Biology
5th Edition
ISBN: 9781260487947
Author: BROOKER
Publisher: MCG
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Chapter 18.6, Problem 2CC
Summary Introduction
To design: A testcross to determine the distance between the al and dp genes.
Introduction: The hereditary characters are determined by DNA sequences called genes. The study of the arrangement of genes and the distance between them on the genome of an organism is called genetic mapping.
Summary Introduction
To determine: The genotypes of P, F1.and F2 generation in the test cross.
Introduction: Genes are the sequences of DNA in the genome that are responsible for the hereditary characters being passed on from generation to generation. With the use of genetic mapping, the arrangement and location of genes on the chromosome can be understood.
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Chapter 18 Solutions
Biology
Ch. 18.2 - Prob. 1CCCh. 18.3 - Prob. 1CCCh. 18.3 - Prob. 1CSCh. 18.5 - Prob. 1CSCh. 18.5 - Prob. 1CCCh. 18.6 - Prob. 1EQCh. 18.6 - Prob. 2EQCh. 18.6 - Prob. 3EQCh. 18.6 - Prob. 1CCCh. 18.6 - Prob. 2CC
Ch. 18 - Which of the following is an example of an...Ch. 18 - Prob. 2TYCh. 18 - A female mouse that is Igf2 Igf2 is crossed to a...Ch. 18 - Prob. 4TYCh. 18 - Prob. 5TYCh. 18 - Prob. 6TYCh. 18 - Prob. 7TYCh. 18 - Prob. 8TYCh. 18 - Based on the ideas proposed by Morgan, which of...Ch. 18 - Extranuclear inheritance occurs because a. certain...Ch. 18 - Define epigenetics. Are all epigenetic changes...Ch. 18 - What is a Barr body? How is its structure...Ch. 18 - Core Concept: Information A core concept of...Ch. 18 - Prob. 1COQCh. 18 - Mendel studied seven traits in garden pea plants,...
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- Part A: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be AB? Enter your answer as a decimal to three places (for example: 0.120). Part B: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be Ab? Part C: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be aB? Part D: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a signal plant will be ab?arrow_forwardYou design Drosophila crosses to provide recombinationdata for gene a, which is located on the chromosome shownin Figure 15.12. Gene a has recombination frequencies of14% with the vestigial wing locus and 26% with the browneye locus. Approximately where is a located along thechromosome?arrow_forwardYou are studying a group of mutations in Drosophila all of which are recessive lethal and map to a small area of chromosome 3. You decide to perform complementation testing to see which mutations are in the same gene. You have bred your flies such that they have the following genotype: +; +; TM3 Sb/mut; + This indicates that chromosomes X, 2, and 4 are wild-type. One copy of chromosome 3 is the balancer TM3 Sb which carries a recessive lethal gene and the dominant marker Stubble (Sb), which causes the flies to have short bristles. The other copy of chromosome 3 carries your recessive lethal mutation. a) Describe how you would maintain each of your individual stocks. What ratios of phenotypes do you expect? b) When you perform your complementation testing, you will have to look at the F1 phenotypes to determine whether two genes complement. Describe what your F1 fly population should look like if the two mutations complement and if they do not. You breed each of your four…arrow_forward
- You are mapping three linked loci in Drosophila melanogaster (the common laboratory fruit fly). You cross flies that are triply mutant for apricot (pale eyes), bristle (extra bristles) and clipped (notched wings) to wild-type flies. The F+ flies are wild-type in appearance. You then backcross the F+ females to pure-breeding (apricot, bristle, clipped) males and score the phenotypes of 1000 F progeny for all three loci. Here are the results: 359 wild-type 361 apricot, bristle, clipped 89 bristle, clipped 91 apricot 42 apricot, bristle 38 clipped 9 apricot, clipped 11 bristle Using these data, first determine what gametes from the F; trihybrid produced each of the eight F2 categories. Note that apricot = aa (recessive to wild-type A); bristle = bb (recessive to wild-type B); and clipped = cc (recessive to wild-type C). Then determine if each gamete is recombinant (R) or nonrecombinant (R) for each pair of alleles (that is, for each genetic interval). Complete the table by dragging the…arrow_forwardThe genes dumpy wings (dp), clot eyes (cl), and apterous wings (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the genetic distances shown below were determined. What is the sequence of the three genes? dp–ap 42 dp–cl 3 ap–cl 39arrow_forwardConsider the following variations in Drosophila melanogaster, relative to the wild-type: White eyes are a recessive trait—the gene of which is found in Chromosome I (X). Vestigial wings are a recessive trait—the gene of which is found in Chromosome II. Aristapedia is a dominant trait—the gene of which is found in Chromosome III. Being homozygous for this condition is lethal. Cross the following mutant females with a wild-type (homozygous) male. Show the Punnett square and obtain the genotypic and phenotypic ratios of the first filial generation (F1). Female with white eyes Q4: Show the Punnett squares and obtain the genotypic and phenotypic ratios of the first filial generation (F1) and second filial generation (F2) of the following crosses. Note: The F2 generation can be obtained by crossing one male and one female from the F1 generation. Female with white eyes and vestigial wings and wild-type malearrow_forward
- In Drosophila, one of the genes controlling wing length is located on the X chromosome. A recessive mutant allele of this gene makes the wings miniature—hence, its symbol m; the wild-type allele of this gene, m_, makes the wings long. One of the genes controlling eye color is located on an autosome. A recessive mutant allele of this gene makes the eyes brown—hence, its symbol bw; the wildtype allele of this gene, bw_, makes the eyes red. Miniature-winged, red-eyed females from one true-breeding strain were crossed to normal-winged, brown-eyed males from another true-breeding strain. 1. Predict the phenotypes of the F1 flies. 2. If these flies are intercrossed with one another, what phenotypes will appear in the F2, and in what proportions?arrow_forwardThe genes dumpy (dp), clot (cl), and apterous (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the following genetic distances were determined. What is the sequence of the three genes? dp–ap 42 dp–cl 3 ap–cl 39arrow_forwardIn Drosophila, the dominant Bar mutation (B, chromosome X, position 57) results in thin bar- shaped eyes, while the recessive singed (sn, chromosome X, position 21) results burnt looking bristles. True breeding, wild type females are mated with true breeding males with Bar eyes and singed bristles. 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
- The following list of four Drosophila mutations indicates the symbol for the mutation, the name of thegene, and the mutant phenotype:Allele symbol Gene name Mutant phenotypedwp dwarp small body, warped wingsrmp rumpled deranged bristlespld pallid pale wingsrv raven dark eyes and bodiesYou perform the following crosses with the indicatedresults:Cross #1: dwarp, rumpled females × pallid, raven males→ dwarp, rumpled males and wild-type femalesCross #2: pallid, raven females × dwarp, rumpled males→ pallid, raven males and wild-type femalesF1 females from cross #1 were crossed to males froma true-breeding dwarp rumpled pallid raven stock.The 1000 progeny obtained were as follows:pallid 3pallid, raven 428pallid, raven, rumpled 48pallid, rumpled 23dwarp, raven 22dwarp, raven, rumpled 2dwarp, rumpled 427dwarp 47Indicate the best map for these four genes, includingall relevant data. Calculate interference values whereappropriate.arrow_forwardFrom a Drosophila testcross, the number of each phenotype obtained was as follows: W+ m f+ 218 W m+ f 236 W+ m+ f 168 W m f+ 178 W+ m f 95 W m+ f+ 101 W+ m+ f+ 3 W m f 1 Total, 1000 Construct a genetic map.arrow_forwardIn Drosophila, the genes for withered wings (whd), smooth abdomen (sm) and speck body (sp) are located on chromosome 2 and are separated by the following map distances:whd ----------(30.5)----------sm-----(15.5)-----spA female with withered wings and a smooth abdomen was mated to a male with a speck body.The resulting phenotypically wild-type females were mated with males that had the mutant phenotype for all three traits, producing 1000 offspring. Calculate the number of expected double crossover progeny. Express your answer to the nearest whole number.arrow_forward
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