CHAPTER 10
Application and Experimental Questions
E1.
Discuss how researchers determined that TMV is a virus that causes damage to
plants.
Answer: First, Adolf Mayer determined that the disease caused by TMV could be
spread by spraying the sap from one plant onto another. By subjecting this sap to
filtration, Ivanovski showed that the causative agent was too small to be a
bacterium. Martinus Beijerinck ruled out that it is a toxin by showing that sap could
continue to transmit the disease after many plant generations.
E2.
What technique must be used to visualize a virus?
Answer: Electron microscopy must be used to visualize a virus.
E3.
Certain drugs to combat human viral diseases affect spike glycoproteins in the
viral envelope. Discuss how you think that such drugs may prevent viral infection.
Answer: Drugs that affect spike glycoproteins can interfere with ability of a virus to bind
to its host cell. This would prevent the virus from entering its host cell.
E4.
If you analyzed a phage plaque from a petri plate, what would it contain?
Answer: A P1 plaque mostly contains P1 bacteriophages that have a phage coat and P1
DNA. On occasion, however, a phage coat contains a segment of the bacterial
chromosome. It would also contain material from the E. coli cells that had been lysed.
E5.
As shown in Figure 10.9, phages with rII mutations cannot produce plaques in E.
coli K12(), but wild-type phages can. From an experimental point of view, explain why
this observation is so significant.
Answer: Benzer could use this observation as a way to evaluate if intragenic
recombination had occurred. If two rII mutations recombined to make a wild-type gene,
the phage would produce plaques in this E. coli K12(λ) strain.
E6.
In the experimental strategy described in Figure 10.11, explain why it was
necessary to dilute the phage preparation used to infect E. coli B so much more than the
phage preparation used to infect E. coli K12().
Answer: In E. coli B, both rII mutants and wild-type strains could produce plaques. In E.
coli K12(λ) only the rare recombinants that produced the wild-type gene would be able to
form plaques. If the phage preparation used to infect E. coli B was not diluted enough,
the entire bacterial lawn would be lysed, so it would be impossible to count the number
of plaques.
E7.
Here are data from several complementation experiments, involving rapid-lysis
mutations in genes rIIA and rIIB. The strain designated L51 is known to have a mutation
in rIIB.
Phage Mixture
Complementation
L91 and L65
No
L65 and L62
No
L33 and L47
Yes
L40 and L51
No
L47 and L92
No
L51 and L47
Yes
L51 and L92
Yes
L33 and L40
No
L91 and L92
Yes
L91 and L33
No
List which groups of mutations are in the rIIA gene and which groups are in the
rIIB gene.
Answer:
rIIA: L47 and L92
rIIB: L33, L40, L51, L62, L65, and L91
E8.
A researcher has several different strains of T4 phage with single mutations in the
same gene. In these strains, the mutations render the phage temperature sensitive. In this
case, this means that temperature-sensitive phages can propagate when the bacterium
(i.e., E. coli) is grown at 32°C but cannot propagate themselves when E. coli is grown at
37°C. Think about Benzer’s strategy for intragenic mapping and propose an experimental
strategy to map the temperature-sensitive mutations.
Answer: You would basically follow the strategy of Benzer (see Figure 10.11), except
that you would not need to use different E. coli strains. Instead, you would grow the cells
at different temperatures. You would coinfect an E. coli strain with two of the
temperature-sensitive phages and then grow at 32°C. You would then reisolate phages, as
shown in step 4 of Figure 10.11. You would take some of the phage preparation, dilute it
greatly (10–8), infect E. coli, and grow at 32°C. This would tell you how many total
phages you had. You would also take some of the phage preparation, dilute it somewhat
(10–6), infect E. coli, and then grow at 37°C. At 37°C, only intragenic recombinants that
produced wild-type phages would form plaques.
E9.
Explain how Benzer’s results indicated that a gene is not an indivisible unit.
Answer: Benzer first determined the individual nature of each gene by showing that
mutations within the same gene did not complement each other. He then could map the
distance between two mutations within the same gene. The map distances defined each
gene as a linear, divisible unit. In this regard, the gene is divisible due to crossing over.
E10. Explain why deletion mapping was used as a step in the intragenic mapping of rII
mutations.
Answer: You can first narrow down a mutation to one of several different regions by
doing pairwise crosses with just a few deletion strains. After that, you make pairwise
crosses only with strains that have also been narrowed down to the same region. You do
not have to make pairwise crosses between all the mutant strains.