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 9PDQ
The genes dumpy (dp), clot (cl), and apterous (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the following genetic distances were determined. What is the sequence of the three genes?
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The genes dumpy (dp), clot (cl), and apterous (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the following genetic distances were determined. What is the sequence of the three genes? dp–ap 42 dp–cl 3 ap–cl 39
Females of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.
Two different strains of Drosophila melanogaster are mated in reciprocal crosses. When strain A males are crossed with strain B females, the progeny are normal. However, when strain A females are crossed with strain B males, there are many mutations and chromosome rearrangements in the gametes of the F1 progeny, and the F1 generation is effectively sterile. Explain these results.
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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- The genes dumpy wings (dp), clot eyes (cl), and apterous wings (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the genetic distances shown below were determined. What is the sequence of the three genes? dp–ap 42 dp–cl 3 ap–cl 39arrow_forwardThe genotype of a Drosophila with a heterozygous translocation between chromosome 2 and chromosome 3 is shown below, where bw = brown eyes and e = ebony body: bw+ bw e Assume there is no crossing over in the female and that alternate = adjacent 1 = adjacent 2 disjunction. [Yes, these assumptions are incorrect, but they make the problem much easier!] Two individuals of the above genotype, i.e. both heterozygous for this 2;3 translocation and both heterozygous for bw (on chromosome 2) and e (on chromosome 3) are crossed. A (Only zygotes with balanced genomes will survive to adulthood. What proportion of F1 zygotes have a balanced genome? B. Draw the genotype(s) and state the phenotype(s) of the surviving progeny you mentioned in part A above. For each different genotype/phenotype indicate it's frequency among the surviving progeny.arrow_forwardIn Drosophila, the brown mutation (bw, chromosome 2, position 104.5) results in brown eyes, while miniature (min, chromosome X, position 36.1) results in wings that are 2/3 the length of wild type. True breeding, wild type females are mated with true breeding males with brown eyes and miniature wings. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1arrow_forward
- In Drosophila, the genes ct(cut wing margin), y (yellow body), and v (vermilion eye color) are X-linked. Females heterozygous for all three markers were mated with wildtype males and the following male progeny were obtained. As is conventional in Drosophila genetics, the wild-type allele of each gene is designated by a “+” sign in the appropriate column. Use the data to (A) create a genetic map of the genes, (B) calculate interference, and (C) interpret the value of interference. Show your work.arrow_forwardMale Drosophila from a true-breeding wild-type stock were irradiated with X-rays and then mated with females from a true-breeding stock carrying the following recessive mutations on the X chromosome: yellow body (y), crossveinless wings (cv), cut wings (ct), singed bristles (sn), and miniature wings (m). These markers are known to map in the order: Recessive alleles: y, cv, ct, sn, m Dominant alleles: y+, cv+, ct+, sn+, m+ y-cv-ct-sn-m у CV ct sn m X-rays х х X ct sn CV у m y+ CV+ ct+ sn+ m+ х X ? Exceptional female: Most of the female progeny of this cross were phenotypically wild type, but one female exhibited ct and sn mutant characteristics. When this exceptional ct sn female was mated with a male from the true-breeding wild-type stock, twice as many females as males appeared among the progeny. a. What is the nature of the X-ray-induced mutation present in the exceptional female? b. Draw the X chromosomes present in the exceptional ct sn female as they would appear during pairing…arrow_forwardIn Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F1, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F2 generation, and 1000 offspring were counted, with the results shown in the following table. Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. (a) Using proper nomenclature, determine the genotypes of the P1 and F1 parents. (b) Determine the sequence of the three genes and the map distances between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence. Does it represent positive or negative interference?arrow_forward
- In Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the F1 cross? X X 3+ cu e + X X e + + + + + cu e + O + ■ 3+ X X X X Y Y + + ■ cu cu cu ' + ■ cu ■ ' + e + e e e e e + cu +arrow_forwardThe genetic map was based on crosses in Drosophila involving the three sex-linked genes a, b and c. “a” gives red eyes, “b” gives normal wings and “c” gives black body. The recombination frequencies between these genes are as follows; a and b is 23.8, b and c is 2.6 and a and c is 28.1, respectively. Could you draw a basic genetic map based on distance between these genes using dots to show distance(s)? Could you make one fundamental comment using these distances based on genetic linkage?arrow_forwardThe normal sequence of markers on a certain Drosophila chromosome is ABCDE*FGHIJK, where the asterisk represents the centromere. Some flies were isolated with a chromosome aberration that has the following structure: ABCDE*FIJK . This represents a O a) deletion of GH segment O b) inversion of GH segment O c) deletion O d) deletion of centromerearrow_forward
- Three recessive mutations in Drosophila melanogaster, roughoid (ru, small rough eyes), javelin (jv, cylindrical bristles), and sepia eyes (se, dark brown eyes) are linked. A three-point cross was carried out and the following progeny obtained: jv+ ru+ se+ 37 jv+ ru+ se 2 jv+ ru se 14 jv+ ru se+ 146 jv ru se+ 2 jv ru se 35 jv ru+ se 154 jv ru+ se+ 10 a. Determine the order of the genes on the chromosome. b. Determine which progeny contain single crossovers and which contain double crossovers and indicate where among the genes the crossovers occurred. c. Calculate the map distances among the genes. d. Calculate the coefficient of coincidence and interference among the genesarrow_forwardDrosophila females heterozygous for three recessive mutations, a, b, and c , were crossed to males homozygous for all three mutations.The cross yielded the following results: in the image Q. Construct a linkage map showing the correct order of these genes and estimate the distances between them.arrow_forwardYou design Drosophila crosses to provide recombinationdata for gene a, which is located on the chromosome shownin Figure 15.12. Gene a has recombination frequencies of14% with the vestigial wing locus and 26% with the browneye locus. Approximately where is a located along thechromosome?arrow_forward
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