E. coli bacterium except cell envelope should use smaller radium as 0.39μm.
V'=π〖r^'〗^2 h=π×〖0.39〗^2×1.98=0.301158π 〖μm〗^3
Cell envelope occupies 1-0.301158π⁄0.32π=5.89%.
(c) E. coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical ribosomes (diameter 18 nm), which carry out protein synthesis. What percentage of the cell volume do the ribosomes occupy?
Ans: We regard ribosome as ball. One ribosome volume is V=4/3 πr^3=4/3 π〖(9nm)〗^3=972π 〖nm〗^3.
The ribosomes volume is 15,000V=15,000×972π 〖nm〗^3=0.0458 〖μm〗^3 . They occupy 15,000V⁄0.32π=4.55625%
1 answer