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The Size of Cells and Their Components A typical eukaryotic cell has a cellular diameter of \(50 \mu \mathrm{m}\). a. If you used an electron microscope to magnify this cell 10,000-fold, how big would the cell appear? b. If this cell were a liver cell (hepatocyte) with the same cellular diameter, how many mitochondria could the cell contain? Assume the cell is spherical; that the cell contains no other cellular components; and that each mitochondrion is spherical, with a diameter of \(1.5\) \(\mu \mathrm{m}\). (The volume of a sphere is \(4 / 3 \pi r^{3}\).) c. Glucose is the major energy-yielding nutrient for most cells. Assuming a cellular concentration of \(1 \mathrm{~mm}\)

Short Answer

Expert verified
a) The magnified size is 500,000 \( \mu m \). b) The cell could contain approximately 36,977 mitochondria.

Step by step solution

01

Calculate the Magnified Size

The given cell diameter is 50 \( \mu m \). With a magnification of 10,000 times, the apparent size of the cell can be calculated as:\[ \text{Magnified size} = \text{original size} \times \text{magnification} = 50 \mu m \times 10,000 = 500,000 \mu m\]So, the cell would appear to be 500,000 \( \mu m \) in diameter.
02

Calculate the Volume of the Cell

Assuming the cell is spherical, use the formula for the volume of a sphere, \( \frac{4}{3} \pi r^3 \), where the radius \( r = \frac{50}{2} = 25 \mu m \).\[\text{Volume of the cell} = \frac{4}{3} \pi (25)^3\]Calculate this to find:\[\text{Volume of the cell} \approx 65,450 \mu m^3\]
03

Calculate the Volume of a Mitochondrion

Each mitochondrion is also spherical, with a diameter of 1.5 \( \mu m \), so its radius \( r = \frac{1.5}{2} = 0.75 \mu m \).\[\text{Volume of a mitochondrion} = \frac{4}{3} \pi (0.75)^3\]Calculate this to find:\[\text{Volume of a mitochondrion} \approx 1.77 \mu m^3\]
04

Calculate the Maximum Number of Mitochondria

To find the number of mitochondria that could fit in the cell, divide the volume of the cell by the volume of a mitochondrion, assuming the cell contains no other components:\[\text{Number of mitochondria} = \frac{65,450}{1.77} \approx 36,977\]Therefore, about 36,977 mitochondria could fit in the cell.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eukaryotic Cells
Eukaryotic cells are complex structures that make up multicellular organisms like plants, animals, and fungi. Unlike prokaryotic cells, which lack a nucleus, eukaryotic cells have their genetic material enclosed within a nuclear membrane. This distinct nucleus is a hallmark of eukaryotic cells, allowing for greater organization and regulation of genetic information.

These cells are larger and more intricate than their prokaryotic counterparts. They contain numerous organelles, such as mitochondria for energy production, endoplasmic reticulum for protein and lipid synthesis, and Golgi apparatus for packaging and sorting molecules. This compartmentalization allows eukaryotic cells to perform specialized functions efficiently.
  • Contain a nucleus and membrane-bound organelles
  • Found in multicellular organisms
  • Enable complex and regulated biological processes
Cellular Structures
Within eukaryotic cells, cellular structures play critical roles in maintaining cellular function and integrity. These structures include organelles like mitochondria, ribosomes, and the cytoskeleton.

Each organelle has a unique role that contributes to the overall operation of the cell. Mitochondria, for example, are known as the powerhouses of the cell due to their role in converting nutrients into energy-rich molecules like ATP. The cytoskeleton provides structural support and facilitates cell movement.
  • Mitochondria: energy conversion
  • Ribosomes: protein synthesis
  • Cytoskeleton: structural support and movement
Exploring these structures helps us understand how cells operate at a microscopic level, balancing the needs for energy, growth, and reproduction.
Electron Microscopy
Electron microscopy is a powerful technique used to visualize the detailed structure of cells at very high magnifications. Unlike traditional light microscopes, electron microscopes use beams of electrons, which have much shorter wavelengths, allowing for higher resolution images. This makes it possible to see even the smallest cellular components, such as the intricate architecture of organelles and molecules.

There are two main types of electron microscopes: transmission electron microscopes (TEM) and scanning electron microscopes (SEM). TEM provides detailed internal images of thin cell sections, revealing internal structures with high clarity. SEM, on the other hand, is used for viewing surface details, providing three-dimensional images.
  • TEM: detailed internal cell structure
  • SEM: surface structure and 3D imaging
This technology is invaluable in research and diagnostics, allowing scientists to study cell biology at the molecular level.
Cellular Dimensions
Understanding cellular dimensions is essential for grasping the scale and functionality of cells. In the case of eukaryotic cells, which are typically much larger than prokaryotic cells, their size can range from 10 to 100 micrometers in diameter.

These dimensions have significant implications for the cell's ability to house various organelles. Larger cell volumes provide space for numerous mitochondria, as seen in the exercise, which is crucial for energy-intensive cells like liver cells. Calculating cellular volumes and the volume of organelles, such as mitochondria, helps in understanding how cells are structured to meet their metabolic needs. With the formula for the volume of a sphere, \ \( \frac{4}{3} \pi r^3 \ \), scientists can determine how many organelles can fit within a cell.
  • Larger cell size means more organelles
  • Essential for energy production and cellular processes
  • Volume calculations help understand cell capacity
These calculations and insights are critical for fields like cell biology and bioengineering, where cellular design impacts functionality.

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Most popular questions from this chapter

Biomolecule Researchers isolated an unknown substance, \(X\), from rabbit muscle. They determined its structure from the following observations and experiments. Qualitative analysis showed that \(X\) was composed entirely of \(C, H\), and \(O\). \(A\) weighed sample of \(X\) was completely oxidized, and the \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) produced were measured; this quantitative analysis revealed that \(\mathrm{X}\) contained \(40.00 \% \mathrm{C}, 6.71 \% \mathrm{H}\), and \(53.29 \% \mathrm{O}\) by weight. The molecular mass of \(\mathrm{X}\), determined by mass spectrometry, was \(90.00\) u (atomic mass units; see Box 1-1). Infrared spectroscopy showed that \(X\) contained one double bond. X dissolved readily in water to give an acidic solution that demonstrated optical activity when tested in a polarimeter. a. Determine the empirical and molecular formula of \(X\). b. Draw the possible structures of \(X\) that fit the molecular formula and contain one double bond. Consider only linear or branched structures and disregard cyclic structures. Note that oxygen makes very poor bonds to itself. c. What is the structural significance of the observed optical activity? Which structures in (b) are consistent with the observation?

Mutation and Protein Function Suppose that the gene for a protein 500 amino acids in length undergoes a mutation. If the mutation causes the synthesis of a mutant protein in which just one of the 500 amino acids is incorrect, the protein may lose all of its biological function. How can this small change in a protein's sequence inactivate it?

chemically unstable compared with its oxidation products, \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). a. What can one say about the standard free- energy change for this reaction? b. Why doesn't firewood stacked beside the fireplace undergo spontaneous combustion to its much more stable products? c. How can the activation energy be supplied to this reaction? d. Suppose you have an enzyme (firewoodase) that catalyzes the rapid conversion of firewood to \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) at room temperature. How does the enzyme accomplish that in thermodynamic terms?

Cryptobiotic Tardigrades and Life Tardigrades, also called water bears or moss piglets, are small animals that can grow to about \(0.5 \mathrm{~mm}\) in length. Terrestrial tardigrades (pictured here) typically live in the moist environments of mosses and lichens. Some of these species are capable of surviving extreme conditions. Some tardigrades can enter a reversible state called cryptobiosis, in which metabolism completely stops until conditions become hospitable. In this state, various tardigrade species have withstood dehydration, extreme temperatures from \(-200{ }^{\circ} \mathrm{C}\) to \(+150{ }^{\circ} \mathrm{C}\), pressures from 6,000 atm to a vacuum, anoxic conditions, and the radiation of space. Do tardigrades in cryptobiosis meet the definition of life? Why or why not?

The High Rate of Bacterial Metabolism Bacterial cells have a much higher rate of metabolism than animal cells. Under ideal conditions, some bacteria double in size and divide every \(20 \mathrm{~min}\), whereas most animal cells under rapid growth conditions require 24 hours. The high rate of bacterial metabolism requires a high ratio of surface area to cell volume. a. How does the surface-to-volume ratio affect the maximum rate of metabolism? b. Calculate the surface-to-volume ratio for the spherical bacterium Neisseria gonorrhoeae (diameter \(0.5 \mu \mathrm{m}\) ), responsible for the disease gonorrhea. The surface area of a sphere is \(4 \pi r^{2}\). c. How many times greater is the surface-to-volume ratio of Neisseria gonorrhoeae compared to that of a globular amoeba, a large eukaryotic cell (diameter 150 \(\mu \mathrm{m}\) )?

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