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Solutions for Chemistry of Life Exam
1. Carbohydrates are one of the four classes of primary metabolites in humans (components that
are required for life). This question will deal with glucose, a product of photosynthesis,
which is consumed during cellular respiration (releasing energy).
a. First, let’s examine the photosynthetic reaction which produces glucose. Complete and
balance the equation below:
(hint: hν is electromagnetic radiation, a.k.a. light)
6
b. Glucose can actually exist in the acyclic/linear (below right) or cyclic (below left) form
when in solution. Both conformations are drawn below.
In solution, cyclic glucose tends to be favored because it consists of a stable
functional group called a hemiacetal.
Hemi = half
acetal =
Using the above definitions, circle the hemiacetal structure on the cyclic form of
glucose.
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c. When chemists think about the linear form of glucose, it is often convenient to use what
are known as Fischer Projections (named after the genius Emil Fischer). Fischer
projections are a shorthand way of representing a 3-dimensional molecule in two
dimensions. There are certain caveats to portraying a molecule in two dimensions – for
example, the molecule can be rotated on the page but not flipped across an axis. The
basic form of a Fischer projection is shown below:
A dashed line represents a bond extending under the plane of the page, and a wedge
represents a bond extending above the plane of the page. Using this basic outline of
Fischer projections as a guide, convert the linear form of glucose to a Fischer projection.
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d. Sugars are classified by the orientation of the second to bottom carbon on the Fischer
projection. The two orientations are mirror images of each other (mirror images are not
identical, just look at your hands for proof!). This is called enantiomerism and can lead
to what is called a chiral molecule. At the most basic level, a chiral molecule is a
molecule that is not super-imposable on its mirror image (like your hands). Generally, a
chiral center is one in which a carbon is connected to four different substituents.
Remember this definition throughout the exam, you will need it!
Typically, one sugar enantiomer is natural and found in biological systems, and the other
isn’t. Using the nomenclature below, your Fischer projection from question c, and your
knowledge of glucose, which enantiomeric form of carbohydrates (D or L) do you
believe is the natural enantiomer? Explain your reasoning.
e. Sugars often get a bad rap in the media due to their high caloric content, promotion of
heart disease, and tendency to lead to obesity when overconsumed. Because of these and
other reasons, artificial sugar substitutes have been researched and become common for
human consumption (think Splenda, stevia). Artificial sugars (like sorbitol) are
metabolized much more slowly in the body than sugars. Why might this make them good
ingredients in diet foods?
Because artificial sugars are not metabolized quickly enough, they do not contribute to
the overall caloric intake.________________________________________________________
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) ] ) is commonly reacted with sugars to determine if the
f. Tollens’ reagent ( [ (
sugar structure has an aldehyde present (in its linear form). Let’s examine the reaction of
Tollens’ reagent with glucose, which oxidizes the sugar.
i.
Complete the chemical equation below.
[
(
) ]
→ _2_ __Ag___ + _4_ __
___ +
Explain why the Tollens’ reagent provides a good visual test for the presence of
sugars.
i.
The reaction of Tollens’ reagent with a sugar will precipitate silver metal, forming an
“silver mirror” on the chamber surface.____________________________________________
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2. Proteins are large and complex molecules made of long chains of amino acids that are very
important for the human body – they do the bulk of the work in cells and help form the
structures of most tissues and organs, regulating and aiding in their functioning. As needed,
please reference the amino acid chart on page 16.
a. Imagine you are a scientist working to engineer a new protein that will span the cell
membrane (the interior of the membrane is hydrophobic) and bind ions to bring them into
the cell. Additionally, you are interested in having the cytoplasmic domain (cytoplasm
interior of cells is hydrophilic) of the protein be relatively unstructured for future
experiments.
With these goals in mind, build a satisfactory protein by choosing amino acids that will
contribute to the properties of the protein. Your protein should have at least three distinct
segments: choose at least three appropriate amino acids that will make up each part of the
protein (9 amino acids total). Be sure to draw your peptide chain in a cell
membrane. You need only write the amino acid letter (ex. N for asparagine). Explain
why you chose each amino acid and the ordering. (Hint: there are many correct answers
for this question)
Protein sequence/diagram:
Any 3 hydrophilic amino acids (by side chain property), followed by any 3
hydrophobic, and then any three hydrophilic.
Explanation:
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b. One particular way that certain polypeptides become more stable is by forming alpha
helices which are stabilized through intermolecular forces between the backbone of
the protein. The backbone of the protein is formed through a series of peptide bonds
(see reference sheets), and does not involve interactions with the side chains. Given
the polypeptide below, answer the following questions.
i.
On the polypeptide structure, circle two groups that contribute to the
intermolecular forces that are responsible for stabilizing the alpha helices.
ii.
How many amino acids make up this polypeptide?
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iii.
Proline is an amino acid known for being an alpha helix breaker. Looking at its
structure below, explain why.
In a polypeptide chain, the proline molecule has no free
amide hydrogen available for H-bonding.
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c. It is often important to be able to chemically synthesize peptide strands (a chain of
amino acids together) – this process is called peptide synthesis. To form a peptide,
the carboxyl group of one amino acid is joined to the amino group of another amino
acid molecule are joined. While this sounds simple in theory, in practice, it is often
quite difficult to prevent lots of side reactions from occurring, leaving the researcher
with only a little bit of the desired synthesized peptide.
One way to prevent side reactions from occurring is to utilize protecting group
chemistry, in which the amino group of one acid is protected (so that it does not
undergo any reactions) and the carboxyl group of the second acid is protected (so that
it does not undergo any reactions). This leads to only one free amino group and one
free carboxyl group to react and form a single product.
Why does using protecting groups in this manner lead to the formation of only one
product?
Protecting the amino group of the first acid makes it so that amino group is inactive and
cannot attack the second acid. Protecting the acid of the second amino acid deactivates
that amino acid acting as a nucleophile and forces it to be an electrophile (brownie points
for mentioning that this protecting group actually helps activate the acid derivative as an
electrophile). Points for noting that there is only one free carboxylic acid and one free
amino group, and that the two need to form a bond.
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d. Solid-phase peptide synthesis (SPPS) is the way many labs synthesize peptides today
– the process is incredibly easy and efficient. In SPPS, peptides are attached to resin
beads, which act as very large protecting groups. These beads are treated with amino
acids, which grow the peptide chain one by one until the chain is removed with
trifluoroacetic acid.
The most important thing to consider in SPPS is the overall yield of your
synthesis. Consider the linear sequence to build a hypothetical peptide below. Using
the yield data for each step, what is the overall yield for this synthesis? Show all
work for your calculation.
yield is (0.92)(0.96)(0.84)(0.99)(0.82) = 0.602
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e. In reality, the yields for each step in solid-phase peptide synthesis are usually at or
above 99.5%. The high fidelity of each step helped transform the entire field of
peptide synthesis. Why do you think solid-phase peptide synthesis leads to faster
peptide syntheses and better yields for each step than liquid-phase peptide synthesis?
Answers that demonstrate comprehension of solid-phase peptide synthesis and a
comparison to liquid-phase (fewer washing steps, faster because physically on a bead,
etc.) most reasonable and well-thought out answer will be accepted._______________
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3. DNA, deoxyribonucleic acid, is perhaps the most important and complex chemical
molecule for all living things’ growth, development, functioning, and reproduction.
a. You know that both DNA and RNA are acidic (think about their names). Given the
abbreviated structure of DNA below, explain why DNA is acidic? As part of your
explanation, circle the acidic hydrogen(s). Note that at biological pH, these protons
would not be a part of the DNA.
OH
OH
OH
The structure of the nucleic acid given contains acidic phosphate groups that at biological pH,
would lose their proton and be acidic.
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b. If you were to put the above molecule into a buffered solution at biological pH, and
applied a voltage so that the negative terminal of a power source is placed in the top
of the solution, and the positive terminal at the bottom, describe how the molecule
would move in the solution.
The molecule at biological pH will be negatively charged due to the phosphate groups, so the
molecule will move down in the solution.
c. A common technique in molecular biology is the extraction of DNA from cells. The
process includes many steps. The cell membranes (hydrophobic) need to be lysed
(burst open), and the DNA has to be separated from other proteins, lipids, and
carbohydrates within the cell. Usually, the first step is to add a detergent, such as
sodium dodecyl sulfate (SDS). The structure of SDS is shown below.
i.
Referencing the structure above, use intermolecular forces to explain what makes
SDS an effective detergent? Why is SDS used in the experimental protocol?
SDS has a hydrophillic head and a hydrophobic tail of hydrocarbons. The
hydrophobic region “sticks” to fats/oils/grease, and the hydrophillic part makes
SDS soluble in water. The SDS is used to break apart the cell membranes, which
are hydrophobic.
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ii.
Once the cells have been lysed, phenol and chloroform (both organic non-polar
solvents) is added to the mixture. The mixture is then centrifuged and then
allowed to separate into two layers (aqueous and organic). Referencing
intermolecular forces, explain why the two layers form, and explain which layer
the DNA is in.
Two layers form because phenol and chloroform are both nonpolar solvents and do not mix with
water. The DNA is in the aqueous layer because it is charged, and water is more polar.
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4. Broadly defined, lipids are just hydrophobic molecules (molecules that are soluble in organic
solvents). Fatty acids are one of the main subtypes of lipids and an important source of
energy in the body. In humans, fatty acids are typically composed of an even number of
carbons, reflecting the pathway for their synthesis.
a. Fatty acids consist of a long-chain hydrocarbon ending in a carboxylic acid. Draw an
example of a fatty acid that contains six carbons.
b. Triglycerides are derived from three fatty acids and a glycerol molecule, forming the
main constituent of body fat in humans and animals. The reaction between fatty acids
and glycerol is a condensation reaction, meaning that water is a product.
Using the structure of glycerol and the three provided fatty acid structures, complete
the below reaction for the formation of triglyceride.
3 H2 O
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c. Triglycerides are esters, meaning they can be saponified to be converted to soap. In
saponification, the triglyceride is treated with a strong base (such as NaOH),
cleaving the ester and releasing the fatty acid salt and glycerol.
Basically, this is the reverse process of part b illustrated above, except the salt of the
fatty acid is formed, rather than the fatty acid itself. Why is this the case?
The salt is formed because saponification happens under basic conditions (the acid
will become deprotonated)
d. In the final part of this problem, steroids, an important class of lipids important for
hormone signaling, will be examined.
i. Using the information that a chiral carbon center is one at which carbon is
attached to four different substituents and the general steroid carbon skeleton,
below, draw an asterisk (*) next to each possible chiral center of testosterone,
shown below.
*
*
*
*
*
ii.
*
The Le Bel-Van’t Hoff rule states that if n is the number of chiral carbon centers
on a molecule, then the maximum number of stereoisomers for the molecule is
equal to 2n. Using your chiral centers above, determine the number of
stereoisomers for testosterone (though the one shown above is the only
stereoisomer of testosterone found in the body).
2^(# of chiral centers) for answer.
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Reference Charts
Amino acid structures and information
Generic amino acid Structure, where R represents the side chain
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Peptide bond formation reaction
Credit: http://schoolbag.info/chemistry/ap_chemistry/ap_chemistry.files/image004.jpg
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Constants and Definitions
Mega (M) =
, Kilo (k) =
, Deci =
, Centi (c) =
, Micro (μ) =
, Milli (m) =
, Nano =
1 kcal = 1 Calorie = 1000 cal = 4184J
⁄
Avogadro’s Number =
1 Ampere = 1 Coulomb (C) / second
(
)
( )
Molarity (M) =
( )
Equations
Equilibrium
[
]
(
)
For an amphiprotic salt with two
Given
( )
(
Given
)
(
[ ]
(
,
] [
[
)
[ ] [ ]
[ ] [ ]
)
]
; If pressures in atm instead of concentration, then
[
Henderson-Hasselbalch Buffer Approximation:
For acid-dissociation,
,
[
][
[
]
[
]
]
]
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√
For
(
)
Physical Properties
√
( )
Absorbance (A) =
, where is the molar absorptivity constant (M-1 cm-1), b is the pathlength in cm,
and c is the concentration in moles per liter.
, where for water,
Thermodynamics and Kinetics
[ ] [ ] , where k is the rate constant
For
Assume that frequency factor (A) is constant at all temperatures. Ea is the activation energy.
Zero Order Reaction: [ ]
[ ]
First Order Reaction: [ ]
[ ]
Second Order Reaction: [
Third Order Reaction: [
]
]
[
Half-life of a first order reaction:
( )
(
[
]
]
( )
)
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, where L is the latent heat of phase change per unit mass
Electrochemistry
Charge of an electron:
( )
(
)
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Credit for this table goes to the Department of Chemistry at Georgetown University. It can be
found at the following url: http://bouman.chem.georgetown.edu/S02/lect25/e6a.gif
Brief Bond-Line Instructions: For these diagrams, unlabeled nodes and ends of line segments
represent carbons. Hydrogens bonded to carbon are not shown, assume they are present
wherever implied by the structure. All heteroatoms are labeled explicitly, along with any
hydrogens bonded to heteroatoms.
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