16Achondroplasia is the most common form of dwarfism in miniature horses. Achondroplasia dwarf horses have normal sized bodies and heads, but noticeably smallerupper legs and ears. This form of dwarfism is an autosomal dominant trait. The pedigree shows the offspring of one horse without dwarfism, Individual 1, and onewith dwarfism, Individual 2. The horses with achondroplasia are shorter.BA Dwarf, Homozygous dominantC7Dwarf, HeterozygousNon-Dwarf, Homozygous recessiveD Non-Dwarf, Heterozygous3Which of the following is the correct pairing of phenotype and genotype of Individual 5 in the pedigree?1114257 of 11?>Submit€

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Description: 16Achondroplasia is the most common form of dwarfism in miniature horses. Achondroplasia dwarf horses have normal sized bodies and heads, but noticeably smallerupper legs and ears. This form of dwarfism is an autosomal dominant trait. The pedigree s

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Which of the following is the correct pairing of phenotype and genotype of Individual 5 in the pedigree? A Dwarf, Homozygous dominant B Dwarf, Heterozygous с Non-Dwarf, Homozygous recessive D Non-Dwarf, Heterozygous

Which of the following is the correct pairing of phenotype and genotype of Individual 5 in the pedigree? A Dwarf, Homozygous dominant B Dwarf, Heterozygous с Non-Dwarf, Homozygous recessive D Non-Dwarf, Heterozygous

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 5QP: Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of...

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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?
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
Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of an individual with achondroplasia is seeking counsel on the likelihood of his being a carrier of the mutant allele. What is the probability that the unaffected client is carrying the achondroplasia allele?
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…
Cystic fibrosis (CF) is an autosomal recessive condition triggered by the overproduction of stickymucus that clogs the lungs and pancreas. It is a life-threatening disease, but medical advances helpedthe afflicted to live through adulthood.The mother of Claudia died from cystic fibrosis, but her father was normal and never had anyrelative with CF. Her fiancé, Marcus, turned out to be a carrier of the CF allele. What are the genotypes of Claudia and Marcus? Claudia: ________________________ Marcus: _____________________ They planned to have four children. What is the probability that: a. all children will be normal b. at least two will be normal
Cystic fibrosis (CF) is an autosomal recessive condition triggered by the overproduction of stickymucus that clogs the lungs and pancreas. It is a life-threatening disease, but medical advances helpedthe afflicted to live through adulthood. The mother of Claudia died from cystic fibrosis, but her father was normal and never had anyrelative with CF. Her fiancé, Marcus, turned out to be a carrier of the CF allele. What are the genotypes of Claudia and Marcus? Claudia: ________________________ Marcus: _____________________ They planned to have four children. What is the probability that:a. all children will be normal b. at least two will be normal
Cystic fibrosis (CF) is an autosomal recessive condition triggered by the overproduction of sticky mucus that clogs the lungs and pancreas. It is a life-threatening disease, but medical advances helped the afflicted to live through adulthood. The mother of Claudia died from cystic fibrosis, but her father was normal and never had any relative with CF. Her fiancé, Marcus, turned out to be a carrier of the CF allele. What are the genotypes of Claudia and Marcus? Claudia: ________________________ Marcus: _____________________ They planned to have four children. What is the probability that: a. all children will be normal b. at least two will be normal PLEASE SHOW COMPLETE SOLUTION
A man is brachydactylous (very short fingers; rare autosomal dominant), and his wife is not. Both can taste the chemical phenylthiocarbamide (autosomal dominant; common allele), but their mothers could not.a. Give the genotypes of the couple.If the genes assort independently and the couple has four children, what is the probability ofb. all of them being brachydactylous?c. none being brachydactylous?d. all of them being tasters?e. all of them being nontasters?f. all of them being brachydactylous tasters?g. none being brachydactylous tasters?h. at least one being a brachydactylous taster?
In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: IA, IB, and i. The IA allele codes for the A sugar, The IB allele codes for the B sugar, and the i allele doesn't code for any sugar. IA and IB are codominant, and i is recessive to both IA and IB. If an individual with type AB blood has a child with an individual with type O blood, what blood types could their children possibly have?
In addition to the allelic pair determining pattern baldness in man (B,b), consider early baldness to be due to another autosomal allele (E) on a different pair of chromosomes and also dominant in males but recessive in females. The phenotype for ee may be late or nonbaldness depending on sex and the genotype for B, b alleles. Two doubly heterozygous persons marry. What is the phenotype of the male parent? What is the phenotype of the female parent? Give the phenotypic ratio expected among male children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio. Give the phenotypic ratio expected among female children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio.
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