Download MCDB 1030

MCDB 1030
1st Hour Exam
Feb. 10, 2005
150 points
Good luck!
Please write legibly!
1. Suppose that a gene encodes a protein that is 100 amino acids long.
a) (5 points) How long would the gene be in a bacterial cell?
300 base pairs encode the protein; the gene including the stop codon would be 303 base
pairs. Either answer is fine. Note that you need to define the units – just saying 300 is
not sufficient.
b) (5 points) Why might the gene be longer in a eukaryotic cell?
Eukaryotic genes often have introns, which are non-coding regions that are
transcribed into RNA, but then spliced out before the mRNA is translated.
c) Describe how the following mutations would affect the structure of the encoded protein.
In each case, discuss whether you expect the mutation to have no effect, a mild effect, or
a severe effect on the function of the protein. For missense mutations, identify the
original and changed amino acids and consider their structures in order to predict the
effects of the mutations.
i)
(5 points) a frameshift mutation at position 35
This would change the reading frame early in the gene and result in production of
a protein that is almost entirely wrong. This would have a severe effect on the
function of the protein.
ii)
(5 points) a frameshift mutation at position 290
This would change the reading frame very close to the end of the gene, and would
change only the last few amino acids. This would likely not have a severe effect on
the function unless those last few amino acids were critical.
iii)
(5 points) nonsense mutation at position 97
This would introduce a stop codon about 1/3 through the gene, and would result
in the production of a severely truncated protein. The effect would be expected to
be severe.
iv)
(5 points) silent mutation at position 12
A silent mutation changes the codon, but not the encoded amino acid, so it
would have no effect on the protein.
v)
(6 points) missense mutation that changes a CTA in the coding strand of the
DNA to CTC
If a CTA in the coding strand is changed to CTC, the mRNA that is
transcribed will change from GAU to GAG. GAU encodes aspartic acid and
GAG encodes glutamic acid. The structures of these amino acids are quite
similar (see your sheet of structures), so this change is not likely to be
important unless it occurs in a particularly critical part of the protein.
vi)
(6 points) missense mutation that changes CCC in the coding strand of the
DNA to TCC
If a CCC in the coding strand is changed to TCC, the mRNA that is
transcribed will change from GGG to AGG. GGG encodes glycine and AGG
encodes arginine. The structures of these amino acids are very different (see
your sheet of structures), so it is likely that this will have a significant effect
on the function of the protein.
2. (6 points) Draw a tri-peptide connecting three amino acids with peptide bonds as would be
found in a protein. Use any of the amino acids on the attached sheet.
R1
O
H 3N
H
N
C
H
R2
O
C
C
R3
H
N
H
C
COO
H
3. (18 points) For each of the following, indicate whether the molecule is polar or non-polar and
explain why.
H
a)
b)
H
O
H
H
O
O
C
C
O
O
H
H
c)
O
H
H
O
H
CH2
O
N
C
C
H
H
O
a) non-polar; C and H have similar electronegativities, so there are no polarized bonds
and therefore no partial charges
b) polar – O is more electronegative than C and H, so every C-O and O-H bond will be
polarized, with partial negative charges on the O and partial positive charges on the C or
H
c) polar – the molecule has a positive and a negative charge, in addition to polarized C-O
and O-H bonds.
4. (9 points) In the GC base pair shown below, mark the partial charges that provide the
attractive force for the H-bonds.
N
N
-
+
O
H
N
R
N
+
H
H
N
N
N
+
N
H
R
O
-
H
5. The following is a mRNA encoding a (very small!) protein.
AUG CCC UAU GAG GGU GAA AAA CAG UAG
(6 points) What is the DNA that would encode this mRNA?
ATG CCC TAT GAG GGT GAA AAA CAG TAG
TAC GGG ATA CTC CCA CTT TTT GTC ATC
(5 points) Mark the coding strand in this DNA.
The coding strand is the one that is copied; therefore it is the bottom one shown above.
(6 points) What is the “protein” that is encoded?
Met – Pro – Tyr – Glu – Gly – Glu - Lys - Gln
6. Contemplate peptidoglycan.
a) (5 points) What is it made of?
Peptidoglycan is made of polysaccharide chains cross-linked with short peptides.
b) (5 points) Where is it found?
It is found in the cell walls of bacteria.
c) (5 points) Why is it important?
It provides structural support that prevents the cells from bursting when they are placed in a
hypotonic environment.
7. (20 points) People of Swedish ancestry have fair skin and never have sickle cell anemia. In
contrast, people of African ancestry have dark skin and often carry a gene encoding Hemoglobin
S (the defective hemoglobin that causes sickle cell anemia in homozygotes). What are the
selective pressures that account for the differences in these two traits in the two different
populations of people?
Skin color is important because skin cells use light to synthesize vitamin D. At higher latitudes
(e.g. Sweden), the skin needs to be lighter in order to let in more light. In Africa, the skin can be
darker and still allow sufficient production of vitamin D.
The presence of malaria provides the selective pressure that maintains the hemoglobin S gene in
the population. Hemoglobin S appears to protect people from malaria, which occurs in Africa
but not in Sweden. Although people who are homozygous for the hemoglobin S gene will have
sickle cell anemia, those who are heterozygotes have relatively little trouble due to the defective
gene, but they are somewhat protected from malaria compared to people who have two normal
genes for hemoglobin.
8. Bacterial and mammalian cells use phospholipids in their cell membranes.
a) (8 points) What is the most important characteristic of phospholipids that makes them
suitable for this role?
Phospholipids have polar head groups and non-polar tails. They form bilayers because the
non-polar tails cluster together in the middle to avoid water, and the polar head groups face
outwards to interact with the water.
b) (15 points) There are actually other possibilities for molecules that could form cell
membranes. For each of the following, consider whether the molecule has properties that
might allow it to form a membrane. Explain your reasoning in each case.
What you are looking for is a molecule with a polar head and a non-polar tail. Only b fits this
description. Molecule a is completely non-polar. Molecule c has a charged group right in the
middle of the non-polar tail that would disrupt the packing of the non-polar tails.
a)
b)
c)
O
H3N
NH3
O
P
O
O
N