Mutations of haplosufficient genes are recessive Homozygous wild type Heterozygote Homozygous recessive mutant +/m m/m +/+ Protein Functional Functional Nonfunctional MRNA and Chromosome Chromosome MRNA Protein Functional Nonfunctional Nonfunctional be distinguished from the heterozygote; that is, at the phenotypic level, A/A = A/a. As mentioned earlier, phenylketonuria (PKU) and many other single-gene human diseases are fully recessive, whereas their wild-type alleles are dominant. Other single-gene diseases such as achondroplasia are fully dominant, whereas, in those cases, the wild-type allele is recessive. How can these dominance relations FIGURE 6-1 In the heterozygote, even though the mutated copy of the gene pro- duces nonfunctional protein, the wild-type copy generates enough functional protein to produce the wild-type phenotype. E E

Mutations of haplosufficient genes are recessive Homozygous wild type Heterozygote Homozygous recessive mutant +/m m/m +/+ Protein Functional Functional Nonfunctional MRNA and Chromosome Chromosome MRNA Protein Functional Nonfunctional Nonfunctional be distinguished from the heterozygote; that is, at the phenotypic level, A/A = A/a. As mentioned earlier, phenylketonuria (PKU) and many other single-gene human diseases are fully recessive, whereas their wild-type alleles are dominant. Other single-gene diseases such as achondroplasia are fully dominant, whereas, in those cases, the wild-type allele is recessive. How can these dominance relations FIGURE 6-1 In the heterozygote, even though the mutated copy of the gene pro- duces nonfunctional protein, the wild-type copy generates enough functional protein to produce the wild-type phenotype. E E

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 2CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

See similar textbooks

Similar questions

The gene controlling ABO blood type and the gene underlying nail-patella syndrome are said to show linkage. What does that mean in terms of their relative locations in the genome? What does it mean in terms of how the two traits are inherited with respect to each other?
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?
In drosophila, a recessive mutation (m-) of a maternal effect gene results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produce by a cross between a heterozygous female and a recessive male. From the offspring, backcross the recessive female with the paternal strain. What are the genotypes and phenotypes of the F2s? Show COMPLETE cross for both cases. If m-m- females produce useless eggs, then how are m-m- produced?
Tay Sachs is a rare autosomal recessive disorder that causes mental and physical disabilities leading to death in infants. Affected individuals are lacking the enzyme hexosaminidase, causing lipids to build up in the brain.The HEXA gene on chromosome 15 codes for hexosaminidase, and a four base pair insertion in the gene results in an altered reading frame and non-functional enzyme being produced. Individuals who are carriers (heterozygotes) of the Tay-Sachs allele are not affected by the disease but appear to have increased protection against tuberculosis.The incidence of Tay-Sachs disease is much higher among Ashkenazi Jews originating from Eastern Europe than the general population of the United States. About 1 in 3 500 babies of Ashkenazi Jewish heritage are born with Tay-Sachs disease and about 1 in 30 Ashkenazi Jews are carriers compared to about 1 in 320 000 babies born with the disease and about 1 in 300 carriers in the general United States population. Ashkenazi Jews living in…
Tay–Sachs disease is caused by loss-of-function mutations ina gene on chromosome 15 that encodes a lysosomal enzyme.Tay–Sachs is inherited as an autosomal recessive condition.Among Ashkenazi Jews of Central European ancestry, about1 in 3600 children is born with the disease. What fraction ofthe individuals in this population are carriers?
Campomelic dysplasia (CMD1) is a congenital humansyndrome featuring malformation of bone and cartilage.It is caused by an autosomal dominant mutation of agene located on chromosome 17. Consider the followingobservations in sequence, and in each case, draw whateverappropriate conclusions are warranted.(a) Of those with the syndrome who are karyotypically46,XY, approximately 75 percent are sex reversed,exhibiting a wide range of female characteristics.(b) The nonmutant form of the gene, called SOX9, isexpressed in the developing gonad of the XY male,but not the XX female.(c) The SOX9 gene shares 71 percent amino acid codingsequence homology with the Y-linked SRY gene.(d) CMD1 patients who exhibit a 46,XX karyotypedevelop as females, with no gonadal abnormalities.
In drosophila, a recessive mutation (m-) of a maternal effect gene results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. (A) Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produce by a cross between a heterozygous female and a recessive male. (B) From the offspring, backcross the recessive female with the paternal strain. What are the genotypes and phenotypes of the F2s? (C) If m-m- females produce useless eggs, then how are m-m- produced?
Take the example of B-thalassemia, an autosomal recessive genetic disease that particularly affects people from around the Mediterranean. This disease is associated with an anomaly of hemoglobin, a protein essential for the transport of oxygen, which is composed of four chains: two alpha (a) and two beta (B). In case of B-thalassemia, the ẞ chains are produced in insufficient or no quantity in an individual homozygous recessive resulting in insufficient production of overall hemoglobin leading to anemia and other physiological challenges. The gene that controls the synthesis of the ẞ chains is located on chromosome 11. Here is part of the coding portion of this gene (which controls a total of 146 amino acids and of which you only see the portion 36 to 41) and one of the targeted mutations: 1. Give the sequence of amino acids from the template and mutated strands. 2. What type of point mutation is it? 3. Using the principles of the theory of evolution, explain briefly and generally why…
Shown below is a pedigree for Phenylketonuria (PKU), an autosomal recessive metabolic disorder. The characteristic feature of PKU is severe mental retardation A) What is the probability that individual II-1 is heterozygous for this gene? B) What is the probability that individual III-4 is heterozygous for this gene? C) If individuals III-3 and III-4 were to marry, what is the probability that their child would express PKU?
Bloom syndrome is an autosomal recessive disease that exhibitshaploinsufficiency. A recent survey showed that people heterozygousfor mutations at the BLM locus are at increased risk of colon cancer.Suppose you are a genetic counselor. A young woman is referred to youwhose mother has Bloom syndrome; the young woman’s father has nofamily history of Bloom syndrome. The young woman asks whether sheis likely to experience any other health problems associated with herfamily history of Bloom syndrome. What advice would you give her?