TA_____________
Solution Name_____________________________________
Section_______ Key - 7.013 Problem Set 1
Question 1
Complete the crossword puzzle by using the following keywords.
Cell junctions
chromosomes
cytoskeleton
cytosol
endoplasmic reticulum (ER)
eukaryotes
nuclear membrane
extracellular matrix (ECM)
mitochondria
flagella
genes
lysosomes
nucleolus
nucleus
organelles
phospholipids
plasma membrane
prokaryotes
ribosomes
D ow n
2. A component of the cell that is involved in the synthesis of ribosomes.
5. A dynamic network of fibers that maintain cell shape and motility and play important roles
in intracellular transport and cell division.
6. Membrane lipids that possess a hydrophobic head and a hydrophilic tail.
7. Filamentous structure that is present between cells, provides them with anchorage and is
involved in intercellular communication. (Note: Use the abbreviation).
8. Organisms that do not contain a nucleus.
9. Location of proteins that form a channel to export mRNA from the nucleus into the cytosol.
11. Locations of the enzymes that digest the toxic substances produced within a cell.
14. Location of a protein that transports water in and out of the cells.
16. Discrete units of hereditary information.
Across
1. An organelle in a eukaryotic cell that is involved in the synthesis of ATP.
3. Cell structures comprised of proteins that are involved in cell-cell adhesion.
4. Part of eukaryotic cell where DNA is synthesized and stored.
10. The assemblies of rRNA and proteins that play a major role in protein synthesis.
12. The assemblies of DNA and proteins that carry hereditary information.
13. One location of protein synthesis in a cell. (Note: Use the abbreviation).
15. Membrane enclosed sacs in eukaryotic cells that have diverse functions.
17. Location of an enzyme that initiates the breakdown of glucose.
18. Organisms whose cells are nucleated.
19. Cellular appendages that propel the movement of sperms.
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Section_______ TA_____________
Question 1 continued
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Question 2
You are given three unlabeled samples and asked to identify the cells in each sample as
bacteria, archaea, or eukaryote. Which cell types will fluoresce if the samples are stained
with
a) A fluorescent dye that recognizes a protein involved in ATP synthesis?
All organisms require ATP as their energy currency in order to live and perform various functions.
Hence a fluorescent dye that recognizes a protein involved in ATP synthesis will stain the cells in all
three samples.
b) A fluorescent dye that recognizes a lysosomal protein?
Lysosomes are a type of organelle present only in eukaryotic cells but absent in bacteria and archaea.
Hence a fluorescent dye that recognizes a lysosomal protein will stain only that sample that contains
eukaryotic cells.
c) The percentage of difference in DNA sequences between organisms is often used as a tool to
classify species into three major groups; bacteria, archaea and eukarya and to devise an
evolutionary tree as shown below.
BACTERIA
ARCHAEA
EUKARYA
Evolutionary time scale
i.
In the evolutionary tree drawn above, indicate the order of origin of the three groups by
filling the boxes.
Bacteria diverged first, and therefore the left-most box must be labeled Bacteria. Archaea and
eukarya diverged later, so the other two boxes should be labeled Archaea and Eukarya, in either
order based on the amount of information provided. It should however be noted that the common
ancestor of archaea and eukarya looked much more similar to modern-day archaea than to
eukaryotes.
ii.
On the schematic above, indicate where the mitochondria (as an organelle) might have
originated using an arrow(s).
An arrow drawn anywhere between the marked region i.e. between the archaea / eukarya common
node and the box labeled Eukarya, is a correct answer based on the information that this question
provides.
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Name_____________________________________
Section_______ TA_____________
Question 3
The major macromolecules present in the cell are deoxyribonucleic acid (DNA), ribonucleic
acid (RNA), carbohydrates, proteins and lipids. Each of these macromolecules is composed of
specific monomers that serve as their building blocks.
a) The following diagram represents a nucleotide that serves as a monomer for nucleic acids.
i.
Would you classify this nucleotide as a monomer for DNA or RNA? Explain.
The nucleotide in the schematic above has a deoxyribose sugar with a 3OH group and no 2OH. This
classifies the nucleotide as a monomer of DNA and not an RNA. Furthermore, the base is Thymine (T),
which is only seen in DNA and is absent in RNA.
ii.
Circle the part(s) of this monomer that will bond so that this will be added to the
growing end of the polymer and name the covalent bond that will be formed.
The covalent bond formed is a phosphodiester bond.
iii.
If the monomer drawn above comprises 28% of a double stranded nucleic acid, predict
the percentages of each of the other nucleotides present in the double stranded nucleic
acid. Explain your answer.
According to the complementary base pairing rule Adenosine (A) pairs with Thymidine (T) by
forming two hydrogen bonds, and Guanosine (G) pairs with Cytosine (C) by forming three
hydrogen bonds. The nucleotide in the above schematic is thymidine. Hence the DNA should contain
28% T, 28% A, 22% G and 22% C.
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Name_____________________________________
Section_______ TA_____________
Question 3 continued
iv.
If you modify the nucleic acid such that this monomer is reduced from 28% to 10%,
would that increase or decrease the amount of energy required to denature the
modified nucleic acid relative to the original polymer? Explain.
If the amount of T is reduced from 28% to 10% in DNA then the altered DNA will have a
composition of 10% T, 10% A, 40%G and 40% C. Because a G/C base - pair involves 3
hydrogen bonds while an A/T base-pair involves only 2 hydrogen bonds, the increased % of G and
C represents an increase in the total number of hydrogen bonds. Therefore the amount of energy
required to break these hydrogen bonds and denature the DNA will be increased.
v.
Identify the primary role of this polymer in a cell.
DNA serves as the hereditary material in a cell.
b) Drawn below is a schematic of a transmembrane protein.
Extracellular
Cell membrane
Cytosolic side
i.
Of the 20 essential amino acids that make a protein, list those that you would expect to
find in a linear stretch that spans the lipid bilayer? Explain why you selected these
amino acids. (Note: A chart of the amino acids is provided as the last page of this problem set).
You would expect amino acids that are non polar, uncharged, hydrophobic : Alanine, Cysteine, Glycine,
Isoleucine, Leucine, Methionine, Phenylalanine, Valine and Tryptophan.
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Section_______ TA_____________
Question 3 continued
ii.
Consider the following amino acid sequence that is a part of the protein.
iii.
Circle one peptide bond that holds together the primary structure of this protein.
Box the groups that provide unique properties to every amino acid drawn above.
Assuming that the mature protein is comprised of a single polypeptide chain what level
of protein structure will not be present in this protein? Explain.
This protein will not have a quaternary structure that is observed only in the proteins that have two or
more than two polypeptide chains.
iv .
Enzymes catalyze biological reactions by binding to specific substrates at their catalytic
site. The catalytic site of each enzyme is comprised of only a few amino acids, although
the protein also includes many amino acids that are not a part of the catalytic site.
Describe one role of the amino acids that are not a part of its catalytic site.
The remaining amino acids are r e quired to provide the enzyme with a specific three dimensional
conformation that is crucial for the functioning of the enzyme. The structur e of the protein is important
to its stability, and may contribute to sub strate re cognition.
They may also be involved in making regulatory sites i.e allosteric sites whe r e specific allosteric
activators and inhibitors can bind and regulate the overall enzyme activity, sites of protein modification,
or sites of protein/protein interactions.
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Name_____________________________________
Section_______ TA_____________
Question 3 continued
c) Assume that the transmembrane protein in the schematic above is a glycoprotein and has a
carbohydrate part as shown below.
i.
ii.
Name the covalent bond marked by a circle in the diagram. Glycosidic bond
What molecule (Molecular weight = 18) is produced as a byproduct following the
making of the bond that you identified?
Water molecule will be produced as a byproduct.
d) Below is the structure of cholesterol that serves as a precursor for the steroid hormones,
Vitamin D and is also an important component of the plasma membrane.
i.
On the diagram above, box the entire hydrophilic region(s) of this molecule.
ii.
On the diagram above, circle the entire hydrophobic region(s) of this molecule.
iii.
What bonds / interaction are most likely to form between cholesterol molecules?
Since cholesterol is highly hydrophobic you will most commonly be observing the hydrophobic
interactions.
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Name_____________________________________
Section_______ TA_____________
Question 4
Phosphofructokinase (PFK) is an enzyme in a multi-step pathway which harvests the energy
from carbohydrates and stores it in ATP. The transfer of a phosphate group from ATP is an
important reaction in a wide variety of biological processes. This enzyme binds to its ligand,
fructose-6-phosphate and catalyses its phosphorylation to fructose-1,6- bisphosphate by
transferring the phosphate group from ATP (ligand bound / active form). PFK exists as a
homotetramer in bacteria and mammals where each monomer possesses 2 similar domains.
One domain has an ATP binding pocket whereas the other houses the substrate-binding site
and the allosteric site (a regulatory binding site distinct from the active site, but that affects
enzyme activity). Deficiency in PFK to leads Tauri's disease that is characterized by severe
nausea, vomiting, muscle cramps in response to bursts of intense or vigorous exercise.
Computer modeling programs that are currently being used by the biologists to view the
structure and function of various proteins both in normal and diseased state. For this problem
you will need to use a computer to view the structure of PFK enzyme. To begin, go to the
site: http://web.mit.edu/star/biochem . This website has several links. In the left menu click
on Supplemental resources. Explore the links marked The Amino Acids, The Peptide
Bond, Secondary Structure, and Nucleotides and DNA. These contain models of different
macromolecules and will help you understand protein and structure. To help you learn how
to use the program, a tutorial is available under StarBiochem User Guide. When you are
ready to begin, click on Start. Once you have Star Biochem open, go to File and choose the
file 2PFK and click open in the dialog box. You will see a schematic of the 2PFK enzyme on
the left, and a menu on the right. Choose STRUCTURE from that menu. You may try
opening these files using Athena Station.
a) Choose Primary under the STRUCTURE menu. You will see two, independent structures
in 2PFK (an artifact of how the structural model was determined); please use one of these to
answer the following questions. How many amino acids comprise this enzyme?
This structure file includes 4 peptide chains, but there are only two chains in each independent enzyme
structure. Looking under Primary Structure, you will see a list of all the amino acids in the file. The
polypeptide chain 1 starts with Met0 and ends with Ile300, so it is composed of 301 amino acids in total.
The polypeptide chain Chain 2 similarly has 301; chain 3 has 302 and chain 4 has 305. Therefore, either
602 or 607 is a correct answer to this question (depending on which structure or pair of chains you
chose to use).
b) Choose Secondary under the STRUCTURE menu. Again, use one of the two separate
structures in this file to answer the following questions.
i.
What are the different types of secondary structures that you can observe?
Alpha-helices, beta-sheets, and coils.
ii.
How many alpha- helices are there in 2PFK?
There are 48 helices in the entire file, and 12 per peptide monomer. Therefore, each dimeric enzyme
has 24 helices.
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Question 4 continued
iii.
How many beta- sheets are there in 2PFK?
There are 4 sheets in each dimmer, 2 from each peptide monomer.
iv .
The secondary structures are predominantly stabilized by hydrogen bonds. For each
type of secondary structure that you found, identify the part of the amino acids that
participates in the formation of the hydrogen bonds. Your choices are the SIDECHAIN
or the BACKBONE? (Note: Explore the STRUCTURE and the VIEW CONTROL menu
to answer this question).
Alpha-helices are stabilized by hydrogen bonds within a chain between the hydrogen of N-H of one
amino acid and the oxygen of the C=O of another amino acid. These atoms belong to the
BACKBONE of the peptide, as opposed to being part of the side-chain or R-group.
Beta-sheets are stabilized by hydrogen bonds between two chains, between the N-H groups on one
chain and the C=O groups on the other: again, this primarily involves interactions between
BACKBONE atoms. (Note: For diagrams of these two structures, see page 45 of your textbook.)
c) Explore the STRUCTURE menu. Again, use one of the two separate structures in this file to
answer the following questions.
i.
How many peptide chains are present in the molecule?
Two.
ii.
What is the highest level of protein structure that you observe?
Quaternary structure , because the enzyme is formed by two interacting peptide chains.
d) To explore the catalytic activity of PFK, open 2PFK, 1PFK and 6PFK in three separate
windows. (Note: You should click on start to open each window). Compare the three forms of
PFK by going through their STRUCTURE menu.
i.
Which of the three molecules represents the resting / ligand-unbound state of PFK?
2PFK
ii.
Now compare the structure of the remaining two molecules that you did not select
while answering the question above. Identify which of the two molecules represent the
active state of PFK and which one represents the inactive state. Explain your choices.
(Note: Carefully compare the ligands that are bound to the enzyme while making your selection,
and review the information given in this problem about PFK. Also apply what you have learned
in lecture).
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Name_____________________________________
Section_______ TA_____________
Question 4 continued
1PFK is the active state of PFK. The enzyme is bound to two different kinds of ligands
(appearing in yellow in the structure), which are its reaction products: fructose 1,6-bisphosphate,
and ADP. 6PFK is the inactive state of PFK. This is the structure of the enzyme bound to an
allosteric inhibitor, not to a reaction substrate or product.
iii.
Deficiency in PFK leads to Tauri's disease, characterized by severe nausea, vomiting,
muscle cramps. Patients with this genetic disorder are advised not to exercise
vigorously. Explain how a reduced physical activity can help these patients.
Physical activity requires the body to use energy. The body uses energy in the form of ATP, but
receives energy in other forms such as carbohydrates. A deficiency of PFK enzyme that acts in
ATP production will result in less efficient ATP production. Therefore, the person with Tauris
disease needs to limit physical activity so that their supplies of ATP will not be depleted.
Question 5
The following diagram represents the active site of an enzyme (E1) that catalyzes the covalent
joining of two peptides (P1 and P2) together to make a large protein (P3) at neutral pH. You
are able to observe this in a test tube (in vitro).
P2
P1
Gln
Gly
Glu
1
Ala
4
3
Asp
2
Arg
His
Ser
E1
Note: The dashed lines represent the bonds (numbered 1-4) that are formed between the
amino acids of the enzyme (E1) and the substrate peptides (P1 and P2).
a) Complete the following table by stating the interactions between the amino acids that are
either located at the catalytic site of the enzyme (E1) or are a part of the two interacting
substrate peptides (P1 or P2).
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Name_____________________________________
Section_______ TA_____________
Question 5 continued
Bond # Amino acids
1
Ala (E1) - Gly (P1)
Interaction / bond (Your choices are hydrogen bond, Van der
Waals forces, ionic bonds, covalent bonds and hydrophobic
interactions.
Hhdrophobic interactions / Van der Waals forces
2
His (E1) Asp (P1)
Ionic bonds
3
Arg (E1) Glu (P2)
Ionic Bonds
4
Ser(E1) Gln (P2)
Hydrogen bonds
b) Explain how the interaction of the enzyme with the substrates will be influenced if the
Arginine at the catalytic site undergoes the following substitutions:
i.
Arg changed to Trp?
Trp unlike Arg is a nonpolar hydrophobic amino acid with a bulky carbon ring that can create steric
hindrance Therefore, if trp substitutes Arg , then it will not form an ionic bond with glu and will also
cause steric hindrance.
ii.
Arg changed to Lys?
.
Lys is positively charged, like Arg, and has a similar structure as well, so it will probably still form an
ionic bond with Glu, and the interaction will not be disrupted.
c) During an experiment, by mistake, you add a mystery compound, M to the original
reaction mixture and find that the reaction is completely inhibited. You then try to make the
reaction work by increasing the concentration of substrates P1 and P2. You find that the
reaction now does work. However as indicated on the diagram below, if you add an excess of
P1 and a normal amount of P2 together with M the reaction works, but if you have a normal
amount of P1 and an excess of P2 the reaction is not restored. Based on this information,
briefly explain how compound M inhibits the enzymatic reaction.
P1+ P2
E1
P3
P1+ P2 + Mystery component
E1
No reaction
E1
P1 (excess)+ P2 (normal) + Mystery component
P3
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Question 5 continued
P1 (normal)+ P2 (excess) + Mystery component
Section_______ TA_____________
E1
No reaction
M is a competitive inhibitor, because excess P1 can out-compete M and allow the reaction to proceed.
d) Draw the energy profile of P1 + P2
i.
P3 in the space provided below if
Free energy (G) is negative.
When free energy is negative, the reactants have higher energy than the products.
Reactants
G
G=-negative
Products
Course of reaction
ii.
Free energy (G) is positive.
In this reaction, the products are higher energy than the reactants.
Reactants
G
G=-positive
Products
Course of reaction
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Section_______ TA_____________
Question 5 continued
iii.
Which of these reactions will require more free energy? Explain.
Reaction (ii), with positive free energy, requires more energy, because its products have higher energy
than its reactants. Reaction (i) on the other hand will release energy.
STRUCTURES OF AMINO ACIDS at pH 7.0
O
O
H
CH3
H
NH3
+
C
CH2CH2CH2 N
O
CH2 SH
H
NH3
+
O
H
N
C
+
C
CH2
NH3
+
O
N
C
H
H
H
H
OC
S
CH3
H
C
O-
OCH
H
H
C
H
CH3
NH3 OH
+
THREONINE
(thr)
C
CH2
H
NH3+
O-
O
C
H C CH2
CH2
H
N
+ CH2
H
H
H
C
CH2
OH
NH3
+
SERINE
(ser)
PROLINE
(pro)
H
N
C
H
H
H
H
O
O
NH3
+
TRYPTOPHAN
(trp)
OC
H
CH2CH2CH2CH2
LYSINE
(lys)
O
H
C
NH3
+
LEUCINE
(leu)
CH2
C
H
CH3
CH3
PHENYLALANINE
(phe)
O
C
C
NH3
+
METHIONINE
(met)
H
CH2
C
CH2CH2
C
H
C
NH3
+
O-
O
NH3
+
O
H
O
O
C
CH2CH3
ISOLEUCINE
(ile)
C
CH
NH3
+
GLYCINE
(gly)
O
O
NH3 CH3
+
H
HISTIDINE
(his)
O
C
H
NH2
GLUTAMINE
(gln)
O
CH
C
C
C
NH3
+
GLUTAMIC ACID
(glu)
C
H
O-
O
O
CH2CH2
O-
ASPARTIC ACID
(asp)
O
C
CH2 C
NH3
+
O-
H
C
NH3
+
CYSTEINE
(cys)
O
CH2CH2
O
C
NH2
C
O
C
H
ASPARAGINE
(asn)
O
C
C
CH2 C
NH3
+
O-
O
C
O
C
NH2
+
ARGININE
(arg)
C
H
C
H
NH2
O-
O
C
H
NH3
+
ALANINE
(ala)
O
O
C
C
O
O
O
C
H
H
H
C
O
O
C
C
CH2
OH
NH3
+
H
TYROSINE
(tyr)
H
H
C
CH3
C
NH3 H
+
CH3
VALINE
(val)
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