Question

We made a simple estimate of the mass of an \(E\). coll cell by assuming that such cells have the same density of water. However, a more reasonable estimate is that the density of the macromolecules of the cell is 1.3 times that of water \((\mathrm{BNID} 101502,104272) .\) As a result, the estimate of the mass of an \(E\) coli cell is off by a bit. Using that two-thirds of the mass is water and that the remaining one-third is macromolecular, compute the percentage error made by treating the macromolecular density as the same as that of water.

Short answer

The percentage error made by treating the macromolecular density the same as that of water is \((\frac{(Mw + 1.3 \times Mw/2)-M}{M}) \times 100 \% \)

Step-by-step solution

Identify the proportions of mass contributed by water and macromolecules

According to the given conditions, two-thirds of the mass is water and one-third is macromolecular. Let's assume the total mass of the E.coli cell is M (which includes the mass contribution from both water and macromolecules).

Compute the mass of macromolecules using the correct density

The density of macromolecules is 1.3 times that of water. Hence, the actual mass of macromolecules (Mm_actual) contributing to the total mass M is 1.3 times the mass considering water density (Mm_water). So, \(Mm_{\text{actual}}=1.3 \times Mm_{\text{water}}\). Let's denote the mass of water as \(Mw\). From the proportions given, we know that \(Mw = 2 \times Mm_{\text{water}}\) or \(Mm_{\text{water}} = \frac{Mw}{2}\). Combining these equations we find \(Mm_{\text{actual}}=1.3 \times \frac{Mw}{2}\).

Calculate the total actual mass and the percentage error

The actual total mass (M_actual) of the E.coli cell is the sum of the actual mass of water and macromolecules, i.e., \(M_{\text{actual}}= Mw + Mm_{\text{actual}}\). Substituting for \(Mm_{\text{actual}}\) from the previous step we find \(M_{\text{actual}}= Mw + 1.3 \times \frac{Mw}{2}\). The total estimated mass Me (using the incorrect assumption that the density of macromolecules equals that of water) is M. To find the percentage error we use the formula \((\frac{M_{\text{actual}}-Me}{Me}) \times 100\%. Since Me=M, replacing this in the formula gives \((\frac{M_{\text{actual}}-M}{M}) \times 100\%\). Replacing \(M_{\text{actual}}\) from our previous calculation completes our solution.

Key concepts

E. coli cell mass estimation

Estimating the mass of an E. coli cell involves understanding the proportions of its components. Typically, an E. coli cell is made up of both water and macromolecules. For estimation, we must consider that, on average:

• Two-thirds of the cell mass is water.
• One-third is composed of macromolecules.

The important factor here is the density of these macromolecules compared to water. Initially, you might assume both have equal densities, misjudging the actual cell mass. However, it's known that the density of macromolecules is about 1.3 times that of water. This affects the calculated mass significantly. To adjust for this, calculations incorporate the real density to ensure accuracy in mass estimation. Underestimating this factor leads to errors, making the newly refined approach much more reliable. This precise estimation is vital in microbiology for understanding cellular functions and responses.

macromolecule density

The density of macromolecules within a cell is a critical metric in the field of cell biology. Macromolecules, such as proteins, DNA, and polysaccharides, contribute to the cell's mass but do not share the same density as water.

The density of these macromolecules is roughly 1.3 times higher than water. This means that for every unit of mass, macromolecules are more compact than water. This property ensures cells maintain structural integrity and optimal functionality. It also affects how cells respond to external stimuli and interact with their environment.
Ignoring this density differential can lead to inaccuracies in scientific calculations. Hence, when measuring or estimating cell mass, it's crucial to use the correct density figures. This understanding plays a fundamental role in a range of biological and chemical processes, emphasizing the critical nature of accurate density measurements in biological science.

water density

Water forms the backbone of cellular life, comprising approximately two-thirds of an E. coli cell's mass. Its density is considered a baseline when estimating and comparing other substances within the cell. Water density is about 1 g/cm³ at standard temperature and pressure (STP), making it a standard reference point in scientific calculations and experiments.

In biological contexts, water density is vital for understanding buoyancy, diffusion, and cellular osmosis. It's straightforward but essential when estimating the mass of a cell. Considering two-thirds of the cell mass comes from water, it highlights water's crucial role in maintaining cell volume and pressure.
By providing a zero-point for relative density studies, understanding water density helps to calculate how much heavier or lighter other cellular components, like macromolecules, are. This basic yet indispensable knowledge facilitates more accurate estimations and deeper insights into cellular biology's complexities.