The great chef and amateur scientist Kahpper Bowles was watching a Nova program
ID: 272378 • Letter: T
Question
The great chef and amateur scientist Kahpper Bowles was watching a Nova program on cell division one day when he wondered what the growth rate was of his new yeast that he bought (Super Fast Rise Yeast). So, he decided to put both normal yeast and SFR yeast in separate sugar and honey solutions and measure the rate at which the cells grow by determining the relative number of cells in each of the two solutions. Calibrating his spectrophotometer for actual number of yeast cells present, he used his spectrophotometer and obtained the following results: Time Number of Cells Present (cells/ml) (days) Regular Yeast SFR Yeast 0 (start) 20 34 0.5 36 62 1 66 112 2 215 366 4 2319 3,942 7 81,920 139,264 10 287,411312,682 1 & 2. What is the doubling-time for cell division (td) for the two types of yeast during the first seven days? 3 & 4. What is the doubling-time for cell division (td) for the two types of yeast during days 7 to 10? Which type of yeast, if either, was growing fastest during the first week (days 1-7)? 5. 6. For each of the cell types, determine if they grew more slowly in days 7-10 than they did during the first week (i.e. was growth during days 7-10 slower than days 1-7)? 7 & 8. If a yeast cell has an approximate volume of 50 um3, at day 7 what percentage of the total volume of the mixture is taken up by the yeast cells in each of the two yeast mixtures? For problems 7 & 8, remember, the numbers given are concentr tions (cells/ml) and that I m|-1 cm3 and 1 ?m-10" cm, so that l ??_ (1 ?m)'= (104 cm)'-10-12 cm3]Explanation / Answer
Regular yeast is considered as yeast 1 and SFR yeast is considered as yeast 2.
1 & 2) t = 7 days
For yeast 1, Nt1 = 81920; N01 = 20;
For yeast 1, Nt2 = 139264; N02 = 34;
k = ln (Nt/N0)/t
k1 = ln (Nt1/N01)/t = ln ( 81920/20) / 7 =ln (4096) / 7 = 8.318 / 7 = 1.188
td1 = ln(2) / k1 = ln (2) /1.188 = 0.583 days
k2 = ln (Nt2/N02)/t = ln ( 139264/34) / 7 =ln (4096) / 7 = 8.318 / 7 = 1.188
td2 = ln(2) / k2 = ln (2) /1.188 = 0.583 days
3&4: 7 to 10 days. or t = 10-7 = 3 days
For yeast 1, Nt1 = 287411; N01 = 81920;
For yeast 1, Nt2 = 312682; N02 = 139264;
k = ln (Nt/N0)/t
k1 = ln (Nt1/N01)/t = ln ( 287411/81920) / 7 =ln (3.508) / 7 = 1.255 / 7 = 0.179
td1 = ln(2) / k1 = ln (2) /0.179 = 3.872 days
k2 = ln (Nt2/N02)/t = ln ( 312682/139264) / 7 =ln (2.245) / 7 = 0.808 / 7 = 0.116
td2 = ln(2) / k2 = ln (2) /0.116 = 5.974 days
5. During 1-7 days, td1 = 0.583 days and td1 = 0.583 days. So td1 = td2. So both yeast are growing at the same speed.
6. More doubling time means that the yeast is growing slow and low doubling time means that the yeast is growing fast. During 7-10 days, td1 = 3.872 days and td1 = 5.972 days. So td1 < td2. So SFR yeast is growing faster than the regular yeast.
7&8: Volume of one cell = 50 µm3 = 50*10-12 cm3 = 50*10-12 ml
At day 7,
Number of yeast 1 cells = 81920 cells/ ml
Number of yeast 2 cells = 139264 cells/ml
Total volume occupied = Number of yeast cells * Volume of one cell
Percentage of volume occupied = Total volume occupied * 100
Total volume occupied by yeast 1 = 81920 * 50*10-12= 4096000*10-12 = 0.4 *10-5 ml
Percentage of volume occupied by yeast 1 = 0.4*10-5 * 100 =0.4*10-3 %
Total volume occupied by yeast2 =139264* 50*10-12= 6963200*10-12 = 0.7 *10-5 ml
Percentage of volume occupied by yeast 2 = 0.7*10-5 * 100 =0.7*10-3 %
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.