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P A S S
F a c i l i t a t o r
B I O L O G Y : G E N E T I C
S E Q U E N C E S
E d i b l e
I n f o ~
B A S E
D N A
A couple of kick off questions (may find a worksheet to use but the questions
may be better to get convo flowing anyhow):
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Does each new cell need DNA to be present?
Where does the extra DNA come from?
Why does DNA replicate?
Does DNA replication end when the old DNA creates a completely new
DNA?
What is the type of replication that DNA goes through called?
Why is it called semi-conservative replication?
Where does translation take place
In what direction do you read DNA?
In which direction do you ‘build’ DNA?
What type of bond exists between nitrogenous bases?
Are these strong or weak bonds?
Materials: coloured mini-marshmallows or other soft sweet, red liquorice ropes
and toothpicks (ALTERNATIVELY: pegs, buttons, string or pipe cleaners can be
used for base sequence pairing )
The idea is to use the liquorice as the phosphorous/ sugar back bone. The
sugar is joined to the nitrogenous base with a toothpick and each
marshmallow/soft sweet represents a base (ie. Different colour for U, C, A, G
and T). Be sure the students remember at which point they must use
Uracil rather than Thymine.
Using a sequence of Amino Acids, the students are to use the bunch of
goodies to play out:
 transcription from the template
 translation to the Amino Acid sequence
Each group will be given a sequence of DNA which will be the template/noncoding strand. There are three variations and in order to determine whether
they got the transcription and translation processes correct the resulting
Amino Acid sequence is available for comparison.
1. TACAAATTCCACCTC
-DNA template / Coding strand
AUGUUUAAGGUGGAG
Met –Phe –Lys –Val –Glu
-mRNA
-Amino Acid sequence
2. TACGGATTCGCAACT
AUGCCUAAGCGUUGA
Met –Pro –Lys –Arg –Ter
3. TACATAACCACGCGG
AUGUAUUGGUGCGCC
Met –Tyr –Trp –Cys –Ala
Second Position of Codon
U
F
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r
s
t
C
P
o
s
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t A
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n
(5’)
G
U
C
A
G
UUU Phe
UUC Phe
UUA Leu
UUG Leu
UCU Ser
UCC Ser
UCA Ser
UCG Ser
UAU Tyr
UAC Tyr
UAA Ter
UAG Ter
UGU Cys
UGC Cys
UGA Ter
UGG Trp
U
C
A
G
CUU Leu
CUC Leu
CUA Leu
CUG Leu
CCU Pro
CCC Pro
CCA Pro
CCG Pro
CAU His
CAC His
CAA Gln
CAG Gln
CGU Arg
CGC Arg
CGA Arg
CGG Arg
U
C
A
G
AUU Ile
AUC Ile
AUA Ile
AUG Met
ACU Thr
ACC Thr
ACA Thr
ACG Thr
AAU Asn
AAC Asn
AAA Lys
AAG Lys
AGU Ser
AGC Ser
AGA Arg
AGG Arg
U
C
A
G
GUU Val
GUC Val
GUA Val
GUG Val
GCU Ala
GCC Ala
GCA Ala
GCG Ala
GAU Asp
GAC Asp
GAA Glu
GAG Glu
GGU Gly
GGC Gly
GGA Gly
GGG Gly
U
C
A
G
T
h
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r
d
P
o
s
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t
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o
n
(3’)
Exam Questions from 2006 BIOL1102:
46. A transcription unit that is 8000 nucleotides long may use 1800 nucleotides to make
a protein consisting of 600 amino acids. This is best explained by the fact that:
1) There are termination exons near the beginning of mRNA.
2) There is redundancy and ambiguity in the genetic code.
3) Many nucleotides are needed to code for each amino acid.
4) Nucleotides break off and are lost during the transcription process.
5) Many non-coding nucleotides are present in mRNA.
47. A particular triplet of bases in the template sequence of DNA is AAA. The anticodon
on the tRNA that binds the mRNA codon is:
1) TTT
2) UUA
3) UUU
4) AAA
48. Which of the following does not occur during the termination phase of translation?
1) A stop codon causes the Acceptor site to accept a release factor.
2) The newly formed polypeptide is released.
3) A tRNA with the next amino acid enters the P site.
4) The ribosome comes off the mRNA and the two ribosomal subunits
separate.
5) Translation stops.
49. Suppose you have an actively growing culture of E. coli to which you add
radioactively labelled guanine (G). You allow the culture to grow for 1 more
generation (ie every cell undergoes one more round of replication). What do you
predict for the resulting cells?
1) All cells would contain radioactive DNA.
2) Half of the cells would have radioactive DNA.
3) All four bases of the DNA would be radioactive
50. Why is the new DNA strand complementary to the 3’ to 5’ strands assembled in short
segments (Okazaki fragments)?
1) The replication forks block the formation of larger strands.
2) DNA polymerases can assemble DNA only in the 3’ to 5’ direction
3) DNA polymerases can assemble DNA only in the 5’ to 3’ direction
4) It is more efficient than assembling complete new strands
5) Only short DNA sequences can extend off the RNA primers.
51. What would the effect of the translation be of a single base pair deletion in an
organism’s DNA?
1) None
2) Changes only to the amino acid where it occurs.
3) Possible changes to the amino acid where it occurs as well as to all
downstream amino acids from that point.
4) Poor replication of the DNA.
5) Immediate death to the cell.