Biology Calculation-is this correct?

Viewing 5 reply threads
  • Author
    Posts
    • #14199
      Kezzer
      Participant

      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 🙂

    • #102508
      pielover1234
      Participant

      i would say so 😀

    • #102514
      JackBean
      Participant

      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…

    • #102549
      Kezzer
      Participant

      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.

    • #102609
      Kezzer
      Participant

      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

    • #102610
      Kezzer
      Participant

      I realised for the first question on coated pits I did need to do 4×3.14×0.05(squared) so got 2165 for how many coated pits there are 🙂

Viewing 5 reply threads
  • You must be logged in to reply to this topic.