Download Practice exam 2 key

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Page 1 of 6
Question 1 (9 points).
a) Does the molecular below come from RNA or DNA? Justify your answer.
RNA. There are two OH groups on the sugar. DNA has one OH group on the sugar.
b) List two other structural differences between RNA and DNA
RNA uses uracil while DNA uses thymine
RNA is usually single-stranded while DNA is usually double-stranded.
Question 2 (11 points)
The following table contains a list of statements that apply to replication, transcription, both, or
neither. In each empty box, put a check mark if that statement applies to replication or
transcription.
In eukaryotes, the process occurs in the nucleus.
A primer is required to initiate synthesis.
The polymerase moves 5’ to 3’ along the template strand.
The polymerase moves 3’ to 5’ along the template strand.
The process requires DNA unwinding
The product is DNA.
The product is RNA.
The template is used only once per cell cycle
Synthesis of the new strand is initiated at a promoter.
The product may be modified after synthesis
Proof-reading and repair are incorporated into this
process.
Replication
√
Transcription
√
√
√
√
√
√
√
√
√
√
√
√
√
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Page 2 of 6
Question 3 (18 points).
The line below represents a prokaryotic gene. The rectangle represents the coding region.
Below the DNA diagram, draw the corresponding RNA (roughly to scale) and polypeptide.
Where appropriate, indicate the location of the following:
start codon
polyA tail
3’UTR
hairpin loop
5’ cap
amino terminal
stop codon
carboxy terminal
5’ UTR
promoter
methionine (met)
intron
DNA
Promoter
5’ UTR
start
stop
coding region
3’ UTR
hairpin loop
RNA
5’ UTR
stop
start
coding region
3’ UTR
Protein
Amino
Terminal
met
protein
Carboxy
Terminal
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Page 3 of 6
Question 4 (16 points).
You are studying an inversion heterozygote.
The order of genes along one homolog is
centromere – A – B – D – E – F
The order of genes along the other homolog is
centromere – A – E – D – B – F
a) Assuming that the first homolog is the normal chromosome, draw an arrow(s) at that
breakpoint(s) that gave rise to the abnormal chromosome
(2 pts) See above. If shown on inverted chromosome (1 pt).
b) The homologs undergo recombination between genes B and D. Draw a clear sketch depicting
the chromosomes of this inversion heterozygote as they align at pachytene. In your sketch, only
include the chromatids undergoing recombination.
D
D
B
B
E
E
centromere
A
F
centromere
A
F
(5 pts) [see figure] No centromeres indicated (-1 pt); no cross-over indicated (-1 pt);
error in placement of gene (-1 pt); attempted loop – but didn't quite get it (3 pts).
c) Draw a clear sketch of the recombinant products from such as cross-over event.
centromere
A
F
B
D
E
A
E
D
B
F
centromere
(4 pts) [see figure] 2 pts for each chromosome. No centromeres indicated (-1 pt). No
deduction if non-recombinant chromosomes shown and ABDEA indicated as bridge
between.
d) Will the recombinant chromatids segregate normally during cell division? Explain.
(5 pts) No (1 pt); decentric will be pulled to opposite sites during anaphase, resulting in
chromosome breakage (2 pts); acentric cannot attach to spindle OR may be lost (2 pts).
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Page 4 of 6
Question 5 (22 points).
You are studying a gene in E. coli that specifies a protein enzyme. Part of the wild-type
sequence is given below. You recover a series of mutants for this gene that show no enzyme
activity. Isolating and sequencing the mutant products, you find the following protein sequences
(assume each mutant is the result of a single nucleotide change):
Wild-type:
- Met - Cys – Ala - Gln - Ile – Tyr –
Mutant 1:
- Met - Cys – Ala - Gln - Thr – Tyr –
Mutant 2:
- Met - Cys – Ala
Mutant 3:
- Met - Val – Pro – Arg – Phe – Thr
a) How may different mRNA sequences could code for this portion of the wild-type protein?
Show your work.
amino acid Met Cys
Ala
Gln Ile
Tyr
# codons
1x
2x
4x
2x
3x
2
= 96 mRNA sequences
(4 pts) 96 possible (4 pts); -1 pt for each component incorrect; -1 for math error.
b) For the FIRST amino acid affected by each mutation, give the original codon(s) and the
mutant codon(s) as specifically as possible. Use all the data and show your work. Clearly
indicate the type of mutation that has occurred.
(18 pts; 6 pts for each mutant)
possible codons are given, actual codons are circles based on analysis of mutant 3
Mutant 1 =
Ile ->
Thr
AUU ->
ACU (4 pts)
C
C
A
A
must include some indication of location or amino acids affected; any one of:
missense/T (or U) to C substitution/transition (2 pts); “point mutation” (1 pt).
Failure to resolve ambiguity (-1 pt); incorrectly resolved ambiguity (-2 pts).
Mutant 2 =
Gln ->
STOP
CAA ->
UAA
CAG
UAG (4 pts)
must include some indication of location or amino acids affected; nonsense
mutation (2 pts) OR stop codon (1 pt) due to transition (1 pt) OR stop codon (1
pt) due to C --> T (U) substitution (1 pt). Failure to resolve ambiguity (-1 pt);
incorrectly resolved ambiguity (-2 pts).
Mutant 3 =
first codon affected
Cys ->
Val
UGU ->
GU_
C
GUG (4 pts)
must include some indication of location or amino acids affected; frameshift (1
pt) due to deletion of 1 bp or indel (1 pt). Failure to resolve ambiguity (-1 pt);
incorrectly resolved ambiguity (-3 pts). If otherwise 0 pts and indicate possible
codons for amino acids involved (1 pt).
Name____________________________________
wild-type protein
mutant 3
Page 5 of 6
Cys
UGU
C
->
Val
GUG
Ala
GCU
GCC
A
G
Pro
CCC
Gln
CAA
CAG
Ile
AUU
C
A
Tyr
UAU
UAC
Arg
AGA
Phe
UUU
Thr
AC_
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Page 6 of 6
Question 6 (24 pts) The following is a list of mutational changes. For each of the specific
changes described, indicate which of the terms listed below could apply, either as a description
of the mutation or as a possible cause. More than one term from the list can apply to each and
individual terms may be used once, more than once, or not at all.
a) a G-C base pair is changed to an A-T base pair (6 pts)
transition, base substitution, SOS system, deamination
b) a G-C base pair is changed to a T-A base pair (6 pts)
transversion, base substitution, SOS system
c) the sequence AAGCTTATCG is changed to AACTTATCG (6 pts)
indel, frameshift, intercalating agent
d) the sequence AACGTCACACATCG is changed to AACGTCACACACACATCG (6 pts)
indel, replication slippage, frameshift
points deducted for incorrect answers. a and d) 4 correct answers: 1 correct (+2 pts), 2 correct
(+3 pts), 3 correct (+5 pts); 4 correct (+ 6 pts); each incorrect answer (-1 pt). b and c) 3
correct answers: each correct (+2 pts); each incorrect (-1 pt).
transition
indel
replication slippage
base substitution
deamination
SOS system
transversion
intercalating agent
photodimer
inversion
frameshift