Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 22, Problem 3CS
CASE STUDY|An unexpected outcome
A newborn screening program identified a baby with a rare autosomal recessive disorder called arginosuccinate aciduria (AGA), which causes high levels of ammonia to accumulate in the blood. Symptoms usually appear in the first week after birth and can progress to include severe liver damage, developmental delay, and mental retardation. AGA occurs with a frequency of about 1 in 70,000 births. There is no history of this disorder in either the father's or mother's family. This case raises several questions:
What are the chances that two heterozygotes will meet and have an affected child?
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Students have asked these similar questions
Niemann Pick Type C disease is a recessive disorder that causes the accumulation of cholesterol and other lipids in lysosomes, ultimately affecting both the liver and the nervous system. Below are the genotypes and phenotypes of offspring of a family with a history of Niemann Pick.
7 NN ( all normal phenotype)
3 Nn (all normal phenotype)
4 nn (1 early onset dementia, 1 mid-life onset dementia, 2 late-onset dementia).
From this information, Niemann-Pick disease is an example of:
A) variable expressivity
B) incomplete dominance
C) incomplete penetrance
D) variable expressivity and incomplete penetrance
E) multiple alleles
Punnet square problems
A=Codominant; B=Codominant; O=Recessive
Mary is homozygous for type A blood. Steve is homozygous for type O blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each?
Mary and Steve have a son, Brad. Brad’s wife, Samantha is heterozygous for type B blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each?
Stella loves roses and decides to cross her red rose with her white rose. All of the resulting offspring of this cross are pink roses. What can you say about the red and white alleles as a result of this cross?
Stella decides to cross two of the pink roses. What are the possible genotypes and phenotypes of the offspring and the probabilities of each?
DNA replication, Transcription and Translation problems
It is S phase of the cell cycle, and time to replicate the cell’s DNA. Using the following strand of DNA…
5) Humans who have an abnormally high level of cholesterol are said to suffer from familialhypercholesterolemia.
The gene for this disorder is dominant (C). A man who isheterozygous for familial hypercholesterolemia marries a woman who is homozygousfor the recessive allele. What is the probability that they will have children that sufferfrom this disorder? please draw the puutnet square and also write the parents and gemtaics top the puutnet square thanks
Chapter 22 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 22 - CASE STUDY |An unexpected outcome A newborn...Ch. 22 - CASE STUDY |Anunexpected outcome A newborn...Ch. 22 - CASE STUDY|An unexpected outcome A newborn...Ch. 22 - HOW DO WE KNOW? Population geneticists study...Ch. 22 - Review the Chapter Concepts on page 441. All these...Ch. 22 -
3. Price et al. (1999. J. Bacteriol. 181:...Ch. 22 -
4. The genetic difference between two Drosophila...Ch. 22 - The use of nucleotide sequence data to measure...Ch. 22 - Calculate the frequencies of the AA, Aa, and aa...Ch. 22 - Prob. 7PDQ
Ch. 22 -
8. What must be assumed in order to validate the...Ch. 22 - In a population where only the total number of...Ch. 22 -
10. If 4 percent of a population in equilibrium...Ch. 22 -
11. Consider a population in which the frequency...Ch. 22 - If the initial allele frequencies are p = 0.5 and...Ch. 22 -
13. Under what circumstances might a lethal...Ch. 22 - Assume that a recessive autosomal disorder occurs...Ch. 22 -
15. One of the first Mendelian traits identified...Ch. 22 -
16. Describe how populations with substantial...Ch. 22 - Achondroplasia is a dominant trait that causes a...Ch. 22 -
18. A recent study examining the mutation rates...Ch. 22 - A form of dwarfism known as Ellis–van Creveld...Ch. 22 -
20. List the barriers that prevent interbreeding...Ch. 22 - What are the two groups of reproductive isolating...Ch. 22 - Prob. 22PDQCh. 22 -
23. In a recent study of cichlid fish inhabiting...Ch. 22 - What genetic changes take place during speciation?Ch. 22 - Some critics have warned that the use of gene...Ch. 22 - Comparisons of Neanderthal mitochondrial DNA with...
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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?arrow_forwardA 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?arrow_forward11. Below is a pedigree chart of an autosomal recessive disorder. Answer the following questions using the correct genetic terminology (do not just write letters like “Ee”). A. What is the genotype of individual 1 in generation II? B: What is the genotype of individual 2 in generation I?C: Is it true that individuals 6, 7, 8, 9, and 10 in generation III all have the same genotype? Why or why not?arrow_forward
- Canavan Disease Canavan disease in an autosomal recessive disorder caused by mutations in the gene that codes for the enzyme aspartoacylase resulting in incomplete synthesis of myelin sheaths. Individuals affected by Canavan disease experience lack of motor development, abnormal muscle tone, and difficulty feeding. Most individuals affected by this disease die before the age of 10. In a particular population in Canada, researchers examined 250 people for Canavan disease and found that 45 of them had the disorder. a) What is the frequency of Canavan disease in the population examined by the researchers? Show all your work. Express your answer as a value between 0 and 1 rounded to two decimal places. b) What is the frequency of the Canavan disease carriers in the population examined by the researchers? Show all your work. Express your answer as a value between 0 and 1 rounded to two decimal places.arrow_forwardCh. 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.arrow_forwardPompe disease is a glycogen storage disorder caused by a lack of ⍺-glucosidase, the enzyme that converts glycogen to glucose in the muscles. Babies born with this disorder die by the age of two. pope disease is a genetic disorder caused by the presence of two recessive autosomal alleles. A. A man and a woman heterozygous for the condition have two female unaffected children. Determine, using a Punnett square, the probability, expressed as a whole number percentage, of a fourth baby being born a male with Pompe disease. B. Iftheman,fromthepreviousquestion,hadtypeAbloodandthewomanhadtypeO blood, what would be probability of them having a daughter with type O blood and affected with Pompe disease? (Show clearly your calculations and express your answer as a frequency between 0 - 1 and rounded to 2 decimal places)arrow_forward
- 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?arrow_forwardShown 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?arrow_forwardTake 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…arrow_forward
- Trait: NeurofibromatosisThe dominant form is neurofibromatosis, caused by the production of an abnormal form of the protein neurofibromin. Affected individuals show spots of abnormal skin pigmentation and non-cancerous tumors that can interfere with the nervous system and cause blindness. Some tumors can convert to a cancerous form.The recessive form is a normal protein - in other words, no neurofibromatosis. Situation:Karina wants to know from whom she got her disorder. Her grandmother Ana has neurofibromatosis. Ana marries Angelo, who does not have the disease. They both have children named Christine (eldest), Monica, Joseph, Marco, and Yna (youngest). Monica, Marco, and Yna have the disorder out of the five children. Christine marries a man with the disorder, and they have two daughters, Kadita and Karina, who both have the disorder, and another daughter, Karrie, who does not have neurofibromatosis. Yna, on the other hand, marries a man with the disorder. They have four children,…arrow_forwardPompe disease is a glycogen storage disorder caused by a lack of ⍺-glucosidase, the enzyme that converts glycogen to glucose in the muscles. Babies born with this disorder die by the age of two. Pompe disease is a genetic disorder caused by the presence of two recessive autosomal alleles. A man and a woman heterozygous for the condition have two female unaffected children. Determine the probability, expressed as a whole number percentage, of a third baby being born a male with Pompe disease. Your answer must include the use of a Punnett square as well as a legend indicating the allele symbols used. (2 marks)arrow_forwardY-linked Inheritance. Holandric genes are inherited by males only. Assuming that a man with hypertrichosis is married to a woman and they had children 1. what is the probability that his sons inherit the trait? 2. what is the probability that his daughters inherit the trait?arrow_forward
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