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 30ESP
In laboratory class, a genetics student was assigned to study an unknown mutation in Drosophila that had a whitish eye. He crossed females from his true-breeding mutant stock to wild-type (brick-red-eyed) males, recovering all wild-type F1 flies. In the F2 generation, the following offspring were recovered in the following proportions:
The student was stumped until the instructor suggested that perhaps the whitish eye in the original stock was the result of homozygosity for a mutation causing brown eyes and a mutation causing bright red eyes, illustrating gene interaction (see Chapter 4). After much thought, the student was able to analyze the data, explain the results, and learn several things about the location of the two genes relative to one another. One key to his understanding was that crossing over occurs in Drosophila females but not in males. Based on his analysis, what did the student learn about the two genes?
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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.
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The phenotype of crooked wings (cw) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive mutant gene for hairy (h) body. Assume that a cross is made between a fly with normal wings and a hairy body and a fly with crooked wings and normal body hair. All F1 flies from this cross were wild-type, and these flies were crossed among each other to produce 288 F2 offspring. Which phenotypes would you expect among the offspring in the F2 generation, and how many of each phenotype would you expect?
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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- In 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_forwardIn 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_forwardIn a cross between a white-eyed female (ww) and a red-eyed male (w+Y), nearly all the progeny were either red-eyed females (w+w) or white-eyed males (wY). However, about 1 in every 2000 F1 flies had an "exceptional phenotype" and was either a white-eyed female or red-eyed male. How did Bridges explain this unexpected result? A) Crossing over B) Incomplete cytokinesis C) Incorrect synapsis D) Nondisjunction E) Pseudoautosomal regionarrow_forward
- In Drosophila melanogaster, vestigial (short) wings (vg) are caused by a recessive mutant gene that independently assorts with a gene pair that influences body hair. Hairy (h ) results in a hairy body. A cross is made between a fly with normal wings and a hairy body and a fly with vestigial wings and a normal body. The phenotypically normal F1 flies were crossed among each other and 1024 F2 flies were reared. What phenotypes would you expect in the F2 and in what actual numbers (not ratio) would you expect to find them?arrow_forwardThe phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly that is homozygous for normal wings and has a hairy body and a fly with vestigial wings that is homozygous for normal body. The wild-type F1 flies were crossed among each other to produce 1024 F2 offspring. Which phenotypes would you expect among the F2 offspring, and how many of each phenotype would you expect? Group of answer choices 192 wild type, 256 vestigial, 64 hairy, and 192 vestigial and hairy All vestigial and hairy. 576 wild type, 192 vestigial, 192 hairy, and 64 vestigial and hairy All wild type 256 wild type; 256 vestigial, 256 hairy, and 256 vestigial and hairyarrow_forwardTwo Drosophila flies that had normal (transparent, long) wings were mated. In the progeny, two new phenotypes appeared, dusky wings (having a semi-opaque appearance) and clipped wings (with squared ends). The progeny were as follows: Females: 179 transparent, long 58 transparent, clipped Males: 92 transparent, long 89 dusky, long 28 transparent, clipped 31 dusky, clipped a) Provide a genetic explanation for these results, showing genotypes of parents and of all progeny classes under your model. b) Design a test for your model.arrow_forward
- In the fruit fly, dumpy wings (d) and purple eyes (p) are encoded by mutant alleles that are recessive to those that produce wild type traits; long wings (d+) and red eyes (p+). These two genes are on the same chromosome. In a particular lab, two researchers Walt and Jesse crossed a fly homozygous for dumpy wings and purple eyes with a fly homozygous for the wild type traits. The F1 progeny, which had long wings and red eyes, was then crossed with flies that had dumpy wings and purple eyes. Unfortunately, the progeny of this cross somehow escaped. To prevent their other projects from contamination, they decided to spend an exceptionally boring hour in the lab catching and counting the progeny and found the following: long wings, red eyes – 482 dumpy wings, purple eyes – 473 long wings, purple eyes – 23 dumpy wings, red eyes - 22 What is the genetic distance between these two loci? a. 4.5 cM b. 55 cM c. 45 cM d. 49.5 cM e. 4.7 cMarrow_forwardAnother cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w) and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F1 females were wild type for all three traits, while the F1 males expressed the yellow-body and white eyes traits. The cross was carried to an F2 progeny and only male offspring were tallied. Based on the data shown here, a genetic map was constructed. a) Diagram the genotypes of the F1 parents. b) Construct a map, assuming the white is at locus 1.5 on the X-chromosome. Phenotype Male offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0arrow_forwardIn fruit flies, you are mapping three genes in a three point cross. The mutants are hairy body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent with a homozygous recessive parent and obtain the following results: Type Number h se g. 5 + se + 450 + se g 27 ++g_ h se + + + + h + g. h + + TOTAL is the gene in the middle and the distance in map units between se and g is Oh; 16.4 se; 7.1 Oh; 7.1 70 82 7 327 32 1000 se; 16.4arrow_forward
- Male 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_forwardRed eyes (r), brown color (b), and curled wings (c) are 3 recessive mutations that occur in house flies. These genes have already been mapped: r-b = 12 mu b-c = 18 mu r-c = 30 mu A fly with red eyes, brown color, and curled wings is crossed with a fly homozygous for the wild-type traits. The F1 males are crossed with females with the three mutant traits and 2000 progeny are produced. Find the number of observed double crossovers if the interference is 0.38. O 43 O 62 O 27 O 16arrow_forwardA Drosophila strain with two mutated phenotypes was crossed with a wild-type Drosophila strain. The wild-type phenotypes were present in all of the F1 flies. The F1 flies were bred with one another. The F2 generation, on the other hand, did not have the predicted 9:3:3:1 phenotypic ratio. Explain why these outcomes occurred.arrow_forward
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