Page 178 - Genetics_From_Genes_to_Genomes_6th_FULL_Part1

170 Chapter 5 Linkage, Recombination, and the Mapping of Genes on Chromosomes

III. The following Neurospora asci were obtained when a between two genes, and the distance between each gene
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wild-type strain (ad leu ) was crossed to a double mu- and the centromere of the chromosome on which the
tant strain that cannot grow in the absence of adenine or gene is located. First designate the five classes of asci
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leucine (ad leu ). Only one member of each spore pair shown. The first class is a PD; the second is an NPD;
produced by the final mitosis is shown because the two the last three are Ts. Next determine if these genes are
cells in a pair have the same genotype. Total asci = 120. linked. The number of PD = number of NPD, so the
genes are not linked. When genes are unlinked, the
Spore pair Ascus type
T asci are generated by a crossover between a gene and
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1–2 ad leu ad leu ad leu + ad leu ad leu + the centromere. Looking at the leu gene, there is an
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3–4 ad leu ad leu ad leu ad leu + ad leu + MII pattern of that gene in the third and fourth asci
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5–6 ad leu ad leu + ad leu + ad leu ad leu − types. Therefore, the fraction of MII asci is:
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7–8 ad leu ad leu + ad leu ad leu + ad leu −
# of asci 30 30 40 2 18 40 + 2
a. What genetic event causes the alleles of two genes to 120 × 100 = 35%.
segregate to different cells at the second meiotic divi-
sion, and when does this event occur? Because only half of the chromatids in the meioses
that generated these T asci were involved in the
b. Provide the best possible map for the two genes and crossover, the map distance between leu and the
the relevant centromere(s). centromere is 35/2, or 17.5 m.u. Asci of the fourth
and fifth types show an MII pattern for the ad gene:
Answer 2 + 18
This problem requires an understanding of tetrad analysis 120 × 100 = 16.6%.
and the process (meiosis) that produces the patterns seen in
ordered asci. Dividing 16.6% by 2 gives the ad gene ↔ centromere
a. A crossover between a gene and the centromere map distance of 8.3 m.u. The map of these two genes
is the following:
causes the segregation of alleles at the second meiotic
division. The crossover event itself occurs during pro- ad
phase of meiosis I.
8.3 m.u.
b. Using ordered tetrads (or ordered octads) you can de- leu
termine whether two genes are linked, the distance 17.5 m.u.
PROBLEMS
Vocabulary i. parental ditype 9. when the two alleles of a gene are segregated into
different cells at the first meiotic division
1. Choose the phrase from the right column that best fits
the term in the left column. j. ascospores 10. an ascus containing only two nonrecombinant
kinds of spores
a. recombination 1. a statistical method for testing the fit between k. first-division 11. structure formed at the spot where crossing-over
observed and expected results segregation occurs between homologs
b. linkage 2. an ascus containing spores of four different l. mosaic 12. fungal spores contained in a sac
genotypes
c. chi-square test 3. one crossover along a chromosome makes a
second nearby crossover less likely Section 5.1
d. chiasma 4. when two loci recombine in less than 50% of 2. a. A Drosophila male from a true-breeding stock
gametes with scabrous eyes was mated with a female from a
e. tetratype 5. the relative chromosomal location of a gene true-breeding stock with javelin bristles. Both sca-
f. locus 6. the ratio of observed double crossovers to brous eyes and javelin bristles are autosomal reces-
expected double crossovers sive mutant traits. The F 1 progeny all had normal
g. coefficient of 7. individual composed of cells with different eyes and bristles. F 1 females from this cross were
coincidence genotypes mated with males with both scabrous eyes and
h. interference 8. formation of new genetic combinations by javelin bristles. Write all the possible phenotypic
exchange of parts between homologs classes of the progeny that could be produced from