ps1answers - Genome 371 Spring 2005 Review of Central Dogma...

Genome 371 Spring, 2005
Review of Central Dogma; Simple Mendelian Inheritance
Problem Set #1 Answers:
1. 5'-ACCGTTATGAC-3'
2. No. You would also need to know if this organism has a double stranded DNA genome.
Assuming that it does have a double stranded genome, G+C = 44%, and A+T = 56%. Thus,
28% of the bases would be adenine.
3. No. tobacco mosaic virus has a single stranded RNA genome and thus, the base-pair
stoichiometry of DNA does not apply.
4. (a). double-stranded DNA
(b) . Double-stranded RNA
(c). Single-stranded RNA
5. (a). False
(b). False
(c). True
(d). True
(e). True
(f). True
(g). True
(h). True
(i). True
(j). False
(k). False
(l). True
(m). True
6. Molecule II will have the higher T
m
. Thermal vibrations from heat can disrupt hydrogen
bonds and base stacking interaction, the major forces that hold the two strands of a DNA
duplex together. Molecule II has more GC base pairs than molecule I: 10 of 15 compared to 4
of 15. GC base pairs have three H-bonds compared to the two of AT base pairs. In addition,
adjacent GC base pairs have greater stacking energy than adjacent AT base pairs. Therefore,
more thermal energy will be required to separate the strands of a DNA molecule with a
higher percent of GC.
7. (a). 2
(b). 10
(c). 8

Genome 371 Spring, 2005
(d). 12
(e). 6
(f). 5
(g). 9
(h). 14
(i). 3
(j). 13
(k). 1
(l). 7
(m). 15
(n). 11
(o). 4
8.
5'
5'
3'
3'
Gene F Gene G
template strand for transcription of gene G
template strand for transcription of gene F
9. Three.
5’C)(TTA)(CAG)(TTT)(ATT)(GAT)(ACG)(GAG)(AAG)(G3’
5’CT)(TAC)(AGT)(TTA)(TTG)(ATA)(CGG)(AGA)(AGG)3’
3’(GAA)(TGT)(CAA)(ATA)(ACT)(ATG)(CCT)(CTT)(CC5’
10. Transcription: Adding the appropriate ribonucleotide complementary to the template.
Translation: codon in mRNA and anticodon in tRNA are complementary.

Genome 371 Spring, 2005
11. (a). Each amino acid is specified by a codon. A codon consists of three nucleotides.
Therefore, there must be at least 477X3=1431nucleotides in the coding part of this yeast gene.
(b). Since this sequence derives from the middle of the coding region it must contain an open
reading frame that extends through the entire sequence. There are six possible reading
frames (including the complementary strand, which we can easily derive) and only one
contains an open reading frame (the complementary strand) that extends through the entire
sequence:
5’A)(CCC)(TGG)(ACT)(AGT)(CGA)(AAG)(TTA)(ACT)(TAC)3’
The corresponding amino acid sequence would be: ?-Pro-Trp-Thr-Ser-Arg-Lys-Leu-Thr-Tyr.
(c). The sequence must correspond to that in which the complete open reading frame exists:
5’ACCCUGGACUAGUCGAAAGUUAACUUAC3’
12. (a). Using the genetic code in the Table, there are eight cases in which knowing the first
two nucleotides does not tell you the specific amino acid.
(b). If you knew the amino acid, you would not know the first two nucleotides in the cases of
Arg, Ser, and Leu.
13. (a). Val: UUU, UUG, UGU, GUU (however, among these, only GUU actually codes for
Val)
Gly: GGG, GGU, GUG, UGG (however, among these, only GGG and GGU actually
code for Gly)
(b). UUU = (0.6)
3
= 21.6%; UUC, UCU, and CUU = (0.6)
2
(0.4) = 14.4% each; UCC, CUC, and
CCU = (0.6)(0.4)
2
= 9.6% each; CCC = (0.4)
3
= 6.4%.