MIT Department of Biology
7.014 Introductory Biology, Spring 2004
7.014 Problem Set 1
Please print out this problem set and record your answers on the printed copy.
Problem sets will not be accepted late.
Question 1
a) Describe the conditions of the atmosphere on prebiotic earth and how these conditions
differ from what is found today.
b) Describe the proposed first organism as it compares to modern organisms; include the type
of genetic material and the source of energy it may have used.
c) Describe the process that led to an O2 atmosphere on earth and how this change in
atmospheric conditions influenced the course of evolution.
Question 2
You are given four test tubes, each contains cells from a different organism. One tube contains
bacterial cells, one contains yeast cells (eukaryotic), one contains human cells and the last
contains dinosaur cells.
Can you identify the cells from each tube if you are given a light microscope? Explain your
answer.
1
Question 3
Growth factor receptors (shown below) are transmembrane proteins found on the cell surface.
extracellular
–
O
– O P O
O
H
H2C
C
O
C O
CH2
CH2
CH2
CH2
CH2
H2C
O
O P O
–
O
–
H
H
C H
O
C OO
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
C O
O
C
H
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
C OO
O
C HH
H
H
C
–
O
O P O
–
O
H
C H2C
O
O
C O C
–
O
O P O
–
O
H
H2C
C
O
C O
CH2
CH2
CH2
CH2
CH2
CH2
H
C H
O
C O
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
C O
O
C
H
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
C O
O
C H
H
CH2 CH2
C O C
O
C
H
O
C H2C
H
O
O P O
O –
–
H2C
O
– O P O
O
–
H
H C
–
O
O P O –
O
H
C H 2C
O
O
O C O C
–
O
O P O
–
O
H
H2C
C
H
C H
O
O
C O C O
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2 CH2
O C O C
O
H C
H
O
C H 2C
H
O
O P O
O –
–
CH2 CH2
C O C O
O
H2C
C
H
O
– O P O
–O
O
C H
H
–
O
O P O –
H
H
O
H C
C H2 C
O
O
O C O C
CH2 CH 2
CH2 CH 2
CH2 CH 2
CH2 CH 2
CH2 CH 2
CH2 CH 2
CH2 CH 2
CH2 CH 2
CH2 CH 2
CH2
CH 2
CH2
CH 2
CH2
CH2
CH 2
CH 2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
O C O
O
H C
H
CH 2
CH 2
CH 2
CH 2
CH 2
CH 2
CH 2
C
O
C H2 C
H
O
O P O–
O
–
intracellular
growth factor receptors
a) The molecules that form the above membrane belong to what class of macromolecules?
Explain the important qualities/properties of these molecules that allow them to form
membranes.
2
Question 3, continued
A smaller schematic of the growth factor receptor is shown here.
transmembrane region
]
b) Which amino acids would you expect to find in the transmembrane region of the receptor?
When a signaling molecule, called the ligand, binds to the extracellular domain of the receptor,
a conformational change occurs in the receptor. Growth factor binding causes dimerization of
two adjacent receptors in the cell membrane. Upon dimerization, the intracellular domains of
the receptors become activated. See schematic below.
ligand
ligand
extracellullar ligandbinding domain
Receptor 1
Receptor 2
plasma
membrane
intracellular domain
inactive domains
active domains
3
Question 3, continued
c) Regions of the two receptors that interact upon dimerization shown below. In parts (i - iv)
below, name the strongest type of interaction (choose from; hydrogen bond, ionic, covalent,
van der Waals) that occurs between the side chains of the amino acids indicated.
Receptor 1
Ala45
Asp68
CH2
CH2 S H
CH2
S
S
CH3
C O
O
Ser53
CH2 OH
+
H3N
CH2
CH2
CH2
Cys75
CH2 CH CH
2
2
H2N
O
Phe50
Receptor 2
Cys82
Cys36
C CH2 CH2
Lys65
Gln12
HC
3
HC
CH
Val98
3
Interacting Side chains
i)
Cys75 : Cys82
ii)
Asp68 : Lys65
iii)
Ser53 : Gln12
iv)
Phe50 : Val98
Type of interaction
d) Explain how Gln12 and Val98, which are far apart in the primary sequence of the protein,
can be close to each other in the region of the protein diagrammed above.
4
Question 4
You have discovered a new enzyme, enzyme E, which breaks down proteins by cleaving
peptide bonds after tyrosine or phenylalanine.
a) Enzyme E is the product of gene G that encodes a protein with the molecular weight of 50
kilodaltons (50 kD). Upon purification of enzyme E, only the expected 50 kD polypeptide is
present. Interestingly, the molecular weight of the active enzyme E is 250 kD, not 50 kD
i) Why might active purified enzyme E be larger than the product encoded by gene G?
ii) Define primary, tertiary, and quaternary structure.
iii) Is the primary structure of the 50 kD protein the same or different than the primary
structure of the 250 kD protein? Explain briefly.
iv) Is the tertiary structure of the 50 kD protein the same or different than the tertiary
structure of the 250 kD protein? Explain breifly.
v) Is the quaternary structure of the 50 kD protein the same or different than the
quaternary structure of the 250 kD protein? Explain breifly.
b) You test the ability of enzyme E to break down a large protein. This large protein is not
broken down by enzyme E. You then treat the large protein with DTT (a compound that
disrupts disulfide bonds) and test the enzyme E activity again. This time the large protein is
broken down by enzyme E.
Why was enzyme E able to cleave the large protein only after the substrate was treated with
DTT?
5
STRUCTURES OF AMINO ACIDS at pH 7.0
O
O
C
H
CH3
H
C
ALANINE
(ala)
CH2 SH
H
NH3
+
O
H
N
C
C
O
+
C
H
H
CH2
NH3
+
N
C
H
O-
O
C
H
C
CH2CH2
S
CH3
H
C
CH2
O-
O
OC H
C
C
CH3
NH3 OH
+
THREONINE
(thr)
C
CH2
H
CH3
H
C
H
H
O-
H
C
H
C
CH2
OH
NH3
+
SERINE
(ser)
PROLINE
(pro)
H
N
C
H
H
H
H
O
O
NH3
+
TRYPTOPHAN
(trp)
O-
O
H C CH2
CH2
H
N
CH2
H +
H
NH3+
LYSINE
(lys)
O
H
CH2CH2CH2CH2
NH3
+
CH3
LEUCINE
(leu)
CH2
C
H
C
C
NH3
+
PHENYLALANINE
(phe)
O
C
H
-
O
C
NH3
+
METHIONINE
(met)
GLYCINE
(gly)
H
C
NH3
+
H
CH2CH3
O-
O
NH3
+
O
C
ISOLEUCINE
(ile)
HISTIDINE
(his)
C H
NH2
O
O
NH3 CH3
+
H
H
GLUTAMINE
(gln)
O
C H
C C
C
NH3
+
GLUTAMIC ACID
(glu)
O
C
O
CH2CH2
O-
ASPARTIC ACID
(asp)
O
C
C
O-
NH3
+
CYSTEINE
(cys)
H
C
NH3
+
O-
H
O
C CH2 C
NH2
ASPARAGINE
(asn)
O
CH2CH2
H
CH2 C
NH3
+
O
C
C
O
C
NH2
+
C
C
H
NH2
O-
O
C
O
C
ARGININE
(arg)
O
H
CH2CH2CH2 N
O
-
O
C
H
NH3
+
NH3
+
O
O
C
C
O
O
O
H
H
C
C
H
C
O
O
CH2
OH
NH3
+
H
TYROSINE
(tyr)
H
H
C
CH3
C
NH3 H
+
CH3
VALINE
(val)
6