A gene is described as X-linked when it occurs on the X chromosome and not the Y. Our convention is to indicate X-linkage by attaching the appropriate gene symbol as a superscript on the letter X. Commonly, the wild-type (+) allele is indicated with only a "+" to avoid having to type a superscript on a superscript. For example, a female that is heterozygous and carrying a recessive mutant allele is indicated as X+Xm. Note the convenience of the shorthand + for m+ in this situation. A mutant male has the genotype XmY. When working with X-linked inheritance, always include the X and Y chromosomes in the descriptions of genotypes, and include the sex (male or female) in the descriptions of the phenotypes (e.g., mutant male, wild-type female, etc.). Here are the genotypes and associated phenotypes for X-linked recessive inheritance: X+X+ Wild-type female X+Xm Wild-type female xmxm Mutant female X+Y Wild-type male xmy Mutant male

A gene is described as X-linked when it occurs on the X chromosome and not the Y. Our convention is to indicate X-linkage by attaching the appropriate gene symbol as a superscript on the letter X. Commonly, the wild-type (+) allele is indicated with only a "+" to avoid having to type a superscript on a superscript. For example, a female that is heterozygous and carrying a recessive mutant allele is indicated as X+Xm. Note the convenience of the shorthand + for m+ in this situation. A mutant male has the genotype XmY. When working with X-linked inheritance, always include the X and Y chromosomes in the descriptions of genotypes, and include the sex (male or female) in the descriptions of the phenotypes (e.g., mutant male, wild-type female, etc.). Here are the genotypes and associated phenotypes for X-linked recessive inheritance: X+X+ Wild-type female X+Xm Wild-type female xmxm Mutant female X+Y Wild-type male xmy Mutant male

Name: X-linked Recessive InheritanceA gene is described as X-linked when it
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Description: X-linked Recessive InheritanceA gene is described as X-linked when it occurs on the X chromosome and not the Y. Our convention is to indicate X-linkage by attaching theappropriate gene symbol as a superscript on the letter X. Commonly, the wild-type

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter4: Pedigree Analysis In Human Genetics

Section: Chapter Questions

Problem 19QP: Analysis of X-Linked Dominant and Recessive Traits Suppose a couple, both phenotypically normal,...

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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 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?
Analysis of X-Linked Dominant and Recessive Traits Suppose a couple, both phenotypically normal, have two children: one unaffected daughter and one son affected with a genetic disorder. The phenotype ratio is 1:1, making it difficult to determine whether the trait is autosomal or X-linked. With your knowledge of genetics, what are the genotypes of the parents and children in the autosomal case? In the X-linked case?
Analysis of X-Linked Dominant and Recessive Traits In the eighteenth century, a young boy with a skin condition known as ichthyosis hystrix gravior was identified. The phenotype of this disorder includes thickening of skin and the formation of loose spines that are sloughed off periodically. This man married and had six sons, all of whom had the same condition. He also had several daughters, all of whom were unaffected. In all succeeding generations, the condition was passed on from father to son. What can you theorize about the location of the gene that causes ichthyosis hystrix gravior?
Analysis of X-Linked Dominant and Recessive Traits A young boy is color-blind. His one brother and five sisters are not. The boy has three maternal uncles and four maternal aunts. None of his uncles children or grandchildren is color-blind. One of the maternal aunts married a color-blind man, and half of her children, both male and female, are color-blind. The other aunts married men who have normal color vision. All their daughters have normal vision, but half of their sons are color-blind. a. Which of the boys four grandparents transmitted the gene for color blindness? b. Are any of the boys aunts or uncles color-blind? c. Is either of the boys parents color-blind?
X-linked ichthyosis is an X-linked recessive trait that manifests in part as dry, scaly skin (“ichthy-” = fish or fish like). Suppose a couple are considering having a child together. Parent A is heterozygous for the ichthyosis allele while Parent B is hemizygous negative for the ichthyosis allele. What is the probability their child would be unafflicted with ichthyosis but be a carrier of the ichthyosis-causing allele? a.0% b.25% c.50% d.75% e.100%
I will be adding two questions here because I asked two of the same questions twice by accident earlier. A man and a woman do not have hemophilioa, but the womans father did. (Hemophilia is X-linked recessive). a) What is the probability that they will have a child with hemophilia? b) Is it possible for any of their daughters to be affected? Explain. A cross between a horse homozygous for red hair and a horse homozygous for white hair results in offspring with the coat colour called roan. When you look at the fur of the roan offspring you see both red and white hairs. What type of inheritance best explains this? a) blending inheritance b)codominance c)incomplete codominance d)multiple alleles
Figure 5.1 presented abbreviated pedigrees looking at the inheritance of X-linked human conditions. El Figure 5.1a initiated with a male who was color blind and had hemophilia A. The grandsons and great-grandsons of this male were either color blind and had hemophilia A, or they had neither condition. a. Which if any individuals in the G Fig. 5.1a pedigree is a double heterozygote? b. Are any of the grandsons or great-grandsons shown in the pedigree the product of recombinant type gametes? c. Do the genes responsible for color blindness and hemophilia A segregate independently? d. If you examined many pedigrees, each starting with a man who was simultaneously color blind and had hemophilia A, do you think any grandsons or great-grandsons would ever have one condition but not the other? L Figure 5.1b initiated with a male who was color blind and had hemophilia B. The grandsons of this male were either only color blind, only had hemophilia B, had both conditions, or they had neither…
Red-green color blindness, in humans, is a sex-linked trait controlled by alleles on the X chromosome. Normal color vision (X+) is dominant to colorblindness (Xc). (NOTE: You could write this as XC for the normal allele and Xc for the recessive colorblind allele. Unfortunately, capital C and lowercase c are hard to tell apart as superscripts, especially in type. I will use C -- "X-plus" -- for the normal allele.) If a colorblind man marries a woman with normal vision and they have a colorblind son, what are the genotypes of the father and the mother?
The Figueroa family has a genetically inherited trait called ocular albinism—lack of pigment in the eye—which is caused by a 516-bp deletion in an exon of a single gene. DNA for this gene was amplified from each member of the family and run on the gel shown below. The thickness of the band indicates the relative amount of DNA. Use the pedigree and the gel to determine the most likely mode of inheritance for this disease. a) x-linked dominant b)x-linked recessive c)autosomal dominant d) autosomal recessive e) cant be determined
Could someone please help me with this grade 11 bio dihybrid cross problem in detail and how to solve this question, using a strategy? Assume that curly hair (C) is dominant to straight hair. Albinism (P ) is recessive to normal skin pigmentation. A woman who is heterozygous for curly hair and albinism has a child. The father is homozygous dominant for curly hair and has albinism. (a) Determine the possible phenotypes for their child. (double-crossing, Puneet square)(b) Calculate the four different probabilities of a child beingboth a male and of each phenotype.(c) What is the probability that the child will expressalbinism and have curly hair like his father?
Could someone please help me with this grade 11 bio dihybrid cross problem on how to solve this question, using a strategy in detail! Assume that curly hair (C) is dominant to straight hair. Albinism (P ) is recessive to normal skin pigmentation. A woman who is heterozygous for curly hair and albinism has a child. The father is homozygous dominant for curly hair and has albinism. (a) Determine the possible phenotypes for their child. (double-crossing, Punnet square) (how to display the traits) (b) Calculate the four different probabilities of a child beingboth a male and of each phenotype. (steps in solving this) (c) What is the probability that the child will expressalbinism and have curly hair like his father? (steps in solving this)