A genetic disorder is caused by a LOF mutation (Bm) and epigenetic imprinting of gene B. Through pedigree analysis of many families, researchers have observed the following results: Female carrier B+Bm x B+B+ --> B+Bm and Mother's genotype Father's genotype Children's genotype Children's phenotype Cross 1: B+Bm B+B+ B+Bm , B+B+ some affected, some unaffected Cross 2: B+B+ B+Bm B+Bm , B+B+ always unaffected Based on these results, gene B is imprinted on the: a. maternal b. paternal

A genetic disorder is caused by a LOF mutation (Bm) and epigenetic imprinting of gene B. Through pedigree analysis of many families, researchers have observed the following results: Female carrier B+Bm x B+B+ --> B+Bm and Mother's genotype Father's genotype Children's genotype Children's phenotype Cross 1: B+Bm B+B+ B+Bm , B+B+ some affected, some unaffected Cross 2: B+B+ B+Bm B+Bm , B+B+ always unaffected Based on these results, gene B is imprinted on the: a. maternal b. paternal

Name: A genetic disorder is caused by a LOF mutation (Bm) and epigenetic imp
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Description: A genetic disorder is caused by a LOF mutation (Bm) and epigenetic imprinting of gene B. Through pedigree analysis of many families, researchers have observed the following results: Female carrier B+Bm x B+B+ --> B+Bm and Mother's genotypeFather's g

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 2QP: Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To...

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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?
Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
A woman has her personal genome analyzed for the BRCA1 mutation after learning that her father is heterozygous and carries one mutant allele. What is her chance of inheriting the mutant allele from her father? 0: men cannot transmit genes affecting breast cancer. 25% 50% 75% 100% Among the progeny of a heterozygous round (Aa) x homozygous wrinkled ( aa) cross, three seeds are chosen at random. What is the probability that all three seeds are round? (1/4)3 1/4 (1/2) 3 1/2 A single gene can produce different proteins. True False
Gene interaction and Epistasis Dog ears may either be drooping or erect. In terms of their barking, some always bark, others sometimes bark, and still other are non-barking. Two dogs from the same species but of different phenotypes were used in the cross: P1 phenotype drooping ears & Non-barking X erect ears & sometimes bark P1 genotype AABB aabb F1 phenotype 100% drooping ears & Non-barking F1 genotype AaBb Mating the male and female from F1 produced: F2 phenotypes F2 genotypes 4 drooping ears & non-barking A_B_ 1 drooping ears & always bark A_bb 2 erect ears & sometimes bark aaB_ aabb Using the same…
Both hemophilia (h) and favism (gd) are inherited as X-linked recessive traits. Hemophilia is an inherited disorder of blood clotting, and favism is an inherited hemolytic anemia caused by absence of the enzyme glucose-6-phosphate dehydrogenase. A phenotypically normal woman is known to have the X chromosome genotype h + / + gd. The frequency of recombination between h and gd is 16%. What proportion of sons born to this woman are expected to be phenotypically normal with respect to both hemophilia and favism?
Help me create a pedigree of this information: Pedigree analysis: Generation 1: Normal parents (AA x AA) Generation 2: Carrier parents (AA x AS) Generation 3: Affected child (AS x AS) Generation 4: Affected grandchild (SS) This pedigree has two normal parents in the first generation. Second generation carriers carry the sickle cell trait from one parent. The disease is 25% more likely to be inherited in the third generation if both parents have the 'S' allele. If both parents have the 'S' allele, their children will have sickle cell anemia in the fourth generation
Arabidopsis thaliana is a diploid plant model organism with 2n 10. Please select the number options to match the following number of copies of each gene an Arabidopsis thaliana 1 leaf cell has number of chromosomes an Arabidopsis thaliana leaf cell contains 1. two 2. five number of chromosomes an Arabidopsis thaliana gamete cell contains 3. ten pairs of homologous chromosomes an Arabidopsis thaliana 2 leaf cell contains
Wild-type mice have brown fur and short tails. Loss of function of a particular gene produces white fur, while loss of function of another gene produces long tails, and loss of function at a third locus produces agitated behavior. Each of these loss of function alleles is recessive. If a wild-type mouse is crossed with a triple mutant, and their F1 progeny is test-crossed, the following recombination frequencies are observed among their progeny. Produce a genetic map for these loci. Brown, short tailed, normal: 955 White, short tailed, normal: 16 Brown, short tailed, agitated: 0 White, short tailed, agitated: 36 Brown, long tailed, normal: White, long tailed, normal: Brown, long tailed, agitated: 46 0 14 White, long tailed, agitated: 933
You are given a Drosophila female that looks wild-type but is heterozygous for mutations intan body (t), miniature wings (m), and white eyes (w). You test cross this female with a tanbodied, miniature winged, and white-eyed homozygous mutant male, and you obtain thefollowing 1400 progeny: Phenotype : number+ + + : 608t m w : 516+ m w : 2t + + : 6+ m + : 39t + w : 46+ + w : 81t m + : 102 Calculate the distance between each pair t-m, m-w, and t-w only using the number ofrecombinants between them (i.e. ignoring the gene in the middle). Draw a linear map with thedistances between genes.

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	Mother's genotype	Father's genotype	Children's genotype	Children's phenotype
Cross 1:	B⁺B^m	B⁺B⁺	B⁺B^m, B⁺B⁺	some affected, some unaffected
Cross 2:	B⁺B⁺	B⁺B^m	B⁺B^m, B⁺B⁺	always unaffected