Biology Reference
In-Depth Information
a) If 100% of the progeny are New and the brood size is fairly large ( > 250), without
evidence of any slow growers or lethals, then the stock is true breeding with
complete penetrance.
b) If a significant fraction of the progeny are sickly or inviable, but all of the
survivors are New, then this is probably due to either the presence of a separate
mutation (''sick-1(*)'') in the strain, or multiple phenotypic effects (pleiotropy)
of the new-1(*) mutation. Although less likely, it is also possible that the strain
has a balanced lethal genotype, for example, new-1(*) sick-2(+)/new-1(+) sick-2
(*), where new-1(*) has a dominant New phenotype, and new-1(*) and sick-2(*)
are each lethal when homozygous. sick-2(*) could be a single gene mutation in a
locus that is fairly tightly linked to the new-1 locus, or it could be a chromosomal
alteration, such as a deletion, inversion, or translocation. It is not essential to
distinguish between these possibilities prior to the initiation of mapping crosses.
c) If there is no evidence of lethality, but < 100% of the progeny exhibit the New
phenotype, then either the mutation has incomplete penetrance or it is not
homozygous (which could occur if new-1(*) is dominant). Clone 12 New her-
maphrodites and three non-New hermaphrodites and examine their progeny. In
the case of a true-breeding stock with incomplete penetrance, each clone should
exhibit the same percentage of New progeny. If the mutation was previously not
homozygous, then a subset of the New clones should exhibit 100% penetrance.
These should then be used for further analysis.
B. Mapping Crosses
Traditional methods for mapping using visible markers and balancer chromo-
somes have been well described. These methods are still necessary in some cases;
however, the overwhelming majority of mutations that have been induced in N2
Bristol-derived strains can be successfully mapped by outcrossing to the polymor-
phic wild-type Hawaiian strain, CB4856 (available from the Caenorhabitis Genetic
Stock Center).
1. Mutations with 100% Penetrances
Set up a cross between CB4856 and your New stock. To do this, use a pick to
transfer a small amount of E. coli ( 3 mm diameter glob) onto the center of an
unseeded plate. Add 8-10 young adult/L4-stage CB4856 males and 3-4 young adult
New hermaphrodites (presumed genotype new-1(*)/new-1(*)). Incubate the plate at
20 for 3 days, and then inspect the progeny. If all of the progeny are New, and males
are not present, then the cross was probably not successful and should be repeated,
adding 20 L4-stage males and 5-6 L4/young adult-stage hermaphrodites.
a) Recessive Autosomal Mutations.
The most likely result is that the progeny will consist of a mixture of New self-
progeny hermaphrodites, plus non-New cross-progeny hermaphrodites and
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