Campbell Biology (11th Edition)
11th Edition
ISBN: 9780134093413
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece
Publisher: PEARSON
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Chapter 15, Problem 6TYU
A wild-type fruit fly (heterozygous for gray body color and led eyes) is mated Willi a black fruit fly wltli purple eye. The offspring are wild-type, 721; black purpic. 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for body color and eye color? Using Information from problem 3, what fruit flies (genotypes and
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A wild-type fruit fly (heterozygous for gray body color andred eyes) is mated with a black fruit fly with purple eyes. Theoffspring are wild-type, 721; black purple, 751; gray purple, 49;black red, 45. What is the recombination frequency betweenthese genes for body color and eye color? Using informationfrom problem 3, what fruit flies (genotypes and phenotypes)would you mate to determine the order of the body color, wingsize, and eye color genes on the chromosome?
A wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wildtype, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?
BLACK VESTIGIAL DIHYBRID TESTCROSS
In the parental generation, you mate a pure-breeding wild-type female (bl+/blt.vg+/vg+) with a pure-breeding black, vestigial (bl/bl vg/vg)
to produce an F1 generation that is all wild-type (bl+/bl vg+/vg). Note that the F1 flies are all dihybrid.
Next, you mate several F1 dihybrid females (bl+/bl vg+/vg) with tester males, which are black, vestigial (bl/bl;vg/vg). The offspring of this
dihybrid testcross are:
Phenotype Genotype Tester Gamete Dihybrid Gamete Number
440
Wild-type
394
108
135
Black, vestigial
Vestigial
Black
Copy the table into your notes and derive the dihybrid gametes following the example in the previous section. The columns in blue
(phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced
from the phenotypes and knowing that the tester contributed bl vg gametes. Finally, you can deduce the dihybrid gametes (green) by
subtracting the tester gamete…
Chapter 15 Solutions
Campbell Biology (11th Edition)
Ch. 15.1 - Which one of Mendel's laws describes the...Ch. 15.1 - MAKE CONNECTIONS Review the description of...Ch. 15.1 - WHAT IF? Propose a possible reason that the first...Ch. 15.2 - A white-eyed female Drosophila is mated with a...Ch. 15.2 - Neither Tim nor Rhoda has Duchenne muscular...Ch. 15.2 - MAKE CONNECTIONS Consider what you learned about...Ch. 15.3 - When two genes are located on the same chromosome,...Ch. 15.3 - VISUAL SKILLS For each type of offspring of the...Ch. 15.3 - Prob. 3CCCh. 15.4 - Prob. 1CC
Ch. 15.4 - Prob. 2CCCh. 15.4 - Prob. 3CCCh. 15.5 - Gene dosagethe number of copies of a gene that are...Ch. 15.5 - Reciprocal crosses between two primrose varieties,...Ch. 15.5 - WHAT IF? Mitochondrial genes are critical to the...Ch. 15 - What characteristic of the sex chromosomes allowed...Ch. 15 - Why are males affected by X-Iinked disorders much...Ch. 15 - Why are specific alleles of two distant genes more...Ch. 15 - Prob. 15.4CRCh. 15 - Explain how genomic imprinting and inheritance of...Ch. 15 - A man with hemophilia (a recessive, sex-linked...Ch. 15 - Pseudohypertrophic muscular dystrophy is an...Ch. 15 - A wild-type fruit fly (heterozygous for gray body...Ch. 15 - A planet is inhabited by creatures that reproduce...Ch. 15 - Using the information from problem 4, scientists...Ch. 15 - A wild-type fruit fly (heterozygous for gray body...Ch. 15 - Assume that genes, A and B are on the same...Ch. 15 - Two genes of a flower, one Controlling blue (B)...Ch. 15 - You design Drosophila crosses to provide...Ch. 15 - Banana plants, which are triploid, are seedless...Ch. 15 - EVOLUTION CONNECTION Crossing over is thought to...Ch. 15 - SCIENTIFIC INQUIRY DRAW IT Assume you are mapping...Ch. 15 - WRITE ABOUT A THEME: INFORMATION The continuity of...Ch. 15 - SYNTHESIZE YOUR KNOWLEDGE Butter flies have an X-Y...
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- (This one is more complicated). Two true breeding fruit flies are allowed to mate. One fly ishomozygous dominant for body color and eye color while the other is homozygous recessive forbody color and eye color. (g+:gray body, g:black body, e+:ebony eyes, e:red), The F1 flies areallowed to mate. What are the following probabilities in the F2 generation? For this problemmake sure to show the Punnett Square for your F2 generation and all of your calculations. Youdo not need to show the Punnett Square for the F1 generation. Answer in percent and round to 2decimals.a. Two of three fruit flies are gray and red.b. The first fruit fly is gray and ebony, the second fruit fly is gray and red, and the third fruit fly is black and ebony.arrow_forwardTwo plants in a cross were each heterozygous for two gene pairs (AB/ab) whose loci are linked and 10 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part D If the two genes are 15 mu apart and the plant is (Ab/aB), what proportion of gametes from a signal plant will be ab? Part E What proportion of the offspring of two plants ( both (Ab/aB)) will be A_B_ if the genes are 15 mu apart? Part F What proportion of the offspring of two plants ( both (Ab/aB)) will be A_bb if the genes are 15 mu apart? Part G What proportion of the offspring of two plants ( both (Ab/aB)) will be aaB_ if the genes are 15 mu apart? Part H What proportion of the offspring of two plants ( both (Ab/aB)) will be aabb if the genes are 15 mu apart? How would I solve these?arrow_forwardPURPLE VESTIGIAL DIHYBRID CROSS In the parental generation, you mate a pure-breeding wild-type female (put/pu+;vg+/vg+) with a pure-breeding purple, vestigial (pu/pu;vg/vg) to produce an F1 generation that is all wild-type (pu*/pu;vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (pu*/pu;vg+/vg) with tester males, which are purple, vestigial (pu/pu;vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number Wild-type 437 417 77 59 Purple, vestigial Vestigial Purple Copy the table into your notes and derive the dihybrid gametes following the example in the first section. The columns in blue (phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed pu vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete contribution…arrow_forward
- Two plants in a cross were each heterozygous for two gene pairs (AB /ab) whose loci are linked and 30 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part E: What proportion of the offspring of two plants (both (AB/ab ) will be A - B- if the genes are 30 mu apart? Part F: What proportion of the offspring of two plants (both (AB/ab)) will be A - bb if the genes are 30 mu apart? Part G: What proportion of the offspring of two plants (both (AB/ab)) will be aaB- If the genes are 30 mu apart? Part H: What proportion of the offspring of two plants (both (AB/ab)) will be aabb if the genes are 30 mu apart?arrow_forwardPURPLE VESTIGIAL DIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (put/put.vg+/vg+) with a pure-breeding purple, vestigial (pu/pu;vg/vg) to produce an F1 generation that is all wild-type (put/pu;vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (put/pu;vg+/vg) with tester males, which are purple, vestigial (pu/pu;vg/vg). The offspring of this dihybrid testcross are: Phenotype Wild-type Purple, vestigial Vestigial Purple Genotype Tester Gamete Dihybrid Gamete Number 437 417 77 59 Copy the table into your notes and derive the dihybrid gametes following the example in the first section. The columns in blue (phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed pu vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete…arrow_forwardTable 1: F1 ebony flies - 0 F1 non-ebony flies - 560 F1 stubble flies - 560 F1 non-stubble flies - 0 The researcher collects and crosses male and female flies from the F1 generation. In the resulting offspring, F2, there are both stubble and ebony flies. Draw a Punnett Square to illustrate the F1 cross for the stubbly phenotype showing the individual gametes of each parent and the combination in the resulting offspring.arrow_forward
- What will be the phenotypic ratio of the offspring of a purebreed male fly with eosin eyes (CCXw-eY) mated to a red-eyed female who is heterozygous for both the cream (C) and eosin eyes (Xw-e) allele.arrow_forwardA fruit fly with a gray body and red eyes (genotype BbPp) is mated with a fly having a black and purple eyes (genotype bbpp). Show diagrammatically a genetic cross between the two flies and the possible genotypes and phenotypes of F1. What ratio of offspring would you expect if the body-colour and eye-colour genes are on different chromosome (unlinked)? When mating is actually carried out, most of the offspring look like the parents, but 3% have a gray body and purple eyes, and 3% have a black body and red eyes. Compare and discuss the observation with your answer in part (arrow_forwardHeterozygous Cc chickens express a condition called creeper, in which the leg and wing bones are shorter than normal (cc). The dominant C allele is lethal when homozygous. Skin color is determined by the alleles W and w. W_ chickens have white skin and ww chickens have yellow skin. In a mating between 2 chickens heterozygous for both of these genes: 1) What phenotypes will be observed in the progeny? 2) What fraction of the offspring will have each phenotype? Assign alleles. List possible phentoypes. Make gametes, draw Punnett square, and list fractions of genotypes and/or phenotypes of offspring,arrow_forward
- The allele b gives Drosophila flies a black body and b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings and wx+ gives non-waxy, the wild-type phenotype. The allele cn of a third gene gives cinnabar eyes and cn+ gives red, the wild-type phenotype. A female heterozygous for these three genes is testcrossed, and 1000 progeny are classified with the following phenotypes. 382 cinnabar 379 black, waxy 69 waxy, cinnabar 67 black 48 waxy 44 black, cinnabar 5 wild type 6 black, waxy, cinnabar Based on this data, what is the correct map of these genes in terms of order and distance?arrow_forwardFruitflies of genotype AABB are bred to fruitflies with genotype aabb. The F1 offspring are then bred to fruitflies of genotype aabb. The F2 offspring have the following genotypes: Aabb 14 flies AaBb 413 flies aabb 419 flies aaBb 15 flies a. Are genes A and B on the same chromosome? Yes No b. Show how you would calculate the genetic distance between genes A and B.arrow_forwardA wild-type fruit fly (heterozygous for gray body color and wings) is mated with a black fruit fly with no wings. The offspring are: gray body with wings: 895 black body with wings: 905 gray body without wings: 110 black body without wings: 90 What is the recombination frequency between the two genes?arrow_forward
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