Biology Resources - Tutorials, Problems, & Answers

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Problem 1
The half-life of radium-226 is 1620 years. If a sample of material contains 16 milligrams of radium-226, how much will it contain in 1620 years? How much will it contain in 3240 years? How long will it take for the sample to contain 1 milligram of radium-226?
- After 1620 (its half-life) half of the radium-226 would be left so:

- After 3240 (two half-lives) half of a half would be left or:

16mg/2=8mg and then 8mg/2=4mg.

- half-life is 1620 years and so it would take 4 half-lives to get the sample down to 1mg or 4x1620=6480:
After 1 half-life (1620 years)–16mg/2=8mg
After 2 half-lives (3240 years)–8mg/2=4mg
After 3 half-lives (4860 years)–4mg/2=2mg
After 4 half-lives (6480 years)–2mg/2=1mg
Problem 2
Scientists find a fossil of a marine animal in the middle of a desert. They use carbon-14 dating to determine how old the fossil is, which would indicate when an ocean covered the desert. There was a 10% (0.10) ratio of 14C compared to 12C in the sample (Nf/No) with a 14C half-life (t1/2) of 5730 years. Using the equation: find the age of the fossil.

An ocean covered the desert 19,041 years ago.

Problem 3
How many molecules of water are used up in the breakdown of a polypeptide 15 amino acids in length?

A polypeptide that has fifteen amino acids is held together by fourteen peptide bonds (count the lines in diagram). One molecule of water is consumed in the breaking of a peptide bond, therefore: 1 molecule of water x number of peptide bonds or: 1 x 14 = 14 water molecules are consumed to breakdown the polypeptide.

Problem 4
Why do you suppose the monosaccharide glucose is circulated in the blood of humans rather than a disaccharide, such as sucrose, which is the transport sugar found in plants?
Humans use glucose in metabolic pathways for the generation of cellular energy. Glucose flows into metabolic pathways directly. A disaccharide, such as sucrose, would require an additional step before entering into a metabolic pathway. Therefore, it is more efficient for humans to transport sugar in the blood as glucose. Plants, on the other hand, use the carbohydrates they produce as building blocks of cellular tissues, therefore, the transportation of carbohydrates as sucrose is more efficient in plants.
Problem 5
If during the action of the sodium-potassium pump, 150 molecules of ATP are used, how many sodium ions are transported across the membrane?

Every time the sodium-potassium pump undergoes its conformational change, 3 sodium ions are transported across the membrane and 1 ATP molecule is required. If 150 ATP molecules have been used by the sodium-potassium pump, then the pump has undergone its conformational change 150 times, and has transported 150 x 3 Na+ ions/time, or 450 total sodium ions.

Problem 6
Cystic fibrosis is a genetic disease that results in thick mucus secretions that clog up air passages in the lungs. Faulty chloride ion channels keep Cl- and Na+ in the cells that line the airways increasing the intracellular ion concentrations. How does this cause the mucus in the airways to become thick?
This is a similar situation of that seen in cholera but in the opposite direction. Normally, chloride ions (Cl-) pass out of the cells lining the airways through chloride ion channels. Sodium ions (Na+) follow the Cl- out of the cells due to the electrochemical gradient created by the movement of Cl-. The movement of ions out of the cells makes the cells hypoosmotic compared to the lumen of the airways. Water passes out of the cells and blood stream through osmosis and into the airways, thinning the mucus that flushes out impurities. In cystic fibrosis, the chloride channel doesn't function properly so Cl- remains in the cells, likewise Na+ remains in the cells. The cells are hyperosmotic compared to the lumen of the airways so water remains in the cells and bloodstream. Without water passing into the airways, the mucus becomes thick and clogs the airways. 
Problem 7
At first glance, the signaling systems that involve cell surface receptors may appear rather complex and indirect, with their use of G proteins, second messengers, and often multiple stages of enzymes. What are the advantages of such seemingly complex response systems?
There are several advantages to response system that is organized in multiple levels. The first is obviously the ability to amplify the signal. With the multiple stages of enzymes, a small amount of a signal is all that is needed to get the response going. This allows for a very high level of sensitivity.
Problem 8
How much energy would be generated in the cells of a person who consumed a diet of pyruvate instead of glucose? Calculate the energy generated on a per molecule basis.

The person would not receive the benefits of energy generated from glycolysis because the pyruvate would enter directly into pyruvate oxidation. For each molecule of pyruvate consumed:

The oxidation of pyruvate to acetyl-CoA would produce:

1 molecule of NADH

One round of the Krebs cycle would produce:

3 molecules of NADH

1 molecule of FADH2
1 molecule of ATP

The electron transport chain would generate:

4 NADH x 2.5 ATP = 9 ATPs
1 FADH2 x 1.5 = 1.5 ATPs

For a total of 11.5 ATP molecules per each pyruvate molecule consumed

Problem 9
Why do plants typically store their excess energy as carbohydrates rather than fat?
Because plants are able to produce their own food, it is not necessary for them to have long-term storage of energy provided by fatty acids. Carbohydrate storage is sufficient for plants. When they need more carbohydrates, they just make them. Animals, on the other hand, need to consume their food therefore their bodies have evolved the ability to store energy in more energy-rich molecules for times of famine.
Problem 10
To reduce six molecules of carbon dioxide to glucose via photosynthesis, how many molecules of NADPH and ATP are required?

For every three molecules of CO2 that enters the Calvin cycle, one molecule of the three carbon glyceraldehyde 3-phosphate (G3P) is produced. Two molecules of G3P are needed to produce one molecule of glucose. Therefore, the Calvin cycle needs to make a total of 6 turns to produce two molecules of G3P. One turn of the Calvin cycle requires 3 molecules of ATP and 2 molecules of NADPH so for 6 turns:

3 ATP X 6 = 18 ATPs
2 NADPH X 6 = 12 NADPHs

Problem 11
What is the advantage of having many pigment molecules in each photosystem but only one reaction center chlorophyll? In other words, why not couple every pigment molecule directly to an electron acceptor?
If every pigment molecule were coupled to an electron acceptor, there would need to be hundreds of electron acceptors present in the photosystem and hundreds of electron transport chains. Maintaining several electron transport chains would consume a lot of energy. It is more energy economical to have the energy from the sun funneling into one electron acceptor and electron transport chain.
Problem 12
An ancient plant called horsetail contains 216 chromosomes. How many homologous pairs of chromosomes does it contain? How many chromosomes are present in its cells during metaphase?
108 homologous pairs (108 unique chromosomes); 432 sister chromatids (during mitosis and meiosis I, each chromosome in the pair will double - 432 is double 216.)
Problem 13

Complete the monohybrid cross below of the two heterozygous parents (recall that heterozygous indicates a parent that has one gene encoding the dominant type of the trait and one gene encoding the recessive type). 'S' represents big offspring (recall that this is the dominant trait symbolized by a capital letter) and 's' represents small offspring (recall that this is the recessive trait symbolized by a small letter).

Ss x Ss


Genotype that resulted from this monohybrid cross (Ss x Ss):

25% homozygous dominant

50% heterozygous

25% homozygous recessive

Problem 14

An organism has 56 chromosomes in its diploid stage. Indicate how many chromosomes are present in the following, and explain your reasoning:     

    a) somatic cells
    b) metaphase (mitosis)
    c) metaphase I (meiosis)
    d) metaphase II (meiosis)
    e) gametes


a) 56 chromosomes would be present in somatic cells, 28 homologous pairs?somatic cells are diploid and would have the full complement of chromosomes.

 b) 112 chromosomes would be present in mitotic cells at metaphase. Prior to metaphase, the chromosomes replicated and so every chromosome is a pair of identical sister chromatids. So, 56 x 2 = 112 chromosomes.

c) 112 chromosomes would be present in meiotic cells in metaphase I. Prior to metaphase I, the chromosomes replicated and so every chromosome is a pair of sister chromatids. So, 56 x 2 = 112 chromosomes. These chromosomes have also undergone crossing over during prophase I so the sister chromatids are no longer identical as in metaphase of mitosis.

d) 56 chromosomes would be present in meiotic cells in metaphase II. During metaphase I and anaphase I, homologous pairs of chromosomes were separated so that the resulting cells contained one chromosome of a homologous pair still paired as sister chromatids.

e) 28 chromosomes would be present in the gametes. Each gamete would contain one chromosome from each homologous pair. During fertilization, the chromosome will be matched up with its homologue.

Problem 15
From an extract of human cells growing in tissue culture, you obtain a white, fibrous substance. How would you distinguish whether it was DNA, RNA, or protein?
First you could test the substance for the presence of amino acids; if present, the substance is a protein. If the substance does not contain amino acids, it is one of the two nucleic acids. To determine which nucleic acid, you can either test the substance for the presence of ribose or deoxyribose or you can test the substance for the presence of thymine or uracil. The presence of thymine or deoxyribose indicates the substance is DNA. The presence of uracil or ribose indicates the substance is RNA.
Problem 16
In 2006 a prominent biomedical scientist was dismissed from his position because of his involvement with fraudulent claims that his laboratory had cloned humans by using a nuclear transplantation technique. Suppose that you were in charge of demonstrating that the cloning claims were false. How would you show this definitively?
A real clone would have the exact same nuclear DNA as its donor. During the cloning process, the nucleus of the donor is fused into an enucleated egg cell. The enucleated egg cell, however, still contains its own cytoplasm, which in most cases contains its own mitochondria. Mitochondria found within the egg are not removed from the cytoplasm. So, to check if the clone is real, one must check if there is a difference in mitochondrial DNA. If there is no difference between the donor and clone’s mitochondrial DNA, then you can assume that it is not clone after all – originating from the exact same source. However, if there is a difference in mitochondrial genome, but have the exact same nucleic DNA, then you can assume it is an authentic clone.
Problem 17

Cells were obtained from a patient with a viral infection. The DNA extracted from these cells consisted of two forms: double-stranded human DNA and single-stranded viral DNA. The base compositions of these two forms of DNA were as follows:

  Adenine Cytosine Guanine Thymine
Form 1 22.1% 27.9% 27.9% 22.1%
Form 2 31.3% 31.3% 18.7% 18.7%
Which form was the viral DNA, and which form was the human DNA? Explain your reasoning
The single-stranded viral DNA does not have complementary base-pairing and so one would not expect the viral DNA to follow the Chargaff's rule of base-pairing. That being the case, Form 2 would be the viral DNA because the adenine and thymine are not present in equal proportions and the cytosine and guanine are not in equal proportions. In Form 1, the adenine and thymine are in equal proportions indicating that they are base-paired and the cytosine and guanine are in equal proportions indicating that they are base-paired. Therefore Form 1 is human DNA.
Problem 18
Why do plants typically store their excess energy as carbohydrates rather than fat?
Because plants are able to produce their own food, it is not necessary for them to have long-term storage of energy provided by fatty acids. Carbohydrate storage is sufficient for plants. When they need more carbohydrates, they just make them. Animals, on the other hand, need to consume their food therefore their bodies have evolved the ability to store energy in more energy-rich molecules for times of famine.
Problem 19

The nucleotide sequence of a hypothetical eukaryotic gene is:


If a mutation in this gene were to change the fifteenth nucleotide (underlined) from guanine to thymine, what effect do you think it might have on the expression of this gene?


The mRNA sequence of this gene is:


The amino acid sequence for this sequence of codons is:


If the underlined guanine was changed to thymine the mRNA sequence for this codon would be UGU and the amino acid encoded by UGU is Cysteine, the same amino acid encoded by UGC, therefore there would be no effect of this mutation on the production of the protein.

Problem 20
Fugu is a vertebrate like humans, with many of the same complex structures and organ systems. However, unlike humans the size of the Fugu genome is a mere 400 Mb compared to the 3000 Mb of humans. What are two likely explanations for this difference? What does this say about genome size?
The difference of genome size is due almost entirely to a reduction, not in the numbers, but in the sizes of the introns, and the sizes of the regions of noncoding DNA that lie between genes. Somehow or other, Fugu has either managed to rid itself of most of the noncoding DNA that clutters the genome of species such as humans (Junk DNA), or has managed to avoid accumulating this DNA in the first place due to less selective pressure in a stable habitat. The reason the two species share similar structures is because of conserved exons and conserved segments of regulatory DNA. Overall, genome size does not necessarily imply that the organism is more or less complex since the bulk of our junk DNA is actually junk.
Problem 21
All human beings have a rich growth of E. coli bacteria in their large intestine. Will the lac operon in the bacteria present in a lactose-intolerant individual who is careful never to consume anything with lactose (milk sugar) be activated or repressed? Explain.
The lac operon will not be needed if there is no lactose and so it will be repressed. The E. coli bacteria don't need lactose to survive. They use lactose as a source of glucose. If there is no lactose present in the intestine but an alternate source of glucose, the bacteria will survive on the alternate glucose source.
Problem 22
You have generated a cell line that expresses an altered form of cadherin. This mutant cadherin has the 110-amino-acid extracellular domain that is required for interaction with other cadherins, but lacks a transmembrane domain. If you were to mix this cell population with other cells expressing a wild-type, or normal, form of cadherin, would you expect these two cell populations aggregate with each other? Why or why not? Would the mutant cells be able to aggregate with other mutant cells?
Although the mutant cells express the cadherin interaction domain, this altered form of the protein would not be present on the cell surface because it lacks a transmembrane domain. Therefore, the mutant cells would be unable to interact and aggregate with each other or with wild-type cells.
Problem 1

Why is it incorrect to think of evolution as progressive (i.e., proceeding from lowest or simplest to highest or most complex as shown in the illustration below):

Although many evolutionary trends noted thus far have been examples of increasing complexity this is not to say that evolution is progressive, pushing in a single direction. When we examine horse evolution, there is a general trend to larger, more complex animals but throughout the fossil record, there are also examples where the animals became smaller, less complex. The evolution of the hemoglobin molecule doesn't show directionality in its evolution; rather just a modification of what is there (i.e., the amino acid sequence). Another example discussed in this chapter is the evolution of the vertebrate eye. Creatures that may be considered "lower" on the evolutionary tree such as mollusks actually have eyes that are more optimally designed for sensing light. Evolution through natural selection occurs more by the appearance of workable solutions rather than the appearance of optimal designs.
Problem 24
Can a human embryo that exhibits polyploidy (3 x 23 = 69 chromosomes/cell) survive until birth? What is the difference between polyploidy and trisomy?
A polyploidy human embryo that is 3n would not be expected to survive to birth. In contrast to polyploidy, trisomy involves the addition of a chromosome to a diploid genome (2n +1). Three forms of trisomy can lead to a viable infant, although the trisomy of chromosome 21, which results in Down syndrome, is the only condition in which a significant number of individuals live longer than a year past birth. The other two viable trisomies, Patau syndrome (trisomy 13) and Edward's syndrome (trisomy 18) have severe birth defects and often die within the first three or four months of birth. It is believed that other forms of trisomy result in spontaneous abortion of the embryo or fetus early in pregnancy. These observations suggest that normal human embryonic development requires a precise diploid number of chromosomes.
Problem 25
Justify the assertion that life on earth could not exist without prokaryotes.
Prokaryotes are essential to all aspects of life on earth. Without bacteria, oxygen levels would not have increased in the atmosphere and although bacterial life would have still existed, we would not see the great diversity that currently inhabits the earth. Also, without prokaryotes, there would be little nitrogen in the soil for plants to use for growth. Prokaryotes decompose dead organic matter so that it recycles back into the environment.
Problem 26
What are the functions of antibiotics in the prokaryotes, such as bacteria, that produce them?
Prokaryotes produce chemicals that protect their environment from competition. In order to reduce the number of competitors for resources, a prokaryote produces a chemical that harms other strains of prokaryotes but not itself. We use these chemicals as antibiotics to treat certain, primarily bacterial, infections. Although effective against some bacteria, these antibiotics are not effective against all prokaryotes, particularly the type of prokaryotes that produced the chemical.
Problem 27
What are endospores? Why would they be an advantage? Would a bacterium with the ability to form endospores have a greater or lesser chance of extinction? Why?
Endospores are thick-walled spores produced by certain bacteria that protect its genome and a small amount of cytoplasm. The bacteria produce these spores under conditions that are harmful to the bacteria, for example extreme heat or cold or arid conditions that could lead to desiccation. Endospores offer an advantage to bacteria that are able to produce them because it protects them from destruction. Bacteria that produce endospores would have a lesser chance of becoming extinct because of this advantage. The endospores are in a state of dormancy and when conditions improve, the endospores germinate and produce live bacteria that can continue to grow and reproduce. Bacteria that cannot form endospores would die off in the extremely harsh environmental conditions.
Problem 28
Protists typically reproduce asexually. However, some protists undergo sexual reproduction during times of environmental stress. What advantage does sexual reproduction give a protist during these times?
Sexual reproduction provides greater genetic variability in the next generation. New gene combinations that are adaptive under the specific environmental stress may enhance the survival and reproduction of offspring with the favorable genotype.
Problem 29
In what ways is an earthworm more complex than a flatworm?
An earthworm, in the Annelida phylum, is a coelomate and a flatworm; in the Platyhelminthes phylum, is an acoelomate. The presence of a body cavity in the earthworm allows for specialization of the internal organs, compared to the flatworm. The coelom also leads to the evolution of a circulatory system in the earthworm that is not present in the flatworm. The reproductive organs and gametes in the earthworm are also larger and more diverse, which expands the reproductive strategies used by earthworms compared to flatworms.
Problem 30
Why is it believed that echinoderms and chordates, which are so dissimilar, are members of the same evolutionary line?
Chordates and echinoderms shared a very key characterization of animal taxonomy—that of embryo growth pattern. These two groups comprise the deuterostomes whose embryological development differs greatly from the protostomes (mollusks, annelids, and arthropods). These embryological developmental patterns are guided by genetic differences in the expression of Hox genes, which suggests that echinoderms and chordates share a key characteristic very distinct from other animal groups. And, although the adult echinoderms look very different from the chordates, an earlier stage in their development looks more closely related to the chordates.
Problem 31
Many fruits can be preserved by candying. The fruit is immersed in a highly concentrated sugar solution; the sugar is then allowed to crystallize. How does the sugar preserve the fruit?

The coating of sugar preserves the fruit by not allowing moisture in the air from interacting with the exposed cells on the fruit; acting as a protecting barrier/layer. The sugar layer also forms a hypertonic environment for bacteria, preventing bacterial cells from carrying out decomposing reactions and spoiling the fruit. This also prevents mould from forming on the fruit because mould requires moisture to start germinating and spreading.

Problem 32
Why does a noncompetitive inhibitor not change the observed Km?

The km stays the same because the inhibitor does not interfere with the binding of substrate to the active site. The Km represents the substrate concentration when the rate is equal to half its maximal value. Thus, the enzyme shows the same Km in the presence or absence of the non-competitive inhibitor.

Problem 33
Explain why using a polypeptide’s primary sequence to determine its three-dimensional shape is problematic.
The protein may undergo drastic structural and chemical changes during the secondary, tertiary, and quaternary structure. During the secondary structure, there is an arrangement in space of the atoms in the peptide backbone; therefore, in the primary phase, whether or not it is an alpha helix or B-pleated sheet (or both) isn’t determined. Hydrogen bonding between the amide N-H and the carbonyl groups of the peptide backbone isn’t complete either. Moreover, the tertiary structure includes the 3-Dimensional arrangement of all the atoms in the protein, including those in the side chains; disulfide bridges will also affect the structure of the nascent protein. Finally, the nascent protein may interact with other proteins in the quaternary phase; this interaction between subunits is mediated by noncovalent interactions, such as hydrogen bonds, electrostatic attractions, and hydrophobic interactions. As a result, a polypeptide’s primary sequence isn’t the best way to determine its 3D shape.
Problem 34

A peptide was cleaved into smaller peptides with cyanogens bromide (CNBr) and into two different peptides by trypsin (Tryp). Their sequences were as follows:

CNBr-1: Gly-Thr-Lys-Ala-Glu

CNBr-2: Ser-Met

Tryp-1: Ser-Met-Gly-Thr-Lys

Tryp-2: Ala-Glu

What was the sequence of the parent peptide? Be sure to explain your reasoning. Full credit will not be given for the peptide sequence without rationale.

In order to distinguish the parent peptide, the different portions must be arranged in sequence where the residues overlap one another:









Therefore, this suggests that the parent peptide sequence is: Ser–Met–Gly–Thr–Lys-Ala-Glu. This is determined simply from the fact that the information provided is redundant, CNBr-2 and Tryp-2 are not required to determine the parent code because 60 per cent of the amino acids from CNBr-1 and Tryp-1 overlap.

It is also known that CNBr hydrolyzes peptide bonds at the c-terminus or carboxyl side of methionine residues, whereas the trypsin hydrolyzes peptide bonds at the c-terminus of the amino acids lysine and arginine, except when either are followed by proline. In this case, arginine and proline are not present.

Problem 35
In which direction will the following proteins move in an electric field [toward the anode, toward the cathode, or toward neither (i.e., remain stationary)]?
a. Egg albumin (pI = 4.6) at pH 5.0 [2]

At the pH of 5.0, the protein will carry a small excess negative charge because 5.0 is above the isoelectric point of 4.6. Since negatively charged anions move towards the anode, the protein will therefore migrate towards the anode.

b. β-lactoglobin (pI = 5.2) at pH 5.0 and pH 7.0 [4]

At the pH of 5.0, the protein β-lactoglobin will be positively charged and move towards the cathode because it is below the isoelectric point of 5.2. The opposite is true at the pH of 7.0 because the protein will carry a negative charge and move toward the anode.
Problem 36
In what order would the following proteins emerge from a gel filtration column of Sephadex G200: myoglobin (Mr=16,000), catalase (Mr=500,000), cytochrome c (Mr=12,000), chymotrypsinogen (Mr=26,000), and haemoglobin (Mr=66,000)? Be sure to include your rationale for full points.
According to the molecular weight provided, the enzyme catalase is the largest protein in the set. It has been noted that substances with a MW greater than 200,000 are supposed to be excluded from the mycell structure of Sephadex G200 and consequently should flow freely, without separation through the gel. This suggests that the catalase enzyme would be excluded completely. Moreover, in this situation, the catalase molecule would dissolve in the fluid outside the Sephadex gel ‘pores’ and not diffuse into the gel since catalase is too large to fit within the gels spaces. The fluid outside the beads is referred to as the void volume (total volume of solvent within the column at any one point). When an eluant of equal volume to the void volume is introduced to the gel column, the catalase enzyme would simply be washed off.

The protein haemoglobin would be next to emerge from a gel filtration column because it is a tetramer consisting of four peptides that does not dissociate into its component sub-units when exposed to Sephadex G200. Therefore, it would not become four single-chained proteins with a molecular weight of 66000 / 4 = 16 500, but instead it behaves as a single complete protein. Since chymotrypsinogen is the third largest protein in the group, it would be the third in the order of elution. Likewise, myoglobin would be fourth since it has the fourth largest molecular weight of 16 000. Finally, the smallest protein to be tested is cytochrome c. This suggests that it will flow freely into the gels perforations. Therefore, it is required that a volume equal to the gels total volume be added for cytochrome c to be successfully eluted from the gel.

Order to which the proteins will emerge from the gel filtration column: Catalase, haemoglobin, chymotrypsinogen, myoglobin, and cytochrome c. (Largest to smallest).
Problem 37
An enzyme was studied by means of gel filtration in aqueous buffer at pH 7.0 and had an apparent Mr of 160,000. When studied by gel electrophoresis in SDS solution, a single band of apparent Mr of 40,000 was formed. Explain these findings. Be specific.

It is possible for enzymes to be can be made up more than one peptide – forming a polypeptide. Gel filtration does not have the ability to dissociate an enzyme or a protein into its subunits on its own; therefore, the analysis indicated an Mr value of 160000 as a whole. Gel electrophoresis in SDS (SDS being an anionic detergent which denatures secondary and non-disulfide linked tertiary structures) may have caused the enzyme to dissociate from its quaternary structure (possibly four identical subunits) and produced a molecular weight figure of its component subunits. In other words, the 160000 figure indicates the enzyme as a whole in its quaternary state while the 40000 indicates that it is made up of four identical subunits, forming a single band.

Problem 38
Which of the following amino acids could be a target for phosphorylation, and why? Phe, Tyr, Ala, Asp, Ser, Cys, Thr.
Protein kinase is a kinase enzyme that modifies other proteins by chemically adding phosphate groups to them (phosphorylation) The chemical activity of a kinase involves removing a phosphate group from ATP and covalently attaching it to one of three amino acids (Tyr, Ser, and Thr) that have a free hydroxyl group.
Problem 39
You are working in a research lab investigating lignin degradation by the bacterium Eatusdetree totalii which was isolated from a rotting piece of birch your supervisor found while strolling in the forest. The genomic sequence of this bacterium has not yet been determined. You found that the addition of a small quantity of hydrogen peroxide induces production of an enzyme which enables the organism to degrade lignin. Zymography (an electrophoretic technique, based on SDS-PAGE) has revealed that a single form of the enzyme is present in cultures grown under solid state fermentation conditions. You go on to purify the enzyme and find that the purified enzyme catalyzes C-C bond cleavage in the side chains of lignin but is unable to do so without added H2O2. When the enzyme preparation is added to pure cellulose under similar reaction conditions, hydrogen peroxide is consumed but to a lesser degree than when lignin is used as substrate. Interestingly, you find that bubbling pure oxygen into the cultures does not result in any enzyme activity against either lignin or cellulose.
We know that the bacterium:

- Produces a single type of active enzyme that can breakdown lignin in the presence of H2O2.
- The catalyzation of lignin carbon-carbon bonds is dependent on H2O2.
- The degree of catalyzation in the presence of cellulose + H2O2 occurs much less.
- In the presence of pure oxygen + enzyme + cellulose or lignin = no enzymatic activity.

o This indicates that the enzyme is not an oxygenase because oxygenases incorporate molecular oxygen into substrates.

The enzyme is quite possibly a heme-containing peroxidase – ligninase – due to its unique requirement for H2O2 as opposed to molecular oxygen to undergo catalytic reactions. We are told that the species produces the active enzyme upon exposure to H2O2; this suggests that the inactive enzyme is already present in the bacteria, but as soon as the H2O2 is present, the iron comprising the heme group is oxidized (losses an electron) causing the iron to become a free radical (less stable): Fe3+ + H2O2 = Fe4+ + H2O. Recall that during this process, the H2O2 gains two electrons, thus producing water. Once a ligninase is active, it is able to cleave Cα-Cβ bonds found in lignin as depicted in the description. This is also the reason why in the presence of cellulose + H2O2, H2O2 was being consumed to a lesser degree because lignin – not cellulose – is required as a substrate for ligninase to revert the oxidized heme iron back to its normal state where it can continue to accept more H2O2. Presumably, once all the enzymes were oxidized, H2O2 would no longer be able to induce free radicals and become converted into water; thus, consumed but to a lesser degree. Finally, it is interesting to point out that cellulose is made up of interconnected β-glucose monomers and do not contain Cα-Cβ bonds to which ligninase can catalyze. If the enzyme were a laccase, a reaction would have taken place when the enzyme was exposed to pure oxygen, since laccases are molecular oxygen-dependent; however, this did not take place.
Problem 40
You are performing site-directed mutagenesis to test predictions about which residues are essential for a protein’s function. Which of each pair of amino acid substitutions listed below would you expect to disrupt protein structure the most? Explain.

a)      Val replaced by Ala or Phe

b)      Lys replaced by Asp or Arg

c)      Gln replaced by Glu or Asn

d)     Pro replaced by His or Gly

(a) Phe. Recall that phenylalanine has a benzene ring, so even though both alanine and phenylalanine are hydrophobic, the larger size of phenylalanine’s R side chain may not fit as well in place of valine in comparison to alanine so it would disrupt the structure – alanine is likely to do a better job because of its smaller R group.

(b) Asp. Recall that Asp (asparagine) is a polar amino acid with no particular charge present in its R group. Lysine on the hand is also polar (basic) but carries a positive charge in its side chain. Altering the proteins charges would ultimately disrupt the structure – Arg is likely a better substitute since it also carries a positive (basic) charge.

(c) Glu. Recall that glutamic acid (Glu) has acidic properties and carries a negative charge in its R group. Gln, although polar, has an amide group as opposed to a carboxylate. Therefore, asparagine’s amide-containing R group would be better suited to substituted for Gln than Glu.

(d) His. Histidine is a hydrophilic amino acid with basic properties and a positive charge due to its side R group. Moreover, proline is hydrophobic and has a constrained geometry due to the heterocyclic ring. Therefore, a protein is better off having an amino acid with no side chain to replace proline than to have an amino acid side chain that is bulkier.
Problem 41
The following observations are made on an unknown membrane protein, pX. It can be extracted from disrupted erythrocyte membranes into a concentrated salt solution, and the isolated pX protein can be cleaved into fragments by proteolytic enzymes. Treatment of erythrocytes with proteolytic enzymes followed by disruption and extraction of membrane components yields intact pX. However, treatment of erythrocyte ‘ghosts’ (which consist of only membranes, produced by disrupting the cells and washing out the hemoglobin) with proteolytic enzymes followed by disruption and extraction yields extensively fragmented pX. What do these observations indicate about the location of pX in the membrane? Do the properties of pX resemble those of an integral or peripheral membrane protein? Explain.
- According to these observations, protein pX is a peripheral membrane protein because it can be extracted from disrupted erythrocyte membranes using a salt treatment.
- When the erythrocytes were treated with proteolytic enzymes (proteases) followed by disruption and extraction of membrane components, it yielded undigested, intact pX. The inability of the protease to digest the protein unless the membrane is disrupted suggests that pX is likely bound to the inner surface of the corresponding cell’s plasma membrane, thus located internally.
Problem 42
You have cloned the gene for a human erythrocyte protein which you suspect is a membrane protein. From the nucleotide sequence of the gene, you know the amino acid sequence. From this sequence alone, how would you evaluate the possibility that your protein is an integral membrane protein? Suppose the protein proves to be an integral membrane protein. How could you experimentally determine whether the external domain is carboxyl- or amino-terminal?

Recall that an integral membrane protein can be divided into two groups: transmembrane proteins, which span the entire membrane) and integral monotropic proteins (which are permanently attached to the membrane from one side). The region spanning the membrane likely consists of hydrophobic amino acids as a means to interact with the phospholipid bilayer. Thus, if we know the amino acid sequence of the gene, we must first construct a protein hydrophobicity plot in order to characterize and analyse the cloned proteins hydrophobic character – this is a useful technique in predicting membrane-spanning domains, potential antigenic sites and regions that are likely exposed on the protein's surface. A window size of 19-21 will make hydrophobic, membrane-spanning domains stand out rather clearly, and so we can assume they are transmembrane segments. Generally, hydrophobicity is determined by the energy required to move an amino acid from a nonpolar to an aqueous environment.

Moreover, membrane-impermeable reagents have been shown to be impermeable to intracytoplasmic membranes of certain bacteria. He shows that when cells are treated with membrane-impermeant reagents, they reacts with proteins on the outer surface of the membrane, all of which have one or more lysine residues on the N-terminal sides of their hydrophobic region (recall that lysine has primary amines located on its R group). Thus, since membrane-impermeable reagents are known to react with primary amines, it can be used to determine whether the external domain is carboxyl- or amino-terminal. If the addition of membrane-impermeant reagents reacts with the external portion of the protein, we can assume that the external domain comprises of an amino terminus. If no reaction occurs, then we can assume that the external domain is a carboxyl-terminus. The specific membrane-impermeant reagent used is guanidinating reagent known as 2–S-[14C] Thiuronium Ethane Sulfonate.
Problem 43
Explain how carbon flux through the pentose phosphate pathway is regulated.
Carbon flux through the pentose phosphate pathway (PPP) is mainly controlled by NADPH on the first enzyme of the irreversible oxidative phase, glucose-6-phosphate dehydrogenase. When [NADPH], it inhibits the enzyme, while low concentrations, which is usually a result of increased lipogenesis (the process by which simple sugars are converted to fatty acids) stimulates glucose-6-phosphate dehydrogenase activity. NADPH is also used by cytochrome P450, a monooxygenase that oxidizes the NADPH to NADP+ during the oxidation of fatty acids in the endoplasmic reticulum, where the first reaction places a hydroxyl group onto the omega carbon – this process reduces [NADPH] concentration. Likewise, the maintenance of reduced glutathione, a powerful antioxidant also requires NADPHRed via the enzyme glutathione reductase to prevent red blood cells from free-radical damage. This process also decreases [NADPH] concentration.
Problem 44
The first patient described to have McArdle disease (type V glycogen storage disease), was a young man who was weak and developed severe muscle pain on doing modest exercise. A biopsy of his muscle showed accumulation of glycogen, and enzyme assays revealed low glycogen phosphorylase activity. Predict the effect of exercise on the lactate concentration in such a patient.

Since glycogen cannot be broken down into glucose-1-phosphate via the enzyme myophosphorylase, glucose molecules as a source of energy flowing through the blood will be very limited, especially during modest exercise. Recall that glucose-1-phosphate is converted to glucose-6-phosphate and subsequently into glucose. Glucose is then used to produce ATP via glycolysis. Under anaerobic conditions, however, pyruvate is converted into lactate. Since there is an insufficient level of glucose, there will be an insufficient level of pyruvate produced and also lactate. Therefore, in McArdle disease, the lactate concentration does not increase during exercise.

Problem 45
Why does the brain use ketone bodies as metabolic fuel during starvation when there are abundant fatty acids available from the blood?

When triacylglycerides are broken down by various lipases in fatty acids, they are insoluble in water due to their hydrophobic tail. Thus, they are transported in the blood by various proteins, including serum albumin. The blood-brain barrier is very specific on the types of proteins it allows to pass from the circulating blood to the interstitial fluid. For instance, cells of the barrier actively transport metabolic products such as glucose across the barrier with specific protein. However, since it generally excludes proteins from entering the cerebrospinal fluid, it limits the efficiency of delivering an amount that is sufficient for adequate energy production via β-oxidation. Ketone bodies, which include acetone, acetoacetate, and β-hydrobutyrate are hydrophilic (water soluble) and are generally small molecules that can pass through certain transporters of the cells lining the blood-brain barrier through specific transport systems (i.e. carrier-mediated transporters, receptor-mediated). This is why the brain uses ketone bodies to harvest energy, simply due to the fact that they are more attainable.

Problem 46
Predict the clinical effect of an obstruction in the bile duct.
The bile duct carries bile salts produced by the liver, such as taurocholate and glycocholate, from the liver to the gallbladder, where they are stored (and subsequently from the gallbladder to the duodenum). If there is an obstruction in the bile duct, the gallbladder will have little to no bile to store (bile also becomes more concentrated in the gallbladder than when it left the liver). When lipids are ingested, they move into the stomach and then into the duodenum; within the duodenum, the fat stimulates the release of Cholecystokinin. Cholecystokinin travels into the bloodstream, reaches the gallbladder, and causes gallbladder contractions. When it contracts, it will release the bile salts produced by the liver through the common bile duct and released into the duodenum. The bile salts will emulsify the lipids into smaller portions where they can be subsequently degraded by pancreatic lipases into monoacylglycerol and 2 fatty acids. This is the form to which they can be absorbed by the enterocytes and, in turn, be distributed throughout the body via chylomicrons, for storage or energy production. Therefore, if the bile duct is obstructed, the triacylglycerides consumed will not be efficiently emulsified. If fats are not emulsified, they will not be enzymatically broken down by the pancreatic lipases into monoacylglycerol and 2 fatty acids, and thus will not be absorbed by the intestinal cells and be distributed into the blood. This will cause one to have digestion complications such as steatorrhea (presence of excess fat in feces), weight loss because ingested fat and cholesterol is not being absorbed and stored, and be less energetic because fat levels used by the body for energy via β-oxidation will decrease. Another direct results could be jaundice due to increased bilirubin levels in the blood, since bilirubin is a main component of bile released to the gallbladder. But if the duct is blocked, it will get concentrated in the blood instead of being excreted. An obstruction of the bile duct usually results when the gall bladder becomes almost filled with crystalline cholesterol, or compounds derived from bilirubin and this forms gallstones which have the potential to rupture the ducts or prevent them from storing bile produced in the liver.
Problem 47
Folic acid deficiency, believed to be the most common vitamin deficiency, causes a type of anemia in which haemoglobin synthesis is impaired and erythrocytes do not mature properly. Explain the metabolic relationship between haemoglobin synthesis and folic acid deficiency. 

Folic acid is itself not biologically active until it has been converted into tetrahydrofolate via the enzyme dihydrofolate reductase in the liver. Therefore, folic acid is a precursor of tetrahydrofolate.


In order to biosynthesize glycine, the amino acid serine is required, where the enzyme serine hydroxymethyltransferase catalyses the reaction between serine and tetrahydrofolate into glycine.


The committed step for porphyrin biosynthesis (the prosthetic group that contains an iron atom contained in the center of a large heterocyclic organic ring) is the formation of D-aminolevulinic acid by the reaction of the amino acid glycine and succinyl-CoA, from the citric acid cycle. Therefore, glycine is a precursor of porphyrin.

[1] Folic Acid -> Tetrahydrofolate

[2] Tetrahydrofolate -> Glycine

[3] Glycine -> Porphyrin

In all, a lack of folic acid impairs hemoglobin synthesis.
Problem 48
Under starvation conditions, organisms can use proteins and amino acids as sources of energy. Deamination of amino acids produces carbon skeletons that can enter the glycolytic pathway and the citric acid cycle to produce energy in the form of ATP. Nucleotides, on the other hand, are not similarly degraded for use as energy-yielding fuels. What observations about cellular physiology support this statement? (i.e. what do you know about nucleotide metabolism that support this statement?). What aspect of nucleotide structure makes them a relatively poor source of energy?
- Generally, nucleotides are not stored in the body to be used as a source of energy; rather, they require constant synthesis, degradation, salvaging, and resynthesis. When nucleotides are ingested, group-specific nucleotidases and non-specific phosphatases degrade nucleotides into nucleosides. They are then further degraded via hydrolysis in-to ribose and their respective base which can be recovered in purine and pyrimidine salvage pathways.

- Since carbon is generally an energy-producing factor, a low carbon to nitrogen ratio would yield a relatively poor energy source; this may explain the reason why the nucleotide structure is not a convincing fuel source for humans (that is, due to its low carbon:nitrogen ratio.) Recall that carbon-carbon bonds are energy rich, such as those found in the fatty acid hydrocarbon tail and carbohydrates.
Problem 49
Allopurinol is an inhibitor of xanthine oxidase and is used to treat chronic gout. Explain the biochemical basis for this treatment. Patients treated with allopurinol sometimes develop xanthine stones in the kidneys, although the incidence of kidney damage is much lower than in untreated gout. Explain this observation considering the following solubilities in urine: uric acid, 0.15 g/L; xanthine, 0.05 g/L; and hypoxanthine, 1.4 g/L.

Xanthine oxidase is responsible for the following chemical reaction:

[1] hypoxanthine + H2O + O2 <-> xanthine + H2O2

[2] xanthine + H2O + O2 <-> uric acid + H2O2

The addition of allopurinol inhibits the conversion of hypoxanthine (a naturally occurring purine derivative) to uric acid. This allows the level of uric acid in the blood to decrease and alleviates the problems associated with AMP degradation. This leads to the accumulation of hypoxanthine, which is more soluble and more readily excreted than uric acid, the chemical responsible for causing gout.

Patients treated with allopurinol also show an increase in xanthine accumulation because guanine cannot be converted to uric acid. Xanthine is less soluble than uric acid, thus causing xanthine stones. The numbers depict that more hypoxanthine is found in the urine because it is more soluble, followed by uric acid, and the least soluble compound xanthine. Because the amount of GMP degradation is low relative to AMP degradation, the kidney damage caused by xanthine stones is less than that caused by untreated gout


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