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 4, Problem 34P

In a breed of domestic cattle, horns can appear on malesand on females. Males and females can also be hornless. The following crosses are performed with parents from pure-breeding lines.

Chapter 4, Problem 34P, In a breed of domestic cattle, horns can appear on malesand on females. Males and females can also

Explain the inheritance of this phenotype in cattle, and assign genotypes to all cattle in Cross I.

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In onion, male sterility is produced when the nuclear genotype is aa and the mitochondrial gene S (sterile) are present. Any other combination of nuclear genotype and mitochondrial gene (including gene F for fertile) will result in a male fertile plant. Give the genotypic ratio and the phenotypic ratio or the percentage of male sterile and male fertile offspring that will be produced in the following crosses. 1. Aa + S male x aa + F female 2. Reciprocal cross of number 1. (Note that when we do reciprocal cross, we interchange/swap the genotypes of the parents (if there is a nuclear gene involved, you interchange the nuclear genotype as well). 3. Aa + S female x Aa + F male 4. Reciprocal cross of number 3.
The A & B antigens in humans may be found in water-soluble forms in secretions, including saliva of individuals with the genotypes E/E& E/e, but not individuals with the genotype e/e.  Thus, the population contains ‘secretors’ & ‘nonsecretors’.  A cross was performed between 2 heterozygous secretors, one with type O blood and the second with type AB blood.  What phenotypic ratios are expected in their offspring?  A) 1/4 type A : 1/4 type B : 1/2 non-secretor B) 3/8 type A : 3/8 type B : 2/8 non-secretor C) 1/2 type A : 1/2 type B D) 3/4 type AB : 1/4 non-secretor
In dogs, dark coat color is dominant over albino, andshort hair is dominant over long hair. Assume that theseeffects are caused by two independently assorting genes.Seven crosses were done as shown below, in which D andA stand for the dark and albino phenotypes, respectively,and S and L stand for the short-hair and long-hairphenotypes.Number of progenyParental phenotypes D, S D, L A, S A, La. D, S × D, S 88 31 29 12b. D, S × D, L 19 18 0 0c. D, S × A, S 21 0 20 0d. A, S × A, S 0 0 29 9e. D, L × D, L 0 31 0 11f. D, S × D, S 45 16 0 0g. D, S × D, L 31 30 10 10Write the genotypes of the parents in each cross. Use thesymbols C and c for the dark and albino coat-color allelesand the symbols H and h for the short-hair and long-hairalleles, respectively. Assume parents are homozygousunless there is evidence otherwise.

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Genetic Analysis: An Integrated Approach (3rd Edition)

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