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Map units between a locus and its centromere in organisms with ordered spores, such as Neurospora, can be calculated as
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map units (1 2) the number of SDS asci total number of asci 100
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Crossing over also occurs during mitosis, but at a much reduced rate. Somatic (mitotic) crossing over can be used to map loci. STUDY OBJECTIVE 3: To learn about analytical tech-
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niques for locating the relative positions of genes on human chromosomes 132 140
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Human chromosomes can be mapped. Recombination distances can be established by pedigrees, and loci can be attributed to speci c chromosomes by synteny and assignment tests in hybrid cell lines.
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PROBLEM 1: A homozygous claret (ca, claret eye color),
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curled uted, claret uted, curled claret, curled uted, claret, curled wild-type 26 24 167 6 298 302
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curled (cu, upcurved wings), uted ( , creased wings) fruit y is crossed with a pure-breeding wild-type y. The F1 females are testcrossed with the following results:
uted claret 4 173
Tamarin: Principles of Genetics, Seventh Edition
II. Mendelism and the Chromosomal Theory
6. Linkage and Mapping in Eukaryotes
The McGraw Hill Companies, 2001
Solved Problems
a. Are the loci linked b. If so, give the gene order, map distances, and coef cient of coincidence. Answer: The pattern of numbers among the eight offspring classes is the pattern we are used to seeing for linkage of three loci. We can tell from the two groups in largest numbers (the nonrecombinants uted, claret, curled and wild-type) that the alleles are in the coupling (cis) arrangement. If we compare either of the nonrecombinant classes with either of the double crossover classes ( uted and claret, curled), we see that the uted locus is in the center. For example, compare uted, a double crossover offspring, with the wild-type, a nonrecombinant; clearly, uted has the odd pattern.Thus the trihybrid female parent had the following arrangement of alleles: ca cu ca+ fl + cu+ A crossover in the ca region produces claret and uted, curled offspring, and a crossover in the cu region produces uted, claret and curled offspring. Counting the crossovers in each region, including the double crossovers in each, and then converting to percentages, yields a claret-to- uted distance of 35.0 map units (173 167 6 4) and a uted-to-curled distance of 6.0 map units (26 24 6 4). We expect 0.35 0.06 1,000 21 double crossovers, but we observed only 6 4 10. Thus, the coef cient of coincidence is 10/21 0.48.
PROBLEM 2: The ad5 locus in Neurospora is a gene for
PROBLEM 3: In yeast, the his5 locus is a gene for an en-
zyme in the synthesis pathway for the amino acid histidine, and the lys11 locus is a gene for an enzyme in the synthesis pathway for the amino acid lysine. A haploid wild-type strain (his5 lys11 ) is crossed with the double mutant (his5 lys11 ). The diploid is allowed to undergo meiosis, and 100 asci are scored with the following results:
his5 lys11 his5 lys11 his5 lys11 his5 lys11 62 his5 his5 his5 his5 lys11 lys11 lys11 lys11 30 his5 his5 his5 his5 8 lys11 lys11 lys11 lys11
What is the linkage arrangement of these loci Answer: Of the 100 asci analyzed, 62 were parental ditypes (PD), 30 were tetratypes (TT), and 8 were nonparental ditypes (NPD). To map the distance between the two loci, we take the percentage of NPD (8%) plus half the percentage of TT (1/2 of 30 15%) 23% or 23 centimorgans between loci.
PROBLEM 4: A particular human enzyme is present only
in clone B.The human chromosomes present in clones A, B, and C appear as pluses in the following table. Determine the probable chromosomal location of the gene for the enzyme.
Human Chromosome Clone A B C 1 2 3 4 5 6 7 8
an enzyme in the synthesis pathway for the DNA base adenine. A wild-type strain (ad5 ) is crossed with an adenine-requiring strain, ad5 . The diploid undergoes meiosis, and one hundred asci are scored for their segregation patterns with the following results:
ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5 40 46 5 3 4 2
What can you say about the linkage arrangements at this locus Answer: You can see that 14 (5 3 4 2) asci are of the second-division segregation type (SDS) and 86 (40 46) are of the rst-division segregation type (FDS). To map the distance of the locus to its centromere, we divide the percentage of SDS types by 2: 14/100 14%; divided by 2 is 7%. Thus, the ad5 locus is 7 map units from its centromere.
Answer: If a gene is located on a chromosome, the gene must be present in the clones with the chromosome ( ). Chromosomes 1, 2, 5, 6 are present in B. If the gene in question were located on chromosome 1, the enzyme should have been present in all three clones. A similar argument holds for chromosome 2, in which the enzyme should have been present in clones A and B, and so on for the rest of the chromosomes. The only chromosome that is unique to clone B is 6.Therefore, the gene is located on chromosome 6.
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