BIOL Name_______________________________ 50 Exam III (100 points total)
Student I.D.______________________ 3/31/04
Signature for honor code
This exam is worth 100 points total. The back page has a copy of the genetic code. You may want to separate this page from the others, but do not write any answers on that page it will NOT be graded.
1. (3 points each) For each of the following terms, write a single sentence definition or description of function:
(a) nucleosome Basic unit of chromatin, consisting of DNA wound around an octamer of histone proteins.
(b) lagging strand DNA strand being synthesized discontinuously, as a series of Okazaki fragments that will later be ligated into one continuous strand.
(c) rRNA Ribosomal RNAs are RNA components of the ribosome, involved in translation.
(d) homodimer Protein in which the quaternary structure consists of two identical polypeptide subunits.
(e) motif Recognizable amino acid sequence within a polypeptide's primary structure.
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Name_______________________________ BIOL 50 Exam III (100 points total)
Student I.D.______________________ 3/31/04
2. (6 points) A single strand of DNA is drawn below.
C O C C C C
O P O O O O O P O O
C O C C C C
O P O O O
C O C C C C
O P O O O
C O C C C C
O P O O O
C O C C C C
O P O O O
C O C C C C
OH
Of the following sequences, only one could be the sequence of this strand. Circle the correct sequence.
5'TATATA3'
5'ATGCAT3'
5'CGCGCG3'
5'CCCGGG3'
(The sequence is RYRYRY, where R=purine and Y=pyrimidine)
3. (2 points each) The UAU codon codes for tyrosine. For each of the following, write the sequence of the new codon and specify which amino acid (if any) it codes for. You need list only one correct answer for each, even if several are possible. (a) A missense mutation CAU, GAU, AAU, UUU, UCU, or UGU his asp asn phe ser cys (b) A nonsense mutation UAA or UAG (no amino acid)
(c) A frameshift mutation Any insertion of one nucleotide.
(d) A silent mutation UAC (tyr)
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Name_______________________________ BIOL 50 Exam III (100 points total)
Student I.D.______________________ 3/31/04
4. (3 points each) Four enzymatic activities are listed below; DNA polymerase uses three of these. Cross out the activity that DNA polymerase does not have. For each of the other three, write a single sentence description of the function of each during replication. (a) 5' to 3' DNA polymerase This is the DNA synthesis activity, through which dNTPs are used to add nucleotides to the 3' end of a growing chain, using complementarity to the template strand as a guide.
(b)
3' to 5' DNA polymerase
(c) 5' to 3' DNA exonuclease
This activity is used to degrade the RNA primer, so that it can be replaced with DNA nucleotides.
(d)
3' to 5' DNA exonuclease
The "proofreading" activity; when DNA polymerase inserts an incorrect nucleotide, 3' to 5' exonuclease cleaves this nucleotide off so that the 5' to 3' DNA polymerase activity can have another go.
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Name_______________________________ BIOL 50 Exam III (100 points total)
Student I.D.______________________ 3/31/04
5. (6 points each) The diagram below is a drawing of transcription. DNA strands are solid and the RNA transcript is dotted.
______stream up
______stream down
non-template / coding strand 5' 3' template strand 5' (a) Label the 5' and 3' ends of each of the three nucleic acid strands. (b) Fill in the blanks to indicate upstream and downstream regions. (c) Label the template and non-template (coding) strands. 3' 3' 5'
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Name_______________________________ BIOL 50 Exam III (100 points total)
Student I.D.______________________ 3/31/04
6. (10 points) Chromatin consists of DNA proteins, and primarily histones. During replication, new histones must be synthesized. The new and old histones may be partitioned among the daughter DNA molecules in a conservative, semi-conservative, or dispersive mechanism. To determine which of these actually happens, you grow cells for several generations in medium with 35S, which labels proteins radioactively. Below is a diagram of a chromosome from one of these cells, through two cell cycles (M = mitosis, S = synthesis). Stippling (dots) indicates that a chromatid is radioactive.
M
S
M
S
G2
G1
G2
G1
G2
Based on these results, what can you conclude about the distribution of newly synthesized histones during DNA replication? Newly synthesized histones are deposited dispersively each daughter DNA molecule is associated with a mix of both old histones and new histones.
7. (5 points each) For each of the following genotypes of the E. coli lac operon, indicate whether -galactosidase activity will be inducible or constitutive? Assume glucose is not present.
+ + + + + C +
a) I P O Z / I P O Z
+ + + + + + C
constitutive ___________________
b) I P O Z / I P O Z
inducible ___________________
c) What property of lacO is demonstrated by these genotypes? Cis-dominance: the allele of lacO that is in cis to lacZ is dominant.
+
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Name_______________________________ BIOL 50 Exam III (100 points total)
Student I.D.______________________ 3/31/04
8. (2 points each) Circle the correct answer for each of the following questions about the carbon atoms in the sugars in nucleotides: (a) Which carbon has OH in ribose and H in deoxyribose? 1' 2' 3' 4' 5' 6'
(b) To which carbon is the base attached? 1' 2' 3' 4' 5' 6'
(c) Which carbon has one or more phosphate groups attached? 1' 2' 3' 4' 5' 6'
(d) Which carbon has an OH in both ribose and deoxyribose? 1' 2' 3' 4' 5' 6'
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Name_______________________________ BIOL 50 Exam III (100 points total)
Student I.D.______________________ 3/31/04
9. (2 points each) The diagram below represents the transcription unit of a gene. The scale is in base pairs, and each feature drawn is an even multiple of 100. The shaded region represents the protein-coding region of the transcript.
0
200
400 AUG
600
800
1000 UGA
1200
1400
(a) (2 points) What will be the length of the primary transcript (pre-mRNA)? 1400 nucleotides
(b) (2 points) Not including the 5' cap or the poly-adenylation tract, how long will the mRNA be? 1200 nucleotides
c. (2 points) How long with the polypeptide produced by this mRNA be? 200 amino acid residues (199 also acceptable)
d. (2 points) This gene undergoes alternative splicing. In the brain, intron 1 is not spliced out. In this case, how long with the mRNA be? 1300 nucleotides
e. (2 points) How long with the polypeptide produced by this brain mRNA be? 200 amino acid residues (199 also acceptable)
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U
UUU UUC U UUA UUG CUU CUC C CUA CUG AUU AUC A AUA AUG GUU GUC G GUA GUG
Second position C A phe leu
UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG
G tyr
UGU UGC UGA UGG CGU CGC CGA CGG AGU AGC AGA AGG GGC GGC GGA GGG
ser
leu
pro
ile
MET
thr
val
ala
UAU UAC UAA UAG CAU CAC CAA CAG AAU AAC AAA AAG GAU GAC GAA GAG
cys
STOP
STOP
trp arg
his gln asn lys asp glu
ser arg
gly
U C A G U C A G U C A G U C A G
Third position (3' end)
A
R
N
D
First position (5' end)
C
Q
E
G
H
I
L
K
M
F
P
S
T
W
Y
V
STOP
ala arg asn asp cys gln glu gly his ile leu lys met phe pro ser thr trp tyr val GCA CGA AAC GAC UGC CAA GAA GGA CAC AUA CUA AAA AUG UUC CCA UCA ACA UGG UAC GUA UAA C U C UAG G U C C U U U G G C U C C C C G G UGA G G G U G G G U U U U U U U U AGC UUA AGA U G G