Genetics – Final Exam Summer 2012 VERSION A Name Multiple

advertisement
Genetics – Final Exam Summer 2012
VERSION A
Name
Multiple Choice
(50 pts.
possible)
Problems
(50 points
possible)
Total
(100 points
possible)
KEY
IF you completed the in-class workshop put a CHECK MARK HERE --->
There are two sections in the exam:
MULTIPLE CHOICE
25 questions worth 2 points each. - Answer on the Scantron Form
PROBLEMS
11 problems worth 5 points each.
If you completed the workshop we will drop the 2 lowest scores.
OR
If you did not complete the workshop we will drop the 1 lowest score.
U
C
A
G
UUU
UUC
UUA
UUG
CUU
CUC
CUA
CUG
AUU
AUC
AUA
AUG
GUU
GUC
GUA
GUG
U
Phe
Phe
Leu
Leu
Leu
Leu
Leu
Leu
Ile
Ile
Ile
Met
Val
Val
Val
Val
UCU
UCC
UCA
UCG
CCU
CCC
CCA
CCG
ACU
ACC
ACA
ACG
GCU
GCC
GCA
GCG
C
Ser
Ser
Ser
Ser
Pro
Pro
Pro
Pro
Thr
Thr
Thr
Thr
Ala
Ala
Ala
Ala
5' Base in Anticodon
G
C
A
U
I
UAU
UAC
UAA
UAG
CAU
CAC
CAA
CAG
AAU
AAC
AAA
AAG
GAU
GAC
GAA
GAG
A
Tyr
Tyr
Stop
Stop
His
His
Gln
Gln
Asn
Asn
Lys
Lys
Asp
Asp
Glu
Glu
UGU
UGC
UGA
UGG
CGU
CGC
CGA
CGG
AGU
AGC
AGA
AGG
GGU
GGC
GGA
GGG
G
Cys
Cys
Stop
Trp
Arg
Arg
Arg
Arg
Ser
Ser
Arg
Arg
Gly
Gly
Gly
Gly
3' Base in Codon
U or C
G
U
A or G
U, C or A
1
1.
In the table below you are given the genotypes of 4 E. coli strains. All 4 strains are
"diploid" for the Lac Operon region. For each of the strains, indicate (using + or -) whether or not
the Lac gene products will be expressed at high levels in the presence or absence of inducer.
Assume that the inducer IPTG is used rather than lactose.
Beta Galactosidase
no inducer
with inducer
GENOTYPE
S +
+ - +
I P O Z Y
+ +
c + I P O Z Y
+ +
c + I P O Z Y
C +
+ + +
I P O Z Y
S +
c + I P O Z Y
+ +
+ - +
I P O Z Y
+ + + +
I P O Z Y
C +
+ - +
I P O Z Y
-
HINTS: I = I
C
c
O =O
Lactose Permease
no inducer
with inducer
+
+
-
-
+
+
-
+
+
+
-
-
-
-
-
+
-
For wild-type E. coli the correct responses would be:
GENOTYPE
+ +
+ + +
I P O Z Y
Beta Galactosidase
no inducer
with inducer
+
Lactose Permease
no inducer
with inducer
+
2.
The diagram below depicts a charged tRNA interacting with a codon in a mRNA. The six
squares represent the nucleotides of the codon paired with the anticodon, and the rectangle
represents the amino acid attached to the acceptor stem of the tRNA. Some of the 7 identities
represented by the squares and the rectangle are given.
Work out the identities for the remaining empty shapes and add them to the diagram. i.e. fill any
remaining empty squares and/or rectangle with the correct nucleotide or amino acid. If there are
any boxes where the identity cannot be unambiguously determined, GIVE ALL THE
POSSIBILITIES.
Ser
tRNA
A
G
U
C
I
U
C
A
mRNA
NOTE: DO NOT ASSUME that the mRNA is shown in the usual orientation. You must determine
the orientation from the information provided.
2
3.
Describe the molecular basis for the fact that GC base pairs in which the C is methylated
(5-methylcytosine) are hotspots for a certain type of mutation.
Deamination of 5-me C creates a G T mismatch. G T mismatches cannot
be repaired by the uracil DNA glycosylase which repairs the normal
cytosine deamination lesion (i.e. G U mismatch).
4.
For a CGA (Arg) codon, how many of the possible single base pair substitution changes
are synonymous, missense, and nonsense?
4 synonomous
4 Missense
1 Nonsense
3
5.
This is a matching problem that involves the mutational processes we considered in
detail. These mechanisms are listed in the left-hand column.
In column 2, write the letter designation only for the pre-mutagenic lesion that is created by each
mechanism.
In column 3 write the letter or letters designation for the repair mechanism/s that can correct the
lesion.
In column 4 write the letter designations to indicate the overall mutational change that results
from each process.
Each choice may be used one or more times. Each box may have one or more appropriate
responses.
MUTAGENIC PROCESS
Tautomeric Shifting of Bases
(e.g. keto/enol)
Ultraviolet (UV) Radiation
Deamination of 5-methyl
Cytosine
Deamination of Cytosine
2.
2.
3.
4.
PREMUTAGENIC
LESION/S
REPAIR
MECHANISM/S
OVERALL
MUTATION/S
A (+ B OK)
H, K, L
M (+N OK)
D
F
N
B
I
N
C
J, G, L
N
3.
PREMUTAGENIC
LESION/S
LETTER
All possible Normal
Base Pair
Mismatches with
incorrect base in
newer DNA strand
G - T Mismatch
A
Photolyase
F
AT ↔ GC
Bidirectional
Transitions
M
B
AP Excision
G
N
G - U Mismatch
C
H
TT Intrastrand
Dimer
T - T Mismatch
D
Methyl-Directed
Mismatch Repair
No specific repair
mechanism
Uracil DNA
Glycosylase
Proofreading by
DNA Polymerase
Single Strand Gap
Repair
GC → AT
Unidirectional
Transition
Transversions
E
REPAIR
MECHANISM/S
4.
Letter
OVERALL
MUTATION/S
LETTER
O
I
J
K
L
4
6.
Imagine that a mutation eliminates expression of the gene for the regulatory protein CAP.
(i.e. As a result of this mutation there is no longer any CAP protein in the cell.)
How would this mutation alter expression of the Lac operon genes? (i.e. How would Lac gene
expression be different in the mutant compared to the wild-type strain where CAP was present?)
Adenyl cyclase is required to synthesize cAMP under conditions of
glucose limitation. Therefore CAP activation of Lac (and other catabolic
operons) would not occur. Production of lac mRNA would still be regulated
by lactose, but even in the absence of glucose would not be produced at
levels as high as in a wild-type strain.
High levels of Lac mRNA transcription would not be observed even in
presence of lactose and absence of glucose because the CAP regulator
would not bind to its operator.
7.
Give the anticodon sequences (with polarities) of ALL possible Tyrosine-specific tRNA's.
Circle the anticodon sequences in your complete set that would represent a minimum set needed
to translate all Tyrosine codons. If they are all required, circle them all. (There may be more than
one minimum set. If so, you need give only one of them.)
Tyrosine = “Tyr”
3' AUA 5'
3' AUG 5'
5
8.
The diagram represents a segment in a larger DNA molecule, where a base mismatch (A
with C) has resulted from tautomeric shifting of a base in the template strand during DNA
replication. The diagram also shows a nearby GATC sequence.
!
1. Which of the bases in the mismatch is erroneous?
C
2. Use an arrow to indicate the approximate location of strand cutting by the endonuclease mutH.
Unmethylated GATC sequence.
3. If the mismatch is not repaired, the overall mutation that would result is: (Circle One)
AT ---> GC
4. What is the identity of the base that experienced the tautmeric shift? (Circle one)
A
9.
Angelman syndrome (AS) is a rare human genetic disorder in which individuals exhibit
unusual seizures and repetitive symmetrical muscle movements and which is due to lack of
expression of the gene UBE3A. Expression of UBE3a is subject to paternal imprinting.
Two normal parents have a female child born with AS who carries a defective copy of UBE3a.
a. Which one of her parents must be heterozygous for the defective UBE3a allele? (Defective
UBE3a alleles are very rare in the human population, so it is unlikely that both parents carry it.)
Female parent is heterozygous for UBE3.
b. If the couple has a son, what is the probability that he will have AS?
1/2
6
10.
Give 3 important and significant differences between bacterial genomes and eukaryotes.
Circular vs Linear DNA
no nucleus vs nucleus
mostly coding DNA vs mostly non-coding DNA
single vs multiple origins of replication
other responses may be acceptable as well
11. In a paragraph, describe the differences between "maternal effect" and "maternal
inheritance". The differences you address should include the inheritance pattern of traits as seen
in genetic crosses, and the cellular/molecular bases of the two phenomena.
If you wrote your answer ahead of time, just staple it to the back of your exam.
Maternal effect genes are transcribed in cells of a female during oogenesis and the mRNA
or protein products of the gene are transported into oocytes where the effect early
development of the embryo. Therefore the phenotype of the offspring reflects the
genotype of the mother.
Maternal inheritance is due to the transmission of mitochondria and chloroplasts, and
their associated genomes, via the egg but not the sperm. Therefore the genotype of the
offspring will be identical to that of the mother.
7
Download