Number 7 need help A A A E E E- E E 1 Teew Roman11AaBbCcAaBbCcNo SpacingAaBbCcO FindU ab x, x' Aa AEE E TNormalHeading 1TrReplaceFontParagraphStylesEditingIn man, normal pigmentation is due to a dominant allele "A" and albinism to its recessive allele "a". A normalman marries an albino woman and their first child is an albino. What are the genotypes of these three people?TA7.

Sir Gregor Mendel was a priest and a teacher who did the famous hybridization experiment on garden…

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Biology

Number 7 need help

Human Biology (MindTap Course List)

11th Edition

ISBN:9781305112100

Author:Cecie Starr, Beverly McMillan

Publisher:Cecie Starr, Beverly McMillan

Chapter19: Introduction To Genetics

Section: Chapter Questions

Problem 7CT

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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 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?
Redgreen color blindness is an X-linked recessive disorder in humans. Your friend is the daughter of a color-blind father. Her mother had normal color vision, but her maternal grandfather was color-blind. What is the probability that your friend is color-blind? (a) 1 (b) (c) (d) (e) 0
A young couple went to see a genetic counselor because each had a sibling with cystic fibrosis. (Cysticfibrosis is a recessive disease, and neither member ofthe couple nor any of their four parents is affected.)a. What is the probability that the female of thiscouple is a carrier?b. What are the chances that their child will havecystic fibrosis?c. What is the probability that their child will be acarrier of the cystic fibrosis disease allele?
Albinism is a recessive autosomal trait for skin pigmentation. Hemophilia is a sex-linked recessivedisorder of the blood. assign alleles to the traitsA – normal skin pigmentation X H – normal blooda – albino X h - hemophilia A double heterozygous woman marries a non-hemophilic man and heterozygous for skinpigmentation.  Double heterozygous means heterozygous for both traits. Aa for skin pigmentation andX H X h for blood trait. Therefore, the genotype of the woman is AaX H X h .  Non-hemophilic man is X H Y and heterozygous for skin pigmentation is Aa. The genotype,therefore, of the man is AaX H Y. What is the probability that they will have:a. a child with normal skin? _____________________b. a child with normal blood? _____________________c. an albino girl? _____________________d. A hemophilic boy? _____________________
A man, Penoy, whose sister died in early childhood from a recessive lethal disease marries a woman Esmae, who has the same family history. Because Penoy has survived beyond childhood, he does not have the disease, but he may be a carrier (i.e. heterozygous, as may also be the case with Esmae). What is the probability that their first child will suffer from the disease? [Hint: first calculate the probability that Penoy is heterozygous; then determine the probability that both parents are carriers. Remember that he has survived to adulthood when calculating this probability].
Two mothers give birth to sons at the same time at a busy urbanhospital. The son of mother 1 has hemophilia, a disease causedby an X-linked recessive allele. Neither parent has the disease.Mother 2 has a son without hemophilia, despite the fact thatthe father has hemophilia. Several years later, couple 1 suesthe hospital, claiming that these two newborns were swappedin the nursery following their birth. As a genetic counselor, youare called to testify. What information can you provide the juryconcerning the allegation?
A couple who are about to get married learn from studying their family histories that, in both their families, theirunaffected grandparents had siblings with cystic fibrosis(a rare autosomal recessive disease).a. If the couple marries and has a child, what is theprobability that the child will have cystic fibrosis?b. If they have four children, what is the chance that thechildren will have the precise Mendelian ratio of 3:1 fornormal:cystic fibrosis?c. If their first child has cystic fibrosis, what is theprobability that their next three children 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 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
Attached are three pedigrees. For each trait, considerwhether it is or is not consistent with X-linked recessiveinheritance. In a sentence or two, indicate why or why not.