Campbell Biology in Focus; Modified Mastering Biology with Pearson eText -- ValuePack Access Card -- for Campbell Biology in Focus (2nd Edition)
2nd Edition
ISBN: 9780134433776
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman
Publisher: PEARSON
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Chapter 11.2, Problem 3CC
Summary Introduction
To determine:
The fraction of off springs with a genotype consisting of at least two alleles in homozygous recessive state from the cross between PpYyIi and ppYyii.
Concept introduction:
A cross in which alleles for three characters are used is called a trihybrid cross. In a trihybrid cross, all the three laws of inheritance given by Mendel are followed which are: law of dominance, law of segregation and law of independent assortment.
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For practice construct a Genetic map with three genes in Arabidopsis plants. A short plant (hy) is recessive to tall plant (HY), a red flower (r) is recessive to white flower (R); and long pollen (po) is recessive to oval shaped pollen (PO). You can come up with arbitrary number of progenies for each class of offspring (follow the numbers like in Fig. 7.14)
( please help, I dont understand what/how to do this)
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Topic: Trihybrid Cross,
Height in merigonias is determined by three unlinked genes that act additively. Each has two alleles, one compltely dominant allele(A,B or C) that makes plants taller and one recessive allele that makes plants shorter. Do the corss AaBbCc X AaBbCc. The fraction of The F1 progenty will be ? Pls explain it with more details. thanks
Chapter 11 Solutions
Campbell Biology in Focus; Modified Mastering Biology with Pearson eText -- ValuePack Access Card -- for Campbell Biology in Focus (2nd Edition)
Ch. 11.1 - DRAW IT Pea plants heterozygous for flower...Ch. 11.1 - List all gametes that could be made by a pea plant...Ch. 11.1 - Prob. 3CCCh. 11.2 - For any gene with a dominant allele A and...Ch. 11.2 - Two organisms, with genotypes BbDD and BBDd, are...Ch. 11.2 - Prob. 3CCCh. 11.3 - Incomplete dominance and epistasis are both terms...Ch. 11.3 - Prob. 2CCCh. 11.3 - WHAT IF? A rooster with gray feathers and a hen of...Ch. 11.4 - Beth and Tom each have a sibling with cystic...
Ch. 11.4 - MAKE CONNECTIONS In Table 11.1, note the...Ch. 11 - DRAW IT Two pea plants heterozygous for the...Ch. 11 - A man with type A blood marries a woman with type...Ch. 11 - A man has six fingers on each hand and six toes on...Ch. 11 - Prob. 4TYUCh. 11 - Flower position, stem length, and seed shape are...Ch. 11 - Prob. 6TYUCh. 11 - Prob. 7TYUCh. 11 - Prob. 8TYUCh. 11 - Prob. 9TYUCh. 11 - Prob. 10TYUCh. 11 - In tigers, a recessive allele that is pleiotropic...Ch. 11 - Prob. 12TYUCh. 11 - Imagine that you are a genetic counselor, and a...Ch. 11 - Prob. 15TYUCh. 11 - Prob. 16TYUCh. 11 - Prob. 17TYU
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- 6arrow_forwardCrossover frequencies for some genes on a tomato plant: • Normal leaves (M) and tall plant (D) 12% • Normal leaves (M) and normal tomato (O) 33% • Normal leaves (M) and simple florescence (S) 64% • Tall plant (D) and normal tomato (0) 21% • Tall plant (D) and simple fluorescence (S) 52% Draw a chromosome map and state the cross-over frequency between genes for a normal tomato and simple florescence.arrow_forwardEXAMPLE PROBLEM A plant with orange, spotted flowers was grown in the greenhouse from a seed collected in the wild. The plant was self-pollinated and gave rise to the following progeny: 88 orange with spots, 34 yellow with spots, 32 orange with no spots, and 8 yellow with no spots. What can you conclude about the dominance relationships of the alleles responsible for the spotted and unspotted phenotypes? What can you conclude about the genotype of the original plant that had orange, spotted flowers?arrow_forward
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- Please help me understandarrow_forward. In nature, the plant Plectritis congesta is dimorphic forfruit shape; that is, individual plants bear either winglessor winged fruits, as shown in the illustration.Wingless fruit Winged fruitPlants were collected from nature before floweringand were crossed or selfed with the following results:Number of progenyPollination Winged WinglessWinged (selfed) 91 1*Winged (selfed) 90 30Wingless (selfed) 4* 80Winged × wingless 161 0Winged × wingless 29 31Winged × wingless 46 0Winged × winged 44 0*Phenotype probably has a nongenetic explanation.Interpret these results, and derive the mode ofinheritance of these fruit-shaped phenotypes. Usesymbols. What do you think is the nongeneticexplanation for the phenotypes marked by asterisks inthe table?arrow_forwardCrossing Pea Plants: Mendels Study of Single Traits Stem length in pea plants is controlled by a single gene. Consider the cross of a true-breeding long-stemmed variety to a true-breeding short-stemmed variety in which long stems are completely dominant. a. If 120 F1 plants are examined, how many plants are expected to be long stemmed? Short stemmed? b. Assign genotypes to both P1 varieties and to all phenotypes listed in (a). c. A long-stemmed F1 plant is self-crossed. Of 300 F2 plants, how many should be long stemmed? Short stemmed? d. For the F2 plants mentioned in (c), what is the expected genotypic ratio?arrow_forward
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