Study Guide for Campbell Biology
11th Edition
ISBN: 9780134443775
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece, Martha R. Taylor, Michael A. Pollock
Publisher: PEARSON
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Chapter 15, Problem 8TYK
Summary Introduction
Introduction: Locus is the specific location of gene in a chromosome. The inheritance of two different characters is affected by the gene linkage. The production of offspring with traits that are combined is studied in recombination. Map units measure genetic linkage. Map units are calculated for linked genes.
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A sex-linked recessive allele rg causes a red-green colorblindness in humans. A normal woman whose father was colorblind marries a colorblind man.
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Chapter 15 Solutions
Study Guide for Campbell Biology
Ch. 15 - Complete the following summary of Morgans crosses...Ch. 15 - Two normal color-sighted individuals have two...Ch. 15 - In a testcross between a heterozygote tall,...Ch. 15 - With unlinked genes, an equal number of parental...Ch. 15 - The following recombination frequencies have been...Ch. 15 - a. What is the difference between an organism with...Ch. 15 - Prob. 7IQCh. 15 - Prob. 8IQCh. 15 - Mendels law of independent assortment applies to...Ch. 15 - You have found a new mutant phenotype in fruit...
Ch. 15 - Prob. 3SYKCh. 15 - Prob. 4SYKCh. 15 - Thomas Hunt Morgan firmly established the location...Ch. 15 - Prob. 2TYKCh. 15 - Sex-linked traits a. are coded for by genes...Ch. 15 - Prob. 4TYKCh. 15 - Prob. 5TYKCh. 15 - Prob. 6TYKCh. 15 - Prob. 7TYKCh. 15 - Prob. 8TYKCh. 15 - Prob. 9TYKCh. 15 - Prob. 10TYKCh. 15 - Consider three genes on the X chromosome: A, B,...Ch. 15 - Prob. 12TYKCh. 15 - Genomic imprinting a. explains cases in which the...Ch. 15 - Prob. 14TYKCh. 15 - Prob. 15TYKCh. 15 - Suppose that alleles for an X-linked character for...Ch. 15 - Some girls who fail to undergo puberty are found...Ch. 15 - Prob. 18TYKCh. 15 - The genetic event that results in Turner syndrome...Ch. 15 - Prob. 20TYKCh. 15 - Prob. 1GPCh. 15 - Prob. 2GPCh. 15 - Prob. 3GPCh. 15 - Prob. 4GPCh. 15 - Prob. 5GPCh. 15 - Red-green color blindness is caused by a...
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- Tomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these in the table below. Probability = Number of Progeny in Phenotype Class ÷ Total Progeny Compare the expected probabilities of each phenotype to…arrow_forwardCross Cross A Cross A Cross B Cross B Phenotype F1 generation F2 generation F1 generation F2 generation Male red eyes 132 150 0 99 Female red eyes 135 295 110 101 Male white eyes 0 147 105 93 Female white eyes 0 0 0 95 Using “+” to indicate the wildtype red-eyed allele and “w” to indicate the mutant white-eyed allele, state the genotypes of the following: Wildtype red-eyed and white-eyed parental flies from cross A and cross B. Males and females from the F1 generation flies from cross A and cross B Males and females, F2 generation flies from cross A and cross B.arrow_forwardThe 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_forward
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- Two 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_forwardTomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Compare the expected probabilities of each phenotype to the observed probabilities. Are the gene for fruit colour and the gene for fruit shape assorting independently? Explain.arrow_forward) Cats also have a locus (gene) for a character called agouti. In cats with agouti fur, pigment is unevenly distributed (see Fig. 1). The agouti allele is dominant to the allele for regular fur. A male cat who long haired and is heterozygous for agouti fur is crossed with a female who has regular, non-agouti fur and is homozygous for short fur. a) What are the genotypes of these cats? b) What are all of the possible allelic combinations in the gametes from each of these cats? c)What are all of the possible phenotypic combinations of these characters in their kittens and in what ratios?arrow_forward
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