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Chapter 1: Problem 12

The genome of the Mycoplasma genitalium consists of 523 genes, encoding 484 proteins, in just 580,074 base pairs (Table 1.6). What fraction of the \(M .\) genitalium genes encode proteins? What do you think the other genes encode? If the fraction of base pairs devoted to protein-coding genes is the same as the fraction of the total genes that they represent, what is the average number of base pairs per protein-coding gene? If it takes 3 base pairs to specify an amino acid in a protein, how many amino acids are found in the average \(M .\) genitalium protein? If each amino acid contributes on average 120 Daltons to the mass of a protein, what is the mass of an average M. genitalium protein?

Short Answer

Expert verified
The fraction of M. Genitalium genes that encode proteins is 0.925. The non-protein-coding genes likely encode functional RNA molecules like rRNAs and tRNAs. The average number of base pairs per protein-encoding gene is nearly 1199. The average number of amino acids in a M. genitalium protein is approximately 400. If each amino acid contributes 120 Daltons to the protein mass, the average mass of a M. genitalium protein is about 48000 Daltons.

Step by step solution

01

Calculate the fraction of genes that encode proteins

The fraction can be calculated by dividing the number of proteins (484) by the total number of genes (523). This will give the fraction of the M. genitalium genes encoding proteins.
02

Identify what non-protein coding genes might encode

Scores of genes encode functional RNA molecules involved in protein synthesis, such as ribosomal RNAs (rRNAs) and transfer RNAs (tRNAs), regulating gene expression, or other cellular functions.
03

Compute the average number of base pairs per protein-coding gene

If the fraction of base pairs assigned to protein-coding genes equals the fraction of the total genes they signify, then the average number of base pairs per protein-coding gene can be calculated as the total number of base pairs (580,074) divided by the number of protein-encoding genes (484).
04

Calculate the number of amino acids in an average M. genitalium protein

If three gene base pairs determine an amino acid in a protein, the previous result should be divided by 3 to get the average number of amino acids per M. genitalium protein.
05

Determine the mass of an average M. genitalium protein

To derive the mass of a M. genitalium protein, each amino acid should be analyzed to assess its contribution to the protein's mass. An estimated scale is previously given as 120 Dalton. Thus, by multiplying the average number of amino acids per protein by 120 Daltons, we can estimate the mass of an average M. genitalium protein.

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Most popular questions from this chapter

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.)

Without consulting the figures in this chapter, sketch the characteristic prokaryotic and eukaryotic cell types and label their pertinent organelle and membrane systems.

Biomolecules interact with one another through molecular surfaces that are structurally complementary. How can various proteins interact with molecules as different as simple ions, hydrophobic lipids, polar but uncharged carbohydrates, and even nucleic acids?

Escherichia coli cells are about \(2 \mu \mathrm{m}\) (microns) long and \(0.8 \mu \mathrm{m}\) in diameter. a. How many \(E\). coli 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 an \(E\). coli cell? (Assume it is a cylinder, with the volume of a cylinder given by \(V=\pi r^{2} h,\) where \(\pi=3.14 .\) c. What is the surface area of an \(E\). colicell? What is the surface-to volume ratio of an \(E .\) colicell? d. Glucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of about \(1 \mathrm{m} M\). What is the concentration of glucose, expressed as \(\mathrm{mg} / \mathrm{mL}\) ? How many glucose molecules are contained in a typical \(E\) coli cell? (Recall that Avogadro's number \(=6.023 \times 10^{23}\).) e. A number of regulatory proteins are present in \(E\). coli at only one or two molecules per cell. If we assume that an \(E\). colicell contains just one molecule of a particular protein, what is the molar concentration of this protein in the cell? If the molecular weight of this protein is \(40 \mathrm{kD},\) what is its concentration, expressed as \(\mathrm{mg} / \mathrm{mL} ?\) f. \(\operatorname{An} E .\) coli cell contains about 15,000 ribosomes, which carry out protein synthesis. Assuming ribosomes are spherical and have a diameter of \(20 \mathrm{nm}\) (nanometers), what fraction of the \(E .\) colicell volume is occupied by ribosomes? g. The \(E\) coli chromosome is a single DNA molecule whose mass is about \(3 \times 10^{9}\) daltons. This macromolecule is actually a linear array of nucleotide pairs. The average molecular weight of a nucleotide pair is \(660,\) and each pair imparts \(0.34 \mathrm{nm}\) to the length of the DNA molecule. What is the total length of the E. coli chromosome? How does this length compare with the overall dimensions of an \(E\). coli cell? How many nucleotide pairs does this DNA contain? The average \(E\). coli protein is a linear chain of 360 amino acids. If three nucleotide pairs in a gene encode one amino acid in a protein, how many different proteins can the E. coli chromosome encode? (The answer to this question is a reasonable approximation of the maximum number of different kinds of proteins that can be expected in bacteria.)

Describe what is meant by the phrase "cells are steady-state systems."
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