Download Genetics BIOL 335 Optional Worksheet 1 solutions 1

GeneticsBIOL335OptionalWorksheet1solutions
1/3
1. What happens if you insert an adenine to the beginning of this coding sequence:
With the addition of A at the beginning of the coding sequence, the reading frame
shifts, and a different set of amino acids is specified by the sequence, as shown here:
(note that a set of nonsense
codons is encountered, which
results in chain termination)
2. Consider the following DNA segment:
5'-GCTTCCCAA-3'
3'-CGAAGGGTT-5'
Assume that the top strand is the template strand used by RNA polymerase.
a. What is the transcribed RNA sequence?
If the top strand is the template, that means the transcribed RNA is complementary
to it. Since polymerases synthesize in a 5'->3' direction, this means the polymerase is
moving to the left. The RNA sequence is therefore 5'-UUGGGAAGC-3'.
Note that this sequence is equivalent to the DNA strand complementary to the
template (the sense strand; the bottom in this case).
b. Translate to protein sequence.
LGS
Assume that the bottom strand is the template strand used by RNA polymerase.
c. What is the transcribed RNA sequence?
If the bottom strand is the template, that means the transcribed RNA is
complementary to it. Since polymerases synthesize in a 5'->3' direction, this means the
polymerase is moving to the right. The RNA sequence is therefore 5'-GCUUCCCAA-3.
Note that this sequence is equivalent to the DNA strand complementary to the
template (the sense strand; the top in this case).
d. Translate to protein sequence.
ASQ
3. Which anticodon would you predict for isoleucine tRNA? Is there more than one possible
answer? If so, state any alternative answers.
There are three codons for ile: AUU, AUC, AUA.
This suggests the complementary anti-codons AAU, GAU, and UAU respectively.
In addition, due to wobble base-pairing permitted at the third codon position, the
tRNA with GAU can recognize both AUU and AUC.
GeneticsBIOL335OptionalWorksheet1solutions
2/3
4. A mutant E coli has no activity for the enzyme isocitrate lyase. Does this result prove that the
mutation is in the gene coding for isocitrate lyase? If not, what other mutations could result in the
same phenotype?
No, it does not. Mutations that affect gene expression could be involved. For
example, promoter mutations that prevent RNA polymerase from binding would have no
activity. A constitutively binding repressor could block expression all the time. Mutations
in a transcription activator protein or its binding site could result in a failure to recruit
RNA polymerase, resulting in no enzyme production.
5. What are the activities (- or +) of the lacZ and lacY proteins when inducer is present and when
inducer is absent for the following merodiploid E coli lac operon genotypes:
a. I-P-OcZ+Y+/I+P+O+Z-YThe first allele listed (I-P-OcZ+Y+) lacks a functional promoter (P-). Therefore, no
transcription of any of the downstream lac genes is possible. The second allele (I+P+O+ZY-) does not have functional lacZ (Z-) or lacY (Y-) genes; therefore, this allele cannot
produce functional protein either.
lacZ
lacY
+inducer
-
-inducer
-
b. I+P+OcZ-Y+/I+P-O+Z+YThe first allele (I+P+OcZ-Y+) has a functional promoter. It also has a constitutive
lac operator (Oc). This results in constitutive expression of downstream lac genes. However,
since lacZ is defective (Z-), only lacY will be produced. The second allele (I+P-O+Z+Y-)
has a non-functional promoter (P-); no transcription of downstream lac genes is possible.
lacZ
lacY
+inducer
+
-inducer
+
6. If we assume that each cycle in a given PCR reaction takes 5 minutes, how much
amplification would be accomplished in 1 hour?
One hour (60 min) has 12 5min cycles. Since each PCR cycle copies each strand
currently in the reaction, the amplification would be 212
GeneticsBIOL335OptionalWorksheet1solutions
3/3
7. Read the following Sanger sequencing gel.
From the bottom:
TGCGGGCTTATCGGGTCTAA
8. The restriction enzyme HindIII cuts DNA at the sequence AAGCTT, and the restriction
enzyme HpaII cuts DNA at the sequence CCGG. On average, how frequently will each enzyme
cut double-stranded DNA?
a. HindIII
Each base (A, T, C, G) has a 1/4 chance of being at any given position. Therefore,
there is a 1/46 or 1/4096 chance of observing AAGCTT, a six-base sequence; you expect a
HindIII site on average every 4096bp.
b. HpaII
There is a 1/44 or 1/256 chance of observing CCGG, a four-base sequence; you
expect a HpaII site on average every 256bp.