First Experiment

1		The purpose of using 32P in this experiment was to
			A)	Label the phage protein
			B)	Label the E. coli
			C)	Label the phage RNA
			D)	Label the phage DNA
2		If the genetic material in T2 had been RNA instead of DNA (as in some other viruses), how would this have affected the results?
			A)	The results would have been the same.
			B)	There would be less radioactivity in the pellet.
			C)	There would be less radioactivity in the supernatant.
			D)	There would be an equal amount of radioactivity in the pellet and supernatant.
3		If the blender treatment step had been omitted, how would the results have been affected?
			A)	The results would have been the same.
			B)	There would be less radioactivity in the pellet.
			C)	There would be less radioactivity in the supernatant.
			D)	There would be an equal amount of radioactivity in the pellet and supernatant.
4		Though this doesn't occur in T2 phage proteins, many cellular proteins are phosphorylated (i.e., have a phosphate group that is covalently attached to an amino acid side chain). If T2 phage proteins had been phosphorylated, how would the results have been affected?
			A)	The results would have been the same.
			B)	There would be more radioactivity in the pellet.
			C)	There would be more radioactivity in the supernatant.
			D)	None of the above
5		The purpose of the centrifugation step was to
			A)	Separate the labeled phages from the unlabeled phages
			B)	Separate the phages from the bacterial cells
			C)	Separate the 32P from the 35S
			D)	Promote the binding of the phages to the bacterial cells

Second Experiment

6		Strong acid treatment breaks
			A)	The hydrogen bonds between bases in opposite strands
			B)	The covalent bonds between bases and deoxyribose
			C)	The hydrogen bonds between bases in the same strand
			D)	The covalent bonds between phosphate and deoxyribose
7		Overall, the data suggest that
			A)	The amount of A = T, and the amount of G = C.
			B)	The amounts of all four bases are equal.
			C)	The base composition among different species has no discernable pattern.
			D)	The amount of A = G, and the amount of C = T.
8		Which data are the least convincing with regard to the AT/GC rule?
			A)	Escherichia coli
			B)	Yeast
			C)	Turtle
			D)	Chicken
9		The purpose of the paper chromatography step is to
			A)	Hydrolyze the bases
			B)	Separate the DNA strands
			C)	Separate the bases
			D)	Quantitate the amount of each base
10		The purpose of the spectroscopy step was to
			A)	Hydrolyze the bases
			B)	Separate the DNA strands
			C)	Separate the bases
			D)	Quantitate the amount of each base