Answers to Quiz 4:
The number of codons that can specify each individual amino acid is indicated in
above the peptide sequence:
2
6
2
2
1
3
1
4
1
6
1
4
2
2
1
Phe-Ser-Glu-Tyr-Met-Ile-Trp-Val-Met-Arg-Tyr-Ala-Gln-Tyr-Met
a) Phe-Ser-Glu-Tyr-Met-Ile-Trp (144); b) Met-Ile-TrpVal-Met-Arg-Tyr (72); c) Glu-Tyr-Met-Ile-Trp-Val-Met
(48); d) Met-Arg-Tyr-Ala-Gln-Tyr-Met (96).
1. The lowest redundancy is found in Glu-Tyr-Met-IleTrp-Val-Met. Ans: (c).
2. This peptide could be encoded by 48 different codon
sequences. Ans: (b).
3. cDNA libraries are derived from mRNAs,
enzymatically converted to DNA using reverse
transcriptase and DNA polymerase. Ans: (c).
4. If we built a probe that was based on the codon sequence, it would be a ‘sense’ probe with
the same nucleotide sequence as the messenger RNA. It therefore would not have sequence
complementarity to the mRNA, and couldn’t bind to mRNA in a Northern blot. Ans: True.
5. Individual I-1 is heterozygous for two different RFLP patterns; one of his homologs
contains a 7 kb fragment that hybridizes
with the probe, and the other contains a
restriction site that allows the probe to
hybridize to a 4 and 3 kb piece: e.g.
7.0 kb
7
I-1
4
*
3
5.3 kb
4.0 kb
3.0 kb
1.7 kb
Individual I-2 also contains one homolog that hybridizes to a
Piece. In this individual, however, the 7 kb fragment that
hybridizes to the probe looks like it can be cleaved into two
fragments due to the presence of a variable internal restriction
site that cleaves the fragment into a 5.3 and 1.7 kb piece: e.g.
7 kb.
7
I-2
5.3
1.7
*
Following the segregation of both patterns through the pedigree, it seems as if the 7 kb
fragment donated by I-2 is always seen when the disease is present, and the 5.3/1.7 pattern
is always seen when the disease is absent, with the exception of individual II-7. This can be
explained if the disease gene locus is linked to the chromosome carrying the 7 kb fragment.
Individual II-7 can be explained by a crossover occurring during meiosis in individual I-2,
that occurs between the RFLP locus and the disease gene locus: e.g.
7
5.3
1.7
D
X
+
5. The 7 kb fragment is linked to the disease gene. Ans: (d).
6. The distance between the RFLP locus and the disease gene is 1/13 or between 7-8 map
units. Ans: (c).
7. The DNA sequence for the STRAND BEING SYNTHESIZED would read from the bottom of
the gel:
5’-CGTGCACGTGCA-3’
The DNA sequence for the TEMPLATE STRAND would be complementary and run
antiparallel to this sequence, i.e.
5’-CGTGCACGTGCA-3’
3’-GCACGTGCACGT-5’
Ans: (c).
8. All of the sequencing reactions need to contain all of the normal components for DNA
synthesis, and deoxyadenosine triphosphate (dATP) is a normal component of DNA. The
individual reactions only contain one chain terminator- a dideoxynucleotide that contains
neither a 3’ or 5’ hydroxyl group on the sugar. So dideoxyATP- ddATP would be present in
the ‘A’ reaction, ddTTP in the ‘T’ reaction, etc. terminating at specific bases as the template
strand is being read. Ans: False.
The pathway would be:
W
6
R
2
V
3
U
4
S
5
T
1
Q
9. The immediate precursor to S would be U. Ans: (d).
10. A mutation in gene 3 would lead to accumulation of precursor V. Ans: (e).