Page 105 - Genetics_From_Genes_to_Genomes_6th_FULL_Part2

264 Chapter 7 Anatomy and Function of a Gene: Dissection Through Mutation

−
of the tests could not be completed because of an How do you think he was able to distinguish rII
accident in the greenhouse. The results of the comple- deletions from point mutations?
mentation tests that could be finished are shown in b. Figure 7.25c shows Benzer’s fine structure map of
the table that follows. point mutations in the rII region. A key feature of
this map is the existence of hotspots, which Benzer
interpreted as nucleotide pairs that were particularly
1 2 3 4 5 6
1 − + − + susceptible to mutation. How could Benzer say that
all of the independent mutations in a hotspot were
2 − − due to mutations of the same nucleotide pair?
3 − − 31. a. You have a test tube containing 5 ml of a solution of
4 − bacteriophages, and you would like to estimate the
5 − + number of bacteriophages in the tube. Assuming the
6 − tube actually contains a total of 15 billion bacterio-
phages, design a serial dilution experiment that
would allow you to estimate this number. Ideally,
a. Exactly what experiment was done to fill in indi- the final plaque-containing plates you count should
vidual boxes in the table with a plus (+) or a contain more than 10 and fewer than 1000 plaques.
minus (−) ? What does + represent? What does −
represent? Why are some boxes in the table filled b. When you count bacteriophages by the serial
in green? dilution method as in part (a), you are assuming a
plating efficiency of 100%; that is, the number of
b. Assuming no complications, what do you expect plaques on the petri plate represents exactly the
for the results of the complementation tests that number of bacteriophages you mixed with the plat-
were not performed? That is, complete the table by ing bacteria. Is there any way to test the possibility
placing a + or a − in each of the blank boxes. that only a certain percentage of bacteriophage
c. How many genes are represented among this particles can form plaques (so that the plating effi-
collection of mutants? Which mutations are in ciency would be less than 100%)? Why is it fair to
which genes? assume that any plaques are initiated by one rather
29. In humans, albinism is normally inherited in an auto- than multiple bacteriophage particles?
somal recessive fashion. Figure 3.24c in Chapter 3 32. You found five T4 rII mutants that will not grow on
−
shows a pedigree in which two albino parents have E. coli K(λ). You mixed together all possible combi-
several children, none of whom is an albino. nations of two mutants (as indicated in the following
a. Interpret this pedigree in terms of a complementa- chart), added the mixtures to E. coli K(λ), and scored
tion test. for the ability of the mixtures to grow and make
b. It is very rare to find examples of human pedigrees plaques (indicated as a + in the chart).
such as Fig. 3.24c that in effect represent a comple-
mentation test. The reason is that most genetic 1 2 3 4 5
conditions in humans are rare, so it is highly unlikely 1 − + + − +
that unrelated people with the same condition would 2 − − + −
mate. In the absence of complementation testing, 3 − + −
what kinds of experiments could be done to deter- 4 − +
mine whether a particular human disease phenotype 5 −
can be caused by mutations at more than one gene?
c. Complementation testing requires that the two mu- a. How many genes were identified by this analysis?
tations to be tested are both recessive to wild type. b. Which mutants belong to the same complementa-
Suppose that two dominant mutations cause similar tion groups?
phenotypes. How could you establish whether these 33. The rosy (ry) gene of Drosophila encodes an enzyme
mutations affected the same gene or different genes? called xanthine dehydrogenase. Flies homozygous for
30. a. Seymour Benzer’s fine structure analysis of the ry mutations exhibit a rosy eye color. Heterozygous
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rII region of bacteriophage T4 depended in large females were made that had ry Sb on one homolog
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41
part on deletion analysis as shown in Fig. 7.25. and Ly ry on the other homolog, where ry and
But to perform such deletion analysis, Benzer ry are two independently isolated alleles of ry. Ly
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−
had to know which rII bacteriophage strains [Lyra (narrow) wings] and Sb [Stubble (short) bris-
were deletions and which were point mutations. tles] are dominant mutant alleles of genes to the left