77 10 9 The following 5 population genetics questions refer to this introduction: "Tay-Sachs disease is inherited as an autosomal recessive, caused by the absence of hexosaminidase-A (Hex-A). Without Hex-A a lipid accumulates in cells, particularly in the brain, resulting in homozygous re individuals regressing in mental and physical function until death in early childhood. Heterozygotes have abnormal Hex-A activity but manifest no disease symptoms. In a large eastern European population, the of Tay-Sachs disease is 3 percent. Using this information answer the following." a) If the population is assumed to be in Hardy-Weinberg equilibrium with respect to Tay-Sachs, what is the frequency of the allele that causes Tay-Sachs (answer in two decimal points [DP])? b) What would be the frequency of heterozygotes (answer in 2 DP)? Asse c) Assuming no assortative mating, what is the probability of two heterozygotes mating (answer in 2DP)? d) In children of such marriages, what would be the frequency of Tay-Sachs disease (answer in 2DP)? e) What proportion of all Tay-Sachs births are produced by such marriages (answer in 2DP)? 10 1 15 16 20 21

77 10 9 The following 5 population genetics questions refer to this introduction: "Tay-Sachs disease is inherited as an autosomal recessive, caused by the absence of hexosaminidase-A (Hex-A). Without Hex-A a lipid accumulates in cells, particularly in the brain, resulting in homozygous re individuals regressing in mental and physical function until death in early childhood. Heterozygotes have abnormal Hex-A activity but manifest no disease symptoms. In a large eastern European population, the of Tay-Sachs disease is 3 percent. Using this information answer the following." a) If the population is assumed to be in Hardy-Weinberg equilibrium with respect to Tay-Sachs, what is the frequency of the allele that causes Tay-Sachs (answer in two decimal points [DP])? b) What would be the frequency of heterozygotes (answer in 2 DP)? Asse c) Assuming no assortative mating, what is the probability of two heterozygotes mating (answer in 2DP)? d) In children of such marriages, what would be the frequency of Tay-Sachs disease (answer in 2DP)? e) What proportion of all Tay-Sachs births are produced by such marriages (answer in 2DP)? 10 1 15 16 20 21

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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Phenylketonuria (PKU) is a human hereditary disease resulting from the inability of the body to process the chemical phenylalanine, which is contained in the protein we eat. PKU is manifest in early infancy and, if it remains untreated, generally leads to cognitive impairment. PKU is caused by a recessive allele with simple Mendelian inheritance. A couple intends to have children but consults a genetic counselor because the man has a sister with PKU and the woman has a brother with PKU. There are no other known cases in their families. They ask the genetic counselor to determine the probability that their first child will have PKU. What is this probability?
Ch. 13-7 Mutations in the TYR gene may render its enzyme product—tyrosinase—nonfunctional.Individuals homozygous for such mutations cannot make the pigment melanin. Albinism, the absence of melanin, results. Humans and many other organisms can have this phenotype (right). Mutated tyrosinase alleles are recessive when paired with the normal allele in heterozygous individuals.In the following situations, what are the probable genotypes of the father, the mother, and their children?a. Both parents have normal phenotypes; some of their children have the albino phenotype and others are unaffected.b. Both parents and children have the albino phenotype.c. The mother and three children are unaffected; the father and one child have the albino phenotype.
Phenylketonuria (PKU) is caused by the absence of the enzyme phenylalanine hydroxylase, which catalyzes the first step of the pathway that breaks down the amino acid phenylalanine, a common component of dietary protein. Explain how environmental intervention is commonly practiced to prevent the development of this human autosomal recessive condition.
If a woman who is a carrier for Tay-Sachs has children with a man who is not a Tay-Sachs carrier, what is the probability that one of their children will be a carrier for Tay-Sachs? (Sketch a Punnett square to help you find the correct answer.) a. None b. 25% c. 50% d. 75% e. 100%
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?
Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. Individuals with PKU have two recessive alleles and have very low levels of an enzyme that is needed to properly break down proteins. If a woman and her husband are both carriers and have three children, what is the probability of each of the following? Show your math. (Hint: You can represent your probabilities as fractions or decimals, but probabilities are always between 0 and 1) a. All three children are of normal phenotype. b. One or more of the three children have the disease. c.All three children have the disease.
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?
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?
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
Neurofibromatosis-1 (NF1) is an autosomal dominant disorder where tumours form in the base layer of the skin or in nerve tissues. What is the probability that individuals II-1 and II-2 will have a genetic son with NF1? Find the image attached.