Loose Leaf For Integrated Principles Of Zoology
18th Edition
ISBN: 9781260411140
Author: Cleveland P Hickman Jr. Emeritus, Susan L. Keen, David J Eisenhour Professor PhD, Allan Larson, Helen I'Anson Associate Professor of Biology
Publisher: McGraw-Hill Education
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Chapter 5, Problem 12RQ
Summary Introduction
To predict: The appearance of offspring in given crosses.
Introduction: Genotype is the genetic makeup of an organism. Offspring receive two sets of chromosomes in which each set belong to each parent.
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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.
A 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.
In Drosophila melanogaster, white eyes (w) is an X-linked recessive mutation.
Normal eyes are red (W). At the autosomal wing shape locus, normal shaped
wings (Vg) are dominant to vestigial wings (vg). The following cross was made:
XWXW Vgvg x XWY Vgvg
What proportion of the progeny from this cross will be males with white eyes
and normal wings? Write your answer as a numerical value rounded properly to 4
decimal digits. Do not write the answer as a fraction or percentage.
Chapter 5 Solutions
Loose Leaf For Integrated Principles Of Zoology
Ch. 5 - What is the relationship between homologous...Ch. 5 - Describe or diagram the sequence of events in...Ch. 5 - What are the designations of the sex chromosomes...Ch. 5 - Prob. 4RQCh. 5 - Diagram by Punnett square a cross between...Ch. 5 - Prob. 6RQCh. 5 - Assuming brown eyes (B) are dominant over blue...Ch. 5 - Prob. 8RQCh. 5 - Prob. 9RQCh. 5 - Prob. 10RQ
Ch. 5 - Assume that right-handedness (R) is genetically...Ch. 5 - Prob. 12RQCh. 5 - Prob. 13RQCh. 5 - Distinguish the following: euploidy, aneuploidy,...Ch. 5 - Prob. 15RQCh. 5 - Prob. 16RQCh. 5 - Prob. 17RQCh. 5 - Prob. 18RQCh. 5 - Prob. 19RQCh. 5 - Prob. 20RQCh. 5 - Prob. 21RQCh. 5 - Why do many mutations have no detectable effects...Ch. 5 - Distinguish between proto-oncogene and oncogene....Ch. 5 - Prob. 24RQCh. 5 - Outline the essential steps in the polymerase...Ch. 5 - Prob. 26RQCh. 5 - Prob. 27RQCh. 5 - Prob. 1FFT
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- In Drosophila melanogaster, white eyes (w) is an X-linked recessive mutation. Normal eyes are red (W). At the autosomal wing shape locus, normal shaped wings (Vg) are dominant to vestigial wings (vg). The following cross was made: X"XW Vgvg x XWY Vgvg What proportion of the progeny from this cross will be males with red eyes and normal wings? Write your answer as a numerical value rounded properly to 4 decimal digits. Do not write the answer as a fraction or percentage. Answer:arrow_forwardIn Drosophila, the white gene located on the X chromosome affects eye color; an autosomal gene, wingless, is on an autosomal chromosome. Use the following allele symbols: Xw+ _ , Xw+Y = wild type red eyes; X-linked dominant allele Xw Xw , XwY = white eyes; X-linked recessive allele Y = Y sex chromosome vg+ = wild type wings; autosomal dominant vg = wingless; autosomal recessive Predict ratios/proportions of genotypes and phenotypes of offspring from the following cross, of a white-eyed male with wild type wings and a wild type red eyed female with wild type wings: indicate sex of offspring along with phenotypes. XwY vg+ vg x Xw+Xw vg+vgarrow_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_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 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_forwardIN DROSOPHILA, AN X-LINKED RECESSIVE MUTATION, Xm CAUSES MINIATURE WINGS. LIST THE F₂ PHENOTYPIC RATIOS IF: A MINIATURE-WINGED FEMALE IS CROSSED WITH A NORMAL MALE AND A MINIATURE-WINGED MALE IS ● ● CROSSED WITH A NORMAL FEMALE. WHAT WOULD THE PHENOTYPIC RATIO FROM (A) BE IF THE MINIATURE- WINGED GENE WERE AUTOSOMAL? ASSUME IN ALL CASES THAT THE P1 INDIVIDUALS ARE TRUE-BREEDING.arrow_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_forward
- In Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. 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 =1arrow_forwardIn Drosophila, an X-linked recessive mutation, Xm causes miniature wings. List the F2 phenotypic ratios if: a miniature-winged female is crossed with a normal male and a miniature-winged male is crossed with a normal female. What would the phenotypic ratio from (a) be if the miniature-winged gene were autosomal? Assume in all cases that the P1 individuals are true-breeding.arrow_forward+ ec +/Y + + w/Y y ec +/Y + ec +/y ec w ++ w/y ec w у ес +у ес и Determine the order in which the three loci y, ec, and w Occur on the chromosome and prepare a linkage map. 7.22 A cross involving X-linked genes was made between yellow, bar, vermilion female fies and wild males, and the F1 females were crossed with y B v males. The following phenotypes were obtained when 1000 progeny were exam- ined: Dra ord ma the 7.2 546 244 160 50 + + + + Bv y Bv y+ + y+v y B+ and an and and and +B + re + + v ge Determine the order in which the three loci occur on the chromosome and prepare a linkage map. 7.23 Female Drosophila heterozygous for ebony (e"le), scarlet (st*/st), and spineless (ss*/ss) were testcrossed, and the following progeny were obtained: PROGENCY PHENOTYPES NUMBER ir Wild type Ebony Ebony, scarlet Ebony, spineless Ebony, scarlet, spineless Scarlet 67 8. 68 347 78 368 Scarlet, spineless Spineless (a) Are these genes linked? Justify your answer. (b) Write the genes given on a…arrow_forward
- The 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_forwardIn 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?arrow_forwarda. In Drosophila, crosses between F1 heterozygotes ofthe form A b / a B always yield the same ratio ofphenotypes in the F2 progeny regardless of the distance between the two genes (assuming completedominance for both autosomal genes). What is thisratio? Would this also be the case if the F1 heterozygotes were A B / a b? (Hint: Remember that inDrosophila, recombination does not take placeduring spermatogenesis.)b. If you intercrossed F1 heterozygotes of the formA b / a B in mice, the phenotypic ratio among the F2progeny would vary with the map distance betweenthe two genes. Is there a simple way to estimate themap distance based on the frequencies of the F2phenotypes, assuming rates of recombination areequal in males and females? Could you estimatemap distances in the same way if the mouse F1heterozygotes were A B / a b?arrow_forward
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