Cell Questions Ch. 1-4

Question 1

How can physics and chemistry account for Life?

Answer 1

Plants and animals obey the laws of thermodynamics; neither cells nor organisms can create energy from nothing. All organisms require an input of energy from the environment to grow and reproduce—even to stay alive. The structures of large proteins can be determined in weeks, and chemists can sequence whole bacterial genomes even more quickly. Organic chemists now understand enzyme- catalyzed reactions as well as any they study. The details of the metabolism of obscure bacteria living at extreme ocean depths on a diet of sulfur and carbon monoxide are well understood. The genes that control the intricate body plans of insects are mapped and sequenced.

Question 2

T or F?: Eukaryotic cells contain either mitochondria or chloroplasts, but not both.

Answer 2

False. Plant cells contain both mitochondria and chloroplasts

Question 3

T or F?: Most of the DNA sequences in a bacterial genome code for proteins, whereas most of the DNA sequences in the human genome do not.

Answer 3

True. Bacterial genomes seem to be pared down to the essentials: most of the DNA sequences encode proteins, a few encode functional RNAs, a small amount of DNA is devoted to regulating gene expression, and there are very few extraneous, nonfunctional sequences. By contrast, only about 1.5% of the DNA sequences in the human genome is thought to code for proteins. Even allowing for large amounts of regulatory DNA, much of the human genome is composed of DNA with no apparent function.

Question 4

T or F?: The only horizontal gene transfer that has occurred in animals is from the mitochondrial genome to the nuclear genome.

Answer 4

False. In addition to transfers from the mitochondrial genome, there are many examples of transfers of viral genomes; for example, some 1% of the mouse genome arose from copies of a sequence that originated as the genome of the mouse mammary tumor virus. What is rare is the transfer of genes from other species.

Question 5

Why False? Plant and animal tissues are considered different because animal tissues do not have conspicuous boundaries that divide the cell.

Answer 5

Initially thought to be true because animal cells do not have cell walls, which made it difficult for early investigators to identify individual cells using the crude microscope available during that time. Schwann (1839) shown that this was incorrect for cartilage cells, which have well defined boundaries of collagen fibers, and later extended to all animal cells.

Question 6

Why False? Living organisms are not governed by the laws of chemistry and physics, but are subject to a “vital force” that is responsible for the formation of organic compounds.

Answer 6

Initially thought to be true because living organisms seems to increase in complexity spontaneously. Synthesis of urea, an organic compound made by all living cells from an inorganic compound by Wohler (1828) finally disproved this concept.

Question 7

Why False? Hereditary materials are of proteins but not DNA, because DNA contains only four nucleotides as monomers.

Answer 7

Originally thought to be true because tetranucleoide in DNA was not complex enough to encode hereditary factors

Question 8

Why False? The fermentation of sugar to alcohol can take place only in the presence of yeast cells.

Answer 8

Initially thought to be true, because of the demonstration by Pasteur that yeast cells were needed for alcoholic fermentation. Later Buchner and Buchner (1897) showed that extracts from yeast cells could substitute for intact cells, an effect which we know now is due to the presence of enzymes that catalyze various steps in fermenting reaction.

Question 9

Why False? The flow of genetic information in a cell is always from DNA to RNA to protein.

Answer 9

Initially thought to be true, however, discovery of the process of reverse transcription demonstrated that in retroviruses genetic information flows from RNA to DNA.

Question 10

Why False? DNA always exists as a duplex of two strands wound together into a right-handed helix.

Answer 10

Initially thought to be true that right-handed helix is the only form of DNA, later on it was discovered the right handed form of DNA as well as single stranded DNA.

Question 11

Why False? The genetic code specifying how the information present in the DNA molecule is used to make proteins is universal in the sense that all organisms use the same code

Answer 11

Genetic code is not universal. In some prokaryotes codons encode different amino acids.

Question 12

Indicate how the property of carbon atom contributes its role as the most important atom in biomolecules:
The carbon atom has a valance of four.

Answer 12

Because of valance of 4, C can form multiple covalent bonds with either other atoms or with C atoms itself to form greater diversity.

Question 13

Indicate how the property of carbon atom contributes its role as the most important atom in biomolecules:
The C-C bond has a bond energy that is above the energy of photons of light in the visible range (400 – 700 nm).

Answer 13

Stability, visible light, cells can survive

Question 14

Indicate how the property of carbon atom contributes its role as the most important atom in biomolecules:
A carbon atom can bond simultaneously to two other carbon atoms

Answer 14

Long chain C atom compounds as well as ring structures

Question 15

Indicate how the property of carbon atom contributes its role as the most important atom in biomolecules:
Carbon atoms can bond readily to H, N, and S atoms.

Answer 15

Diversity

Question 16

Indicate how the property of carbon atom contributes its role as the most important atom in biomolecules:
Carbon-containing compounds can contain asymmetric carbon atoms.

Answer 16

Structural diversity in the form of stereoisomers

Question 17

T or F? and what benefit?

Water is a polar molecule, hence an excellent solvent for polar compounds,

Answer 17

T, Living organisms are essentially aqueous solutions containing many kinds of molecules, most of which are polar and are readily soluble in water.

Question 18

T or F? and what benefit?

Water can be formed by the reduction of molecular oxygen (O2.)

Answer 18

T, Oxygen is the ultimate electron acceptor in cellular respiration with H20 as product

Question 19

T or F? and what benefit?

Water does not absorb visible light.

Answer 19

T, this property of water allows light to penetrate readily so that submerged photosynthetic organisms can survive.

Question 20

Carefully state a reasonable conclusion that can be drawn from the experimental findings concerning the reassembly of tobacco mosaic virus (TMV) virions from TMV RNA and coat protein subunits.

When RNA from a specific strain of TMV is mixed with coat protein from the same strain, infectious virions are formed.

Answer 20

TMV virions self-assemble spontaneously without the input of energy or information, which means that all of the information necessary to direct their assembly must be already present RNA and/or proteins

Question 21

Carefully state a reasonable conclusion that can be drawn from the experimental findings concerning the reassembly of tobacco mosaic virus (TMV) virions from TMV RNA and coat protein subunits.

When RNA from strain A of TMV is mixed with coat protein from strain B, the reassembled virions are infectious, giving rise to strain A virus particles in the infected tobacco cells.

Answer 21

The strain specific assembly of TMV in vivo is determined by the RNA, not the coat protein.

Question 22

Carefully state a reasonable conclusion that can be drawn from the experimental findings concerning the reassembly of tobacco mosaic virus (TMV) virions from TMV RNA and coat protein subunits.

Isolated coat protein monomers can polymerize into a virus like helix in the absence of RNA.

Answer 22

The information necessary to direct self-assembly in TMV virions appears to be reside in the coat protein monomers.

Question 23

Carefully state a reasonable conclusion that can be drawn from the experimental findings concerning the reassembly of tobacco mosaic virus (TMV) virions from TMV RNA and coat protein subunits.

In infected plant cells, the TMV virions that form contain only TMV RNA and never any of the various kinds of cellular RNAs present in the host cell.

Answer 23

The self-assembly of TMV virions is specific for TMV RNA.

Question 24

Carefully state a reasonable conclusion that can be drawn from the experimental findings concerning the reassembly of tobacco mosaic virus (TMV) virions from TMV RNA and coat protein subunits.

Regardless of the ratio of RNA to coat protein in the starting mixture, the reassembled virions always contain RNA and coat protein in the ratio of three nucleotides of RNA per coat protein monomer.

Answer 24

The most stable conformation for TMV virions is achieved by the 3:1 ratio of nucleotides and coat protein monomers and is therefore the product formed upon self-assembly regardless of the starting ratio of nucleotides and monomers.

Question 25

Suppose life has been discovered on Mars and shown to contain a new type of macromolecule, named marsalive. You have been hired to study this new compound and want to determine whether marsalive is a structural or an informational macromolecule. What features would you look for?

Answer 25

Determine whether the macromolecule is made up of a series of monomeric units, if so, determine the arrangement of monomers. A repetitive pattern consisting of one or two different monomers suggests that it is structural, otherwise, it is informational.

Question 26

Your company has developed a new anticancer drug, but it will not cross through the membrane of target cells because it is large and contains many polar functional groups. Can you design a strategy to get this new drug into the cancer cells?

Answer 26

Encapsulate within a nonpolar lipid soluble vescicle (liposome) that will readily absorbed by the cell.

Question 27

Why wouldn’t your blood and other fluids become significantly acidic after you drink a half gallon of orange juice or soda?

Answer 27

Although the orange juice is acidic, your body contains buffers that absorb these excess hydrogen ions and preserve the neutral pH of your blood

Question 28

For each of the following pairs of amino acids, choose the one that is more likely to be found in the interior of a globular protein and explain why:
Alanine; Glycine

Answer 28

Alanine; (Interior, More Nonpolar)

Question 29

For each of the following pairs of amino acids, choose the one that is more likely to be found in the interior of a globular protein and explain why:
Tyrosine; Phenylalanine

Answer 29

Phenylalanine (Interior, Hydrphobic and nonpolar)

Question 30

For each of the following pairs of amino acids, choose the one that is more likely to be found in the interior of a globular protein and explain why:
Glutamate; Aspartate

Answer 30

Glutamate; (Interior, More hydrophibic, because of the extra methyl group)

Question 31

For each of the following pairs of amino acids, choose the one that is more likely to be found in the interior of a globular protein and explain why:
Methionine; Cysteine

Answer 31

Methionine;- (Interior, More hydrophibic, because of internal “S”)

Question 32

Given the chemical nature of glutamate and valine, why substitution of valine at position 6 of the b chain of hemoglobin is deleterious in sickle cell?

Answer 32

The amino acid glutamate is hydrophilic and ionizes at cellular pH, whereas valine is hydrophobic and nonionic. Substitution of the latter for the former is likely to change the chemical and structural nature of the part of the protein to affect the function.

Question 33

What are other amino acids that would be much less likely than valine to cause impairment of hemoglobin function if substituted for glutamate at position 6 of the b chain?

Answer 33

Aspartate is another acidic amino acid and is therefore conserving the charge, structure and function of the hemoglobin. Others that are unlikely to have major effects are the polar but uncharged amino acids serine, threonine, tyrosine and cysteine.

Question 34

Explain why, in some cases, two proteins could differ at several points in their amino acid sequence and still be very similar in structure and function?

Answer 34

Yes, if substitutions of amino acids are always of like-for-like amino acids in terms of chemical properties, it will become a chemically conservative changes, therefore, polypeptide could maintain the structure.

Question 35

What is the chemical basis of permanent hair styling? Be sure to include the use of a reducing and a oxidizing agent in your explanation.

Answer 35

Hair proteins are first treated with a sulfhydryl reducing agent to break disulfide bonds and thereby destroy much of the natural tertiary structure and shape of the hair. After being “set” in the desired shape, the hair is treated with an oxidizing agent to allow disulfide bonds to re-form, but now between different cysteine groups, as determined by the positioning imposed by the curlers. These unnatural disulfide bonds then stabilize the desired configuration.

Question 36

Why do you suppose a permanent isn’t permanent? (Explain why the wave or curl is gradually lost after permanent.)

Answer 36

There are two reasons for the lack of permanence: (1) Disulfide bonds occasionally break and re-form spontaneously, allowing the hair proteins to return gradually to their original, thermodynamically more favorable shape. (2) Hair continues to grow, and the new α-keratin molecules will have the natural (correct) disulfide bonds.

Question 37

Can you suggest an explanation for naturally curly hair?

Answer 37

There is probably a genetic difference in the positioning of cysteine groups and hence in the formation of disulfide bonds.

Question 38

Why False? Proteins, nucleic acids, polysaccharides, and lipids are polymers synthesized by condensation of monomer units.

Answer 38

Proteins, nucleic acids, polysaccharides are synthesized from monomers through condensation reaction, but not lipids.

Question 39

Why False? The amino acid proline is not found in a helices because its R group too large to fit into the a helix.

Answer 39

The amino acid proline is not found in a helices because its R group is covalently bonded to the amino nitrogen. Once the peptide bond is formed involving proline, its amino nitrogen has no available hydrogen for forming a hydrogen bond within an a-helix.

Question 40

Why False? While a protein can be denatured by high temperature, extremes of pH generally have no effect on tertiary structure.

Answer 40

Both high temperature and extreme pH denature protein and hence the tertiary structure.

Question 41

Why False? a-D-glucose and b-D-glucose are stereoisomers.

Answer 41

a-D-glucose and b-D-glucose are isomers differing in the position of the hydroxyl group attached to carbon atom 1, but they are not stereoisomers (mirror images). D-glucose and L-glucose are stereo isomers.

Question 42

Why false? Fatty acids are important component of all cellular lipids.

Answer 42

Fatty acids are important components of the phospholipids found in all cellular membranes. However, there are many other cellular lipids that do not contain fatty acids.

Question 43

Why False? It is easy to predict the final folded structure of a protein from its amino acid sequence using today’s powerful super computers.

Answer 43

Although a polypeptide’s primary sequence determine its final folded (tertiary) structure, because there are limitless ways in which even a small protein can fold, it is still not possible to predict this structure from primary sequence.

Question 44

What is the advantage of storage polysaccharides to have a branched-chain structure instead of a linear structure?

Answer 44

Compared to a linear molecule, a branched-chain polymer has more termini for addition or hydrolysis of glucose units per unit volume of polymer, thereby facilitating both the deposition and mobilization of glucose by providing more sites for enzymatic activity.

Question 45

Can you foresee any metabolic complications in the process of glycogen degradation? How cell handle this?

Answer 45

Every branch point will have an α(1→6) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the α(1→6) bond.

Question 46

Why cells that degrade amylose instead of amylopectin have enzymes capable of both exolytic and endolytic (internal) cleavage of glycosidic bonds?

Answer 46

Endolytic cleavage breaks the molecule infernally, creating additional ends for exolytic attack and thereby allowing the mobilization of more glucose per unit time.

Question 47

Why the structural polysaccharide cellulose does not contain branches?

Answer 47

Cellulose molecules are rigid, linear rods that aggregate laterally into microfibrils. Branches in the molecule would generate side chains which would almost certainly make it difficult to pack the cellulose molecules into microfibrils, thereby decreasing the rigidity and strength of the microfibrils.

Question 48

What does partial hydrogenation of cooking oils accomplish chemically? and physically?

Answer 48

Partial hydrogenation results in the partial reduction of C=C double bonds to C–C single bonds in the fatty acyl chains of the triglycerides. Become more solid at room temperature.

Question 49

Why would it be misleading to say that the shortening is “made from 100% polyunsaturated oils”?

Answer 49

Although the ingredients for making the shortening may be 100% polyunsaturated oils, to convert them into solid, they must be hydrogenated and thus become saturated fats in the final product.

Question 50

Why DNA has Thymine instead of Uracil as its pyrimidine base?

Answer 50

Suppose if uracil is incorporated in DNA, a mutation (deamination of cytosine will create uracil) would confuse DNA repairing enzymes, whether the original nucleotide was either Uracil or Cytosine. The additional CH3 residue on Thymine would avoid this confusion.

Question 51

Why are bacteria often a poor choice for the production of proteins for therapeutic purposes?

Answer 51

Bacteria is prokaryotic as a result can’t modify proteins as in eukaryotic cells, therefore, may affect the structure of overexpressed proteins to affect their function. Protein aggregation resulting in from protein misfolding is a common problem in the overexpression of eukaryotic proteins in bacteria.

Question 52

Biological Relevance? When you need a burst of energy, the hormones epinephrine and glucagon are secreted into your bloodstream and circulated to your muscle cells, where they initiate a cascade of reactions that leads to the phosphorylation of the inactive b form of glycogen phosphorylase, thereby converting the enzyme into the active a form.

Answer 52

The burst of energy needed for muscle contraction is provided by increased break down of glucose and ATP synthesis by the muscle cells. Since muscle cell store glucose in the form of glycogen, the enzyme glycogen phosphorylase must be activated to breakdown glycogen to glucose. When epinephrine (adrenal gland) and glucagon (pancrease) secreted into blood stream, they reach into mitochondria and convert glycogen phosphorylase into active form (convert it from “b” to “a” form) to breakdown glycogen.

Question 53

Biological Relevance? Even in the a form, glycogen phosphorylase is allosterically inhibited by a high concentration of glucose or ATP within a specific liver cell.

Answer 53

When there is high concentration of ATP or simple sugar glucose, there is no need of further production of glucose for energy purpose. Therefore, ATP allosterically inhibit phosphorylase enzyme.

Question 54

Biological Relevance? Your pancreas synthesizes and secretes the proteolytic enzyme carboxypeptidase in the form of an inactive precursor called procarboxypeptidase, which is activated as a result of proteolytic cleavage by the enzyme trypsin in the duodenum of your small intestine.

Answer 54

It will allow better regulation of carboxypeptidase, signal amplification, and kept its activity suppressed when it is not needed.

Question 55

To what ligand do you suppose antifreeze proteins bind to keep Antarctic fish from freezing? Or do you think this might be an example of a protein that functions in the absence of any molecular interaction?

Answer 55

Antifreeze proteins function by binding to tiny ice crystals and arresting their growth, thereby preventing the fish from freezing. Ice crystals that form in the presence of antifreeze proteins are abnormal in that their sur- faces are curved instead of straight. The various forms of the antifreeze proteins in these fishes are all composed of repeats of a simple glycotripeptide (Thr-Ala/Pro-Ala) with a disaccharide attached to each threonine. The genes for these antifreeze proteins were apparently derived by repeated duplication of a small segment of a protease gene.

Question 56

Why do you think that only L-amino acids and not a random mixture of L- and D-amino acids are used to make proteins?

Answer 56

The synthesis of a macromolecule with a unique structure with a specific function requires that a single stereoisomer is used at each position. Changing one amino acid from its L to its D form would result in a different protein. Thus, if a random mixture of the D and L forms were used to build a protein, its amino acid sequence would not specify a single structure, but rather many different structures (2n different structures would be formed, where n is the number of amino acids in the protein).
Why L-amino acids were selected in evolution as the exclusive building blocks of proteins is a mystery.

Question 57

When egg white is heated, it hardens which is irreversible, but hard-boiled egg white can be dissolved by heating it in a solution containing a strong detergent (such as sodium dodecyl sulfate) together with a reducing agent, like 2-mercaptoethanol. Neither reagent alone has any effect.

Why does boiling an egg white cause it to harden?

Answer 57

Heating egg-white proteins denatures them, allowing them to interact with one another in ways that were not possible at the lower temperature of the hen’s oviduct. This process forms a tangled meshwork of polypeptide chains. In addition to these interactions, interchain disulfide bonds also form, so that hard-boiled egg white becomes one giant macromolecule.

Question 58

Why does it require both a detergent and a reducing agent to dissolve the hard-boiled egg white?

Answer 58

Dissolving hard-boiled egg white requires a strong detergent to overcome the noncovalent interchain bonds and mercaptoethanol to break the covalent disulfide bonds. Together, but not separately, these two reagents eliminate the bonds that hold the tangled protein chains in place.

Try it for yourself

Question 59

Polypeptides can be translated in vitro. Would a bacterial mRNA be translated in vitro by eukaryotic ribosomes? Would a eukaryotic mRNA be translated in vitro by bacterial ribosomes? Why or why not?

Answer 59

Both can translate but with very poor efficiency. The initiation phase of translation is very different in bacteria and in eukaryotes. Bacterial mRNA would not translate very efficiently in eukaryotic system because they lacks cap structure at 5’ end of the mRNA. Similarly ekaryotic mRNA lacks Shine-Dalgerno sequence to be translated efficiently in bacterial system.

Question 60

Polycistronic mRNAs are common in prokaryotes but extremely rare in eukaryotes. Describe the key differences in protein synthesis that underlie this observation.

Answer 60

In eukaryotic cells, protein synthesis is normally initiated by scanning from the 5ʹ end of the mRNA until the first AUG codon is found. (Some- times the second or third AUG codon may be used instead—a phenomenon known as leaky scanning.) This mechanism of initiation ensures that ribosomes will all start translating near the 5ʹ end of the mRNA. When the ribosomes complete synthesis of the protein, they fall off the mRNA and must reinitiate by scanning from the 5ʹ end. By contrast, in prokary- otic cells, protein synthesis is initiated by base-pairing between mRNA sequences adjacent to an initiation AUG codon and sequences in the 16S rRNA of the small ribosomal subunit. The prokaryotic initiation strategy allows ribosomes to recognize several start sites in the same mRNA. This key difference in mechanism underlies their ability to make several pro- teins from a single polycistronic mRNA.