Assume that liver cells are cuboidal in shape, \(20 \mu \mathrm{m}\) on a side. a. How many liver cells laid end to end would fit across the diameter of a pinhead? (Assume a pinhead diameter of \(0.5 \mathrm{mm} .\) ) b. What is the volume of a liver cell? (Assume it is a cube.) c. What is the surface area of a liver cell? What is the surface to-volume ratio of a liver cell? How does this compare to the surface-to-volume ratio of an \(E\) coli cell (compare this answer with that of problem \(3 c\) )? What problems must cells with low surface to-volume ratios confront that do not occur in cells with high surface-to-volume ratios? A. A human liver cell contains two sets of 23 chromosomes, each set being roughly equivalent in information content. The total mass of DNA contained in these 46 enormous DNA molecules is \(4 \times 10^{12}\) daltons. Because each nucleotide pair contributes 660 daltons to the mass of DNA and 0.34 nm to the length of DNA, what is the total number of nucleotide pairs and the complete length of the DNA in a liver cell? How does this length compare with the overall dimensions of a liver cell? The maximal information in each set of liver cell chromosomes should be related to the number of nucleotide pairs in the chromosome set's DNA. This number can be obtained by dividing the total number of nucleotide pairs just calculated by 2 . What is this value? If this information is expressed in proteins that average 400 amino acids in length and three nucleotide pairs encode one amino acid in a protein, how many different kinds of proteins might a liver cell be able to produce? (In reality, liver cell DNA encodes approximately 20,000 different proteins. Thus, a large discrepancy exists between the theoretical information content of DNA in liver cells and the amount of information actually expressed.)

Step by step solution

01

Calculate how many liver cells fit across the diameter of a pinhead

First convert the pinhead diameter to \(\mu m\) since the liver cell size is given in \(\mu m\). Hence the conversion is \(0.5mm = 500\mu m\). Then, divide the diameter of the pinhead by the size of the liver cell, \(500\mu m / 20\mu m = 25\) cells.

02

Calculate the volume of a liver cell

If we assume the liver cell is a perfect cube, the calculation of its volume in \(µm^3\) is a simple cube of its side length. Hence, \(Volume = 20\mu m × 20\mu m × 20\mu m = 8000\mu m^3\).

03

Calculate the surface area of a liver cell

The surface area of a cube is given by the formula \(6a^2\), where \(a\) represents the length of a side. So, following this formula, the surface area of a liver cell is \(6 × (20\mu m)² = 2400\mu m²\).

04

Calculate the surface-to-volume ratio of a liver cell

The surface-to-volume ratio is simply calculated as the surface area divided by the volume. Therefore, \(Surface to Volume Ratio = 2400\mu m² / 8000\mu m^3 = 0.3\mu m^{-1}\). This result allows comparing with the surface-to-volume ratio of another cell, such as the E.Coli mentioned in the problem.

05

Calculate the total number of nucleotide pairs and the complete length of the DNA in a liver cell

Given the total mass of DNA and the mass contribution of each nucleotide pair, we can calculate the total number of nucleotide pairs by dividing the total mass by the mass of each pair. \(4 × 10^{12}\) daltons/ 660 daltons = approx 6.06 × 10^{9} pairs. To find the total length, we multiply it by the length each nucleotide pair contributes. Hence, \(6.06 × 10^{9} pairs × 0.34 nm/pair = approx 2.06 × 10^{9} nm = approx 2.06 m\). We can see this length is much longer than the dimensions of the liver cell.

06

Calculate the maximum number of different proteins a liver cell can produce

The amount of information in each liver cell chromosome is equivalent to half the total number of nucleotide pairs, which is \(6.06 × 10^{9} pairs / 2 = approx 3.03 × 10^{9} pairs\). As each amino acid in a protein is encoded by three pairs, we divide \(3.03 × 10^{9} pairs / 3 = approx 1.01 × 10^{9} amino acids\). If a protein has an average of 400 amino acids, then the liver cell can theoretically produce \(1.01 × 10^9 amino acids / 400 amino acids/protein = 2.52 × 10^6\) different proteins.

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Key Concepts

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

Surface-to-Volume Ratio in Liver Cells

The surface-to-volume ratio is a critical concept in cell biology, especially important for understanding how cells function and interact with their environment. In the context of liver cells, this concept illustrates how the cell's surface area relates to its overall volume. When considering the geometry of a cell, like the liver cell assumed to be a cube in the exercise, this ratio fundamentally affects how substances move in and out of the cell.

Surface area represents the part of the cell that is exposed to the external environment and through which gas exchange, nutrient absorption, and waste expulsion occur. Volume, on the other hand, indicates the space available for these materials once inside the cell, as well as space for intracellular structures. A higher surface-to-volume ratio means that a cell has more surface area relative to its volume, which generally enhances its ability to quickly exchange materials with its surroundings.

A liver cell, like many other cells, must maintain an optimal surface-to-volume ratio. A low ratio can lead to challenges such as difficulties in efficiently exchanging gases and nutrients due to limited surface area. In contrast, bacteria like E. coli tend to have a higher surface-to-volume ratio, aiding in faster exchange rates suitable for their metabolic needs.

Understanding DNA Nucleotide Pairs in Liver Cells

The liver cell's genome is comprised of DNA, where the fundamental unit is the nucleotide pair. This pair is part of a double helix structure that makes up the chromosomes found in the nucleus of the cell. The exercise highlights that human liver cells contain two sets of 23 chromosomes with a total mass for these large DNA molecules. Nucleotide pairs contribute to both the mass and length of the DNA.

Each nucleotide pair has a mass of 660 daltons. By knowing the total mass of DNA within a liver cell, students can calculate the total number of nucleotide pairs. This provides insights into the enormous amount of genetic information contained within a single cell, which is then translated into a diversity of proteins necessary for the cell to perform its functions. The total length of the DNA strands compared to the dimensions of the liver cell also emphasizes how efficiently DNA is packed within the cell nucleus.

Protein Synthesis in Liver Cells

Protein synthesis is a vital function within liver cells and is a process that translates the genetic code into functional proteins. Proteins are essential for virtually all cell functions, from structure to enzymes catalyzing biochemical reactions.

In the case of the liver cell, the vast amount of DNA holds the genetic code for a theoretically huge number of different proteins. Amino acids, the building blocks of proteins, are encoded by sequences of three nucleotide pairs known as codons. Given the average length of proteins and the understanding that three nucleotide pairs encode a single amino acid, we can estimate the theoretical number of different proteins a cell can produce.

Despite the high theoretical protein production capacity, the real number of proteins expressed by liver cells is much lower. This discrepancy reminds students that not all of the genetic information is used for protein synthesis - some regions of DNA serve regulatory functions, while others may be silent or non-coding, showcasing the complexity of genetic expression in cells.

Cell Dimension Calculations

Cell dimension calculations help students understand the scale at which cellular processes take place. By assuming that liver cells are perfect cubes with a side length, students learn to calculate cell volume and surface area, which in turn are used to understand the cell's surface-to-volume ratio.

In the provided exercise, the size of a liver cell is compared to the diameter of a pinhead, prompting students to visualize and compare sizes in a real-world context. This not only aids in understanding the microscale but also demonstrates the precision with which cells must operate. Accurately calculating cell dimensions enables students to appreciate the intricate balance between a cell's structure and its function, such as how size impacts its efficiency in processing substances and communicating with its environment.