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?
Answer
- After 1620 (its half-life)
half of the radium-226 would be left so:
16mg/2=8mg.
- 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.
Answer
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?
Answer
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?
Answer
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?
Answer
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?
Answer
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?
Answer
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.
Answer
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?
Answer
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?
Answer
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?
Answer
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?
Answer
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
Answer
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
Answer
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?
Answer
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?
Answer
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
Answer
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?
Answer
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:
TACATACTAGTTACGTCGCCCGGAAATATC
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?
Answer
The mRNA sequence of this gene is:
AUG UAU GAU CAA UGC AGC GGG
CCU UUA UAG
The amino acid sequence for this
sequence of codons is:
Met-Tyr-Asp-Glu-Cys-Ser-Gly-Pro-Leu-Stop
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?
Answer
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.
Answer
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?
Answer
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):
Answer
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?
Answer
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.
Answer
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?
Answer
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?
Answer
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?
Answer
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?
Answer
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?
Answer
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?
Answer
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?
Answer
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.
Answer
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.
Answer
In order to distinguish the parent peptide, the different
portions must be arranged in sequence where the residues
overlap one another:
CNBr-2
Ser-Met
Tryp-1
Ser-Met-Gly-Thr-Lys
CNBr-1
Gly-Thr-Lys-Ala-Glu
Tryp-2
Ala-Glu
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)]?
Answer
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.
Answer
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.
Answer
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.
Answer
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.
Answer
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
Answer
(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.
Answer
- 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?
Answer
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.
Answer
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.
Answer
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?
Answer
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.
Answer
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.
Answer
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?
Answer
- 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.
Answer
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
