GAGAG-HW

Chapter 1
1.6-Since the phenol destroys only proteins, not DNA, transformation can still occur (the DNA is still intact and thus can be incorporated into R cells transforming them into S cells). A strong alkali would destroy the DNA so transformation cannot occur.

1.8-Viruses are considered non-living mainly because they are unable to replicate and synthesize proteins on their own. Therefore, since a DNA virus was used in the Hershey-Chase experiment, the progeny phage left over at the end for analysis contains only DNA and protein (capsid made of proteins) and thus no RNA. So only labeled DNA and/or protein can be seen in the progeny phages and therefore labeled RNA does not compromise the results (radioactive phosphorus gets incorporated into RNA only inside the infected cell).

1.9-One can conclude that this DNA is double stranded because of complimentary base pairs (equal amounts of A to T and G to C).

1.10-Chargaff's rule is broken. The ratios between A's and T's and between C's and G's are not 1:1. The structure of this DNA is single stranded.

1.18-3'-AUAGCAUA-5'

1.20-Since the sequence ends in A with previous nucleotides all being of U than the end codon must be UUA. Therefore the codon for leucine is UUA

1.21-There could be 3 reading frames for an RNA molecule (one reading frame would start with C, one with G, one with C, which are the first 3 nucleotides of this sequence). In vivo there would be 1 reading frame.

1.29-a. Methionine-Serine-Threonine-Alanine-Valine-Leucine-Glutamate-Asparagine-Proline-Glycine
b. Valine would replace methionine at amino acid 1 (M1V mutation)
c. Would result in same sequence (silent S2S mutation)
d. Alanine would replace valine at amino acid 5 (V5A mutation)
e. The amino acid sequence would be terminated right after leucine resulting in a 6 amino acid sequence.

1.30- a. Methionine-Valine-Histidine-Leucine-Threonine-Proline-Glutamate-Glutamate-Lysine-Serine
b. Methionine-Valine-Histidine-Leucine-Threonine-Proline-Arginine-Arginine-Serine
c. Methionine-Valine-Histidine-Leucine-Threonine-Proline-Termination-Glycine-Glutamate-Valine
Note: The proline in b. and c. are same as a. because of redundancy
In b. the result is an aa short compared to the original (b4 framshifting).
In c. the result is that the aas chain is terminal after the #6 amino acid.

Challenge Problem 1-a. 3'-TAC CAC GTG GAC TGA GGA CTC CTC TTC AGA-5'
5'-ATG GTG CAC CTG ACT CCT GAG GAG AAG TCT-3'

b. 3'-TAC CAC GTG GAC TGA GGC TCC TCT TCA GA-5'
5'-ATG GTG CAC CTG ACT CCG AGG AGA AGT CT-3'

c. 3'-TAC CAC GTG GAC TGA GGC ACT CCT CTT CAG A-5'
5'- ATG GTG CAC CTG ACT CCG TGA GGA GAA GTC T-3'

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