Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 22, Problem 15P
Explain the similarities and differences between the procedures used for the coarse-scale mapping and fine-scale mapping of QTLs through controlled crosses, as was accomplished for QTLs involved with tomato size.
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ISSR is generally a dominant STS DNA marker. Nonetheless, with validated experimental evidence (e.g. laboratory and population genetics data), the marker can be used in codominance marker genotyping. Briefly explain each case below:
a) Codominant marker targets specific locus and reveals allelic variations in that locus among DNA samples.
b) Dominant marker: primers can complement other repeat sequences or in multiple loci thereby non-specificity in sampled genomes.
What advantages do anonymous DNA markers afford for genetic mapping as opposed to traditional allelic markers associated with visible phenotypes? What are the disadvantages of anonymous DNA markers for mapping?
How can linkage disequilibrium mapping sometimes provide a much higher resolution of gene location than classical linkage mapping?
Chapter 22 Solutions
Genetics: From Genes to Genomes
Ch. 22 - Choose the best matching phrase in the right...Ch. 22 - Suppose you grew genetically identical dandelion...Ch. 22 - How can each of the following be used in...Ch. 22 - Two different groups of scientists studying a rare...Ch. 22 - Which of the following statements would be true of...Ch. 22 - Studies have indicated that for pairs of twins...Ch. 22 - Prob. 7PCh. 22 - Prob. 8PCh. 22 - Table 22.2 lists concordance values for MZ and DZ...Ch. 22 - Prob. 10P
Ch. 22 - Prob. 11PCh. 22 - Two alleles at one locus produce three distinct...Ch. 22 - In a certain plant, leaf size is determined by...Ch. 22 - Compare and contrast the use of SNP genotyping: i...Ch. 22 - Explain the similarities and differences between...Ch. 22 - In Fig. 22.14c, the fw2.2 causal gene was...Ch. 22 - Among the most prevalent pathologies that afflict...Ch. 22 - Human geneticists have found the Finnish...Ch. 22 - Canavan disease, caused by homozygosity for a...Ch. 22 - In GWAS analysis, because of the existence of LD...Ch. 22 - In Fig. 22.15: a. Why do some chromosomes in the...Ch. 22 - Consider the triangle diagram shown in Fig. 22.17....Ch. 22 - Prob. 23PCh. 22 - You conduct a Case/Control study comparing the...Ch. 22 - Prob. 25PCh. 22 - ALS amyotrophic lateral sclerosis is a rare, fatal...Ch. 22 - Through GWAS explorations, scientists have...Ch. 22 - In domesticated dogs, size has a high...Ch. 22 - Suppose a GWAS investigation found a particular LD...Ch. 22 - In 2008, Time magazine named as its invention of...
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- Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forward(a) Explain sensitivity and selectivity in terms of true and false positives. (b) What is the advantage of a Needleman-Wunsch alignment compared to a seeded alignment? (c) What does the expectation parameter mean in local alignment? What will happen if the expectation value is increased from its default value of 10 to a 100?arrow_forwardBelow is the mapping data from three different crosses of Hfr strains with separate samples of an F- strain. Appearance of Genes in F- cells Hfrl: Genes d+ c+ ft et g+ Time (min) 4 15 26 44 58 Hfr2: Genes b+ d+ c+ f+ g+ 59 Time (min) 3 5 16 27 Hfr3: Genes et f+ c+ d+ b+ Time (min) 6 24 35 46 48 Use the mapping data from these interrupted conjugation crosses to construct a genetic map for these genes, indicating their order on the bacterial chromosome and the distances between them.arrow_forward
- Is there any difference between the alignments, alignment scores, identity percentages, positives and gaps? If so, what is your explanation for the differences?arrow_forwardCompare and contrast the use of SNP genotyping:(i) in the positional cloning of Mendelian diseasegenes, (ii) in direct QTL mapping, and (iii) in GWAS.arrow_forwardConsider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PCR was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PCR products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forward
- How do you use qPCR to confirm the phenotype of dpy-5 and skn-1. Describe the experiment.arrow_forwardDNA from a strain of Bacillus subtilis with the genotype trp+ tyr+ was used to transform a recipient strain with the genotype trp− tyr−. The following numbers of transformed cells were recovered: Genotype Number of transformed cells trp+ tyr− 154 trp− tyr+ 312 trp+ tyr+ 354 What do these results suggest about the linkage of the trp and tyr genes?arrow_forwardYou would like to isolate loci in the genome that influence plant height in tomato plants. Describe how you would achieve this through QTL mapping. Choose a true breeding tall strain and a true breeding short strain and compare the genomic sequences. Choose a true breeding tall strain and true breeding short strain and type them for molecular markers. Mate them together to produce F1 offspring and evaluate whether the F1 are tall or short. Choose two true breeding tall strains and compare the genomic sequences. Choose a true breeding tall strain and true breeding short strain and type them for molecular markers. Mate them together to produce F1 offspring. Genotype the F1 to identify makers that are associated with differences in plant height. Choose a true breeding tall strain and true breeding short strain and type them for molecular markers. Mate them together to produce F1 offspring, then backeross the F1 offspring to produce F2 offspring carrying recombinant chromosomes from tall…arrow_forward
- DNA obtalned from the indicated donor strains is used to TRANSFORM the indicated recipient strains. The resulting progeny are plated on minimal medium so that only wild- type recombinants are scored. The number of wild-types for each cross is given in the chart below. What is the order of the genes? Donor Recipient wid type colonies a-b- c+ a+ b+ c- 273 a- b+ c- a+ b-c+ 462 a- b+ C+ a+ b-c- 2 a b- C+ a- b+ c+ O 4-b-c a-o-b b-a-c cannot be determined Question 18 of 26arrow_forwardConsider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion. What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+? Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.arrow_forwardmapping three linked genes, the heterozygous AaBbCc was crossed to aabbcc individual the following phenotypes were obtained: ABC 4 abc 2 ABc 150 abC 141 aBc 18 AbC 20 Abc 30 aBC 35 A) What are the alleles found on each parental chromosome(genotypes of the parents)? B) How many double recombinants(cross over) were there? c) what is the linear arrangement of loci on the chromosome?arrow_forward
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