Loose Leaf For Integrated Principles Of Zoology
18th Edition
ISBN: 9781260411140
Author: Cleveland P Hickman Jr. Emeritus, Susan L. Keen, David J Eisenhour Professor PhD, Allan Larson, Helen I'Anson Associate Professor of Biology
Publisher: McGraw-Hill Education
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Chapter 5, Problem 7RQ
Assuming brown eyes (B) are dominant over blue eyes (b), determine the genotypes of all the following individuals. The blue-eyed son of two brown-eyed parents marries a brown-eyed woman whose mother was brown eyed and whose father was blue eyed. Their child is blue eyed.
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Assuming brown eyes (B) are dominant over blue eyes (b), determine the genotypes of all the following individuals. The blue-eyed son of two brown-eyed parents marries a brown-eyed woman whose mother was brown eyed and whose father was blue eyed. Their child is blue eyed.
i) Assuming brown eyes (E) are dominant over blue eyes (e), determine the genotypes of all the following individuals. The blue-eyed son of two brown-eyed parents marries a brown eyed woman whose mother was brown eyed and whose father was blue eyed. Their child is blue eyed.
In humans, brown eyes (B) are dominant over blue (b)*. A brown-eyed man marries a blue-eyed woman and they have three children, two of whom are brown-eyed and one of whom is blue-eyed. Draw the Punnett square that illustrates this marriage. What is the man’s genotype? What are the genotypes of the children?
Chapter 5 Solutions
Loose Leaf For Integrated Principles Of Zoology
Ch. 5 - What is the relationship between homologous...Ch. 5 - Describe or diagram the sequence of events in...Ch. 5 - What are the designations of the sex chromosomes...Ch. 5 - Prob. 4RQCh. 5 - Diagram by Punnett square a cross between...Ch. 5 - Prob. 6RQCh. 5 - Assuming brown eyes (B) are dominant over blue...Ch. 5 - Prob. 8RQCh. 5 - Prob. 9RQCh. 5 - Prob. 10RQ
Ch. 5 - Assume that right-handedness (R) is genetically...Ch. 5 - Prob. 12RQCh. 5 - Prob. 13RQCh. 5 - Distinguish the following: euploidy, aneuploidy,...Ch. 5 - Prob. 15RQCh. 5 - Prob. 16RQCh. 5 - Prob. 17RQCh. 5 - Prob. 18RQCh. 5 - Prob. 19RQCh. 5 - Prob. 20RQCh. 5 - Prob. 21RQCh. 5 - Why do many mutations have no detectable effects...Ch. 5 - Distinguish between proto-oncogene and oncogene....Ch. 5 - Prob. 24RQCh. 5 - Outline the essential steps in the polymerase...Ch. 5 - Prob. 26RQCh. 5 - Prob. 27RQCh. 5 - Prob. 1FFT
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- In addition to the allelic pair determining pattern baldness in man (B,b), consider early baldness to be due to another autosomal allele (E) on a different pair of chromosomes and also dominant in males but recessive in females. The phenotype for ee may be late or nonbaldness depending on sex and the genotype for B, b alleles. Two doubly heterozygous persons marry. What is the phenotype of the male parent? What is the phenotype of the female parent? Give the phenotypic ratio expected among male children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio. Give the phenotypic ratio expected among female children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio.arrow_forwardin humans brown eyes (B) are dominant over blue (b).A brown eye man marries a blue-eyed woman and they have three children , two of whom are brown eyed and one of whom is blue eyed.Draw the punnett square that illustrate this marrage.what is the man's genotype ? what are the genotype of the children?arrow_forwardSuppose a man is heterozygous for heterochromia, an autosomal dominant disorder which causes two different‑colored eyes in an individual, produced 25‑offspring with his normal‑eyed wife. Of their children, 16 were heterochromatic and 9 were‑normal. Calculate the chi‑square value for this observation.arrow_forward
- In humans, Widow’s peak (W) is dominant over a continuous hairline (w), and short fingers (F) are dominant over long fingers (f). Two individuals with widow’s peaks and short fingers have a child with a continuous hairline and long fingers. Determine the genotype of the parents.arrow_forwardOn the basis of Mendel’s observations (i.e. Mendelian patterns), predict the results from the following crosses with peas. Show the results in terms of genotypes, phenotypes, and their proportions in the offspring. The progeny of a tall (dominant and homozygous) variety crossed with a dwarf variety. The progeny of (a) self-fertilized. The progeny from (a) crossed with the original tall parent. The progeny of (a) crossed with the original dwarf parent.arrow_forwardIn basic eye color in humans, brown (B) is dominant to blue (b). A brown-eyed man, whose father was brown-eyed and whose mother was blue-eyed, married a blue-eyed woman whose father and mother were both brown-eyed' The couple has a blue-eyed son. For which of the individuals mentioned can you be sure of the genotypes? What genotypes are possible for the others?arrow_forward
- In a dihybrid cross of two bi-allelic Mendelian genes, A (two alleles – completely dominant A and recessive a), and B (two alleles – completely dominant B and recessive b), such that the parental generation comprises of pure-bred homozygotes (i.e. AABB with aabb), what are the expected genotype and phenotype ratios in the F1 and F2, if the F1 cross is (a) an intercross and if the F1 cross is (b) a testcrossarrow_forwardIn human beings, the gene for red‑green colorblindness (r) is sex‑linked and recessive to its allele for normal vision (R), while the gene for freckles (F) is autosomal and dominant over its allele for nonfreckled (f). A nonfreckled, normal‑visioned woman whose father was freckled and colorblind, marries a freckled, colorblind man whose mother was nonfreckled. What is the genotype of the woman's father? What is the probability that the couple's first child will be a non-freckled, normal visioned girl? What is the probability that the first two children born to the couple will be freckled and colorblind girls?arrow_forwardBlack eye (B) is dominant over brown eye (b). A man that is a homozygous dominant black-eyed marries a woman that is a heterozygous black-eyed. The offspring resulted to 50% homozygous black and 50% heterozygous black. If one of their heterozygous children marries another heterozygous partner. Regardless of its gender, what are the genotypes of their children? a. 25% BB, 50% Bb, 25% bb b. 50% BB, 25% Bb, 25% bb c. 25% BB, 25% Bb, 50% bb d. 0% BB, 75% Bb, 25% bb In Law of Codominance, a pure line dominant trait crossed with a recessive trait will result to the appearance of both dominant and recessive trait in the offspring. What is the percentage of having a red-haired offspring if the parental genotype is both a heterozygous roan haired? Red hair is a dominant trait and white hair is a recessive trait. a. 100% b. 75% c. 50% d. 25% e. 0%arrow_forward
- In human beings, the gene for red‑green colorblindness (r) is sex‑linked and recessive to its allele for normal vision (R), while the gene for freckles (F) is autosomal and dominant over its allele for nonfreckled (f). A nonfreckled, normal‑visioned woman whose father was freckled and colorblind, marries a freckled, colorblind man whose mother was nonfreckled. What is the probability that the first child born to the couple will either be a freckled, colorblind boy or a non‑freckled, normal visioned girl or a non-freckled, normal visioned boy? What is the probability that the first four children born to the couple will be freckled and normal visioned girls?arrow_forwardIn humans, brown eyes (B) are dominant over blue. A brown eyed man marries a blue-eyed (b) woman and they have three children two of whom are brown-eyed and one of whom is blue-eyed. what are the possible genotypes of the children?arrow_forwardA light-haired man has blood type O, his wife has dark hair and blood type AB, but her father had light hair. Assume dark hair (H) is dominant to light (h). What is the probability they will have a child with dark hair and blood type A? What is the probability they will have a child with light hair and blood type B? What is the total number of phenotypes their children could show?arrow_forward
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