SOLUTION KEY- 7.

013 EXAM 1 (2 / 23 / 15)

Question 1 (20 points)
a) Name an organelle in a prokaryotic cell: Ribosomes (2pts)

b) List three subcellular regions in human skin cells where ribosomes function:
RER, Mitochondria and cytoplasm (3pts, 1 for each)

c) Psoriasis is a skin disorder that is caused by the increased proliferation (division) of skin cells. You
develop a mouse model to study psoriasis. You isolate skin cells from this mouse and culture them in
Plate 1 with Colchicine and Plate 2 with Latrunculin.

Colchicine disrupts the microtubules by inhibiting the polymerization of tubulin protein monomers.
Latrunculin disrupts the microfilaments by inhibiting the polymerization of actin protein monomers.

Which inhibitor (Colchicine or Latrunculin) could potentially be used to treat psoriasis? Explain why
you selected this inhibitor and not the other.
You would colchicine that disrupts the microtubules formation by disrupting the assembly of the tubulin
monomers. In the absence of the microtubules the chromosomes duplicated during the S phase will not
be able to align at the metaphase plate. So the cell division will halt during the late prophase or
metaphase of mitosis. (2pts for colchicine and 3pts for explanation)

d) The assembly or disassembly of actin monomers regulates the formation of microfilaments.

i. Circle the highest level of protein structure for an actin monomer from the choices below.

Primary Secondary Tertiary (2pts) Quaternary

ii. Circle is the highest level of protein structure that is changing during assembly and disassembly
of actin monomers?

Primary Secondary Tertiary Quaternary (2pts)

e) The following is a drawing of a triglyceride molecule in which glycerol is condensed with three fatty
acid chains. O

O
O

O O

i. Although the fatty acid chains of this triglyceride are similar to the lipids that make the plasma
membrane, this triglyceride is almost never found in the plasma membrane. What property of this
triglyceride prevents it from forming a plasma membrane?
Triglycerides lack a phosphate group, they are not amphipathic since they do not have polar head
group. This prevents them from forming a lipid bilayer. (3pts)

ii. You analyze the plasma membrane of two cell types and see that the lipids in the membrane of
Cell type A pack more tightly than those that make the membrane of Cell type B. Which cell type
(Type A or Type B) has higher concentration of fatty acid chains that are shown in the drawing
above? Explain why you selected this cell type.
Unsaturated fatty acids have one or more than one double bonds between the carbon atoms of their
fatty acid chains. This prevents their close packaging. In comparison, saturated fatty acids (as shown in
the drawing) have no double bonds so they closely pack and have higher melting points. So the
membrane of Cell type A will have higher concentration of saturated fatty acids as shown in the
drawing, these will show close packaging of their fatty acid chains and hence exhibit less mobility or
fluidity compared to the membrane of Cell type B which are rich in unsaturated fatty acids. (3pts)
1
Question 2 (22 points)
a) Following is the structure of adenosine nucleoside.

i. Number the carbon atoms of the sugar in the schematic

(2pts).

ii. Circle the carbon atom position to which you would add
5 phosphates such that it becomes a nucleotide. (2pts)
You should circle the 5C of the ribose sugar.
4 1
3 2 iii. Would the nucleotide you generate in part (ii) be a monomer
of deoxyribonucleic acid (DNA)? Why or why not? (3pts)
No, it would be a monomer of RNA since it has an OH group at the
2C position of the ribose sugar.

b) You are working with two double helical DNA molecules that are of the same length (in base pairs):
DNA double helix 1 contains 25% Guanine (G) whereas DNA double helix 2 has 25% Adenine (A).

i. Give the % of Thymine (T) in

DNA helix 1: 25% (2pts) DNA helix 2: 25% (2pts)

ii. Would the sequence of nucleotides in DNA helix 1 be identical to that in DNA helix 2? Why or
why not?
No, although they have the same % of As, Ts, Gs and Cs, the order of the bases in the two helices may
differ resulting in different sequences (3pts)

c) The following schematic shows the last four amino acids at the C- terminus of a protein.

i. On the schematic, circle the side-chain

of the amino acid(s) that is hydrophilic.
(3pts)

ii. You plan to add Molecule 1 to the growing end of the protein sequence above in vitro (in a test
tube).
H
NH 2 C CH3 On Molecule 1, circle the group that participates in forming a covalent
CH3 peptide bond. (2pts)
Molecule 1

Once Molecule 1 is added to the growing end of the protein sequence, could it form a peptide
bond with an incoming amino acid? Explain your answer.
No since its lacks the carboxyl group that is involved in forming the peptide bond. (3pts)

2
Question 3 (24 points)
This schematic shows the binding of [Substrate A], to the active site of Enzyme E1. Note: For
simplicity, only the groups of amino acids X90, Y51 and Z23 that interact with [Substrate A] are shown and
the interactions are circled and labeled as (i), (ii) and (iii).

C N a) Classify the amino acids, X, Y and Z as

Substrate A ! 51!
90 (i) polar or nonpolar. (3pts, 1pt for each)
i. Amino acid X90: Polar or nonpolar
ii. Amino acid Y51: Polar or nonpolar
(iii)
iii. Amino acid Z23: Polar or nonpolar
(ii)
Active site of 51
23
Enzyme E1

b) List the strongest non-covalent bonds (choose from ionic bonds, hydrogen bonds, hydrophobic
interactions) between Substrate A and the side-chains of amino acids X90, Y51 and Z23 at regions (i), (ii)
and (iii)).
i. Strongest non- covalent bonds at Region (i): Hydrogen bond (2pts)
ii. Strongest non- covalent bonds at Region (ii): Hydrogen bond (2pts)
iii. Strongest non- covalent bonds at Region (iii): Hydrogen bond (2pts)

c) You create two mutants (Mutant 1 and Mutant 2) of Enzyme E1, each of which has amino acid
substitutions at its active site as shown in the table below.

E1 Amino acids at the Which mutant (Mutant 1 or Mutant 2) is likely to bind to

active site [Substrate A] similar to the wild-type version of E1?
Wild- type X90 Y51 Z23 Explain why the other mutant form of E1 cannot bind to
[Substrate A].
Mutant 1 X90 Y51 Asn23 Mutant 1 can still bind to Substrate A unlike Mutant 2
(2pts). In mutant 1 the Asn23 can still hydrogen bond with
Mutant 2 Asp90 Y51 Z23 the substrate similar to Z23. In contrast, in mutant 2 the
Asp90 has a negatively charged side-chain unlike X90. So
Asn90 will repel the like charged group of the substrate
thus preventing/ destabilizing the formation of Enzyme
substrate A complex. (3pts)

d) Further analysis shows that hydrolysis of [Substrate A]

by Enzyme E1 is an endergonic reaction. This means
that the (2pts each or 6pts total)

i. Free energy of the substrate is less than / more than / the same as that of the product.
ii. Activation energy of E1 catalyzed reaction is less than/ more than/ same as that of E1 uncatalzed
reaction.
iii. Reaction rate of E1 catalyzed reaction is less than/ more than/ same as that of E1 uncatalyzed
reaction

3
Question 3 continued
e) The E1 catalyzed reaction is a part of the following multi-step reaction cascade in a cell where
each step is catalyzed by a specific enzyme (E1-E3).

E1 E2 E3
Substrate A B C D

Product D non-covalently binds to E1 and alters its 3- dimensional conformation. In this altered
conformation, E1 cannot bind to [Substrate A]. Based on this information, circle one or more option(s)
from the choices below that best describes the effect of Product D on E1 activity and explain why you
selected this option(s) (4pts)

This is a Competitive, reversible inhibitor Its non-covalent binding to a site on E1

other than the active site makes it
This is an allosteric, reversible inhibitor
reversible and allosteric. (4pts, 2pts for
This is a competitive, irreversible inhibitor circling and 2pts for explain, 1 pt if
This is an allosteric, irreversible inhibitor competitive with proper explanation)

Question 4 (14 points)

BD bd
You are following the segregation of three genes; A, B and D in Heidis cells (genotype: AAX X ).
Note: Gene A is located on the autosome whereas Genes B and D are on the X chromosome.

a) Give the possible genotype(s) of a cell from Heidis dad for A, B and D genes. Note: Assume that
Heidis dad is homozygous for Gene A.
BD bd
AAX Y (2pts) or AAX Y (2pts)

b) Assume that a somatic cell (shown below) from Heidi undergoes cell division. Draw the
arrangement of A, B and D genes at the metaphase plate of this dividing cell. Note: The centrioles at
the two ends are drawn for you. (4pts total for correct alignment)

A A
Centriole
Centriole A A
B B
D D
b b
d d

c) Assume that the somatic cell drawn above undergoes nondisjunction of one X chromosome
carrying B and D genes. All other chromosomes separate normally. Give the genotypes of all possible
daughter cells in terms of A, B and D genes. (4pts or 1pt for each)

BD BD bd bd bd bd BD BD
AAX X X and AAX or AAX X X and AAX

BD bd bd
d) Heidi has a daughter whose genotype is AAX X X . Could the presence of the extra X
chromosome be explained by nondisjunction in part (c)? Why or why not?
No, the daughter can have this genotype only if there was nondisjunction event that resulted in
Heidi producing a gamete that had an extra X chromosome. Non-disjunctions in somatic cell of
an individual are not passed on to the subsequent generations. (2pts)
4
Question 5 (20 points)
You are studying three traits in mice; fur color (A or a), size (B or b) and temperament (D or d).
Note: Use the uppercase letters for the alleles associated with the dominant phenotypes and the
lowercase letters for the alleles associated with the recessive phenotypes. Assume that each trait is
regulated by one autosomal gene.

a) You mate a mouse having black fur (P1) with a mouse having white fur (P2) and obtain 100 mice in
F1, of which 50 have black fur and 50 have white fur. Assuming that black fur color is the dominant
phenotype, give the genotypes of the following mice. (4pts total or 1pt each)
i. P1 mouse: Aa
ii. P2 mouse: aa
iii. F1 mice with black fur: Aa
iv. F1 mice with white fur: aa
b) You mate two F1 mice both having black fur to get 100 mice in F2 generation. Fill in the table below
for the F2 mice. (4pts total; 2pts for correct phenotype and corresponding numbers together and 2pts
for correct genotypes and corresponding numbers together)

100 F2 mice

Phenotypes Corresponding Genotypes Corresponding

numbers or ratio numbers or ratio

Black : White 75: 25 AA 25

(3:1) Aa 50
aa 25 or (1: 2: 1)

c) In a separate experiment, you mate a true breeding small & aggressive mouse (P3) with a true
breeding large & mild- tempered mouse (P4). All of the resulting F1 mice are large & aggressive.
Give the genotypes of P3 & P4 mice.

Parental generation Phenotype Corresponding genotype

P3 Small & aggressive bbDD (2pts)

P4 Large & mild- tempered BBdd (2pts)

d) You mate an F1 mouse with a double homozygous recessive mouse and obtain four different
classes of phenotypes in F2. If the map distance between the two genes is 20cM, complete the table
below by filling in the genotypes and corresponding % for each class of mice obtained in F2. (8pts total
or 2pts for each correct genotype and corresponding %)

Phenotypes in F2 Corresponding genotype Corresponding %

Small & aggressive bDbd 40%

Large & mild- tempered Bdbd 40%

Large & aggressive bdbd 10%

Small & mild-tempered BDbd 10%