CAMPBELL BIOLOGY-MASTERINGBIO.>CUSTOM<
17th Edition
ISBN: 9781323739488
Author: Central Texas
Publisher: PEARSON C
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Textbook Question
Chapter 15, Problem 5TYU
Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.
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Chapter 15 Solutions
CAMPBELL BIOLOGY-MASTERINGBIO.>CUSTOM<
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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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- A homozygous tomato plant with orange fruit and white flowers was crossed with a homozygous tomato plant with red fruit and red flowers. The F1 all had orange fruit and white flowers. The F1 were testcrossed by crossing them to homozygous recessive individuals and the following offspring were obtained: Orange fruit and white flowers- 64 Red fruit and red flowers- 69 Orange fruit and red flowers- 14 Red fruit and white flowers- 13 What is the recombination frequency of these two genes?arrow_forwardHemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?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_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. 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_forwardHemophilia and color blindness are both recessive conditions caused by genes on the X chromosome . To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness , their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters’ sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia 24 have neither of the traits , 1 has color blindness only and 1 has hemophilia only . how many centimorgans seperate the hemophilia locus from the locus of the color blindness.arrow_forwardCalculate the recombination frequency from the following data collected from a breeding experiment with butterflies. Parent 1: dihybrid, long legs and blue wings. Parent 2: double recessive, short legs and white wings Offspring: 50 long legs and white wings 200 long legs and blue wings 50 short legs and blue wings 200 short legs and white wings Recombination frequency is: Select one: a. 0.25 (25%) b. 0.10 (10%) c. 0.20 (20%) d. 0.33 (33%) e. 0.50 (50%)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 interference value for this data set?arrow_forwardTwo SNPs are located in the short arm of chromosome 12. SNP1 has allele frequencies of 60% T and 40% C, and SNP2 has allele frequencies of 40% A and 60% G. What is the frequency of the haplotype T-G in a population if the two SNPs are in absolute linkage disequilibrium and if they are in linkage equilibrium?arrow_forwardA. Part 1: F1 Generation Yellow Sampaguita- Homozygous recessive (cc) White Sampaguita- Homozygous dominant (CC) a. With the given sampaguita flowers, you will be crossing these two samples and identify the offspring using a punnet square. Indicate the ratio between each offspring B. Part 2: F2 generation a. With the results you have from Part 1 and using a punnet square, what are the offspring of your heterozygous parents to identify the F2 generation. Indicate the ratio between each offspring. Note: Please provide the punnet square and a detailed explanationarrow_forward
- Genes X, Y, and Z are linked. Crossover gametes between genes X and Y are observed with a frequency of 20%, and crossover gametes between genes Y and Z are observed with a frequency of 10%. What is the expected frequency of double crossover gametes among these genes?arrow_forwardThe 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.arrow_forwardFor linkage analysis, a test cross is used rather than a hybrid cross. Why is this essential? Why would a hybrid cross result in incorrect estimates of genetic distance?arrow_forward
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