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Genetics 320 Fall 2007, Midterm Exam 1. Sept 21

• Expression, mutation, consequences (25pt)
• Gene expression/domains (25pt)
• Mitotic and meiotic recombination (25pt)
• T/F short answer (25pt)

TOTAL
2

• Gene expression, mutation and consequences. The
  top strand of a DNA sequence of a gene is shown.
  The complementary sequence on the bottom strand
  is not shown. The promoter sequence is in
  capital letters and boxed in. 1 of the mRNA is
  a C. There is no splicing. The Atg is the start
  codon and is circled. Each amino acid encoded in
  the protein is essential for protein function.

d
e
f
g
11
1
gcgTATAttgaccgctaCgtaggctttAtgccggttaaatttcgcgggta
gcccgcaag
53

• Label (5, 3) the ends of the strands with the
  correct polarity. (4pt)
• Put an arrow to the template strand used by RNA
  polymerase to make mRNA.(2pt)
• Circle the stop codon. How many amino acids are
  encoded in this gene? (4 pt) 7
• d-g. Fill in the table below. Use less, same or
  more to indicate the AMOUNT PER CELL in the
  mutant compared to the original gene. NOTE in
  diagram above d,e,f,g are indicated for your
  convenience! Need to say something-write on
  back(9pt)

d. TATAcc to TggAcc less
less e. Delete t at 6 same
same f. Insert g between 10g and
11g same less g. Change
cgc to cgt same
same
Amount of Transcription
Functioning Protein (per cell)
h. Explain briefly the change, if any in and
kind of amino acids for parts f and g. (6pt) In
f, inserting a g introduces a frameshift
mutation, and a stop codon after a total of 3
amino acids. In g, the basepair change generates
a syonomyous codon--they both encode arginine. No
change to amino acid sequence.
3
2. Consider the domain structure of the protein
below. The protein has three domains (amino acids
10-100 for the rectangle, 120-150 for the
triangle, and 180-210 for the oblong). The
corresponding gene is shown above the protein.
There is one exon. The triangle is
required for normal hearing and the oblong domain
for normal learning.
Gene
Protein
Amino 1 10 100 120 150 180
210 250 Acid

• A polypeptide backbone is shown. Draw a box
  around one peptide subunit (3pt).
• Draw the R groups for alanine and for lysine in
  the space below (6pt).
• A mutation changes an alanine to a lysine at
  codon 130. What is the likely phenotype of an
  organism with this mutant protein (and no wild
  type protein) and why? (5pt) Triangle domain
  defective because alanine (hydrophobic) to lysine
  ( charged) is dramatic change in a domain.
  Hearing defective but learning functional because
  the oblong domain should be unaffected.
• What is the likely molecular and phenotypic
  consequence of a frameshift mutation in the
  rectangle domain? Explain briefly.(5pt)
  Frameshift causes wrong reading frame for rest of
  protein, no domains made, complete defect for
  hearing and learning.
• The G-C base pair at the 3rd base of the ATG
  start codon is deaminated, and replication occurs
  before repair is complete. Explain briefly what
  happens to the genes sequence, transcription and
  translation. (6pt) Deamination without repair
  results in one normal cell and one cell with a
  ATA mutation that will not alter transcription
  but will eliminate translation of the protein.

O
R
H
O
R
H
O
R
H
O
R
H
O
R
H
N
C
N
C
N
C
N
C
N
C
C
C
C
C
C
H
H
H
H
H
Pts given for peptide bond too
H

C
NH
H
lysine
alanine
4
3. Recombination
Mitotic recombination, part a.
a. Draw a recombination and segregation event
that forms an Aa-b-b-cell. Label centromeres!
EACH LINE IS A DOUBLESTRAND CHROMOSOME. (6pt)
b-
a-
b- a-
1
b-
A
2
1 3 cosegregate
3
4
B A
Meiotic recombination, part b,c,d
b. Draw a double Holliday structure intermediate
(after strand invasion and D-loop formation and
before replication). NEATLY! Indicate the
polarity of all 4 strands. Circle each 3 site
where DNA polymerase will synthesize DNA. Place a
box around ONE region that will become a Holliday
structure. (10 pt)
53 35
Full pts given even if replication included
c. Why is it important that Spo11 cuts only one,
and not all four, of the chromosomes at a
specific DNA sequence? (4pt) The DSB needs to
be repaired using an intact template if all four
sequences were cut by Spo11 there would be no
intact template!
d. In the G2 diploid cell below, place an A at
a sequence that is allelicto X. Put an N at
ANY ONE sequence that is non-allelic to X. EACH
LINE IS A DOUBLESTRAND CHROMOSOME. AND What
chemical principle determines that RecA pairs the
correct sequences? (5pt)
Hydrogen bonding between complementary base
pairs. Or some other verbage..!
X
A
NA
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• 4. True/False, short relevant comment. Full
  credit requires comment.
• RNA polymerase stops transcription when it comes
  to a stop codon.(5pt)
• False. Stop codons stop the ribosome/RNA
  polymerase stops at special sequences after the
  stop codon.
• The Philadelphia chromosome is an example of how
  allelic recombination leads to a drastic genetic
  change. (5pt)
• False. The Philly chromosome joins Chr 9 and
  Ch22, its a translocation, and thus by
  definition two non-allelic sequences.
• The Protein Interaction Network suggests one
  reason why a drug that targets one protein in one
  process may affect many processes. (5pt)
• True drug inactivation of a protein in a
  network will affect proteins in the
  subnetwork,and some of those proteins also act in
  subnetworks so those other subnetworks could be
  affected. Some mixed domains in here toopartial
  credit.
• RecA probably did not initially evolve to
  generate genetic diversity. (5pt)
• True, Rec A probably evolved to repair DSB in an
  error-free way, to limit mutation and thus limit
  genetic diversity.
• A mutation that deletes several exons and introns
  of the CCR5 gene will be as useful in gene
  therapy as a ccr5delta32.(.(5pt)
• False. The ccr5delta32 mutation is a tried and
  true mutation present in 1 of the human
  population, and thus has known effects on people.
  A more draconian mutation removing multiple
  exons and introns may unwittingly alter other
  non-CCR5 gene functions sometimes introns encode
  a function, for example.