Name _______________KEY_____________________
Biology 201 (Genetics)
Exam #3
24 November 2003

Read the question carefully before answering. Think before you write. Be concise.

You will have up to one hour to take this exam. After that, you MUST stop no matter where you
are in the exam.

If I can not read your handwriting, I will count the question wrong.

Sign the honor pledge if applicable.

Good luck!
I pledge that I have neither given nor received unauthorized assistance during the completion of this
work.
Signature: _________________________________________________________
10pts
.
10pts
.
1. Oculocutaneous albinism type IA is an autosomal recessive disorder characterized by absence of
pigment in hair, skin, and eyes. People with this disease have a complete lack of tyrosinase activity
in their pigment cells (tyrosinase is an enzyme required for pigment production). BRIEFLY,
explain how a missense mutation might cause this genetic disease.
A missense mutation is a change in one nucleotide in the DNA (in this case in the tyrosinase gene) that
alters one codon to that of another amino acid,. The change in one amino acid alters the primary
structure of the tyrosinase protein. A change in primary structure of a protein may affect folding of the
protein into the proper secondary and/or terteriary structure, so the tyrosinase protein does not function
properly to make pigment. Alternatively, the change in one amino acid may eliminate the activity of the
tyrosinase protein that results in pigment formation because that one amino acid is critical to function.
See notes from Nature of the Gene (last half of 13) and slide 26 from Ch 14 (for missense)
2. Spontaneous tautomerization and addition of base analogues both cause mutations in the DNA by a
common mechanism. BRIEFLY explain how these events cause mutations to occur?
Spontaneous tautomerization is a transient shift in a proton on a nitrogenous base from one atom to
another. This shift alters the hydrogen bonding between bases which results in improper basepairing,
allowing the tautomerized base to pair with bases other than the one it is normally paired with during
DNA replication. Base analogues are compounds sufficiently similar to basepair with the correct base
during DNA replication; however, they tend to tautomerize even more that normal bases (as described
above). In either case, improper hydrogen bonding during DNA replication results in the wrong
nucleotide being incorporated into the new DNA strand. After one more round of DNA replication, the
mutation will be permanently established (see Figure 14-6).
See question 5 in Ch 14 and Fig 14-6
5 pts.
3. A mutant allele of the CFTR gene was isolated, yet the change in nucleotide sequence was not in the
coding sequence that encodes the actual CFTR protein. Give two possible locations of the mutation.
promoter
promoter element (such as TATA box)
enhancer
polyA adenylation site
splicing sites (in introns)
ribosome binding site
stop codon
note the question said mutant allele, so the mutation had to be in CFTR (not another gene whose product
regulates or is required for CFTR expression)
See slide 29 from Ch 14 notes
15pts
.
4. You recover a cloned DNA segment of interest and determine that the insert is 13 kb in length. To
characterize this cloned insert, you isolate just the insert and construct a restriction map. Using
EcoRI (E) and BamHI (B) and gel electrophoresis, you obtain the following results.
Restriction enzyme
E
B
E+B
None
DNA fragment size (kb)
2, 11
3.5, 9.5
1.5, 2, 9.5
13
See slide 14 from Ch 16 notes, Figure 16-15, and question 9 in Ch 16
A. On the gel below:
(i.) draw in the approximate location of the 10 DNA fragments for the 1 kb marker (band sizes
are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 kb). Label just the 1kb and the 10 kb DNA fragments.
(ii.) draw what the DNA fragments from the table would look like on the gel, making sure you put
each band in the appropriately labeled lane
-
uncut
E
B
E+B
1 kb Marker
wells
10 kb
1 kb
+
B. Label next to the gel where the positive and negative electrodes would be if this gel were in a gel
box with a + for positive and a – for negative.
C. Construct a restriction enzyme map from these data, showing the positions of the restriction sites
relative to each other and the distance between them in basepairs.
E
2 kb
B
1.5 kb
9.5 kb
15pts
.
5. The partial DNA sequence for the normal and mutant alleles of a recessive inherited disease are
shown here:
Normal (B)
5’...TTTTTGATCCCGAAATTCGATTTGAGAGGCC...3’
3’...AAAAACTAGGGCTTTAAGCTAAACTCTCCGG...5’
Mutant (b)
5’...TTTTTGATCCCGAAATACGATTTGAGAGGCC...3’
3’...AAAAACTAGGGCTTTATGCTAAACTCTCCGG...5’
DNA from three individuals is denatured, spotted to a filter, and hybridized to an added radioactive ASO
(allele specific oligonucleotide) with the following sequence:
ASO = 5’GATCCCGAAATACGATTTGAGA3’
The following pattern of hybridization is observed:
bb
Bb
BB
a) Indicate the genotype of the DNA in each spot in the blank provided below the figure.
15pts
.
b) Box the exact sequence to which entire the ASO shown in this problem binds on the appropriate
DNA sequence above.
See slide 10 from Ch 19 notes, Figure 19-7, and question 8 in Ch 19
20pts
.
6. The table below lists several genotypes associated with the lac operon in E. coli. For each, indicate
with a "+" or a "-" whether B-galactosidase would be expected to be produced in the indicated media.
See slide 12, 14, and 17 from Ch 15 notes, Table 15.1, and question 4,5 in Ch 15
Genotype (chromosome/plasmid)
I+
III+
I+
Is
Is
Is
O+
O+
O+
Oc
Oc
O+
O+
OC
Z+
Z+
Z+/p I+ O+ Z+
Z+
Z+/p I+ O+ ZZ+
Z+/p I+
Z+
I+ = wild type repressor
I- = no repressor
Is = mutant repressor that can not bind lactose
O+ = wild type operator
Oc = mutant operator
B-galactosidase production in media containing:
NO Lactose
With Lactose
+
+
+
+
+
+
+
+
+
+
Multiple choice section: (25 points total – 5 points per question): Write your answer in the blank provided to
the left. If you want to explain your answer, you can do so next to the question.
A
7. You have obtained an E. coli with a mutation in the uvrA gene. There is no UvrA protein (part of the
nucleotide excision repair system) made in this strain. You UV irradiate this uvrA mutant strain and a
wildtype strain with normal uvrA gene. After exposure to the UV light, you incubate the cells in the dark
and then measure the percent of cells that survived the UV irradiation. Which of the following results do
you expect based on what you know about DNA repair?
a. Some cell death for both strains, but less cell survival for the uvr mutant relative to wildtype E. coli
b. Some cell death for both strains, but more cell survival for the uvr mutant relative to wildtype E. coli
c. Some cell death for both strains, but equal cell survival for the uvr mutant and the wildtype E. coli
because the PRE system is functional in both strains.
d. No cell death for either the uvr mutant or the wildtype E. coli because the PRE system is functional in
both strains.
e. None of the above.
C
D
A
8. The araB gene allows bacteria to use arabinose as a carbon source (bacteria need carbon for growth). Some
biochemical mutations in the araB gene are “leaky”. This means that the bacteria with the leaky mutation
show a very small amount of growth in media where arabinose is the only source of carbon. Which of the
following mutation types would most likely lead to a leaky mutation?
a. Frameshift
b. Nonsense
c. Missense
d. A and B equally
e. Insertion of a transposon
9. The best library for cloning the human growth hormone gene (hgh) for production of human growth
hormone in E. coli would be see question 11 and notes from slide 2 ch 19
a. genomic DNA library of E. coli DNA
b. genomic DNA library of human DNA
c. cDNA library of E. coli DNA
d. cDNA library of human DNA
e. chromosome DNA library of human DNA (with the chromosome that contains the hgh gene)
10. The steps that occur at each temperature in a standard PCR reaction are:
a. annealing of the primer at 58C; copying the DNA by the Pfu polymerase at 72C; denaturation of the
DNA at 95C
b. annealing of the primer at 58C; denaturation of the DNA at 72C; copying the DNA by the Pfu
polymerase at 95C
c. denaturation of the DNA at 58C; annealing of the primer at 72C; copying the DNA by the Pfu
polymerase at 95C
d. copying the DNA by the Pfu polymerase at 58C; denaturation of the DNA at 72C; annealing of the
primer at 95C
e. None of the above because the polymerase would be inactivated at 95C.
D
11. Expression of the trp genes is repressed by tryptophan, which activates the TrpR repressor protein. An E.
coli mutant constitutively expressed the tryptophan synthesis genes. The mutation could be:
a. a deletion of the trp repressor gene
b. a deletion of the trp operator
c. a deletion of the trp promoter
d. A and B
e. All of the above