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Description: 2. Write the genotype of the individuals in the pedigree based on the mode of inheritance given.Pedigree 1710 1112131415

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2. Write the genotype of the individuals in the pedigree based on the mode of inheritance given. Pedigree 1 10 11 12 13 14 15

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter14: Chromosomes And Human Inheritance

Section: Chapter Questions

Problem 14SQ

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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?
A testcross is a way to determine ________. a. phenotype b. genotype c. dominance
9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
2. Create a pedigree chart based on the situation given. IV Many years ago, when I was 23 My middle child chose not to I was married to a lovely lady, as pretty as can be Maybe she's not that into "I like your beard" said she, on the day that she Her husband agrees to it too met me So I told her on that day, "will you marry me?" I wonder why they don't want to? II V The youngest fell in love early And had five children that she raised fairly We wed and had three daughters All of which has found their partners My daughters are now carriers Her only son, Harry, also wed early I'm sure the beard will have them prosper I guess his beard also got him his Daisy III VI My eldest had twins Now that we have Gregory Both boys yes it seems Came from Harry and her Daisy I guess now they've not stayed as babies Will a beard be shown in his daily? Because they've been chased after by more ladies If his mother also has a history.
A. HUMAN PEDIGREE CASE ANALYSIS1. One couple has three children with the following sexes and ages: one son (40 y.o.) and two daughters (35 y.o. and 33 y.o.), all of them have normal pigmentation. Another couple has a son (35 y.o) and a daughter (20 y.o.) and all of them also have normal pigmentation. Both couples have normal pigmentation. The younger daughter from the first couple married the son of the second couple and they had three children. Their eldest daughter (5 y.o.) has normal pigmentation while their only son (3 y.o.) and one daughter (1 y.o.) have albinism. a. Draw the pedigree of this family. Follow protocols in making a pedigree. Provide the genotype of all individuals in the pedigree. Please provide also the gene notation. b. What is the mode of inheritance of this trait? c. Justify your answer in letter (b).d. For their normal daughter, what is the probability that she is a carrier? Show solution. e. If they will have a fourth child, what is the probability that the…
Double click on the image below to complete Question 5. 5. Examine the pedigree below. Write the genotype for each individual in the family. 6. The two children of the original couple are both carriers. How do you know?
2. The pedigree below shows a family with a history of a recessive disease. DETERMINE the genotype of each individual for an autosomal trait 13 5 1 O= 14 201 N insowed up 15 O. O 8 16 17 3 Hist 10 4 11 18 2sion 199 www.c Oc 12 a brea 3. If individual 12 has a child with someone who has the disease, DETERMINE the probability that their first child will not have the disease 4. The pedigree of a family with a history of a particular genetic disease is shown below. Squares represent males and circles represent females. Shaded symbols represent those who have the disease.
1 pts The pedigree below shows the expression of Huntington's disease in a family. Huntington's disease is a fatal genetic disorder that causes the progressive breakdown of nerve cells in the brain. It deteriorates a person's physical and mental abilities usually during their prime working years and has no cure. Huntington's is inherited as a dominant allele (H). Examine the pedigree below and determine the genotype for individual "2" DODO0O 2. O HH O Hh О h O either HH or Hh
3) The most likely inheritance pattern in the pedigree above is:
2) 0,0 0,0 1 The most likely inheritance pattern in the pedigree abolve s: