Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 4, Problem 9PDQ
Given the inheritance pattern of coat color in rats described in Problem 8, predict the genotype and
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A standard three-point mapping is conducted for recessive mutations in autosomal genes purple eye (pr), curved wing ( c) and black body (b). Their wild type alleles are also used for genetic mapping. An F1 Drosophila female heterozygous for purple eye (pr), curved wing (c) and black (b) is crossed to a triply homozygous mutant male. The observed numbers and phenotypes of the offspring are as follows:
360 pr c b
380 pr+ c+ b+
104 pr c+ b
96 pr+ c b+
30 pr c b+
20 pr+ c+ b
6 pr c+ b+
4 pr+ c b
PROVIDE THE FOLLOWING:
A) State the order of genes on this chromosome.
B) Calculate map distances between the gene pairs: pr-c, pr-b, c-b. Show calculations, state the number of map units and which gene pairs they refer to.
In a testcross of a female Drosophila heterozygous for 3 linked, recessive genes the following phenotypes were observed and used to make a map of these genes. Calculate the interference.
A---(3.0)---D-----(4.8)-----B
A) There is negative interference.
B) 1.44
C) 2.08
D) 1
E) 0
Three recessive mutations in Drosophila melanogaster, roughoid (ru, small rough eyes), javelin (jv, cylindrical bristles), and sepia eyes (se, dark brown eyes) are linked. A three-point cross was carried out and the following progeny obtained: jv+ ru+ se+ 37 jv+ ru+ se 2 jv+ ru se 14 jv+ ru se+ 146 jv ru se+ 2 jv ru se 35 jv ru+ se 154 jv ru+ se+ 10 a. Determine the order of the genes on the chromosome. b. Determine which progeny contain single crossovers and which contain double crossovers and indicate where among the genes the crossovers occurred. c. Calculate the map distances among the genes. d. Calculate the coefficient of coincidence and interference among the genes
Chapter 4 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 4 - CASE STUDY | But he isn't deaf Researching their...Ch. 4 -
CASE STUDY | But he isn’t deaf
Researching...Ch. 4 - CASE STUDY | But he isn't deaf Researching their...Ch. 4 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 4 - Review the Chapter Concepts list on page 53. These...Ch. 4 - In Shorthorn cattle, coat color may be red, white,...Ch. 4 -
4. With regard to the ABO blood types in humans,...Ch. 4 - In foxes, two alleles of a single gene, P and p,...Ch. 4 - Three gene pairs located on separate autosomes...Ch. 4 - As in the plants of Problem 6, color may be red,...
Ch. 4 -
8. The following genotypes of two independently...Ch. 4 - Given the inheritance pattern of coat color in...Ch. 4 - A husband and wife have normal vision, although...Ch. 4 - In humans, the ABO blood type is under the control...Ch. 4 - In goals, development of the beard is due to a...Ch. 4 -
13. In cats, orange coal color is determined by...Ch. 4 - In Drosophila, an X-linked recessive mutation,...Ch. 4 - Another recessive mutation in Drosophila, ebony...Ch. 4 - While vermilion is X-linked in Drosophila and...Ch. 4 - In pigs, coat color may be sandy, red, or white. A...Ch. 4 - A geneticist from an alien planet that prohibits...Ch. 4 - In another cross, the frog geneticist from Problem...Ch. 4 - In cattle, coats may be solid white, solid black,...Ch. 4 - Consider the following three pedigrees, all...Ch. 4 - Labrador retrievers may be black, brown, or golden...Ch. 4 - Three autosomal recessive mutations in Drosophila,...Ch. 4 -
24. Horses can be cremello (a light cream...Ch. 4 - Pigment in the mouse is produced only when the C...Ch. 4 - Five human matings numbered 1–5 are shown in the...Ch. 4 - Two mothers give birth to sons at the same time at...Ch. 4 - In Dexter and Kerry cattle, animals may be polled...Ch. 4 - What genetic criteria distinguish a case of...Ch. 4 -
30. The specification of the anterior-posterior...Ch. 4 - The maternal-effect mutation bicoid(bcd)is...Ch. 4 -
32. Students taking a genetics exam were...Ch. 4 - In four o'clock plants, many flower colors are...Ch. 4 - Prob. 34PDQ
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- Miniature 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_forwardThe wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated that have either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes. Carefully examine the data from the following five crosses shown below (running across both columns). (a) Identify each mutation as either dominant or recessive. In each case, indicate which crosses support your answer. (b) Assign gene symbols and, for each cross, determine the genotypes of the parents.arrow_forwardIn a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.arrow_forward
- Gene mapping using the Three-point Testcross a) Given the following alleles that control seed traits:W = wrinkled G = green R = roundw = smooth g = yellow r = oval b) Results of a cross with a triple heterozygote revealed the following phenotypes:30 smooth yellow round4 smooth green round958 wrinkled green round2 wrinkled yellow oval18 wrinkled yellow round946 smooth yellow oval16 smooth green oval26 wrinkled green oval c) Determine the order of the genes and the distance between them in centiMorgan (cM).Construct a gene map to show your results. TIP: Based on the phenotypes, determine the alleles in the gametesarrow_forwardA researcher crosses mice with brown eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 6 apricot, short : 30 brown, long : 15 brown, short : 9 apricot, long You know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these results. a) Making use of the appropriate genetic convention for naming alleles, give the genotype of the male parent in this cross. b) What is your null hypothesis for the chi-square test? c) Give the expected number of individuals in the "brown, long" class. d) You obtain a value of 3.47 for the chi-square test. What conclusion can you make from the results of the chi-square test? P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357…arrow_forwardYour internship with Dr. Nefario at Gru Industries is going well so far, but a recent mission to obtain a shrink ray requires a special kind of Minion. You identify that the ideal Minion phenotypes for this mission are two eyes, pale yellow, and short. From your previous work, you’ve found that:Two-eyed (D) is completely dominant over one-eyed (d)Yellow (Y) is incompletely dominant with white (y)Tall (T) is completely dominant over short (t) a. The only reproductive individuals you have at the moment (it was surprise mission!) are a true breeding two-eyed, yellow, short Minion and a one heterozygous for each trait. What genotypic ratios would you expect for this crossassuming that each locus is on a different chromosome? What total proportion would you expect to be the desired phenotype for the mission? b. The mission has since been delayed due to lack of financial support, so you’ve got some time. All the individuals from the F1 cross above are now reproductive. Which genotypes would…arrow_forward
- Miniature 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 231 long, 250 miniature Long Long 560 long O male: Xm/Y and female X* /X* male: X* / Y and female Xm /x O male: X* /Y and female X* /X* male: Xm/Y and female Xm /x+arrow_forwardGiven the following testcross data for corn in which the genes for fine stripe (f), bronze aleurone (bz) and knotted leaf (Kn) are involved: PhenotypeNumber Kn + +451 Kn f +134 + + + 97 + f bz436 Kn + bz18 + + bz119 + f +24 Kn f bz86 Total:1,365 Maize geneticists tend to be like Drosophila geneticists utilizing a + to indicate the wild type allele (which also is dominant) and a lower case letter for the mutant allele (which is recessive). The first thing I suggest in attacking this problem is to use the typical convention of upper case letters for dominant alleles and lower case letters for recessive alleles as follows: F = wild type, f = fine stripe; B = wild type, b = bronze aleurone K = wild type, k = knotted leaf Then we can rewrite the phenotypes in a “language” that is more easily understood: PhenotypeNumber k F B451 k f B134 K F B97 K f b436 k F b18 K F b119 K f B24 k f b86 Total:1,365 a) Determine the sequence (order) of the…arrow_forwardThree autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progenyarrow_forward
- Wing shape (A/a) and body color (B/b) in Drosophila are controlled by two independently assorting gene pairs. Flies with curved wings and gray body were crossed with each other and produced the following offspring: 36 curved wings, gray body 12 curved wings, black body 18 straight wings, gray body 6 straight wings, black body Questions: 1. Describe the interactions that will explain the observed phenotypic ratio. Indicate the phenotype specified by each allele, and the interaction(s) between the alleles.arrow_forwardConsider the following three autosomal recessive mutations in Drosophila:vestigial wings (v); wild type is long (v+)black body color (b); wildtype is gray (b+)plum eyes (p); wildtype is red (p+)A vestigal, gray, red female (homozygous for all three genes) is crossed with a long wing, black, plum male (homozygous for all three genes). The F1 female progeny are mated with triple homozygous recessive males. Here is the phenotypic data for the F2 progeny:vestigal; gray; red 580long wings; black; plum 592vestigal; black; red 45long; gray; plum 40vestigal; black; plum 89long; gray; red 94vestigal; gray; plum 3long; black; red 5A total of 1448 progeny were counted.Which one of the following values is the approximate distance between the plum eye color and black body color loci?arrow_forward3.13 In Drosophila, the mutant allele bwdts causing brown eyes (normal eyes are red) is temperature sensitive. In flies reared at 29°C the mutant allele is dominant, but in flies reared at 22°C the mutant allele is recessive. In a cross of bwdts/+ X bwdts/+, where the + sign denotes the wild-type allele of bwdts, what is the expected ratio of brown-eyed flies to red-eyed flies if the progeny are reared at 29°C? At 22°C?arrow_forward
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