6. ) In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (a*). At an independently assorting locus, an allele for tan (t) coat color is recessive to an allele for black (t*) coat color. A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F1 are intercrossed to produce the F2. In a litter of eight F2 mice, what is the probability that exactly two will have apricot eyes and tan coats? Use three decimal places for the answer.

6. ) In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (a). At an independently assorting locus, an allele for tan (t) coat color is recessive to an allele for black (t) coat color. A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F1 are intercrossed to produce the F2. In a litter of eight F2 mice, what is the probability that exactly two will have apricot eyes and tan coats? Use three decimal places for the answer.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 5QP: As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women....

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As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
A pedigree analysis was performed on the family of a man with schizophrenia. Based on the known concordance statistics, would his MZ twin be at high risk for the disease? Would the twins risk decrease if he were raised in an environment different from that of his schizophrenic brother?
I'm having trouble with my genetics study guide, and am stuck on this question. If someone could explain it with the work it would help me so much. Thank you! In humans, the inheritance of the ABO blood group system exhibits both complete dominance (alleles IA and IB are dominant to allele i) and codominance (alleles IA and IB). In addition, thalassemia shows incomplete dominance; heterozygotes (Tt) exhibit a mild form of the disease (thalassemia minor) and homozygotes (tt) have a more severe form (thalassemia major). A man has blood type A and his mother has blood type O. His wife has blood type AB. Both members of the couple have thalassemia minor. What is the probability that they will have a child with: a. thalassemia minor and blood type A? b. no anemia and blood type AB? c. thalassemia major and blood type B?
A. Look at the pedigree, and DISREGARD individual II-8 for the moment. Is the pattern of inheritance of Unetan syndrome dominant or recessive? You may assume that the gene is FULLY-PENETRANT in this family. Please give two specific reasons that support your conclusion. B. Now, looking at BOTH the pedigree AND at the Southern blot, is this trait autosomal, X-linked, or Y-linked? Please give two specific reasons that support your conclusion. Once again, disregard II-8 for the moment. One of your two reasons must refer specifically to evidence present in the Southern blot. C. Define the gene alleles associated with Unetan syndrome. Your alleles MUST be consistent with the pattern of inheritance, AND your genetic notation must be consistent with that used throughout the course. Unetan syndrome allele: ________ Normal allele: ________
A recessive maternal effect mutant in zebrafish, called ichabod, results in embryos lacking heads that are non-viable. You have been instructed to identify females that are homozygous for the ichabod mutant allele. At your disposal are a tank of wild-type fish (males and females), a tank of male and female parental fish that are all heterozygous for the ichabod mutant allele (ichabod/+), and a tank of F1 fish derived from a cross between a heterozygous male and heterozygous female (ichabod/+). Which of the following would be a way to identify females that are homozygous mutant, i.e. ichabod/ichabod? Select all answers that would work.
From the pedigree shown here, answer the following questions with regard to individual VII-1. A. Who are the common ancestors of her parents? B. What is the inbreeding coefficient for this individual?
1. For each question, use allele symbols and text to explain your answer. A. Is it possible for a man with type A blood to have a child with type B? B. Is it possible for a woman with type O blood to have a child with type A? C. Is it possible for a man with type O blood to have a child with type AB? D. Is it possible for a person with AB blood to have a full sibling with type O? 2. In a strain of chickens, there are birds with white, black, or blue feathers. In crosses between ten white hens and ten blue roosters, there are 100 offspring: 24 hens with blue feathers, 26 roosters with blue feathers, 23 hens with white feathers, 27 roosters with white feathers. When crossing blue hens and white roosters, there are 100 offspring: 25 hens with blue feathers, 25 roosters with blue feathers, 26 hens with white feathers, 24 roosters with white feathers. A. Using allele symbols, create a plausible explanation for these traits. Why is this trait not sex-linked? When crossing ten black…
In rabbits grey hair is dominant to white & black eyes are dominant to red. -Cross 2 parents who are both heterozygous for both traits. 1st Question to answer ---- How many of the offspring could have grey hair & black eyes? 2nd Question to answer ---- How many of the offspring could have grey hair & red eyes? 3rd Question to answer ---- How many of the offspring could have white hair & black eyes? 4th Question to answer ---- How many of the offspring could have white hair & red eyes?
In rabbits, the gene for fur color is multiallelic. Agouti (C) is dominant to chinchilla (Cch), chinchillas is dominant to himalaya (Ch), and himalaya is dominant to albino (c). Determine the genotypes of each corresponding phenotypes. Please show the solution.