Page 100 - Genetics_From_Genes_to_Genomes_6th_FULL_Part2

Solved Problems 259

SOLVED PROBLEMS

I. Imagine that 10 independently isolated recessive same alleles resulted in no complementation because
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lethal mutations (l , l , l , etc.) map to chromosome homozygotes for the recessive lethal mutation were gener-
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7 in mice. You perform complementation testing by ated.) The three complementation groups consist of (1) l ,
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mating all pairwise combinations of heterozygotes l , l ; (2) l , l ; and (3) l , l , l , l , l .
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bearing these lethal mutations, and you score the
absence of complementation by examining pregnant II. W, X, and Y are the intermediates (in that order) in a
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females for dead fetuses. A + in the chart means that biochemical pathway whose product is Z. Z mutants
the two lethals complemented, and dead embryos are found in five different complementation groups.
were not found. A − indicates that dead embryos were Z1 mutants will grow on Y or Z, but not W or X. Z2
found, at the rate of about one in four conceptions. mutants will grow on X, Y, or Z. Z3 mutants will only
(The crosses between heterozygous mice would be grow on Z. Z4 mutants will grow on Y or Z. Finally,
expected to yield the homozygous recessive showing Z5 mutants will grow on W, X, Y, or Z.
the lethal phenotype in 1/4 of the embryos.) The a. Order the five complementation groups in terms of
lethal mutation in the parental heterozygotes for each the steps they block.
cross are listed across the top and down the left side b. What does this genetic information reveal about
of the chart (that is, l indicates a heterozygote in the nature of the enzyme that carries out the
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which one chromosome bears the l mutation and the conversion of X to Y?
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homologous chromosome is wild type).
Answer
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l l l l l l l l l l 10 This problem requires that you understand complementa-
l 1 − + + + + − − + + + tion and the connection between genes and enzymes in a
l 2 − + + + + + + + − biochemical pathway.
l 3 − − − + + − − +
l 4 − − + + − − + a. A biochemical pathway represents an ordered set of
l 5 − + + − − + reactions that must occur to produce a product. This
l 6 − − + + + problem gives the order of intermediates in a pathway
l 7 − + + + for producing product Z. The lack of any enzyme
l 8 − − + along the way will cause the phenotype of Z , but the
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l 9 − +
l 10 − block can occur at different places along the pathway.
If the mutant grows when given an intermediate com-
How many genes do the 10 lethal mutations repre- pound, the enzymatic (and hence gene) defect must be
sent? What are the complementation groups? before production of that intermediate compound.
The Z1 mutants that grow on Y or Z (but not on
W or X) must have a defect in the enzyme that pro-
Answer duces Y. Z2 mutants have a defect prior to X; Z3 mu-
This problem involves the application of the complementa- tants have a defect prior to Z; Z4 mutants have a defect
tion concept to a set of data. There are two ways to analyze prior to Y; Z5 mutants have a defect prior to W. The
these results. You can focus on the mutations that do com- five complementation groups can be placed in order of
plement each other, conclude that they are in different activity within the biochemical pathway as follows:
genes, and begin to create a list of mutations in separate
genes. Alternatively, you can focus on mutations that do Z5 Z2 Z1, Z4 Z3
not complement each other and therefore are alleles of the W X Y Z
same genes. The latter approach is more efficient when
several mutations are involved. For example, l does not b. Mutants Z1 and Z4 affect the same step, but because
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complement l and l . These three alleles are in one com- they are in different complementation groups, we
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plementation group. l does not complement l ; they are in know they are in different genes. Mutations Z1 and
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a second complementation group. l does not complement Z4 are probably in genes that encode subunits of a
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l , l , l , or l , so they form a third complementation group. multisubunit enzyme that carries out the conversion
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Three complementation groups exist. (Note also that for of X to Y. Alternatively, a currently unknown addi-
each mutant, the cross between individuals carrying the tional intermediate step between X and Y could exist.