A cross in Drosophila melanogaster involved the recessive X-linked genes for white eye (w), yellow body (y), and cut wings (ct). A wild-type tri-hybrid female was crossed with white eye, yellow body, cut wings males. On the basis of the results shown below, which of the choices shown best represents the genetic map of the three loci on the X-chromosome? Phenotype + y ct W + + Male Offspring 21 33 wy ct 111 + + + 109 + + ct 424 w y + 394 + y + LO 5 W + ct 3 20.7 mu 5.6 mu W ct 20 mu 4.9 mu ct W 5.6 mu 20.7 mu ct W 5.6 mu 20.7 mu ct W 4.9 mu 20 mu ct W

A cross in Drosophila melanogaster involved the recessive X-linked genes for white eye (w), yellow body (y), and cut wings (ct). A wild-type tri-hybrid female was crossed with white eye, yellow body, cut wings males. On the basis of the results shown below, which of the choices shown best represents the genetic map of the three loci on the X-chromosome? Phenotype + y ct W + + Male Offspring 21 33 wy ct 111 + + + 109 + + ct 424 w y + 394 + y + LO 5 W + ct 3 20.7 mu 5.6 mu W ct 20 mu 4.9 mu ct W 5.6 mu 20.7 mu ct W 5.6 mu 20.7 mu ct W 4.9 mu 20 mu ct W

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter11: The Basic Principles Of Heredity

Section: Chapter Questions

Problem 11TYU: Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were...

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A cross in Drosophila melanogaster involved the recessive X-linked genes for white eye (w), yellow body (y), and cut wings (c). A wild-type tri-hybrid female was crossed with wild-type males and only the male offspring were tallied. On the basis of the results shown below, which of the choices shown best represents the genetic map of the three loci on the X-chromosome? Phenotype Male Offspring + у t 494 394 + + ct 28 w y + 35 + у + 105 w + ct 101 w y ct 5 5.4 mu 17.6 mu ct 6.1 mu 18.3 mu w ct 17.6 mu 5.4 mu ct 18.3 mu 6.1 mu ct 6.1 mu 18.3 mu ct Submit Request Answer de Feedback Next >
A cross in Drosophila melanogaster involved the recessive X-linked genes for white eye (w), yellow body (y), and cut wings (c). A wild-type tri-hybrid female was crossed with wild-type males and only the male offspring were tallied. On the basis of the results shown below, which of the choices shown best represents the genetic map of the three loci on the X-chromosome? Phenotype Male Offspring + y ct 494 w + + 394 + + ct 28 w y + 35 + y + 105 w + ct 101 w y ct 5 + + + 3
In 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?
The 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?
In flies, assume black (B) is dominant and green (b) is recessive for abdomen color, and straight (S) is dominant and bent (s) is recessive for antenna shape. Assume also that these genes are found on the same chromosome. If single mutants for each of these genes (green abdomen x bent antenna) are crossed with one another, and you then testcross the resulting offspring, you get the following numbers of each phenotype: black, straight 17 green, bent 12 black, bent 337 green, straight 364 What is the map distance between B and S? You must show and clearly label ALL work
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 +
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, 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 =1
Wild-type mice have brown fur and short tails. Loss of function of a particular gene produces white fur, while loss of function of another gene produces long tails, and loss of function at a third locus produces agitated behavior. Each of these loss of function alleles is recessive. If a wild-type mouse is crossed with a triple mutant, and their F1 progeny is test-crossed, the following recombination frequencies are observed among their progeny. Produce a genetic map for these loci. Brown, short tailed, normal: 955 White, short tailed, normal: 16 Brown, short tailed, agitated: 0 White, short tailed, agitated: 36 Brown, long tailed, normal: White, long tailed, normal: Brown, long tailed, agitated: 46 0 14 White, long tailed, agitated: 933
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-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?
Drosophila 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.
In fruit flies, red eyes (pr+_) are dominant to purple eyes (prpr) and normal wings (vg+_) are dominant to vestigial wings (vgvg). The genes are located on the same chromosome. A purebreeding red-eyed fly with vestigial wings was crossed with a pure-breeding purple-eyed fly with normal wings. All of the F1 progeny had a WT phenotype. The recombination frequency between the two genes is 15%. If an F1 individual was test crossed, what percentage of the progeny would you expect to have the WT phenotype