Answer:
1) What is the mitochondrion length in nanometers? Show all your working out.
The mitochondrion length in nanometers is 85,000 nm.
Working out:
1.3 cm = 1300 mm
1 mm = 1000 μm
1 μm = 1000 nm
Therefore, 1300 mm x 1000 μm x 1000 nm = 1,300,000,000 nm
1,300,000,000 nm / 65 nm = 20,000,000
20,000,000 x 65 nm = 1,300,000,000 nm
1,300,000,000 nm - 1,300,000,000 nm = 0
Therefore, 1,300,000,000 nm / 20,000,000 = 65 nm
65 nm x 1.3 cm = 85,000 nm
2) Calculate the magnification. Show all your working out.
The magnification is 65,000.
Working out:
1.3 cm = 1300 mm
1 mm = 1000 μm
1 μm = 1000 nm
Therefore, 1300 mm x 1000 μm x 1000 nm = 1,300,000,000 nm
1,300,000,000 nm / 65 nm = 20,000,000
20,000,000 x 65 nm = 1,300,000,000 nm
1,300,000,000 nm - 1,300,000,000 nm = 0
Therefore, 1,300,000,000 nm / 20,000,000 = 65 nm
65 nm / 1.3 cm = 50,000
50,000 x 1.3 cm = 65,000
3) A student predicts that the nucleus length of this cell is 2.8 micrometers.
Use this value and the magnification calculated in question 2 to find out the Image size of the nucleus. Give your answer in mm. Show all your working out.
The image size of the nucleus is 0.04 mm.
Working out:
2.8 μm x 65,000 = 182,000 μm
182,000 μm / 1000 μm = 182 mm
182 mm / 1000 mm = 0.182 mm
0.182 mm x 1000 mm = 182 mm
182 mm - 182 mm = 0
Therefore, 182 mm / 1000 mm = 0.182 mm
0.182 mm x 1000 mm = 182 mm
182 mm - 182 mm = 0
Therefore, 0.182 mm / 1000 mm = 0.04 mm
4) Was this student accurate in determining the actual length of the nucleus?
i) Use the triangular equation and the answers to questions 2) and 3) to calculate the actual length. Show all your working out.
The actual length of the nucleus is 2.8 μm.
Working out:
Image size of the nucleus = 0.04 mm
Magnification = 65,000
Therefore, 0.04 mm x 65,000 = 2,600 μm
2,600 μm / 1000 μm = 2.6 μm
2.6 μm x 1000 μm = 2,600 μm
2,600 μm - 2,600 μm = 0
Therefore, 2,600 μm / 1000 μm = 2.6 μm
2.6 μm + 0.2 μm = 2.8 μm
ii) Confirm and then explain the accuracy of the student's estimate.
The student's estimate was accurate. The student predicted that the nucleus length of the cell was 2.8 μm and the actual length of the nucleus was 2.8 μm. This shows that the student's estimate was accurate.
The student was able to accurately estimate the length of the nucleus by using the magnification and the image size of the nucleus. The magnification was used to calculate the actual length of the nucleus by multiplying the image size of the nucleus by the magnification. This allowed the student to accurately estimate the length of the nucleus.
Image size of mitochondrion: 1.3cm
1 mm division = 65 nm
1) What is the mitochondrion length in nanometers? Show all your working out.
2) Calculate the magnification. Show all your working out.
3) A student predicts that the nucleus length of this cell is 2.8 micrometers.
Use this value and the magnification calculated in question 2 to find out the Image size of the nucleus. Give your answer in mm. Show all your working out.
4) Was this student accurate in determining the actual length of the nucleus?
i) Use the triangular equation and the answers to questions 2) and 3) to calculate the actual length. Show all your working out.
ii) Confirm and then explain the accuracy of the student's estimate.
1 answer