BIOL 3301
EXAM 3
Section 07870
GENETICS
Tuesday April 1 (no kidding), 2003, 5.30-7.00 pm
NOTE: MAKE SURE YOU READ THE QUESTIONS CAREFULLY!!!
________________________________________________________________________
Name:
SSN:
________________________________________________________________________
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. Protein synthesis is terminated by a rho-dependent mechanism. F
7. High glucose prevents formation of the CAP-cAMP complex. T
8. The lac repressor binds to DNA via a helix-turn-helix motif. T
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. A bacterial promoter has the transcription start site at position:
a. -35
b. +20
c. +1
d. -10
e. has no transcription start
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. Beta galactosidase converts
a. glucose to galactose
b. galactose to lactose
c. galactose to glucose and lactose
d. lactose to galactose and glucose
e. lactose to two glucose molecules
12. A super-repressor mutation (IS) affects the Lac operon. It results in
a. no transcription
b. inducible transcription
c. transcription but no translation
d. no translation
e. constitutive transcription
13. A partial diploid of genotype I-P+OCZ-Y+/ISP-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. constitutive 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, Cytosine and Thymine 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. Basic leucine zippers, helix-loop-helix, zinc fingers, and homeodomains are different
DNA.-binding domains.
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 all components necessary for and explain RNA transcription (initiation,
elongation, termination) and protein translation initiation in prokaryotes.
RNA TRANSCRIPTION IN PROKARYOTES.
Transcription initiation.
Components:
-RNA polymerase consists of the following proteins:
2' form the core enzyme
-The core enzyme plusform the holoenzyme.
-The promoter in prokaryotes has conserved sequences at -35, at -10, and has the
transcription start site at +1.
Initiation:
- RNA polymerase holoenzyme scans for the promoter
- RNA polymerase holoenzyme binds to the promoter and forms a closed complex
- The promoter is unwound and an open complex is formed.
- Transcription is initiated.
-  factor is released.
Transcription elongation.
Components: ribonucleotides and RNA polymerase core enzyme
Elongation:
(NMP)n + NTP ----> (NMP)n+1 + Ppi
catalyzed by the RNA polymerase core enzyme
Transcription termination.
1. Intrinsic termination makes use of a hairpin structure that forms in the transcribed
RNA which leads to steric hindrance and a dissolution of the RNA-DNA-RNA
polymerase complex = transcription termination.
2. Rho-dependent termination. A hexamer of the bacterially encodes Rho protein binds
to the rut sequence in the RNA, moves towards the transcription bubble and pulls the
RNA of the template = transcription termination.
PROTEIN TRANSLATION INITIATION IN PROKARYOTES.
1. The complex of IF3 and the 30S small subunit of the ribosome binds to the mRNA.
Correct positioning of the 30S small subunit on the initiator Met codon, depends on the
basepairing between the Shine-Dalgarno sequence and the 3' end of the 16S rRNA.
2. formyl-methionine-tRNA-IF2-GTP complex associates with the IF3-30S-mRNA
complex. The formyl-methionine-tRNA is located in the P-site of the ribosome.
3. GTP is cleaved, the 50S large subunit of the ribosome associates with the mRNA-30Sformyl-methionine-tRNA complex and IF2 and IF3 are released.