CAMP BIO L/M&MOD MSTG (HARDBOUND BUNDLE
11th Edition
ISBN: 9780134810133
Author: Urry
Publisher: PEARSON
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Chapter 15, Problem 3TYU
A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following
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A wild-type fruit fly (heterozygous for gray body color andnormal wings) is mated with a black fly with vestigial wings.The offspring have the following phenotypic distribution: wildtype, 778; black vestigial, 785; black normal, 158; gray vestigial,162. What is the recombination frequency between these genesfor body color and wing size? Is this consistent with the resultsof the experiment in Figure 15.9?
A 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?
The following results were obtained from a dihybrid cross between two pea plants:
315 plants with round seeds and yellow flowers
101 plants with wrinkled seeds and yellow flowers
108 plants with round seeds and white flowers
32 plants with wrinkled seeds and white flowers
Assume that round seeds and yellow flowers are dominant and both parent plants were heterozygous for both traits.
Chapter 15 Solutions
CAMP BIO L/M&MOD MSTG (HARDBOUND BUNDLE
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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- Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?arrow_forwardTwo true-breeding varieties of maize, one 11 cm high and the other 47 cm high were crossed and the resultant F1 hybrids were then crossed to generate the F2 . In the F2 there were a total of 13,923 plants with a continuous variation in heights between the two extremes and with only 3 plants as large as 47 cm high and 5 plants of 11 cm high. How many genes and how many alleles are involved in determining height in this plant?arrow_forwardA cross was performed using Drosophila melanogaster involving a female known to be heterozygous for both ebony body and sepia eyes and a male known to be homozygous wild type male. The resulting progeny were allowed to mate with one another to produce the data set. Three repetitions of the experiment were conducted. The following data were produced from the crosses. Test these data to determine if they are significantly different from the expected phenotypic ratio. Use the 5% level of significance. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject for each experiment. Wild eye Wild body – 112, Wild eye Ebony body – 40, Sepia eye Wild body – 35, Sepia eye Ebony body – 11arrow_forward
- 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?arrow_forwardPhenotypically wild-type female Drosophila, whose mothers had spineless bristles (ss) and fathers had ebony bodies (e), produced the following offspring when crossed to homozygous males with ebony bodies and spineless bristles: 43 ebony bodies and spineless bristles 49 wild type 336 ebony bodies, wild type bristles 328 spineless bristles and wild type body color What is the map distance between the ebony body and spineless bristles loci?arrow_forwardPhenotypically wild-type female Drosophila, whose mothers had spineless bristles (ss) and fathers had ebony bodies (e), produced the following offspring when crossed to homozygous males with ebony bodies and spineless bristles: 43 ebony bodies and spineless bristles 49 wild type 336 ebony bodies, wild type bristles 328 spineless bristles and wild type body color What is the map distance between the ebony body and spineless bristles loci? Group of answer choices 12.2 43.9 24.4 87.8 6.1arrow_forward
- A series of three-point testcrosses is made to determine the genetic map order of seven linked allele pairs: A/a, B/b, G/g, H/h, Q/q, R/r, and Y/y.From each cross between a triply heterozygous parent listed below, two recombinant classes were noticed as the least frequent among all 8 progeny classes, and are listed at the right in the table. A. For each testcross write the genotype of the F1 heterozygous parent. F1 Parental Phenotype Least frequent F2 Phenotype 1.AHB&ahb AHb & ahB 2.RYh&ryH RYH & ryh 3.BhY&bHy Bhy & bHY 4.qYB&Qyb qYb & QyB 5.AbQ&aBq Abq & aBQ 6.ghR&GHr ghr & GHR B. Write the unified map order of these genes, showing your reasoning.arrow_forwardTwo pure-breeding parents produced a heterozygous female offspring (AaBb) that was then testcrossed with an aabb male. The offspring produced from the testcross included 50 AaBb, 450 Aabb, 450 aaBb, 50 aabb individuals. Describe how you can tell if these two genes are linked or unlinked (What ratio would you expect to see from the testcross if they were not linked?). What were the genotypes of the original parents that produced the heterozygous female? What is the genetic map distance between the two genes?arrow_forwardIn a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. Why is the recombination frequency for the outside pair of genes not equal to the sum of the recombination between the adjacent gene pairs?arrow_forward
- In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the gene order of these linked genes?arrow_forwardIn corn, a colored aleurone is due to the presence of an R allele; r/r is colorless. Another gene controls the color of the plant, with g/g being yellow and G_being green. A plant of unknown genotype is test-crossed, and the following progeny plants were obtained. Colored green 89 Colored yellow 13 Colorless green 9 Colorless yellow 92 What is the recombination frequency between the R locus and the G locus? A. 45.6% B. 9.85% C. 91.15% D. 4.93% E. 6.4%arrow_forwardWe are interested in detremining the genetic map of three different house finch traits: beak color (dark vs light), song length (short vs long) and patch size (reduced vs normal). We have done a large experiment involving the usual PO crossed, with the F1 test crossed. The following are the data for the offspring from the test cross: Beak Shade Patch Size Song Length Number Observed Dark Normal Long 67 Light Normal Long 3 Light Normal Short 498 Dark Normal Short 43 Dark Reduced Short 13 Light Reduced Short 80 Light Reduced Long 50 Dark Reduced Long 389 Choose the map below that best fits the data Beak Shade- 14.26CM- Patch Size- 9.54CM-Song Length O Patch Size-9.54cM- Beak Shade- 14.26CM-Song Length O Patch Size- 14.26cM- Beak Shade- 9.54CM-Song Length O Beak Shade-9.54cM-Patch Size- 14.26CM-Song Length O Patch Size-9.54CM- Song Length- 14.26CM-Beak Shade O Patch Size- 14.26cM-Song Length- 9.54cM-Beak Shadearrow_forward
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