Yeast cells can respire aerobically or anaerobically - AQA - A-Level Biology - Question 3 - 2018 - Paper 2

The movement of the coloured liquid to the right indicates that carbon dioxide was being produced as a by-product of cellular respiration. As yeast metabolizes the glucose present in the culture, it releases carbon dioxide, which increases the pressure in the apparatus, causing the liquid to move.

To calculate the volume of gas produced, we use the formula for the volume of a cylinder:

$V = ext{Area} imes ext{Length}$

The area of the capillary tubing's cross-section (lumen) is calculated as:

A = rac{ ext{π} d^2}{4} = rac{3.14 imes (0.1 ext{ cm})^2}{4} = 0.00785 ext{ cm}^2

Then, since the liquid moved 1.5 cm in 24 hours:

$V = 0.00785 ext{ cm}^2 imes 1.5 ext{ cm} = 0.011775 ext{ cm}^3$

To find the volume produced per hour:

ext{Volume per hour} = rac{0.011775 ext{ cm}^3}{24 ext{ hours}} imes 60 ext{ minutes} = 0.0294 ext{ cm}^3 ext{ hour}^{-1}

A log scale is utilized to accommodate the exponential growth patterns observed in population studies. This makes it easier to visualize large ranges of cell numbers and highlight the relative changes in population size over time, particularly during rapid growth phases.

One reason many yeast cells die during the death phase is the depletion of essential nutrients in the medium, which leads to increased competition for limited resources and ultimately results in cell death.

Using the equation:

$X_t = X_0 e^{rt}$

where:

• $X_0 = 2000$ (initial population)
• $r = 0.5$ (growth rate)
• $t = 10$ (time in hours)

We calculate:

$X_t = 2000 e^{0.5 imes 10} = 2000 e^{5}$

Substituting $e^{5} ext{ (approximately 148.41)}$:

$X_t = 2000 imes 148.41 hickapprox 296820$

Thus, the predicted size of the population after 10 hours is approximately 296820 yeast cells.