Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 2, Problem 45P
A pure-breeding fruit fly with the recessive mutation cut wing, caused by the homozygous cc genotype, is crossed to a pure-breeding fly with normal wings, genotype CC. Their
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The recessive allele s causes Drosophila to have small wings, and the s+ allele causes normal wings. This gene is known to be X linked. If a small-winged male is crossed with a homozygous wild-type female, what ratio of normal to small-winged flies can be expected in each sex in the F1? If F1 flies are intercrossed, what F2 progeny ratios are expected? What progeny ratios are predicted if F1 females are backcrossed with their father?
Two pure-breeding strains of flies are mated, and the F1 are intercrossed. The first strain has curled wings and black bodies. The second strain has straight wings and brown bodies. The F2 progeny are 271 straight wings with brown bodies, 31 curled wings with black bodies, 94 curled wings with brown bodies and 90 straight wings with black bodies.
If instead of the above, assume the wing shape gene and the body color gene are completely linked. From parents that are curled winged with brown bodies mated to straight winged with black bodies, what would be the outcome of an F1 intercross? (Specify the phenotypes and the frequency of each expected).
. In fruit flies (Drosophila melanogaster), the bn+ allele for normal dull red eyes is dominant to the bn allele that gives brown eyes. Another gene affects wing shape; for this gene, the ct+ allele for normal wings is dominant to the ct allele, which gives “cut” wings, with jagged edges. A fly with dull red eyes and normal wings was crossed with a fly that had dull red eyes and cut wings, and the following progeny were obtained:
16 dull red eyes, normal wings
14 dull red eyes, cut wings
5 brown eyes, normal wings
5 brown eyes, cut wings
What were the genotypes of the parents?
Chapter 2 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 2 -
1. Compare and contrast the following terms:
a....Ch. 2 - For the cross , what is the expected genotype...Ch. 2 - 37. Galactosemia is an autosomal recessive...Ch. 2 - In mice, black coat color is dominant to white...Ch. 2 - Two parents plan to have three children. What is...Ch. 2 - Consider the cross AaBbCCAABbCc. a. How many...Ch. 2 - If a chi-square test produces a chi-square value...Ch. 2 -
8. Determine whether the statements below are...Ch. 2 - In the datura plant, purple flower color is...Ch. 2 - 10. The dorsal pigment pattern of frogs can be...
Ch. 2 - 11. Black skin color is dominant to pink skin...Ch. 2 - A male mouse with brown fur color is mated to two...Ch. 2 - 13. Figure 2.12 shows the results of Mendel’s...Ch. 2 - 14. An experienced goldfish breeder receives two...Ch. 2 -
15. The accompanying pedigree shows the...Ch. 2 -
16. A geneticist crosses a pure-breeding strain...Ch. 2 - Suppose an F1 plant from Problem 16 is crossed to...Ch. 2 - 18. In pea plants, the appearance of flowers along...Ch. 2 - 19. If two six-sided dice are rolled, what is the...Ch. 2 - Experimental Insight 2.1 describes data, collected...Ch. 2 -
21. The accompanying pedigree shows the...Ch. 2 - 22. The seeds in bush bean pods are each the...Ch. 2 - List all the different gametes that are possible...Ch. 2 - Organisms with the genotypes AABbCcDd and AaBbCcDd...Ch. 2 -
52. In humans, the ability to bend the thumb...Ch. 2 - In the fruit fly Drosophila, a rudimentary wing...Ch. 2 - In pea plants, plant height, seed shape, and seed...Ch. 2 - A variety of pea plant called Blue Persian...Ch. 2 - 29. In tomato plants, the production of red fruit...Ch. 2 - A male and a female are each heterozygous for both...Ch. 2 - 31. In a sample of families with children each,...Ch. 2 - Prob. 32PCh. 2 - A woman expressing a dominant phenotype is...Ch. 2 - Two parents who are each known to be carriers of...Ch. 2 - 33. An organism having the genotype AaBbCcDdEe is...Ch. 2 - 34. A man and a woman are each heterozygous...Ch. 2 - For a single dice roll, there is a 16 chance that...Ch. 2 - You have four guinea pigs for a genetic study. One...Ch. 2 - 37. Galactosemia is an autosomal recessive...Ch. 2 - Sweet yellow tomatoes with a pear shape bring a...Ch. 2 - A cross between a spicy variety of Capsicum annum...Ch. 2 - Alkaptonuria is an infrequent autosomal recessive...Ch. 2 - 41. Humans vary in many ways from one another....Ch. 2 - 42. In chickens, the presence of feathers on the...Ch. 2 -
43. A pure-breeding fruit fly with the...Ch. 2 - 44. Situs inversus is a congenital condition in...Ch. 2 - 45. Domestic dogs evolved from ancestral grey...Ch. 2 - Alleles of the IGF-1 gene in dogs, encoding...Ch. 2 - 49. The Basalt Seed Lending Library run by the...
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- Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?arrow_forwardFigure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?arrow_forwardHemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?arrow_forward
- Two pure-breeding strains of flies are mated, and the F1 are intercrossed. The first strain has curled wings and black bodies. The second strain has straight wings and brown bodies. The F2 progeny are 271 straight wings with brown bodies, 31 curled wings with black bodies, 94 curled wings with brown bodies and 90 straight wings with black bodies. If the F1 were backcrossed to the straight, wing brown bodied parent, what phenotypes would be produced among the progeny? What would be the proportion of each phenotype?arrow_forwardYou have been given a virgin Drosophila female. You notice that the bristles on her thorax are much shorter than normal. You mate her with a normal male (with long bristles) and obtain the following F1 progeny1 3 short-bristled females, 1 3 long-bristled females, and 1 3 long-bristled males. A cross of the F1 long-bristled females with their brothers gives only long-bristled F2. A cross of short-bristled females with their brothers gives 1 3 short-bristled females, 1 3 long-bristled females, and 1 3 long-bristled males. Provide a genetic hypothesis to account for all these results, showing genotypes in every cross.arrow_forwardA cross was made between 2 Drosophila that have the wild phenotype. Their progenies were found to have the same phenotype. A sample was taken from 2000 progenies, each of which was crossed with a fly having purple eye color. Half of the cross gave only wild type flies and the other half gave 50% purple-eyed progeny. What were the genotypes of the original pairs of wild type flies?arrow_forward
- In roses, purple flower color is determined by the dominant P allele, while pphomozygotes are white. The presence of long stems is determined by the dominant S allele, while ss homozygotes have short stems. Both mutations are completely penetrant. A test cross was performed between a rose plant of unknown genotype with a white flowered, short stemmed rose plant (pp ss) and the following 200 progeny plants were obtained: 84 white flowers, long stems 16 purple flowers, long stems 82 purple flowers, short stems 18 white flowers, short stems Select two statements below that are TRUE. options: The P and S genes independently assort during meiosis. The map distance between P and S is 17 cM. The genotype of the progeny plants with purple flowers and short stems is PP ss. The map distance between P and S is 83 cM. The homologs in the plant with…arrow_forwardIn Drosophila, white eyes (w) are recessive to red eyes (w+) at one locus and black body (b) is recessive to gray body (b+). A homozygous white eyes, gray bodied female is crossed with a homozygous red eyes, black bodied male to produce the F1 progeny. The F1 progeny are testcrossed and produce the following progeny: White eyes, black body: 212 White eyes, gray body: 288 Red eyes, black body: 308 Red eyes, gray body: 192 Does the evidence indicate that w and b loci are linked? Explain why or why not? If they are linked, what is the map distance between the two loci? If they are not linked, what is the map distance between the two loci? If they are linked, are the allels in the F1 in coupling or repulsion? How do you know? Draw the genotypes of all individuals described in the problem (original parents, F1, testcross, and F2 progeny) using the appropriate notation.arrow_forwardIn the fruit fly Drosophila melanogaster, the trait of black body is due to a gene on chromosome 2 and black body b is recessive to wild type body b + . The trait of purple eyes is controlled by a gene that is also on chromosome 2 and purple eyes p is recessive to wild type eyes p + . A true-breeding wild type strain is crossed with a true breeding strain that has black bodies and purple eyes. The F1 generation is then testcrossed to the black body, purple eye strain and 500 progeny are produced as follows: 224 wild type for both body and eye 236 black body and purple eye 18 wild type body and purple eye 22 black body and wild type eye. What is the recombination frequency and genetic map distance between the two genes?arrow_forward
- Three autosomal genes are linked along the same chromosome.The distance between gene A and B is 7 mu, the distance betweenB and C is 11 mu, and the distance between A and C is 4 mu. Anindividual that is AA bb CC was crossed to an individual that is aaBB cc to produce heterozygous F1 offspring. The F1 offspring werethen crossed to homozygous aa bb cc individuals to produce F2offspring.If we assume that no double crossovers occur, what percentageof F2 offspring is likely to be homozygous for all three genes?arrow_forwardThe 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?arrow_forwardWhen a white guinea pig is crossed to a yellow guinea pig, all the progeny are cream-colored. From a cross of two such F1’s, the F2 showed 32 white, 66 cream, and 30 yellow guinea pigs. How is this trait inherited? _______________ Identify the genotype of the white guinea pig. ________ Identify the genotype of the yellow guinea pig. ________ Identify the genotype of the cream-colored guinea pig. ________ Identify the genotypes and the corresponding phenotypes of the F2 progeny. Give the genotypic ratio. __________________ Give the phenotypic ratio. ___________________arrow_forward
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