Introduction to Genetic Analysis
Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 4, Problem 27P

a.

Summary Introduction

To determine:  The pedigree chart for the whole family.

Introduction. The human consists of 23 pairs of chromosomes, condensed form of chromatids which divide during cell division into daughter cells. The human has 22 autosomes and one pair of sex chromosomes. The sex chromosomes determine the sex in an individual based on the type of sex chromosomes that are present in the fusing gametes.

b.

Summary Introduction

To determine:  The agreement of the pedigree with the hypothesis that Rh+ is dominant allele and Rh- is recessive.

Introduction. The pedigree chart helps to depict the complete history of the pattern of inheritance of genetic disease from one generation to another generation. The inheritable genetic disease is either autosomal dominant or recessive or sex-linked based on which the probability of disease in a future generation can be predicted.

c.

Summary Introduction

To determine:  The mechanism of the transmission of elliptocytosis.

Introduction. The dominant allele masks the expression of the recessive allele. Therefore the dominant allele is expressed in homozygous and heterozygous genotype while the recessive phenotype is only expressed under homozygous condition.

d.

Summary Introduction

To determine:  The presence of the genes of same chromosome that govern E and Rh phenotypes.

Introduction. Recombination is the process that is exclusive to the meiotic division as it allows the exchange of genetic material between the non-homologous chromosomes. The recombination process is responsible for the shuffling of the characters and producing a zygote that is different from both the parents but have the chromosomes from both the parents.

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Question #3: In the KeyGene paper, the authors state that it would be useful if pollen from an apomict would transmit apomixis-inducing genes to the female in the cross (assuming the pollen is viable). Assuming there was just one gene conferring gametophytic obligate apomixis, and that the two parents are inbreds, what would be the consequences of such a cross if: a) The apomixis was a dominant trait? Indicate the genotypes and phenotypes (apomict or non- apomict) of the parents, F1 and F2 generations. Remember to include the expected genotypic and phenotypic ratios (or percentages) in the F1 and F2 generations, and to position the female first (left side) in the parental cross. b) The apomixis was a recessive trait? Indicate the genotypes and phenotypes (apomict or non- apomict) of the parents, F1 and F2 generations. Remember to include the expected genotypic and phenotypic ratios (or percentages) in the F1 and F2 generations, and to position the female first (left side) in the…
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Chapter 4 Solutions

Introduction to Genetic Analysis

Ch. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 21.1PCh. 4 - Prob. 21.2PCh. 4 - Prob. 21.3PCh. 4 - Prob. 21.4PCh. 4 - Prob. 21.5PCh. 4 - Prob. 21.6PCh. 4 - Prob. 21.7PCh. 4 - Prob. 21.8PCh. 4 - Prob. 21.9PCh. 4 - Prob. 21.10PCh. 4 - Prob. 21.11PCh. 4 - Prob. 21.12PCh. 4 - Prob. 21.13PCh. 4 - Prob. 21.14PCh. 4 - Prob. 21.15PCh. 4 - Prob. 21.16PCh. 4 - Prob. 21.17PCh. 4 - Prob. 21.18PCh. 4 - Prob. 21.19PCh. 4 - Prob. 21.20PCh. 4 - Prob. 21.21PCh. 4 - Prob. 21.22PCh. 4 - Prob. 21.23PCh. 4 - Prob. 21.24PCh. 4 - Prob. 21.25PCh. 4 - Prob. 21.26PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 38.1PCh. 4 - Prob. 38.2PCh. 4 - Prob. 38.3PCh. 4 - Prob. 38.4PCh. 4 - Prob. 38.5PCh. 4 - Prob. 38.6PCh. 4 - Prob. 38.7PCh. 4 - Prob. 38.8PCh. 4 - Prob. 38.9PCh. 4 - Prob. 38.10PCh. 4 - Prob. 38.11PCh. 4 - Prob. 38.12PCh. 4 - Prob. 38.13PCh. 4 - Prob. 38.14PCh. 4 - Prob. 38.15PCh. 4 - Prob. 38.16PCh. 4 - Prob. 38.17PCh. 4 - Prob. 38.18PCh. 4 - Prob. 38.19PCh. 4 - Prob. 38.20PCh. 4 - Prob. 38.21PCh. 4 - Prob. 38.22PCh. 4 - Prob. 38.23PCh. 4 - Prob. 38.24PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69P
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