Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 5, Problem 2PDQ
Review the Chapter Concepts list on page 94. Most of these center around the process of crossing over between linked genes. Write a short essay that discusses how crossing over can be detected and how the resultant data provide the basis of chromosome mapping.
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Chapter 5 Solutions
Concepts of Genetics (12th Edition)
Ch. 5 - In a family with one autistic child the risk for...Ch. 5 - Given that the prenatal test can provide only a...Ch. 5 - Consider two hypothetical recessive autosomal...Ch. 5 - With two pairs of genes involved (P/p and Z/z), a...Ch. 5 - In Drosophila, a heterozygous female for the...Ch. 5 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 5 - Review the Chapter Concepts list on page 94. Most...Ch. 5 - Describe the cytological observation that suggests...Ch. 5 - Why does more crossing over occur between two...Ch. 5 - Explain why a 50 percent recovery of...
Ch. 5 - Why are double-crossover events expected less...Ch. 5 - What is the proposed basis for positive...Ch. 5 - What two essential criteria must be met in order...Ch. 5 - The genes dumpy (dp), clot (cl), and apterous (ap)...Ch. 5 - Colored aleurone in the kernels of com is due to...Ch. 5 - In the cross shown here, involving two linked...Ch. 5 - In a series of two-point mapping crosses involving...Ch. 5 - Two different female Drosophila were isolated,...Ch. 5 - In Drosophila, a cross was made between femalesall...Ch. 5 - Another cross in Drosophila involved the...Ch. 5 - In Drosophila, Dichaete (D) is a mutation on...Ch. 5 - Drosophila females homozygous for the third...Ch. 5 - In Drosophila, two mutations, Stubble (Sb) and...Ch. 5 - If the cross described in Problem 18 were made,...Ch. 5 - Are mitotic recombinations and sister chromatid...Ch. 5 - What possible conclusions can be drawn from the...Ch. 5 - An organism of the genotype AaBbCc was testcrossed...Ch. 5 - Based on our discussion of the potential...Ch. 5 - Traditional gene mapping has been applied...Ch. 5 - DNA markers have greatly enhanced the mapping of...Ch. 5 - In a certain plant, fruit is either red or yellow,...Ch. 5 - Two plants in a cross were each heterozygous for...Ch. 5 - A number of humanmouse somatic cell hybrid clones...Ch. 5 - A female of genotype produces 100 meiotic tetrads....Ch. 5 - In laboratory class, a genetics student was...Ch. 5 - Drosophila melanogaster has one pair of sex...Ch. 5 - In Drosophila, a female fly is heterozygous for...Ch. 5 - The gene controlling the Xg blood group alleles...Ch. 5 - Prob. 34ESP
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- Genes Q, R, and S are on the same chromosome. The crossoverfrequency between S and Q is 5%, the crossover frequency betweenQ and R is 30%, and the crossover frequency between R and S is35%. Use this information to create a linkage map for thechromosomearrow_forwardIn an electrophoretic gel across which is applied a powerful electrical alternating pulsed field, the DNA of the haploid fungus Neurospora crassa (n = 7) moves slowly but eventually forms seven bands, which represent DNA fractions that are of different sizes and hence have moved at different speeds. These bands are presumed to be the seven chromosomes. How would you show which band corresponds to which chromosome?arrow_forwardWhat is FISH? Briefly describe how it works How is FISH used to characterize chromosomal translocations associated with certain genetic disorders and specific types of cancers?arrow_forward
- For questions 22 through 30, please refer to the pictures below: Normal B B B C D D D D' E E 22 23 24 B D G. Y D M H. B C E 25 26 27 28 B D Break through centromere B B A D 29 30 Identify the type of chromosomal aberration in each number above. Please refer to the choices below. Reciprocal translocation Robertsonian translocation Isochromosome Terminal deletion Interstitial deletion Duplication Pericentric inversion Ring chromosome Marker chromosome Paracentric inversionarrow_forwardThe mutations called bobbed in Drosophila result from variable reductions (deletions) in the number of amplified genes coding for rRNA. Researchers trying to maintain bobbed stocks have often documented their tendency to revert to wild type in successive generations. Propose a mechanism based on meiotic recombination which could account for this reversion phenomenon. Why would wild-type flies become more prevalent in Drosophila cultures?arrow_forwardIn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines Using the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringarrow_forward
- crossing over within a pericentric inversion produces chromosomes that have extra copies of some genes and no copies of other genes. The fertilization of gametes containing chromosomes with duplications or deletions often result in children with syndromes characterized by developmental delay, intellectual disability, abnormal development of organ systems, and early death. Maarit Jaarola and colleagues examined individual sperm cells of a male who was heterozygous for a pericentric inversion on chromosome 8 and determined that crossing over took place within the pericentric inversion in 26% of the meiotic divisions (M. Jaarola, R. H. Martin, and T. Ashley. 1998. American Journal of Human Genetics 63:218–224). Assume that you are a genetic counselor and that a couple seeks counseling from you. Both the man and the woman are phenotypically normal, but the woman is heterozygous for a pericentric inversion on chromosome 8. The man is karyotypically normal. What is the probability that this…arrow_forwardIn Drosophila the genes A B D are linked. You cross a strain that is homozygous for d with a strain that is homozygous for ab. Below are the phenotypes and progeny of a test cross. Give the correct order of the genes in the chromosome and determine the map distances between genes. Phenotype Number ABd 389 abD 413 ABD 60 abd 68 aBd 29 AbD 34 Abd 3 aBD 4arrow_forwardIn corn, the genes v (virescent seedlings), pr (red aleurone), and bm (brown midrib) are all on chromosome 5, but not necessarily in the order given. The cross: v+ pr bm/ v pr+ bm+ with v pr bm/ v pr bm produces 1000 progeny with the following phenotypes: v+ pr bm 226 v pr+ bm+ 229 v+ pr bm+ 153 v pr+ bm 185 v+ pr+ bm 59 v pr bm+ 71 v+ pr+ bm+ 36 v pr bm 41 What is the gene order, and the (b) genetic map of these three genes?arrow_forward
- For this particular family, what is the recombination rate between the D17S74 marker and the breast cancer gene? Lets say that at age 45 the third granddaughter III-3, is diagnosed with breast cancer. Now recalculate the recombination rate between the marker locus D17S74 and the breast cancer gene? Use only the 5 women in your analysis:the two daugthers in generation II and the three granddaughters in generation IIarrow_forwardConsider the following "pericentric" ("around the center") inversion. In one an individual, a simplified 8-gene sequence along one chromosome is 1234 5678 while a pericentric inversion occurred on their other homologous chromosome, resulting in the sequence 1265 4378 (the dot represents the centromere). Draw these homologous chromosomes lined up (as during crossing over, in Prophase l). If a crossover event occurred between gene locus 3 and 4 (slicing between them) on the original chromosome, what would the crossover products look like? What would be the "problem" with each of the chromosomes that result?arrow_forwardConsider the following two parents that are real chromosome encoded. Then, draw and calculate the result of performing a complete arithmetic recombination with the subtraction operation and a = 0.6, as well as complement mutation on the following parents, to produce the first child, with details? P1 .3 .6 .1 .5 .8 .7 .9 .4 .2 P2 .9 .5 .8 .1 .2 .3 .7 .4 .6arrow_forward
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