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BIOL 3301
EXAM 3
Section 07869
GENETICS
Wednesday April 2, 2003, 1.00-2.30 pm
NOTE: MAKE SURE YOU READ THE QUESTIONS CAREFULLY!!!
________________________________________________________________________
Name:
SSN:
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True-false statements (write True or False in front of the statement)
(2 points each).
1. The experiments by Griffith on Streptococcus pneumoniae were the first demonstration
of bacterial transformation. T
2. DNA polymerase III is the main activity during DNA replication. T
3. Helicase is required for synthesis of the RNA primer during DNA replication. F
4. Topoisomerase is a reverse transcriptase that maintains chromosome ends. F
5. DNA ligase catalyzes the formation of hydrogen bonds. F
6. The RNA polymerase core enzyme is sufficient for correct initiation of transcription. F
7. Protein synthesis is terminated by a rho-dependent mechanism. F
8. IF2 stimulates binding of the 30S ribosome subunit to mRNA. F
9. A homeodomain is a cis-regulatory sequence. F
10. Enhancers can only operate upstream of promoters they are enhancing. F
Multiple choice (circle the correct answer) (2 points each).
1. A sample of normal double-stranded DNA was found to have a thymine content of
27%. What is the expected proportion of guanine ?
a. 9%
b. 23%
c. 32%
d. 36%
e. 73%
2. The figure shown below depicts the structure of:
a. dATP
b. dCTP
c. ddATP
d. UTP
e. ddTTP
3. Hershey and Chase proved that DNA is the transforming principle. They observed that
the isotope accumulated in the bacteria was:
a. 31P
b. 35S
c. 32P
d. 15N
e. none of the above
4. In a DNA double-helix, nucleotides on complementary strands are held together by
a. covalent bonds
b. salt bridges
c. hydrogen bonds
d. phosphodiester bonds
e. none of the above
5. Meselsohn and Stahl demonstrated that DNA replication is:
a. conservative
b. dispersive
c. funky
d. semi-conservative
e. none of the above
6. rII- mutants are grown on the restrictive host E. coli K (plate 1) and on the permissive
host E. coli B (plate 2). The plaques appear as follows:
a. Plate 1: no plaques
Plate 2: large plaques
b. Plate 1: large plaques
Plate 2: large plaques
c. Plate 1: small plaques
Plate 2: small plaques
d. Plate 1: no plaques
Plate 2: small plaques
e. Plate 1: no plaques
Plate 2: no plaques
7. Mutations that do not complement, are:
a. allelic
b. non-allelic
c. intergenic
d. null mutants
e. missense mutations
8. A tRNA with the anticodon 3´UGG 5´would carry
a. phenylalanine
b. tyrosine
c. tryptophane
d. serine
e. threonine
9. The E. coli RNA polymerase holoenzyme consists of the following subunits:
a. 2, 1, 1', 1
b. 2, 2, 1'
c. 2, 1, 1'
d. 1, 2, 1', 1
e. 2, 2', 1
10. The following diagram shows a fragment of transcribed DNA, and the upper strand is
the template strand:
5´ ATTGCC 3´
3´ TAACGG 5´
The transcribed RNA can be represented by:
a. 5´ AUUGCC 3´
b. 5´ TAACGG 3´
c. 5´ AUUGCC 3´
d. 5´ UAACGG 3´
e. 5´ GGCAAU 3´
11. EF-Tu is required for:
a. loading aminoacyl-tRNA into the A-site of the ribosome
b.
c.
d.
e.
regenerating EF-Ts
translocation of the peptidylchain
translocation of the ribosome
releasing empty tRNAs from the ribosome
12. Stop codons are not recognized by tRNAs, but by:
a. ribosomes
b. the Shine-Dalgarno sequence
c. release factors
d. EF-Tu
e. IF3
13. A partial diploid of genotype I-P+O+Z-Y+/I+P-O+Z+Y+ will show
a. inducible production of beta galactosidase
b. inducible production of beta galactosidase and permease
c. constitutive production of beta galactosidase and permease
d. inducible production of permease
e. no beta galactosidase or acetylase production at all
14. Which of the following statements is true (for eukaryotic RNA polymerases)?
a. RNA polymerase I synthesizes mRNA
b. RNA polymerase II synthesizes snRNA and rRNA
c. RNA polymerase I synthesizes rRNA
d. RNA polymerase III synthesizes mRNA
e. none of the above
15. Transcription termination in eukaryotes depends on:
a. a hairpin structure
b. Rho termination factor
c. recognition of a AAUAAA sequence and endonuclease cleavage
d. polyadenylation
e. guanyltransferase
Fill in (2 points each).
1. Two of the bases, Adenine and Guanine., are similar in structure and are called purines
.
The other two bases, Thymine and Cytosine are called pyrimidines.
2. B-DNA is the most common form of DNA. The spacing between basepairs is 3.4 A.
3. Yanofsky studied the trp synthetase gene and demonstrated the colinearity of gene and
protein.
4. E. coli DNA replication begins from a fixed origin, but then proceeds bidirectionally ,
ending at a site called the terminus .
5. Translation initiation depends on recognition of the Shine-Dalgarno.sequence by the 3'
end of 16S rRNA.
Homework.
1. (A) We are studying some of the Neurospora mutants described by Beadle and Tatum.
These mutants (numbered 1 to 7) are all involved in the same biosynthetic pathway. The
compounds generated in the different biosynthetic steps are A, B, C, D, F, G, H, I.
Reconstruct the biosynthetic pathway (starting on the left with the compound that comes
first in the biosynthetic pathway)! (7 points)
Compound tested (+= growth; -+ no growth)
A
B
C
D
F
G
H
I
1
-
-
+
-
-
-
-
-
2
-
+
+
+
+
-
-
-
3
+
+
+
+
+
-
-
-
4
+
+
+
+
+
+
-
+
5
-
+
+
-
+
-
-
-
6
-
+
+
-
-
-
-
-
7
+
+
+
+
+
-
-
+
H -> G -> I -> A -> D -> F -> B -> C
4
7
3 2
5
6
1
(B) What are auxotrophic mutants ? (3 points)
Mutants that require addition of a supplement to grow on minimal medium.
2. You are studying a gene in E. coli that specifies a protein. A part of its sequence is:
-Phe-Tyr-Ser-Trp-Glu-Ala-Met-ThrYou recover a series of mutants for this gene that show no enzymatic activity. By
isolating the mutant enzyme products, you find the following sequences:
Mutant 1:
-Phe-Asn-Ser-Trp-Glu-Ala-Met-Thr-
Mutant 2:
-Phe-Tyr-Ser-Trp
Mutant 3:
-Phe-His-Cys-Leu-Pro-Ala-Val-Thr-
Mutant 4:
-Phe-Tyr-Met-Leu-Gly-Ala-Met-Thr-
What is the DNA sequence that specifies this part of the protein ? What is the molecular
nature for each mutation? (10 points)
Wildtype
TTC TAT TCN TGG GAA GCN ATG ACN
C
C AGT
G
C
Mutant 1
TTC AAT TCN TGG GAA GCN ATG ACN
C
C AGT
G
C
Point mutation leads to amino acid substitution.
Mutant 2
TTC AAT TCN TGG TAA
C
C AGT
C
Point mutation leads to stop codon.
Mutant 3
TTC CAT TGT CTN CCN GCN GTN ACN
C
C
C TTA
G
Inversion.
Mutant 4
TTC TAT ATG CTN GGN GCN ATG ACN
C
C
TTA
G
Insertion (single underlined) and deletion (double underlined).
Open questions (20 points).
NOTE: If you would run out of space, you can also use the back of this page to continue
your answer.
Describe and explain the lac operon and its regulation (including catabolite repression) !
LAC OPERON STRUCTURE.
The lac operon consists of POZYA.
P & O are cis-regulatory sequences
P: promoter
O: operator
ZYA are structural genes
lacZ encodes beta-galactosidase
lacY encodes permease
lacA encodes transacetylase
Beta-galactosidase converts lactose into glucose and galactose.
Permease is required for entry of lactose into the cell.
Transacetylase is probably involved in detoxification.
LAC OPERON REGULATION BY THE LAC REPRESSOR.
RNA polymerase and lac repressor are the trans-acting factors that interact with P & O.
- RNA polymerase holoenzyme binds to the promoter and eventually leads to
transcription initiation.
- lac repressor is encoded by the lacI gene. This gene has a promoter that is always on, so
in wildtype E. coli lac repressor is always present in the cell.
The lac repressor binds to the operator sequences and results in repression of the lac
operon, i.e. no transcription.
When lactose enters the cell, it will bind to the lac repressor. Lactose binding to the
repressor results in a conformational change and the repressor can no longer bind to the
operator DNA. Consequently, transcription of the lac operon is initiated.
Mutants:
P- : no or defective promoter -> no transcription
OC : mutant operator - > lac repressor can not bind to the operator -> constitutive
transcription
I- : mutant lacI gene -> no functional lac repressor is encoded
IS: mutant lacI -> encodes a repressor protein to which lactose can no longer bind -> the
repressor remains bound to the operator -> constitutive transcriptional repression
Note: I+ is the wildtype lacI gene and encodes a normal repressor protein
Order of dominance: P- > OC > IS > I+ > ILac repressor structure:
The lac repressor contains three protein domains:
- a multimerization domain
- a ligand-binding domain
- a DNA-binding domain with 3 alpha-helices and a helix-turn-helix DNA-binding
motif.
LAC OPERON REGULATION: CATABOLITE REPRESSION.
E. coli preferentially uses glucose to help provide in its energy requirements. However, it
can metabolize other sugars, but will only express the genes necessary for metabolization
at maximum levels when no glucose is present. The fine-tuned regulation is referred to as
catabolite repression. We essentially distinguish 3 situations.
1. Glucose, no lactose. The lac repressor is bound to the operator and the lac operon is not
transcribed.
2. Glucose and lactose are present. The lac repressor has been bound by lactose and
undergone a conformational change, so it is no longer bound to the operator.
Consequently, there is transcription of the lac operon. However, this transcription is at
relatively low levels. This is caused by the presence of glucose and its indirect effect on
DNA-binding of the CAP protein. CAP only binds DNA as a cAMP-CAP complex. In
the presence of glucose, there are insufficient levels of cAMP in the cell to make the
cAMP-CAP complex, so no DNA binding.
3. Lactose, but no glucose. Since lactose is present the lac repressor is not bound to the
operator so we have transcription of the lac operon. Since there is no glucose, there will
be high levels of cAMP in the cell, and the cAMP-CAP complex is efficiently formed.
The cAMP-CAP complex binds to the promoter region and bends the DNA. This in turn
facilitates transcription initiation by RNA polymerase holoenzyme. The end result is high
level transcription.