Question

2. Consider four spherical cells with the diameters given in the table below. Based on estimates from the scientific literature, the cytoplasm in cells can require 100,000 molecules of glucose per per second. These cells use a protein on the cell surface called the glucose transporter (Glu) to transport glucose from the outside of the cell into the cytoplasm. Scientific measurements indicate that under normal conditions each Glutr has a transport rate capacity of around 400 molecules per second. Each Glutr occupies approximately 12 nm of membrane surface. Glucose 200 um Glucose carrier protein Inside of cell SA/V Surface Volume Area () 1.2 x 10° 4.0 x 10' (4) Diameter 20 pm Glucose needed every second 4.0 x 10 # of Guy needed 1.0 x 10 Area of Glue (H) 1.2 x 10' % of SA covered by Glu 1.0% 0.3 4.0 x 10" 200 um 2.0 mm 2.0 cm 1.2 x 10° 4.0 x 100.03 1.2 x 107 4.0 x 10° 0.003 1.2 x 10° 4.0 x 100.0003 4.0 x 10" 1.0 x 10° 1.2 x 10 1.0 x 10" 1.2 x 10' 1.0 x 10" 1.2 x 10" 4.0 x 10" *Use a value of 3 for it to simplify calculations. Remember: SA-4rr' and V-(4/3)

Answer 1

As per your table all the values and calculation done as per surface area and volume. Now come to your % of SA covered by Glutr we have to divide the Area of Glutr by Surface area and your answer are:

Cell Diameter 20 pm Surface Volume SAN Area (H) (u) 1.2 x 104.0 x 100.3 Glucose needed every second 4.0 x 108 # of Gluy needed | 1.0 x 10° Area of Glu, (?) 1.2 x 10 % of SA covered by Glu, 1.0% 200 m 1.2 x 104.0 x 10° 0.03 4.0 x 10" 10% 1.0 x 10° 1.2 x 10 110 x 1012 121 2.0 mm 4.0 x 10" 100% 1.2 x 10' 4.0 x 10' 0.003 1.2 x 10° 4.0 x 100.0003 2.0 cm 4.0 x 107 1.0 x 10" 1.2 x 100 1000%