Chapter 1 Test

FACTS OF lIFE

 What properties must proteins possess to interact with mol-ecules as different as simple ions, hydrophobic lipids, polar but uncharged carbohydrates, and nucleic acids? (Hint: Molecular Recognition through Structural Complementarity)

The amino acid side chains of proteins provide a range of shapes, polarity, and chemical features that allow a protein to be tailored to fit almost any possible molecular surface in a complementary way.

Consult Table 1.1 to compare the size of alanine (an amino acid) and immunoglobulin (a protein). Assume this protein can inter- act with alanine. What conclusions can you draw about the rela- tive surface area of the protein and the parts of it that interact with alanine. (Section 1.4)

(Hint: Molecular Recognition through Structural Complementarity)

Immunoglobulin is hundreds of times bigger than alanine. It would take only a very small fraction of the surface area of IgG to accommodate alanine.

What organizational features allow biological polymers to be informational macromolecules? Hint:Properties of informational Molecule

Biopolymers may be informational molecules because they are constructed of different monomeric units (“letters”) joined head to tail in a particular order (“words, sentences”). Polysaccharides are often linear polymers composed of only one (or two repeating) monosaccharide unit(s) and thus display little information content. Polysaccharides with a variety of monosaccharide units may convey information through specific recognition by other biomolecules. Also, most monosaccharide units are typically capable of forming branched polysaccharide structures that are potentially very rich in information content (as in cell surface molecules that act as the unique labels displayed by different cell types in multicellular organisms).

What are the two essential criteria for a linear polymer to be an informa- tional macromolecule?

Informational polymers must have “sense” or direction, and they must be composed of more than one kind of monomer unit.

DNA molecules adopt the famous double- helix shape (see Figure 1.4). Proteins adopt a variety of shapes (see Figures 1.10 and 1.16). Based on your understanding of the central dogma of molecular biology (see section 1.4b), provide a possible explanation for why DNA molecules adopt one shape whereas pro- teins have multiple different shapes?

DNA molecules essentially have only one function: “store infor- mation on how to make RNA and protein molecules.” Proteins are the “workhorses” of our biochemistry; the multitude of bio- chemical events happening in our cells are mediated by a multi- tude of different proteins; requiring them to adopt a multitude of different shapes.

Why is it important that weak forces, not strong forces, mediate biomolecular recognition? (Section 1.

Molecular recognition is based on structural complementarity. If complementary interactions involved covalent bonds (strong forces), stable structures would be formed that would be less responsive to the continually changing dynamic interactions that characterize living processes.

1. 4)How can you change the kinetic energy in a system of interacting molecules? (Hint: Weak forces in Biomolecular Recognition)

2. Predict the effect on such interactions of

   1. 1)  an increase in kinetic energy.

   2. 2)  a decrease in kinetic energy.

Raise the temperature.

b. The kinetic energy may overwhelm the energy of the weak forces and disrupt interactions between molecules.

c. Generally speaking, the opposite of b.

Biomacromolecules are often built by linking monomeric units through a process that removes a water molecule from between the monomers (see Figure 1.9). If one ton (1000 kg) of a monomer with a molecular mass of 200 g/mol is linked into a macromolecular structure, what volume (in L) of water would be released (1 mL water 5 1 g)?

1,000 kg of monomer would correspond to 5,000 mol monomer. (5,000–1) mol linkages would have been made with release
of (5,000–1) mol of water. 4,999 mol water corresponds to 89,982 g water. 89,982 g water would correspond to 89.982 L water.

Macromolecules are often formed
by linking simple monomeric units through repeated chemical bonds between such monomers (see Figure 1.8). Consider the structure shown below and draw the structure of a macromolecule that could be created from this monomer based on the chemistry shown in Figure 1.8.

H2N-CH2-O-CH2-COOH

As this compound has both an amine and a carboxylic acid functional group, it can make repeated amide bonds (as shown in Figure 1.9) with other molecules of its kind.

Calcium stearate, the calcium salt of stearic acid, is poorly soluble in water despite being an ionic compound. Explain why using your understanding of the weak forces described in Section 1.4. (it has long linkage change oxygens and ca

The long-chain alkanes that make up the structure of calcium stearate do not contain any functional groups that could interact favorably with water. This inability to interact with water will lead to a decreased solubility and increased tendency to separate from water (the hydrophobic effect).

Interatomic Distances in Weak Forces versus Chemical Bonds What is the distance between the centers of two carbon atoms (their limit of approach) that are interacting through van der Waals forces? What
is the distance between the centers of two carbon atoms joined in a covalent bond? (See Table 1.4.) (Section 1.4)

Two carbon atoms interacting through van der Waals forces are 0.34 nm apart;

Interatomic Distances in Weak Forces versus Chemical Bonds What is the distance between the centers of two carbon atoms (their limit of approach) that are interacting through van der Waals forces? What
is the distance between the centers of two carbon atoms joined in a covalent bond? (See Table 1.4.) (Section 1.4)

two carbon atoms joined in a covalent bond are 0.154 nm apart.

Cells as Steady-State Systems Describe what is meant by the phrase “cells are steady-state systems.” (Section 1.4)

Living systems are maintained by a continuous flow of matter and energy through them. Despite the ongoing transformations of matter and energy by these highly organized, dynamic systems, no overt changes seem to occur in them: They are in a steady state.

The Biosynthetic Capacity of Cells The nutritional requirements
of Escherichia coli cells are far simpler than those of humans, yet
the macromolecules found in bacteria are as complex as those of animals. Because bacteria can make all their essential biomolecules while subsisting on a simpler diet, do you think bacteria may have more biosynthetic capacity and thus more metabolic complexity than animals? Organize your thoughts on this question, pro and con, into a rational argument. (Section 1.5)

Because bacteria (compared with humans) have simple nutri- tional requirements, their cells obviously contain enzyme systems that allow them to convert rudimentary precursors (even inor- ganic substances such as NH4+, NO3−, N2, and CO2) into com- plex biomolecules—proteins, nucleic acids, polysaccharides, and complex lipids. On the other hand, animals have an assortment of different cell types designed for specific physiological func- tions; these cells possess a correspondingly greater repertoire of complex biomolecules to accomplish their intricate physiology.

Cell Structure Without consulting the figures in this chapter, sketch the characteristic prokaryotic and eukaryotic cell types and label their organelles and membranes. (Section 1.5)

Consult Figures 1.19 to 1.21 to confirm your answer.

11. The Dimensions of Prokaryotic Cells and Their Constituents Escherichia coli cells are about 2 μm long and 0.8 μm in diameter. (Section 1.5)

    1. How many E. coli cells laid end to end would fit across the diam- eter of a pinhead? (Assume a pinhead diameter of 0.5 mm.)

Laid end to end, 250 E. coli cells would span the head of a pin.

1. What is the volume of an E. coli cell? (Assume it is a cylinder, with the volume of a cylinder given by V 5 pr2h, where p 5 3.14.)

The volume of an E. coli cell is about 10215 L.

1. What is the surface area of an E. coli cell? What is the surface-to- volume ratio of an E. coli cell?

The surface area of an E. coli cell is about 6.3 3 10212 m2. Its surface-to-volume ratio is 6.3 3 106 m21.

1. Glucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of about 1 mM.

   What is the concentration of glucose, expressed as mg/mL? How many glucose molecules are contained in a typical E. coli

   cell? (Recall that Avogadro’s number 5 6.023 3 1023.)

600,000 molecules.

1. A number of regulatory proteins are present in E. coli at only one or two molecules per cell. If we assume that an E. coli cell 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 kDa, what is its con- centration, expressed as mg/mL?

5. 1.7 nM.

1. An E. coli cell contains about 15,000 ribosomes, supramolecular complexes that carry out protein synthesis. Assuming ribosomes are spherical and have a diameter of 20 nm (nanometers), what fraction of the E. coli cell volume is occupied by ribosomes?

5. Because we can calculate the volume of one ribosome to be

   4.2 3 10224 m3 (or 4.2 3 10221 L), 15,000 ribosomes would occupy 6.3 3 10217 L, or 6.3% of the total cell volume.

1. The E. coli chromosome is a single DNA molecule whose mass is about 3 3 109 Da. 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 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?

5. Because the E. coli chromosome contains 4600 kilobase pairs (4.6 3 106 bp) of DNA, its total length would be 1.6 mm— approximately 800 times the length of an E. coli cell.

1. 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.)

5. This DNA would encode 4300 different proteins, each 360 amino acids long.

11. The Dimensions of Mitochondria and Their Constituents Assume that mitochondria are cylinders 1.5 mm in length and 0.6 mm in diameter. (Section 1.5)

    1. What is the volume of a single mitochondrion?

The volume of a single mitochondrion is about 4.2 3 10216 L
(about 40% the volume calculated for an E. coli cell in problem 3).

11. The Dimensions of Mitochondria and Their Constituents Assume that mitochondria are cylinders 1.5 mm in length and 0.6 mm in diameter. Oxaloacetate is an intermediate in the citric acid cycle, an important metabolic pathway localized in the mitochondria of eukaryotic cells. The concentration of oxaloacetate in mitochondria is about 0.03 mM.

1. How many molecules of oxaloacetate are in a single mitochondrion?

A mitochondrion would contain on average fewer than eight molecules of oxaloacetate.

The Dimensions of Eukaryotic Cells and Their Constituents Assume that liver cells are cuboidal in shape, 20 mm on a side. (Section 1.5)

a. How many liver cells laid end to end would fit across the diam- eter of a pinhead? (Assume a pinhead diameter of 0.5 mm.)

Laid end to end 25 liver cells would spain the head of a pin

What is the volume of a liver cell?

Laid end to end, 25 liver cells would span the head of a pin.

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volume of an E. coli cell)

What is the surface area of a liver cell?

The surface area of a liver cell is 2.4 3 1029 m2.

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 3c.)

Cells exchange matter and communicate with the environment through their surface membranes.

What problems do cells with low surface-to-volume ratios face that cells with high surface-to-volume ratios don’t face?

4. Its surface-to-volume ratio is 3 3 105 m21, or about 0.05 (1/20) that of an E. coli cell.

   Cells with lower surface-to-volume ratios are limited in their exchange of materials with the environment.

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 3 1012 Da. If each nucleotide pair has a mass of 660 Da, what is the total number of nucleotide pairs in these 46 DNA molecules?

4. The number of base pairs in the DNA of a liver cell is 6.06 3 109 bp.

If each nucleotide pair contributes 0.34 nm to the length of DNA, what would be the total length of the 46 liver-cell DNA molecules if laid end-to-end?

How does this length compare with the overall dimensions of a liver cell?

4. The total length of the DNA in liver cell is 2.06 m (or more than 6 feet of DNA!) contained within a cell that is only
   20 mm on a side.

The maximal information in each set of liver cell chromosomes should be related to the number of nucleotide pairs in the chro- mosome 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.)

4. Maximal information content of liver cell DNA = 3.03 3 109 bp, which, expressed in proteins 400 amino acids in length, could encode 2.5 3 106 proteins.

A Simple Genome and Its Protein-Encoding Capacity The Mycoplasma genitalium genome consists of 523 genes, encoding 484 proteins, in just 580,074 base pairs (see Table 1.5).

a. What fraction of the M. genitalium genes encode proteins?

The fraction of the M. genitalium genes encoding proteins = 0.925.

A Simple Genome and Its Protein-Encoding Capacity The Mycoplasma genitalium genome consists of 523 genes, encoding 484 proteins, in just 580,074 base pairs (see Table 1.5).

b. What do you think the other genes encode?

Genes not encoding proteins encode RNA molecules.

A Simple Genome and Its Protein-Encoding Capacity The Mycoplasma genitalium genome consists of 523 genes, encoding 484 proteins, in just 580,074 base pairs (see Table 1.5).

c. 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?

(0.925)(580,074 base pairs) = 536,820 base pairs devoted to protein-coding genes.

A Simple Genome and Its Protein-Encoding Capacity The Mycoplasma genitalium genome consists of 523 genes, encoding 484 proteins, in just 580,074 base pairs (see Table 1.5).

d. If it takes three base pairs to specify an amino acid in a protein, how many amino acids are found in the average M. genitalium protein?

Since three base pairs specify an amino acid in a protein,
369 amino acids are found in the average M. genitalium protein.

A Simple Genome and Its Protein-Encoding Capacity The Mycoplasma genitalium genome consists of 523 genes, encoding 484 proteins, in just 580,074 base pairs (see Table 1.5).

e. If each amino acid contributes, on average, 120 Da to the mass of a protein, what is the mass of an average M. genitalium protein? (Section 1.5)

If each amino acid contributes on average 120 D to the mass of a protein, the mass of an average M. genitalium protein is 44,280 D.

AnEstimationofMinimalGenomeSizeforaLivingCell Oneprominent study of existing cells to determine the minimum number of genes needed by a living cell has suggested that 206 genes are sufficient.

If the ratio of protein-coding genes to non–protein-coding genes is the same in this minimal organism as the genes of Mycoplasma genitalium, how many proteins are represented in these 206 genes?

How many base pairs would be required to form the genome of this minimal organism if the genes are the same size as M. genitalium genes? (Section 1.5)

(0.925)(206) = 191 proteins. Assuming its genes are the same size as M. genitalium, the minimal genome would be 228,480 base pairs.

An Estimation of the Number of Genes in a Virus Virus genomes range in size from approximately 3500 nucleotides to 280,000 base pairs. The genome of SARS-CoV-2, the virus responsible for the COVID-19 pandemic, consists of 29,811 nucleotides.

If viral genes are about the same size as M. genitalium genes, what is the minimum and maximum number of genes in viruses? (Section 1.6)

Given 1109 nucleotides (or base pairs) per gene, the minimal virus, with a 3500-nucleotide genome, would have only three genes; the maximal virus, with a 280,000-bp genome, would have 252 genes.

1. Intracellular Protein Transport Proteins made by ribosomes associated with the endoplasmic reticulum (ER) may pass into the ER mem- brane or enter the lumen of the ER. Devise pathways by which:

1. a plasma membrane protein may reach the plasma membrane.

19. Fate of proteins synthesized by the rough ER:

    1. Membrane proteins would enter the ER membrane, and, as part of the membrane, be passed to the Golgi, from which vesicles depart and fuse with the plasma membrane.

1. Intracellular Protein Transport Proteins made by ribosomes associated with the endoplasmic reticulum (ER) may pass into the ER mem- brane or enter the lumen of the ER. Devise pathways by which:

1. a protein destined for secretion is deposited outside the cell. (Section 1.5)

A secreted protein would enter the ER lumen and be trans- ferred as a luminal protein to the Golgi, from which vesicles depart. When the vesicle fuses with the plasma membrane, the protein would be deposited outside the cell.