Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN: 9781305389892
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Cengage Learning
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Chapter 13, Problem 8TYK
Another gene in Drosophila determines wing length. The dominant wild-type allele of this gene produces long wings; a recessive allele produces vestigial (short) wings. A female that is true- breeding for red eyes and long wings is mated with a male that has purple eyes and vestigial wings. F1 females are then crossed with purple-eyed, vestigial-winged males. From this second cross, a total of 600 offspring are obtained with the following combinations of traits:
252 with red eyes and long wings
276 with purple eyes and vestigial wings
42 with red eyes and vestigial wings
30 with purple eyes and long wings
Are the genes linked, unlinked, or sex-linked? If they are linked, how many map units separate them on the chromosome?
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In Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up,
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Considering Drosophila notation, which of the following correctly diagrams the F1 cross?
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In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows:
genotype
number
sn ct
15
sn ct+
34
sn+ ct
33
sn+ct+
18
What is the map distance between sn and ct?
In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows
sn ct
36
sn ct+
13
sn+ ct
12
sn+ ct+
39
What is the map distance between sn and ct?
Chapter 13 Solutions
Biology: The Dynamic Science (MindTap Course List)
Ch. 13.1 - You want to determine whether genes a and b are...Ch. 13.2 - You have a true-breeding strain of...Ch. 13.3 - What mechanisms are responsible for: (a)...Ch. 13.4 - A man has Simpson syndrome, an addiction to a...Ch. 13.4 - Prob. 2SBCh. 13.5 - Prob. 1SBCh. 13 - In humans, redgreen color blindness is an X-linked...Ch. 13 - The following pedigree shows the pattern of...Ch. 13 - Individuals affected by a condition known as...Ch. 13 - A number of genes carried on the same chromosome...
Ch. 13 - Prob. 5TYKCh. 13 - Discuss Concepts Can a linkage map be made for a...Ch. 13 - In Drosophila, two genes, one for body color and...Ch. 13 - Another gene in Drosophila determines wing length....Ch. 13 - Prob. 9TYKCh. 13 - You conduct a cross in Drosophila that produces...Ch. 13 - Discuss Concepts Crossing-over does not occur...Ch. 13 - Prob. 12TYKCh. 13 - Prob. 13TYKCh. 13 - Prob. 14TYKCh. 13 - Prob. 1ITDCh. 13 - Prob. 2ITDCh. 13 - Prob. 3ITDCh. 13 - Prob. 4ITD
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- 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 P1 cross? X X ++ e + + + O+ X + X + ■ + X + + + 3+ X X X X + + Y Y cu cu cu + cu cu J e e e e e (D e + cu cu (Darrow_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 mutant sex-linked trait called “notched” (N) is deadly in Drosophila when homozygous in females. Males who have a single N allele will also die. The heterozygous condition (Nn) causes small notches on the wing. The normal condition in both male and females is represented by the allele n. a) Indicate the phenotypes of the F1 generation from the following cross: XNXn x XnY b) Explain why dead females are never found in the F1 generation no matter which parents are crossed. c) Explain why the mating of female XNXn and a male XNy is unlikely.arrow_forward
- Another 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 *******ANSWER PART B NOT PART A!!!! Phenotype Male offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0arrow_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 silkmoths (Bombyx mori), red eyes (re) and white-banded wings (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have wild-type eyes and wild-type wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are wild-type eyes, wild-type wings red eyes, wild-type wings wild-type eyes, white-banded wings red eyes, white-banded wings a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes? b. What is the rate of recombination between the gene for red eyes and the gene for white-banded wings?arrow_forward
- In Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/Xw V/v×Xw/Y v/v, Xw/Xw V/v × XW/Y V/v.arrow_forwardMiniature wings (Xm) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X*). Give the genotypes of the parents in the following cross: Male parent Female parent Male offspring Female offspring Long Miniature 750 miniature 761 long O male: X* / X* and female X™ /x+ O male: X*/Xt and female Xm /xm O male: X*/ Y and female Xm /xm O male: Xm/ Y and female Xm /xmarrow_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, a cross was made between a yellow-bodied male with vestigial wings and a wild-type (WT) female(brown body and normal wings). The F1 generation consisted of WT males and WT females. The F1 males and females were crossed, and the F2 progeny consisted of 16 yellow males with vestigial wings, 48 yellow males with WT wings, 15 brown males with vestigial wings, 49 WT males, 31 brown females with vestigial wings, and 97 WT females. Based on these results, explain the inheritance of the two genes (i.e. autosomal or sex-linked, dominant or recessive).arrow_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-eye traits. The cross was carried to an F2 progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed. Phenotype Male Offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0 (a) Diagram the genotypes of the F1 parents. (b) Construct a map, assuming that white is at locus 1.5 on the X chromosome. (c) Were any double-crossover offspring expected? (d) Could the F2 female offspring be used to construct the map? Why or why not?arrow_forwardIn Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/XwV/v×Xw/Y v/v, Xw/XwV/v×XW/Y V/v.arrow_forward
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