Biology Calculation-is this correct?
Moderators: honeev, Leonid, amiradm, BioTeam
Biology Calculation-is this correct?
Hey just wondering if anyone can tell me whether I have done this calculation right or not.
A baby hamster kidney (BHK) cell has a surface area of 3,400 micrometres². Assume that the endocyte clathrin-coated pits account for 2% of this area and that the surface area of a single coated pit is the same of that of a 100nm diameter clathrin coated vesicle that will be formed from it. If the cell surface has 100,000 transferrin receptors, around 70% are localised within coated pits: work out how many coated pits there are.
Ive got:
3,400 * 2% = 68 micrometers(squared)
every pit has a 100 nanometer diameter. area of a circle 2*pi*r(squared)
100nanometers=0.1micrometer
r=radius 2r=d
r=0.05micrometers
2*pi*0.05(squared)=0.0157micromters(squared)
68/0.0157=4329.014
is this correct or would I do pir2 rather than 2*pir2 which would give me an answer of 8658? Or do I need to work out the surface area of a sphere? 4*pi*r2?
Please help, thanks
A baby hamster kidney (BHK) cell has a surface area of 3,400 micrometres². Assume that the endocyte clathrin-coated pits account for 2% of this area and that the surface area of a single coated pit is the same of that of a 100nm diameter clathrin coated vesicle that will be formed from it. If the cell surface has 100,000 transferrin receptors, around 70% are localised within coated pits: work out how many coated pits there are.
Ive got:
3,400 * 2% = 68 micrometers(squared)
every pit has a 100 nanometer diameter. area of a circle 2*pi*r(squared)
100nanometers=0.1micrometer
r=radius 2r=d
r=0.05micrometers
2*pi*0.05(squared)=0.0157micromters(squared)
68/0.0157=4329.014
is this correct or would I do pir2 rather than 2*pir2 which would give me an answer of 8658? Or do I need to work out the surface area of a sphere? 4*pi*r2?
Please help, thanks
BSc (Hons) Forensic Science,
MSc Biological Sciences
MSc Biological Sciences
1) the vesicle is 3D, not 2D 
2) why do you calculate two surfaces and divide them?
3) I think you should calaculate the number of the receptors on pits and from that calculate the number of pits. But I think there is missing some info...
2) why do you calculate two surfaces and divide them?
3) I think you should calaculate the number of the receptors on pits and from that calculate the number of pits. But I think there is missing some info...
http://www.biolib.cz/en/main/
Cis or trans? That's what matters.
Cis or trans? That's what matters.
Re: Biology Calculation-is this correct?
so I need to calculate the surface area of a sphere then, 4*pir(squared)?
I have done the division at the end because they only account for 2% of the total suface area which is 68micrometers squared. The next question is to then calculate how many transferrin receptors there are in each coated pit, but I just can't get my head round how to do that.
I have done the division at the end because they only account for 2% of the total suface area which is 68micrometers squared. The next question is to then calculate how many transferrin receptors there are in each coated pit, but I just can't get my head round how to do that.
BSc (Hons) Forensic Science,
MSc Biological Sciences
MSc Biological Sciences
Hey, for the number of transferrin receptors in each coated pit I've got:
100 000 receptors, 70% localised within the coated pits so:
100 000 x 70% = 70 000 receptors in the coated pits.
There are 70 000 receptors and 2165 coated pits so:
70 000/2165 = 32 transferrin receptors in each coated pit.
Does think look right to anyone? Not very confident when it comes to maths lol
Thanks
K
100 000 receptors, 70% localised within the coated pits so:
100 000 x 70% = 70 000 receptors in the coated pits.
There are 70 000 receptors and 2165 coated pits so:
70 000/2165 = 32 transferrin receptors in each coated pit.
Does think look right to anyone? Not very confident when it comes to maths lol
Thanks
K
BSc (Hons) Forensic Science,
MSc Biological Sciences
MSc Biological Sciences
Who is online
Users browsing this forum: Bing [Bot] and 10 guests