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Question 1
You are studying an enzyme called shinase. Its substrate is the tripeptide, Ala-Lys-Thr,
with an unusual molecule on its C terminus, the GLOW molecule. When this GLOW
molecule is leaved from the tripeptide, it acquires the ability to glow green. Pictured here
is the substrate bound to the active site of shinase, with important amino acid side chains
depicted.
a) In the chart below, indicate the strongest intermolecular force that occurs between the
amino acid side chains of the shinase enzyme and the tripeptide substrate.
Choose one from the following: ionic bond, van der Waals force, covalent bond,
hydrogen bond.
3 points each
Amino Acid in
Shinase Enzyme
Amino Acid in
Substrate
Strongest Force
Asn 74
Thr
Hydrogen
Glu 110
Lys
Ionic
Val 150
Ala
Van der Waals
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Question 1 continued
In order to study the enzymatic activity of shinase, you create mutant versions of the
shinase enzyme and measure kinetic constants.
b) Based on the data above, which mutant enzyme has the highest affinity for the
substrate? Circle one.
3 points
c) Based on the data above, which mutant enzyme has the highest catalytic activity?
Circle one.
2 points
d) Based on the data above, which amino acid residue in the wildtype shinase enzyme
appears to be directly involved in cleaving the peptide bond? Circle one.
3 points
e) When Glu 110 of shinase is converted to Lys, the resulting mutant enzyme no longer
binds nor cleaves the tripeptide substrate. Which of the following tripeptides might be a
substrate for this mutant enzyme? Circle one.
4 points
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Question 2
You are studying fly genetics, and you find a mutant strain of fly that has blue legs. You
cross a pure-breeding blue-legged female fly with a wild-type (WT) pure-breeding
black-legged male. Resulting phenotypes are listed below.
a) Provide the likely genotypes of all of these F0, F1, and F2 flies with respect to leg color
in the adjacent boxes.
2 points each
F0
F1
Curious, you cross the F1 females and males, and obtain the following phenotypes.
1 Point each
F1
F2 files
Now you study genes that control wing size, wing shape and wing color –
wild-type flies have large, round, transparent wings. You study three mutations, shown
below, that give the following phenotypes, which are all recessive to the wild-type allele.
Wing Genotypes and Phenotypes
Wild type Mutants
+/+: Large t/t: Tiny wings.
+/+: Round n/n: Notched wings.
+/+: Transparent b/b: Black wings
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Question 2 continued
You cross a fly that is homozygous for the mutant wing alleles (t/t, n/n, b/b) with a
wild-type fly to obtain flies that are all heterozygous in all three genes. (Their genotype
shown here.)
You cross one of these triple heterozygotes back to the pure-breeding triple mutant strain
and see the following phenotypes in 1000 F2 progeny:
F2:
You chart the F2 data as shown below.
b) Complete the genetic map below by placing the genes in the correct location.
(Distances shown are recombination frequencies.) SHOW YOUR WORK BELOW.
6 points
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Question 3
You discovered a new species of purple yeast that secrete a substance “Thera”. Thera is
required for growth on minimal medium and is also responsible for the purple color of the
yeast.
a) You have isolated several haploid mutant strains unable to synthesize Thera. You cross
each haploid mutant with both wild type and every other mutant, and score the ability of
the resulting diploid strain to grow on minimal medium. The results are shown below.
(+) = growth on minimal medium
(-) = unable to grow on minimal medium
b) Are you able to place all of the mutants into complementation groups?
3 Points
(Circle one). If not, why not? Mutant 2 has a dominant phenotype
c) Place the mutants into complementation groups.
8 Points
c) What is the minimum number of different genes you have identified?
2 points
4_
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Question 3 continued
After more experimentation you harvest more recessive mutants and make the striking
observation that the distinct haploid mutant yeast strains are each a different color from
the wild-type yeast making purple Thera. For instance:
Wildtype Purple
Mutant A Black
Mutant B: Silver
Mutant C Gold
Mutant D Green
Mutant E Turquoise
To determine the order in which these mutants’ gene products are acting, you construct
haploid yeast strains each with 2 mutations and observe the color of the resulting double
mutant. The results of this analysis are shown below.
Double
Mutants
Color
A,B
Black
C,D
Green
E,F
Turquoise
A,D
Black
A,E
Turquoise
B,C
Silver
B,D
Green
d) Draw the Thera intermediate pathway below:
Example:
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Question 4
A well-respected molecular biologist, Dr. Ivanna A. Plus discovers an interesting gene,
“phr”, shown below, and determines its sequence.
a) Draw a schematic of the fully processed mature mRNA transcript.
5 points
200 +100+200+200=700
b) Calculate the size of the coding region in this gene (bp).
3 points
700-50, -100 = 550
Dr. Plus’s colleague, Dr. Igetta Bee, obtains several mutants. The DNA for the coding
sequence
starting at the start codon is shown for 4 mutants.
c) For each mutation, indicate the type of mutation: nonsense, missense, silent, or
frameshift.
The genetic code is provided at the back of the quiz.
3 points-one for type, 2 for effect
Strain
Sequence
Type of
mutation
Effect on protein
produced
Wildtype
ATGTGGGCTAGACATCAA..
WT
Phr1
ATGTGGGCTAGACATCAA..
ATGTGGGCTTGACATCAA..
Nonsense
Arg Stop, protein
truncated after 3 aa
WT
Phr2
ATGTGGGCTAGACATCAA..
ATGTGGGCTAGACACCAA..
Silent
His His
No chang
WT
Phr3
ATGTGGGCTAGACATCAA..
ATGTGGGCTAGTCATCAA..
Missense
Arg Ser, 1 aa
different
WT
Phr 4
ATGTGGGCTAGACATCAA..
ATGTGGCTAGACATCAAA.
Deletion
All diff aa’s after
Frameshift first
2 aa’s.
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Question 5
a) Below is a picture of replication forks with four sites, A-D, highlighted with sequence.
Indicate below whether the primer: 5'-GCGAC-3' can bind these sites to initiate
replication.
IF the primer binds, indicate:
• the direction of elongation of the daughter DNA strand (left or right).
• whether DNA synthesis would be performed in a continuous or a discontinuous
fashion relative to the nearest replication fork.
8 points
b) Enter a check () in the box next to the statement if it’s true of Prokaryotes (Proks)
and /or
Eukaryotes (Euks).
12 points
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