Genetic Analysis: An Integrated Approach (3rd Edition)
Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter E, Problem 10P
Summary Introduction

To analyze:

The four allelic frequencies contributed to the child by probable father F1 in Problem 7 are 0.18, 0.23, 0.13, and 0.14.

From the given information,

a. Determine the combined paternity index for the four genes mentioned in this analysis.

b. Formulate a statement about the probable paternity of F1 based on this analysis.

Introduction:

DNA analysis can be used for identification of parents in sexually reproducing animals including humans. A child receives exactly one half of genes from the mother. Genetic testing can be used to reveal that one half of the genetic markers in the child match those of the mother. Genetic markers that are not maternal should have come from the father. To identify paternity, every non-maternal marker carried by a child must be carried by the father.

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B What is the genotype ratio? What is the phenotype ratio? 3. What are the expected genotype and phenotype ratios in the following genetic conditions? Use scratch paper to do the Punnett square if needed but you do not need to draw it on the worksheet. a. Monohybrid cross between 2 heterozygous individuals (Aa x Aa) Genotype ratio: 1:2:1 Phenotype ratio: 3:1 G q 8:32 a b. Dihybrid cross between 2 heterozygous individuals (AaBb x AaBb) Genotype ratio: Phenotype ratio: 4. Both Mrs. Smith and Mrs. Jones had babies the same day in the same hospital. Mrs. Smith took home a baby girl, whom she named Sharon. Mrs. Jones took home a girl, whom she named Jane. Mrs. Jones began to suspect, however, that the child had been accidentally switched with Mrs. Smith baby in the nursery. Blood test were made; Mr. Smith was type A, Mrs. Smith was type B, Mr. Jones was type A, Mrs. Jones was type A. Sharon was type O, and Jane was type. B. Had a mix-up occurred? Use scratch paper, you do not need to…
Phenotypic ratio Type A : Type O = 2:2 or 1:1 Let's Try This! I1. Solve for the genetic problems below. One point each for: a. genotype of the parents b. Punnett square C. genotypic ratio d. phenotypic ratio e. type of Mendelian or non-Mendelian principle
A. 0.0025   B. 0.01   C. 0.95   D. 0.99   E. None of the above   The common morning glory (Ipomoea purpurea) is highly polymorphic for flower color. Part of this phenotypic variation is due to allelic variation at the P locus. This locus determines whether flowers will be in purple or white color. Further genetic studies showed that the purple color is dominant. Based on a survey, you found that out of a total of 1000 individuals in the population, 922 are purple and 78 are white. Assuming that this population is in Hardy-Weinberg equilibrium, what is the allele frequency for the purple allele?   A. 0.078   B. 0.279   C. 0.721   D. 0.922   E. 0.960
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